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MRD:01409469

2019-05-10T16:05:31Z

Stc1917: /* Polanyi's empirical rules */

== Exercise 1: H + H2 system ==

=== Definition of a transition state ===

A potential energy profile results from plotting the potential versus reaction path.

[[File:Surface_Plot01409469_pic1.jpg|thumb|center|400px|'''Figure 1.''' A potential energy surface plot]]

The transition state is defined as the maximum on the minimum energy path linking reactants and products. This means that the transition state is a stationary point on the potential energy surface. Thus, at the transition state, the derivative of the potential energy is zero with respect to each Cartesian coordinate.

<math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>

However, a local minimum on the potential energy surface corresponding to a well would have the same <math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>. Therefore, to distinguish between a local minimum on the potential energy surface and the transition state, the second derivative, <math>\tfrac{\partial^2 V(r_i)}{\partial r_i ^2}\ </math> is determined. This can be calculated by finding the determinant of the Hessian matrix shown below.

:<math>\mathbf H =
\begin{bmatrix}
\dfrac{\partial^2 \ V(r_i)}{\partial \ x^2} & \dfrac{\partial^2 \ V(r_i)}{\partial \ x \partial \mathbf y} \\
\dfrac{\partial^2 \ V(r_i)}{\partial \ y \partial \mathbf x} & \dfrac{\partial^2 \ V(r_i)}{\partial \mathbf y^2}
\end{bmatrix}</math>

The Hessian matrix of a convex function is positive semi-definite. Refining this property allows one to test if a critical point (x,y) is a local maximum, local minimum or saddle point.<ref name="ref1" />

If <math>f</math> < 0 (exactly one negative eigenvalue), then (x,y) is an isolated local maximum.

If <math>f</math> > 0 (exactly one positive eigenvalue), then (x,y) is an isolated local minimum.

If the Hessian has both positive and negative eigenvalues then (x,y) is a saddle point.

=== Locating the transition state ===

[[File:01409469 combined pic 2 and 3.jpg|thumb|center|800px|'''Figure 2.''' A surface and contour plot of the trajectory when rAB=rBC=0.907743 Å and pA=pBC=0]]

When the molecules are not at the transition state, a surface plot of the conditions shows a trajectory being plotted. This means that the trajectory is "rolling" towards the products or reactants. However, if a trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

[[File:01409469_internuclearpic3.png|thumb|center|400px|'''Figure 3.''' A plot of internuclear distance against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

An initial estimate for the position of the transition state was made by observing the potential energy surface and was found to be 0.90 Å. At 0.90 Å, the internuclear distance against time plot showed that the internuclear distance was constantly oscillating. Therefore, the position of the transition state was further refined by increasing the number of decimal places until a horizontal line in the internuclear distance againt time plot was obtained. This means that the internuclear distance is constant and that the molecules are simply undergoing periodic symmetric vibration at the transition state. Therefore, the position of the transition state was determined to be 0.907743 Å.

[[File:01409469 combined energy pic.png|thumb|center|400px|'''Figure 4.''' A plot of energy against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

When rAB=rBC=0.907743 Å and pA=pBC=0, the kinetic energy is 0. This means that HA and HBC are stationary, which is as predicted when the trajectory is started at the transition state with no initial momentum. Furthermore, the forces acting on HA and HBC are 0 as should be the case in the transition state.

=== Calculation type: Dynamic and minimum energy path ===

The minimum energy path, is the lowest energy path from the reactants to the products. The minimum energy path is determined by following the mode corresponding to the negative Hessian eigenvalue in both directions along the steepest descent direction.<ref name="ref1" /> Following this path allows one to approach two valleys of stable structures. The resulting path generated is the minimum energy path. The reaction paths were chartered for a dynamic and a minimum energy path. The initial conditions used were that of the transition state except that the internuclear distance between HA and HB had increased to 0.917743 Å. The reaction path obtained using the two different calculation methods was different.

[[File:01409469 mep and dynamics.jpg|thumb|center|800px|'''Figure 5.''' A surface plot of the trajectory using minimum energy path '''Figure 6.''' A surface plot of the trajectory using dynamics (Left-right)]]

On the surface plot, the reaction path is depicted by the black line. Following the minimum energy path, the trajectory simply follows the valley floor to Ha+Hb-Hc. The reaction path appears to be a relatively straight line indicating that the atoms are not vibrating. In the minimum energy path, the reactants change into the products in an infinitely slow motion. This is done by having the momenta/velocities reset to zero in each time step. While the minimum energy path is useful in characterising a chemical reaction, it does not provide a realistic representation of the motion of the atoms during the reaction. This is because the mass and inertia of the atoms have been ignored.

Since the atoms simply follow the minimum energy path, they do not undergo slight displacements into surrounding valleys of higher potential energy. Furthermore, because the atoms are assumed to have no inertia, they do not undergo vibrations even if they were to be displaced to valleys of higher potential energy.

However, when dynamic calculation is performed, the reaction path is wavy indicating that the atoms are oscillating. This is because the atoms are displaced to regions of higher potential on the path from reactants to products. This leads to atom oscillation due to the inertia of the atoms.

'''Internuclear distance against time plot'''

Setting the initial system conditions such that rAB= 0.907743, rBC= 0.917743 and pA= pBC= 0:

[[File:01409469 Internuclear distance mep and dynamic.jpg|thumb|center|800px|'''Figure 6.''' A graph of internuclear distance against steps using minimum energy path '''Figure 7.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In the minimum energy path, the distance between HA and HB decreases from 0.90 Å to approximately 0.79 Å during the first 50 steps before plateauing off to a value of 0.74 Å which is the approximate length of a hydrogen bond. This means that a bond was formed between HA and HB. In addition, the plateau indicates that the bond is static. On the other hand, the distance between HB and HC has increased to 1.91 Å at large step values. This shows that HC has been ejected off following a successful collision between HA and HBC.

Similarly, in the dynamic trajectory, the distance between HA and HB decreases to approximately 0.75 Å after 0.3 s. This is indicative of hydrogen bond formation between HA and HB. In addition, the distance between HB and HC increases to 3.44 Å after HC is ejected following a successful collision.

One obvious difference between the two plots is that in the minimum energy path trajectory, the internuclear distance between HA and HB is constant indicating that the bond is static. However, in the dynamic trajectory the internuclear distance is oscillating showing the the bond is vibrating. This difference could be attributed to the fact that minimum energy path does not take oscillatory motion of atoms into account while dynamic trajectory does.

However, changing the initial conditions of the system such that rAB = 0.917743 Å, rBC = 0.907743 Å and pA=pBC=0:
[[File:01409469 01409469 Reversed values dynamic mep.jpg|thumb|center|800px|'''Figure 8.''' A graph of internuclear distance against steps using minimum energy path '''Figure 9.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In this case, a hydrogen bond of length 0.77-0.79 Å is formed between HB and HC while HA is repelled away to beyond 3.45 Å.

Additionally, from the internuclear distance plot it can be seen that the distance between HB and HC increases in a rectangular hyperbola fashion when calculated using the minimum energy path. On the other hand, the internuclear distance between HB and HC increases in a linear manner when dynamic calculation is applied. A possible reason for this difference is that in the minimum energy path, the atoms are not vibrating and as such reach the maximum distance and plateau at a faster rate. However, in the dynamic calculation, the atoms have momentum and as such move further away from each other.

=== Reactive and unreactive trajectories ===

With the initial positions r'''1''' = 0.74 and r'''2''' = 2.0, trajectories were run with the momenta combination shown below.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Contour plot !! Reactivity !! Description of the dynamics
|-
| -1.25 || -2.5 ||-99.018 ||[[File:01409469 trajectory1.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. There are no oscillations observed in HA-HB initially, indicating that the diatomic molecule is stable. A new bond is formed between HB and HC. Oscillatory behaviour is observed indicating that the bond is vibrating. On the other hand, HA is ejected away to beyond 10 Å.
|-
| -1.5 || -2.0 ||-100.456 ||[[File:01409469 trajectory2.jpg|thumb|center|400px]] ||Unreactive ||HC collides with the vibrating diatomic HA-HB. However, no bond is formed between HB and HC as the collision is unsuccessful. This is because the molecules lack sufficient kinetic energy to overcome the activation energy barrier and so no reaction occurs. HC is subsequently repelled to beyond 8 Å.
|-
| -1.5 || -2.5 ||-98.956 ||[[File:01409469 trajectory3.jpg|thumb|center|400px]] ||Reactive |||HC collides with HA-HB. The collision is successful and a bond is formed between HB and HC. This is because the atoms have sufficient kinetic energy to overcome the activation energy barrier and form a new HB-HC bond. HA is ejected away to beyond 10 Å. The new hydrogen bond formed is vibrating as seen by the oscillatory behaviour.
|-
| -2.5 || -5.0 ||-84.956 ||[[File:01409469 trajectory4.jpg|thumb|center|400px]] ||Unreactive ||HC collides with HA-HB. Initially the HA-HB bond is elongating, indicating that the hydrogen bond is weakening and going to break. However, after 0.5 s, the HA-HB bond length decreases and the HAHB bond is not broken. HC is repelled beyond 6 Å with very strong vibrations.
A reaction did not occur even though the atoms have sufficient kinetic energy to cross the activation energy barrier. This is because the vibrations are too strong. The system crosses the transition state region and a bond is formed in the product temporarily but then the system reverts back to the reactants. This is an example of barrier recrossing.
|-
| -2.5 || -5.2 ||-83.416 ||[[File:01409469 trajectory5.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. The collision is successful and a new hydrogen bond is formed between HB and HC with HA being ejected away. The products formed experience very strong vibrations as evidenced by the oscillatory motion which occur with a very high frequency.
|}

From the table, it can be seen that the chance of a reaction successfully occurring is not solely dependent on the amount of kinetic energy the atoms possess. Even if the atoms have sufficient kinetic energy to cross the activation energy barrier, barrier recrossing can occur if the atoms are too vibrationally excited.

=== Transition State Theory ===

Transition state theory provided the first theoretical attempt to determine absolute reaction rates. In this theory, a transition state separating the reactants from products is used to find an expression for the thermal rate constant.

The main assumptions of transition state theory are<ref name="ref2" />:

1. Electronic and nuclear motions are separated (equivalent to the Born-Oppenheimer approximation in quantum mechanics)

2. Reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution

3. Molecular systems that have crossed the transition state in the direction of products cannot turn around and reform reactants

4. In the transition state, motion along the reaction coordinate may be separated from the other motions and treated classically as a translation

5. Even in the absence of an equilibrium between reactant and product molecules, the transition states that are becoming products are distributed among their states according to the Maxwell-Boltzmann laws

The goal of transition state theory is to predict the rate constant, ''k'', of a reaction given a potential energy surface for the reaction.

However, transition state theory suffers from a set of limitations due to the assumptions stated above not holding true in reality.<ref name="ref3" />

Firstly, transition state theory assumes that barrier recrossing is negligible. However, as seen from the fourth trajectory in table 1, barrier recrossing occurred because the atoms were too vibrationally excited and the system reverted back to the reactants even though the system had already crossed the transition state region. In this situation, transition state theory would predict a faster rate of reaction compared to what was obtained experimentally.

Secondly, the transition state theory breaks down for systems with light atoms such as hydrogens. In such systems, quantum effects become crucial. Quantum effects tend to increase the rate of the reaction for two reasons. First, particles lie in energy levels which are higher than the bottom of the potential energy well (there is additional zero-point energy). Secondly, the particles can tunnel through the barrier. These two effects effectively lower the activation energy barrier. As a result, the rate of reaction predicted by transition state theory would be less than what is obtained experimentally.

Lastly, transition state theory does not hold true for systems which are far from equilibrium. In such cases, the assumption of a starting Boltzmann distribution breaks down.

== Exercise 2: F - H - H system ==

=== Potential energy surface inspection===

F + HA-HB → HA-F + HB

[[File:01409469 Surface_Plot F 3.jpg|thumb|center|400px|'''Figure 10.''' Potential energy surface for the reaction between F and H2]]

The reaction between F and H2 is an exothermic reaction. This means that the reactants are at a higher energy than the products. The overall reaction involves the breaking of a hydrogen bond and the formation of a hydrogen-fluorine bond. The HF bond formed is stronger than the hydrogen bond that was broken. This is because fluorine is more electronegative than hydrogen resulting in a polar HF bond. The negative enthalpy of H-F bond formation is greater than the positive enthalpy of hydrogen bond dissociation. Therefore, the reaction between F and H2 is an exothermic reaction.

HA + HB-F → HA-HB + F

[[File:01409469 Surface_Plot F 4.jpg|thumb|center|400px|'''Figure 11.''' Potential energy surface for the reaction between H and HF]]

The reaction between H and HF is an endothermic reaction. The products are at a higher energy level than the reactants. The overall reaction involves the breaking of a HF bond and the formation of a hydrogen bond. The formation of a homonuclear H2 molecule results in a non-polar hydrogen bond being formed. The HF bond broken is stronger than the hydrogen bond that was formed. The negative enthalpy of hydrogen bond formation is less than the positive enthalpy of hydrogen-fluorine dissociation. Therefore, the reaction between H and HF is endothermic.

=== Locating the transition state ===

A = HA, B = HB, C = F:

[[File:01409469 Surface_Plot F TS4.jpg|thumb|center|800px|'''Figure 12.''' Determining the position of the transition state for the reaction between F and H2]]

As mentioned previously, when the molecules are not at the transition state, a reaction path will be plotted. However, if the initial conditions of the system are set such that the trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

An initial estimate for the transition state was made by setting rAB = 0.74 which is the length of a hydrogen bond. The position of the transition state was further refined by increasing the number of decimal places such that the final value of rAB was found to be 0.74478 and rBC = 1.81079. This set of distances gave a reaction path that was represented by a point indicating that no trajectory was plotted and that the molecules remained at the transition state since their initial momentum was 0.

Moreover, the kinetic energy as well as the forces acting on the atoms was calculated to be 0, which is consistent with the transition state.

Therefore, the position of the transition state was found to be at rAB = 0.74478 and rBC = 1.81079 when pA = pBC = 0.

=== Determining the activation energy ===

[[File:01409469 activation energy r1.1 combined.jpg|thumb|center|800px|'''Figure 13.''' Determining the activation energy for the reaction between F and H2]]

By reducing the value of rBC by 0.01, the atoms are no longer at the transition state and would move towards the formation of products. The activation energy was found by calculating the change in total energy after the reaction. The activation energy of F + H2 → HF + H was found to be +23.279 kcal/ mol.

[[File:01409469 activation energy r2.1 combined.jpg|thumb|center|800px|'''Figure 14.''' Determining the activation energy for the reaction between FH and H]]

The activation energy of FH + H → F + H2 was found by calculating the small dip in potential energy. The activation energy was found to be +0.280 kcal/ mol.

=== Reaction dynamics of F + HA-HB → HA-F + HB ===

The set of initial conditions that result in a reactive trajectory for F + H2 is as follows: rHH = 0.74 Å, rHF = 1.79 Å and pHH = pHF = 0.

[[File:01409469 momentum r1.jpg|thumb|center|400px|'''Figure 15.''' Graph of momentum against time for the reaction between F and H2]]

Initially, the H2 molecule is travelling in a relatively straight line with very little oscillations as evidenced by the low amplitude of momentum between 0-2 s. Following successful collision between F and H2, the amplitude of momentum of the product, HF, increases dramatically. This is because the reaction between F and H2 is exothermic and the excess energy released causes the molecule to become vibrationally excited and the amplitude of oscillations in the product to increase in the form of H-F vibrations.

[[File:01409469 energy1 r1.jpg|thumb|center|400px|'''Figure 16.''' Graph of energy against time for the reaction between F and H2]]

The law of conservation of energy states that energy cannot be created or destroyed. It simply changes from one form to another. As can be seen from the graph in figure 16, total energy remains constant. In this reaction, potential energy is converted to kinetic energy. A decrease in potential energy of the atoms following a successful collision between F and H2 results in an equal increase in kinetic energy which manifests in the form of increased product (HF) oscillations.

This can be confirmed experimentally using infrared chemiluminescence.<ref name="ref2" /> Observation of radiation from excited product molecules is a clear indication that these molecules are produced in the excited state. If the excitation is primarily vibrational, the radiation will appear in the infrared region of the spectrum (between 3-15 μm). This technique has been used extensively in hydrogen halogen reactions which produce HX molecules in the excited state.

One such experiment involves the use of the arrested relaxation method.<ref name="ref2" /> In this technique, the reaction is carried out in a chamber whose walls are maintained at a very low temperature (20 K). This results in cryopumping of the chamber in that all the products and unreacted starting materials are condensed on the walls and only those product molecules that radiate before being pumped away are observed.

=== Polanyi's empirical rules ===

In a typical chemical reaction, there is an energetic barrier — a saddle point that the reactants must surmount to become products. Which form of energy is initially deposited in the reactants, translational or vibrational, is more efficacious in surmounting the activation energy barrier is one of the core topics in the area of reaction dynamics. The Polyani rules state that vibrational energy is more effective at promoting a late-barrier reaction (transition state resembles products). Conversely, translational energy is better at promoting an early-barrier reaction (transition state resembles the reactants).<ref name="ref4" />

The forward reaction between H2 and F is exothermic. Therefore, the transition state is closer to the reactants than to the products in energy. Since the transition state occurs early in an exothermic reaction, translational energy is more effective at promoting an exothermic reaction than vibrational energy.

Conversely, the backward reaction between H and HF is endothermic. According to Hammond's postulate, the transition state resembles the products more than it resembles the reactants. Therefore, in an endothermic reaction, a late transition state is observed and vibrational energy is more effective at promoting a reaction than translational energy.

=== Promoting the forward reaction: F + HA-HB → HA-F + HB ===

1. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 2.95 and pHF = -0.5.

[[File:01409469 polyani 1+2.jpg|thumb|center|800px|'''Figure 17.''' Contour plot of the reaction between F and H2 when pHH = 2.95 '''Figure 18.''' Graph of momentum against time for the reaction between F and H2 when pHH = 2.95 (Left-right)]]

Vibrational energy is represented by pHH while translational energy is represented by pHF.

A molecule of H2 collides with F. However, the collision is unsuccessful and no reaction occurs. A reaction did not occur even though the molecules have sufficient kinetic energy to overcome the activation energy barrier. This is because most of the energy is deposited in the reactants as H-H vibrations as seen by the large oscillations in H2. Barrier recrossing has occured as a result of the large vibrational energy causing the system to revert back to the reactants even though the system has already crossed the transition state and a bond was formed between HB and F temporarily.

This appears to run counter to what one would expect to happen when more energy is put into the system. It is predicted that when more energy is put into the system, the reactants would possess more kinetic energy to overcome the activation energy barrier. However, as can be seen from the simulation above, an increase in the vibrational energy of the reactants is not effective at promoting the occurrence of an exothermic reaction with an early transition state. This is in accordance to Polyani's rules.

2. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 0.1 and pHF = -0.8.

[[File:01409469 polyani 3+4 2.jpg|thumb|center|800px|'''Figure 19.''' Contour plot of the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 '''Figure 20.''' Graph of momentum against time for the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 (Left-right)]]

The collision between a molecule of H2 and F is successful and a chemical reaction has occurred. A reaction has occurred even though the overall energy of the system has been significantly reduced by lowering the amount of vibrational energy in the reactants but slightly increasing the amount of translational energy deposited in the reactants. Therefore, H2 and F have more translational energy to move towards each other and form a H-F bond. The product formed is vibrationally excited as seen by the large amplitude of oscillations in the product HF which correspond to the strong H-F vibrations.

From the second simulation, it can be seen that an increase in translational energy in the reactants is more effective at promoting an exothermic reaction with an early transition state to occur, which again follows Polyani's rules.

=== Promoting the backward reaction: HA + HB-F → HA-HB + F ===

1. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 0.2 and pHH = -3.0.

[[File:01409469 polyani 5+6 .jpg|thumb|center|800px|'''Figure 21.''' Contour plot of the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 '''Figure 22.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 (Left-right)]]

An atom of HA collides with a molecule of HBF. However, the collision is unsuccessful and no chemical reaction occurs. This is because the reactants lack sufficient vibrational energy to surmount the activation energy barrier. Therefore, following an unsuccessful collision with HBF, HA simply bounces off and moves away at constant momentum. This is in agreement with Polyani's rules which state that vibrational energy is more effective than translational energy at promoting reactions with a late transition state, for example, endothermic chemical reactions.

2. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 7.1 and pHH = -0.5.

[[File:01409469 polyani 7+8 .jpg|thumb|center|800px|'''Figure 23.''' Contour plot of the reaction between HA + HB-F when pFH = 7.1 and pHH = -0.5 '''Figure 24.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 7.1 and pHH = -0.5 (Left-right)]]

Increasing the amount of vibrational energy in the reactants by increasing pFH resulted in a chemical reaction successfully occurring. This is because the reactants now have sufficient kinetic energy in the form of H-F vibrations to overcome the activation energy barrier. In addition, because the reaction between HA + HB-F is endothermic, energy is taken in from the surroundings during the reaction and the products are less vibrationally excited than the reactants. This can be seen in the lower amplitude of oscillations in the product formed.

In conclusion, the vibrational energy of the reactants should be increased to promote an endothermic reaction with a late transition state while the translational energy of the reactants should be increased to promote an exothermic reaction with an early transition state. All 4 simulations shown above obey Polyani's rules.

== References ==
<references>
<ref name="ref1">A. C. Vaucher, M. Reiher, ''J. Chem. Theory. Comput.'', 2018, '''14''', 3091-3099.</ref>
<ref name="ref2">J. I. Steinfield, J. S. Francisco, W. L. Hase, ''Chemical Kinetics and Dynamics'', Pearson, 1998.</ref>
<ref name="ref3">D. C. Elton, ''Transition State Theory for Physicists'', 2013.</ref>
<ref name="ref4">Z. Zhang, Y. Zhou, D. H. Zhang, ''J. Phys. Chem. Lett.'', 2012, '''3''', 3416-3419.</ref>
</references>

MRD:01409469

2019-05-10T16:01:20Z

Stc1917: /* Locating the transition state */

== Exercise 1: H + H2 system ==

=== Definition of a transition state ===

A potential energy profile results from plotting the potential versus reaction path.

[[File:Surface_Plot01409469_pic1.jpg|thumb|center|400px|'''Figure 1.''' A potential energy surface plot]]

The transition state is defined as the maximum on the minimum energy path linking reactants and products. This means that the transition state is a stationary point on the potential energy surface. Thus, at the transition state, the derivative of the potential energy is zero with respect to each Cartesian coordinate.

<math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>

However, a local minimum on the potential energy surface corresponding to a well would have the same <math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>. Therefore, to distinguish between a local minimum on the potential energy surface and the transition state, the second derivative, <math>\tfrac{\partial^2 V(r_i)}{\partial r_i ^2}\ </math> is determined. This can be calculated by finding the determinant of the Hessian matrix shown below.

:<math>\mathbf H =
\begin{bmatrix}
\dfrac{\partial^2 \ V(r_i)}{\partial \ x^2} & \dfrac{\partial^2 \ V(r_i)}{\partial \ x \partial \mathbf y} \\
\dfrac{\partial^2 \ V(r_i)}{\partial \ y \partial \mathbf x} & \dfrac{\partial^2 \ V(r_i)}{\partial \mathbf y^2}
\end{bmatrix}</math>

The Hessian matrix of a convex function is positive semi-definite. Refining this property allows one to test if a critical point (x,y) is a local maximum, local minimum or saddle point.<ref name="ref1" />

If <math>f</math> < 0 (exactly one negative eigenvalue), then (x,y) is an isolated local maximum.

If <math>f</math> > 0 (exactly one positive eigenvalue), then (x,y) is an isolated local minimum.

If the Hessian has both positive and negative eigenvalues then (x,y) is a saddle point.

=== Locating the transition state ===

[[File:01409469 combined pic 2 and 3.jpg|thumb|center|800px|'''Figure 2.''' A surface and contour plot of the trajectory when rAB=rBC=0.907743 Å and pA=pBC=0]]

When the molecules are not at the transition state, a surface plot of the conditions shows a trajectory being plotted. This means that the trajectory is "rolling" towards the products or reactants. However, if a trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

[[File:01409469_internuclearpic3.png|thumb|center|400px|'''Figure 3.''' A plot of internuclear distance against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

An initial estimate for the position of the transition state was made by observing the potential energy surface and was found to be 0.90 Å. At 0.90 Å, the internuclear distance against time plot showed that the internuclear distance was constantly oscillating. Therefore, the position of the transition state was further refined by increasing the number of decimal places until a horizontal line in the internuclear distance againt time plot was obtained. This means that the internuclear distance is constant and that the molecules are simply undergoing periodic symmetric vibration at the transition state. Therefore, the position of the transition state was determined to be 0.907743 Å.

[[File:01409469 combined energy pic.png|thumb|center|400px|'''Figure 4.''' A plot of energy against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

When rAB=rBC=0.907743 Å and pA=pBC=0, the kinetic energy is 0. This means that HA and HBC are stationary, which is as predicted when the trajectory is started at the transition state with no initial momentum. Furthermore, the forces acting on HA and HBC are 0 as should be the case in the transition state.

=== Calculation type: Dynamic and minimum energy path ===

The minimum energy path, is the lowest energy path from the reactants to the products. The minimum energy path is determined by following the mode corresponding to the negative Hessian eigenvalue in both directions along the steepest descent direction.<ref name="ref1" /> Following this path allows one to approach two valleys of stable structures. The resulting path generated is the minimum energy path. The reaction paths were chartered for a dynamic and a minimum energy path. The initial conditions used were that of the transition state except that the internuclear distance between HA and HB had increased to 0.917743 Å. The reaction path obtained using the two different calculation methods was different.

[[File:01409469 mep and dynamics.jpg|thumb|center|800px|'''Figure 5.''' A surface plot of the trajectory using minimum energy path '''Figure 6.''' A surface plot of the trajectory using dynamics (Left-right)]]

On the surface plot, the reaction path is depicted by the black line. Following the minimum energy path, the trajectory simply follows the valley floor to Ha+Hb-Hc. The reaction path appears to be a relatively straight line indicating that the atoms are not vibrating. In the minimum energy path, the reactants change into the products in an infinitely slow motion. This is done by having the momenta/velocities reset to zero in each time step. While the minimum energy path is useful in characterising a chemical reaction, it does not provide a realistic representation of the motion of the atoms during the reaction. This is because the mass and inertia of the atoms have been ignored.

Since the atoms simply follow the minimum energy path, they do not undergo slight displacements into surrounding valleys of higher potential energy. Furthermore, because the atoms are assumed to have no inertia, they do not undergo vibrations even if they were to be displaced to valleys of higher potential energy.

However, when dynamic calculation is performed, the reaction path is wavy indicating that the atoms are oscillating. This is because the atoms are displaced to regions of higher potential on the path from reactants to products. This leads to atom oscillation due to the inertia of the atoms.

'''Internuclear distance against time plot'''

Setting the initial system conditions such that rAB= 0.907743, rBC= 0.917743 and pA= pBC= 0:

[[File:01409469 Internuclear distance mep and dynamic.jpg|thumb|center|800px|'''Figure 6.''' A graph of internuclear distance against steps using minimum energy path '''Figure 7.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In the minimum energy path, the distance between HA and HB decreases from 0.90 Å to approximately 0.79 Å during the first 50 steps before plateauing off to a value of 0.74 Å which is the approximate length of a hydrogen bond. This means that a bond was formed between HA and HB. In addition, the plateau indicates that the bond is static. On the other hand, the distance between HB and HC has increased to 1.91 Å at large step values. This shows that HC has been ejected off following a successful collision between HA and HBC.

Similarly, in the dynamic trajectory, the distance between HA and HB decreases to approximately 0.75 Å after 0.3 s. This is indicative of hydrogen bond formation between HA and HB. In addition, the distance between HB and HC increases to 3.44 Å after HC is ejected following a successful collision.

One obvious difference between the two plots is that in the minimum energy path trajectory, the internuclear distance between HA and HB is constant indicating that the bond is static. However, in the dynamic trajectory the internuclear distance is oscillating showing the the bond is vibrating. This difference could be attributed to the fact that minimum energy path does not take oscillatory motion of atoms into account while dynamic trajectory does.

However, changing the initial conditions of the system such that rAB = 0.917743 Å, rBC = 0.907743 Å and pA=pBC=0:
[[File:01409469 01409469 Reversed values dynamic mep.jpg|thumb|center|800px|'''Figure 8.''' A graph of internuclear distance against steps using minimum energy path '''Figure 9.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In this case, a hydrogen bond of length 0.77-0.79 Å is formed between HB and HC while HA is repelled away to beyond 3.45 Å.

Additionally, from the internuclear distance plot it can be seen that the distance between HB and HC increases in a rectangular hyperbola fashion when calculated using the minimum energy path. On the other hand, the internuclear distance between HB and HC increases in a linear manner when dynamic calculation is applied. A possible reason for this difference is that in the minimum energy path, the atoms are not vibrating and as such reach the maximum distance and plateau at a faster rate. However, in the dynamic calculation, the atoms have momentum and as such move further away from each other.

=== Reactive and unreactive trajectories ===

With the initial positions r'''1''' = 0.74 and r'''2''' = 2.0, trajectories were run with the momenta combination shown below.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Contour plot !! Reactivity !! Description of the dynamics
|-
| -1.25 || -2.5 ||-99.018 ||[[File:01409469 trajectory1.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. There are no oscillations observed in HA-HB initially, indicating that the diatomic molecule is stable. A new bond is formed between HB and HC. Oscillatory behaviour is observed indicating that the bond is vibrating. On the other hand, HA is ejected away to beyond 10 Å.
|-
| -1.5 || -2.0 ||-100.456 ||[[File:01409469 trajectory2.jpg|thumb|center|400px]] ||Unreactive ||HC collides with the vibrating diatomic HA-HB. However, no bond is formed between HB and HC as the collision is unsuccessful. This is because the molecules lack sufficient kinetic energy to overcome the activation energy barrier and so no reaction occurs. HC is subsequently repelled to beyond 8 Å.
|-
| -1.5 || -2.5 ||-98.956 ||[[File:01409469 trajectory3.jpg|thumb|center|400px]] ||Reactive |||HC collides with HA-HB. The collision is successful and a bond is formed between HB and HC. This is because the atoms have sufficient kinetic energy to overcome the activation energy barrier and form a new HB-HC bond. HA is ejected away to beyond 10 Å. The new hydrogen bond formed is vibrating as seen by the oscillatory behaviour.
|-
| -2.5 || -5.0 ||-84.956 ||[[File:01409469 trajectory4.jpg|thumb|center|400px]] ||Unreactive ||HC collides with HA-HB. Initially the HA-HB bond is elongating, indicating that the hydrogen bond is weakening and going to break. However, after 0.5 s, the HA-HB bond length decreases and the HAHB bond is not broken. HC is repelled beyond 6 Å with very strong vibrations.
A reaction did not occur even though the atoms have sufficient kinetic energy to cross the activation energy barrier. This is because the vibrations are too strong. The system crosses the transition state region and a bond is formed in the product temporarily but then the system reverts back to the reactants. This is an example of barrier recrossing.
|-
| -2.5 || -5.2 ||-83.416 ||[[File:01409469 trajectory5.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. The collision is successful and a new hydrogen bond is formed between HB and HC with HA being ejected away. The products formed experience very strong vibrations as evidenced by the oscillatory motion which occur with a very high frequency.
|}

From the table, it can be seen that the chance of a reaction successfully occurring is not solely dependent on the amount of kinetic energy the atoms possess. Even if the atoms have sufficient kinetic energy to cross the activation energy barrier, barrier recrossing can occur if the atoms are too vibrationally excited.

=== Transition State Theory ===

Transition state theory provided the first theoretical attempt to determine absolute reaction rates. In this theory, a transition state separating the reactants from products is used to find an expression for the thermal rate constant.

The main assumptions of transition state theory are<ref name="ref2" />:

1. Electronic and nuclear motions are separated (equivalent to the Born-Oppenheimer approximation in quantum mechanics)

2. Reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution

3. Molecular systems that have crossed the transition state in the direction of products cannot turn around and reform reactants

4. In the transition state, motion along the reaction coordinate may be separated from the other motions and treated classically as a translation

5. Even in the absence of an equilibrium between reactant and product molecules, the transition states that are becoming products are distributed among their states according to the Maxwell-Boltzmann laws

The goal of transition state theory is to predict the rate constant, ''k'', of a reaction given a potential energy surface for the reaction.

However, transition state theory suffers from a set of limitations due to the assumptions stated above not holding true in reality.<ref name="ref3" />

Firstly, transition state theory assumes that barrier recrossing is negligible. However, as seen from the fourth trajectory in table 1, barrier recrossing occurred because the atoms were too vibrationally excited and the system reverted back to the reactants even though the system had already crossed the transition state region. In this situation, transition state theory would predict a faster rate of reaction compared to what was obtained experimentally.

Secondly, the transition state theory breaks down for systems with light atoms such as hydrogens. In such systems, quantum effects become crucial. Quantum effects tend to increase the rate of the reaction for two reasons. First, particles lie in energy levels which are higher than the bottom of the potential energy well (there is additional zero-point energy). Secondly, the particles can tunnel through the barrier. These two effects effectively lower the activation energy barrier. As a result, the rate of reaction predicted by transition state theory would be less than what is obtained experimentally.

Lastly, transition state theory does not hold true for systems which are far from equilibrium. In such cases, the assumption of a starting Boltzmann distribution breaks down.

== Exercise 2: F - H - H system ==

=== Potential energy surface inspection===

F + HA-HB → HA-F + HB

[[File:01409469 Surface_Plot F 3.jpg|thumb|center|400px|'''Figure 10.''' Potential energy surface for the reaction between F and H2]]

The reaction between F and H2 is an exothermic reaction. This means that the reactants are at a higher energy than the products. The overall reaction involves the breaking of a hydrogen bond and the formation of a hydrogen-fluorine bond. The HF bond formed is stronger than the hydrogen bond that was broken. This is because fluorine is more electronegative than hydrogen resulting in a polar HF bond. The negative enthalpy of H-F bond formation is greater than the positive enthalpy of hydrogen bond dissociation. Therefore, the reaction between F and H2 is an exothermic reaction.

HA + HB-F → HA-HB + F

[[File:01409469 Surface_Plot F 4.jpg|thumb|center|400px|'''Figure 11.''' Potential energy surface for the reaction between H and HF]]

The reaction between H and HF is an endothermic reaction. The products are at a higher energy level than the reactants. The overall reaction involves the breaking of a HF bond and the formation of a hydrogen bond. The formation of a homonuclear H2 molecule results in a non-polar hydrogen bond being formed. The HF bond broken is stronger than the hydrogen bond that was formed. The negative enthalpy of hydrogen bond formation is less than the positive enthalpy of hydrogen-fluorine dissociation. Therefore, the reaction between H and HF is endothermic.

=== Locating the transition state ===

A = HA, B = HB, C = F:

[[File:01409469 Surface_Plot F TS4.jpg|thumb|center|800px|'''Figure 12.''' Determining the position of the transition state for the reaction between F and H2]]

As mentioned previously, when the molecules are not at the transition state, a reaction path will be plotted. However, if the initial conditions of the system are set such that the trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

An initial estimate for the transition state was made by setting rAB = 0.74 which is the length of a hydrogen bond. The position of the transition state was further refined by increasing the number of decimal places such that the final value of rAB was found to be 0.74478 and rBC = 1.81079. This set of distances gave a reaction path that was represented by a point indicating that no trajectory was plotted and that the molecules remained at the transition state since their initial momentum was 0.

Moreover, the kinetic energy as well as the forces acting on the atoms was calculated to be 0, which is consistent with the transition state.

Therefore, the position of the transition state was found to be at rAB = 0.74478 and rBC = 1.81079 when pA = pBC = 0.

=== Determining the activation energy ===

[[File:01409469 activation energy r1.1 combined.jpg|thumb|center|800px|'''Figure 13.''' Determining the activation energy for the reaction between F and H2]]

By reducing the value of rBC by 0.01, the atoms are no longer at the transition state and would move towards the formation of products. The activation energy was found by calculating the change in total energy after the reaction. The activation energy of F + H2 → HF + H was found to be +23.279 kcal/ mol.

[[File:01409469 activation energy r2.1 combined.jpg|thumb|center|800px|'''Figure 14.''' Determining the activation energy for the reaction between FH and H]]

The activation energy of FH + H → F + H2 was found by calculating the small dip in potential energy. The activation energy was found to be +0.280 kcal/ mol.

=== Reaction dynamics of F + HA-HB → HA-F + HB ===

The set of initial conditions that result in a reactive trajectory for F + H2 is as follows: rHH = 0.74 Å, rHF = 1.79 Å and pHH = pHF = 0.

[[File:01409469 momentum r1.jpg|thumb|center|400px|'''Figure 15.''' Graph of momentum against time for the reaction between F and H2]]

Initially, the H2 molecule is travelling in a relatively straight line with very little oscillations as evidenced by the low amplitude of momentum between 0-2 s. Following successful collision between F and H2, the amplitude of momentum of the product, HF, increases dramatically. This is because the reaction between F and H2 is exothermic and the excess energy released causes the molecule to become vibrationally excited and the amplitude of oscillations in the product to increase in the form of H-F vibrations.

[[File:01409469 energy1 r1.jpg|thumb|center|400px|'''Figure 16.''' Graph of energy against time for the reaction between F and H2]]

The law of conservation of energy states that energy cannot be created or destroyed. It simply changes from one form to another. As can be seen from the graph in figure 16, total energy remains constant. In this reaction, potential energy is converted to kinetic energy. A decrease in potential energy of the atoms following a successful collision between F and H2 results in an equal increase in kinetic energy which manifests in the form of increased product (HF) oscillations.

This can be confirmed experimentally using infrared chemiluminescence.<ref name="ref2" /> Observation of radiation from excited product molecules is a clear indication that these molecules are produced in the excited state. If the excitation is primarily vibrational, the radiation will appear in the infrared region of the spectrum (between 3-15 μm). This technique has been used extensively in hydrogen halogen reactions which produce HX molecules in the excited state.

One such experiment involves the use of the arrested relaxation method.<ref name="ref2" /> In this technique, the reaction is carried out in a chamber whose walls are maintained at a very low temperature (20 K). This results in cryopumping of the chamber in that all the products and unreacted starting materials are condensed on the walls and only those product molecules that radiate before being pumped away are observed.

=== Polanyi's empirical rules ===

In a typical chemical reaction, there is an energetic barrier — a saddle point that the reactants must surmount to become products. Which form of energy is initially deposited in the reactants, translational or vibrational, is more efficacious in surmounting the activation energy barrier is one of the core topics in the area of reaction dynamics. The Polyani rules state that vibrational energy is more effective at promoting a late-barrier reaction (transition state resembles products). Conversely, translational energy is better at promoting an early-barrier reaction (transition state resembles the reactants).<ref name="ref4" />

The forward reaction between H2 and F is exothermic. Therefore, the transition state is closer to the reactants than to the products in energy. Since the transition state occurs early in an exothermic reaction, translational energy is more effective at promoting a chemical reaction than vibrational energy.

Conversely, the backward reaction between H and HF is endothermic. According to Hammond's postulate, the transition state resembles the products more than it resembles the reactants. Therefore, in an endothermic reaction, a late transition state is observed and vibrational energy is more effective at promoting a reaction than translational energy.

=== Promoting the forward reaction: F + HA-HB → HA-F + HB ===

1. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 2.95 and pHF = -0.5.

[[File:01409469 polyani 1+2.jpg|thumb|center|800px|'''Figure 17.''' Contour plot of the reaction between F and H2 when pHH = 2.95 '''Figure 18.''' Graph of momentum against time for the reaction between F and H2 when pHH = 2.95 (Left-right)]]

Vibrational energy is represented by pHH while translational energy is represented by pHF.

A molecule of H2 collides with F. However, the collision is unsuccessful and no reaction occurs. A reaction did not occur even though the molecules have sufficient kinetic energy to overcome the activation energy barrier. This is because most of the energy is deposited in the reactants as H-H vibrations as seen by the large oscillations in H2. Barrier recrossing has occured as a result of the large vibrational energy causing the system to revert back to the reactants even though the system has already crossed the transition state and a bond was formed between HB and F temporarily.

This appears to run counter to what one would expect to happen when more energy is put into the system. It is predicted that when more energy is put into the system, the reactants would possess more kinetic energy to overcome the activation energy barrier. However, as can be seen from the simulation above, an increase in the vibrational energy of the reactants is not effective at promoting the occurrence of an exothermic reaction with an early transition state. This is in accordance to Polyani's rules.

2. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 0.1 and pHF = -0.8.

[[File:01409469 polyani 3+4 2.jpg|thumb|center|800px|'''Figure 19.''' Contour plot of the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 '''Figure 20.''' Graph of momentum against time for the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 (Left-right)]]

The collision between a molecule of H2 and F is successful and a chemical reaction has occurred. A reaction has occurred even though the overall energy of the system has been significantly reduced by lowering the amount of vibrational energy in the reactants but slightly increasing the amount of translational energy deposited in the reactants. Therefore, H2 and F have more translational energy to move towards each other and form a H-F bond. The product formed is vibrationally excited as seen by the large amplitude of oscillations in the product HF which correspond to the strong H-F vibrations.

From the second simulation, it can be seen that an increase in translational energy in the reactants is more effective at promoting an exothermic reaction with an early transition state to occur, which again follows Polyani's rules.

=== Promoting the backward reaction: HA + HB-F → HA-HB + F ===

1. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 0.2 and pHH = -3.0.

[[File:01409469 polyani 5+6 .jpg|thumb|center|800px|'''Figure 21.''' Contour plot of the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 '''Figure 22.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 (Left-right)]]

An atom of HA collides with a molecule of HBF. However, the collision is unsuccessful and no chemical reaction occurs. This is because the reactants lack sufficient vibrational energy to surmount the activation energy barrier. Therefore, following an unsuccessful collision with HBF, HA simply bounces off and moves away at constant momentum. This is in agreement with Polyani's rules which state that vibrational energy is more effective than translational energy at promoting reactions with a late transition state, for example, endothermic chemical reactions.

2. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 7.1 and pHH = -0.5.

[[File:01409469 polyani 7+8 .jpg|thumb|center|800px|'''Figure 23.''' Contour plot of the reaction between HA + HB-F when pFH = 7.1 and pHH = -0.5 '''Figure 24.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 7.1 and pHH = -0.5 (Left-right)]]

Increasing the amount of vibrational energy in the reactants by increasing pFH resulted in a chemical reaction successfully occurring. This is because the reactants now have sufficient kinetic energy in the form of H-F vibrations to overcome the activation energy barrier. In addition, because the reaction between HA + HB-F is endothermic, energy is taken in from the surroundings during the reaction and the products are less vibrationally excited than the reactants. This can be seen in the lower amplitude of oscillations in the product formed.

In conclusion, the vibrational energy of the reactants should be increased to promote an endothermic reaction with a late transition state while the translational energy of the reactants should be increased to promote an exothermic reaction with an early transition state. All 4 simulations shown above obey Polyani's rules.

== References ==
<references>
<ref name="ref1">A. C. Vaucher, M. Reiher, ''J. Chem. Theory. Comput.'', 2018, '''14''', 3091-3099.</ref>
<ref name="ref2">J. I. Steinfield, J. S. Francisco, W. L. Hase, ''Chemical Kinetics and Dynamics'', Pearson, 1998.</ref>
<ref name="ref3">D. C. Elton, ''Transition State Theory for Physicists'', 2013.</ref>
<ref name="ref4">Z. Zhang, Y. Zhou, D. H. Zhang, ''J. Phys. Chem. Lett.'', 2012, '''3''', 3416-3419.</ref>
</references>

MRD:01409469

2019-05-10T16:00:50Z

Stc1917: /* Locating the transition state */

== Exercise 1: H + H2 system ==

=== Definition of a transition state ===

A potential energy profile results from plotting the potential versus reaction path.

[[File:Surface_Plot01409469_pic1.jpg|thumb|center|400px|'''Figure 1.''' A potential energy surface plot]]

The transition state is defined as the maximum on the minimum energy path linking reactants and products. This means that the transition state is a stationary point on the potential energy surface. Thus, at the transition state, the derivative of the potential energy is zero with respect to each Cartesian coordinate.

<math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>

However, a local minimum on the potential energy surface corresponding to a well would have the same <math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>. Therefore, to distinguish between a local minimum on the potential energy surface and the transition state, the second derivative, <math>\tfrac{\partial^2 V(r_i)}{\partial r_i ^2}\ </math> is determined. This can be calculated by finding the determinant of the Hessian matrix shown below.

:<math>\mathbf H =
\begin{bmatrix}
\dfrac{\partial^2 \ V(r_i)}{\partial \ x^2} & \dfrac{\partial^2 \ V(r_i)}{\partial \ x \partial \mathbf y} \\
\dfrac{\partial^2 \ V(r_i)}{\partial \ y \partial \mathbf x} & \dfrac{\partial^2 \ V(r_i)}{\partial \mathbf y^2}
\end{bmatrix}</math>

The Hessian matrix of a convex function is positive semi-definite. Refining this property allows one to test if a critical point (x,y) is a local maximum, local minimum or saddle point.<ref name="ref1" />

If <math>f</math> < 0 (exactly one negative eigenvalue), then (x,y) is an isolated local maximum.

If <math>f</math> > 0 (exactly one positive eigenvalue), then (x,y) is an isolated local minimum.

If the Hessian has both positive and negative eigenvalues then (x,y) is a saddle point.

=== Locating the transition state ===

[[File:01409469 combined pic 2 and 3.jpg|thumb|center|800px|'''Figure 2.''' A surface and contour plot of the trajectory when rAB=rBC=0.907743 Å and pA=pBC=0]]

When the molecules are not at the transition state, a surface plot of the conditions shows a trajectory being plotted. This means that the trajectory is "rolling" towards the products or reactants. However, if a trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

[[File:01409469_internuclearpic3.png|thumb|center|400px|'''Figure 3.''' A plot of internuclear distance against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

An initial estimate for the position of the transition state was made by observing the potential energy surface and was found to be 0.90 Å. At 0.90 Å, the internuclear distance against time plot showed that the internuclear distance was constantly oscillating. Therefore, the position of the transition state was further refined by increasing the number of decimal places until a horizontal line in the internuclear distance againt time plot was obtained. This means that the internuclear distance is constant and that the molecules are simply undergoing periodic symmetric vibration at the transition state. Therefore, the position of the transition state was determined to be 0.907743 Å.

[[File:01409469 combined energy pic.png|thumb|center|400px|'''Figure 4.''' A plot of energy against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

When rAB=rBC=0.907743 Å and pA=pBC=0, the kinetic energy is 0. This means that HA and HBC are stationary, which is as predicted when the trajectory is started at the transition state with no initial momentum. Furthermore, the forces acting on HA and HBC are 0 as should be the case in the transition state.

=== Calculation type: Dynamic and minimum energy path ===

The minimum energy path, is the lowest energy path from the reactants to the products. The minimum energy path is determined by following the mode corresponding to the negative Hessian eigenvalue in both directions along the steepest descent direction.<ref name="ref1" /> Following this path allows one to approach two valleys of stable structures. The resulting path generated is the minimum energy path. The reaction paths were chartered for a dynamic and a minimum energy path. The initial conditions used were that of the transition state except that the internuclear distance between HA and HB had increased to 0.917743 Å. The reaction path obtained using the two different calculation methods was different.

[[File:01409469 mep and dynamics.jpg|thumb|center|800px|'''Figure 5.''' A surface plot of the trajectory using minimum energy path '''Figure 6.''' A surface plot of the trajectory using dynamics (Left-right)]]

On the surface plot, the reaction path is depicted by the black line. Following the minimum energy path, the trajectory simply follows the valley floor to Ha+Hb-Hc. The reaction path appears to be a relatively straight line indicating that the atoms are not vibrating. In the minimum energy path, the reactants change into the products in an infinitely slow motion. This is done by having the momenta/velocities reset to zero in each time step. While the minimum energy path is useful in characterising a chemical reaction, it does not provide a realistic representation of the motion of the atoms during the reaction. This is because the mass and inertia of the atoms have been ignored.

Since the atoms simply follow the minimum energy path, they do not undergo slight displacements into surrounding valleys of higher potential energy. Furthermore, because the atoms are assumed to have no inertia, they do not undergo vibrations even if they were to be displaced to valleys of higher potential energy.

However, when dynamic calculation is performed, the reaction path is wavy indicating that the atoms are oscillating. This is because the atoms are displaced to regions of higher potential on the path from reactants to products. This leads to atom oscillation due to the inertia of the atoms.

'''Internuclear distance against time plot'''

Setting the initial system conditions such that rAB= 0.907743, rBC= 0.917743 and pA= pBC= 0:

[[File:01409469 Internuclear distance mep and dynamic.jpg|thumb|center|800px|'''Figure 6.''' A graph of internuclear distance against steps using minimum energy path '''Figure 7.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In the minimum energy path, the distance between HA and HB decreases from 0.90 Å to approximately 0.79 Å during the first 50 steps before plateauing off to a value of 0.74 Å which is the approximate length of a hydrogen bond. This means that a bond was formed between HA and HB. In addition, the plateau indicates that the bond is static. On the other hand, the distance between HB and HC has increased to 1.91 Å at large step values. This shows that HC has been ejected off following a successful collision between HA and HBC.

Similarly, in the dynamic trajectory, the distance between HA and HB decreases to approximately 0.75 Å after 0.3 s. This is indicative of hydrogen bond formation between HA and HB. In addition, the distance between HB and HC increases to 3.44 Å after HC is ejected following a successful collision.

One obvious difference between the two plots is that in the minimum energy path trajectory, the internuclear distance between HA and HB is constant indicating that the bond is static. However, in the dynamic trajectory the internuclear distance is oscillating showing the the bond is vibrating. This difference could be attributed to the fact that minimum energy path does not take oscillatory motion of atoms into account while dynamic trajectory does.

However, changing the initial conditions of the system such that rAB = 0.917743 Å, rBC = 0.907743 Å and pA=pBC=0:
[[File:01409469 01409469 Reversed values dynamic mep.jpg|thumb|center|800px|'''Figure 8.''' A graph of internuclear distance against steps using minimum energy path '''Figure 9.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In this case, a hydrogen bond of length 0.77-0.79 Å is formed between HB and HC while HA is repelled away to beyond 3.45 Å.

Additionally, from the internuclear distance plot it can be seen that the distance between HB and HC increases in a rectangular hyperbola fashion when calculated using the minimum energy path. On the other hand, the internuclear distance between HB and HC increases in a linear manner when dynamic calculation is applied. A possible reason for this difference is that in the minimum energy path, the atoms are not vibrating and as such reach the maximum distance and plateau at a faster rate. However, in the dynamic calculation, the atoms have momentum and as such move further away from each other.

=== Reactive and unreactive trajectories ===

With the initial positions r'''1''' = 0.74 and r'''2''' = 2.0, trajectories were run with the momenta combination shown below.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Contour plot !! Reactivity !! Description of the dynamics
|-
| -1.25 || -2.5 ||-99.018 ||[[File:01409469 trajectory1.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. There are no oscillations observed in HA-HB initially, indicating that the diatomic molecule is stable. A new bond is formed between HB and HC. Oscillatory behaviour is observed indicating that the bond is vibrating. On the other hand, HA is ejected away to beyond 10 Å.
|-
| -1.5 || -2.0 ||-100.456 ||[[File:01409469 trajectory2.jpg|thumb|center|400px]] ||Unreactive ||HC collides with the vibrating diatomic HA-HB. However, no bond is formed between HB and HC as the collision is unsuccessful. This is because the molecules lack sufficient kinetic energy to overcome the activation energy barrier and so no reaction occurs. HC is subsequently repelled to beyond 8 Å.
|-
| -1.5 || -2.5 ||-98.956 ||[[File:01409469 trajectory3.jpg|thumb|center|400px]] ||Reactive |||HC collides with HA-HB. The collision is successful and a bond is formed between HB and HC. This is because the atoms have sufficient kinetic energy to overcome the activation energy barrier and form a new HB-HC bond. HA is ejected away to beyond 10 Å. The new hydrogen bond formed is vibrating as seen by the oscillatory behaviour.
|-
| -2.5 || -5.0 ||-84.956 ||[[File:01409469 trajectory4.jpg|thumb|center|400px]] ||Unreactive ||HC collides with HA-HB. Initially the HA-HB bond is elongating, indicating that the hydrogen bond is weakening and going to break. However, after 0.5 s, the HA-HB bond length decreases and the HAHB bond is not broken. HC is repelled beyond 6 Å with very strong vibrations.
A reaction did not occur even though the atoms have sufficient kinetic energy to cross the activation energy barrier. This is because the vibrations are too strong. The system crosses the transition state region and a bond is formed in the product temporarily but then the system reverts back to the reactants. This is an example of barrier recrossing.
|-
| -2.5 || -5.2 ||-83.416 ||[[File:01409469 trajectory5.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. The collision is successful and a new hydrogen bond is formed between HB and HC with HA being ejected away. The products formed experience very strong vibrations as evidenced by the oscillatory motion which occur with a very high frequency.
|}

From the table, it can be seen that the chance of a reaction successfully occurring is not solely dependent on the amount of kinetic energy the atoms possess. Even if the atoms have sufficient kinetic energy to cross the activation energy barrier, barrier recrossing can occur if the atoms are too vibrationally excited.

=== Transition State Theory ===

Transition state theory provided the first theoretical attempt to determine absolute reaction rates. In this theory, a transition state separating the reactants from products is used to find an expression for the thermal rate constant.

The main assumptions of transition state theory are<ref name="ref2" />:

1. Electronic and nuclear motions are separated (equivalent to the Born-Oppenheimer approximation in quantum mechanics)

2. Reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution

3. Molecular systems that have crossed the transition state in the direction of products cannot turn around and reform reactants

4. In the transition state, motion along the reaction coordinate may be separated from the other motions and treated classically as a translation

5. Even in the absence of an equilibrium between reactant and product molecules, the transition states that are becoming products are distributed among their states according to the Maxwell-Boltzmann laws

The goal of transition state theory is to predict the rate constant, ''k'', of a reaction given a potential energy surface for the reaction.

However, transition state theory suffers from a set of limitations due to the assumptions stated above not holding true in reality.<ref name="ref3" />

Firstly, transition state theory assumes that barrier recrossing is negligible. However, as seen from the fourth trajectory in table 1, barrier recrossing occurred because the atoms were too vibrationally excited and the system reverted back to the reactants even though the system had already crossed the transition state region. In this situation, transition state theory would predict a faster rate of reaction compared to what was obtained experimentally.

Secondly, the transition state theory breaks down for systems with light atoms such as hydrogens. In such systems, quantum effects become crucial. Quantum effects tend to increase the rate of the reaction for two reasons. First, particles lie in energy levels which are higher than the bottom of the potential energy well (there is additional zero-point energy). Secondly, the particles can tunnel through the barrier. These two effects effectively lower the activation energy barrier. As a result, the rate of reaction predicted by transition state theory would be less than what is obtained experimentally.

Lastly, transition state theory does not hold true for systems which are far from equilibrium. In such cases, the assumption of a starting Boltzmann distribution breaks down.

== Exercise 2: F - H - H system ==

=== Potential energy surface inspection===

F + HA-HB → HA-F + HB

[[File:01409469 Surface_Plot F 3.jpg|thumb|center|400px|'''Figure 10.''' Potential energy surface for the reaction between F and H2]]

The reaction between F and H2 is an exothermic reaction. This means that the reactants are at a higher energy than the products. The overall reaction involves the breaking of a hydrogen bond and the formation of a hydrogen-fluorine bond. The HF bond formed is stronger than the hydrogen bond that was broken. This is because fluorine is more electronegative than hydrogen resulting in a polar HF bond. The negative enthalpy of H-F bond formation is greater than the positive enthalpy of hydrogen bond dissociation. Therefore, the reaction between F and H2 is an exothermic reaction.

HA + HB-F → HA-HB + F

[[File:01409469 Surface_Plot F 4.jpg|thumb|center|400px|'''Figure 11.''' Potential energy surface for the reaction between H and HF]]

The reaction between H and HF is an endothermic reaction. The products are at a higher energy level than the reactants. The overall reaction involves the breaking of a HF bond and the formation of a hydrogen bond. The formation of a homonuclear H2 molecule results in a non-polar hydrogen bond being formed. The HF bond broken is stronger than the hydrogen bond that was formed. The negative enthalpy of hydrogen bond formation is less than the positive enthalpy of hydrogen-fluorine dissociation. Therefore, the reaction between H and HF is endothermic.

=== Locating the transition state ===

A = HA, B = A = HB, C = F:

[[File:01409469 Surface_Plot F TS4.jpg|thumb|center|800px|'''Figure 12.''' Determining the position of the transition state for the reaction between F and H2]]

As mentioned previously, when the molecules are not at the transition state, a reaction path will be plotted. However, if the initial conditions of the system are set such that the trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

An initial estimate for the transition state was made by setting rAB = 0.74 which is the length of a hydrogen bond. The position of the transition state was further refined by increasing the number of decimal places such that the final value of rAB was found to be 0.74478 and rBC = 1.81079. This set of distances gave a reaction path that was represented by a point indicating that no trajectory was plotted and that the molecules remained at the transition state since their initial momentum was 0.

Moreover, the kinetic energy as well as the forces acting on the atoms was calculated to be 0, which is consistent with the transition state.

Therefore, the position of the transition state was found to be at rAB = 0.74478 and rBC = 1.81079 when pA = pBC = 0.

=== Determining the activation energy ===

[[File:01409469 activation energy r1.1 combined.jpg|thumb|center|800px|'''Figure 13.''' Determining the activation energy for the reaction between F and H2]]

By reducing the value of rBC by 0.01, the atoms are no longer at the transition state and would move towards the formation of products. The activation energy was found by calculating the change in total energy after the reaction. The activation energy of F + H2 → HF + H was found to be +23.279 kcal/ mol.

[[File:01409469 activation energy r2.1 combined.jpg|thumb|center|800px|'''Figure 14.''' Determining the activation energy for the reaction between FH and H]]

The activation energy of FH + H → F + H2 was found by calculating the small dip in potential energy. The activation energy was found to be +0.280 kcal/ mol.

=== Reaction dynamics of F + HA-HB → HA-F + HB ===

The set of initial conditions that result in a reactive trajectory for F + H2 is as follows: rHH = 0.74 Å, rHF = 1.79 Å and pHH = pHF = 0.

[[File:01409469 momentum r1.jpg|thumb|center|400px|'''Figure 15.''' Graph of momentum against time for the reaction between F and H2]]

Initially, the H2 molecule is travelling in a relatively straight line with very little oscillations as evidenced by the low amplitude of momentum between 0-2 s. Following successful collision between F and H2, the amplitude of momentum of the product, HF, increases dramatically. This is because the reaction between F and H2 is exothermic and the excess energy released causes the molecule to become vibrationally excited and the amplitude of oscillations in the product to increase in the form of H-F vibrations.

[[File:01409469 energy1 r1.jpg|thumb|center|400px|'''Figure 16.''' Graph of energy against time for the reaction between F and H2]]

The law of conservation of energy states that energy cannot be created or destroyed. It simply changes from one form to another. As can be seen from the graph in figure 16, total energy remains constant. In this reaction, potential energy is converted to kinetic energy. A decrease in potential energy of the atoms following a successful collision between F and H2 results in an equal increase in kinetic energy which manifests in the form of increased product (HF) oscillations.

This can be confirmed experimentally using infrared chemiluminescence.<ref name="ref2" /> Observation of radiation from excited product molecules is a clear indication that these molecules are produced in the excited state. If the excitation is primarily vibrational, the radiation will appear in the infrared region of the spectrum (between 3-15 μm). This technique has been used extensively in hydrogen halogen reactions which produce HX molecules in the excited state.

One such experiment involves the use of the arrested relaxation method.<ref name="ref2" /> In this technique, the reaction is carried out in a chamber whose walls are maintained at a very low temperature (20 K). This results in cryopumping of the chamber in that all the products and unreacted starting materials are condensed on the walls and only those product molecules that radiate before being pumped away are observed.

=== Polanyi's empirical rules ===

In a typical chemical reaction, there is an energetic barrier — a saddle point that the reactants must surmount to become products. Which form of energy is initially deposited in the reactants, translational or vibrational, is more efficacious in surmounting the activation energy barrier is one of the core topics in the area of reaction dynamics. The Polyani rules state that vibrational energy is more effective at promoting a late-barrier reaction (transition state resembles products). Conversely, translational energy is better at promoting an early-barrier reaction (transition state resembles the reactants).<ref name="ref4" />

The forward reaction between H2 and F is exothermic. Therefore, the transition state is closer to the reactants than to the products in energy. Since the transition state occurs early in an exothermic reaction, translational energy is more effective at promoting a chemical reaction than vibrational energy.

Conversely, the backward reaction between H and HF is endothermic. According to Hammond's postulate, the transition state resembles the products more than it resembles the reactants. Therefore, in an endothermic reaction, a late transition state is observed and vibrational energy is more effective at promoting a reaction than translational energy.

=== Promoting the forward reaction: F + HA-HB → HA-F + HB ===

1. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 2.95 and pHF = -0.5.

[[File:01409469 polyani 1+2.jpg|thumb|center|800px|'''Figure 17.''' Contour plot of the reaction between F and H2 when pHH = 2.95 '''Figure 18.''' Graph of momentum against time for the reaction between F and H2 when pHH = 2.95 (Left-right)]]

Vibrational energy is represented by pHH while translational energy is represented by pHF.

A molecule of H2 collides with F. However, the collision is unsuccessful and no reaction occurs. A reaction did not occur even though the molecules have sufficient kinetic energy to overcome the activation energy barrier. This is because most of the energy is deposited in the reactants as H-H vibrations as seen by the large oscillations in H2. Barrier recrossing has occured as a result of the large vibrational energy causing the system to revert back to the reactants even though the system has already crossed the transition state and a bond was formed between HB and F temporarily.

This appears to run counter to what one would expect to happen when more energy is put into the system. It is predicted that when more energy is put into the system, the reactants would possess more kinetic energy to overcome the activation energy barrier. However, as can be seen from the simulation above, an increase in the vibrational energy of the reactants is not effective at promoting the occurrence of an exothermic reaction with an early transition state. This is in accordance to Polyani's rules.

2. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 0.1 and pHF = -0.8.

[[File:01409469 polyani 3+4 2.jpg|thumb|center|800px|'''Figure 19.''' Contour plot of the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 '''Figure 20.''' Graph of momentum against time for the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 (Left-right)]]

The collision between a molecule of H2 and F is successful and a chemical reaction has occurred. A reaction has occurred even though the overall energy of the system has been significantly reduced by lowering the amount of vibrational energy in the reactants but slightly increasing the amount of translational energy deposited in the reactants. Therefore, H2 and F have more translational energy to move towards each other and form a H-F bond. The product formed is vibrationally excited as seen by the large amplitude of oscillations in the product HF which correspond to the strong H-F vibrations.

From the second simulation, it can be seen that an increase in translational energy in the reactants is more effective at promoting an exothermic reaction with an early transition state to occur, which again follows Polyani's rules.

=== Promoting the backward reaction: HA + HB-F → HA-HB + F ===

1. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 0.2 and pHH = -3.0.

[[File:01409469 polyani 5+6 .jpg|thumb|center|800px|'''Figure 21.''' Contour plot of the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 '''Figure 22.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 (Left-right)]]

An atom of HA collides with a molecule of HBF. However, the collision is unsuccessful and no chemical reaction occurs. This is because the reactants lack sufficient vibrational energy to surmount the activation energy barrier. Therefore, following an unsuccessful collision with HBF, HA simply bounces off and moves away at constant momentum. This is in agreement with Polyani's rules which state that vibrational energy is more effective than translational energy at promoting reactions with a late transition state, for example, endothermic chemical reactions.

2. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 7.1 and pHH = -0.5.

[[File:01409469 polyani 7+8 .jpg|thumb|center|800px|'''Figure 23.''' Contour plot of the reaction between HA + HB-F when pFH = 7.1 and pHH = -0.5 '''Figure 24.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 7.1 and pHH = -0.5 (Left-right)]]

Increasing the amount of vibrational energy in the reactants by increasing pFH resulted in a chemical reaction successfully occurring. This is because the reactants now have sufficient kinetic energy in the form of H-F vibrations to overcome the activation energy barrier. In addition, because the reaction between HA + HB-F is endothermic, energy is taken in from the surroundings during the reaction and the products are less vibrationally excited than the reactants. This can be seen in the lower amplitude of oscillations in the product formed.

In conclusion, the vibrational energy of the reactants should be increased to promote an endothermic reaction with a late transition state while the translational energy of the reactants should be increased to promote an exothermic reaction with an early transition state. All 4 simulations shown above obey Polyani's rules.

== References ==
<references>
<ref name="ref1">A. C. Vaucher, M. Reiher, ''J. Chem. Theory. Comput.'', 2018, '''14''', 3091-3099.</ref>
<ref name="ref2">J. I. Steinfield, J. S. Francisco, W. L. Hase, ''Chemical Kinetics and Dynamics'', Pearson, 1998.</ref>
<ref name="ref3">D. C. Elton, ''Transition State Theory for Physicists'', 2013.</ref>
<ref name="ref4">Z. Zhang, Y. Zhou, D. H. Zhang, ''J. Phys. Chem. Lett.'', 2012, '''3''', 3416-3419.</ref>
</references>

MRD:01409469

2019-05-10T15:56:12Z

Stc1917: /* Potential energy surface inspection */

== Exercise 1: H + H2 system ==

=== Definition of a transition state ===

A potential energy profile results from plotting the potential versus reaction path.

[[File:Surface_Plot01409469_pic1.jpg|thumb|center|400px|'''Figure 1.''' A potential energy surface plot]]

The transition state is defined as the maximum on the minimum energy path linking reactants and products. This means that the transition state is a stationary point on the potential energy surface. Thus, at the transition state, the derivative of the potential energy is zero with respect to each Cartesian coordinate.

<math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>

However, a local minimum on the potential energy surface corresponding to a well would have the same <math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>. Therefore, to distinguish between a local minimum on the potential energy surface and the transition state, the second derivative, <math>\tfrac{\partial^2 V(r_i)}{\partial r_i ^2}\ </math> is determined. This can be calculated by finding the determinant of the Hessian matrix shown below.

:<math>\mathbf H =
\begin{bmatrix}
\dfrac{\partial^2 \ V(r_i)}{\partial \ x^2} & \dfrac{\partial^2 \ V(r_i)}{\partial \ x \partial \mathbf y} \\
\dfrac{\partial^2 \ V(r_i)}{\partial \ y \partial \mathbf x} & \dfrac{\partial^2 \ V(r_i)}{\partial \mathbf y^2}
\end{bmatrix}</math>

The Hessian matrix of a convex function is positive semi-definite. Refining this property allows one to test if a critical point (x,y) is a local maximum, local minimum or saddle point.<ref name="ref1" />

If <math>f</math> < 0 (exactly one negative eigenvalue), then (x,y) is an isolated local maximum.

If <math>f</math> > 0 (exactly one positive eigenvalue), then (x,y) is an isolated local minimum.

If the Hessian has both positive and negative eigenvalues then (x,y) is a saddle point.

=== Locating the transition state ===

[[File:01409469 combined pic 2 and 3.jpg|thumb|center|800px|'''Figure 2.''' A surface and contour plot of the trajectory when rAB=rBC=0.907743 Å and pA=pBC=0]]

When the molecules are not at the transition state, a surface plot of the conditions shows a trajectory being plotted. This means that the trajectory is "rolling" towards the products or reactants. However, if a trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

[[File:01409469_internuclearpic3.png|thumb|center|400px|'''Figure 3.''' A plot of internuclear distance against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

An initial estimate for the position of the transition state was made by observing the potential energy surface and was found to be 0.90 Å. At 0.90 Å, the internuclear distance against time plot showed that the internuclear distance was constantly oscillating. Therefore, the position of the transition state was further refined by increasing the number of decimal places until a horizontal line in the internuclear distance againt time plot was obtained. This means that the internuclear distance is constant and that the molecules are simply undergoing periodic symmetric vibration at the transition state. Therefore, the position of the transition state was determined to be 0.907743 Å.

[[File:01409469 combined energy pic.png|thumb|center|400px|'''Figure 4.''' A plot of energy against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

When rAB=rBC=0.907743 Å and pA=pBC=0, the kinetic energy is 0. This means that HA and HBC are stationary, which is as predicted when the trajectory is started at the transition state with no initial momentum. Furthermore, the forces acting on HA and HBC are 0 as should be the case in the transition state.

=== Calculation type: Dynamic and minimum energy path ===

The minimum energy path, is the lowest energy path from the reactants to the products. The minimum energy path is determined by following the mode corresponding to the negative Hessian eigenvalue in both directions along the steepest descent direction.<ref name="ref1" /> Following this path allows one to approach two valleys of stable structures. The resulting path generated is the minimum energy path. The reaction paths were chartered for a dynamic and a minimum energy path. The initial conditions used were that of the transition state except that the internuclear distance between HA and HB had increased to 0.917743 Å. The reaction path obtained using the two different calculation methods was different.

[[File:01409469 mep and dynamics.jpg|thumb|center|800px|'''Figure 5.''' A surface plot of the trajectory using minimum energy path '''Figure 6.''' A surface plot of the trajectory using dynamics (Left-right)]]

On the surface plot, the reaction path is depicted by the black line. Following the minimum energy path, the trajectory simply follows the valley floor to Ha+Hb-Hc. The reaction path appears to be a relatively straight line indicating that the atoms are not vibrating. In the minimum energy path, the reactants change into the products in an infinitely slow motion. This is done by having the momenta/velocities reset to zero in each time step. While the minimum energy path is useful in characterising a chemical reaction, it does not provide a realistic representation of the motion of the atoms during the reaction. This is because the mass and inertia of the atoms have been ignored.

Since the atoms simply follow the minimum energy path, they do not undergo slight displacements into surrounding valleys of higher potential energy. Furthermore, because the atoms are assumed to have no inertia, they do not undergo vibrations even if they were to be displaced to valleys of higher potential energy.

However, when dynamic calculation is performed, the reaction path is wavy indicating that the atoms are oscillating. This is because the atoms are displaced to regions of higher potential on the path from reactants to products. This leads to atom oscillation due to the inertia of the atoms.

'''Internuclear distance against time plot'''

Setting the initial system conditions such that rAB= 0.907743, rBC= 0.917743 and pA= pBC= 0:

[[File:01409469 Internuclear distance mep and dynamic.jpg|thumb|center|800px|'''Figure 6.''' A graph of internuclear distance against steps using minimum energy path '''Figure 7.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In the minimum energy path, the distance between HA and HB decreases from 0.90 Å to approximately 0.79 Å during the first 50 steps before plateauing off to a value of 0.74 Å which is the approximate length of a hydrogen bond. This means that a bond was formed between HA and HB. In addition, the plateau indicates that the bond is static. On the other hand, the distance between HB and HC has increased to 1.91 Å at large step values. This shows that HC has been ejected off following a successful collision between HA and HBC.

Similarly, in the dynamic trajectory, the distance between HA and HB decreases to approximately 0.75 Å after 0.3 s. This is indicative of hydrogen bond formation between HA and HB. In addition, the distance between HB and HC increases to 3.44 Å after HC is ejected following a successful collision.

One obvious difference between the two plots is that in the minimum energy path trajectory, the internuclear distance between HA and HB is constant indicating that the bond is static. However, in the dynamic trajectory the internuclear distance is oscillating showing the the bond is vibrating. This difference could be attributed to the fact that minimum energy path does not take oscillatory motion of atoms into account while dynamic trajectory does.

However, changing the initial conditions of the system such that rAB = 0.917743 Å, rBC = 0.907743 Å and pA=pBC=0:
[[File:01409469 01409469 Reversed values dynamic mep.jpg|thumb|center|800px|'''Figure 8.''' A graph of internuclear distance against steps using minimum energy path '''Figure 9.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In this case, a hydrogen bond of length 0.77-0.79 Å is formed between HB and HC while HA is repelled away to beyond 3.45 Å.

Additionally, from the internuclear distance plot it can be seen that the distance between HB and HC increases in a rectangular hyperbola fashion when calculated using the minimum energy path. On the other hand, the internuclear distance between HB and HC increases in a linear manner when dynamic calculation is applied. A possible reason for this difference is that in the minimum energy path, the atoms are not vibrating and as such reach the maximum distance and plateau at a faster rate. However, in the dynamic calculation, the atoms have momentum and as such move further away from each other.

=== Reactive and unreactive trajectories ===

With the initial positions r'''1''' = 0.74 and r'''2''' = 2.0, trajectories were run with the momenta combination shown below.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Contour plot !! Reactivity !! Description of the dynamics
|-
| -1.25 || -2.5 ||-99.018 ||[[File:01409469 trajectory1.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. There are no oscillations observed in HA-HB initially, indicating that the diatomic molecule is stable. A new bond is formed between HB and HC. Oscillatory behaviour is observed indicating that the bond is vibrating. On the other hand, HA is ejected away to beyond 10 Å.
|-
| -1.5 || -2.0 ||-100.456 ||[[File:01409469 trajectory2.jpg|thumb|center|400px]] ||Unreactive ||HC collides with the vibrating diatomic HA-HB. However, no bond is formed between HB and HC as the collision is unsuccessful. This is because the molecules lack sufficient kinetic energy to overcome the activation energy barrier and so no reaction occurs. HC is subsequently repelled to beyond 8 Å.
|-
| -1.5 || -2.5 ||-98.956 ||[[File:01409469 trajectory3.jpg|thumb|center|400px]] ||Reactive |||HC collides with HA-HB. The collision is successful and a bond is formed between HB and HC. This is because the atoms have sufficient kinetic energy to overcome the activation energy barrier and form a new HB-HC bond. HA is ejected away to beyond 10 Å. The new hydrogen bond formed is vibrating as seen by the oscillatory behaviour.
|-
| -2.5 || -5.0 ||-84.956 ||[[File:01409469 trajectory4.jpg|thumb|center|400px]] ||Unreactive ||HC collides with HA-HB. Initially the HA-HB bond is elongating, indicating that the hydrogen bond is weakening and going to break. However, after 0.5 s, the HA-HB bond length decreases and the HAHB bond is not broken. HC is repelled beyond 6 Å with very strong vibrations.
A reaction did not occur even though the atoms have sufficient kinetic energy to cross the activation energy barrier. This is because the vibrations are too strong. The system crosses the transition state region and a bond is formed in the product temporarily but then the system reverts back to the reactants. This is an example of barrier recrossing.
|-
| -2.5 || -5.2 ||-83.416 ||[[File:01409469 trajectory5.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. The collision is successful and a new hydrogen bond is formed between HB and HC with HA being ejected away. The products formed experience very strong vibrations as evidenced by the oscillatory motion which occur with a very high frequency.
|}

From the table, it can be seen that the chance of a reaction successfully occurring is not solely dependent on the amount of kinetic energy the atoms possess. Even if the atoms have sufficient kinetic energy to cross the activation energy barrier, barrier recrossing can occur if the atoms are too vibrationally excited.

=== Transition State Theory ===

Transition state theory provided the first theoretical attempt to determine absolute reaction rates. In this theory, a transition state separating the reactants from products is used to find an expression for the thermal rate constant.

The main assumptions of transition state theory are<ref name="ref2" />:

1. Electronic and nuclear motions are separated (equivalent to the Born-Oppenheimer approximation in quantum mechanics)

2. Reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution

3. Molecular systems that have crossed the transition state in the direction of products cannot turn around and reform reactants

4. In the transition state, motion along the reaction coordinate may be separated from the other motions and treated classically as a translation

5. Even in the absence of an equilibrium between reactant and product molecules, the transition states that are becoming products are distributed among their states according to the Maxwell-Boltzmann laws

The goal of transition state theory is to predict the rate constant, ''k'', of a reaction given a potential energy surface for the reaction.

However, transition state theory suffers from a set of limitations due to the assumptions stated above not holding true in reality.<ref name="ref3" />

Firstly, transition state theory assumes that barrier recrossing is negligible. However, as seen from the fourth trajectory in table 1, barrier recrossing occurred because the atoms were too vibrationally excited and the system reverted back to the reactants even though the system had already crossed the transition state region. In this situation, transition state theory would predict a faster rate of reaction compared to what was obtained experimentally.

Secondly, the transition state theory breaks down for systems with light atoms such as hydrogens. In such systems, quantum effects become crucial. Quantum effects tend to increase the rate of the reaction for two reasons. First, particles lie in energy levels which are higher than the bottom of the potential energy well (there is additional zero-point energy). Secondly, the particles can tunnel through the barrier. These two effects effectively lower the activation energy barrier. As a result, the rate of reaction predicted by transition state theory would be less than what is obtained experimentally.

Lastly, transition state theory does not hold true for systems which are far from equilibrium. In such cases, the assumption of a starting Boltzmann distribution breaks down.

== Exercise 2: F - H - H system ==

=== Potential energy surface inspection===

F + HA-HB → HA-F + HB

[[File:01409469 Surface_Plot F 3.jpg|thumb|center|400px|'''Figure 10.''' Potential energy surface for the reaction between F and H2]]

The reaction between F and H2 is an exothermic reaction. This means that the reactants are at a higher energy than the products. The overall reaction involves the breaking of a hydrogen bond and the formation of a hydrogen-fluorine bond. The HF bond formed is stronger than the hydrogen bond that was broken. This is because fluorine is more electronegative than hydrogen resulting in a polar HF bond. The negative enthalpy of H-F bond formation is greater than the positive enthalpy of hydrogen bond dissociation. Therefore, the reaction between F and H2 is an exothermic reaction.

HA + HB-F → HA-HB + F

[[File:01409469 Surface_Plot F 4.jpg|thumb|center|400px|'''Figure 11.''' Potential energy surface for the reaction between H and HF]]

The reaction between H and HF is an endothermic reaction. The products are at a higher energy level than the reactants. The overall reaction involves the breaking of a HF bond and the formation of a hydrogen bond. The formation of a homonuclear H2 molecule results in a non-polar hydrogen bond being formed. The HF bond broken is stronger than the hydrogen bond that was formed. The negative enthalpy of hydrogen bond formation is less than the positive enthalpy of hydrogen-fluorine dissociation. Therefore, the reaction between H and HF is endothermic.

=== Locating the transition state ===

[[File:01409469 Surface_Plot F TS4.jpg|thumb|center|800px|'''Figure 12.''' Determining the position of the transition state for the reaction between F and H2]]

As mentioned previously, when the molecules are not at the transition state, a reaction path will be plotted. However, if the initial conditions of the system are set such that the trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

An initial estimate for the transition state was made by setting rAB = 0.74 which is the length of a hydrogen bond. The position of the transition state was further refined by increasing the number of decimal places such that the final value of rAB was found to be 0.74478 and rBC = 1.81079. This set of distances gave a reaction path that was represented by a point indicating that no trajectory was plotted and that the molecules remained at the transition state since their initial momentum was 0.

Moreover, the kinetic energy as well as the forces acting on the atoms was calculated to be 0, which is consistent with the transition state.

Therefore, the position of the transition state was found to be at rAB = 0.74478 and rBC = 1.81079 when pA = pBC = 0.

=== Determining the activation energy ===

[[File:01409469 activation energy r1.1 combined.jpg|thumb|center|800px|'''Figure 13.''' Determining the activation energy for the reaction between F and H2]]

By reducing the value of rBC by 0.01, the atoms are no longer at the transition state and would move towards the formation of products. The activation energy was found by calculating the change in total energy after the reaction. The activation energy of F + H2 → HF + H was found to be +23.279 kcal/ mol.

[[File:01409469 activation energy r2.1 combined.jpg|thumb|center|800px|'''Figure 14.''' Determining the activation energy for the reaction between FH and H]]

The activation energy of FH + H → F + H2 was found by calculating the small dip in potential energy. The activation energy was found to be +0.280 kcal/ mol.

=== Reaction dynamics of F + HA-HB → HA-F + HB ===

The set of initial conditions that result in a reactive trajectory for F + H2 is as follows: rHH = 0.74 Å, rHF = 1.79 Å and pHH = pHF = 0.

[[File:01409469 momentum r1.jpg|thumb|center|400px|'''Figure 15.''' Graph of momentum against time for the reaction between F and H2]]

Initially, the H2 molecule is travelling in a relatively straight line with very little oscillations as evidenced by the low amplitude of momentum between 0-2 s. Following successful collision between F and H2, the amplitude of momentum of the product, HF, increases dramatically. This is because the reaction between F and H2 is exothermic and the excess energy released causes the molecule to become vibrationally excited and the amplitude of oscillations in the product to increase in the form of H-F vibrations.

[[File:01409469 energy1 r1.jpg|thumb|center|400px|'''Figure 16.''' Graph of energy against time for the reaction between F and H2]]

The law of conservation of energy states that energy cannot be created or destroyed. It simply changes from one form to another. As can be seen from the graph in figure 16, total energy remains constant. In this reaction, potential energy is converted to kinetic energy. A decrease in potential energy of the atoms following a successful collision between F and H2 results in an equal increase in kinetic energy which manifests in the form of increased product (HF) oscillations.

This can be confirmed experimentally using infrared chemiluminescence.<ref name="ref2" /> Observation of radiation from excited product molecules is a clear indication that these molecules are produced in the excited state. If the excitation is primarily vibrational, the radiation will appear in the infrared region of the spectrum (between 3-15 μm). This technique has been used extensively in hydrogen halogen reactions which produce HX molecules in the excited state.

One such experiment involves the use of the arrested relaxation method.<ref name="ref2" /> In this technique, the reaction is carried out in a chamber whose walls are maintained at a very low temperature (20 K). This results in cryopumping of the chamber in that all the products and unreacted starting materials are condensed on the walls and only those product molecules that radiate before being pumped away are observed.

=== Polanyi's empirical rules ===

In a typical chemical reaction, there is an energetic barrier — a saddle point that the reactants must surmount to become products. Which form of energy is initially deposited in the reactants, translational or vibrational, is more efficacious in surmounting the activation energy barrier is one of the core topics in the area of reaction dynamics. The Polyani rules state that vibrational energy is more effective at promoting a late-barrier reaction (transition state resembles products). Conversely, translational energy is better at promoting an early-barrier reaction (transition state resembles the reactants).<ref name="ref4" />

The forward reaction between H2 and F is exothermic. Therefore, the transition state is closer to the reactants than to the products in energy. Since the transition state occurs early in an exothermic reaction, translational energy is more effective at promoting a chemical reaction than vibrational energy.

Conversely, the backward reaction between H and HF is endothermic. According to Hammond's postulate, the transition state resembles the products more than it resembles the reactants. Therefore, in an endothermic reaction, a late transition state is observed and vibrational energy is more effective at promoting a reaction than translational energy.

=== Promoting the forward reaction: F + HA-HB → HA-F + HB ===

1. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 2.95 and pHF = -0.5.

[[File:01409469 polyani 1+2.jpg|thumb|center|800px|'''Figure 17.''' Contour plot of the reaction between F and H2 when pHH = 2.95 '''Figure 18.''' Graph of momentum against time for the reaction between F and H2 when pHH = 2.95 (Left-right)]]

Vibrational energy is represented by pHH while translational energy is represented by pHF.

A molecule of H2 collides with F. However, the collision is unsuccessful and no reaction occurs. A reaction did not occur even though the molecules have sufficient kinetic energy to overcome the activation energy barrier. This is because most of the energy is deposited in the reactants as H-H vibrations as seen by the large oscillations in H2. Barrier recrossing has occured as a result of the large vibrational energy causing the system to revert back to the reactants even though the system has already crossed the transition state and a bond was formed between HB and F temporarily.

This appears to run counter to what one would expect to happen when more energy is put into the system. It is predicted that when more energy is put into the system, the reactants would possess more kinetic energy to overcome the activation energy barrier. However, as can be seen from the simulation above, an increase in the vibrational energy of the reactants is not effective at promoting the occurrence of an exothermic reaction with an early transition state. This is in accordance to Polyani's rules.

2. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 0.1 and pHF = -0.8.

[[File:01409469 polyani 3+4 2.jpg|thumb|center|800px|'''Figure 19.''' Contour plot of the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 '''Figure 20.''' Graph of momentum against time for the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 (Left-right)]]

The collision between a molecule of H2 and F is successful and a chemical reaction has occurred. A reaction has occurred even though the overall energy of the system has been significantly reduced by lowering the amount of vibrational energy in the reactants but slightly increasing the amount of translational energy deposited in the reactants. Therefore, H2 and F have more translational energy to move towards each other and form a H-F bond. The product formed is vibrationally excited as seen by the large amplitude of oscillations in the product HF which correspond to the strong H-F vibrations.

From the second simulation, it can be seen that an increase in translational energy in the reactants is more effective at promoting an exothermic reaction with an early transition state to occur, which again follows Polyani's rules.

=== Promoting the backward reaction: HA + HB-F → HA-HB + F ===

1. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 0.2 and pHH = -3.0.

[[File:01409469 polyani 5+6 .jpg|thumb|center|800px|'''Figure 21.''' Contour plot of the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 '''Figure 22.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 (Left-right)]]

An atom of HA collides with a molecule of HBF. However, the collision is unsuccessful and no chemical reaction occurs. This is because the reactants lack sufficient vibrational energy to surmount the activation energy barrier. Therefore, following an unsuccessful collision with HBF, HA simply bounces off and moves away at constant momentum. This is in agreement with Polyani's rules which state that vibrational energy is more effective than translational energy at promoting reactions with a late transition state, for example, endothermic chemical reactions.

2. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 7.1 and pHH = -0.5.

[[File:01409469 polyani 7+8 .jpg|thumb|center|800px|'''Figure 23.''' Contour plot of the reaction between HA + HB-F when pFH = 7.1 and pHH = -0.5 '''Figure 24.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 7.1 and pHH = -0.5 (Left-right)]]

Increasing the amount of vibrational energy in the reactants by increasing pFH resulted in a chemical reaction successfully occurring. This is because the reactants now have sufficient kinetic energy in the form of H-F vibrations to overcome the activation energy barrier. In addition, because the reaction between HA + HB-F is endothermic, energy is taken in from the surroundings during the reaction and the products are less vibrationally excited than the reactants. This can be seen in the lower amplitude of oscillations in the product formed.

In conclusion, the vibrational energy of the reactants should be increased to promote an endothermic reaction with a late transition state while the translational energy of the reactants should be increased to promote an exothermic reaction with an early transition state. All 4 simulations shown above obey Polyani's rules.

== References ==
<references>
<ref name="ref1">A. C. Vaucher, M. Reiher, ''J. Chem. Theory. Comput.'', 2018, '''14''', 3091-3099.</ref>
<ref name="ref2">J. I. Steinfield, J. S. Francisco, W. L. Hase, ''Chemical Kinetics and Dynamics'', Pearson, 1998.</ref>
<ref name="ref3">D. C. Elton, ''Transition State Theory for Physicists'', 2013.</ref>
<ref name="ref4">Z. Zhang, Y. Zhou, D. H. Zhang, ''J. Phys. Chem. Lett.'', 2012, '''3''', 3416-3419.</ref>
</references>

MRD:01409469

2019-05-10T15:51:34Z

Stc1917: /* Transition State Theory */

== Exercise 1: H + H2 system ==

=== Definition of a transition state ===

A potential energy profile results from plotting the potential versus reaction path.

[[File:Surface_Plot01409469_pic1.jpg|thumb|center|400px|'''Figure 1.''' A potential energy surface plot]]

The transition state is defined as the maximum on the minimum energy path linking reactants and products. This means that the transition state is a stationary point on the potential energy surface. Thus, at the transition state, the derivative of the potential energy is zero with respect to each Cartesian coordinate.

<math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>

However, a local minimum on the potential energy surface corresponding to a well would have the same <math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>. Therefore, to distinguish between a local minimum on the potential energy surface and the transition state, the second derivative, <math>\tfrac{\partial^2 V(r_i)}{\partial r_i ^2}\ </math> is determined. This can be calculated by finding the determinant of the Hessian matrix shown below.

:<math>\mathbf H =
\begin{bmatrix}
\dfrac{\partial^2 \ V(r_i)}{\partial \ x^2} & \dfrac{\partial^2 \ V(r_i)}{\partial \ x \partial \mathbf y} \\
\dfrac{\partial^2 \ V(r_i)}{\partial \ y \partial \mathbf x} & \dfrac{\partial^2 \ V(r_i)}{\partial \mathbf y^2}
\end{bmatrix}</math>

The Hessian matrix of a convex function is positive semi-definite. Refining this property allows one to test if a critical point (x,y) is a local maximum, local minimum or saddle point.<ref name="ref1" />

If <math>f</math> < 0 (exactly one negative eigenvalue), then (x,y) is an isolated local maximum.

If <math>f</math> > 0 (exactly one positive eigenvalue), then (x,y) is an isolated local minimum.

If the Hessian has both positive and negative eigenvalues then (x,y) is a saddle point.

=== Locating the transition state ===

[[File:01409469 combined pic 2 and 3.jpg|thumb|center|800px|'''Figure 2.''' A surface and contour plot of the trajectory when rAB=rBC=0.907743 Å and pA=pBC=0]]

When the molecules are not at the transition state, a surface plot of the conditions shows a trajectory being plotted. This means that the trajectory is "rolling" towards the products or reactants. However, if a trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

[[File:01409469_internuclearpic3.png|thumb|center|400px|'''Figure 3.''' A plot of internuclear distance against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

An initial estimate for the position of the transition state was made by observing the potential energy surface and was found to be 0.90 Å. At 0.90 Å, the internuclear distance against time plot showed that the internuclear distance was constantly oscillating. Therefore, the position of the transition state was further refined by increasing the number of decimal places until a horizontal line in the internuclear distance againt time plot was obtained. This means that the internuclear distance is constant and that the molecules are simply undergoing periodic symmetric vibration at the transition state. Therefore, the position of the transition state was determined to be 0.907743 Å.

[[File:01409469 combined energy pic.png|thumb|center|400px|'''Figure 4.''' A plot of energy against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

When rAB=rBC=0.907743 Å and pA=pBC=0, the kinetic energy is 0. This means that HA and HBC are stationary, which is as predicted when the trajectory is started at the transition state with no initial momentum. Furthermore, the forces acting on HA and HBC are 0 as should be the case in the transition state.

=== Calculation type: Dynamic and minimum energy path ===

The minimum energy path, is the lowest energy path from the reactants to the products. The minimum energy path is determined by following the mode corresponding to the negative Hessian eigenvalue in both directions along the steepest descent direction.<ref name="ref1" /> Following this path allows one to approach two valleys of stable structures. The resulting path generated is the minimum energy path. The reaction paths were chartered for a dynamic and a minimum energy path. The initial conditions used were that of the transition state except that the internuclear distance between HA and HB had increased to 0.917743 Å. The reaction path obtained using the two different calculation methods was different.

[[File:01409469 mep and dynamics.jpg|thumb|center|800px|'''Figure 5.''' A surface plot of the trajectory using minimum energy path '''Figure 6.''' A surface plot of the trajectory using dynamics (Left-right)]]

On the surface plot, the reaction path is depicted by the black line. Following the minimum energy path, the trajectory simply follows the valley floor to Ha+Hb-Hc. The reaction path appears to be a relatively straight line indicating that the atoms are not vibrating. In the minimum energy path, the reactants change into the products in an infinitely slow motion. This is done by having the momenta/velocities reset to zero in each time step. While the minimum energy path is useful in characterising a chemical reaction, it does not provide a realistic representation of the motion of the atoms during the reaction. This is because the mass and inertia of the atoms have been ignored.

Since the atoms simply follow the minimum energy path, they do not undergo slight displacements into surrounding valleys of higher potential energy. Furthermore, because the atoms are assumed to have no inertia, they do not undergo vibrations even if they were to be displaced to valleys of higher potential energy.

However, when dynamic calculation is performed, the reaction path is wavy indicating that the atoms are oscillating. This is because the atoms are displaced to regions of higher potential on the path from reactants to products. This leads to atom oscillation due to the inertia of the atoms.

'''Internuclear distance against time plot'''

Setting the initial system conditions such that rAB= 0.907743, rBC= 0.917743 and pA= pBC= 0:

[[File:01409469 Internuclear distance mep and dynamic.jpg|thumb|center|800px|'''Figure 6.''' A graph of internuclear distance against steps using minimum energy path '''Figure 7.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In the minimum energy path, the distance between HA and HB decreases from 0.90 Å to approximately 0.79 Å during the first 50 steps before plateauing off to a value of 0.74 Å which is the approximate length of a hydrogen bond. This means that a bond was formed between HA and HB. In addition, the plateau indicates that the bond is static. On the other hand, the distance between HB and HC has increased to 1.91 Å at large step values. This shows that HC has been ejected off following a successful collision between HA and HBC.

Similarly, in the dynamic trajectory, the distance between HA and HB decreases to approximately 0.75 Å after 0.3 s. This is indicative of hydrogen bond formation between HA and HB. In addition, the distance between HB and HC increases to 3.44 Å after HC is ejected following a successful collision.

One obvious difference between the two plots is that in the minimum energy path trajectory, the internuclear distance between HA and HB is constant indicating that the bond is static. However, in the dynamic trajectory the internuclear distance is oscillating showing the the bond is vibrating. This difference could be attributed to the fact that minimum energy path does not take oscillatory motion of atoms into account while dynamic trajectory does.

However, changing the initial conditions of the system such that rAB = 0.917743 Å, rBC = 0.907743 Å and pA=pBC=0:
[[File:01409469 01409469 Reversed values dynamic mep.jpg|thumb|center|800px|'''Figure 8.''' A graph of internuclear distance against steps using minimum energy path '''Figure 9.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In this case, a hydrogen bond of length 0.77-0.79 Å is formed between HB and HC while HA is repelled away to beyond 3.45 Å.

Additionally, from the internuclear distance plot it can be seen that the distance between HB and HC increases in a rectangular hyperbola fashion when calculated using the minimum energy path. On the other hand, the internuclear distance between HB and HC increases in a linear manner when dynamic calculation is applied. A possible reason for this difference is that in the minimum energy path, the atoms are not vibrating and as such reach the maximum distance and plateau at a faster rate. However, in the dynamic calculation, the atoms have momentum and as such move further away from each other.

=== Reactive and unreactive trajectories ===

With the initial positions r'''1''' = 0.74 and r'''2''' = 2.0, trajectories were run with the momenta combination shown below.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Contour plot !! Reactivity !! Description of the dynamics
|-
| -1.25 || -2.5 ||-99.018 ||[[File:01409469 trajectory1.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. There are no oscillations observed in HA-HB initially, indicating that the diatomic molecule is stable. A new bond is formed between HB and HC. Oscillatory behaviour is observed indicating that the bond is vibrating. On the other hand, HA is ejected away to beyond 10 Å.
|-
| -1.5 || -2.0 ||-100.456 ||[[File:01409469 trajectory2.jpg|thumb|center|400px]] ||Unreactive ||HC collides with the vibrating diatomic HA-HB. However, no bond is formed between HB and HC as the collision is unsuccessful. This is because the molecules lack sufficient kinetic energy to overcome the activation energy barrier and so no reaction occurs. HC is subsequently repelled to beyond 8 Å.
|-
| -1.5 || -2.5 ||-98.956 ||[[File:01409469 trajectory3.jpg|thumb|center|400px]] ||Reactive |||HC collides with HA-HB. The collision is successful and a bond is formed between HB and HC. This is because the atoms have sufficient kinetic energy to overcome the activation energy barrier and form a new HB-HC bond. HA is ejected away to beyond 10 Å. The new hydrogen bond formed is vibrating as seen by the oscillatory behaviour.
|-
| -2.5 || -5.0 ||-84.956 ||[[File:01409469 trajectory4.jpg|thumb|center|400px]] ||Unreactive ||HC collides with HA-HB. Initially the HA-HB bond is elongating, indicating that the hydrogen bond is weakening and going to break. However, after 0.5 s, the HA-HB bond length decreases and the HAHB bond is not broken. HC is repelled beyond 6 Å with very strong vibrations.
A reaction did not occur even though the atoms have sufficient kinetic energy to cross the activation energy barrier. This is because the vibrations are too strong. The system crosses the transition state region and a bond is formed in the product temporarily but then the system reverts back to the reactants. This is an example of barrier recrossing.
|-
| -2.5 || -5.2 ||-83.416 ||[[File:01409469 trajectory5.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. The collision is successful and a new hydrogen bond is formed between HB and HC with HA being ejected away. The products formed experience very strong vibrations as evidenced by the oscillatory motion which occur with a very high frequency.
|}

From the table, it can be seen that the chance of a reaction successfully occurring is not solely dependent on the amount of kinetic energy the atoms possess. Even if the atoms have sufficient kinetic energy to cross the activation energy barrier, barrier recrossing can occur if the atoms are too vibrationally excited.

=== Transition State Theory ===

Transition state theory provided the first theoretical attempt to determine absolute reaction rates. In this theory, a transition state separating the reactants from products is used to find an expression for the thermal rate constant.

The main assumptions of transition state theory are<ref name="ref2" />:

1. Electronic and nuclear motions are separated (equivalent to the Born-Oppenheimer approximation in quantum mechanics)

2. Reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution

3. Molecular systems that have crossed the transition state in the direction of products cannot turn around and reform reactants

4. In the transition state, motion along the reaction coordinate may be separated from the other motions and treated classically as a translation

5. Even in the absence of an equilibrium between reactant and product molecules, the transition states that are becoming products are distributed among their states according to the Maxwell-Boltzmann laws

The goal of transition state theory is to predict the rate constant, ''k'', of a reaction given a potential energy surface for the reaction.

However, transition state theory suffers from a set of limitations due to the assumptions stated above not holding true in reality.<ref name="ref3" />

Firstly, transition state theory assumes that barrier recrossing is negligible. However, as seen from the fourth trajectory in table 1, barrier recrossing occurred because the atoms were too vibrationally excited and the system reverted back to the reactants even though the system had already crossed the transition state region. In this situation, transition state theory would predict a faster rate of reaction compared to what was obtained experimentally.

Secondly, the transition state theory breaks down for systems with light atoms such as hydrogens. In such systems, quantum effects become crucial. Quantum effects tend to increase the rate of the reaction for two reasons. First, particles lie in energy levels which are higher than the bottom of the potential energy well (there is additional zero-point energy). Secondly, the particles can tunnel through the barrier. These two effects effectively lower the activation energy barrier. As a result, the rate of reaction predicted by transition state theory would be less than what is obtained experimentally.

Lastly, transition state theory does not hold true for systems which are far from equilibrium. In such cases, the assumption of a starting Boltzmann distribution breaks down.

== Exercise 2: F - H - H system ==

=== Potential energy surface inspection===

F + HA-HB → HA-F + HB

[[File:01409469 Surface_Plot F 3.jpg|thumb|center|400px|'''Figure 10.''' Potential energy surface for the reaction between F and H2]]

The reaction between F and H2 is an exothermic reaction. This means that the reactants are of a higher energy than the products. The overall reaction involves the breaking of a hydrogen bond and the formation of a hydrogen-fluorine bond. The HF bond formed is stronger than the hydrogen bond that was broken. This is because fluorine is more electronegative than hydrogen resulting in a polar HF bond. The negative enthalpy of H-F bond formation is greater than the positive enthalpy of hydrogen bond dissociation. Therefore, the reaction between F and H2 is an exothermic reaction.

HA + HB-F → HA-HB + F

[[File:01409469 Surface_Plot F 4.jpg|thumb|center|400px|'''Figure 11.''' Potential energy surface for the reaction between H and HF]]

The reaction between H and HF is an endothermic reaction. The products are at a higher energy level than the reactants. The overall reaction involves the breaking of a HF bond and the formation of a hydrogen bond. The formation of a homonuclear H2 molecule results in a non-polar hydrogen bond being formed. The HF bond broken is stronger than the hydrogen bond that was formed. The negative enthalpy of hydrogen bond formation is less than the positive enthalpy of hydrogen-fluorine dissociation. Therefore, the reaction between H and HF is endothermic.

=== Locating the transition state ===

[[File:01409469 Surface_Plot F TS4.jpg|thumb|center|800px|'''Figure 12.''' Determining the position of the transition state for the reaction between F and H2]]

As mentioned previously, when the molecules are not at the transition state, a reaction path will be plotted. However, if the initial conditions of the system are set such that the trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

An initial estimate for the transition state was made by setting rAB = 0.74 which is the length of a hydrogen bond. The position of the transition state was further refined by increasing the number of decimal places such that the final value of rAB was found to be 0.74478 and rBC = 1.81079. This set of distances gave a reaction path that was represented by a point indicating that no trajectory was plotted and that the molecules remained at the transition state since their initial momentum was 0.

Moreover, the kinetic energy as well as the forces acting on the atoms was calculated to be 0, which is consistent with the transition state.

Therefore, the position of the transition state was found to be at rAB = 0.74478 and rBC = 1.81079 when pA = pBC = 0.

=== Determining the activation energy ===

[[File:01409469 activation energy r1.1 combined.jpg|thumb|center|800px|'''Figure 13.''' Determining the activation energy for the reaction between F and H2]]

By reducing the value of rBC by 0.01, the atoms are no longer at the transition state and would move towards the formation of products. The activation energy was found by calculating the change in total energy after the reaction. The activation energy of F + H2 → HF + H was found to be +23.279 kcal/ mol.

[[File:01409469 activation energy r2.1 combined.jpg|thumb|center|800px|'''Figure 14.''' Determining the activation energy for the reaction between FH and H]]

The activation energy of FH + H → F + H2 was found by calculating the small dip in potential energy. The activation energy was found to be +0.280 kcal/ mol.

=== Reaction dynamics of F + HA-HB → HA-F + HB ===

The set of initial conditions that result in a reactive trajectory for F + H2 is as follows: rHH = 0.74 Å, rHF = 1.79 Å and pHH = pHF = 0.

[[File:01409469 momentum r1.jpg|thumb|center|400px|'''Figure 15.''' Graph of momentum against time for the reaction between F and H2]]

Initially, the H2 molecule is travelling in a relatively straight line with very little oscillations as evidenced by the low amplitude of momentum between 0-2 s. Following successful collision between F and H2, the amplitude of momentum of the product, HF, increases dramatically. This is because the reaction between F and H2 is exothermic and the excess energy released causes the molecule to become vibrationally excited and the amplitude of oscillations in the product to increase in the form of H-F vibrations.

[[File:01409469 energy1 r1.jpg|thumb|center|400px|'''Figure 16.''' Graph of energy against time for the reaction between F and H2]]

The law of conservation of energy states that energy cannot be created or destroyed. It simply changes from one form to another. As can be seen from the graph in figure 16, total energy remains constant. In this reaction, potential energy is converted to kinetic energy. A decrease in potential energy of the atoms following a successful collision between F and H2 results in an equal increase in kinetic energy which manifests in the form of increased product (HF) oscillations.

This can be confirmed experimentally using infrared chemiluminescence.<ref name="ref2" /> Observation of radiation from excited product molecules is a clear indication that these molecules are produced in the excited state. If the excitation is primarily vibrational, the radiation will appear in the infrared region of the spectrum (between 3-15 μm). This technique has been used extensively in hydrogen halogen reactions which produce HX molecules in the excited state.

One such experiment involves the use of the arrested relaxation method.<ref name="ref2" /> In this technique, the reaction is carried out in a chamber whose walls are maintained at a very low temperature (20 K). This results in cryopumping of the chamber in that all the products and unreacted starting materials are condensed on the walls and only those product molecules that radiate before being pumped away are observed.

=== Polanyi's empirical rules ===

In a typical chemical reaction, there is an energetic barrier — a saddle point that the reactants must surmount to become products. Which form of energy is initially deposited in the reactants, translational or vibrational, is more efficacious in surmounting the activation energy barrier is one of the core topics in the area of reaction dynamics. The Polyani rules state that vibrational energy is more effective at promoting a late-barrier reaction (transition state resembles products). Conversely, translational energy is better at promoting an early-barrier reaction (transition state resembles the reactants).<ref name="ref4" />

The forward reaction between H2 and F is exothermic. Therefore, the transition state is closer to the reactants than to the products in energy. Since the transition state occurs early in an exothermic reaction, translational energy is more effective at promoting a chemical reaction than vibrational energy.

Conversely, the backward reaction between H and HF is endothermic. According to Hammond's postulate, the transition state resembles the products more than it resembles the reactants. Therefore, in an endothermic reaction, a late transition state is observed and vibrational energy is more effective at promoting a reaction than translational energy.

=== Promoting the forward reaction: F + HA-HB → HA-F + HB ===

1. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 2.95 and pHF = -0.5.

[[File:01409469 polyani 1+2.jpg|thumb|center|800px|'''Figure 17.''' Contour plot of the reaction between F and H2 when pHH = 2.95 '''Figure 18.''' Graph of momentum against time for the reaction between F and H2 when pHH = 2.95 (Left-right)]]

Vibrational energy is represented by pHH while translational energy is represented by pHF.

A molecule of H2 collides with F. However, the collision is unsuccessful and no reaction occurs. A reaction did not occur even though the molecules have sufficient kinetic energy to overcome the activation energy barrier. This is because most of the energy is deposited in the reactants as H-H vibrations as seen by the large oscillations in H2. Barrier recrossing has occured as a result of the large vibrational energy causing the system to revert back to the reactants even though the system has already crossed the transition state and a bond was formed between HB and F temporarily.

This appears to run counter to what one would expect to happen when more energy is put into the system. It is predicted that when more energy is put into the system, the reactants would possess more kinetic energy to overcome the activation energy barrier. However, as can be seen from the simulation above, an increase in the vibrational energy of the reactants is not effective at promoting the occurrence of an exothermic reaction with an early transition state. This is in accordance to Polyani's rules.

2. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 0.1 and pHF = -0.8.

[[File:01409469 polyani 3+4 2.jpg|thumb|center|800px|'''Figure 19.''' Contour plot of the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 '''Figure 20.''' Graph of momentum against time for the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 (Left-right)]]

The collision between a molecule of H2 and F is successful and a chemical reaction has occurred. A reaction has occurred even though the overall energy of the system has been significantly reduced by lowering the amount of vibrational energy in the reactants but slightly increasing the amount of translational energy deposited in the reactants. Therefore, H2 and F have more translational energy to move towards each other and form a H-F bond. The product formed is vibrationally excited as seen by the large amplitude of oscillations in the product HF which correspond to the strong H-F vibrations.

From the second simulation, it can be seen that an increase in translational energy in the reactants is more effective at promoting an exothermic reaction with an early transition state to occur, which again follows Polyani's rules.

=== Promoting the backward reaction: HA + HB-F → HA-HB + F ===

1. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 0.2 and pHH = -3.0.

[[File:01409469 polyani 5+6 .jpg|thumb|center|800px|'''Figure 21.''' Contour plot of the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 '''Figure 22.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 (Left-right)]]

An atom of HA collides with a molecule of HBF. However, the collision is unsuccessful and no chemical reaction occurs. This is because the reactants lack sufficient vibrational energy to surmount the activation energy barrier. Therefore, following an unsuccessful collision with HBF, HA simply bounces off and moves away at constant momentum. This is in agreement with Polyani's rules which state that vibrational energy is more effective than translational energy at promoting reactions with a late transition state, for example, endothermic chemical reactions.

2. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 7.1 and pHH = -0.5.

[[File:01409469 polyani 7+8 .jpg|thumb|center|800px|'''Figure 23.''' Contour plot of the reaction between HA + HB-F when pFH = 7.1 and pHH = -0.5 '''Figure 24.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 7.1 and pHH = -0.5 (Left-right)]]

Increasing the amount of vibrational energy in the reactants by increasing pFH resulted in a chemical reaction successfully occurring. This is because the reactants now have sufficient kinetic energy in the form of H-F vibrations to overcome the activation energy barrier. In addition, because the reaction between HA + HB-F is endothermic, energy is taken in from the surroundings during the reaction and the products are less vibrationally excited than the reactants. This can be seen in the lower amplitude of oscillations in the product formed.

In conclusion, the vibrational energy of the reactants should be increased to promote an endothermic reaction with a late transition state while the translational energy of the reactants should be increased to promote an exothermic reaction with an early transition state. All 4 simulations shown above obey Polyani's rules.

== References ==
<references>
<ref name="ref1">A. C. Vaucher, M. Reiher, ''J. Chem. Theory. Comput.'', 2018, '''14''', 3091-3099.</ref>
<ref name="ref2">J. I. Steinfield, J. S. Francisco, W. L. Hase, ''Chemical Kinetics and Dynamics'', Pearson, 1998.</ref>
<ref name="ref3">D. C. Elton, ''Transition State Theory for Physicists'', 2013.</ref>
<ref name="ref4">Z. Zhang, Y. Zhou, D. H. Zhang, ''J. Phys. Chem. Lett.'', 2012, '''3''', 3416-3419.</ref>
</references>

MRD:01409469

2019-05-10T15:48:10Z

Stc1917: /* Reactive and unreactive trajectories */

== Exercise 1: H + H2 system ==

=== Definition of a transition state ===

A potential energy profile results from plotting the potential versus reaction path.

[[File:Surface_Plot01409469_pic1.jpg|thumb|center|400px|'''Figure 1.''' A potential energy surface plot]]

The transition state is defined as the maximum on the minimum energy path linking reactants and products. This means that the transition state is a stationary point on the potential energy surface. Thus, at the transition state, the derivative of the potential energy is zero with respect to each Cartesian coordinate.

<math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>

However, a local minimum on the potential energy surface corresponding to a well would have the same <math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>. Therefore, to distinguish between a local minimum on the potential energy surface and the transition state, the second derivative, <math>\tfrac{\partial^2 V(r_i)}{\partial r_i ^2}\ </math> is determined. This can be calculated by finding the determinant of the Hessian matrix shown below.

:<math>\mathbf H =
\begin{bmatrix}
\dfrac{\partial^2 \ V(r_i)}{\partial \ x^2} & \dfrac{\partial^2 \ V(r_i)}{\partial \ x \partial \mathbf y} \\
\dfrac{\partial^2 \ V(r_i)}{\partial \ y \partial \mathbf x} & \dfrac{\partial^2 \ V(r_i)}{\partial \mathbf y^2}
\end{bmatrix}</math>

The Hessian matrix of a convex function is positive semi-definite. Refining this property allows one to test if a critical point (x,y) is a local maximum, local minimum or saddle point.<ref name="ref1" />

If <math>f</math> < 0 (exactly one negative eigenvalue), then (x,y) is an isolated local maximum.

If <math>f</math> > 0 (exactly one positive eigenvalue), then (x,y) is an isolated local minimum.

If the Hessian has both positive and negative eigenvalues then (x,y) is a saddle point.

=== Locating the transition state ===

[[File:01409469 combined pic 2 and 3.jpg|thumb|center|800px|'''Figure 2.''' A surface and contour plot of the trajectory when rAB=rBC=0.907743 Å and pA=pBC=0]]

When the molecules are not at the transition state, a surface plot of the conditions shows a trajectory being plotted. This means that the trajectory is "rolling" towards the products or reactants. However, if a trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

[[File:01409469_internuclearpic3.png|thumb|center|400px|'''Figure 3.''' A plot of internuclear distance against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

An initial estimate for the position of the transition state was made by observing the potential energy surface and was found to be 0.90 Å. At 0.90 Å, the internuclear distance against time plot showed that the internuclear distance was constantly oscillating. Therefore, the position of the transition state was further refined by increasing the number of decimal places until a horizontal line in the internuclear distance againt time plot was obtained. This means that the internuclear distance is constant and that the molecules are simply undergoing periodic symmetric vibration at the transition state. Therefore, the position of the transition state was determined to be 0.907743 Å.

[[File:01409469 combined energy pic.png|thumb|center|400px|'''Figure 4.''' A plot of energy against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

When rAB=rBC=0.907743 Å and pA=pBC=0, the kinetic energy is 0. This means that HA and HBC are stationary, which is as predicted when the trajectory is started at the transition state with no initial momentum. Furthermore, the forces acting on HA and HBC are 0 as should be the case in the transition state.

=== Calculation type: Dynamic and minimum energy path ===

The minimum energy path, is the lowest energy path from the reactants to the products. The minimum energy path is determined by following the mode corresponding to the negative Hessian eigenvalue in both directions along the steepest descent direction.<ref name="ref1" /> Following this path allows one to approach two valleys of stable structures. The resulting path generated is the minimum energy path. The reaction paths were chartered for a dynamic and a minimum energy path. The initial conditions used were that of the transition state except that the internuclear distance between HA and HB had increased to 0.917743 Å. The reaction path obtained using the two different calculation methods was different.

[[File:01409469 mep and dynamics.jpg|thumb|center|800px|'''Figure 5.''' A surface plot of the trajectory using minimum energy path '''Figure 6.''' A surface plot of the trajectory using dynamics (Left-right)]]

On the surface plot, the reaction path is depicted by the black line. Following the minimum energy path, the trajectory simply follows the valley floor to Ha+Hb-Hc. The reaction path appears to be a relatively straight line indicating that the atoms are not vibrating. In the minimum energy path, the reactants change into the products in an infinitely slow motion. This is done by having the momenta/velocities reset to zero in each time step. While the minimum energy path is useful in characterising a chemical reaction, it does not provide a realistic representation of the motion of the atoms during the reaction. This is because the mass and inertia of the atoms have been ignored.

Since the atoms simply follow the minimum energy path, they do not undergo slight displacements into surrounding valleys of higher potential energy. Furthermore, because the atoms are assumed to have no inertia, they do not undergo vibrations even if they were to be displaced to valleys of higher potential energy.

However, when dynamic calculation is performed, the reaction path is wavy indicating that the atoms are oscillating. This is because the atoms are displaced to regions of higher potential on the path from reactants to products. This leads to atom oscillation due to the inertia of the atoms.

'''Internuclear distance against time plot'''

Setting the initial system conditions such that rAB= 0.907743, rBC= 0.917743 and pA= pBC= 0:

[[File:01409469 Internuclear distance mep and dynamic.jpg|thumb|center|800px|'''Figure 6.''' A graph of internuclear distance against steps using minimum energy path '''Figure 7.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In the minimum energy path, the distance between HA and HB decreases from 0.90 Å to approximately 0.79 Å during the first 50 steps before plateauing off to a value of 0.74 Å which is the approximate length of a hydrogen bond. This means that a bond was formed between HA and HB. In addition, the plateau indicates that the bond is static. On the other hand, the distance between HB and HC has increased to 1.91 Å at large step values. This shows that HC has been ejected off following a successful collision between HA and HBC.

Similarly, in the dynamic trajectory, the distance between HA and HB decreases to approximately 0.75 Å after 0.3 s. This is indicative of hydrogen bond formation between HA and HB. In addition, the distance between HB and HC increases to 3.44 Å after HC is ejected following a successful collision.

One obvious difference between the two plots is that in the minimum energy path trajectory, the internuclear distance between HA and HB is constant indicating that the bond is static. However, in the dynamic trajectory the internuclear distance is oscillating showing the the bond is vibrating. This difference could be attributed to the fact that minimum energy path does not take oscillatory motion of atoms into account while dynamic trajectory does.

However, changing the initial conditions of the system such that rAB = 0.917743 Å, rBC = 0.907743 Å and pA=pBC=0:
[[File:01409469 01409469 Reversed values dynamic mep.jpg|thumb|center|800px|'''Figure 8.''' A graph of internuclear distance against steps using minimum energy path '''Figure 9.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In this case, a hydrogen bond of length 0.77-0.79 Å is formed between HB and HC while HA is repelled away to beyond 3.45 Å.

Additionally, from the internuclear distance plot it can be seen that the distance between HB and HC increases in a rectangular hyperbola fashion when calculated using the minimum energy path. On the other hand, the internuclear distance between HB and HC increases in a linear manner when dynamic calculation is applied. A possible reason for this difference is that in the minimum energy path, the atoms are not vibrating and as such reach the maximum distance and plateau at a faster rate. However, in the dynamic calculation, the atoms have momentum and as such move further away from each other.

=== Reactive and unreactive trajectories ===

With the initial positions r'''1''' = 0.74 and r'''2''' = 2.0, trajectories were run with the momenta combination shown below.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Contour plot !! Reactivity !! Description of the dynamics
|-
| -1.25 || -2.5 ||-99.018 ||[[File:01409469 trajectory1.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. There are no oscillations observed in HA-HB initially, indicating that the diatomic molecule is stable. A new bond is formed between HB and HC. Oscillatory behaviour is observed indicating that the bond is vibrating. On the other hand, HA is ejected away to beyond 10 Å.
|-
| -1.5 || -2.0 ||-100.456 ||[[File:01409469 trajectory2.jpg|thumb|center|400px]] ||Unreactive ||HC collides with the vibrating diatomic HA-HB. However, no bond is formed between HB and HC as the collision is unsuccessful. This is because the molecules lack sufficient kinetic energy to overcome the activation energy barrier and so no reaction occurs. HC is subsequently repelled to beyond 8 Å.
|-
| -1.5 || -2.5 ||-98.956 ||[[File:01409469 trajectory3.jpg|thumb|center|400px]] ||Reactive |||HC collides with HA-HB. The collision is successful and a bond is formed between HB and HC. This is because the atoms have sufficient kinetic energy to overcome the activation energy barrier and form a new HB-HC bond. HA is ejected away to beyond 10 Å. The new hydrogen bond formed is vibrating as seen by the oscillatory behaviour.
|-
| -2.5 || -5.0 ||-84.956 ||[[File:01409469 trajectory4.jpg|thumb|center|400px]] ||Unreactive ||HC collides with HA-HB. Initially the HA-HB bond is elongating, indicating that the hydrogen bond is weakening and going to break. However, after 0.5 s, the HA-HB bond length decreases and the HAHB bond is not broken. HC is repelled beyond 6 Å with very strong vibrations.
A reaction did not occur even though the atoms have sufficient kinetic energy to cross the activation energy barrier. This is because the vibrations are too strong. The system crosses the transition state region and a bond is formed in the product temporarily but then the system reverts back to the reactants. This is an example of barrier recrossing.
|-
| -2.5 || -5.2 ||-83.416 ||[[File:01409469 trajectory5.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. The collision is successful and a new hydrogen bond is formed between HB and HC with HA being ejected away. The products formed experience very strong vibrations as evidenced by the oscillatory motion which occur with a very high frequency.
|}

From the table, it can be seen that the chance of a reaction successfully occurring is not solely dependent on the amount of kinetic energy the atoms possess. Even if the atoms have sufficient kinetic energy to cross the activation energy barrier, barrier recrossing can occur if the atoms are too vibrationally excited.

=== Transition State Theory ===

Transition state theory provided the first theoretical attempt to determine absolute reaction rates. In this theory, a transition state separating the reactants from products is used to find an expression for the thermal rate constant.

The main assumptions of transition state theory are<ref name="ref2" />:

1. Electronic and nuclear motions are separated (equivalent to the Born-Oppenheimer approximation in quantum mechanics)

2. Reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution

3. Molecular systems that have crossed the transition state in the direction of products cannot turn around and reform reactants

4. In the transition state, motion along the reaction coordinate may be separated from the other motions and treated classically as a translation

5. Even in the absence of an equilibrium between reactant and product molecules, the transition states that are becoming products are distributed among their states according to the Maxwell-Boltzmann laws

The goal of transition state theory is to predict the rate constant, ''k'', of a reaction given a potential energy surface for the reaction.

However, transition state theory suffers from a set of limitations due to the assumptions stated above not holding true in reality.<ref name="ref3" />

Firstly, transition state theory assumes that barrier recrossing is negligible. However, as seen from the fourth trajectory in table 1, barrier recrossing occurred because the atoms were too vibrationally excited and the system reverted back to the reactants even though the system had already crossed the transition state region. In this situation, transition state theory would predict a faster rate of reaction compared to what was obtained experimentally.

Secondly, the transition state theory breaks down for systems with light atoms such as hydrogens. In such systems, quantum effects become crucial. Quantum effects tend to increase the rate of the reaction for two reasons. First, particles lie in energy levels which are higher then the bottom of the potential energy well (there is additional zero-point energy). Secondly, the particles can tunnel through the barrier. These two effects effectively lower the activation energy barrier. As a result, the rate of reaction predicted by transition state theory would be less than what is obtained experimentally.

Lastly, transition state theory does not hold true for systems which are far from equilibrium. In such cases, the assumption of a starting Boltzmann distribution breaks down.

== Exercise 2: F - H - H system ==

=== Potential energy surface inspection===

F + HA-HB → HA-F + HB

[[File:01409469 Surface_Plot F 3.jpg|thumb|center|400px|'''Figure 10.''' Potential energy surface for the reaction between F and H2]]

The reaction between F and H2 is an exothermic reaction. This means that the reactants are of a higher energy than the products. The overall reaction involves the breaking of a hydrogen bond and the formation of a hydrogen-fluorine bond. The HF bond formed is stronger than the hydrogen bond that was broken. This is because fluorine is more electronegative than hydrogen resulting in a polar HF bond. The negative enthalpy of H-F bond formation is greater than the positive enthalpy of hydrogen bond dissociation. Therefore, the reaction between F and H2 is an exothermic reaction.

HA + HB-F → HA-HB + F

[[File:01409469 Surface_Plot F 4.jpg|thumb|center|400px|'''Figure 11.''' Potential energy surface for the reaction between H and HF]]

The reaction between H and HF is an endothermic reaction. The products are at a higher energy level than the reactants. The overall reaction involves the breaking of a HF bond and the formation of a hydrogen bond. The formation of a homonuclear H2 molecule results in a non-polar hydrogen bond being formed. The HF bond broken is stronger than the hydrogen bond that was formed. The negative enthalpy of hydrogen bond formation is less than the positive enthalpy of hydrogen-fluorine dissociation. Therefore, the reaction between H and HF is endothermic.

=== Locating the transition state ===

[[File:01409469 Surface_Plot F TS4.jpg|thumb|center|800px|'''Figure 12.''' Determining the position of the transition state for the reaction between F and H2]]

As mentioned previously, when the molecules are not at the transition state, a reaction path will be plotted. However, if the initial conditions of the system are set such that the trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

An initial estimate for the transition state was made by setting rAB = 0.74 which is the length of a hydrogen bond. The position of the transition state was further refined by increasing the number of decimal places such that the final value of rAB was found to be 0.74478 and rBC = 1.81079. This set of distances gave a reaction path that was represented by a point indicating that no trajectory was plotted and that the molecules remained at the transition state since their initial momentum was 0.

Moreover, the kinetic energy as well as the forces acting on the atoms was calculated to be 0, which is consistent with the transition state.

Therefore, the position of the transition state was found to be at rAB = 0.74478 and rBC = 1.81079 when pA = pBC = 0.

=== Determining the activation energy ===

[[File:01409469 activation energy r1.1 combined.jpg|thumb|center|800px|'''Figure 13.''' Determining the activation energy for the reaction between F and H2]]

By reducing the value of rBC by 0.01, the atoms are no longer at the transition state and would move towards the formation of products. The activation energy was found by calculating the change in total energy after the reaction. The activation energy of F + H2 → HF + H was found to be +23.279 kcal/ mol.

[[File:01409469 activation energy r2.1 combined.jpg|thumb|center|800px|'''Figure 14.''' Determining the activation energy for the reaction between FH and H]]

The activation energy of FH + H → F + H2 was found by calculating the small dip in potential energy. The activation energy was found to be +0.280 kcal/ mol.

=== Reaction dynamics of F + HA-HB → HA-F + HB ===

The set of initial conditions that result in a reactive trajectory for F + H2 is as follows: rHH = 0.74 Å, rHF = 1.79 Å and pHH = pHF = 0.

[[File:01409469 momentum r1.jpg|thumb|center|400px|'''Figure 15.''' Graph of momentum against time for the reaction between F and H2]]

Initially, the H2 molecule is travelling in a relatively straight line with very little oscillations as evidenced by the low amplitude of momentum between 0-2 s. Following successful collision between F and H2, the amplitude of momentum of the product, HF, increases dramatically. This is because the reaction between F and H2 is exothermic and the excess energy released causes the molecule to become vibrationally excited and the amplitude of oscillations in the product to increase in the form of H-F vibrations.

[[File:01409469 energy1 r1.jpg|thumb|center|400px|'''Figure 16.''' Graph of energy against time for the reaction between F and H2]]

The law of conservation of energy states that energy cannot be created or destroyed. It simply changes from one form to another. As can be seen from the graph in figure 16, total energy remains constant. In this reaction, potential energy is converted to kinetic energy. A decrease in potential energy of the atoms following a successful collision between F and H2 results in an equal increase in kinetic energy which manifests in the form of increased product (HF) oscillations.

This can be confirmed experimentally using infrared chemiluminescence.<ref name="ref2" /> Observation of radiation from excited product molecules is a clear indication that these molecules are produced in the excited state. If the excitation is primarily vibrational, the radiation will appear in the infrared region of the spectrum (between 3-15 μm). This technique has been used extensively in hydrogen halogen reactions which produce HX molecules in the excited state.

One such experiment involves the use of the arrested relaxation method.<ref name="ref2" /> In this technique, the reaction is carried out in a chamber whose walls are maintained at a very low temperature (20 K). This results in cryopumping of the chamber in that all the products and unreacted starting materials are condensed on the walls and only those product molecules that radiate before being pumped away are observed.

=== Polanyi's empirical rules ===

In a typical chemical reaction, there is an energetic barrier — a saddle point that the reactants must surmount to become products. Which form of energy is initially deposited in the reactants, translational or vibrational, is more efficacious in surmounting the activation energy barrier is one of the core topics in the area of reaction dynamics. The Polyani rules state that vibrational energy is more effective at promoting a late-barrier reaction (transition state resembles products). Conversely, translational energy is better at promoting an early-barrier reaction (transition state resembles the reactants).<ref name="ref4" />

The forward reaction between H2 and F is exothermic. Therefore, the transition state is closer to the reactants than to the products in energy. Since the transition state occurs early in an exothermic reaction, translational energy is more effective at promoting a chemical reaction than vibrational energy.

Conversely, the backward reaction between H and HF is endothermic. According to Hammond's postulate, the transition state resembles the products more than it resembles the reactants. Therefore, in an endothermic reaction, a late transition state is observed and vibrational energy is more effective at promoting a reaction than translational energy.

=== Promoting the forward reaction: F + HA-HB → HA-F + HB ===

1. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 2.95 and pHF = -0.5.

[[File:01409469 polyani 1+2.jpg|thumb|center|800px|'''Figure 17.''' Contour plot of the reaction between F and H2 when pHH = 2.95 '''Figure 18.''' Graph of momentum against time for the reaction between F and H2 when pHH = 2.95 (Left-right)]]

Vibrational energy is represented by pHH while translational energy is represented by pHF.

A molecule of H2 collides with F. However, the collision is unsuccessful and no reaction occurs. A reaction did not occur even though the molecules have sufficient kinetic energy to overcome the activation energy barrier. This is because most of the energy is deposited in the reactants as H-H vibrations as seen by the large oscillations in H2. Barrier recrossing has occured as a result of the large vibrational energy causing the system to revert back to the reactants even though the system has already crossed the transition state and a bond was formed between HB and F temporarily.

This appears to run counter to what one would expect to happen when more energy is put into the system. It is predicted that when more energy is put into the system, the reactants would possess more kinetic energy to overcome the activation energy barrier. However, as can be seen from the simulation above, an increase in the vibrational energy of the reactants is not effective at promoting the occurrence of an exothermic reaction with an early transition state. This is in accordance to Polyani's rules.

2. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 0.1 and pHF = -0.8.

[[File:01409469 polyani 3+4 2.jpg|thumb|center|800px|'''Figure 19.''' Contour plot of the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 '''Figure 20.''' Graph of momentum against time for the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 (Left-right)]]

The collision between a molecule of H2 and F is successful and a chemical reaction has occurred. A reaction has occurred even though the overall energy of the system has been significantly reduced by lowering the amount of vibrational energy in the reactants but slightly increasing the amount of translational energy deposited in the reactants. Therefore, H2 and F have more translational energy to move towards each other and form a H-F bond. The product formed is vibrationally excited as seen by the large amplitude of oscillations in the product HF which correspond to the strong H-F vibrations.

From the second simulation, it can be seen that an increase in translational energy in the reactants is more effective at promoting an exothermic reaction with an early transition state to occur, which again follows Polyani's rules.

=== Promoting the backward reaction: HA + HB-F → HA-HB + F ===

1. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 0.2 and pHH = -3.0.

[[File:01409469 polyani 5+6 .jpg|thumb|center|800px|'''Figure 21.''' Contour plot of the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 '''Figure 22.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 (Left-right)]]

An atom of HA collides with a molecule of HBF. However, the collision is unsuccessful and no chemical reaction occurs. This is because the reactants lack sufficient vibrational energy to surmount the activation energy barrier. Therefore, following an unsuccessful collision with HBF, HA simply bounces off and moves away at constant momentum. This is in agreement with Polyani's rules which state that vibrational energy is more effective than translational energy at promoting reactions with a late transition state, for example, endothermic chemical reactions.

2. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 7.1 and pHH = -0.5.

[[File:01409469 polyani 7+8 .jpg|thumb|center|800px|'''Figure 23.''' Contour plot of the reaction between HA + HB-F when pFH = 7.1 and pHH = -0.5 '''Figure 24.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 7.1 and pHH = -0.5 (Left-right)]]

Increasing the amount of vibrational energy in the reactants by increasing pFH resulted in a chemical reaction successfully occurring. This is because the reactants now have sufficient kinetic energy in the form of H-F vibrations to overcome the activation energy barrier. In addition, because the reaction between HA + HB-F is endothermic, energy is taken in from the surroundings during the reaction and the products are less vibrationally excited than the reactants. This can be seen in the lower amplitude of oscillations in the product formed.

In conclusion, the vibrational energy of the reactants should be increased to promote an endothermic reaction with a late transition state while the translational energy of the reactants should be increased to promote an exothermic reaction with an early transition state. All 4 simulations shown above obey Polyani's rules.

== References ==
<references>
<ref name="ref1">A. C. Vaucher, M. Reiher, ''J. Chem. Theory. Comput.'', 2018, '''14''', 3091-3099.</ref>
<ref name="ref2">J. I. Steinfield, J. S. Francisco, W. L. Hase, ''Chemical Kinetics and Dynamics'', Pearson, 1998.</ref>
<ref name="ref3">D. C. Elton, ''Transition State Theory for Physicists'', 2013.</ref>
<ref name="ref4">Z. Zhang, Y. Zhou, D. H. Zhang, ''J. Phys. Chem. Lett.'', 2012, '''3''', 3416-3419.</ref>
</references>

MRD:01409469

2019-05-10T15:45:08Z

Stc1917: /* Calculation type: Dynamic and minimum energy path */

== Exercise 1: H + H2 system ==

=== Definition of a transition state ===

A potential energy profile results from plotting the potential versus reaction path.

[[File:Surface_Plot01409469_pic1.jpg|thumb|center|400px|'''Figure 1.''' A potential energy surface plot]]

The transition state is defined as the maximum on the minimum energy path linking reactants and products. This means that the transition state is a stationary point on the potential energy surface. Thus, at the transition state, the derivative of the potential energy is zero with respect to each Cartesian coordinate.

<math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>

However, a local minimum on the potential energy surface corresponding to a well would have the same <math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>. Therefore, to distinguish between a local minimum on the potential energy surface and the transition state, the second derivative, <math>\tfrac{\partial^2 V(r_i)}{\partial r_i ^2}\ </math> is determined. This can be calculated by finding the determinant of the Hessian matrix shown below.

:<math>\mathbf H =
\begin{bmatrix}
\dfrac{\partial^2 \ V(r_i)}{\partial \ x^2} & \dfrac{\partial^2 \ V(r_i)}{\partial \ x \partial \mathbf y} \\
\dfrac{\partial^2 \ V(r_i)}{\partial \ y \partial \mathbf x} & \dfrac{\partial^2 \ V(r_i)}{\partial \mathbf y^2}
\end{bmatrix}</math>

The Hessian matrix of a convex function is positive semi-definite. Refining this property allows one to test if a critical point (x,y) is a local maximum, local minimum or saddle point.<ref name="ref1" />

If <math>f</math> < 0 (exactly one negative eigenvalue), then (x,y) is an isolated local maximum.

If <math>f</math> > 0 (exactly one positive eigenvalue), then (x,y) is an isolated local minimum.

If the Hessian has both positive and negative eigenvalues then (x,y) is a saddle point.

=== Locating the transition state ===

[[File:01409469 combined pic 2 and 3.jpg|thumb|center|800px|'''Figure 2.''' A surface and contour plot of the trajectory when rAB=rBC=0.907743 Å and pA=pBC=0]]

When the molecules are not at the transition state, a surface plot of the conditions shows a trajectory being plotted. This means that the trajectory is "rolling" towards the products or reactants. However, if a trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

[[File:01409469_internuclearpic3.png|thumb|center|400px|'''Figure 3.''' A plot of internuclear distance against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

An initial estimate for the position of the transition state was made by observing the potential energy surface and was found to be 0.90 Å. At 0.90 Å, the internuclear distance against time plot showed that the internuclear distance was constantly oscillating. Therefore, the position of the transition state was further refined by increasing the number of decimal places until a horizontal line in the internuclear distance againt time plot was obtained. This means that the internuclear distance is constant and that the molecules are simply undergoing periodic symmetric vibration at the transition state. Therefore, the position of the transition state was determined to be 0.907743 Å.

[[File:01409469 combined energy pic.png|thumb|center|400px|'''Figure 4.''' A plot of energy against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

When rAB=rBC=0.907743 Å and pA=pBC=0, the kinetic energy is 0. This means that HA and HBC are stationary, which is as predicted when the trajectory is started at the transition state with no initial momentum. Furthermore, the forces acting on HA and HBC are 0 as should be the case in the transition state.

=== Calculation type: Dynamic and minimum energy path ===

The minimum energy path, is the lowest energy path from the reactants to the products. The minimum energy path is determined by following the mode corresponding to the negative Hessian eigenvalue in both directions along the steepest descent direction.<ref name="ref1" /> Following this path allows one to approach two valleys of stable structures. The resulting path generated is the minimum energy path. The reaction paths were chartered for a dynamic and a minimum energy path. The initial conditions used were that of the transition state except that the internuclear distance between HA and HB had increased to 0.917743 Å. The reaction path obtained using the two different calculation methods was different.

[[File:01409469 mep and dynamics.jpg|thumb|center|800px|'''Figure 5.''' A surface plot of the trajectory using minimum energy path '''Figure 6.''' A surface plot of the trajectory using dynamics (Left-right)]]

On the surface plot, the reaction path is depicted by the black line. Following the minimum energy path, the trajectory simply follows the valley floor to Ha+Hb-Hc. The reaction path appears to be a relatively straight line indicating that the atoms are not vibrating. In the minimum energy path, the reactants change into the products in an infinitely slow motion. This is done by having the momenta/velocities reset to zero in each time step. While the minimum energy path is useful in characterising a chemical reaction, it does not provide a realistic representation of the motion of the atoms during the reaction. This is because the mass and inertia of the atoms have been ignored.

Since the atoms simply follow the minimum energy path, they do not undergo slight displacements into surrounding valleys of higher potential energy. Furthermore, because the atoms are assumed to have no inertia, they do not undergo vibrations even if they were to be displaced to valleys of higher potential energy.

However, when dynamic calculation is performed, the reaction path is wavy indicating that the atoms are oscillating. This is because the atoms are displaced to regions of higher potential on the path from reactants to products. This leads to atom oscillation due to the inertia of the atoms.

'''Internuclear distance against time plot'''

Setting the initial system conditions such that rAB= 0.907743, rBC= 0.917743 and pA= pBC= 0:

[[File:01409469 Internuclear distance mep and dynamic.jpg|thumb|center|800px|'''Figure 6.''' A graph of internuclear distance against steps using minimum energy path '''Figure 7.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In the minimum energy path, the distance between HA and HB decreases from 0.90 Å to approximately 0.79 Å during the first 50 steps before plateauing off to a value of 0.74 Å which is the approximate length of a hydrogen bond. This means that a bond was formed between HA and HB. In addition, the plateau indicates that the bond is static. On the other hand, the distance between HB and HC has increased to 1.91 Å at large step values. This shows that HC has been ejected off following a successful collision between HA and HBC.

Similarly, in the dynamic trajectory, the distance between HA and HB decreases to approximately 0.75 Å after 0.3 s. This is indicative of hydrogen bond formation between HA and HB. In addition, the distance between HB and HC increases to 3.44 Å after HC is ejected following a successful collision.

One obvious difference between the two plots is that in the minimum energy path trajectory, the internuclear distance between HA and HB is constant indicating that the bond is static. However, in the dynamic trajectory the internuclear distance is oscillating showing the the bond is vibrating. This difference could be attributed to the fact that minimum energy path does not take oscillatory motion of atoms into account while dynamic trajectory does.

However, changing the initial conditions of the system such that rAB = 0.917743 Å, rBC = 0.907743 Å and pA=pBC=0:
[[File:01409469 01409469 Reversed values dynamic mep.jpg|thumb|center|800px|'''Figure 8.''' A graph of internuclear distance against steps using minimum energy path '''Figure 9.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In this case, a hydrogen bond of length 0.77-0.79 Å is formed between HB and HC while HA is repelled away to beyond 3.45 Å.

Additionally, from the internuclear distance plot it can be seen that the distance between HB and HC increases in a rectangular hyperbola fashion when calculated using the minimum energy path. On the other hand, the internuclear distance between HB and HC increases in a linear manner when dynamic calculation is applied. A possible reason for this difference is that in the minimum energy path, the atoms are not vibrating and as such reach the maximum distance and plateau at a faster rate. However, in the dynamic calculation, the atoms have momentum and as such move further away from each other.

=== Reactive and unreactive trajectories ===

With the initial positions r'''1''' = 0.74 and r'''2''' = 2.0, trajectories were run with the momenta combination shown below.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Contour plot !! Reactivity !! Description of the dynamics
|-
| -1.25 || -2.5 ||-99.018 ||[[File:01409469 trajectory1.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. There are no oscillations observed in HA-HB initially, indicating that the diatomic molecule is stable. A new bond is formed between HB and HC. Oscillatory behaviour is observed indicating that the bond is vibrating. On the other hand, HA is ejected away to beyond 10 Å.
|-
| -1.5 || -2.0 ||-100.456 ||[[File:01409469 trajectory2.jpg|thumb|center|400px]] ||Unreactive ||HC collides with the vibrating diatomic HA-HB. However, no bond is formed between HB and HC as the collision is unsuccessful. This is because the molecules lack sufficient kinetic energy to overcome the activation energy barrier and so no reaction occurs. HC is subsequently repelled to beyond 8 Å.
|-
| -1.5 || -2.5 ||-98.956 ||[[File:01409469 trajectory3.jpg|thumb|center|400px]] ||Reactive |||HC collides with HA-HB. The collision is successful and a bond is formed between HB and HC. This is because the atoms have sufficient kinetic energy to overcome the activation energy barrier and form a new HB-HC bond. HA is ejected away to beyond 10 Å. The new hydrogen bond formed is vibrating as seen by the oscillatory behaviour.
|-
| -2.5 || -5.0 ||-84.956 ||[[File:01409469 trajectory4.jpg|thumb|center|400px]] ||Unreactive ||HC collides with HA-HB. Initially the HA-HB bond is elongating, indicating that the hydrogen bond is weakening and going to break. However, after 0.5 s, the HA-HB bond length decreases and the HAHB is not broken. HC is repelled beyond 6 Å with very strong vibrations.
A reaction did not occur even though the atoms have sufficient kinetic energy to cross the activation energy barrier. This is because the vibrations are too strong. The system crosses the transition state region and a bond is formed in the product temporarily but then the system reverts back to the reactants. This is an example of barrier recrossing.
|-
| -2.5 || -5.2 ||-83.416 ||[[File:01409469 trajectory5.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. The collision is successful and a new hydrogen bond is formed between HB and HC with HA being ejected away. The products formed experience very strong vibrations as evidenced by the oscillatory motion which occur with a very high frequency.
|}

From the table, it can be seen that the chance of a reaction successfully occurring is not solely dependent on the amount of kinetic energy the atoms possess. Even if the atoms have sufficient kinetic energy to cross the activation energy barrier, barrier recrossing can occur if the atoms are too vibrationally excited.

=== Transition State Theory ===

Transition state theory provided the first theoretical attempt to determine absolute reaction rates. In this theory, a transition state separating the reactants from products is used to find an expression for the thermal rate constant.

The main assumptions of transition state theory are<ref name="ref2" />:

1. Electronic and nuclear motions are separated (equivalent to the Born-Oppenheimer approximation in quantum mechanics)

2. Reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution

3. Molecular systems that have crossed the transition state in the direction of products cannot turn around and reform reactants

4. In the transition state, motion along the reaction coordinate may be separated from the other motions and treated classically as a translation

5. Even in the absence of an equilibrium between reactant and product molecules, the transition states that are becoming products are distributed among their states according to the Maxwell-Boltzmann laws

The goal of transition state theory is to predict the rate constant, ''k'', of a reaction given a potential energy surface for the reaction.

However, transition state theory suffers from a set of limitations due to the assumptions stated above not holding true in reality.<ref name="ref3" />

Firstly, transition state theory assumes that barrier recrossing is negligible. However, as seen from the fourth trajectory in table 1, barrier recrossing occurred because the atoms were too vibrationally excited and the system reverted back to the reactants even though the system had already crossed the transition state region. In this situation, transition state theory would predict a faster rate of reaction compared to what was obtained experimentally.

Secondly, the transition state theory breaks down for systems with light atoms such as hydrogens. In such systems, quantum effects become crucial. Quantum effects tend to increase the rate of the reaction for two reasons. First, particles lie in energy levels which are higher then the bottom of the potential energy well (there is additional zero-point energy). Secondly, the particles can tunnel through the barrier. These two effects effectively lower the activation energy barrier. As a result, the rate of reaction predicted by transition state theory would be less than what is obtained experimentally.

Lastly, transition state theory does not hold true for systems which are far from equilibrium. In such cases, the assumption of a starting Boltzmann distribution breaks down.

== Exercise 2: F - H - H system ==

=== Potential energy surface inspection===

F + HA-HB → HA-F + HB

[[File:01409469 Surface_Plot F 3.jpg|thumb|center|400px|'''Figure 10.''' Potential energy surface for the reaction between F and H2]]

The reaction between F and H2 is an exothermic reaction. This means that the reactants are of a higher energy than the products. The overall reaction involves the breaking of a hydrogen bond and the formation of a hydrogen-fluorine bond. The HF bond formed is stronger than the hydrogen bond that was broken. This is because fluorine is more electronegative than hydrogen resulting in a polar HF bond. The negative enthalpy of H-F bond formation is greater than the positive enthalpy of hydrogen bond dissociation. Therefore, the reaction between F and H2 is an exothermic reaction.

HA + HB-F → HA-HB + F

[[File:01409469 Surface_Plot F 4.jpg|thumb|center|400px|'''Figure 11.''' Potential energy surface for the reaction between H and HF]]

The reaction between H and HF is an endothermic reaction. The products are at a higher energy level than the reactants. The overall reaction involves the breaking of a HF bond and the formation of a hydrogen bond. The formation of a homonuclear H2 molecule results in a non-polar hydrogen bond being formed. The HF bond broken is stronger than the hydrogen bond that was formed. The negative enthalpy of hydrogen bond formation is less than the positive enthalpy of hydrogen-fluorine dissociation. Therefore, the reaction between H and HF is endothermic.

=== Locating the transition state ===

[[File:01409469 Surface_Plot F TS4.jpg|thumb|center|800px|'''Figure 12.''' Determining the position of the transition state for the reaction between F and H2]]

As mentioned previously, when the molecules are not at the transition state, a reaction path will be plotted. However, if the initial conditions of the system are set such that the trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

An initial estimate for the transition state was made by setting rAB = 0.74 which is the length of a hydrogen bond. The position of the transition state was further refined by increasing the number of decimal places such that the final value of rAB was found to be 0.74478 and rBC = 1.81079. This set of distances gave a reaction path that was represented by a point indicating that no trajectory was plotted and that the molecules remained at the transition state since their initial momentum was 0.

Moreover, the kinetic energy as well as the forces acting on the atoms was calculated to be 0, which is consistent with the transition state.

Therefore, the position of the transition state was found to be at rAB = 0.74478 and rBC = 1.81079 when pA = pBC = 0.

=== Determining the activation energy ===

[[File:01409469 activation energy r1.1 combined.jpg|thumb|center|800px|'''Figure 13.''' Determining the activation energy for the reaction between F and H2]]

By reducing the value of rBC by 0.01, the atoms are no longer at the transition state and would move towards the formation of products. The activation energy was found by calculating the change in total energy after the reaction. The activation energy of F + H2 → HF + H was found to be +23.279 kcal/ mol.

[[File:01409469 activation energy r2.1 combined.jpg|thumb|center|800px|'''Figure 14.''' Determining the activation energy for the reaction between FH and H]]

The activation energy of FH + H → F + H2 was found by calculating the small dip in potential energy. The activation energy was found to be +0.280 kcal/ mol.

=== Reaction dynamics of F + HA-HB → HA-F + HB ===

The set of initial conditions that result in a reactive trajectory for F + H2 is as follows: rHH = 0.74 Å, rHF = 1.79 Å and pHH = pHF = 0.

[[File:01409469 momentum r1.jpg|thumb|center|400px|'''Figure 15.''' Graph of momentum against time for the reaction between F and H2]]

Initially, the H2 molecule is travelling in a relatively straight line with very little oscillations as evidenced by the low amplitude of momentum between 0-2 s. Following successful collision between F and H2, the amplitude of momentum of the product, HF, increases dramatically. This is because the reaction between F and H2 is exothermic and the excess energy released causes the molecule to become vibrationally excited and the amplitude of oscillations in the product to increase in the form of H-F vibrations.

[[File:01409469 energy1 r1.jpg|thumb|center|400px|'''Figure 16.''' Graph of energy against time for the reaction between F and H2]]

The law of conservation of energy states that energy cannot be created or destroyed. It simply changes from one form to another. As can be seen from the graph in figure 16, total energy remains constant. In this reaction, potential energy is converted to kinetic energy. A decrease in potential energy of the atoms following a successful collision between F and H2 results in an equal increase in kinetic energy which manifests in the form of increased product (HF) oscillations.

This can be confirmed experimentally using infrared chemiluminescence.<ref name="ref2" /> Observation of radiation from excited product molecules is a clear indication that these molecules are produced in the excited state. If the excitation is primarily vibrational, the radiation will appear in the infrared region of the spectrum (between 3-15 μm). This technique has been used extensively in hydrogen halogen reactions which produce HX molecules in the excited state.

One such experiment involves the use of the arrested relaxation method.<ref name="ref2" /> In this technique, the reaction is carried out in a chamber whose walls are maintained at a very low temperature (20 K). This results in cryopumping of the chamber in that all the products and unreacted starting materials are condensed on the walls and only those product molecules that radiate before being pumped away are observed.

=== Polanyi's empirical rules ===

In a typical chemical reaction, there is an energetic barrier — a saddle point that the reactants must surmount to become products. Which form of energy is initially deposited in the reactants, translational or vibrational, is more efficacious in surmounting the activation energy barrier is one of the core topics in the area of reaction dynamics. The Polyani rules state that vibrational energy is more effective at promoting a late-barrier reaction (transition state resembles products). Conversely, translational energy is better at promoting an early-barrier reaction (transition state resembles the reactants).<ref name="ref4" />

The forward reaction between H2 and F is exothermic. Therefore, the transition state is closer to the reactants than to the products in energy. Since the transition state occurs early in an exothermic reaction, translational energy is more effective at promoting a chemical reaction than vibrational energy.

Conversely, the backward reaction between H and HF is endothermic. According to Hammond's postulate, the transition state resembles the products more than it resembles the reactants. Therefore, in an endothermic reaction, a late transition state is observed and vibrational energy is more effective at promoting a reaction than translational energy.

=== Promoting the forward reaction: F + HA-HB → HA-F + HB ===

1. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 2.95 and pHF = -0.5.

[[File:01409469 polyani 1+2.jpg|thumb|center|800px|'''Figure 17.''' Contour plot of the reaction between F and H2 when pHH = 2.95 '''Figure 18.''' Graph of momentum against time for the reaction between F and H2 when pHH = 2.95 (Left-right)]]

Vibrational energy is represented by pHH while translational energy is represented by pHF.

A molecule of H2 collides with F. However, the collision is unsuccessful and no reaction occurs. A reaction did not occur even though the molecules have sufficient kinetic energy to overcome the activation energy barrier. This is because most of the energy is deposited in the reactants as H-H vibrations as seen by the large oscillations in H2. Barrier recrossing has occured as a result of the large vibrational energy causing the system to revert back to the reactants even though the system has already crossed the transition state and a bond was formed between HB and F temporarily.

This appears to run counter to what one would expect to happen when more energy is put into the system. It is predicted that when more energy is put into the system, the reactants would possess more kinetic energy to overcome the activation energy barrier. However, as can be seen from the simulation above, an increase in the vibrational energy of the reactants is not effective at promoting the occurrence of an exothermic reaction with an early transition state. This is in accordance to Polyani's rules.

2. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 0.1 and pHF = -0.8.

[[File:01409469 polyani 3+4 2.jpg|thumb|center|800px|'''Figure 19.''' Contour plot of the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 '''Figure 20.''' Graph of momentum against time for the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 (Left-right)]]

The collision between a molecule of H2 and F is successful and a chemical reaction has occurred. A reaction has occurred even though the overall energy of the system has been significantly reduced by lowering the amount of vibrational energy in the reactants but slightly increasing the amount of translational energy deposited in the reactants. Therefore, H2 and F have more translational energy to move towards each other and form a H-F bond. The product formed is vibrationally excited as seen by the large amplitude of oscillations in the product HF which correspond to the strong H-F vibrations.

From the second simulation, it can be seen that an increase in translational energy in the reactants is more effective at promoting an exothermic reaction with an early transition state to occur, which again follows Polyani's rules.

=== Promoting the backward reaction: HA + HB-F → HA-HB + F ===

1. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 0.2 and pHH = -3.0.

[[File:01409469 polyani 5+6 .jpg|thumb|center|800px|'''Figure 21.''' Contour plot of the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 '''Figure 22.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 (Left-right)]]

An atom of HA collides with a molecule of HBF. However, the collision is unsuccessful and no chemical reaction occurs. This is because the reactants lack sufficient vibrational energy to surmount the activation energy barrier. Therefore, following an unsuccessful collision with HBF, HA simply bounces off and moves away at constant momentum. This is in agreement with Polyani's rules which state that vibrational energy is more effective than translational energy at promoting reactions with a late transition state, for example, endothermic chemical reactions.

2. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 7.1 and pHH = -0.5.

[[File:01409469 polyani 7+8 .jpg|thumb|center|800px|'''Figure 23.''' Contour plot of the reaction between HA + HB-F when pFH = 7.1 and pHH = -0.5 '''Figure 24.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 7.1 and pHH = -0.5 (Left-right)]]

Increasing the amount of vibrational energy in the reactants by increasing pFH resulted in a chemical reaction successfully occurring. This is because the reactants now have sufficient kinetic energy in the form of H-F vibrations to overcome the activation energy barrier. In addition, because the reaction between HA + HB-F is endothermic, energy is taken in from the surroundings during the reaction and the products are less vibrationally excited than the reactants. This can be seen in the lower amplitude of oscillations in the product formed.

In conclusion, the vibrational energy of the reactants should be increased to promote an endothermic reaction with a late transition state while the translational energy of the reactants should be increased to promote an exothermic reaction with an early transition state. All 4 simulations shown above obey Polyani's rules.

== References ==
<references>
<ref name="ref1">A. C. Vaucher, M. Reiher, ''J. Chem. Theory. Comput.'', 2018, '''14''', 3091-3099.</ref>
<ref name="ref2">J. I. Steinfield, J. S. Francisco, W. L. Hase, ''Chemical Kinetics and Dynamics'', Pearson, 1998.</ref>
<ref name="ref3">D. C. Elton, ''Transition State Theory for Physicists'', 2013.</ref>
<ref name="ref4">Z. Zhang, Y. Zhou, D. H. Zhang, ''J. Phys. Chem. Lett.'', 2012, '''3''', 3416-3419.</ref>
</references>

MRD:01409469

2019-05-10T15:42:55Z

Stc1917: /* Calculation type: Dynamic and minimum energy path */

== Exercise 1: H + H2 system ==

=== Definition of a transition state ===

A potential energy profile results from plotting the potential versus reaction path.

[[File:Surface_Plot01409469_pic1.jpg|thumb|center|400px|'''Figure 1.''' A potential energy surface plot]]

The transition state is defined as the maximum on the minimum energy path linking reactants and products. This means that the transition state is a stationary point on the potential energy surface. Thus, at the transition state, the derivative of the potential energy is zero with respect to each Cartesian coordinate.

<math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>

However, a local minimum on the potential energy surface corresponding to a well would have the same <math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>. Therefore, to distinguish between a local minimum on the potential energy surface and the transition state, the second derivative, <math>\tfrac{\partial^2 V(r_i)}{\partial r_i ^2}\ </math> is determined. This can be calculated by finding the determinant of the Hessian matrix shown below.

:<math>\mathbf H =
\begin{bmatrix}
\dfrac{\partial^2 \ V(r_i)}{\partial \ x^2} & \dfrac{\partial^2 \ V(r_i)}{\partial \ x \partial \mathbf y} \\
\dfrac{\partial^2 \ V(r_i)}{\partial \ y \partial \mathbf x} & \dfrac{\partial^2 \ V(r_i)}{\partial \mathbf y^2}
\end{bmatrix}</math>

The Hessian matrix of a convex function is positive semi-definite. Refining this property allows one to test if a critical point (x,y) is a local maximum, local minimum or saddle point.<ref name="ref1" />

If <math>f</math> < 0 (exactly one negative eigenvalue), then (x,y) is an isolated local maximum.

If <math>f</math> > 0 (exactly one positive eigenvalue), then (x,y) is an isolated local minimum.

If the Hessian has both positive and negative eigenvalues then (x,y) is a saddle point.

=== Locating the transition state ===

[[File:01409469 combined pic 2 and 3.jpg|thumb|center|800px|'''Figure 2.''' A surface and contour plot of the trajectory when rAB=rBC=0.907743 Å and pA=pBC=0]]

When the molecules are not at the transition state, a surface plot of the conditions shows a trajectory being plotted. This means that the trajectory is "rolling" towards the products or reactants. However, if a trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

[[File:01409469_internuclearpic3.png|thumb|center|400px|'''Figure 3.''' A plot of internuclear distance against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

An initial estimate for the position of the transition state was made by observing the potential energy surface and was found to be 0.90 Å. At 0.90 Å, the internuclear distance against time plot showed that the internuclear distance was constantly oscillating. Therefore, the position of the transition state was further refined by increasing the number of decimal places until a horizontal line in the internuclear distance againt time plot was obtained. This means that the internuclear distance is constant and that the molecules are simply undergoing periodic symmetric vibration at the transition state. Therefore, the position of the transition state was determined to be 0.907743 Å.

[[File:01409469 combined energy pic.png|thumb|center|400px|'''Figure 4.''' A plot of energy against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

When rAB=rBC=0.907743 Å and pA=pBC=0, the kinetic energy is 0. This means that HA and HBC are stationary, which is as predicted when the trajectory is started at the transition state with no initial momentum. Furthermore, the forces acting on HA and HBC are 0 as should be the case in the transition state.

=== Calculation type: Dynamic and minimum energy path ===

The minimum energy path, is the lowest energy path from the reactants to the products. The minimum energy path is determined by following the mode corresponding to the negative Hessian eigenvalue in both directions along the steepest descent direction.<ref name="ref1" /> Following this path allows one to approach two valleys of stable structures. The resulting path generated is the minimum energy path. The reaction paths were chartered for a dynamic and a minimum energy path. The initial conditions used were that of the transition state except that the internuclear distance between HA and HB had increased to 0.917743 Å. The reaction path obtained using the two different calculation methods was different.

[[File:01409469 mep and dynamics.jpg|thumb|center|800px|'''Figure 5.''' A surface plot of the trajectory using minimum energy path '''Figure 6.''' A surface plot of the trajectory using dynamics (Left-right)]]

On the surface plot, the reaction path is depicted by the black line. Following the minimum energy path, the trajectory simply follows the valley floor to Ha+Hb-Hc. The reaction path appears to be a relatively straight line indicating that the atoms are not vibrating. In the minimum energy path, the reactants change into the products in an infinitely slow motion. This is done by having the momenta/velocities reset to zero in each time step. While the minimum energy path is useful in characterising a chemical reaction, it does not provide a realistic representation of the motion of the atoms during the reaction. This is because the mass and inertia of the atoms have been ignored.

Since the atoms simply follow the minimum energy path, they do not undergo slight displacements into surrounding valleys of higher potential energy. Furthermore, because the atoms are assumed to have no inertia, they do not undergo vibrations even if they were to be displaced to valleys of higher potential energy.

However, when dynamic calculation is performed, the reaction path is wavy indicating that the atoms are oscillating. This is because the atoms are displaced to regions of higher potential on the path from reactants to products. This leads to atom oscillation due to the inertia of the atoms.

'''Internuclear distance against time plot'''

Setting the initial system conditions such that rAB= 0.907743, rBC= 0.917743 and pA= pBC= 0:

[[File:01409469 Internuclear distance mep and dynamic.jpg|thumb|center|800px|'''Figure 6.''' A graph of internuclear distance against steps using minimum energy path '''Figure 7.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In the minimum energy path, the distance between HA and HB decreases from 0.90 Å to approximately 0.79 Å during the first 50 steps before plateauing off to a value of 0.74 Å which is the approximate length of a hydrogen bond. This means that a bond was formed between HA and HB. In addition, the plateau indicates that the bond is static. On the other hand, the distance between HB and HC has increased to 1.91 Å at large step values. This shows that HC has been ejected off following a successful collision between HA and HBC.

Similarly, in the dynamic trajectory, the distance between HA and HB decreases to approximately 0.75 Å after 0.3 s. This is indicative of hydrogen bond formation between HA and HB. In addition, the distance between HB and HC increases to 3.44 Å after HC is ejected following a successful collision.

One obvious difference between the two plots is that in the minimum energy path trajectory, the internuclear distance between HA and HB is constant indicating that the bond is static. However, in the dynamic trajectory the internuclear distance is oscillating showing the the bond is vibrating. This difference could be attributed to the fact that minimum energy path does not take oscillatory motion of atoms into account while dynamic trajectory does.

However, changing the initial conditions of the system such that rAB = 0.917743 Å, rBC = 0.907743 Å and pA=pBC=0:
[[File:01409469 01409469 Reversed values dynamic mep.jpg|thumb|center|800px|'''Figure 8.''' A graph of internuclear distance against steps using minimum energy path '''Figure 9.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In this case, a hydrogen bond of length 0.77-0.79 Å is formed between HB and HC while HA is repelled away to beyond 3.45 Å.

Additionally, from the internuclear distance plot it can be seen that the distance between HB and HC increases in a rectangular hyperbola fashion when calculated using the minimum energy path. On the other hand, the internuclear distance between HB and HC increases in a linear manner when dynamic calculation is applied. A possible reason for this difference is that in the minimum energy path, the atoms are not vibrating and as such the reach the maximum distance and plateau at a faster rate. However, in the dynamic calculation, the atoms have momentum and as such move further away from each other.

=== Reactive and unreactive trajectories ===

With the initial positions r'''1''' = 0.74 and r'''2''' = 2.0, trajectories were run with the momenta combination shown below.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Contour plot !! Reactivity !! Description of the dynamics
|-
| -1.25 || -2.5 ||-99.018 ||[[File:01409469 trajectory1.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. There are no oscillations observed in HA-HB initially, indicating that the diatomic molecule is stable. A new bond is formed between HB and HC. Oscillatory behaviour is observed indicating that the bond is vibrating. On the other hand, HA is ejected away to beyond 10 Å.
|-
| -1.5 || -2.0 ||-100.456 ||[[File:01409469 trajectory2.jpg|thumb|center|400px]] ||Unreactive ||HC collides with the vibrating diatomic HA-HB. However, no bond is formed between HB and HC as the collision is unsuccessful. This is because the molecules lack sufficient kinetic energy to overcome the activation energy barrier and so no reaction occurs. HC is subsequently repelled to beyond 8 Å.
|-
| -1.5 || -2.5 ||-98.956 ||[[File:01409469 trajectory3.jpg|thumb|center|400px]] ||Reactive |||HC collides with HA-HB. The collision is successful and a bond is formed between HB and HC. This is because the atoms have sufficient kinetic energy to overcome the activation energy barrier and form a new HB-HC bond. HA is ejected away to beyond 10 Å. The new hydrogen bond formed is vibrating as seen by the oscillatory behaviour.
|-
| -2.5 || -5.0 ||-84.956 ||[[File:01409469 trajectory4.jpg|thumb|center|400px]] ||Unreactive ||HC collides with HA-HB. Initially the HA-HB bond is elongating, indicating that the hydrogen bond is weakening and going to break. However, after 0.5 s, the HA-HB bond length decreases and the HAHB is not broken. HC is repelled beyond 6 Å with very strong vibrations.
A reaction did not occur even though the atoms have sufficient kinetic energy to cross the activation energy barrier. This is because the vibrations are too strong. The system crosses the transition state region and a bond is formed in the product temporarily but then the system reverts back to the reactants. This is an example of barrier recrossing.
|-
| -2.5 || -5.2 ||-83.416 ||[[File:01409469 trajectory5.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. The collision is successful and a new hydrogen bond is formed between HB and HC with HA being ejected away. The products formed experience very strong vibrations as evidenced by the oscillatory motion which occur with a very high frequency.
|}

From the table, it can be seen that the chance of a reaction successfully occurring is not solely dependent on the amount of kinetic energy the atoms possess. Even if the atoms have sufficient kinetic energy to cross the activation energy barrier, barrier recrossing can occur if the atoms are too vibrationally excited.

=== Transition State Theory ===

Transition state theory provided the first theoretical attempt to determine absolute reaction rates. In this theory, a transition state separating the reactants from products is used to find an expression for the thermal rate constant.

The main assumptions of transition state theory are<ref name="ref2" />:

1. Electronic and nuclear motions are separated (equivalent to the Born-Oppenheimer approximation in quantum mechanics)

2. Reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution

3. Molecular systems that have crossed the transition state in the direction of products cannot turn around and reform reactants

4. In the transition state, motion along the reaction coordinate may be separated from the other motions and treated classically as a translation

5. Even in the absence of an equilibrium between reactant and product molecules, the transition states that are becoming products are distributed among their states according to the Maxwell-Boltzmann laws

The goal of transition state theory is to predict the rate constant, ''k'', of a reaction given a potential energy surface for the reaction.

However, transition state theory suffers from a set of limitations due to the assumptions stated above not holding true in reality.<ref name="ref3" />

Firstly, transition state theory assumes that barrier recrossing is negligible. However, as seen from the fourth trajectory in table 1, barrier recrossing occurred because the atoms were too vibrationally excited and the system reverted back to the reactants even though the system had already crossed the transition state region. In this situation, transition state theory would predict a faster rate of reaction compared to what was obtained experimentally.

Secondly, the transition state theory breaks down for systems with light atoms such as hydrogens. In such systems, quantum effects become crucial. Quantum effects tend to increase the rate of the reaction for two reasons. First, particles lie in energy levels which are higher then the bottom of the potential energy well (there is additional zero-point energy). Secondly, the particles can tunnel through the barrier. These two effects effectively lower the activation energy barrier. As a result, the rate of reaction predicted by transition state theory would be less than what is obtained experimentally.

Lastly, transition state theory does not hold true for systems which are far from equilibrium. In such cases, the assumption of a starting Boltzmann distribution breaks down.

== Exercise 2: F - H - H system ==

=== Potential energy surface inspection===

F + HA-HB → HA-F + HB

[[File:01409469 Surface_Plot F 3.jpg|thumb|center|400px|'''Figure 10.''' Potential energy surface for the reaction between F and H2]]

The reaction between F and H2 is an exothermic reaction. This means that the reactants are of a higher energy than the products. The overall reaction involves the breaking of a hydrogen bond and the formation of a hydrogen-fluorine bond. The HF bond formed is stronger than the hydrogen bond that was broken. This is because fluorine is more electronegative than hydrogen resulting in a polar HF bond. The negative enthalpy of H-F bond formation is greater than the positive enthalpy of hydrogen bond dissociation. Therefore, the reaction between F and H2 is an exothermic reaction.

HA + HB-F → HA-HB + F

[[File:01409469 Surface_Plot F 4.jpg|thumb|center|400px|'''Figure 11.''' Potential energy surface for the reaction between H and HF]]

The reaction between H and HF is an endothermic reaction. The products are at a higher energy level than the reactants. The overall reaction involves the breaking of a HF bond and the formation of a hydrogen bond. The formation of a homonuclear H2 molecule results in a non-polar hydrogen bond being formed. The HF bond broken is stronger than the hydrogen bond that was formed. The negative enthalpy of hydrogen bond formation is less than the positive enthalpy of hydrogen-fluorine dissociation. Therefore, the reaction between H and HF is endothermic.

=== Locating the transition state ===

[[File:01409469 Surface_Plot F TS4.jpg|thumb|center|800px|'''Figure 12.''' Determining the position of the transition state for the reaction between F and H2]]

As mentioned previously, when the molecules are not at the transition state, a reaction path will be plotted. However, if the initial conditions of the system are set such that the trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

An initial estimate for the transition state was made by setting rAB = 0.74 which is the length of a hydrogen bond. The position of the transition state was further refined by increasing the number of decimal places such that the final value of rAB was found to be 0.74478 and rBC = 1.81079. This set of distances gave a reaction path that was represented by a point indicating that no trajectory was plotted and that the molecules remained at the transition state since their initial momentum was 0.

Moreover, the kinetic energy as well as the forces acting on the atoms was calculated to be 0, which is consistent with the transition state.

Therefore, the position of the transition state was found to be at rAB = 0.74478 and rBC = 1.81079 when pA = pBC = 0.

=== Determining the activation energy ===

[[File:01409469 activation energy r1.1 combined.jpg|thumb|center|800px|'''Figure 13.''' Determining the activation energy for the reaction between F and H2]]

By reducing the value of rBC by 0.01, the atoms are no longer at the transition state and would move towards the formation of products. The activation energy was found by calculating the change in total energy after the reaction. The activation energy of F + H2 → HF + H was found to be +23.279 kcal/ mol.

[[File:01409469 activation energy r2.1 combined.jpg|thumb|center|800px|'''Figure 14.''' Determining the activation energy for the reaction between FH and H]]

The activation energy of FH + H → F + H2 was found by calculating the small dip in potential energy. The activation energy was found to be +0.280 kcal/ mol.

=== Reaction dynamics of F + HA-HB → HA-F + HB ===

The set of initial conditions that result in a reactive trajectory for F + H2 is as follows: rHH = 0.74 Å, rHF = 1.79 Å and pHH = pHF = 0.

[[File:01409469 momentum r1.jpg|thumb|center|400px|'''Figure 15.''' Graph of momentum against time for the reaction between F and H2]]

Initially, the H2 molecule is travelling in a relatively straight line with very little oscillations as evidenced by the low amplitude of momentum between 0-2 s. Following successful collision between F and H2, the amplitude of momentum of the product, HF, increases dramatically. This is because the reaction between F and H2 is exothermic and the excess energy released causes the molecule to become vibrationally excited and the amplitude of oscillations in the product to increase in the form of H-F vibrations.

[[File:01409469 energy1 r1.jpg|thumb|center|400px|'''Figure 16.''' Graph of energy against time for the reaction between F and H2]]

The law of conservation of energy states that energy cannot be created or destroyed. It simply changes from one form to another. As can be seen from the graph in figure 16, total energy remains constant. In this reaction, potential energy is converted to kinetic energy. A decrease in potential energy of the atoms following a successful collision between F and H2 results in an equal increase in kinetic energy which manifests in the form of increased product (HF) oscillations.

This can be confirmed experimentally using infrared chemiluminescence.<ref name="ref2" /> Observation of radiation from excited product molecules is a clear indication that these molecules are produced in the excited state. If the excitation is primarily vibrational, the radiation will appear in the infrared region of the spectrum (between 3-15 μm). This technique has been used extensively in hydrogen halogen reactions which produce HX molecules in the excited state.

One such experiment involves the use of the arrested relaxation method.<ref name="ref2" /> In this technique, the reaction is carried out in a chamber whose walls are maintained at a very low temperature (20 K). This results in cryopumping of the chamber in that all the products and unreacted starting materials are condensed on the walls and only those product molecules that radiate before being pumped away are observed.

=== Polanyi's empirical rules ===

In a typical chemical reaction, there is an energetic barrier — a saddle point that the reactants must surmount to become products. Which form of energy is initially deposited in the reactants, translational or vibrational, is more efficacious in surmounting the activation energy barrier is one of the core topics in the area of reaction dynamics. The Polyani rules state that vibrational energy is more effective at promoting a late-barrier reaction (transition state resembles products). Conversely, translational energy is better at promoting an early-barrier reaction (transition state resembles the reactants).<ref name="ref4" />

The forward reaction between H2 and F is exothermic. Therefore, the transition state is closer to the reactants than to the products in energy. Since the transition state occurs early in an exothermic reaction, translational energy is more effective at promoting a chemical reaction than vibrational energy.

Conversely, the backward reaction between H and HF is endothermic. According to Hammond's postulate, the transition state resembles the products more than it resembles the reactants. Therefore, in an endothermic reaction, a late transition state is observed and vibrational energy is more effective at promoting a reaction than translational energy.

=== Promoting the forward reaction: F + HA-HB → HA-F + HB ===

1. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 2.95 and pHF = -0.5.

[[File:01409469 polyani 1+2.jpg|thumb|center|800px|'''Figure 17.''' Contour plot of the reaction between F and H2 when pHH = 2.95 '''Figure 18.''' Graph of momentum against time for the reaction between F and H2 when pHH = 2.95 (Left-right)]]

Vibrational energy is represented by pHH while translational energy is represented by pHF.

A molecule of H2 collides with F. However, the collision is unsuccessful and no reaction occurs. A reaction did not occur even though the molecules have sufficient kinetic energy to overcome the activation energy barrier. This is because most of the energy is deposited in the reactants as H-H vibrations as seen by the large oscillations in H2. Barrier recrossing has occured as a result of the large vibrational energy causing the system to revert back to the reactants even though the system has already crossed the transition state and a bond was formed between HB and F temporarily.

This appears to run counter to what one would expect to happen when more energy is put into the system. It is predicted that when more energy is put into the system, the reactants would possess more kinetic energy to overcome the activation energy barrier. However, as can be seen from the simulation above, an increase in the vibrational energy of the reactants is not effective at promoting the occurrence of an exothermic reaction with an early transition state. This is in accordance to Polyani's rules.

2. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 0.1 and pHF = -0.8.

[[File:01409469 polyani 3+4 2.jpg|thumb|center|800px|'''Figure 19.''' Contour plot of the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 '''Figure 20.''' Graph of momentum against time for the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 (Left-right)]]

The collision between a molecule of H2 and F is successful and a chemical reaction has occurred. A reaction has occurred even though the overall energy of the system has been significantly reduced by lowering the amount of vibrational energy in the reactants but slightly increasing the amount of translational energy deposited in the reactants. Therefore, H2 and F have more translational energy to move towards each other and form a H-F bond. The product formed is vibrationally excited as seen by the large amplitude of oscillations in the product HF which correspond to the strong H-F vibrations.

From the second simulation, it can be seen that an increase in translational energy in the reactants is more effective at promoting an exothermic reaction with an early transition state to occur, which again follows Polyani's rules.

=== Promoting the backward reaction: HA + HB-F → HA-HB + F ===

1. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 0.2 and pHH = -3.0.

[[File:01409469 polyani 5+6 .jpg|thumb|center|800px|'''Figure 21.''' Contour plot of the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 '''Figure 22.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 (Left-right)]]

An atom of HA collides with a molecule of HBF. However, the collision is unsuccessful and no chemical reaction occurs. This is because the reactants lack sufficient vibrational energy to surmount the activation energy barrier. Therefore, following an unsuccessful collision with HBF, HA simply bounces off and moves away at constant momentum. This is in agreement with Polyani's rules which state that vibrational energy is more effective than translational energy at promoting reactions with a late transition state, for example, endothermic chemical reactions.

2. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 7.1 and pHH = -0.5.

[[File:01409469 polyani 7+8 .jpg|thumb|center|800px|'''Figure 23.''' Contour plot of the reaction between HA + HB-F when pFH = 7.1 and pHH = -0.5 '''Figure 24.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 7.1 and pHH = -0.5 (Left-right)]]

Increasing the amount of vibrational energy in the reactants by increasing pFH resulted in a chemical reaction successfully occurring. This is because the reactants now have sufficient kinetic energy in the form of H-F vibrations to overcome the activation energy barrier. In addition, because the reaction between HA + HB-F is endothermic, energy is taken in from the surroundings during the reaction and the products are less vibrationally excited than the reactants. This can be seen in the lower amplitude of oscillations in the product formed.

In conclusion, the vibrational energy of the reactants should be increased to promote an endothermic reaction with a late transition state while the translational energy of the reactants should be increased to promote an exothermic reaction with an early transition state. All 4 simulations shown above obey Polyani's rules.

== References ==
<references>
<ref name="ref1">A. C. Vaucher, M. Reiher, ''J. Chem. Theory. Comput.'', 2018, '''14''', 3091-3099.</ref>
<ref name="ref2">J. I. Steinfield, J. S. Francisco, W. L. Hase, ''Chemical Kinetics and Dynamics'', Pearson, 1998.</ref>
<ref name="ref3">D. C. Elton, ''Transition State Theory for Physicists'', 2013.</ref>
<ref name="ref4">Z. Zhang, Y. Zhou, D. H. Zhang, ''J. Phys. Chem. Lett.'', 2012, '''3''', 3416-3419.</ref>
</references>

MRD:01409469

2019-05-10T15:42:13Z

Stc1917: /* Calculation type: Dynamic and minimum energy path */

== Exercise 1: H + H2 system ==

=== Definition of a transition state ===

A potential energy profile results from plotting the potential versus reaction path.

[[File:Surface_Plot01409469_pic1.jpg|thumb|center|400px|'''Figure 1.''' A potential energy surface plot]]

The transition state is defined as the maximum on the minimum energy path linking reactants and products. This means that the transition state is a stationary point on the potential energy surface. Thus, at the transition state, the derivative of the potential energy is zero with respect to each Cartesian coordinate.

<math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>

However, a local minimum on the potential energy surface corresponding to a well would have the same <math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>. Therefore, to distinguish between a local minimum on the potential energy surface and the transition state, the second derivative, <math>\tfrac{\partial^2 V(r_i)}{\partial r_i ^2}\ </math> is determined. This can be calculated by finding the determinant of the Hessian matrix shown below.

:<math>\mathbf H =
\begin{bmatrix}
\dfrac{\partial^2 \ V(r_i)}{\partial \ x^2} & \dfrac{\partial^2 \ V(r_i)}{\partial \ x \partial \mathbf y} \\
\dfrac{\partial^2 \ V(r_i)}{\partial \ y \partial \mathbf x} & \dfrac{\partial^2 \ V(r_i)}{\partial \mathbf y^2}
\end{bmatrix}</math>

The Hessian matrix of a convex function is positive semi-definite. Refining this property allows one to test if a critical point (x,y) is a local maximum, local minimum or saddle point.<ref name="ref1" />

If <math>f</math> < 0 (exactly one negative eigenvalue), then (x,y) is an isolated local maximum.

If <math>f</math> > 0 (exactly one positive eigenvalue), then (x,y) is an isolated local minimum.

If the Hessian has both positive and negative eigenvalues then (x,y) is a saddle point.

=== Locating the transition state ===

[[File:01409469 combined pic 2 and 3.jpg|thumb|center|800px|'''Figure 2.''' A surface and contour plot of the trajectory when rAB=rBC=0.907743 Å and pA=pBC=0]]

When the molecules are not at the transition state, a surface plot of the conditions shows a trajectory being plotted. This means that the trajectory is "rolling" towards the products or reactants. However, if a trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

[[File:01409469_internuclearpic3.png|thumb|center|400px|'''Figure 3.''' A plot of internuclear distance against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

An initial estimate for the position of the transition state was made by observing the potential energy surface and was found to be 0.90 Å. At 0.90 Å, the internuclear distance against time plot showed that the internuclear distance was constantly oscillating. Therefore, the position of the transition state was further refined by increasing the number of decimal places until a horizontal line in the internuclear distance againt time plot was obtained. This means that the internuclear distance is constant and that the molecules are simply undergoing periodic symmetric vibration at the transition state. Therefore, the position of the transition state was determined to be 0.907743 Å.

[[File:01409469 combined energy pic.png|thumb|center|400px|'''Figure 4.''' A plot of energy against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

When rAB=rBC=0.907743 Å and pA=pBC=0, the kinetic energy is 0. This means that HA and HBC are stationary, which is as predicted when the trajectory is started at the transition state with no initial momentum. Furthermore, the forces acting on HA and HBC are 0 as should be the case in the transition state.

=== Calculation type: Dynamic and minimum energy path ===

The minimum energy path, is the lowest energy path from the reactants to the products. The minimum energy path is determined by following the mode corresponding to the negative Hessian eigenvalue in both directions along the steepest descent direction.<ref name="ref1" /> Following this path allows one to approach two valleys of stable structures. The resulting path generated is the minimum energy path. The reaction paths were chartered for a dynamic and a minimum energy path. The initial conditions used were that of the transition state except that the internuclear distance between HA and HB had increased to 0.917743 Å. The reaction path obtained using the two different calculation methods was different.

[[File:01409469 mep and dynamics.jpg|thumb|center|800px|'''Figure 5.''' A surface plot of the trajectory using minimum energy path '''Figure 6.''' A surface plot of the trajectory using dynamics (Left-right)]]

On the surface plot, the reaction path is depicted by the black line. Following the minimum energy path, the trajectory simply follows the valley floor to Ha+Hb-Hc. The reaction path appears to be a relatively straight line indicating that the atoms are not vibrating. In the minimum energy path, the reactants change into the products in an infinitely slow motion. This is done by having the momenta/velocities reset to zero in each time step. While the minimum energy path is useful in characterising a chemical reaction, it does not provide a realistic representation of the motion of the atoms during the reaction. This is because the mass and inertia of the atoms have been ignored.

Since the atoms simply follow the minimum energy path, they do not undergo slight displacements into surrounding valleys of higher potential energy. Furthermore, because the atoms are assumed to have no inertia, they do not undergo vibrations even if they were to be displaced to valleys of higher potential energy.

However, when dynamic calculation is performed, the reaction path is wavy indicating that the atoms are oscillating. This is because the atoms are displaced to regions of higher potential on the path from reactants to products. This leads to atom oscillation due to the inertia of the atoms.

'''Internuclear distance against time plot'''

Setting the initial system conditions such that rAB= 0.907743, rBC= 0.917743 and pA= pBC= 0:

[[File:01409469 Internuclear distance mep and dynamic.jpg|thumb|center|800px|'''Figure 6.''' A graph of internuclear distance against steps using minimum energy path '''Figure 7.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In the minimum energy path, the distance between HA and HB decreases from 0.90 Å to approximately 0.79 Å during the first 50 steps before plateauing off to a value of 0.74 Å which is the approximate length of a hydrogen bond. This means that a bond was formed between HA and HB. In addition, the plateau indicates that the bond is static. On the other hand, the distance between HB and HC has increased to 1.91 Å at large step values. This shows that HC has been ejected off following a successful collision between HA and HBC.

Similarly, in the dynamic trajectory, the distance between HA and HB decreases to approximately 0.75 Å after 0.3 s. This is indicative of hydrogen bond formation between HA and HB. In addition, the distance between HB and HC increases to 3.44 Å after HC is ejected following a successful collision.

One obvious difference between the two plots is that in the minimum energy path trajectory, the internuclear distance between HA and HB is constant indicating that the bond is static. However, in the dynamic trajectory the internuclear distance is oscillating showing the the bond is vibrating. This difference could be attributed to the fact that minimum energy path does not take oscillatory motion into account while dynamic trajectory does.

However, changing the initial conditions of the system such that rAB = 0.917743 Å, rBC = 0.907743 Å and pA=pBC=0:
[[File:01409469 01409469 Reversed values dynamic mep.jpg|thumb|center|800px|'''Figure 8.''' A graph of internuclear distance against steps using minimum energy path '''Figure 9.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In this case, a hydrogen bond of length 0.77-0.79 Å is formed between HB and HC while HA is repelled away to beyond 3.45 Å.

Additionally, from the internuclear distance plot it can be seen that the distance between HB and HC increases in a rectangular hyperbola fashion when calculated using the minimum energy path. On the other hand, the internuclear distance between HB and HC increases in a linear manner when dynamic calculation is applied. A possible reason for this difference is that in the minimum energy path, the atoms are not vibrating and as such the reach the maximum distance and plateau at a faster rate. However, in the dynamic calculation, the atoms have momentum and as such move further away from each other.

=== Reactive and unreactive trajectories ===

With the initial positions r'''1''' = 0.74 and r'''2''' = 2.0, trajectories were run with the momenta combination shown below.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Contour plot !! Reactivity !! Description of the dynamics
|-
| -1.25 || -2.5 ||-99.018 ||[[File:01409469 trajectory1.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. There are no oscillations observed in HA-HB initially, indicating that the diatomic molecule is stable. A new bond is formed between HB and HC. Oscillatory behaviour is observed indicating that the bond is vibrating. On the other hand, HA is ejected away to beyond 10 Å.
|-
| -1.5 || -2.0 ||-100.456 ||[[File:01409469 trajectory2.jpg|thumb|center|400px]] ||Unreactive ||HC collides with the vibrating diatomic HA-HB. However, no bond is formed between HB and HC as the collision is unsuccessful. This is because the molecules lack sufficient kinetic energy to overcome the activation energy barrier and so no reaction occurs. HC is subsequently repelled to beyond 8 Å.
|-
| -1.5 || -2.5 ||-98.956 ||[[File:01409469 trajectory3.jpg|thumb|center|400px]] ||Reactive |||HC collides with HA-HB. The collision is successful and a bond is formed between HB and HC. This is because the atoms have sufficient kinetic energy to overcome the activation energy barrier and form a new HB-HC bond. HA is ejected away to beyond 10 Å. The new hydrogen bond formed is vibrating as seen by the oscillatory behaviour.
|-
| -2.5 || -5.0 ||-84.956 ||[[File:01409469 trajectory4.jpg|thumb|center|400px]] ||Unreactive ||HC collides with HA-HB. Initially the HA-HB bond is elongating, indicating that the hydrogen bond is weakening and going to break. However, after 0.5 s, the HA-HB bond length decreases and the HAHB is not broken. HC is repelled beyond 6 Å with very strong vibrations.
A reaction did not occur even though the atoms have sufficient kinetic energy to cross the activation energy barrier. This is because the vibrations are too strong. The system crosses the transition state region and a bond is formed in the product temporarily but then the system reverts back to the reactants. This is an example of barrier recrossing.
|-
| -2.5 || -5.2 ||-83.416 ||[[File:01409469 trajectory5.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. The collision is successful and a new hydrogen bond is formed between HB and HC with HA being ejected away. The products formed experience very strong vibrations as evidenced by the oscillatory motion which occur with a very high frequency.
|}

From the table, it can be seen that the chance of a reaction successfully occurring is not solely dependent on the amount of kinetic energy the atoms possess. Even if the atoms have sufficient kinetic energy to cross the activation energy barrier, barrier recrossing can occur if the atoms are too vibrationally excited.

=== Transition State Theory ===

Transition state theory provided the first theoretical attempt to determine absolute reaction rates. In this theory, a transition state separating the reactants from products is used to find an expression for the thermal rate constant.

The main assumptions of transition state theory are<ref name="ref2" />:

1. Electronic and nuclear motions are separated (equivalent to the Born-Oppenheimer approximation in quantum mechanics)

2. Reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution

3. Molecular systems that have crossed the transition state in the direction of products cannot turn around and reform reactants

4. In the transition state, motion along the reaction coordinate may be separated from the other motions and treated classically as a translation

5. Even in the absence of an equilibrium between reactant and product molecules, the transition states that are becoming products are distributed among their states according to the Maxwell-Boltzmann laws

The goal of transition state theory is to predict the rate constant, ''k'', of a reaction given a potential energy surface for the reaction.

However, transition state theory suffers from a set of limitations due to the assumptions stated above not holding true in reality.<ref name="ref3" />

Firstly, transition state theory assumes that barrier recrossing is negligible. However, as seen from the fourth trajectory in table 1, barrier recrossing occurred because the atoms were too vibrationally excited and the system reverted back to the reactants even though the system had already crossed the transition state region. In this situation, transition state theory would predict a faster rate of reaction compared to what was obtained experimentally.

Secondly, the transition state theory breaks down for systems with light atoms such as hydrogens. In such systems, quantum effects become crucial. Quantum effects tend to increase the rate of the reaction for two reasons. First, particles lie in energy levels which are higher then the bottom of the potential energy well (there is additional zero-point energy). Secondly, the particles can tunnel through the barrier. These two effects effectively lower the activation energy barrier. As a result, the rate of reaction predicted by transition state theory would be less than what is obtained experimentally.

Lastly, transition state theory does not hold true for systems which are far from equilibrium. In such cases, the assumption of a starting Boltzmann distribution breaks down.

== Exercise 2: F - H - H system ==

=== Potential energy surface inspection===

F + HA-HB → HA-F + HB

[[File:01409469 Surface_Plot F 3.jpg|thumb|center|400px|'''Figure 10.''' Potential energy surface for the reaction between F and H2]]

The reaction between F and H2 is an exothermic reaction. This means that the reactants are of a higher energy than the products. The overall reaction involves the breaking of a hydrogen bond and the formation of a hydrogen-fluorine bond. The HF bond formed is stronger than the hydrogen bond that was broken. This is because fluorine is more electronegative than hydrogen resulting in a polar HF bond. The negative enthalpy of H-F bond formation is greater than the positive enthalpy of hydrogen bond dissociation. Therefore, the reaction between F and H2 is an exothermic reaction.

HA + HB-F → HA-HB + F

[[File:01409469 Surface_Plot F 4.jpg|thumb|center|400px|'''Figure 11.''' Potential energy surface for the reaction between H and HF]]

The reaction between H and HF is an endothermic reaction. The products are at a higher energy level than the reactants. The overall reaction involves the breaking of a HF bond and the formation of a hydrogen bond. The formation of a homonuclear H2 molecule results in a non-polar hydrogen bond being formed. The HF bond broken is stronger than the hydrogen bond that was formed. The negative enthalpy of hydrogen bond formation is less than the positive enthalpy of hydrogen-fluorine dissociation. Therefore, the reaction between H and HF is endothermic.

=== Locating the transition state ===

[[File:01409469 Surface_Plot F TS4.jpg|thumb|center|800px|'''Figure 12.''' Determining the position of the transition state for the reaction between F and H2]]

As mentioned previously, when the molecules are not at the transition state, a reaction path will be plotted. However, if the initial conditions of the system are set such that the trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

An initial estimate for the transition state was made by setting rAB = 0.74 which is the length of a hydrogen bond. The position of the transition state was further refined by increasing the number of decimal places such that the final value of rAB was found to be 0.74478 and rBC = 1.81079. This set of distances gave a reaction path that was represented by a point indicating that no trajectory was plotted and that the molecules remained at the transition state since their initial momentum was 0.

Moreover, the kinetic energy as well as the forces acting on the atoms was calculated to be 0, which is consistent with the transition state.

Therefore, the position of the transition state was found to be at rAB = 0.74478 and rBC = 1.81079 when pA = pBC = 0.

=== Determining the activation energy ===

[[File:01409469 activation energy r1.1 combined.jpg|thumb|center|800px|'''Figure 13.''' Determining the activation energy for the reaction between F and H2]]

By reducing the value of rBC by 0.01, the atoms are no longer at the transition state and would move towards the formation of products. The activation energy was found by calculating the change in total energy after the reaction. The activation energy of F + H2 → HF + H was found to be +23.279 kcal/ mol.

[[File:01409469 activation energy r2.1 combined.jpg|thumb|center|800px|'''Figure 14.''' Determining the activation energy for the reaction between FH and H]]

The activation energy of FH + H → F + H2 was found by calculating the small dip in potential energy. The activation energy was found to be +0.280 kcal/ mol.

=== Reaction dynamics of F + HA-HB → HA-F + HB ===

The set of initial conditions that result in a reactive trajectory for F + H2 is as follows: rHH = 0.74 Å, rHF = 1.79 Å and pHH = pHF = 0.

[[File:01409469 momentum r1.jpg|thumb|center|400px|'''Figure 15.''' Graph of momentum against time for the reaction between F and H2]]

Initially, the H2 molecule is travelling in a relatively straight line with very little oscillations as evidenced by the low amplitude of momentum between 0-2 s. Following successful collision between F and H2, the amplitude of momentum of the product, HF, increases dramatically. This is because the reaction between F and H2 is exothermic and the excess energy released causes the molecule to become vibrationally excited and the amplitude of oscillations in the product to increase in the form of H-F vibrations.

[[File:01409469 energy1 r1.jpg|thumb|center|400px|'''Figure 16.''' Graph of energy against time for the reaction between F and H2]]

The law of conservation of energy states that energy cannot be created or destroyed. It simply changes from one form to another. As can be seen from the graph in figure 16, total energy remains constant. In this reaction, potential energy is converted to kinetic energy. A decrease in potential energy of the atoms following a successful collision between F and H2 results in an equal increase in kinetic energy which manifests in the form of increased product (HF) oscillations.

This can be confirmed experimentally using infrared chemiluminescence.<ref name="ref2" /> Observation of radiation from excited product molecules is a clear indication that these molecules are produced in the excited state. If the excitation is primarily vibrational, the radiation will appear in the infrared region of the spectrum (between 3-15 μm). This technique has been used extensively in hydrogen halogen reactions which produce HX molecules in the excited state.

One such experiment involves the use of the arrested relaxation method.<ref name="ref2" /> In this technique, the reaction is carried out in a chamber whose walls are maintained at a very low temperature (20 K). This results in cryopumping of the chamber in that all the products and unreacted starting materials are condensed on the walls and only those product molecules that radiate before being pumped away are observed.

=== Polanyi's empirical rules ===

In a typical chemical reaction, there is an energetic barrier — a saddle point that the reactants must surmount to become products. Which form of energy is initially deposited in the reactants, translational or vibrational, is more efficacious in surmounting the activation energy barrier is one of the core topics in the area of reaction dynamics. The Polyani rules state that vibrational energy is more effective at promoting a late-barrier reaction (transition state resembles products). Conversely, translational energy is better at promoting an early-barrier reaction (transition state resembles the reactants).<ref name="ref4" />

The forward reaction between H2 and F is exothermic. Therefore, the transition state is closer to the reactants than to the products in energy. Since the transition state occurs early in an exothermic reaction, translational energy is more effective at promoting a chemical reaction than vibrational energy.

Conversely, the backward reaction between H and HF is endothermic. According to Hammond's postulate, the transition state resembles the products more than it resembles the reactants. Therefore, in an endothermic reaction, a late transition state is observed and vibrational energy is more effective at promoting a reaction than translational energy.

=== Promoting the forward reaction: F + HA-HB → HA-F + HB ===

1. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 2.95 and pHF = -0.5.

[[File:01409469 polyani 1+2.jpg|thumb|center|800px|'''Figure 17.''' Contour plot of the reaction between F and H2 when pHH = 2.95 '''Figure 18.''' Graph of momentum against time for the reaction between F and H2 when pHH = 2.95 (Left-right)]]

Vibrational energy is represented by pHH while translational energy is represented by pHF.

A molecule of H2 collides with F. However, the collision is unsuccessful and no reaction occurs. A reaction did not occur even though the molecules have sufficient kinetic energy to overcome the activation energy barrier. This is because most of the energy is deposited in the reactants as H-H vibrations as seen by the large oscillations in H2. Barrier recrossing has occured as a result of the large vibrational energy causing the system to revert back to the reactants even though the system has already crossed the transition state and a bond was formed between HB and F temporarily.

This appears to run counter to what one would expect to happen when more energy is put into the system. It is predicted that when more energy is put into the system, the reactants would possess more kinetic energy to overcome the activation energy barrier. However, as can be seen from the simulation above, an increase in the vibrational energy of the reactants is not effective at promoting the occurrence of an exothermic reaction with an early transition state. This is in accordance to Polyani's rules.

2. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 0.1 and pHF = -0.8.

[[File:01409469 polyani 3+4 2.jpg|thumb|center|800px|'''Figure 19.''' Contour plot of the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 '''Figure 20.''' Graph of momentum against time for the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 (Left-right)]]

The collision between a molecule of H2 and F is successful and a chemical reaction has occurred. A reaction has occurred even though the overall energy of the system has been significantly reduced by lowering the amount of vibrational energy in the reactants but slightly increasing the amount of translational energy deposited in the reactants. Therefore, H2 and F have more translational energy to move towards each other and form a H-F bond. The product formed is vibrationally excited as seen by the large amplitude of oscillations in the product HF which correspond to the strong H-F vibrations.

From the second simulation, it can be seen that an increase in translational energy in the reactants is more effective at promoting an exothermic reaction with an early transition state to occur, which again follows Polyani's rules.

=== Promoting the backward reaction: HA + HB-F → HA-HB + F ===

1. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 0.2 and pHH = -3.0.

[[File:01409469 polyani 5+6 .jpg|thumb|center|800px|'''Figure 21.''' Contour plot of the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 '''Figure 22.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 (Left-right)]]

An atom of HA collides with a molecule of HBF. However, the collision is unsuccessful and no chemical reaction occurs. This is because the reactants lack sufficient vibrational energy to surmount the activation energy barrier. Therefore, following an unsuccessful collision with HBF, HA simply bounces off and moves away at constant momentum. This is in agreement with Polyani's rules which state that vibrational energy is more effective than translational energy at promoting reactions with a late transition state, for example, endothermic chemical reactions.

2. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 7.1 and pHH = -0.5.

[[File:01409469 polyani 7+8 .jpg|thumb|center|800px|'''Figure 23.''' Contour plot of the reaction between HA + HB-F when pFH = 7.1 and pHH = -0.5 '''Figure 24.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 7.1 and pHH = -0.5 (Left-right)]]

Increasing the amount of vibrational energy in the reactants by increasing pFH resulted in a chemical reaction successfully occurring. This is because the reactants now have sufficient kinetic energy in the form of H-F vibrations to overcome the activation energy barrier. In addition, because the reaction between HA + HB-F is endothermic, energy is taken in from the surroundings during the reaction and the products are less vibrationally excited than the reactants. This can be seen in the lower amplitude of oscillations in the product formed.

In conclusion, the vibrational energy of the reactants should be increased to promote an endothermic reaction with a late transition state while the translational energy of the reactants should be increased to promote an exothermic reaction with an early transition state. All 4 simulations shown above obey Polyani's rules.

== References ==
<references>
<ref name="ref1">A. C. Vaucher, M. Reiher, ''J. Chem. Theory. Comput.'', 2018, '''14''', 3091-3099.</ref>
<ref name="ref2">J. I. Steinfield, J. S. Francisco, W. L. Hase, ''Chemical Kinetics and Dynamics'', Pearson, 1998.</ref>
<ref name="ref3">D. C. Elton, ''Transition State Theory for Physicists'', 2013.</ref>
<ref name="ref4">Z. Zhang, Y. Zhou, D. H. Zhang, ''J. Phys. Chem. Lett.'', 2012, '''3''', 3416-3419.</ref>
</references>

MRD:01409469

2019-05-10T15:24:53Z

Stc1917: /* Reaction dynamics of F + HA-HB → HA-F + HB */

== Exercise 1: H + H2 system ==

=== Definition of a transition state ===

A potential energy profile results from plotting the potential versus reaction path.

[[File:Surface_Plot01409469_pic1.jpg|thumb|center|400px|'''Figure 1.''' A potential energy surface plot]]

The transition state is defined as the maximum on the minimum energy path linking reactants and products. This means that the transition state is a stationary point on the potential energy surface. Thus, at the transition state, the derivative of the potential energy is zero with respect to each Cartesian coordinate.

<math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>

However, a local minimum on the potential energy surface corresponding to a well would have the same <math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>. Therefore, to distinguish between a local minimum on the potential energy surface and the transition state, the second derivative, <math>\tfrac{\partial^2 V(r_i)}{\partial r_i ^2}\ </math> is determined. This can be calculated by finding the determinant of the Hessian matrix shown below.

:<math>\mathbf H =
\begin{bmatrix}
\dfrac{\partial^2 \ V(r_i)}{\partial \ x^2} & \dfrac{\partial^2 \ V(r_i)}{\partial \ x \partial \mathbf y} \\
\dfrac{\partial^2 \ V(r_i)}{\partial \ y \partial \mathbf x} & \dfrac{\partial^2 \ V(r_i)}{\partial \mathbf y^2}
\end{bmatrix}</math>

The Hessian matrix of a convex function is positive semi-definite. Refining this property allows one to test if a critical point (x,y) is a local maximum, local minimum or saddle point.<ref name="ref1" />

If <math>f</math> < 0 (exactly one negative eigenvalue), then (x,y) is an isolated local maximum.

If <math>f</math> > 0 (exactly one positive eigenvalue), then (x,y) is an isolated local minimum.

If the Hessian has both positive and negative eigenvalues then (x,y) is a saddle point.

=== Locating the transition state ===

[[File:01409469 combined pic 2 and 3.jpg|thumb|center|800px|'''Figure 2.''' A surface and contour plot of the trajectory when rAB=rBC=0.907743 Å and pA=pBC=0]]

When the molecules are not at the transition state, a surface plot of the conditions shows a trajectory being plotted. This means that the trajectory is "rolling" towards the products or reactants. However, if a trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

[[File:01409469_internuclearpic3.png|thumb|center|400px|'''Figure 3.''' A plot of internuclear distance against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

An initial estimate for the position of the transition state was made by observing the potential energy surface and was found to be 0.90 Å. At 0.90 Å, the internuclear distance against time plot showed that the internuclear distance was constantly oscillating. Therefore, the position of the transition state was further refined by increasing the number of decimal places until a horizontal line in the internuclear distance againt time plot was obtained. This means that the internuclear distance is constant and that the molecules are simply undergoing periodic symmetric vibration at the transition state. Therefore, the position of the transition state was determined to be 0.907743 Å.

[[File:01409469 combined energy pic.png|thumb|center|400px|'''Figure 4.''' A plot of energy against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

When rAB=rBC=0.907743 Å and pA=pBC=0, the kinetic energy is 0. This means that HA and HBC are stationary, which is as predicted when the trajectory is started at the transition state with no initial momentum. Furthermore, the forces acting on HA and HBC are 0 as should be the case in the transition state.

=== Calculation type: Dynamic and minimum energy path ===

The minimum energy path, is the lowest energy path from the reactants to the products. The minimum energy path is determined by following the mode corresponding to the negative Hessian eigenvalue in both directions along the steepest descent direction.<ref name="ref1" /> Following this path allows one to approach two valleys of stable structures. The resulting path generated is the minimum energy path. The reaction paths were chartered for a dynamic and a minimum energy path. The initial conditions used were that of the transition state except that the internuclear distance between HA and HB had increased to 0.917743 Å. The reaction path obtained using the two different calculation methods was different.

[[File:01409469 mep and dynamics.jpg|thumb|center|800px|'''Figure 5.''' A surface plot of the trajectory using minimum energy path '''Figure 6.''' A surface plot of the trajectory using dynamics (Left-right)]]

On the surface plot, the reaction path is depicted by the black line. Following the minimum energy path, the trajectory simply follows the valley floor to Ha+Hb-Hc. The reaction path appears to be a relatively straight line indicating that the atoms are not vibrating. In the minimum energy path, the reactants change into the products in an infinitely slow motion. This is done by having the momenta/velocities reset to zero in each time step. While the minimum energy path is useful in characterising a chemical reaction, it does not provide a realistic representation of the motion of the atoms during the reaction. This is because the mass and inertia of the atoms have been ignored.

Since the atoms simply follow the minimum energy path, they do not undergo slight displacements into surrounding valleys of higher potential energy. Furthermore, because the atoms are assumed to have no inertia, they do not undergo vibrations even if they were to be displaced to valleys of higher potential energy.

However, when dynamic calculation is performed, the reaction path is wavy indicating that the atoms are oscillating. This is because the atoms are displaced to regions of higher potential on the path from reactants to products. This leads to atom oscillation due to the inertia of the atoms.

'''Internuclear distance against time plot'''

Setting the initial system conditions such that rAB= 0.907743, rBC= 0.917743 and pA= pBC= 0:

[[File:01409469 Internuclear distance mep and dynamic.jpg|thumb|center|800px|'''Figure 6.''' A graph of internuclear distance against steps using minimum energy path '''Figure 7.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In the minimum energy path, the distance between HA and HB decreases from 0.90 Å to approximately 0.79 Å during the first 50 steps before plateauing off to a value of 0.74 Å which is the approximate length of a hydrogen bond. This means that a bond was formed between HA and HB. In addition, the plateau indicates that the bond is static. On the other hand, the distance between HB and HC has increased to 1.91 Å at large step values. This shows that HC has been ejected off following a successful collision between HA and HBC.

Similarly, in the dynamic trajectory, the distance between HA and HB decreases to approximately 0.75 Å after 0.3 s. This is indicative of hydrogen bond formation between HA and HB. In addition, the distance between HB and HC increases to 3.44 Å after HC is ejected following a successful collision.

One obvious distance between the two plots is that in the minimum energy path trajectory, the internuclear distance between HA and HB is constant indicating that the bond is static. However, in the dynamic trajectory the internuclear distance is oscillating showing the the bond is vibrating. This difference could be attributed to the fact that minimum energy path does not take oscillatory motion into account while dynamic trajectory does.

However, changing the initial conditions of the system such that rAB = 0.917743 Å, rBC = 0.907743 Å and pA=pBC=0:
[[File:01409469 01409469 Reversed values dynamic mep.jpg|thumb|center|800px|'''Figure 8.''' A graph of internuclear distance against steps using minimum energy path '''Figure 9.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In this case, a hydrogen bond of length 0.77-0.79 Å is formed between HB and HC while HA is repelled away to beyond 3.45 Å.

Additionally, from the internuclear distance plot it can be seen that the distance between HB and HC increases in a rectangular hyperbola fashion when calculated using the minimum energy path. On the other hand, the internuclear distance between HB and HC increases in a linear manner when dynamic calculation is applied. A possible reason for this difference is that in the minimum energy path, the atoms are not vibrating and as such the reach the maximum distance and plateau at a faster rate. However, in the dynamic calculation, the atoms have momentum and as such move further away from each other.

=== Reactive and unreactive trajectories ===

With the initial positions r'''1''' = 0.74 and r'''2''' = 2.0, trajectories were run with the momenta combination shown below.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Contour plot !! Reactivity !! Description of the dynamics
|-
| -1.25 || -2.5 ||-99.018 ||[[File:01409469 trajectory1.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. There are no oscillations observed in HA-HB initially, indicating that the diatomic molecule is stable. A new bond is formed between HB and HC. Oscillatory behaviour is observed indicating that the bond is vibrating. On the other hand, HA is ejected away to beyond 10 Å.
|-
| -1.5 || -2.0 ||-100.456 ||[[File:01409469 trajectory2.jpg|thumb|center|400px]] ||Unreactive ||HC collides with the vibrating diatomic HA-HB. However, no bond is formed between HB and HC as the collision is unsuccessful. This is because the molecules lack sufficient kinetic energy to overcome the activation energy barrier and so no reaction occurs. HC is subsequently repelled to beyond 8 Å.
|-
| -1.5 || -2.5 ||-98.956 ||[[File:01409469 trajectory3.jpg|thumb|center|400px]] ||Reactive |||HC collides with HA-HB. The collision is successful and a bond is formed between HB and HC. This is because the atoms have sufficient kinetic energy to overcome the activation energy barrier and form a new HB-HC bond. HA is ejected away to beyond 10 Å. The new hydrogen bond formed is vibrating as seen by the oscillatory behaviour.
|-
| -2.5 || -5.0 ||-84.956 ||[[File:01409469 trajectory4.jpg|thumb|center|400px]] ||Unreactive ||HC collides with HA-HB. Initially the HA-HB bond is elongating, indicating that the hydrogen bond is weakening and going to break. However, after 0.5 s, the HA-HB bond length decreases and the HAHB is not broken. HC is repelled beyond 6 Å with very strong vibrations.
A reaction did not occur even though the atoms have sufficient kinetic energy to cross the activation energy barrier. This is because the vibrations are too strong. The system crosses the transition state region and a bond is formed in the product temporarily but then the system reverts back to the reactants. This is an example of barrier recrossing.
|-
| -2.5 || -5.2 ||-83.416 ||[[File:01409469 trajectory5.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. The collision is successful and a new hydrogen bond is formed between HB and HC with HA being ejected away. The products formed experience very strong vibrations as evidenced by the oscillatory motion which occur with a very high frequency.
|}

From the table, it can be seen that the chance of a reaction successfully occurring is not solely dependent on the amount of kinetic energy the atoms possess. Even if the atoms have sufficient kinetic energy to cross the activation energy barrier, barrier recrossing can occur if the atoms are too vibrationally excited.

=== Transition State Theory ===

Transition state theory provided the first theoretical attempt to determine absolute reaction rates. In this theory, a transition state separating the reactants from products is used to find an expression for the thermal rate constant.

The main assumptions of transition state theory are<ref name="ref2" />:

1. Electronic and nuclear motions are separated (equivalent to the Born-Oppenheimer approximation in quantum mechanics)

2. Reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution

3. Molecular systems that have crossed the transition state in the direction of products cannot turn around and reform reactants

4. In the transition state, motion along the reaction coordinate may be separated from the other motions and treated classically as a translation

5. Even in the absence of an equilibrium between reactant and product molecules, the transition states that are becoming products are distributed among their states according to the Maxwell-Boltzmann laws

The goal of transition state theory is to predict the rate constant, ''k'', of a reaction given a potential energy surface for the reaction.

However, transition state theory suffers from a set of limitations due to the assumptions stated above not holding true in reality.<ref name="ref3" />

Firstly, transition state theory assumes that barrier recrossing is negligible. However, as seen from the fourth trajectory in table 1, barrier recrossing occurred because the atoms were too vibrationally excited and the system reverted back to the reactants even though the system had already crossed the transition state region. In this situation, transition state theory would predict a faster rate of reaction compared to what was obtained experimentally.

Secondly, the transition state theory breaks down for systems with light atoms such as hydrogens. In such systems, quantum effects become crucial. Quantum effects tend to increase the rate of the reaction for two reasons. First, particles lie in energy levels which are higher then the bottom of the potential energy well (there is additional zero-point energy). Secondly, the particles can tunnel through the barrier. These two effects effectively lower the activation energy barrier. As a result, the rate of reaction predicted by transition state theory would be less than what is obtained experimentally.

Lastly, transition state theory does not hold true for systems which are far from equilibrium. In such cases, the assumption of a starting Boltzmann distribution breaks down.

== Exercise 2: F - H - H system ==

=== Potential energy surface inspection===

F + HA-HB → HA-F + HB

[[File:01409469 Surface_Plot F 3.jpg|thumb|center|400px|'''Figure 10.''' Potential energy surface for the reaction between F and H2]]

The reaction between F and H2 is an exothermic reaction. This means that the reactants are of a higher energy than the products. The overall reaction involves the breaking of a hydrogen bond and the formation of a hydrogen-fluorine bond. The HF bond formed is stronger than the hydrogen bond that was broken. This is because fluorine is more electronegative than hydrogen resulting in a polar HF bond. The negative enthalpy of H-F bond formation is greater than the positive enthalpy of hydrogen bond dissociation. Therefore, the reaction between F and H2 is an exothermic reaction.

HA + HB-F → HA-HB + F

[[File:01409469 Surface_Plot F 4.jpg|thumb|center|400px|'''Figure 11.''' Potential energy surface for the reaction between H and HF]]

The reaction between H and HF is an endothermic reaction. The products are at a higher energy level than the reactants. The overall reaction involves the breaking of a HF bond and the formation of a hydrogen bond. The formation of a homonuclear H2 molecule results in a non-polar hydrogen bond being formed. The HF bond broken is stronger than the hydrogen bond that was formed. The negative enthalpy of hydrogen bond formation is less than the positive enthalpy of hydrogen-fluorine dissociation. Therefore, the reaction between H and HF is endothermic.

=== Locating the transition state ===

[[File:01409469 Surface_Plot F TS4.jpg|thumb|center|800px|'''Figure 12.''' Determining the position of the transition state for the reaction between F and H2]]

As mentioned previously, when the molecules are not at the transition state, a reaction path will be plotted. However, if the initial conditions of the system are set such that the trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

An initial estimate for the transition state was made by setting rAB = 0.74 which is the length of a hydrogen bond. The position of the transition state was further refined by increasing the number of decimal places such that the final value of rAB was found to be 0.74478 and rBC = 1.81079. This set of distances gave a reaction path that was represented by a point indicating that no trajectory was plotted and that the molecules remained at the transition state since their initial momentum was 0.

Moreover, the kinetic energy as well as the forces acting on the atoms was calculated to be 0, which is consistent with the transition state.

Therefore, the position of the transition state was found to be at rAB = 0.74478 and rBC = 1.81079 when pA = pBC = 0.

=== Determining the activation energy ===

[[File:01409469 activation energy r1.1 combined.jpg|thumb|center|800px|'''Figure 13.''' Determining the activation energy for the reaction between F and H2]]

By reducing the value of rBC by 0.01, the atoms are no longer at the transition state and would move towards the formation of products. The activation energy was found by calculating the change in total energy after the reaction. The activation energy of F + H2 → HF + H was found to be +23.279 kcal/ mol.

[[File:01409469 activation energy r2.1 combined.jpg|thumb|center|800px|'''Figure 14.''' Determining the activation energy for the reaction between FH and H]]

The activation energy of FH + H → F + H2 was found by calculating the small dip in potential energy. The activation energy was found to be +0.280 kcal/ mol.

=== Reaction dynamics of F + HA-HB → HA-F + HB ===

The set of initial conditions that result in a reactive trajectory for F + H2 is as follows: rHH = 0.74 Å, rHF = 1.79 Å and pHH = pHF = 0.

[[File:01409469 momentum r1.jpg|thumb|center|400px|'''Figure 15.''' Graph of momentum against time for the reaction between F and H2]]

Initially, the H2 molecule is travelling in a relatively straight line with very little oscillations as evidenced by the low amplitude of momentum between 0-2 s. Following successful collision between F and H2, the amplitude of momentum of the product, HF, increases dramatically. This is because the reaction between F and H2 is exothermic and the excess energy released causes the molecule to become vibrationally excited and the amplitude of oscillations in the product to increase in the form of H-F vibrations.

[[File:01409469 energy1 r1.jpg|thumb|center|400px|'''Figure 16.''' Graph of energy against time for the reaction between F and H2]]

The law of conservation of energy states that energy cannot be created or destroyed. It simply changes from one form to another. As can be seen from the graph in figure 16, total energy remains constant. In this reaction, potential energy is converted to kinetic energy. A decrease in potential energy of the atoms following a successful collision between F and H2 results in an equal increase in kinetic energy which manifests in the form of increased product (HF) oscillations.

This can be confirmed experimentally using infrared chemiluminescence.<ref name="ref2" /> Observation of radiation from excited product molecules is a clear indication that these molecules are produced in the excited state. If the excitation is primarily vibrational, the radiation will appear in the infrared region of the spectrum (between 3-15 μm). This technique has been used extensively in hydrogen halogen reactions which produce HX molecules in the excited state.

One such experiment involves the use of the arrested relaxation method.<ref name="ref2" /> In this technique, the reaction is carried out in a chamber whose walls are maintained at a very low temperature (20 K). This results in cryopumping of the chamber in that all the products and unreacted starting materials are condensed on the walls and only those product molecules that radiate before being pumped away are observed.

=== Polanyi's empirical rules ===

In a typical chemical reaction, there is an energetic barrier — a saddle point that the reactants must surmount to become products. Which form of energy is initially deposited in the reactants, translational or vibrational, is more efficacious in surmounting the activation energy barrier is one of the core topics in the area of reaction dynamics. The Polyani rules state that vibrational energy is more effective at promoting a late-barrier reaction (transition state resembles products). Conversely, translational energy is better at promoting an early-barrier reaction (transition state resembles the reactants).<ref name="ref4" />

The forward reaction between H2 and F is exothermic. Therefore, the transition state is closer to the reactants than to the products in energy. Since the transition state occurs early in an exothermic reaction, translational energy is more effective at promoting a chemical reaction than vibrational energy.

Conversely, the backward reaction between H and HF is endothermic. According to Hammond's postulate, the transition state resembles the products more than it resembles the reactants. Therefore, in an endothermic reaction, a late transition state is observed and vibrational energy is more effective at promoting a reaction than translational energy.

=== Promoting the forward reaction: F + HA-HB → HA-F + HB ===

1. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 2.95 and pHF = -0.5.

[[File:01409469 polyani 1+2.jpg|thumb|center|800px|'''Figure 17.''' Contour plot of the reaction between F and H2 when pHH = 2.95 '''Figure 18.''' Graph of momentum against time for the reaction between F and H2 when pHH = 2.95 (Left-right)]]

Vibrational energy is represented by pHH while translational energy is represented by pHF.

A molecule of H2 collides with F. However, the collision is unsuccessful and no reaction occurs. A reaction did not occur even though the molecules have sufficient kinetic energy to overcome the activation energy barrier. This is because most of the energy is deposited in the reactants as H-H vibrations as seen by the large oscillations in H2. Barrier recrossing has occured as a result of the large vibrational energy causing the system to revert back to the reactants even though the system has already crossed the transition state and a bond was formed between HB and F temporarily.

This appears to run counter to what one would expect to happen when more energy is put into the system. It is predicted that when more energy is put into the system, the reactants would possess more kinetic energy to overcome the activation energy barrier. However, as can be seen from the simulation above, an increase in the vibrational energy of the reactants is not effective at promoting the occurrence of an exothermic reaction with an early transition state. This is in accordance to Polyani's rules.

2. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 0.1 and pHF = -0.8.

[[File:01409469 polyani 3+4 2.jpg|thumb|center|800px|'''Figure 19.''' Contour plot of the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 '''Figure 20.''' Graph of momentum against time for the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 (Left-right)]]

The collision between a molecule of H2 and F is successful and a chemical reaction has occurred. A reaction has occurred even though the overall energy of the system has been significantly reduced by lowering the amount of vibrational energy in the reactants but slightly increasing the amount of translational energy deposited in the reactants. Therefore, H2 and F have more translational energy to move towards each other and form a H-F bond. The product formed is vibrationally excited as seen by the large amplitude of oscillations in the product HF which correspond to the strong H-F vibrations.

From the second simulation, it can be seen that an increase in translational energy in the reactants is more effective at promoting an exothermic reaction with an early transition state to occur, which again follows Polyani's rules.

=== Promoting the backward reaction: HA + HB-F → HA-HB + F ===

1. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 0.2 and pHH = -3.0.

[[File:01409469 polyani 5+6 .jpg|thumb|center|800px|'''Figure 21.''' Contour plot of the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 '''Figure 22.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 (Left-right)]]

An atom of HA collides with a molecule of HBF. However, the collision is unsuccessful and no chemical reaction occurs. This is because the reactants lack sufficient vibrational energy to surmount the activation energy barrier. Therefore, following an unsuccessful collision with HBF, HA simply bounces off and moves away at constant momentum. This is in agreement with Polyani's rules which state that vibrational energy is more effective than translational energy at promoting reactions with a late transition state, for example, endothermic chemical reactions.

2. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 7.1 and pHH = -0.5.

[[File:01409469 polyani 7+8 .jpg|thumb|center|800px|'''Figure 23.''' Contour plot of the reaction between HA + HB-F when pFH = 7.1 and pHH = -0.5 '''Figure 24.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 7.1 and pHH = -0.5 (Left-right)]]

Increasing the amount of vibrational energy in the reactants by increasing pFH resulted in a chemical reaction successfully occurring. This is because the reactants now have sufficient kinetic energy in the form of H-F vibrations to overcome the activation energy barrier. In addition, because the reaction between HA + HB-F is endothermic, energy is taken in from the surroundings during the reaction and the products are less vibrationally excited than the reactants. This can be seen in the lower amplitude of oscillations in the product formed.

In conclusion, the vibrational energy of the reactants should be increased to promote an endothermic reaction with a late transition state while the translational energy of the reactants should be increased to promote an exothermic reaction with an early transition state. All 4 simulations shown above obey Polyani's rules.

== References ==
<references>
<ref name="ref1">A. C. Vaucher, M. Reiher, ''J. Chem. Theory. Comput.'', 2018, '''14''', 3091-3099.</ref>
<ref name="ref2">J. I. Steinfield, J. S. Francisco, W. L. Hase, ''Chemical Kinetics and Dynamics'', Pearson, 1998.</ref>
<ref name="ref3">D. C. Elton, ''Transition State Theory for Physicists'', 2013.</ref>
<ref name="ref4">Z. Zhang, Y. Zhou, D. H. Zhang, ''J. Phys. Chem. Lett.'', 2012, '''3''', 3416-3419.</ref>
</references>

MRD:01409469

2019-05-10T15:12:08Z

Stc1917: /* References */

== Exercise 1: H + H2 system ==

=== Definition of a transition state ===

A potential energy profile results from plotting the potential versus reaction path.

[[File:Surface_Plot01409469_pic1.jpg|thumb|center|400px|'''Figure 1.''' A potential energy surface plot]]

The transition state is defined as the maximum on the minimum energy path linking reactants and products. This means that the transition state is a stationary point on the potential energy surface. Thus, at the transition state, the derivative of the potential energy is zero with respect to each Cartesian coordinate.

<math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>

However, a local minimum on the potential energy surface corresponding to a well would have the same <math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>. Therefore, to distinguish between a local minimum on the potential energy surface and the transition state, the second derivative, <math>\tfrac{\partial^2 V(r_i)}{\partial r_i ^2}\ </math> is determined. This can be calculated by finding the determinant of the Hessian matrix shown below.

:<math>\mathbf H =
\begin{bmatrix}
\dfrac{\partial^2 \ V(r_i)}{\partial \ x^2} & \dfrac{\partial^2 \ V(r_i)}{\partial \ x \partial \mathbf y} \\
\dfrac{\partial^2 \ V(r_i)}{\partial \ y \partial \mathbf x} & \dfrac{\partial^2 \ V(r_i)}{\partial \mathbf y^2}
\end{bmatrix}</math>

The Hessian matrix of a convex function is positive semi-definite. Refining this property allows one to test if a critical point (x,y) is a local maximum, local minimum or saddle point.<ref name="ref1" />

If <math>f</math> < 0 (exactly one negative eigenvalue), then (x,y) is an isolated local maximum.

If <math>f</math> > 0 (exactly one positive eigenvalue), then (x,y) is an isolated local minimum.

If the Hessian has both positive and negative eigenvalues then (x,y) is a saddle point.

=== Locating the transition state ===

[[File:01409469 combined pic 2 and 3.jpg|thumb|center|800px|'''Figure 2.''' A surface and contour plot of the trajectory when rAB=rBC=0.907743 Å and pA=pBC=0]]

When the molecules are not at the transition state, a surface plot of the conditions shows a trajectory being plotted. This means that the trajectory is "rolling" towards the products or reactants. However, if a trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

[[File:01409469_internuclearpic3.png|thumb|center|400px|'''Figure 3.''' A plot of internuclear distance against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

An initial estimate for the position of the transition state was made by observing the potential energy surface and was found to be 0.90 Å. At 0.90 Å, the internuclear distance against time plot showed that the internuclear distance was constantly oscillating. Therefore, the position of the transition state was further refined by increasing the number of decimal places until a horizontal line in the internuclear distance againt time plot was obtained. This means that the internuclear distance is constant and that the molecules are simply undergoing periodic symmetric vibration at the transition state. Therefore, the position of the transition state was determined to be 0.907743 Å.

[[File:01409469 combined energy pic.png|thumb|center|400px|'''Figure 4.''' A plot of energy against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

When rAB=rBC=0.907743 Å and pA=pBC=0, the kinetic energy is 0. This means that HA and HBC are stationary, which is as predicted when the trajectory is started at the transition state with no initial momentum. Furthermore, the forces acting on HA and HBC are 0 as should be the case in the transition state.

=== Calculation type: Dynamic and minimum energy path ===

The minimum energy path, is the lowest energy path from the reactants to the products. The minimum energy path is determined by following the mode corresponding to the negative Hessian eigenvalue in both directions along the steepest descent direction.<ref name="ref1" /> Following this path allows one to approach two valleys of stable structures. The resulting path generated is the minimum energy path. The reaction paths were chartered for a dynamic and a minimum energy path. The initial conditions used were that of the transition state except that the internuclear distance between HA and HB had increased to 0.917743 Å. The reaction path obtained using the two different calculation methods was different.

[[File:01409469 mep and dynamics.jpg|thumb|center|800px|'''Figure 5.''' A surface plot of the trajectory using minimum energy path '''Figure 6.''' A surface plot of the trajectory using dynamics (Left-right)]]

On the surface plot, the reaction path is depicted by the black line. Following the minimum energy path, the trajectory simply follows the valley floor to Ha+Hb-Hc. The reaction path appears to be a relatively straight line indicating that the atoms are not vibrating. In the minimum energy path, the reactants change into the products in an infinitely slow motion. This is done by having the momenta/velocities reset to zero in each time step. While the minimum energy path is useful in characterising a chemical reaction, it does not provide a realistic representation of the motion of the atoms during the reaction. This is because the mass and inertia of the atoms have been ignored.

Since the atoms simply follow the minimum energy path, they do not undergo slight displacements into surrounding valleys of higher potential energy. Furthermore, because the atoms are assumed to have no inertia, they do not undergo vibrations even if they were to be displaced to valleys of higher potential energy.

However, when dynamic calculation is performed, the reaction path is wavy indicating that the atoms are oscillating. This is because the atoms are displaced to regions of higher potential on the path from reactants to products. This leads to atom oscillation due to the inertia of the atoms.

'''Internuclear distance against time plot'''

Setting the initial system conditions such that rAB= 0.907743, rBC= 0.917743 and pA= pBC= 0:

[[File:01409469 Internuclear distance mep and dynamic.jpg|thumb|center|800px|'''Figure 6.''' A graph of internuclear distance against steps using minimum energy path '''Figure 7.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In the minimum energy path, the distance between HA and HB decreases from 0.90 Å to approximately 0.79 Å during the first 50 steps before plateauing off to a value of 0.74 Å which is the approximate length of a hydrogen bond. This means that a bond was formed between HA and HB. In addition, the plateau indicates that the bond is static. On the other hand, the distance between HB and HC has increased to 1.91 Å at large step values. This shows that HC has been ejected off following a successful collision between HA and HBC.

Similarly, in the dynamic trajectory, the distance between HA and HB decreases to approximately 0.75 Å after 0.3 s. This is indicative of hydrogen bond formation between HA and HB. In addition, the distance between HB and HC increases to 3.44 Å after HC is ejected following a successful collision.

One obvious distance between the two plots is that in the minimum energy path trajectory, the internuclear distance between HA and HB is constant indicating that the bond is static. However, in the dynamic trajectory the internuclear distance is oscillating showing the the bond is vibrating. This difference could be attributed to the fact that minimum energy path does not take oscillatory motion into account while dynamic trajectory does.

However, changing the initial conditions of the system such that rAB = 0.917743 Å, rBC = 0.907743 Å and pA=pBC=0:
[[File:01409469 01409469 Reversed values dynamic mep.jpg|thumb|center|800px|'''Figure 8.''' A graph of internuclear distance against steps using minimum energy path '''Figure 9.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In this case, a hydrogen bond of length 0.77-0.79 Å is formed between HB and HC while HA is repelled away to beyond 3.45 Å.

Additionally, from the internuclear distance plot it can be seen that the distance between HB and HC increases in a rectangular hyperbola fashion when calculated using the minimum energy path. On the other hand, the internuclear distance between HB and HC increases in a linear manner when dynamic calculation is applied. A possible reason for this difference is that in the minimum energy path, the atoms are not vibrating and as such the reach the maximum distance and plateau at a faster rate. However, in the dynamic calculation, the atoms have momentum and as such move further away from each other.

=== Reactive and unreactive trajectories ===

With the initial positions r'''1''' = 0.74 and r'''2''' = 2.0, trajectories were run with the momenta combination shown below.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Contour plot !! Reactivity !! Description of the dynamics
|-
| -1.25 || -2.5 ||-99.018 ||[[File:01409469 trajectory1.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. There are no oscillations observed in HA-HB initially, indicating that the diatomic molecule is stable. A new bond is formed between HB and HC. Oscillatory behaviour is observed indicating that the bond is vibrating. On the other hand, HA is ejected away to beyond 10 Å.
|-
| -1.5 || -2.0 ||-100.456 ||[[File:01409469 trajectory2.jpg|thumb|center|400px]] ||Unreactive ||HC collides with the vibrating diatomic HA-HB. However, no bond is formed between HB and HC as the collision is unsuccessful. This is because the molecules lack sufficient kinetic energy to overcome the activation energy barrier and so no reaction occurs. HC is subsequently repelled to beyond 8 Å.
|-
| -1.5 || -2.5 ||-98.956 ||[[File:01409469 trajectory3.jpg|thumb|center|400px]] ||Reactive |||HC collides with HA-HB. The collision is successful and a bond is formed between HB and HC. This is because the atoms have sufficient kinetic energy to overcome the activation energy barrier and form a new HB-HC bond. HA is ejected away to beyond 10 Å. The new hydrogen bond formed is vibrating as seen by the oscillatory behaviour.
|-
| -2.5 || -5.0 ||-84.956 ||[[File:01409469 trajectory4.jpg|thumb|center|400px]] ||Unreactive ||HC collides with HA-HB. Initially the HA-HB bond is elongating, indicating that the hydrogen bond is weakening and going to break. However, after 0.5 s, the HA-HB bond length decreases and the HAHB is not broken. HC is repelled beyond 6 Å with very strong vibrations.
A reaction did not occur even though the atoms have sufficient kinetic energy to cross the activation energy barrier. This is because the vibrations are too strong. The system crosses the transition state region and a bond is formed in the product temporarily but then the system reverts back to the reactants. This is an example of barrier recrossing.
|-
| -2.5 || -5.2 ||-83.416 ||[[File:01409469 trajectory5.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. The collision is successful and a new hydrogen bond is formed between HB and HC with HA being ejected away. The products formed experience very strong vibrations as evidenced by the oscillatory motion which occur with a very high frequency.
|}

From the table, it can be seen that the chance of a reaction successfully occurring is not solely dependent on the amount of kinetic energy the atoms possess. Even if the atoms have sufficient kinetic energy to cross the activation energy barrier, barrier recrossing can occur if the atoms are too vibrationally excited.

=== Transition State Theory ===

Transition state theory provided the first theoretical attempt to determine absolute reaction rates. In this theory, a transition state separating the reactants from products is used to find an expression for the thermal rate constant.

The main assumptions of transition state theory are<ref name="ref2" />:

1. Electronic and nuclear motions are separated (equivalent to the Born-Oppenheimer approximation in quantum mechanics)

2. Reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution

3. Molecular systems that have crossed the transition state in the direction of products cannot turn around and reform reactants

4. In the transition state, motion along the reaction coordinate may be separated from the other motions and treated classically as a translation

5. Even in the absence of an equilibrium between reactant and product molecules, the transition states that are becoming products are distributed among their states according to the Maxwell-Boltzmann laws

The goal of transition state theory is to predict the rate constant, ''k'', of a reaction given a potential energy surface for the reaction.

However, transition state theory suffers from a set of limitations due to the assumptions stated above not holding true in reality.<ref name="ref3" />

Firstly, transition state theory assumes that barrier recrossing is negligible. However, as seen from the fourth trajectory in table 1, barrier recrossing occurred because the atoms were too vibrationally excited and the system reverted back to the reactants even though the system had already crossed the transition state region. In this situation, transition state theory would predict a faster rate of reaction compared to what was obtained experimentally.

Secondly, the transition state theory breaks down for systems with light atoms such as hydrogens. In such systems, quantum effects become crucial. Quantum effects tend to increase the rate of the reaction for two reasons. First, particles lie in energy levels which are higher then the bottom of the potential energy well (there is additional zero-point energy). Secondly, the particles can tunnel through the barrier. These two effects effectively lower the activation energy barrier. As a result, the rate of reaction predicted by transition state theory would be less than what is obtained experimentally.

Lastly, transition state theory does not hold true for systems which are far from equilibrium. In such cases, the assumption of a starting Boltzmann distribution breaks down.

== Exercise 2: F - H - H system ==

=== Potential energy surface inspection===

F + HA-HB → HA-F + HB

[[File:01409469 Surface_Plot F 3.jpg|thumb|center|400px|'''Figure 10.''' Potential energy surface for the reaction between F and H2]]

The reaction between F and H2 is an exothermic reaction. This means that the reactants are of a higher energy than the products. The overall reaction involves the breaking of a hydrogen bond and the formation of a hydrogen-fluorine bond. The HF bond formed is stronger than the hydrogen bond that was broken. This is because fluorine is more electronegative than hydrogen resulting in a polar HF bond. The negative enthalpy of H-F bond formation is greater than the positive enthalpy of hydrogen bond dissociation. Therefore, the reaction between F and H2 is an exothermic reaction.

HA + HB-F → HA-HB + F

[[File:01409469 Surface_Plot F 4.jpg|thumb|center|400px|'''Figure 11.''' Potential energy surface for the reaction between H and HF]]

The reaction between H and HF is an endothermic reaction. The products are at a higher energy level than the reactants. The overall reaction involves the breaking of a HF bond and the formation of a hydrogen bond. The formation of a homonuclear H2 molecule results in a non-polar hydrogen bond being formed. The HF bond broken is stronger than the hydrogen bond that was formed. The negative enthalpy of hydrogen bond formation is less than the positive enthalpy of hydrogen-fluorine dissociation. Therefore, the reaction between H and HF is endothermic.

=== Locating the transition state ===

[[File:01409469 Surface_Plot F TS4.jpg|thumb|center|800px|'''Figure 12.''' Determining the position of the transition state for the reaction between F and H2]]

As mentioned previously, when the molecules are not at the transition state, a reaction path will be plotted. However, if the initial conditions of the system are set such that the trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

An initial estimate for the transition state was made by setting rAB = 0.74 which is the length of a hydrogen bond. The position of the transition state was further refined by increasing the number of decimal places such that the final value of rAB was found to be 0.74478 and rBC = 1.81079. This set of distances gave a reaction path that was represented by a point indicating that no trajectory was plotted and that the molecules remained at the transition state since their initial momentum was 0.

Moreover, the kinetic energy as well as the forces acting on the atoms was calculated to be 0, which is consistent with the transition state.

Therefore, the position of the transition state was found to be at rAB = 0.74478 and rBC = 1.81079 when pA = pBC = 0.

=== Determining the activation energy ===

[[File:01409469 activation energy r1.1 combined.jpg|thumb|center|800px|'''Figure 13.''' Determining the activation energy for the reaction between F and H2]]

By reducing the value of rBC by 0.01, the atoms are no longer at the transition state and would move towards the formation of products. The activation energy was found by calculating the change in total energy after the reaction. The activation energy of F + H2 → HF + H was found to be +23.279 kcal/ mol.

[[File:01409469 activation energy r2.1 combined.jpg|thumb|center|800px|'''Figure 14.''' Determining the activation energy for the reaction between FH and H]]

The activation energy of FH + H → F + H2 was found by calculating the small dip in potential energy. The activation energy was found to be +0.280 kcal/ mol.

=== Reaction dynamics of F + HA-HB → HA-F + HB ===

The set of initial conditions that result in a reactive trajectory for F + H2 is as follows: rHH = 0.74 Å, rHF = 1.79 Å and pHH = pHF = 0.

[[File:01409469 momentum r1.jpg|thumb|center|400px|'''Figure 15.''' Graph of momentum against time for the reaction between F and H2]]

Initially, the H2 molecule is travelling in a relatively straight line with very little oscillations as evidenced by the low amplitude of momentum between 0-2 s. Following successful collision between F and H2, the amplitude of momentum of the product, HF, increases dramatically. This is because the reaction between F and H2 is exothermic and the excess energy released causes the molecule to become vibrationally excited and the amplitude of oscillations in the product to increase.

[[File:01409469 energy1 r1.jpg|thumb|center|400px|'''Figure 16.''' Graph of energy against time for the reaction between F and H2]]

The law of conservation of energy states that energy cannot be created or destroyed. It simply changes from one form to another. As can be seen from the graph in figure 16, total energy remains constant. In this reaction, potential energy is converted to kinetic energy. A decrease in potential energy of the atoms following a successful collision between F and H2 results in an equal increase in kinetic energy which manifests in the form of increased product (HF) oscillations.

This can be confirmed experimentally using infrared chemiluminescence.<ref name="ref2" /> Observation of radiation from excited product molecules is a clear indication that these molecules are produced in the excited state. If the excitation is primarily vibrational, the radiation will appear in the infrared region of the spectrum (between 3-15 μm). This technique has been used extensively in hydrogen halogen reactions which produce HX molecules in the excited state.

One such experiment involves the use of the arrested relaxation method.<ref name="ref2" /> In this technique, the reaction is carried out in a chamber whose walls are maintained at a very low temperature (20 K). This results in cryopumping of the chamber in that all the products and unreacted starting materials are condensed on the walls and only those product molecules that radiate before being pumped away are observed.

=== Polanyi's empirical rules ===

In a typical chemical reaction, there is an energetic barrier — a saddle point that the reactants must surmount to become products. Which form of energy is initially deposited in the reactants, translational or vibrational, is more efficacious in surmounting the activation energy barrier is one of the core topics in the area of reaction dynamics. The Polyani rules state that vibrational energy is more effective at promoting a late-barrier reaction (transition state resembles products). Conversely, translational energy is better at promoting an early-barrier reaction (transition state resembles the reactants).<ref name="ref4" />

The forward reaction between H2 and F is exothermic. Therefore, the transition state is closer to the reactants than to the products in energy. Since the transition state occurs early in an exothermic reaction, translational energy is more effective at promoting a chemical reaction than vibrational energy.

Conversely, the backward reaction between H and HF is endothermic. According to Hammond's postulate, the transition state resembles the products more than it resembles the reactants. Therefore, in an endothermic reaction, a late transition state is observed and vibrational energy is more effective at promoting a reaction than translational energy.

=== Promoting the forward reaction: F + HA-HB → HA-F + HB ===

1. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 2.95 and pHF = -0.5.

[[File:01409469 polyani 1+2.jpg|thumb|center|800px|'''Figure 17.''' Contour plot of the reaction between F and H2 when pHH = 2.95 '''Figure 18.''' Graph of momentum against time for the reaction between F and H2 when pHH = 2.95 (Left-right)]]

Vibrational energy is represented by pHH while translational energy is represented by pHF.

A molecule of H2 collides with F. However, the collision is unsuccessful and no reaction occurs. A reaction did not occur even though the molecules have sufficient kinetic energy to overcome the activation energy barrier. This is because most of the energy is deposited in the reactants as H-H vibrations as seen by the large oscillations in H2. Barrier recrossing has occured as a result of the large vibrational energy causing the system to revert back to the reactants even though the system has already crossed the transition state and a bond was formed between HB and F temporarily.

This appears to run counter to what one would expect to happen when more energy is put into the system. It is predicted that when more energy is put into the system, the reactants would possess more kinetic energy to overcome the activation energy barrier. However, as can be seen from the simulation above, an increase in the vibrational energy of the reactants is not effective at promoting the occurrence of an exothermic reaction with an early transition state. This is in accordance to Polyani's rules.

2. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 0.1 and pHF = -0.8.

[[File:01409469 polyani 3+4 2.jpg|thumb|center|800px|'''Figure 19.''' Contour plot of the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 '''Figure 20.''' Graph of momentum against time for the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 (Left-right)]]

The collision between a molecule of H2 and F is successful and a chemical reaction has occurred. A reaction has occurred even though the overall energy of the system has been significantly reduced by lowering the amount of vibrational energy in the reactants but slightly increasing the amount of translational energy deposited in the reactants. Therefore, H2 and F have more translational energy to move towards each other and form a H-F bond. The product formed is vibrationally excited as seen by the large amplitude of oscillations in the product HF which correspond to the strong H-F vibrations.

From the second simulation, it can be seen that an increase in translational energy in the reactants is more effective at promoting an exothermic reaction with an early transition state to occur, which again follows Polyani's rules.

=== Promoting the backward reaction: HA + HB-F → HA-HB + F ===

1. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 0.2 and pHH = -3.0.

[[File:01409469 polyani 5+6 .jpg|thumb|center|800px|'''Figure 21.''' Contour plot of the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 '''Figure 22.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 (Left-right)]]

An atom of HA collides with a molecule of HBF. However, the collision is unsuccessful and no chemical reaction occurs. This is because the reactants lack sufficient vibrational energy to surmount the activation energy barrier. Therefore, following an unsuccessful collision with HBF, HA simply bounces off and moves away at constant momentum. This is in agreement with Polyani's rules which state that vibrational energy is more effective than translational energy at promoting reactions with a late transition state, for example, endothermic chemical reactions.

2. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 7.1 and pHH = -0.5.

[[File:01409469 polyani 7+8 .jpg|thumb|center|800px|'''Figure 23.''' Contour plot of the reaction between HA + HB-F when pFH = 7.1 and pHH = -0.5 '''Figure 24.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 7.1 and pHH = -0.5 (Left-right)]]

Increasing the amount of vibrational energy in the reactants by increasing pFH resulted in a chemical reaction successfully occurring. This is because the reactants now have sufficient kinetic energy in the form of H-F vibrations to overcome the activation energy barrier. In addition, because the reaction between HA + HB-F is endothermic, energy is taken in from the surroundings during the reaction and the products are less vibrationally excited than the reactants. This can be seen in the lower amplitude of oscillations in the product formed.

In conclusion, the vibrational energy of the reactants should be increased to promote an endothermic reaction with a late transition state while the translational energy of the reactants should be increased to promote an exothermic reaction with an early transition state. All 4 simulations shown above obey Polyani's rules.

== References ==
<references>
<ref name="ref1">A. C. Vaucher, M. Reiher, ''J. Chem. Theory. Comput.'', 2018, '''14''', 3091-3099.</ref>
<ref name="ref2">J. I. Steinfield, J. S. Francisco, W. L. Hase, ''Chemical Kinetics and Dynamics'', Pearson, 1998.</ref>
<ref name="ref3">D. C. Elton, ''Transition State Theory for Physicists'', 2013.</ref>
<ref name="ref4">Z. Zhang, Y. Zhou, D. H. Zhang, ''J. Phys. Chem. Lett.'', 2012, '''3''', 3416-3419.</ref>
</references>

MRD:01409469

2019-05-10T15:10:01Z

Stc1917: /* Polanyi's empirical rules */

== Exercise 1: H + H2 system ==

=== Definition of a transition state ===

A potential energy profile results from plotting the potential versus reaction path.

[[File:Surface_Plot01409469_pic1.jpg|thumb|center|400px|'''Figure 1.''' A potential energy surface plot]]

The transition state is defined as the maximum on the minimum energy path linking reactants and products. This means that the transition state is a stationary point on the potential energy surface. Thus, at the transition state, the derivative of the potential energy is zero with respect to each Cartesian coordinate.

<math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>

However, a local minimum on the potential energy surface corresponding to a well would have the same <math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>. Therefore, to distinguish between a local minimum on the potential energy surface and the transition state, the second derivative, <math>\tfrac{\partial^2 V(r_i)}{\partial r_i ^2}\ </math> is determined. This can be calculated by finding the determinant of the Hessian matrix shown below.

:<math>\mathbf H =
\begin{bmatrix}
\dfrac{\partial^2 \ V(r_i)}{\partial \ x^2} & \dfrac{\partial^2 \ V(r_i)}{\partial \ x \partial \mathbf y} \\
\dfrac{\partial^2 \ V(r_i)}{\partial \ y \partial \mathbf x} & \dfrac{\partial^2 \ V(r_i)}{\partial \mathbf y^2}
\end{bmatrix}</math>

The Hessian matrix of a convex function is positive semi-definite. Refining this property allows one to test if a critical point (x,y) is a local maximum, local minimum or saddle point.<ref name="ref1" />

If <math>f</math> < 0 (exactly one negative eigenvalue), then (x,y) is an isolated local maximum.

If <math>f</math> > 0 (exactly one positive eigenvalue), then (x,y) is an isolated local minimum.

If the Hessian has both positive and negative eigenvalues then (x,y) is a saddle point.

=== Locating the transition state ===

[[File:01409469 combined pic 2 and 3.jpg|thumb|center|800px|'''Figure 2.''' A surface and contour plot of the trajectory when rAB=rBC=0.907743 Å and pA=pBC=0]]

When the molecules are not at the transition state, a surface plot of the conditions shows a trajectory being plotted. This means that the trajectory is "rolling" towards the products or reactants. However, if a trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

[[File:01409469_internuclearpic3.png|thumb|center|400px|'''Figure 3.''' A plot of internuclear distance against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

An initial estimate for the position of the transition state was made by observing the potential energy surface and was found to be 0.90 Å. At 0.90 Å, the internuclear distance against time plot showed that the internuclear distance was constantly oscillating. Therefore, the position of the transition state was further refined by increasing the number of decimal places until a horizontal line in the internuclear distance againt time plot was obtained. This means that the internuclear distance is constant and that the molecules are simply undergoing periodic symmetric vibration at the transition state. Therefore, the position of the transition state was determined to be 0.907743 Å.

[[File:01409469 combined energy pic.png|thumb|center|400px|'''Figure 4.''' A plot of energy against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

When rAB=rBC=0.907743 Å and pA=pBC=0, the kinetic energy is 0. This means that HA and HBC are stationary, which is as predicted when the trajectory is started at the transition state with no initial momentum. Furthermore, the forces acting on HA and HBC are 0 as should be the case in the transition state.

=== Calculation type: Dynamic and minimum energy path ===

The minimum energy path, is the lowest energy path from the reactants to the products. The minimum energy path is determined by following the mode corresponding to the negative Hessian eigenvalue in both directions along the steepest descent direction.<ref name="ref1" /> Following this path allows one to approach two valleys of stable structures. The resulting path generated is the minimum energy path. The reaction paths were chartered for a dynamic and a minimum energy path. The initial conditions used were that of the transition state except that the internuclear distance between HA and HB had increased to 0.917743 Å. The reaction path obtained using the two different calculation methods was different.

[[File:01409469 mep and dynamics.jpg|thumb|center|800px|'''Figure 5.''' A surface plot of the trajectory using minimum energy path '''Figure 6.''' A surface plot of the trajectory using dynamics (Left-right)]]

On the surface plot, the reaction path is depicted by the black line. Following the minimum energy path, the trajectory simply follows the valley floor to Ha+Hb-Hc. The reaction path appears to be a relatively straight line indicating that the atoms are not vibrating. In the minimum energy path, the reactants change into the products in an infinitely slow motion. This is done by having the momenta/velocities reset to zero in each time step. While the minimum energy path is useful in characterising a chemical reaction, it does not provide a realistic representation of the motion of the atoms during the reaction. This is because the mass and inertia of the atoms have been ignored.

Since the atoms simply follow the minimum energy path, they do not undergo slight displacements into surrounding valleys of higher potential energy. Furthermore, because the atoms are assumed to have no inertia, they do not undergo vibrations even if they were to be displaced to valleys of higher potential energy.

However, when dynamic calculation is performed, the reaction path is wavy indicating that the atoms are oscillating. This is because the atoms are displaced to regions of higher potential on the path from reactants to products. This leads to atom oscillation due to the inertia of the atoms.

'''Internuclear distance against time plot'''

Setting the initial system conditions such that rAB= 0.907743, rBC= 0.917743 and pA= pBC= 0:

[[File:01409469 Internuclear distance mep and dynamic.jpg|thumb|center|800px|'''Figure 6.''' A graph of internuclear distance against steps using minimum energy path '''Figure 7.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In the minimum energy path, the distance between HA and HB decreases from 0.90 Å to approximately 0.79 Å during the first 50 steps before plateauing off to a value of 0.74 Å which is the approximate length of a hydrogen bond. This means that a bond was formed between HA and HB. In addition, the plateau indicates that the bond is static. On the other hand, the distance between HB and HC has increased to 1.91 Å at large step values. This shows that HC has been ejected off following a successful collision between HA and HBC.

Similarly, in the dynamic trajectory, the distance between HA and HB decreases to approximately 0.75 Å after 0.3 s. This is indicative of hydrogen bond formation between HA and HB. In addition, the distance between HB and HC increases to 3.44 Å after HC is ejected following a successful collision.

One obvious distance between the two plots is that in the minimum energy path trajectory, the internuclear distance between HA and HB is constant indicating that the bond is static. However, in the dynamic trajectory the internuclear distance is oscillating showing the the bond is vibrating. This difference could be attributed to the fact that minimum energy path does not take oscillatory motion into account while dynamic trajectory does.

However, changing the initial conditions of the system such that rAB = 0.917743 Å, rBC = 0.907743 Å and pA=pBC=0:
[[File:01409469 01409469 Reversed values dynamic mep.jpg|thumb|center|800px|'''Figure 8.''' A graph of internuclear distance against steps using minimum energy path '''Figure 9.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In this case, a hydrogen bond of length 0.77-0.79 Å is formed between HB and HC while HA is repelled away to beyond 3.45 Å.

Additionally, from the internuclear distance plot it can be seen that the distance between HB and HC increases in a rectangular hyperbola fashion when calculated using the minimum energy path. On the other hand, the internuclear distance between HB and HC increases in a linear manner when dynamic calculation is applied. A possible reason for this difference is that in the minimum energy path, the atoms are not vibrating and as such the reach the maximum distance and plateau at a faster rate. However, in the dynamic calculation, the atoms have momentum and as such move further away from each other.

=== Reactive and unreactive trajectories ===

With the initial positions r'''1''' = 0.74 and r'''2''' = 2.0, trajectories were run with the momenta combination shown below.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Contour plot !! Reactivity !! Description of the dynamics
|-
| -1.25 || -2.5 ||-99.018 ||[[File:01409469 trajectory1.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. There are no oscillations observed in HA-HB initially, indicating that the diatomic molecule is stable. A new bond is formed between HB and HC. Oscillatory behaviour is observed indicating that the bond is vibrating. On the other hand, HA is ejected away to beyond 10 Å.
|-
| -1.5 || -2.0 ||-100.456 ||[[File:01409469 trajectory2.jpg|thumb|center|400px]] ||Unreactive ||HC collides with the vibrating diatomic HA-HB. However, no bond is formed between HB and HC as the collision is unsuccessful. This is because the molecules lack sufficient kinetic energy to overcome the activation energy barrier and so no reaction occurs. HC is subsequently repelled to beyond 8 Å.
|-
| -1.5 || -2.5 ||-98.956 ||[[File:01409469 trajectory3.jpg|thumb|center|400px]] ||Reactive |||HC collides with HA-HB. The collision is successful and a bond is formed between HB and HC. This is because the atoms have sufficient kinetic energy to overcome the activation energy barrier and form a new HB-HC bond. HA is ejected away to beyond 10 Å. The new hydrogen bond formed is vibrating as seen by the oscillatory behaviour.
|-
| -2.5 || -5.0 ||-84.956 ||[[File:01409469 trajectory4.jpg|thumb|center|400px]] ||Unreactive ||HC collides with HA-HB. Initially the HA-HB bond is elongating, indicating that the hydrogen bond is weakening and going to break. However, after 0.5 s, the HA-HB bond length decreases and the HAHB is not broken. HC is repelled beyond 6 Å with very strong vibrations.
A reaction did not occur even though the atoms have sufficient kinetic energy to cross the activation energy barrier. This is because the vibrations are too strong. The system crosses the transition state region and a bond is formed in the product temporarily but then the system reverts back to the reactants. This is an example of barrier recrossing.
|-
| -2.5 || -5.2 ||-83.416 ||[[File:01409469 trajectory5.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. The collision is successful and a new hydrogen bond is formed between HB and HC with HA being ejected away. The products formed experience very strong vibrations as evidenced by the oscillatory motion which occur with a very high frequency.
|}

From the table, it can be seen that the chance of a reaction successfully occurring is not solely dependent on the amount of kinetic energy the atoms possess. Even if the atoms have sufficient kinetic energy to cross the activation energy barrier, barrier recrossing can occur if the atoms are too vibrationally excited.

=== Transition State Theory ===

Transition state theory provided the first theoretical attempt to determine absolute reaction rates. In this theory, a transition state separating the reactants from products is used to find an expression for the thermal rate constant.

The main assumptions of transition state theory are<ref name="ref2" />:

1. Electronic and nuclear motions are separated (equivalent to the Born-Oppenheimer approximation in quantum mechanics)

2. Reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution

3. Molecular systems that have crossed the transition state in the direction of products cannot turn around and reform reactants

4. In the transition state, motion along the reaction coordinate may be separated from the other motions and treated classically as a translation

5. Even in the absence of an equilibrium between reactant and product molecules, the transition states that are becoming products are distributed among their states according to the Maxwell-Boltzmann laws

The goal of transition state theory is to predict the rate constant, ''k'', of a reaction given a potential energy surface for the reaction.

However, transition state theory suffers from a set of limitations due to the assumptions stated above not holding true in reality.<ref name="ref3" />

Firstly, transition state theory assumes that barrier recrossing is negligible. However, as seen from the fourth trajectory in table 1, barrier recrossing occurred because the atoms were too vibrationally excited and the system reverted back to the reactants even though the system had already crossed the transition state region. In this situation, transition state theory would predict a faster rate of reaction compared to what was obtained experimentally.

Secondly, the transition state theory breaks down for systems with light atoms such as hydrogens. In such systems, quantum effects become crucial. Quantum effects tend to increase the rate of the reaction for two reasons. First, particles lie in energy levels which are higher then the bottom of the potential energy well (there is additional zero-point energy). Secondly, the particles can tunnel through the barrier. These two effects effectively lower the activation energy barrier. As a result, the rate of reaction predicted by transition state theory would be less than what is obtained experimentally.

Lastly, transition state theory does not hold true for systems which are far from equilibrium. In such cases, the assumption of a starting Boltzmann distribution breaks down.

== Exercise 2: F - H - H system ==

=== Potential energy surface inspection===

F + HA-HB → HA-F + HB

[[File:01409469 Surface_Plot F 3.jpg|thumb|center|400px|'''Figure 10.''' Potential energy surface for the reaction between F and H2]]

The reaction between F and H2 is an exothermic reaction. This means that the reactants are of a higher energy than the products. The overall reaction involves the breaking of a hydrogen bond and the formation of a hydrogen-fluorine bond. The HF bond formed is stronger than the hydrogen bond that was broken. This is because fluorine is more electronegative than hydrogen resulting in a polar HF bond. The negative enthalpy of H-F bond formation is greater than the positive enthalpy of hydrogen bond dissociation. Therefore, the reaction between F and H2 is an exothermic reaction.

HA + HB-F → HA-HB + F

[[File:01409469 Surface_Plot F 4.jpg|thumb|center|400px|'''Figure 11.''' Potential energy surface for the reaction between H and HF]]

The reaction between H and HF is an endothermic reaction. The products are at a higher energy level than the reactants. The overall reaction involves the breaking of a HF bond and the formation of a hydrogen bond. The formation of a homonuclear H2 molecule results in a non-polar hydrogen bond being formed. The HF bond broken is stronger than the hydrogen bond that was formed. The negative enthalpy of hydrogen bond formation is less than the positive enthalpy of hydrogen-fluorine dissociation. Therefore, the reaction between H and HF is endothermic.

=== Locating the transition state ===

[[File:01409469 Surface_Plot F TS4.jpg|thumb|center|800px|'''Figure 12.''' Determining the position of the transition state for the reaction between F and H2]]

As mentioned previously, when the molecules are not at the transition state, a reaction path will be plotted. However, if the initial conditions of the system are set such that the trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

An initial estimate for the transition state was made by setting rAB = 0.74 which is the length of a hydrogen bond. The position of the transition state was further refined by increasing the number of decimal places such that the final value of rAB was found to be 0.74478 and rBC = 1.81079. This set of distances gave a reaction path that was represented by a point indicating that no trajectory was plotted and that the molecules remained at the transition state since their initial momentum was 0.

Moreover, the kinetic energy as well as the forces acting on the atoms was calculated to be 0, which is consistent with the transition state.

Therefore, the position of the transition state was found to be at rAB = 0.74478 and rBC = 1.81079 when pA = pBC = 0.

=== Determining the activation energy ===

[[File:01409469 activation energy r1.1 combined.jpg|thumb|center|800px|'''Figure 13.''' Determining the activation energy for the reaction between F and H2]]

By reducing the value of rBC by 0.01, the atoms are no longer at the transition state and would move towards the formation of products. The activation energy was found by calculating the change in total energy after the reaction. The activation energy of F + H2 → HF + H was found to be +23.279 kcal/ mol.

[[File:01409469 activation energy r2.1 combined.jpg|thumb|center|800px|'''Figure 14.''' Determining the activation energy for the reaction between FH and H]]

The activation energy of FH + H → F + H2 was found by calculating the small dip in potential energy. The activation energy was found to be +0.280 kcal/ mol.

=== Reaction dynamics of F + HA-HB → HA-F + HB ===

The set of initial conditions that result in a reactive trajectory for F + H2 is as follows: rHH = 0.74 Å, rHF = 1.79 Å and pHH = pHF = 0.

[[File:01409469 momentum r1.jpg|thumb|center|400px|'''Figure 15.''' Graph of momentum against time for the reaction between F and H2]]

Initially, the H2 molecule is travelling in a relatively straight line with very little oscillations as evidenced by the low amplitude of momentum between 0-2 s. Following successful collision between F and H2, the amplitude of momentum of the product, HF, increases dramatically. This is because the reaction between F and H2 is exothermic and the excess energy released causes the molecule to become vibrationally excited and the amplitude of oscillations in the product to increase.

[[File:01409469 energy1 r1.jpg|thumb|center|400px|'''Figure 16.''' Graph of energy against time for the reaction between F and H2]]

The law of conservation of energy states that energy cannot be created or destroyed. It simply changes from one form to another. As can be seen from the graph in figure 16, total energy remains constant. In this reaction, potential energy is converted to kinetic energy. A decrease in potential energy of the atoms following a successful collision between F and H2 results in an equal increase in kinetic energy which manifests in the form of increased product (HF) oscillations.

This can be confirmed experimentally using infrared chemiluminescence.<ref name="ref2" /> Observation of radiation from excited product molecules is a clear indication that these molecules are produced in the excited state. If the excitation is primarily vibrational, the radiation will appear in the infrared region of the spectrum (between 3-15 μm). This technique has been used extensively in hydrogen halogen reactions which produce HX molecules in the excited state.

One such experiment involves the use of the arrested relaxation method.<ref name="ref2" /> In this technique, the reaction is carried out in a chamber whose walls are maintained at a very low temperature (20 K). This results in cryopumping of the chamber in that all the products and unreacted starting materials are condensed on the walls and only those product molecules that radiate before being pumped away are observed.

=== Polanyi's empirical rules ===

In a typical chemical reaction, there is an energetic barrier — a saddle point that the reactants must surmount to become products. Which form of energy is initially deposited in the reactants, translational or vibrational, is more efficacious in surmounting the activation energy barrier is one of the core topics in the area of reaction dynamics. The Polyani rules state that vibrational energy is more effective at promoting a late-barrier reaction (transition state resembles products). Conversely, translational energy is better at promoting an early-barrier reaction (transition state resembles the reactants).<ref name="ref4" />

The forward reaction between H2 and F is exothermic. Therefore, the transition state is closer to the reactants than to the products in energy. Since the transition state occurs early in an exothermic reaction, translational energy is more effective at promoting a chemical reaction than vibrational energy.

Conversely, the backward reaction between H and HF is endothermic. According to Hammond's postulate, the transition state resembles the products more than it resembles the reactants. Therefore, in an endothermic reaction, a late transition state is observed and vibrational energy is more effective at promoting a reaction than translational energy.

=== Promoting the forward reaction: F + HA-HB → HA-F + HB ===

1. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 2.95 and pHF = -0.5.

[[File:01409469 polyani 1+2.jpg|thumb|center|800px|'''Figure 17.''' Contour plot of the reaction between F and H2 when pHH = 2.95 '''Figure 18.''' Graph of momentum against time for the reaction between F and H2 when pHH = 2.95 (Left-right)]]

Vibrational energy is represented by pHH while translational energy is represented by pHF.

A molecule of H2 collides with F. However, the collision is unsuccessful and no reaction occurs. A reaction did not occur even though the molecules have sufficient kinetic energy to overcome the activation energy barrier. This is because most of the energy is deposited in the reactants as H-H vibrations as seen by the large oscillations in H2. Barrier recrossing has occured as a result of the large vibrational energy causing the system to revert back to the reactants even though the system has already crossed the transition state and a bond was formed between HB and F temporarily.

This appears to run counter to what one would expect to happen when more energy is put into the system. It is predicted that when more energy is put into the system, the reactants would possess more kinetic energy to overcome the activation energy barrier. However, as can be seen from the simulation above, an increase in the vibrational energy of the reactants is not effective at promoting the occurrence of an exothermic reaction with an early transition state. This is in accordance to Polyani's rules.

2. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 0.1 and pHF = -0.8.

[[File:01409469 polyani 3+4 2.jpg|thumb|center|800px|'''Figure 19.''' Contour plot of the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 '''Figure 20.''' Graph of momentum against time for the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 (Left-right)]]

The collision between a molecule of H2 and F is successful and a chemical reaction has occurred. A reaction has occurred even though the overall energy of the system has been significantly reduced by lowering the amount of vibrational energy in the reactants but slightly increasing the amount of translational energy deposited in the reactants. Therefore, H2 and F have more translational energy to move towards each other and form a H-F bond. The product formed is vibrationally excited as seen by the large amplitude of oscillations in the product HF which correspond to the strong H-F vibrations.

From the second simulation, it can be seen that an increase in translational energy in the reactants is more effective at promoting an exothermic reaction with an early transition state to occur, which again follows Polyani's rules.

=== Promoting the backward reaction: HA + HB-F → HA-HB + F ===

1. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 0.2 and pHH = -3.0.

[[File:01409469 polyani 5+6 .jpg|thumb|center|800px|'''Figure 21.''' Contour plot of the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 '''Figure 22.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 (Left-right)]]

An atom of HA collides with a molecule of HBF. However, the collision is unsuccessful and no chemical reaction occurs. This is because the reactants lack sufficient vibrational energy to surmount the activation energy barrier. Therefore, following an unsuccessful collision with HBF, HA simply bounces off and moves away at constant momentum. This is in agreement with Polyani's rules which state that vibrational energy is more effective than translational energy at promoting reactions with a late transition state, for example, endothermic chemical reactions.

2. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 7.1 and pHH = -0.5.

[[File:01409469 polyani 7+8 .jpg|thumb|center|800px|'''Figure 23.''' Contour plot of the reaction between HA + HB-F when pFH = 7.1 and pHH = -0.5 '''Figure 24.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 7.1 and pHH = -0.5 (Left-right)]]

Increasing the amount of vibrational energy in the reactants by increasing pFH resulted in a chemical reaction successfully occurring. This is because the reactants now have sufficient kinetic energy in the form of H-F vibrations to overcome the activation energy barrier. In addition, because the reaction between HA + HB-F is endothermic, energy is taken in from the surroundings during the reaction and the products are less vibrationally excited than the reactants. This can be seen in the lower amplitude of oscillations in the product formed.

In conclusion, the vibrational energy of the reactants should be increased to promote an endothermic reaction with a late transition state while the translational energy of the reactants should be increased to promote an exothermic reaction with an early transition state. All 4 simulations shown above obey Polyani's rules.

== References ==
<references>
<ref name="ref1">A. C. Vaucher, M. Reiher, ''J. Chem. Theory. Comput.'', 2018, '''14''', 3091-3099.</ref>
<ref name="ref2">J. I. Steinfield, J. S. Francisco, W. L. Hase, ''Chemical Kinetics and Dynamics'', Pearson, 1998.</ref>
<ref name="ref3">D. C. Elton, ''Transition State Theory for Physicists'', 2013.</ref>
</references>

MRD:01409469

2019-05-10T15:09:18Z

Stc1917: /* Reaction dynamics of F + HA-HB → HA-F + HB */

== Exercise 1: H + H2 system ==

=== Definition of a transition state ===

A potential energy profile results from plotting the potential versus reaction path.

[[File:Surface_Plot01409469_pic1.jpg|thumb|center|400px|'''Figure 1.''' A potential energy surface plot]]

The transition state is defined as the maximum on the minimum energy path linking reactants and products. This means that the transition state is a stationary point on the potential energy surface. Thus, at the transition state, the derivative of the potential energy is zero with respect to each Cartesian coordinate.

<math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>

However, a local minimum on the potential energy surface corresponding to a well would have the same <math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>. Therefore, to distinguish between a local minimum on the potential energy surface and the transition state, the second derivative, <math>\tfrac{\partial^2 V(r_i)}{\partial r_i ^2}\ </math> is determined. This can be calculated by finding the determinant of the Hessian matrix shown below.

:<math>\mathbf H =
\begin{bmatrix}
\dfrac{\partial^2 \ V(r_i)}{\partial \ x^2} & \dfrac{\partial^2 \ V(r_i)}{\partial \ x \partial \mathbf y} \\
\dfrac{\partial^2 \ V(r_i)}{\partial \ y \partial \mathbf x} & \dfrac{\partial^2 \ V(r_i)}{\partial \mathbf y^2}
\end{bmatrix}</math>

The Hessian matrix of a convex function is positive semi-definite. Refining this property allows one to test if a critical point (x,y) is a local maximum, local minimum or saddle point.<ref name="ref1" />

If <math>f</math> < 0 (exactly one negative eigenvalue), then (x,y) is an isolated local maximum.

If <math>f</math> > 0 (exactly one positive eigenvalue), then (x,y) is an isolated local minimum.

If the Hessian has both positive and negative eigenvalues then (x,y) is a saddle point.

=== Locating the transition state ===

[[File:01409469 combined pic 2 and 3.jpg|thumb|center|800px|'''Figure 2.''' A surface and contour plot of the trajectory when rAB=rBC=0.907743 Å and pA=pBC=0]]

When the molecules are not at the transition state, a surface plot of the conditions shows a trajectory being plotted. This means that the trajectory is "rolling" towards the products or reactants. However, if a trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

[[File:01409469_internuclearpic3.png|thumb|center|400px|'''Figure 3.''' A plot of internuclear distance against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

An initial estimate for the position of the transition state was made by observing the potential energy surface and was found to be 0.90 Å. At 0.90 Å, the internuclear distance against time plot showed that the internuclear distance was constantly oscillating. Therefore, the position of the transition state was further refined by increasing the number of decimal places until a horizontal line in the internuclear distance againt time plot was obtained. This means that the internuclear distance is constant and that the molecules are simply undergoing periodic symmetric vibration at the transition state. Therefore, the position of the transition state was determined to be 0.907743 Å.

[[File:01409469 combined energy pic.png|thumb|center|400px|'''Figure 4.''' A plot of energy against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

When rAB=rBC=0.907743 Å and pA=pBC=0, the kinetic energy is 0. This means that HA and HBC are stationary, which is as predicted when the trajectory is started at the transition state with no initial momentum. Furthermore, the forces acting on HA and HBC are 0 as should be the case in the transition state.

=== Calculation type: Dynamic and minimum energy path ===

The minimum energy path, is the lowest energy path from the reactants to the products. The minimum energy path is determined by following the mode corresponding to the negative Hessian eigenvalue in both directions along the steepest descent direction.<ref name="ref1" /> Following this path allows one to approach two valleys of stable structures. The resulting path generated is the minimum energy path. The reaction paths were chartered for a dynamic and a minimum energy path. The initial conditions used were that of the transition state except that the internuclear distance between HA and HB had increased to 0.917743 Å. The reaction path obtained using the two different calculation methods was different.

[[File:01409469 mep and dynamics.jpg|thumb|center|800px|'''Figure 5.''' A surface plot of the trajectory using minimum energy path '''Figure 6.''' A surface plot of the trajectory using dynamics (Left-right)]]

On the surface plot, the reaction path is depicted by the black line. Following the minimum energy path, the trajectory simply follows the valley floor to Ha+Hb-Hc. The reaction path appears to be a relatively straight line indicating that the atoms are not vibrating. In the minimum energy path, the reactants change into the products in an infinitely slow motion. This is done by having the momenta/velocities reset to zero in each time step. While the minimum energy path is useful in characterising a chemical reaction, it does not provide a realistic representation of the motion of the atoms during the reaction. This is because the mass and inertia of the atoms have been ignored.

Since the atoms simply follow the minimum energy path, they do not undergo slight displacements into surrounding valleys of higher potential energy. Furthermore, because the atoms are assumed to have no inertia, they do not undergo vibrations even if they were to be displaced to valleys of higher potential energy.

However, when dynamic calculation is performed, the reaction path is wavy indicating that the atoms are oscillating. This is because the atoms are displaced to regions of higher potential on the path from reactants to products. This leads to atom oscillation due to the inertia of the atoms.

'''Internuclear distance against time plot'''

Setting the initial system conditions such that rAB= 0.907743, rBC= 0.917743 and pA= pBC= 0:

[[File:01409469 Internuclear distance mep and dynamic.jpg|thumb|center|800px|'''Figure 6.''' A graph of internuclear distance against steps using minimum energy path '''Figure 7.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In the minimum energy path, the distance between HA and HB decreases from 0.90 Å to approximately 0.79 Å during the first 50 steps before plateauing off to a value of 0.74 Å which is the approximate length of a hydrogen bond. This means that a bond was formed between HA and HB. In addition, the plateau indicates that the bond is static. On the other hand, the distance between HB and HC has increased to 1.91 Å at large step values. This shows that HC has been ejected off following a successful collision between HA and HBC.

Similarly, in the dynamic trajectory, the distance between HA and HB decreases to approximately 0.75 Å after 0.3 s. This is indicative of hydrogen bond formation between HA and HB. In addition, the distance between HB and HC increases to 3.44 Å after HC is ejected following a successful collision.

One obvious distance between the two plots is that in the minimum energy path trajectory, the internuclear distance between HA and HB is constant indicating that the bond is static. However, in the dynamic trajectory the internuclear distance is oscillating showing the the bond is vibrating. This difference could be attributed to the fact that minimum energy path does not take oscillatory motion into account while dynamic trajectory does.

However, changing the initial conditions of the system such that rAB = 0.917743 Å, rBC = 0.907743 Å and pA=pBC=0:
[[File:01409469 01409469 Reversed values dynamic mep.jpg|thumb|center|800px|'''Figure 8.''' A graph of internuclear distance against steps using minimum energy path '''Figure 9.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In this case, a hydrogen bond of length 0.77-0.79 Å is formed between HB and HC while HA is repelled away to beyond 3.45 Å.

Additionally, from the internuclear distance plot it can be seen that the distance between HB and HC increases in a rectangular hyperbola fashion when calculated using the minimum energy path. On the other hand, the internuclear distance between HB and HC increases in a linear manner when dynamic calculation is applied. A possible reason for this difference is that in the minimum energy path, the atoms are not vibrating and as such the reach the maximum distance and plateau at a faster rate. However, in the dynamic calculation, the atoms have momentum and as such move further away from each other.

=== Reactive and unreactive trajectories ===

With the initial positions r'''1''' = 0.74 and r'''2''' = 2.0, trajectories were run with the momenta combination shown below.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Contour plot !! Reactivity !! Description of the dynamics
|-
| -1.25 || -2.5 ||-99.018 ||[[File:01409469 trajectory1.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. There are no oscillations observed in HA-HB initially, indicating that the diatomic molecule is stable. A new bond is formed between HB and HC. Oscillatory behaviour is observed indicating that the bond is vibrating. On the other hand, HA is ejected away to beyond 10 Å.
|-
| -1.5 || -2.0 ||-100.456 ||[[File:01409469 trajectory2.jpg|thumb|center|400px]] ||Unreactive ||HC collides with the vibrating diatomic HA-HB. However, no bond is formed between HB and HC as the collision is unsuccessful. This is because the molecules lack sufficient kinetic energy to overcome the activation energy barrier and so no reaction occurs. HC is subsequently repelled to beyond 8 Å.
|-
| -1.5 || -2.5 ||-98.956 ||[[File:01409469 trajectory3.jpg|thumb|center|400px]] ||Reactive |||HC collides with HA-HB. The collision is successful and a bond is formed between HB and HC. This is because the atoms have sufficient kinetic energy to overcome the activation energy barrier and form a new HB-HC bond. HA is ejected away to beyond 10 Å. The new hydrogen bond formed is vibrating as seen by the oscillatory behaviour.
|-
| -2.5 || -5.0 ||-84.956 ||[[File:01409469 trajectory4.jpg|thumb|center|400px]] ||Unreactive ||HC collides with HA-HB. Initially the HA-HB bond is elongating, indicating that the hydrogen bond is weakening and going to break. However, after 0.5 s, the HA-HB bond length decreases and the HAHB is not broken. HC is repelled beyond 6 Å with very strong vibrations.
A reaction did not occur even though the atoms have sufficient kinetic energy to cross the activation energy barrier. This is because the vibrations are too strong. The system crosses the transition state region and a bond is formed in the product temporarily but then the system reverts back to the reactants. This is an example of barrier recrossing.
|-
| -2.5 || -5.2 ||-83.416 ||[[File:01409469 trajectory5.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. The collision is successful and a new hydrogen bond is formed between HB and HC with HA being ejected away. The products formed experience very strong vibrations as evidenced by the oscillatory motion which occur with a very high frequency.
|}

From the table, it can be seen that the chance of a reaction successfully occurring is not solely dependent on the amount of kinetic energy the atoms possess. Even if the atoms have sufficient kinetic energy to cross the activation energy barrier, barrier recrossing can occur if the atoms are too vibrationally excited.

=== Transition State Theory ===

Transition state theory provided the first theoretical attempt to determine absolute reaction rates. In this theory, a transition state separating the reactants from products is used to find an expression for the thermal rate constant.

The main assumptions of transition state theory are<ref name="ref2" />:

1. Electronic and nuclear motions are separated (equivalent to the Born-Oppenheimer approximation in quantum mechanics)

2. Reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution

3. Molecular systems that have crossed the transition state in the direction of products cannot turn around and reform reactants

4. In the transition state, motion along the reaction coordinate may be separated from the other motions and treated classically as a translation

5. Even in the absence of an equilibrium between reactant and product molecules, the transition states that are becoming products are distributed among their states according to the Maxwell-Boltzmann laws

The goal of transition state theory is to predict the rate constant, ''k'', of a reaction given a potential energy surface for the reaction.

However, transition state theory suffers from a set of limitations due to the assumptions stated above not holding true in reality.<ref name="ref3" />

Firstly, transition state theory assumes that barrier recrossing is negligible. However, as seen from the fourth trajectory in table 1, barrier recrossing occurred because the atoms were too vibrationally excited and the system reverted back to the reactants even though the system had already crossed the transition state region. In this situation, transition state theory would predict a faster rate of reaction compared to what was obtained experimentally.

Secondly, the transition state theory breaks down for systems with light atoms such as hydrogens. In such systems, quantum effects become crucial. Quantum effects tend to increase the rate of the reaction for two reasons. First, particles lie in energy levels which are higher then the bottom of the potential energy well (there is additional zero-point energy). Secondly, the particles can tunnel through the barrier. These two effects effectively lower the activation energy barrier. As a result, the rate of reaction predicted by transition state theory would be less than what is obtained experimentally.

Lastly, transition state theory does not hold true for systems which are far from equilibrium. In such cases, the assumption of a starting Boltzmann distribution breaks down.

== Exercise 2: F - H - H system ==

=== Potential energy surface inspection===

F + HA-HB → HA-F + HB

[[File:01409469 Surface_Plot F 3.jpg|thumb|center|400px|'''Figure 10.''' Potential energy surface for the reaction between F and H2]]

The reaction between F and H2 is an exothermic reaction. This means that the reactants are of a higher energy than the products. The overall reaction involves the breaking of a hydrogen bond and the formation of a hydrogen-fluorine bond. The HF bond formed is stronger than the hydrogen bond that was broken. This is because fluorine is more electronegative than hydrogen resulting in a polar HF bond. The negative enthalpy of H-F bond formation is greater than the positive enthalpy of hydrogen bond dissociation. Therefore, the reaction between F and H2 is an exothermic reaction.

HA + HB-F → HA-HB + F

[[File:01409469 Surface_Plot F 4.jpg|thumb|center|400px|'''Figure 11.''' Potential energy surface for the reaction between H and HF]]

The reaction between H and HF is an endothermic reaction. The products are at a higher energy level than the reactants. The overall reaction involves the breaking of a HF bond and the formation of a hydrogen bond. The formation of a homonuclear H2 molecule results in a non-polar hydrogen bond being formed. The HF bond broken is stronger than the hydrogen bond that was formed. The negative enthalpy of hydrogen bond formation is less than the positive enthalpy of hydrogen-fluorine dissociation. Therefore, the reaction between H and HF is endothermic.

=== Locating the transition state ===

[[File:01409469 Surface_Plot F TS4.jpg|thumb|center|800px|'''Figure 12.''' Determining the position of the transition state for the reaction between F and H2]]

As mentioned previously, when the molecules are not at the transition state, a reaction path will be plotted. However, if the initial conditions of the system are set such that the trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

An initial estimate for the transition state was made by setting rAB = 0.74 which is the length of a hydrogen bond. The position of the transition state was further refined by increasing the number of decimal places such that the final value of rAB was found to be 0.74478 and rBC = 1.81079. This set of distances gave a reaction path that was represented by a point indicating that no trajectory was plotted and that the molecules remained at the transition state since their initial momentum was 0.

Moreover, the kinetic energy as well as the forces acting on the atoms was calculated to be 0, which is consistent with the transition state.

Therefore, the position of the transition state was found to be at rAB = 0.74478 and rBC = 1.81079 when pA = pBC = 0.

=== Determining the activation energy ===

[[File:01409469 activation energy r1.1 combined.jpg|thumb|center|800px|'''Figure 13.''' Determining the activation energy for the reaction between F and H2]]

By reducing the value of rBC by 0.01, the atoms are no longer at the transition state and would move towards the formation of products. The activation energy was found by calculating the change in total energy after the reaction. The activation energy of F + H2 → HF + H was found to be +23.279 kcal/ mol.

[[File:01409469 activation energy r2.1 combined.jpg|thumb|center|800px|'''Figure 14.''' Determining the activation energy for the reaction between FH and H]]

The activation energy of FH + H → F + H2 was found by calculating the small dip in potential energy. The activation energy was found to be +0.280 kcal/ mol.

=== Reaction dynamics of F + HA-HB → HA-F + HB ===

The set of initial conditions that result in a reactive trajectory for F + H2 is as follows: rHH = 0.74 Å, rHF = 1.79 Å and pHH = pHF = 0.

[[File:01409469 momentum r1.jpg|thumb|center|400px|'''Figure 15.''' Graph of momentum against time for the reaction between F and H2]]

Initially, the H2 molecule is travelling in a relatively straight line with very little oscillations as evidenced by the low amplitude of momentum between 0-2 s. Following successful collision between F and H2, the amplitude of momentum of the product, HF, increases dramatically. This is because the reaction between F and H2 is exothermic and the excess energy released causes the molecule to become vibrationally excited and the amplitude of oscillations in the product to increase.

[[File:01409469 energy1 r1.jpg|thumb|center|400px|'''Figure 16.''' Graph of energy against time for the reaction between F and H2]]

The law of conservation of energy states that energy cannot be created or destroyed. It simply changes from one form to another. As can be seen from the graph in figure 16, total energy remains constant. In this reaction, potential energy is converted to kinetic energy. A decrease in potential energy of the atoms following a successful collision between F and H2 results in an equal increase in kinetic energy which manifests in the form of increased product (HF) oscillations.

This can be confirmed experimentally using infrared chemiluminescence.<ref name="ref2" /> Observation of radiation from excited product molecules is a clear indication that these molecules are produced in the excited state. If the excitation is primarily vibrational, the radiation will appear in the infrared region of the spectrum (between 3-15 μm). This technique has been used extensively in hydrogen halogen reactions which produce HX molecules in the excited state.

One such experiment involves the use of the arrested relaxation method.<ref name="ref2" /> In this technique, the reaction is carried out in a chamber whose walls are maintained at a very low temperature (20 K). This results in cryopumping of the chamber in that all the products and unreacted starting materials are condensed on the walls and only those product molecules that radiate before being pumped away are observed.

=== Polanyi's empirical rules ===

In a typical chemical reaction, there is an energetic barrier — a saddle point that the reactants must surmount to become products. Which form of energy is initially deposited in the reactants, translational or vibrational, is more efficacious in surmounting the activation energy barrier is one of the core topics in the area of reaction dynamics. The Polyani rules state that vibrational energy is more effective at promoting a late-barrier reaction (transition state resembles products). Conversely, translational energy is better at promoting an early-barrier reaction (transition state resembles the reactants).

The forward reaction between H2 and F is exothermic. Therefore, the transition state is closer to the reactants than to the products in energy. Since the transition state occurs early in an exothermic reaction, translational energy is more effective at promoting a chemical reaction than vibrational energy.

Conversely, the backward reaction between H and HF is endothermic. According to Hammond's postulate, the transition state resembles the products more than it resembles the reactants. Therefore, in an endothermic reaction, a late transition state is observed and vibrational energy is more effective at promoting a reaction than translational energy.

=== Promoting the forward reaction: F + HA-HB → HA-F + HB ===

1. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 2.95 and pHF = -0.5.

[[File:01409469 polyani 1+2.jpg|thumb|center|800px|'''Figure 17.''' Contour plot of the reaction between F and H2 when pHH = 2.95 '''Figure 18.''' Graph of momentum against time for the reaction between F and H2 when pHH = 2.95 (Left-right)]]

Vibrational energy is represented by pHH while translational energy is represented by pHF.

A molecule of H2 collides with F. However, the collision is unsuccessful and no reaction occurs. A reaction did not occur even though the molecules have sufficient kinetic energy to overcome the activation energy barrier. This is because most of the energy is deposited in the reactants as H-H vibrations as seen by the large oscillations in H2. Barrier recrossing has occured as a result of the large vibrational energy causing the system to revert back to the reactants even though the system has already crossed the transition state and a bond was formed between HB and F temporarily.

This appears to run counter to what one would expect to happen when more energy is put into the system. It is predicted that when more energy is put into the system, the reactants would possess more kinetic energy to overcome the activation energy barrier. However, as can be seen from the simulation above, an increase in the vibrational energy of the reactants is not effective at promoting the occurrence of an exothermic reaction with an early transition state. This is in accordance to Polyani's rules.

2. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 0.1 and pHF = -0.8.

[[File:01409469 polyani 3+4 2.jpg|thumb|center|800px|'''Figure 19.''' Contour plot of the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 '''Figure 20.''' Graph of momentum against time for the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 (Left-right)]]

The collision between a molecule of H2 and F is successful and a chemical reaction has occurred. A reaction has occurred even though the overall energy of the system has been significantly reduced by lowering the amount of vibrational energy in the reactants but slightly increasing the amount of translational energy deposited in the reactants. Therefore, H2 and F have more translational energy to move towards each other and form a H-F bond. The product formed is vibrationally excited as seen by the large amplitude of oscillations in the product HF which correspond to the strong H-F vibrations.

From the second simulation, it can be seen that an increase in translational energy in the reactants is more effective at promoting an exothermic reaction with an early transition state to occur, which again follows Polyani's rules.

=== Promoting the backward reaction: HA + HB-F → HA-HB + F ===

1. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 0.2 and pHH = -3.0.

[[File:01409469 polyani 5+6 .jpg|thumb|center|800px|'''Figure 21.''' Contour plot of the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 '''Figure 22.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 (Left-right)]]

An atom of HA collides with a molecule of HBF. However, the collision is unsuccessful and no chemical reaction occurs. This is because the reactants lack sufficient vibrational energy to surmount the activation energy barrier. Therefore, following an unsuccessful collision with HBF, HA simply bounces off and moves away at constant momentum. This is in agreement with Polyani's rules which state that vibrational energy is more effective than translational energy at promoting reactions with a late transition state, for example, endothermic chemical reactions.

2. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 7.1 and pHH = -0.5.

[[File:01409469 polyani 7+8 .jpg|thumb|center|800px|'''Figure 23.''' Contour plot of the reaction between HA + HB-F when pFH = 7.1 and pHH = -0.5 '''Figure 24.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 7.1 and pHH = -0.5 (Left-right)]]

Increasing the amount of vibrational energy in the reactants by increasing pFH resulted in a chemical reaction successfully occurring. This is because the reactants now have sufficient kinetic energy in the form of H-F vibrations to overcome the activation energy barrier. In addition, because the reaction between HA + HB-F is endothermic, energy is taken in from the surroundings during the reaction and the products are less vibrationally excited than the reactants. This can be seen in the lower amplitude of oscillations in the product formed.

In conclusion, the vibrational energy of the reactants should be increased to promote an endothermic reaction with a late transition state while the translational energy of the reactants should be increased to promote an exothermic reaction with an early transition state. All 4 simulations shown above obey Polyani's rules.

== References ==
<references>
<ref name="ref1">A. C. Vaucher, M. Reiher, ''J. Chem. Theory. Comput.'', 2018, '''14''', 3091-3099.</ref>
<ref name="ref2">J. I. Steinfield, J. S. Francisco, W. L. Hase, ''Chemical Kinetics and Dynamics'', Pearson, 1998.</ref>
<ref name="ref3">D. C. Elton, ''Transition State Theory for Physicists'', 2013.</ref>
</references>

MRD:01409469

2019-05-10T15:07:54Z

Stc1917: /* References */

== Exercise 1: H + H2 system ==

=== Definition of a transition state ===

A potential energy profile results from plotting the potential versus reaction path.

[[File:Surface_Plot01409469_pic1.jpg|thumb|center|400px|'''Figure 1.''' A potential energy surface plot]]

The transition state is defined as the maximum on the minimum energy path linking reactants and products. This means that the transition state is a stationary point on the potential energy surface. Thus, at the transition state, the derivative of the potential energy is zero with respect to each Cartesian coordinate.

<math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>

However, a local minimum on the potential energy surface corresponding to a well would have the same <math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>. Therefore, to distinguish between a local minimum on the potential energy surface and the transition state, the second derivative, <math>\tfrac{\partial^2 V(r_i)}{\partial r_i ^2}\ </math> is determined. This can be calculated by finding the determinant of the Hessian matrix shown below.

:<math>\mathbf H =
\begin{bmatrix}
\dfrac{\partial^2 \ V(r_i)}{\partial \ x^2} & \dfrac{\partial^2 \ V(r_i)}{\partial \ x \partial \mathbf y} \\
\dfrac{\partial^2 \ V(r_i)}{\partial \ y \partial \mathbf x} & \dfrac{\partial^2 \ V(r_i)}{\partial \mathbf y^2}
\end{bmatrix}</math>

The Hessian matrix of a convex function is positive semi-definite. Refining this property allows one to test if a critical point (x,y) is a local maximum, local minimum or saddle point.<ref name="ref1" />

If <math>f</math> < 0 (exactly one negative eigenvalue), then (x,y) is an isolated local maximum.

If <math>f</math> > 0 (exactly one positive eigenvalue), then (x,y) is an isolated local minimum.

If the Hessian has both positive and negative eigenvalues then (x,y) is a saddle point.

=== Locating the transition state ===

[[File:01409469 combined pic 2 and 3.jpg|thumb|center|800px|'''Figure 2.''' A surface and contour plot of the trajectory when rAB=rBC=0.907743 Å and pA=pBC=0]]

When the molecules are not at the transition state, a surface plot of the conditions shows a trajectory being plotted. This means that the trajectory is "rolling" towards the products or reactants. However, if a trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

[[File:01409469_internuclearpic3.png|thumb|center|400px|'''Figure 3.''' A plot of internuclear distance against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

An initial estimate for the position of the transition state was made by observing the potential energy surface and was found to be 0.90 Å. At 0.90 Å, the internuclear distance against time plot showed that the internuclear distance was constantly oscillating. Therefore, the position of the transition state was further refined by increasing the number of decimal places until a horizontal line in the internuclear distance againt time plot was obtained. This means that the internuclear distance is constant and that the molecules are simply undergoing periodic symmetric vibration at the transition state. Therefore, the position of the transition state was determined to be 0.907743 Å.

[[File:01409469 combined energy pic.png|thumb|center|400px|'''Figure 4.''' A plot of energy against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

When rAB=rBC=0.907743 Å and pA=pBC=0, the kinetic energy is 0. This means that HA and HBC are stationary, which is as predicted when the trajectory is started at the transition state with no initial momentum. Furthermore, the forces acting on HA and HBC are 0 as should be the case in the transition state.

=== Calculation type: Dynamic and minimum energy path ===

The minimum energy path, is the lowest energy path from the reactants to the products. The minimum energy path is determined by following the mode corresponding to the negative Hessian eigenvalue in both directions along the steepest descent direction.<ref name="ref1" /> Following this path allows one to approach two valleys of stable structures. The resulting path generated is the minimum energy path. The reaction paths were chartered for a dynamic and a minimum energy path. The initial conditions used were that of the transition state except that the internuclear distance between HA and HB had increased to 0.917743 Å. The reaction path obtained using the two different calculation methods was different.

[[File:01409469 mep and dynamics.jpg|thumb|center|800px|'''Figure 5.''' A surface plot of the trajectory using minimum energy path '''Figure 6.''' A surface plot of the trajectory using dynamics (Left-right)]]

On the surface plot, the reaction path is depicted by the black line. Following the minimum energy path, the trajectory simply follows the valley floor to Ha+Hb-Hc. The reaction path appears to be a relatively straight line indicating that the atoms are not vibrating. In the minimum energy path, the reactants change into the products in an infinitely slow motion. This is done by having the momenta/velocities reset to zero in each time step. While the minimum energy path is useful in characterising a chemical reaction, it does not provide a realistic representation of the motion of the atoms during the reaction. This is because the mass and inertia of the atoms have been ignored.

Since the atoms simply follow the minimum energy path, they do not undergo slight displacements into surrounding valleys of higher potential energy. Furthermore, because the atoms are assumed to have no inertia, they do not undergo vibrations even if they were to be displaced to valleys of higher potential energy.

However, when dynamic calculation is performed, the reaction path is wavy indicating that the atoms are oscillating. This is because the atoms are displaced to regions of higher potential on the path from reactants to products. This leads to atom oscillation due to the inertia of the atoms.

'''Internuclear distance against time plot'''

Setting the initial system conditions such that rAB= 0.907743, rBC= 0.917743 and pA= pBC= 0:

[[File:01409469 Internuclear distance mep and dynamic.jpg|thumb|center|800px|'''Figure 6.''' A graph of internuclear distance against steps using minimum energy path '''Figure 7.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In the minimum energy path, the distance between HA and HB decreases from 0.90 Å to approximately 0.79 Å during the first 50 steps before plateauing off to a value of 0.74 Å which is the approximate length of a hydrogen bond. This means that a bond was formed between HA and HB. In addition, the plateau indicates that the bond is static. On the other hand, the distance between HB and HC has increased to 1.91 Å at large step values. This shows that HC has been ejected off following a successful collision between HA and HBC.

Similarly, in the dynamic trajectory, the distance between HA and HB decreases to approximately 0.75 Å after 0.3 s. This is indicative of hydrogen bond formation between HA and HB. In addition, the distance between HB and HC increases to 3.44 Å after HC is ejected following a successful collision.

One obvious distance between the two plots is that in the minimum energy path trajectory, the internuclear distance between HA and HB is constant indicating that the bond is static. However, in the dynamic trajectory the internuclear distance is oscillating showing the the bond is vibrating. This difference could be attributed to the fact that minimum energy path does not take oscillatory motion into account while dynamic trajectory does.

However, changing the initial conditions of the system such that rAB = 0.917743 Å, rBC = 0.907743 Å and pA=pBC=0:
[[File:01409469 01409469 Reversed values dynamic mep.jpg|thumb|center|800px|'''Figure 8.''' A graph of internuclear distance against steps using minimum energy path '''Figure 9.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In this case, a hydrogen bond of length 0.77-0.79 Å is formed between HB and HC while HA is repelled away to beyond 3.45 Å.

Additionally, from the internuclear distance plot it can be seen that the distance between HB and HC increases in a rectangular hyperbola fashion when calculated using the minimum energy path. On the other hand, the internuclear distance between HB and HC increases in a linear manner when dynamic calculation is applied. A possible reason for this difference is that in the minimum energy path, the atoms are not vibrating and as such the reach the maximum distance and plateau at a faster rate. However, in the dynamic calculation, the atoms have momentum and as such move further away from each other.

=== Reactive and unreactive trajectories ===

With the initial positions r'''1''' = 0.74 and r'''2''' = 2.0, trajectories were run with the momenta combination shown below.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Contour plot !! Reactivity !! Description of the dynamics
|-
| -1.25 || -2.5 ||-99.018 ||[[File:01409469 trajectory1.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. There are no oscillations observed in HA-HB initially, indicating that the diatomic molecule is stable. A new bond is formed between HB and HC. Oscillatory behaviour is observed indicating that the bond is vibrating. On the other hand, HA is ejected away to beyond 10 Å.
|-
| -1.5 || -2.0 ||-100.456 ||[[File:01409469 trajectory2.jpg|thumb|center|400px]] ||Unreactive ||HC collides with the vibrating diatomic HA-HB. However, no bond is formed between HB and HC as the collision is unsuccessful. This is because the molecules lack sufficient kinetic energy to overcome the activation energy barrier and so no reaction occurs. HC is subsequently repelled to beyond 8 Å.
|-
| -1.5 || -2.5 ||-98.956 ||[[File:01409469 trajectory3.jpg|thumb|center|400px]] ||Reactive |||HC collides with HA-HB. The collision is successful and a bond is formed between HB and HC. This is because the atoms have sufficient kinetic energy to overcome the activation energy barrier and form a new HB-HC bond. HA is ejected away to beyond 10 Å. The new hydrogen bond formed is vibrating as seen by the oscillatory behaviour.
|-
| -2.5 || -5.0 ||-84.956 ||[[File:01409469 trajectory4.jpg|thumb|center|400px]] ||Unreactive ||HC collides with HA-HB. Initially the HA-HB bond is elongating, indicating that the hydrogen bond is weakening and going to break. However, after 0.5 s, the HA-HB bond length decreases and the HAHB is not broken. HC is repelled beyond 6 Å with very strong vibrations.
A reaction did not occur even though the atoms have sufficient kinetic energy to cross the activation energy barrier. This is because the vibrations are too strong. The system crosses the transition state region and a bond is formed in the product temporarily but then the system reverts back to the reactants. This is an example of barrier recrossing.
|-
| -2.5 || -5.2 ||-83.416 ||[[File:01409469 trajectory5.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. The collision is successful and a new hydrogen bond is formed between HB and HC with HA being ejected away. The products formed experience very strong vibrations as evidenced by the oscillatory motion which occur with a very high frequency.
|}

From the table, it can be seen that the chance of a reaction successfully occurring is not solely dependent on the amount of kinetic energy the atoms possess. Even if the atoms have sufficient kinetic energy to cross the activation energy barrier, barrier recrossing can occur if the atoms are too vibrationally excited.

=== Transition State Theory ===

Transition state theory provided the first theoretical attempt to determine absolute reaction rates. In this theory, a transition state separating the reactants from products is used to find an expression for the thermal rate constant.

The main assumptions of transition state theory are<ref name="ref2" />:

1. Electronic and nuclear motions are separated (equivalent to the Born-Oppenheimer approximation in quantum mechanics)

2. Reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution

3. Molecular systems that have crossed the transition state in the direction of products cannot turn around and reform reactants

4. In the transition state, motion along the reaction coordinate may be separated from the other motions and treated classically as a translation

5. Even in the absence of an equilibrium between reactant and product molecules, the transition states that are becoming products are distributed among their states according to the Maxwell-Boltzmann laws

The goal of transition state theory is to predict the rate constant, ''k'', of a reaction given a potential energy surface for the reaction.

However, transition state theory suffers from a set of limitations due to the assumptions stated above not holding true in reality.<ref name="ref3" />

Firstly, transition state theory assumes that barrier recrossing is negligible. However, as seen from the fourth trajectory in table 1, barrier recrossing occurred because the atoms were too vibrationally excited and the system reverted back to the reactants even though the system had already crossed the transition state region. In this situation, transition state theory would predict a faster rate of reaction compared to what was obtained experimentally.

Secondly, the transition state theory breaks down for systems with light atoms such as hydrogens. In such systems, quantum effects become crucial. Quantum effects tend to increase the rate of the reaction for two reasons. First, particles lie in energy levels which are higher then the bottom of the potential energy well (there is additional zero-point energy). Secondly, the particles can tunnel through the barrier. These two effects effectively lower the activation energy barrier. As a result, the rate of reaction predicted by transition state theory would be less than what is obtained experimentally.

Lastly, transition state theory does not hold true for systems which are far from equilibrium. In such cases, the assumption of a starting Boltzmann distribution breaks down.

== Exercise 2: F - H - H system ==

=== Potential energy surface inspection===

F + HA-HB → HA-F + HB

[[File:01409469 Surface_Plot F 3.jpg|thumb|center|400px|'''Figure 10.''' Potential energy surface for the reaction between F and H2]]

The reaction between F and H2 is an exothermic reaction. This means that the reactants are of a higher energy than the products. The overall reaction involves the breaking of a hydrogen bond and the formation of a hydrogen-fluorine bond. The HF bond formed is stronger than the hydrogen bond that was broken. This is because fluorine is more electronegative than hydrogen resulting in a polar HF bond. The negative enthalpy of H-F bond formation is greater than the positive enthalpy of hydrogen bond dissociation. Therefore, the reaction between F and H2 is an exothermic reaction.

HA + HB-F → HA-HB + F

[[File:01409469 Surface_Plot F 4.jpg|thumb|center|400px|'''Figure 11.''' Potential energy surface for the reaction between H and HF]]

The reaction between H and HF is an endothermic reaction. The products are at a higher energy level than the reactants. The overall reaction involves the breaking of a HF bond and the formation of a hydrogen bond. The formation of a homonuclear H2 molecule results in a non-polar hydrogen bond being formed. The HF bond broken is stronger than the hydrogen bond that was formed. The negative enthalpy of hydrogen bond formation is less than the positive enthalpy of hydrogen-fluorine dissociation. Therefore, the reaction between H and HF is endothermic.

=== Locating the transition state ===

[[File:01409469 Surface_Plot F TS4.jpg|thumb|center|800px|'''Figure 12.''' Determining the position of the transition state for the reaction between F and H2]]

As mentioned previously, when the molecules are not at the transition state, a reaction path will be plotted. However, if the initial conditions of the system are set such that the trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

An initial estimate for the transition state was made by setting rAB = 0.74 which is the length of a hydrogen bond. The position of the transition state was further refined by increasing the number of decimal places such that the final value of rAB was found to be 0.74478 and rBC = 1.81079. This set of distances gave a reaction path that was represented by a point indicating that no trajectory was plotted and that the molecules remained at the transition state since their initial momentum was 0.

Moreover, the kinetic energy as well as the forces acting on the atoms was calculated to be 0, which is consistent with the transition state.

Therefore, the position of the transition state was found to be at rAB = 0.74478 and rBC = 1.81079 when pA = pBC = 0.

=== Determining the activation energy ===

[[File:01409469 activation energy r1.1 combined.jpg|thumb|center|800px|'''Figure 13.''' Determining the activation energy for the reaction between F and H2]]

By reducing the value of rBC by 0.01, the atoms are no longer at the transition state and would move towards the formation of products. The activation energy was found by calculating the change in total energy after the reaction. The activation energy of F + H2 → HF + H was found to be +23.279 kcal/ mol.

[[File:01409469 activation energy r2.1 combined.jpg|thumb|center|800px|'''Figure 14.''' Determining the activation energy for the reaction between FH and H]]

The activation energy of FH + H → F + H2 was found by calculating the small dip in potential energy. The activation energy was found to be +0.280 kcal/ mol.

=== Reaction dynamics of F + HA-HB → HA-F + HB ===

The set of initial conditions that result in a reactive trajectory for F + H2 is as follows: rHH = 0.74 Å, rHF = 1.79 Å and pHH = pHF = 0.

[[File:01409469 momentum r1.jpg|thumb|center|400px|'''Figure 15.''' Graph of momentum against time for the reaction between F and H2]]

Initially, the H2 molecule is travelling in a relatively straight line with very little oscillations as evidenced by the low amplitude of momentum between 0-2 s. Following successful collision between F and H2, the amplitude of momentum of the product, HF, increases dramatically. This is because the reaction between F and H2 is exothermic and the excess energy released causes the molecule to become vibrationally excited and the amplitude of oscillations in the product to increase.

[[File:01409469 energy1 r1.jpg|thumb|center|400px|'''Figure 16.''' Graph of energy against time for the reaction between F and H2]]

The law of conservation of energy states that energy cannot be created or destroyed. It simply changes from one form to another. As can be seen from the graph in figure 16, total energy remains constant. In this reaction, potential energy is converted to kinetic energy. A decrease in potential energy of the atoms following a successful collision between F and H2 results in an equal increase in kinetic energy which manifests in the form of increased product (HF) oscillations.

This can be confirmed experimentally using infrared chemiluminescence. Observation of radiation from excited product molecules is a clear indication that these molecules are produced in the excited state. If the excitation is primarily vibrational, the radiation will appear in the infrared region of the spectrum (between 3-15 μm). This technique has been used extensively in hydrogen halogen reactions which produce HX molecules in the excited state.

One such experiment involves the use of the arrested relaxation method. In this technique, the reaction is carried out in a chamber whose walls are maintained at a very low temperature (20 K). This results in cryopumping of the chamber in that all the products and unreacted starting materials are condensed on the walls and only those product molecules that radiate before being pumped away are observed.

=== Polanyi's empirical rules ===

In a typical chemical reaction, there is an energetic barrier — a saddle point that the reactants must surmount to become products. Which form of energy is initially deposited in the reactants, translational or vibrational, is more efficacious in surmounting the activation energy barrier is one of the core topics in the area of reaction dynamics. The Polyani rules state that vibrational energy is more effective at promoting a late-barrier reaction (transition state resembles products). Conversely, translational energy is better at promoting an early-barrier reaction (transition state resembles the reactants).

The forward reaction between H2 and F is exothermic. Therefore, the transition state is closer to the reactants than to the products in energy. Since the transition state occurs early in an exothermic reaction, translational energy is more effective at promoting a chemical reaction than vibrational energy.

Conversely, the backward reaction between H and HF is endothermic. According to Hammond's postulate, the transition state resembles the products more than it resembles the reactants. Therefore, in an endothermic reaction, a late transition state is observed and vibrational energy is more effective at promoting a reaction than translational energy.

=== Promoting the forward reaction: F + HA-HB → HA-F + HB ===

1. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 2.95 and pHF = -0.5.

[[File:01409469 polyani 1+2.jpg|thumb|center|800px|'''Figure 17.''' Contour plot of the reaction between F and H2 when pHH = 2.95 '''Figure 18.''' Graph of momentum against time for the reaction between F and H2 when pHH = 2.95 (Left-right)]]

Vibrational energy is represented by pHH while translational energy is represented by pHF.

A molecule of H2 collides with F. However, the collision is unsuccessful and no reaction occurs. A reaction did not occur even though the molecules have sufficient kinetic energy to overcome the activation energy barrier. This is because most of the energy is deposited in the reactants as H-H vibrations as seen by the large oscillations in H2. Barrier recrossing has occured as a result of the large vibrational energy causing the system to revert back to the reactants even though the system has already crossed the transition state and a bond was formed between HB and F temporarily.

This appears to run counter to what one would expect to happen when more energy is put into the system. It is predicted that when more energy is put into the system, the reactants would possess more kinetic energy to overcome the activation energy barrier. However, as can be seen from the simulation above, an increase in the vibrational energy of the reactants is not effective at promoting the occurrence of an exothermic reaction with an early transition state. This is in accordance to Polyani's rules.

2. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 0.1 and pHF = -0.8.

[[File:01409469 polyani 3+4 2.jpg|thumb|center|800px|'''Figure 19.''' Contour plot of the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 '''Figure 20.''' Graph of momentum against time for the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 (Left-right)]]

The collision between a molecule of H2 and F is successful and a chemical reaction has occurred. A reaction has occurred even though the overall energy of the system has been significantly reduced by lowering the amount of vibrational energy in the reactants but slightly increasing the amount of translational energy deposited in the reactants. Therefore, H2 and F have more translational energy to move towards each other and form a H-F bond. The product formed is vibrationally excited as seen by the large amplitude of oscillations in the product HF which correspond to the strong H-F vibrations.

From the second simulation, it can be seen that an increase in translational energy in the reactants is more effective at promoting an exothermic reaction with an early transition state to occur, which again follows Polyani's rules.

=== Promoting the backward reaction: HA + HB-F → HA-HB + F ===

1. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 0.2 and pHH = -3.0.

[[File:01409469 polyani 5+6 .jpg|thumb|center|800px|'''Figure 21.''' Contour plot of the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 '''Figure 22.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 (Left-right)]]

An atom of HA collides with a molecule of HBF. However, the collision is unsuccessful and no chemical reaction occurs. This is because the reactants lack sufficient vibrational energy to surmount the activation energy barrier. Therefore, following an unsuccessful collision with HBF, HA simply bounces off and moves away at constant momentum. This is in agreement with Polyani's rules which state that vibrational energy is more effective than translational energy at promoting reactions with a late transition state, for example, endothermic chemical reactions.

2. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 7.1 and pHH = -0.5.

[[File:01409469 polyani 7+8 .jpg|thumb|center|800px|'''Figure 23.''' Contour plot of the reaction between HA + HB-F when pFH = 7.1 and pHH = -0.5 '''Figure 24.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 7.1 and pHH = -0.5 (Left-right)]]

Increasing the amount of vibrational energy in the reactants by increasing pFH resulted in a chemical reaction successfully occurring. This is because the reactants now have sufficient kinetic energy in the form of H-F vibrations to overcome the activation energy barrier. In addition, because the reaction between HA + HB-F is endothermic, energy is taken in from the surroundings during the reaction and the products are less vibrationally excited than the reactants. This can be seen in the lower amplitude of oscillations in the product formed.

In conclusion, the vibrational energy of the reactants should be increased to promote an endothermic reaction with a late transition state while the translational energy of the reactants should be increased to promote an exothermic reaction with an early transition state. All 4 simulations shown above obey Polyani's rules.

== References ==
<references>
<ref name="ref1">A. C. Vaucher, M. Reiher, ''J. Chem. Theory. Comput.'', 2018, '''14''', 3091-3099.</ref>
<ref name="ref2">J. I. Steinfield, J. S. Francisco, W. L. Hase, ''Chemical Kinetics and Dynamics'', Pearson, 1998.</ref>
<ref name="ref3">D. C. Elton, ''Transition State Theory for Physicists'', 2013.</ref>
</references>

MRD:01409469

2019-05-10T15:05:17Z

Stc1917: /* Transition State Theory */

== Exercise 1: H + H2 system ==

=== Definition of a transition state ===

A potential energy profile results from plotting the potential versus reaction path.

[[File:Surface_Plot01409469_pic1.jpg|thumb|center|400px|'''Figure 1.''' A potential energy surface plot]]

The transition state is defined as the maximum on the minimum energy path linking reactants and products. This means that the transition state is a stationary point on the potential energy surface. Thus, at the transition state, the derivative of the potential energy is zero with respect to each Cartesian coordinate.

<math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>

However, a local minimum on the potential energy surface corresponding to a well would have the same <math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>. Therefore, to distinguish between a local minimum on the potential energy surface and the transition state, the second derivative, <math>\tfrac{\partial^2 V(r_i)}{\partial r_i ^2}\ </math> is determined. This can be calculated by finding the determinant of the Hessian matrix shown below.

:<math>\mathbf H =
\begin{bmatrix}
\dfrac{\partial^2 \ V(r_i)}{\partial \ x^2} & \dfrac{\partial^2 \ V(r_i)}{\partial \ x \partial \mathbf y} \\
\dfrac{\partial^2 \ V(r_i)}{\partial \ y \partial \mathbf x} & \dfrac{\partial^2 \ V(r_i)}{\partial \mathbf y^2}
\end{bmatrix}</math>

The Hessian matrix of a convex function is positive semi-definite. Refining this property allows one to test if a critical point (x,y) is a local maximum, local minimum or saddle point.<ref name="ref1" />

If <math>f</math> < 0 (exactly one negative eigenvalue), then (x,y) is an isolated local maximum.

If <math>f</math> > 0 (exactly one positive eigenvalue), then (x,y) is an isolated local minimum.

If the Hessian has both positive and negative eigenvalues then (x,y) is a saddle point.

=== Locating the transition state ===

[[File:01409469 combined pic 2 and 3.jpg|thumb|center|800px|'''Figure 2.''' A surface and contour plot of the trajectory when rAB=rBC=0.907743 Å and pA=pBC=0]]

When the molecules are not at the transition state, a surface plot of the conditions shows a trajectory being plotted. This means that the trajectory is "rolling" towards the products or reactants. However, if a trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

[[File:01409469_internuclearpic3.png|thumb|center|400px|'''Figure 3.''' A plot of internuclear distance against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

An initial estimate for the position of the transition state was made by observing the potential energy surface and was found to be 0.90 Å. At 0.90 Å, the internuclear distance against time plot showed that the internuclear distance was constantly oscillating. Therefore, the position of the transition state was further refined by increasing the number of decimal places until a horizontal line in the internuclear distance againt time plot was obtained. This means that the internuclear distance is constant and that the molecules are simply undergoing periodic symmetric vibration at the transition state. Therefore, the position of the transition state was determined to be 0.907743 Å.

[[File:01409469 combined energy pic.png|thumb|center|400px|'''Figure 4.''' A plot of energy against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

When rAB=rBC=0.907743 Å and pA=pBC=0, the kinetic energy is 0. This means that HA and HBC are stationary, which is as predicted when the trajectory is started at the transition state with no initial momentum. Furthermore, the forces acting on HA and HBC are 0 as should be the case in the transition state.

=== Calculation type: Dynamic and minimum energy path ===

The minimum energy path, is the lowest energy path from the reactants to the products. The minimum energy path is determined by following the mode corresponding to the negative Hessian eigenvalue in both directions along the steepest descent direction.<ref name="ref1" /> Following this path allows one to approach two valleys of stable structures. The resulting path generated is the minimum energy path. The reaction paths were chartered for a dynamic and a minimum energy path. The initial conditions used were that of the transition state except that the internuclear distance between HA and HB had increased to 0.917743 Å. The reaction path obtained using the two different calculation methods was different.

[[File:01409469 mep and dynamics.jpg|thumb|center|800px|'''Figure 5.''' A surface plot of the trajectory using minimum energy path '''Figure 6.''' A surface plot of the trajectory using dynamics (Left-right)]]

On the surface plot, the reaction path is depicted by the black line. Following the minimum energy path, the trajectory simply follows the valley floor to Ha+Hb-Hc. The reaction path appears to be a relatively straight line indicating that the atoms are not vibrating. In the minimum energy path, the reactants change into the products in an infinitely slow motion. This is done by having the momenta/velocities reset to zero in each time step. While the minimum energy path is useful in characterising a chemical reaction, it does not provide a realistic representation of the motion of the atoms during the reaction. This is because the mass and inertia of the atoms have been ignored.

Since the atoms simply follow the minimum energy path, they do not undergo slight displacements into surrounding valleys of higher potential energy. Furthermore, because the atoms are assumed to have no inertia, they do not undergo vibrations even if they were to be displaced to valleys of higher potential energy.

However, when dynamic calculation is performed, the reaction path is wavy indicating that the atoms are oscillating. This is because the atoms are displaced to regions of higher potential on the path from reactants to products. This leads to atom oscillation due to the inertia of the atoms.

'''Internuclear distance against time plot'''

Setting the initial system conditions such that rAB= 0.907743, rBC= 0.917743 and pA= pBC= 0:

[[File:01409469 Internuclear distance mep and dynamic.jpg|thumb|center|800px|'''Figure 6.''' A graph of internuclear distance against steps using minimum energy path '''Figure 7.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In the minimum energy path, the distance between HA and HB decreases from 0.90 Å to approximately 0.79 Å during the first 50 steps before plateauing off to a value of 0.74 Å which is the approximate length of a hydrogen bond. This means that a bond was formed between HA and HB. In addition, the plateau indicates that the bond is static. On the other hand, the distance between HB and HC has increased to 1.91 Å at large step values. This shows that HC has been ejected off following a successful collision between HA and HBC.

Similarly, in the dynamic trajectory, the distance between HA and HB decreases to approximately 0.75 Å after 0.3 s. This is indicative of hydrogen bond formation between HA and HB. In addition, the distance between HB and HC increases to 3.44 Å after HC is ejected following a successful collision.

One obvious distance between the two plots is that in the minimum energy path trajectory, the internuclear distance between HA and HB is constant indicating that the bond is static. However, in the dynamic trajectory the internuclear distance is oscillating showing the the bond is vibrating. This difference could be attributed to the fact that minimum energy path does not take oscillatory motion into account while dynamic trajectory does.

However, changing the initial conditions of the system such that rAB = 0.917743 Å, rBC = 0.907743 Å and pA=pBC=0:
[[File:01409469 01409469 Reversed values dynamic mep.jpg|thumb|center|800px|'''Figure 8.''' A graph of internuclear distance against steps using minimum energy path '''Figure 9.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In this case, a hydrogen bond of length 0.77-0.79 Å is formed between HB and HC while HA is repelled away to beyond 3.45 Å.

Additionally, from the internuclear distance plot it can be seen that the distance between HB and HC increases in a rectangular hyperbola fashion when calculated using the minimum energy path. On the other hand, the internuclear distance between HB and HC increases in a linear manner when dynamic calculation is applied. A possible reason for this difference is that in the minimum energy path, the atoms are not vibrating and as such the reach the maximum distance and plateau at a faster rate. However, in the dynamic calculation, the atoms have momentum and as such move further away from each other.

=== Reactive and unreactive trajectories ===

With the initial positions r'''1''' = 0.74 and r'''2''' = 2.0, trajectories were run with the momenta combination shown below.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Contour plot !! Reactivity !! Description of the dynamics
|-
| -1.25 || -2.5 ||-99.018 ||[[File:01409469 trajectory1.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. There are no oscillations observed in HA-HB initially, indicating that the diatomic molecule is stable. A new bond is formed between HB and HC. Oscillatory behaviour is observed indicating that the bond is vibrating. On the other hand, HA is ejected away to beyond 10 Å.
|-
| -1.5 || -2.0 ||-100.456 ||[[File:01409469 trajectory2.jpg|thumb|center|400px]] ||Unreactive ||HC collides with the vibrating diatomic HA-HB. However, no bond is formed between HB and HC as the collision is unsuccessful. This is because the molecules lack sufficient kinetic energy to overcome the activation energy barrier and so no reaction occurs. HC is subsequently repelled to beyond 8 Å.
|-
| -1.5 || -2.5 ||-98.956 ||[[File:01409469 trajectory3.jpg|thumb|center|400px]] ||Reactive |||HC collides with HA-HB. The collision is successful and a bond is formed between HB and HC. This is because the atoms have sufficient kinetic energy to overcome the activation energy barrier and form a new HB-HC bond. HA is ejected away to beyond 10 Å. The new hydrogen bond formed is vibrating as seen by the oscillatory behaviour.
|-
| -2.5 || -5.0 ||-84.956 ||[[File:01409469 trajectory4.jpg|thumb|center|400px]] ||Unreactive ||HC collides with HA-HB. Initially the HA-HB bond is elongating, indicating that the hydrogen bond is weakening and going to break. However, after 0.5 s, the HA-HB bond length decreases and the HAHB is not broken. HC is repelled beyond 6 Å with very strong vibrations.
A reaction did not occur even though the atoms have sufficient kinetic energy to cross the activation energy barrier. This is because the vibrations are too strong. The system crosses the transition state region and a bond is formed in the product temporarily but then the system reverts back to the reactants. This is an example of barrier recrossing.
|-
| -2.5 || -5.2 ||-83.416 ||[[File:01409469 trajectory5.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. The collision is successful and a new hydrogen bond is formed between HB and HC with HA being ejected away. The products formed experience very strong vibrations as evidenced by the oscillatory motion which occur with a very high frequency.
|}

From the table, it can be seen that the chance of a reaction successfully occurring is not solely dependent on the amount of kinetic energy the atoms possess. Even if the atoms have sufficient kinetic energy to cross the activation energy barrier, barrier recrossing can occur if the atoms are too vibrationally excited.

=== Transition State Theory ===

Transition state theory provided the first theoretical attempt to determine absolute reaction rates. In this theory, a transition state separating the reactants from products is used to find an expression for the thermal rate constant.

The main assumptions of transition state theory are<ref name="ref2" />:

1. Electronic and nuclear motions are separated (equivalent to the Born-Oppenheimer approximation in quantum mechanics)

2. Reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution

3. Molecular systems that have crossed the transition state in the direction of products cannot turn around and reform reactants

4. In the transition state, motion along the reaction coordinate may be separated from the other motions and treated classically as a translation

5. Even in the absence of an equilibrium between reactant and product molecules, the transition states that are becoming products are distributed among their states according to the Maxwell-Boltzmann laws

The goal of transition state theory is to predict the rate constant, ''k'', of a reaction given a potential energy surface for the reaction.

However, transition state theory suffers from a set of limitations due to the assumptions stated above not holding true in reality.<ref name="ref3" />

Firstly, transition state theory assumes that barrier recrossing is negligible. However, as seen from the fourth trajectory in table 1, barrier recrossing occurred because the atoms were too vibrationally excited and the system reverted back to the reactants even though the system had already crossed the transition state region. In this situation, transition state theory would predict a faster rate of reaction compared to what was obtained experimentally.

Secondly, the transition state theory breaks down for systems with light atoms such as hydrogens. In such systems, quantum effects become crucial. Quantum effects tend to increase the rate of the reaction for two reasons. First, particles lie in energy levels which are higher then the bottom of the potential energy well (there is additional zero-point energy). Secondly, the particles can tunnel through the barrier. These two effects effectively lower the activation energy barrier. As a result, the rate of reaction predicted by transition state theory would be less than what is obtained experimentally.

Lastly, transition state theory does not hold true for systems which are far from equilibrium. In such cases, the assumption of a starting Boltzmann distribution breaks down.

== Exercise 2: F - H - H system ==

=== Potential energy surface inspection===

F + HA-HB → HA-F + HB

[[File:01409469 Surface_Plot F 3.jpg|thumb|center|400px|'''Figure 10.''' Potential energy surface for the reaction between F and H2]]

The reaction between F and H2 is an exothermic reaction. This means that the reactants are of a higher energy than the products. The overall reaction involves the breaking of a hydrogen bond and the formation of a hydrogen-fluorine bond. The HF bond formed is stronger than the hydrogen bond that was broken. This is because fluorine is more electronegative than hydrogen resulting in a polar HF bond. The negative enthalpy of H-F bond formation is greater than the positive enthalpy of hydrogen bond dissociation. Therefore, the reaction between F and H2 is an exothermic reaction.

HA + HB-F → HA-HB + F

[[File:01409469 Surface_Plot F 4.jpg|thumb|center|400px|'''Figure 11.''' Potential energy surface for the reaction between H and HF]]

The reaction between H and HF is an endothermic reaction. The products are at a higher energy level than the reactants. The overall reaction involves the breaking of a HF bond and the formation of a hydrogen bond. The formation of a homonuclear H2 molecule results in a non-polar hydrogen bond being formed. The HF bond broken is stronger than the hydrogen bond that was formed. The negative enthalpy of hydrogen bond formation is less than the positive enthalpy of hydrogen-fluorine dissociation. Therefore, the reaction between H and HF is endothermic.

=== Locating the transition state ===

[[File:01409469 Surface_Plot F TS4.jpg|thumb|center|800px|'''Figure 12.''' Determining the position of the transition state for the reaction between F and H2]]

As mentioned previously, when the molecules are not at the transition state, a reaction path will be plotted. However, if the initial conditions of the system are set such that the trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

An initial estimate for the transition state was made by setting rAB = 0.74 which is the length of a hydrogen bond. The position of the transition state was further refined by increasing the number of decimal places such that the final value of rAB was found to be 0.74478 and rBC = 1.81079. This set of distances gave a reaction path that was represented by a point indicating that no trajectory was plotted and that the molecules remained at the transition state since their initial momentum was 0.

Moreover, the kinetic energy as well as the forces acting on the atoms was calculated to be 0, which is consistent with the transition state.

Therefore, the position of the transition state was found to be at rAB = 0.74478 and rBC = 1.81079 when pA = pBC = 0.

=== Determining the activation energy ===

[[File:01409469 activation energy r1.1 combined.jpg|thumb|center|800px|'''Figure 13.''' Determining the activation energy for the reaction between F and H2]]

By reducing the value of rBC by 0.01, the atoms are no longer at the transition state and would move towards the formation of products. The activation energy was found by calculating the change in total energy after the reaction. The activation energy of F + H2 → HF + H was found to be +23.279 kcal/ mol.

[[File:01409469 activation energy r2.1 combined.jpg|thumb|center|800px|'''Figure 14.''' Determining the activation energy for the reaction between FH and H]]

The activation energy of FH + H → F + H2 was found by calculating the small dip in potential energy. The activation energy was found to be +0.280 kcal/ mol.

=== Reaction dynamics of F + HA-HB → HA-F + HB ===

The set of initial conditions that result in a reactive trajectory for F + H2 is as follows: rHH = 0.74 Å, rHF = 1.79 Å and pHH = pHF = 0.

[[File:01409469 momentum r1.jpg|thumb|center|400px|'''Figure 15.''' Graph of momentum against time for the reaction between F and H2]]

Initially, the H2 molecule is travelling in a relatively straight line with very little oscillations as evidenced by the low amplitude of momentum between 0-2 s. Following successful collision between F and H2, the amplitude of momentum of the product, HF, increases dramatically. This is because the reaction between F and H2 is exothermic and the excess energy released causes the molecule to become vibrationally excited and the amplitude of oscillations in the product to increase.

[[File:01409469 energy1 r1.jpg|thumb|center|400px|'''Figure 16.''' Graph of energy against time for the reaction between F and H2]]

The law of conservation of energy states that energy cannot be created or destroyed. It simply changes from one form to another. As can be seen from the graph in figure 16, total energy remains constant. In this reaction, potential energy is converted to kinetic energy. A decrease in potential energy of the atoms following a successful collision between F and H2 results in an equal increase in kinetic energy which manifests in the form of increased product (HF) oscillations.

This can be confirmed experimentally using infrared chemiluminescence. Observation of radiation from excited product molecules is a clear indication that these molecules are produced in the excited state. If the excitation is primarily vibrational, the radiation will appear in the infrared region of the spectrum (between 3-15 μm). This technique has been used extensively in hydrogen halogen reactions which produce HX molecules in the excited state.

One such experiment involves the use of the arrested relaxation method. In this technique, the reaction is carried out in a chamber whose walls are maintained at a very low temperature (20 K). This results in cryopumping of the chamber in that all the products and unreacted starting materials are condensed on the walls and only those product molecules that radiate before being pumped away are observed.

=== Polanyi's empirical rules ===

In a typical chemical reaction, there is an energetic barrier — a saddle point that the reactants must surmount to become products. Which form of energy is initially deposited in the reactants, translational or vibrational, is more efficacious in surmounting the activation energy barrier is one of the core topics in the area of reaction dynamics. The Polyani rules state that vibrational energy is more effective at promoting a late-barrier reaction (transition state resembles products). Conversely, translational energy is better at promoting an early-barrier reaction (transition state resembles the reactants).

The forward reaction between H2 and F is exothermic. Therefore, the transition state is closer to the reactants than to the products in energy. Since the transition state occurs early in an exothermic reaction, translational energy is more effective at promoting a chemical reaction than vibrational energy.

Conversely, the backward reaction between H and HF is endothermic. According to Hammond's postulate, the transition state resembles the products more than it resembles the reactants. Therefore, in an endothermic reaction, a late transition state is observed and vibrational energy is more effective at promoting a reaction than translational energy.

=== Promoting the forward reaction: F + HA-HB → HA-F + HB ===

1. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 2.95 and pHF = -0.5.

[[File:01409469 polyani 1+2.jpg|thumb|center|800px|'''Figure 17.''' Contour plot of the reaction between F and H2 when pHH = 2.95 '''Figure 18.''' Graph of momentum against time for the reaction between F and H2 when pHH = 2.95 (Left-right)]]

Vibrational energy is represented by pHH while translational energy is represented by pHF.

A molecule of H2 collides with F. However, the collision is unsuccessful and no reaction occurs. A reaction did not occur even though the molecules have sufficient kinetic energy to overcome the activation energy barrier. This is because most of the energy is deposited in the reactants as H-H vibrations as seen by the large oscillations in H2. Barrier recrossing has occured as a result of the large vibrational energy causing the system to revert back to the reactants even though the system has already crossed the transition state and a bond was formed between HB and F temporarily.

This appears to run counter to what one would expect to happen when more energy is put into the system. It is predicted that when more energy is put into the system, the reactants would possess more kinetic energy to overcome the activation energy barrier. However, as can be seen from the simulation above, an increase in the vibrational energy of the reactants is not effective at promoting the occurrence of an exothermic reaction with an early transition state. This is in accordance to Polyani's rules.

2. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 0.1 and pHF = -0.8.

[[File:01409469 polyani 3+4 2.jpg|thumb|center|800px|'''Figure 19.''' Contour plot of the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 '''Figure 20.''' Graph of momentum against time for the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 (Left-right)]]

The collision between a molecule of H2 and F is successful and a chemical reaction has occurred. A reaction has occurred even though the overall energy of the system has been significantly reduced by lowering the amount of vibrational energy in the reactants but slightly increasing the amount of translational energy deposited in the reactants. Therefore, H2 and F have more translational energy to move towards each other and form a H-F bond. The product formed is vibrationally excited as seen by the large amplitude of oscillations in the product HF which correspond to the strong H-F vibrations.

From the second simulation, it can be seen that an increase in translational energy in the reactants is more effective at promoting an exothermic reaction with an early transition state to occur, which again follows Polyani's rules.

=== Promoting the backward reaction: HA + HB-F → HA-HB + F ===

1. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 0.2 and pHH = -3.0.

[[File:01409469 polyani 5+6 .jpg|thumb|center|800px|'''Figure 21.''' Contour plot of the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 '''Figure 22.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 (Left-right)]]

An atom of HA collides with a molecule of HBF. However, the collision is unsuccessful and no chemical reaction occurs. This is because the reactants lack sufficient vibrational energy to surmount the activation energy barrier. Therefore, following an unsuccessful collision with HBF, HA simply bounces off and moves away at constant momentum. This is in agreement with Polyani's rules which state that vibrational energy is more effective than translational energy at promoting reactions with a late transition state, for example, endothermic chemical reactions.

2. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 7.1 and pHH = -0.5.

[[File:01409469 polyani 7+8 .jpg|thumb|center|800px|'''Figure 23.''' Contour plot of the reaction between HA + HB-F when pFH = 7.1 and pHH = -0.5 '''Figure 24.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 7.1 and pHH = -0.5 (Left-right)]]

Increasing the amount of vibrational energy in the reactants by increasing pFH resulted in a chemical reaction successfully occurring. This is because the reactants now have sufficient kinetic energy in the form of H-F vibrations to overcome the activation energy barrier. In addition, because the reaction between HA + HB-F is endothermic, energy is taken in from the surroundings during the reaction and the products are less vibrationally excited than the reactants. This can be seen in the lower amplitude of oscillations in the product formed.

In conclusion, the vibrational energy of the reactants should be increased to promote an endothermic reaction with a late transition state while the translational energy of the reactants should be increased to promote an exothermic reaction with an early transition state. All 4 simulations shown above obey Polyani's rules.

== References ==
<references>
<ref name="ref1">A. C. Vaucher, M. Reiher, ''J. Chem. Theory. Comput.'', 2018, '''14''', 3091-3099.</ref>
<ref name="ref2">J. I. Steinfield, J. S. Francisco, W. L. Hase, ''Chemical Kinetics and Dynamics'', Pearson, 1998.</ref>
</references>

MRD:01409469

2019-05-10T15:04:47Z

Stc1917: /* References */

== Exercise 1: H + H2 system ==

=== Definition of a transition state ===

A potential energy profile results from plotting the potential versus reaction path.

[[File:Surface_Plot01409469_pic1.jpg|thumb|center|400px|'''Figure 1.''' A potential energy surface plot]]

The transition state is defined as the maximum on the minimum energy path linking reactants and products. This means that the transition state is a stationary point on the potential energy surface. Thus, at the transition state, the derivative of the potential energy is zero with respect to each Cartesian coordinate.

<math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>

However, a local minimum on the potential energy surface corresponding to a well would have the same <math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>. Therefore, to distinguish between a local minimum on the potential energy surface and the transition state, the second derivative, <math>\tfrac{\partial^2 V(r_i)}{\partial r_i ^2}\ </math> is determined. This can be calculated by finding the determinant of the Hessian matrix shown below.

:<math>\mathbf H =
\begin{bmatrix}
\dfrac{\partial^2 \ V(r_i)}{\partial \ x^2} & \dfrac{\partial^2 \ V(r_i)}{\partial \ x \partial \mathbf y} \\
\dfrac{\partial^2 \ V(r_i)}{\partial \ y \partial \mathbf x} & \dfrac{\partial^2 \ V(r_i)}{\partial \mathbf y^2}
\end{bmatrix}</math>

The Hessian matrix of a convex function is positive semi-definite. Refining this property allows one to test if a critical point (x,y) is a local maximum, local minimum or saddle point.<ref name="ref1" />

If <math>f</math> < 0 (exactly one negative eigenvalue), then (x,y) is an isolated local maximum.

If <math>f</math> > 0 (exactly one positive eigenvalue), then (x,y) is an isolated local minimum.

If the Hessian has both positive and negative eigenvalues then (x,y) is a saddle point.

=== Locating the transition state ===

[[File:01409469 combined pic 2 and 3.jpg|thumb|center|800px|'''Figure 2.''' A surface and contour plot of the trajectory when rAB=rBC=0.907743 Å and pA=pBC=0]]

When the molecules are not at the transition state, a surface plot of the conditions shows a trajectory being plotted. This means that the trajectory is "rolling" towards the products or reactants. However, if a trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

[[File:01409469_internuclearpic3.png|thumb|center|400px|'''Figure 3.''' A plot of internuclear distance against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

An initial estimate for the position of the transition state was made by observing the potential energy surface and was found to be 0.90 Å. At 0.90 Å, the internuclear distance against time plot showed that the internuclear distance was constantly oscillating. Therefore, the position of the transition state was further refined by increasing the number of decimal places until a horizontal line in the internuclear distance againt time plot was obtained. This means that the internuclear distance is constant and that the molecules are simply undergoing periodic symmetric vibration at the transition state. Therefore, the position of the transition state was determined to be 0.907743 Å.

[[File:01409469 combined energy pic.png|thumb|center|400px|'''Figure 4.''' A plot of energy against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

When rAB=rBC=0.907743 Å and pA=pBC=0, the kinetic energy is 0. This means that HA and HBC are stationary, which is as predicted when the trajectory is started at the transition state with no initial momentum. Furthermore, the forces acting on HA and HBC are 0 as should be the case in the transition state.

=== Calculation type: Dynamic and minimum energy path ===

The minimum energy path, is the lowest energy path from the reactants to the products. The minimum energy path is determined by following the mode corresponding to the negative Hessian eigenvalue in both directions along the steepest descent direction.<ref name="ref1" /> Following this path allows one to approach two valleys of stable structures. The resulting path generated is the minimum energy path. The reaction paths were chartered for a dynamic and a minimum energy path. The initial conditions used were that of the transition state except that the internuclear distance between HA and HB had increased to 0.917743 Å. The reaction path obtained using the two different calculation methods was different.

[[File:01409469 mep and dynamics.jpg|thumb|center|800px|'''Figure 5.''' A surface plot of the trajectory using minimum energy path '''Figure 6.''' A surface plot of the trajectory using dynamics (Left-right)]]

On the surface plot, the reaction path is depicted by the black line. Following the minimum energy path, the trajectory simply follows the valley floor to Ha+Hb-Hc. The reaction path appears to be a relatively straight line indicating that the atoms are not vibrating. In the minimum energy path, the reactants change into the products in an infinitely slow motion. This is done by having the momenta/velocities reset to zero in each time step. While the minimum energy path is useful in characterising a chemical reaction, it does not provide a realistic representation of the motion of the atoms during the reaction. This is because the mass and inertia of the atoms have been ignored.

Since the atoms simply follow the minimum energy path, they do not undergo slight displacements into surrounding valleys of higher potential energy. Furthermore, because the atoms are assumed to have no inertia, they do not undergo vibrations even if they were to be displaced to valleys of higher potential energy.

However, when dynamic calculation is performed, the reaction path is wavy indicating that the atoms are oscillating. This is because the atoms are displaced to regions of higher potential on the path from reactants to products. This leads to atom oscillation due to the inertia of the atoms.

'''Internuclear distance against time plot'''

Setting the initial system conditions such that rAB= 0.907743, rBC= 0.917743 and pA= pBC= 0:

[[File:01409469 Internuclear distance mep and dynamic.jpg|thumb|center|800px|'''Figure 6.''' A graph of internuclear distance against steps using minimum energy path '''Figure 7.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In the minimum energy path, the distance between HA and HB decreases from 0.90 Å to approximately 0.79 Å during the first 50 steps before plateauing off to a value of 0.74 Å which is the approximate length of a hydrogen bond. This means that a bond was formed between HA and HB. In addition, the plateau indicates that the bond is static. On the other hand, the distance between HB and HC has increased to 1.91 Å at large step values. This shows that HC has been ejected off following a successful collision between HA and HBC.

Similarly, in the dynamic trajectory, the distance between HA and HB decreases to approximately 0.75 Å after 0.3 s. This is indicative of hydrogen bond formation between HA and HB. In addition, the distance between HB and HC increases to 3.44 Å after HC is ejected following a successful collision.

One obvious distance between the two plots is that in the minimum energy path trajectory, the internuclear distance between HA and HB is constant indicating that the bond is static. However, in the dynamic trajectory the internuclear distance is oscillating showing the the bond is vibrating. This difference could be attributed to the fact that minimum energy path does not take oscillatory motion into account while dynamic trajectory does.

However, changing the initial conditions of the system such that rAB = 0.917743 Å, rBC = 0.907743 Å and pA=pBC=0:
[[File:01409469 01409469 Reversed values dynamic mep.jpg|thumb|center|800px|'''Figure 8.''' A graph of internuclear distance against steps using minimum energy path '''Figure 9.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In this case, a hydrogen bond of length 0.77-0.79 Å is formed between HB and HC while HA is repelled away to beyond 3.45 Å.

Additionally, from the internuclear distance plot it can be seen that the distance between HB and HC increases in a rectangular hyperbola fashion when calculated using the minimum energy path. On the other hand, the internuclear distance between HB and HC increases in a linear manner when dynamic calculation is applied. A possible reason for this difference is that in the minimum energy path, the atoms are not vibrating and as such the reach the maximum distance and plateau at a faster rate. However, in the dynamic calculation, the atoms have momentum and as such move further away from each other.

=== Reactive and unreactive trajectories ===

With the initial positions r'''1''' = 0.74 and r'''2''' = 2.0, trajectories were run with the momenta combination shown below.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Contour plot !! Reactivity !! Description of the dynamics
|-
| -1.25 || -2.5 ||-99.018 ||[[File:01409469 trajectory1.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. There are no oscillations observed in HA-HB initially, indicating that the diatomic molecule is stable. A new bond is formed between HB and HC. Oscillatory behaviour is observed indicating that the bond is vibrating. On the other hand, HA is ejected away to beyond 10 Å.
|-
| -1.5 || -2.0 ||-100.456 ||[[File:01409469 trajectory2.jpg|thumb|center|400px]] ||Unreactive ||HC collides with the vibrating diatomic HA-HB. However, no bond is formed between HB and HC as the collision is unsuccessful. This is because the molecules lack sufficient kinetic energy to overcome the activation energy barrier and so no reaction occurs. HC is subsequently repelled to beyond 8 Å.
|-
| -1.5 || -2.5 ||-98.956 ||[[File:01409469 trajectory3.jpg|thumb|center|400px]] ||Reactive |||HC collides with HA-HB. The collision is successful and a bond is formed between HB and HC. This is because the atoms have sufficient kinetic energy to overcome the activation energy barrier and form a new HB-HC bond. HA is ejected away to beyond 10 Å. The new hydrogen bond formed is vibrating as seen by the oscillatory behaviour.
|-
| -2.5 || -5.0 ||-84.956 ||[[File:01409469 trajectory4.jpg|thumb|center|400px]] ||Unreactive ||HC collides with HA-HB. Initially the HA-HB bond is elongating, indicating that the hydrogen bond is weakening and going to break. However, after 0.5 s, the HA-HB bond length decreases and the HAHB is not broken. HC is repelled beyond 6 Å with very strong vibrations.
A reaction did not occur even though the atoms have sufficient kinetic energy to cross the activation energy barrier. This is because the vibrations are too strong. The system crosses the transition state region and a bond is formed in the product temporarily but then the system reverts back to the reactants. This is an example of barrier recrossing.
|-
| -2.5 || -5.2 ||-83.416 ||[[File:01409469 trajectory5.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. The collision is successful and a new hydrogen bond is formed between HB and HC with HA being ejected away. The products formed experience very strong vibrations as evidenced by the oscillatory motion which occur with a very high frequency.
|}

From the table, it can be seen that the chance of a reaction successfully occurring is not solely dependent on the amount of kinetic energy the atoms possess. Even if the atoms have sufficient kinetic energy to cross the activation energy barrier, barrier recrossing can occur if the atoms are too vibrationally excited.

=== Transition State Theory ===

Transition state theory provided the first theoretical attempt to determine absolute reaction rates. In this theory, a transition state separating the reactants from products is used to find an expression for the thermal rate constant.

The main assumptions of transition state theory are<ref name="ref2" />:

1. Electronic and nuclear motions are separated (equivalent to the Born-Oppenheimer approximation in quantum mechanics)

2. Reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution

3. Molecular systems that have crossed the transition state in the direction of products cannot turn around and reform reactants

4. In the transition state, motion along the reaction coordinate may be separated from the other motions and treated classically as a translation

5. Even in the absence of an equilibrium between reactant and product molecules, the transition states that are becoming products are distributed among their states according to the Maxwell-Boltzmann laws

The goal of transition state theory is to predict the rate constant, ''k'', of a reaction given a potential energy surface for the reaction.

However, transition state theory suffers from a set of limitations due to the assumptions stated above not holding true in reality.

Firstly, transition state theory assumes that barrier recrossing is negligible. However, as seen from the fourth trajectory in table 1, barrier recrossing occurred because the atoms were too vibrationally excited and the system reverted back to the reactants even though the system had already crossed the transition state region. In this situation, transition state theory would predict a faster rate of reaction compared to what was obtained experimentally.

Secondly, the transition state theory breaks down for systems with light atoms such as hydrogens. In such systems, quantum effects become crucial. Quantum effects tend to increase the rate of the reaction for two reasons. First, particles lie in energy levels which are higher then the bottom of the potential energy well (there is additional zero-point energy). Secondly, the particles can tunnel through the barrier. These two effects effectively lower the activation energy barrier. As a result, the rate of reaction predicted by transition state theory would be less than what is obtained experimentally.

Lastly, transition state theory does not hold true for systems which are far from equilibrium. In such cases, the assumption of a starting Boltzmann distribution breaks down.

== Exercise 2: F - H - H system ==

=== Potential energy surface inspection===

F + HA-HB → HA-F + HB

[[File:01409469 Surface_Plot F 3.jpg|thumb|center|400px|'''Figure 10.''' Potential energy surface for the reaction between F and H2]]

The reaction between F and H2 is an exothermic reaction. This means that the reactants are of a higher energy than the products. The overall reaction involves the breaking of a hydrogen bond and the formation of a hydrogen-fluorine bond. The HF bond formed is stronger than the hydrogen bond that was broken. This is because fluorine is more electronegative than hydrogen resulting in a polar HF bond. The negative enthalpy of H-F bond formation is greater than the positive enthalpy of hydrogen bond dissociation. Therefore, the reaction between F and H2 is an exothermic reaction.

HA + HB-F → HA-HB + F

[[File:01409469 Surface_Plot F 4.jpg|thumb|center|400px|'''Figure 11.''' Potential energy surface for the reaction between H and HF]]

The reaction between H and HF is an endothermic reaction. The products are at a higher energy level than the reactants. The overall reaction involves the breaking of a HF bond and the formation of a hydrogen bond. The formation of a homonuclear H2 molecule results in a non-polar hydrogen bond being formed. The HF bond broken is stronger than the hydrogen bond that was formed. The negative enthalpy of hydrogen bond formation is less than the positive enthalpy of hydrogen-fluorine dissociation. Therefore, the reaction between H and HF is endothermic.

=== Locating the transition state ===

[[File:01409469 Surface_Plot F TS4.jpg|thumb|center|800px|'''Figure 12.''' Determining the position of the transition state for the reaction between F and H2]]

As mentioned previously, when the molecules are not at the transition state, a reaction path will be plotted. However, if the initial conditions of the system are set such that the trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

An initial estimate for the transition state was made by setting rAB = 0.74 which is the length of a hydrogen bond. The position of the transition state was further refined by increasing the number of decimal places such that the final value of rAB was found to be 0.74478 and rBC = 1.81079. This set of distances gave a reaction path that was represented by a point indicating that no trajectory was plotted and that the molecules remained at the transition state since their initial momentum was 0.

Moreover, the kinetic energy as well as the forces acting on the atoms was calculated to be 0, which is consistent with the transition state.

Therefore, the position of the transition state was found to be at rAB = 0.74478 and rBC = 1.81079 when pA = pBC = 0.

=== Determining the activation energy ===

[[File:01409469 activation energy r1.1 combined.jpg|thumb|center|800px|'''Figure 13.''' Determining the activation energy for the reaction between F and H2]]

By reducing the value of rBC by 0.01, the atoms are no longer at the transition state and would move towards the formation of products. The activation energy was found by calculating the change in total energy after the reaction. The activation energy of F + H2 → HF + H was found to be +23.279 kcal/ mol.

[[File:01409469 activation energy r2.1 combined.jpg|thumb|center|800px|'''Figure 14.''' Determining the activation energy for the reaction between FH and H]]

The activation energy of FH + H → F + H2 was found by calculating the small dip in potential energy. The activation energy was found to be +0.280 kcal/ mol.

=== Reaction dynamics of F + HA-HB → HA-F + HB ===

The set of initial conditions that result in a reactive trajectory for F + H2 is as follows: rHH = 0.74 Å, rHF = 1.79 Å and pHH = pHF = 0.

[[File:01409469 momentum r1.jpg|thumb|center|400px|'''Figure 15.''' Graph of momentum against time for the reaction between F and H2]]

Initially, the H2 molecule is travelling in a relatively straight line with very little oscillations as evidenced by the low amplitude of momentum between 0-2 s. Following successful collision between F and H2, the amplitude of momentum of the product, HF, increases dramatically. This is because the reaction between F and H2 is exothermic and the excess energy released causes the molecule to become vibrationally excited and the amplitude of oscillations in the product to increase.

[[File:01409469 energy1 r1.jpg|thumb|center|400px|'''Figure 16.''' Graph of energy against time for the reaction between F and H2]]

The law of conservation of energy states that energy cannot be created or destroyed. It simply changes from one form to another. As can be seen from the graph in figure 16, total energy remains constant. In this reaction, potential energy is converted to kinetic energy. A decrease in potential energy of the atoms following a successful collision between F and H2 results in an equal increase in kinetic energy which manifests in the form of increased product (HF) oscillations.

This can be confirmed experimentally using infrared chemiluminescence. Observation of radiation from excited product molecules is a clear indication that these molecules are produced in the excited state. If the excitation is primarily vibrational, the radiation will appear in the infrared region of the spectrum (between 3-15 μm). This technique has been used extensively in hydrogen halogen reactions which produce HX molecules in the excited state.

One such experiment involves the use of the arrested relaxation method. In this technique, the reaction is carried out in a chamber whose walls are maintained at a very low temperature (20 K). This results in cryopumping of the chamber in that all the products and unreacted starting materials are condensed on the walls and only those product molecules that radiate before being pumped away are observed.

=== Polanyi's empirical rules ===

In a typical chemical reaction, there is an energetic barrier — a saddle point that the reactants must surmount to become products. Which form of energy is initially deposited in the reactants, translational or vibrational, is more efficacious in surmounting the activation energy barrier is one of the core topics in the area of reaction dynamics. The Polyani rules state that vibrational energy is more effective at promoting a late-barrier reaction (transition state resembles products). Conversely, translational energy is better at promoting an early-barrier reaction (transition state resembles the reactants).

The forward reaction between H2 and F is exothermic. Therefore, the transition state is closer to the reactants than to the products in energy. Since the transition state occurs early in an exothermic reaction, translational energy is more effective at promoting a chemical reaction than vibrational energy.

Conversely, the backward reaction between H and HF is endothermic. According to Hammond's postulate, the transition state resembles the products more than it resembles the reactants. Therefore, in an endothermic reaction, a late transition state is observed and vibrational energy is more effective at promoting a reaction than translational energy.

=== Promoting the forward reaction: F + HA-HB → HA-F + HB ===

1. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 2.95 and pHF = -0.5.

[[File:01409469 polyani 1+2.jpg|thumb|center|800px|'''Figure 17.''' Contour plot of the reaction between F and H2 when pHH = 2.95 '''Figure 18.''' Graph of momentum against time for the reaction between F and H2 when pHH = 2.95 (Left-right)]]

Vibrational energy is represented by pHH while translational energy is represented by pHF.

A molecule of H2 collides with F. However, the collision is unsuccessful and no reaction occurs. A reaction did not occur even though the molecules have sufficient kinetic energy to overcome the activation energy barrier. This is because most of the energy is deposited in the reactants as H-H vibrations as seen by the large oscillations in H2. Barrier recrossing has occured as a result of the large vibrational energy causing the system to revert back to the reactants even though the system has already crossed the transition state and a bond was formed between HB and F temporarily.

This appears to run counter to what one would expect to happen when more energy is put into the system. It is predicted that when more energy is put into the system, the reactants would possess more kinetic energy to overcome the activation energy barrier. However, as can be seen from the simulation above, an increase in the vibrational energy of the reactants is not effective at promoting the occurrence of an exothermic reaction with an early transition state. This is in accordance to Polyani's rules.

2. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 0.1 and pHF = -0.8.

[[File:01409469 polyani 3+4 2.jpg|thumb|center|800px|'''Figure 19.''' Contour plot of the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 '''Figure 20.''' Graph of momentum against time for the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 (Left-right)]]

The collision between a molecule of H2 and F is successful and a chemical reaction has occurred. A reaction has occurred even though the overall energy of the system has been significantly reduced by lowering the amount of vibrational energy in the reactants but slightly increasing the amount of translational energy deposited in the reactants. Therefore, H2 and F have more translational energy to move towards each other and form a H-F bond. The product formed is vibrationally excited as seen by the large amplitude of oscillations in the product HF which correspond to the strong H-F vibrations.

From the second simulation, it can be seen that an increase in translational energy in the reactants is more effective at promoting an exothermic reaction with an early transition state to occur, which again follows Polyani's rules.

=== Promoting the backward reaction: HA + HB-F → HA-HB + F ===

1. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 0.2 and pHH = -3.0.

[[File:01409469 polyani 5+6 .jpg|thumb|center|800px|'''Figure 21.''' Contour plot of the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 '''Figure 22.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 (Left-right)]]

An atom of HA collides with a molecule of HBF. However, the collision is unsuccessful and no chemical reaction occurs. This is because the reactants lack sufficient vibrational energy to surmount the activation energy barrier. Therefore, following an unsuccessful collision with HBF, HA simply bounces off and moves away at constant momentum. This is in agreement with Polyani's rules which state that vibrational energy is more effective than translational energy at promoting reactions with a late transition state, for example, endothermic chemical reactions.

2. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 7.1 and pHH = -0.5.

[[File:01409469 polyani 7+8 .jpg|thumb|center|800px|'''Figure 23.''' Contour plot of the reaction between HA + HB-F when pFH = 7.1 and pHH = -0.5 '''Figure 24.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 7.1 and pHH = -0.5 (Left-right)]]

Increasing the amount of vibrational energy in the reactants by increasing pFH resulted in a chemical reaction successfully occurring. This is because the reactants now have sufficient kinetic energy in the form of H-F vibrations to overcome the activation energy barrier. In addition, because the reaction between HA + HB-F is endothermic, energy is taken in from the surroundings during the reaction and the products are less vibrationally excited than the reactants. This can be seen in the lower amplitude of oscillations in the product formed.

In conclusion, the vibrational energy of the reactants should be increased to promote an endothermic reaction with a late transition state while the translational energy of the reactants should be increased to promote an exothermic reaction with an early transition state. All 4 simulations shown above obey Polyani's rules.

== References ==
<references>
<ref name="ref1">A. C. Vaucher, M. Reiher, ''J. Chem. Theory. Comput.'', 2018, '''14''', 3091-3099.</ref>
<ref name="ref2">J. I. Steinfield, J. S. Francisco, W. L. Hase, ''Chemical Kinetics and Dynamics'', Pearson, 1998.</ref>
</references>

MRD:01409469

2019-05-10T15:00:57Z

Stc1917: /* Transition State Theory */

== Exercise 1: H + H2 system ==

=== Definition of a transition state ===

A potential energy profile results from plotting the potential versus reaction path.

[[File:Surface_Plot01409469_pic1.jpg|thumb|center|400px|'''Figure 1.''' A potential energy surface plot]]

The transition state is defined as the maximum on the minimum energy path linking reactants and products. This means that the transition state is a stationary point on the potential energy surface. Thus, at the transition state, the derivative of the potential energy is zero with respect to each Cartesian coordinate.

<math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>

However, a local minimum on the potential energy surface corresponding to a well would have the same <math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>. Therefore, to distinguish between a local minimum on the potential energy surface and the transition state, the second derivative, <math>\tfrac{\partial^2 V(r_i)}{\partial r_i ^2}\ </math> is determined. This can be calculated by finding the determinant of the Hessian matrix shown below.

:<math>\mathbf H =
\begin{bmatrix}
\dfrac{\partial^2 \ V(r_i)}{\partial \ x^2} & \dfrac{\partial^2 \ V(r_i)}{\partial \ x \partial \mathbf y} \\
\dfrac{\partial^2 \ V(r_i)}{\partial \ y \partial \mathbf x} & \dfrac{\partial^2 \ V(r_i)}{\partial \mathbf y^2}
\end{bmatrix}</math>

The Hessian matrix of a convex function is positive semi-definite. Refining this property allows one to test if a critical point (x,y) is a local maximum, local minimum or saddle point.<ref name="ref1" />

If <math>f</math> < 0 (exactly one negative eigenvalue), then (x,y) is an isolated local maximum.

If <math>f</math> > 0 (exactly one positive eigenvalue), then (x,y) is an isolated local minimum.

If the Hessian has both positive and negative eigenvalues then (x,y) is a saddle point.

=== Locating the transition state ===

[[File:01409469 combined pic 2 and 3.jpg|thumb|center|800px|'''Figure 2.''' A surface and contour plot of the trajectory when rAB=rBC=0.907743 Å and pA=pBC=0]]

When the molecules are not at the transition state, a surface plot of the conditions shows a trajectory being plotted. This means that the trajectory is "rolling" towards the products or reactants. However, if a trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

[[File:01409469_internuclearpic3.png|thumb|center|400px|'''Figure 3.''' A plot of internuclear distance against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

An initial estimate for the position of the transition state was made by observing the potential energy surface and was found to be 0.90 Å. At 0.90 Å, the internuclear distance against time plot showed that the internuclear distance was constantly oscillating. Therefore, the position of the transition state was further refined by increasing the number of decimal places until a horizontal line in the internuclear distance againt time plot was obtained. This means that the internuclear distance is constant and that the molecules are simply undergoing periodic symmetric vibration at the transition state. Therefore, the position of the transition state was determined to be 0.907743 Å.

[[File:01409469 combined energy pic.png|thumb|center|400px|'''Figure 4.''' A plot of energy against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

When rAB=rBC=0.907743 Å and pA=pBC=0, the kinetic energy is 0. This means that HA and HBC are stationary, which is as predicted when the trajectory is started at the transition state with no initial momentum. Furthermore, the forces acting on HA and HBC are 0 as should be the case in the transition state.

=== Calculation type: Dynamic and minimum energy path ===

The minimum energy path, is the lowest energy path from the reactants to the products. The minimum energy path is determined by following the mode corresponding to the negative Hessian eigenvalue in both directions along the steepest descent direction.<ref name="ref1" /> Following this path allows one to approach two valleys of stable structures. The resulting path generated is the minimum energy path. The reaction paths were chartered for a dynamic and a minimum energy path. The initial conditions used were that of the transition state except that the internuclear distance between HA and HB had increased to 0.917743 Å. The reaction path obtained using the two different calculation methods was different.

[[File:01409469 mep and dynamics.jpg|thumb|center|800px|'''Figure 5.''' A surface plot of the trajectory using minimum energy path '''Figure 6.''' A surface plot of the trajectory using dynamics (Left-right)]]

On the surface plot, the reaction path is depicted by the black line. Following the minimum energy path, the trajectory simply follows the valley floor to Ha+Hb-Hc. The reaction path appears to be a relatively straight line indicating that the atoms are not vibrating. In the minimum energy path, the reactants change into the products in an infinitely slow motion. This is done by having the momenta/velocities reset to zero in each time step. While the minimum energy path is useful in characterising a chemical reaction, it does not provide a realistic representation of the motion of the atoms during the reaction. This is because the mass and inertia of the atoms have been ignored.

Since the atoms simply follow the minimum energy path, they do not undergo slight displacements into surrounding valleys of higher potential energy. Furthermore, because the atoms are assumed to have no inertia, they do not undergo vibrations even if they were to be displaced to valleys of higher potential energy.

However, when dynamic calculation is performed, the reaction path is wavy indicating that the atoms are oscillating. This is because the atoms are displaced to regions of higher potential on the path from reactants to products. This leads to atom oscillation due to the inertia of the atoms.

'''Internuclear distance against time plot'''

Setting the initial system conditions such that rAB= 0.907743, rBC= 0.917743 and pA= pBC= 0:

[[File:01409469 Internuclear distance mep and dynamic.jpg|thumb|center|800px|'''Figure 6.''' A graph of internuclear distance against steps using minimum energy path '''Figure 7.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In the minimum energy path, the distance between HA and HB decreases from 0.90 Å to approximately 0.79 Å during the first 50 steps before plateauing off to a value of 0.74 Å which is the approximate length of a hydrogen bond. This means that a bond was formed between HA and HB. In addition, the plateau indicates that the bond is static. On the other hand, the distance between HB and HC has increased to 1.91 Å at large step values. This shows that HC has been ejected off following a successful collision between HA and HBC.

Similarly, in the dynamic trajectory, the distance between HA and HB decreases to approximately 0.75 Å after 0.3 s. This is indicative of hydrogen bond formation between HA and HB. In addition, the distance between HB and HC increases to 3.44 Å after HC is ejected following a successful collision.

One obvious distance between the two plots is that in the minimum energy path trajectory, the internuclear distance between HA and HB is constant indicating that the bond is static. However, in the dynamic trajectory the internuclear distance is oscillating showing the the bond is vibrating. This difference could be attributed to the fact that minimum energy path does not take oscillatory motion into account while dynamic trajectory does.

However, changing the initial conditions of the system such that rAB = 0.917743 Å, rBC = 0.907743 Å and pA=pBC=0:
[[File:01409469 01409469 Reversed values dynamic mep.jpg|thumb|center|800px|'''Figure 8.''' A graph of internuclear distance against steps using minimum energy path '''Figure 9.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In this case, a hydrogen bond of length 0.77-0.79 Å is formed between HB and HC while HA is repelled away to beyond 3.45 Å.

Additionally, from the internuclear distance plot it can be seen that the distance between HB and HC increases in a rectangular hyperbola fashion when calculated using the minimum energy path. On the other hand, the internuclear distance between HB and HC increases in a linear manner when dynamic calculation is applied. A possible reason for this difference is that in the minimum energy path, the atoms are not vibrating and as such the reach the maximum distance and plateau at a faster rate. However, in the dynamic calculation, the atoms have momentum and as such move further away from each other.

=== Reactive and unreactive trajectories ===

With the initial positions r'''1''' = 0.74 and r'''2''' = 2.0, trajectories were run with the momenta combination shown below.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Contour plot !! Reactivity !! Description of the dynamics
|-
| -1.25 || -2.5 ||-99.018 ||[[File:01409469 trajectory1.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. There are no oscillations observed in HA-HB initially, indicating that the diatomic molecule is stable. A new bond is formed between HB and HC. Oscillatory behaviour is observed indicating that the bond is vibrating. On the other hand, HA is ejected away to beyond 10 Å.
|-
| -1.5 || -2.0 ||-100.456 ||[[File:01409469 trajectory2.jpg|thumb|center|400px]] ||Unreactive ||HC collides with the vibrating diatomic HA-HB. However, no bond is formed between HB and HC as the collision is unsuccessful. This is because the molecules lack sufficient kinetic energy to overcome the activation energy barrier and so no reaction occurs. HC is subsequently repelled to beyond 8 Å.
|-
| -1.5 || -2.5 ||-98.956 ||[[File:01409469 trajectory3.jpg|thumb|center|400px]] ||Reactive |||HC collides with HA-HB. The collision is successful and a bond is formed between HB and HC. This is because the atoms have sufficient kinetic energy to overcome the activation energy barrier and form a new HB-HC bond. HA is ejected away to beyond 10 Å. The new hydrogen bond formed is vibrating as seen by the oscillatory behaviour.
|-
| -2.5 || -5.0 ||-84.956 ||[[File:01409469 trajectory4.jpg|thumb|center|400px]] ||Unreactive ||HC collides with HA-HB. Initially the HA-HB bond is elongating, indicating that the hydrogen bond is weakening and going to break. However, after 0.5 s, the HA-HB bond length decreases and the HAHB is not broken. HC is repelled beyond 6 Å with very strong vibrations.
A reaction did not occur even though the atoms have sufficient kinetic energy to cross the activation energy barrier. This is because the vibrations are too strong. The system crosses the transition state region and a bond is formed in the product temporarily but then the system reverts back to the reactants. This is an example of barrier recrossing.
|-
| -2.5 || -5.2 ||-83.416 ||[[File:01409469 trajectory5.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. The collision is successful and a new hydrogen bond is formed between HB and HC with HA being ejected away. The products formed experience very strong vibrations as evidenced by the oscillatory motion which occur with a very high frequency.
|}

From the table, it can be seen that the chance of a reaction successfully occurring is not solely dependent on the amount of kinetic energy the atoms possess. Even if the atoms have sufficient kinetic energy to cross the activation energy barrier, barrier recrossing can occur if the atoms are too vibrationally excited.

=== Transition State Theory ===

Transition state theory provided the first theoretical attempt to determine absolute reaction rates. In this theory, a transition state separating the reactants from products is used to find an expression for the thermal rate constant.

The main assumptions of transition state theory are<ref name="ref2" />:

1. Electronic and nuclear motions are separated (equivalent to the Born-Oppenheimer approximation in quantum mechanics)

2. Reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution

3. Molecular systems that have crossed the transition state in the direction of products cannot turn around and reform reactants

4. In the transition state, motion along the reaction coordinate may be separated from the other motions and treated classically as a translation

5. Even in the absence of an equilibrium between reactant and product molecules, the transition states that are becoming products are distributed among their states according to the Maxwell-Boltzmann laws

The goal of transition state theory is to predict the rate constant, ''k'', of a reaction given a potential energy surface for the reaction.

However, transition state theory suffers from a set of limitations due to the assumptions stated above not holding true in reality.

Firstly, transition state theory assumes that barrier recrossing is negligible. However, as seen from the fourth trajectory in table 1, barrier recrossing occurred because the atoms were too vibrationally excited and the system reverted back to the reactants even though the system had already crossed the transition state region. In this situation, transition state theory would predict a faster rate of reaction compared to what was obtained experimentally.

Secondly, the transition state theory breaks down for systems with light atoms such as hydrogens. In such systems, quantum effects become crucial. Quantum effects tend to increase the rate of the reaction for two reasons. First, particles lie in energy levels which are higher then the bottom of the potential energy well (there is additional zero-point energy). Secondly, the particles can tunnel through the barrier. These two effects effectively lower the activation energy barrier. As a result, the rate of reaction predicted by transition state theory would be less than what is obtained experimentally.

Lastly, transition state theory does not hold true for systems which are far from equilibrium. In such cases, the assumption of a starting Boltzmann distribution breaks down.

== Exercise 2: F - H - H system ==

=== Potential energy surface inspection===

F + HA-HB → HA-F + HB

[[File:01409469 Surface_Plot F 3.jpg|thumb|center|400px|'''Figure 10.''' Potential energy surface for the reaction between F and H2]]

The reaction between F and H2 is an exothermic reaction. This means that the reactants are of a higher energy than the products. The overall reaction involves the breaking of a hydrogen bond and the formation of a hydrogen-fluorine bond. The HF bond formed is stronger than the hydrogen bond that was broken. This is because fluorine is more electronegative than hydrogen resulting in a polar HF bond. The negative enthalpy of H-F bond formation is greater than the positive enthalpy of hydrogen bond dissociation. Therefore, the reaction between F and H2 is an exothermic reaction.

HA + HB-F → HA-HB + F

[[File:01409469 Surface_Plot F 4.jpg|thumb|center|400px|'''Figure 11.''' Potential energy surface for the reaction between H and HF]]

The reaction between H and HF is an endothermic reaction. The products are at a higher energy level than the reactants. The overall reaction involves the breaking of a HF bond and the formation of a hydrogen bond. The formation of a homonuclear H2 molecule results in a non-polar hydrogen bond being formed. The HF bond broken is stronger than the hydrogen bond that was formed. The negative enthalpy of hydrogen bond formation is less than the positive enthalpy of hydrogen-fluorine dissociation. Therefore, the reaction between H and HF is endothermic.

=== Locating the transition state ===

[[File:01409469 Surface_Plot F TS4.jpg|thumb|center|800px|'''Figure 12.''' Determining the position of the transition state for the reaction between F and H2]]

As mentioned previously, when the molecules are not at the transition state, a reaction path will be plotted. However, if the initial conditions of the system are set such that the trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

An initial estimate for the transition state was made by setting rAB = 0.74 which is the length of a hydrogen bond. The position of the transition state was further refined by increasing the number of decimal places such that the final value of rAB was found to be 0.74478 and rBC = 1.81079. This set of distances gave a reaction path that was represented by a point indicating that no trajectory was plotted and that the molecules remained at the transition state since their initial momentum was 0.

Moreover, the kinetic energy as well as the forces acting on the atoms was calculated to be 0, which is consistent with the transition state.

Therefore, the position of the transition state was found to be at rAB = 0.74478 and rBC = 1.81079 when pA = pBC = 0.

=== Determining the activation energy ===

[[File:01409469 activation energy r1.1 combined.jpg|thumb|center|800px|'''Figure 13.''' Determining the activation energy for the reaction between F and H2]]

By reducing the value of rBC by 0.01, the atoms are no longer at the transition state and would move towards the formation of products. The activation energy was found by calculating the change in total energy after the reaction. The activation energy of F + H2 → HF + H was found to be +23.279 kcal/ mol.

[[File:01409469 activation energy r2.1 combined.jpg|thumb|center|800px|'''Figure 14.''' Determining the activation energy for the reaction between FH and H]]

The activation energy of FH + H → F + H2 was found by calculating the small dip in potential energy. The activation energy was found to be +0.280 kcal/ mol.

=== Reaction dynamics of F + HA-HB → HA-F + HB ===

The set of initial conditions that result in a reactive trajectory for F + H2 is as follows: rHH = 0.74 Å, rHF = 1.79 Å and pHH = pHF = 0.

[[File:01409469 momentum r1.jpg|thumb|center|400px|'''Figure 15.''' Graph of momentum against time for the reaction between F and H2]]

Initially, the H2 molecule is travelling in a relatively straight line with very little oscillations as evidenced by the low amplitude of momentum between 0-2 s. Following successful collision between F and H2, the amplitude of momentum of the product, HF, increases dramatically. This is because the reaction between F and H2 is exothermic and the excess energy released causes the molecule to become vibrationally excited and the amplitude of oscillations in the product to increase.

[[File:01409469 energy1 r1.jpg|thumb|center|400px|'''Figure 16.''' Graph of energy against time for the reaction between F and H2]]

The law of conservation of energy states that energy cannot be created or destroyed. It simply changes from one form to another. As can be seen from the graph in figure 16, total energy remains constant. In this reaction, potential energy is converted to kinetic energy. A decrease in potential energy of the atoms following a successful collision between F and H2 results in an equal increase in kinetic energy which manifests in the form of increased product (HF) oscillations.

This can be confirmed experimentally using infrared chemiluminescence. Observation of radiation from excited product molecules is a clear indication that these molecules are produced in the excited state. If the excitation is primarily vibrational, the radiation will appear in the infrared region of the spectrum (between 3-15 μm). This technique has been used extensively in hydrogen halogen reactions which produce HX molecules in the excited state.

One such experiment involves the use of the arrested relaxation method. In this technique, the reaction is carried out in a chamber whose walls are maintained at a very low temperature (20 K). This results in cryopumping of the chamber in that all the products and unreacted starting materials are condensed on the walls and only those product molecules that radiate before being pumped away are observed.

=== Polanyi's empirical rules ===

In a typical chemical reaction, there is an energetic barrier — a saddle point that the reactants must surmount to become products. Which form of energy is initially deposited in the reactants, translational or vibrational, is more efficacious in surmounting the activation energy barrier is one of the core topics in the area of reaction dynamics. The Polyani rules state that vibrational energy is more effective at promoting a late-barrier reaction (transition state resembles products). Conversely, translational energy is better at promoting an early-barrier reaction (transition state resembles the reactants).

The forward reaction between H2 and F is exothermic. Therefore, the transition state is closer to the reactants than to the products in energy. Since the transition state occurs early in an exothermic reaction, translational energy is more effective at promoting a chemical reaction than vibrational energy.

Conversely, the backward reaction between H and HF is endothermic. According to Hammond's postulate, the transition state resembles the products more than it resembles the reactants. Therefore, in an endothermic reaction, a late transition state is observed and vibrational energy is more effective at promoting a reaction than translational energy.

=== Promoting the forward reaction: F + HA-HB → HA-F + HB ===

1. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 2.95 and pHF = -0.5.

[[File:01409469 polyani 1+2.jpg|thumb|center|800px|'''Figure 17.''' Contour plot of the reaction between F and H2 when pHH = 2.95 '''Figure 18.''' Graph of momentum against time for the reaction between F and H2 when pHH = 2.95 (Left-right)]]

Vibrational energy is represented by pHH while translational energy is represented by pHF.

A molecule of H2 collides with F. However, the collision is unsuccessful and no reaction occurs. A reaction did not occur even though the molecules have sufficient kinetic energy to overcome the activation energy barrier. This is because most of the energy is deposited in the reactants as H-H vibrations as seen by the large oscillations in H2. Barrier recrossing has occured as a result of the large vibrational energy causing the system to revert back to the reactants even though the system has already crossed the transition state and a bond was formed between HB and F temporarily.

This appears to run counter to what one would expect to happen when more energy is put into the system. It is predicted that when more energy is put into the system, the reactants would possess more kinetic energy to overcome the activation energy barrier. However, as can be seen from the simulation above, an increase in the vibrational energy of the reactants is not effective at promoting the occurrence of an exothermic reaction with an early transition state. This is in accordance to Polyani's rules.

2. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 0.1 and pHF = -0.8.

[[File:01409469 polyani 3+4 2.jpg|thumb|center|800px|'''Figure 19.''' Contour plot of the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 '''Figure 20.''' Graph of momentum against time for the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 (Left-right)]]

The collision between a molecule of H2 and F is successful and a chemical reaction has occurred. A reaction has occurred even though the overall energy of the system has been significantly reduced by lowering the amount of vibrational energy in the reactants but slightly increasing the amount of translational energy deposited in the reactants. Therefore, H2 and F have more translational energy to move towards each other and form a H-F bond. The product formed is vibrationally excited as seen by the large amplitude of oscillations in the product HF which correspond to the strong H-F vibrations.

From the second simulation, it can be seen that an increase in translational energy in the reactants is more effective at promoting an exothermic reaction with an early transition state to occur, which again follows Polyani's rules.

=== Promoting the backward reaction: HA + HB-F → HA-HB + F ===

1. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 0.2 and pHH = -3.0.

[[File:01409469 polyani 5+6 .jpg|thumb|center|800px|'''Figure 21.''' Contour plot of the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 '''Figure 22.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 (Left-right)]]

An atom of HA collides with a molecule of HBF. However, the collision is unsuccessful and no chemical reaction occurs. This is because the reactants lack sufficient vibrational energy to surmount the activation energy barrier. Therefore, following an unsuccessful collision with HBF, HA simply bounces off and moves away at constant momentum. This is in agreement with Polyani's rules which state that vibrational energy is more effective than translational energy at promoting reactions with a late transition state, for example, endothermic chemical reactions.

2. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 7.1 and pHH = -0.5.

[[File:01409469 polyani 7+8 .jpg|thumb|center|800px|'''Figure 23.''' Contour plot of the reaction between HA + HB-F when pFH = 7.1 and pHH = -0.5 '''Figure 24.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 7.1 and pHH = -0.5 (Left-right)]]

Increasing the amount of vibrational energy in the reactants by increasing pFH resulted in a chemical reaction successfully occurring. This is because the reactants now have sufficient kinetic energy in the form of H-F vibrations to overcome the activation energy barrier. In addition, because the reaction between HA + HB-F is endothermic, energy is taken in from the surroundings during the reaction and the products are less vibrationally excited than the reactants. This can be seen in the lower amplitude of oscillations in the product formed.

In conclusion, the vibrational energy of the reactants should be increased to promote an endothermic reaction with a late transition state while the translational energy of the reactants should be increased to promote an exothermic reaction with an early transition state. All 4 simulations shown above obey Polyani's rules.

== References ==
<references>
<ref name="ref1">A. C. Vaucher, M. Reiher, ''J. Chem. Theory. Comput.'', 2018, '''14''', 3091-3099.</ref>
</references>

MRD:01409469

2019-05-10T14:59:31Z

Stc1917: /* Calculation type: Dynamic and minimum energy path */

== Exercise 1: H + H2 system ==

=== Definition of a transition state ===

A potential energy profile results from plotting the potential versus reaction path.

[[File:Surface_Plot01409469_pic1.jpg|thumb|center|400px|'''Figure 1.''' A potential energy surface plot]]

The transition state is defined as the maximum on the minimum energy path linking reactants and products. This means that the transition state is a stationary point on the potential energy surface. Thus, at the transition state, the derivative of the potential energy is zero with respect to each Cartesian coordinate.

<math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>

However, a local minimum on the potential energy surface corresponding to a well would have the same <math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>. Therefore, to distinguish between a local minimum on the potential energy surface and the transition state, the second derivative, <math>\tfrac{\partial^2 V(r_i)}{\partial r_i ^2}\ </math> is determined. This can be calculated by finding the determinant of the Hessian matrix shown below.

:<math>\mathbf H =
\begin{bmatrix}
\dfrac{\partial^2 \ V(r_i)}{\partial \ x^2} & \dfrac{\partial^2 \ V(r_i)}{\partial \ x \partial \mathbf y} \\
\dfrac{\partial^2 \ V(r_i)}{\partial \ y \partial \mathbf x} & \dfrac{\partial^2 \ V(r_i)}{\partial \mathbf y^2}
\end{bmatrix}</math>

The Hessian matrix of a convex function is positive semi-definite. Refining this property allows one to test if a critical point (x,y) is a local maximum, local minimum or saddle point.<ref name="ref1" />

If <math>f</math> < 0 (exactly one negative eigenvalue), then (x,y) is an isolated local maximum.

If <math>f</math> > 0 (exactly one positive eigenvalue), then (x,y) is an isolated local minimum.

If the Hessian has both positive and negative eigenvalues then (x,y) is a saddle point.

=== Locating the transition state ===

[[File:01409469 combined pic 2 and 3.jpg|thumb|center|800px|'''Figure 2.''' A surface and contour plot of the trajectory when rAB=rBC=0.907743 Å and pA=pBC=0]]

When the molecules are not at the transition state, a surface plot of the conditions shows a trajectory being plotted. This means that the trajectory is "rolling" towards the products or reactants. However, if a trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

[[File:01409469_internuclearpic3.png|thumb|center|400px|'''Figure 3.''' A plot of internuclear distance against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

An initial estimate for the position of the transition state was made by observing the potential energy surface and was found to be 0.90 Å. At 0.90 Å, the internuclear distance against time plot showed that the internuclear distance was constantly oscillating. Therefore, the position of the transition state was further refined by increasing the number of decimal places until a horizontal line in the internuclear distance againt time plot was obtained. This means that the internuclear distance is constant and that the molecules are simply undergoing periodic symmetric vibration at the transition state. Therefore, the position of the transition state was determined to be 0.907743 Å.

[[File:01409469 combined energy pic.png|thumb|center|400px|'''Figure 4.''' A plot of energy against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

When rAB=rBC=0.907743 Å and pA=pBC=0, the kinetic energy is 0. This means that HA and HBC are stationary, which is as predicted when the trajectory is started at the transition state with no initial momentum. Furthermore, the forces acting on HA and HBC are 0 as should be the case in the transition state.

=== Calculation type: Dynamic and minimum energy path ===

The minimum energy path, is the lowest energy path from the reactants to the products. The minimum energy path is determined by following the mode corresponding to the negative Hessian eigenvalue in both directions along the steepest descent direction.<ref name="ref1" /> Following this path allows one to approach two valleys of stable structures. The resulting path generated is the minimum energy path. The reaction paths were chartered for a dynamic and a minimum energy path. The initial conditions used were that of the transition state except that the internuclear distance between HA and HB had increased to 0.917743 Å. The reaction path obtained using the two different calculation methods was different.

[[File:01409469 mep and dynamics.jpg|thumb|center|800px|'''Figure 5.''' A surface plot of the trajectory using minimum energy path '''Figure 6.''' A surface plot of the trajectory using dynamics (Left-right)]]

On the surface plot, the reaction path is depicted by the black line. Following the minimum energy path, the trajectory simply follows the valley floor to Ha+Hb-Hc. The reaction path appears to be a relatively straight line indicating that the atoms are not vibrating. In the minimum energy path, the reactants change into the products in an infinitely slow motion. This is done by having the momenta/velocities reset to zero in each time step. While the minimum energy path is useful in characterising a chemical reaction, it does not provide a realistic representation of the motion of the atoms during the reaction. This is because the mass and inertia of the atoms have been ignored.

Since the atoms simply follow the minimum energy path, they do not undergo slight displacements into surrounding valleys of higher potential energy. Furthermore, because the atoms are assumed to have no inertia, they do not undergo vibrations even if they were to be displaced to valleys of higher potential energy.

However, when dynamic calculation is performed, the reaction path is wavy indicating that the atoms are oscillating. This is because the atoms are displaced to regions of higher potential on the path from reactants to products. This leads to atom oscillation due to the inertia of the atoms.

'''Internuclear distance against time plot'''

Setting the initial system conditions such that rAB= 0.907743, rBC= 0.917743 and pA= pBC= 0:

[[File:01409469 Internuclear distance mep and dynamic.jpg|thumb|center|800px|'''Figure 6.''' A graph of internuclear distance against steps using minimum energy path '''Figure 7.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In the minimum energy path, the distance between HA and HB decreases from 0.90 Å to approximately 0.79 Å during the first 50 steps before plateauing off to a value of 0.74 Å which is the approximate length of a hydrogen bond. This means that a bond was formed between HA and HB. In addition, the plateau indicates that the bond is static. On the other hand, the distance between HB and HC has increased to 1.91 Å at large step values. This shows that HC has been ejected off following a successful collision between HA and HBC.

Similarly, in the dynamic trajectory, the distance between HA and HB decreases to approximately 0.75 Å after 0.3 s. This is indicative of hydrogen bond formation between HA and HB. In addition, the distance between HB and HC increases to 3.44 Å after HC is ejected following a successful collision.

One obvious distance between the two plots is that in the minimum energy path trajectory, the internuclear distance between HA and HB is constant indicating that the bond is static. However, in the dynamic trajectory the internuclear distance is oscillating showing the the bond is vibrating. This difference could be attributed to the fact that minimum energy path does not take oscillatory motion into account while dynamic trajectory does.

However, changing the initial conditions of the system such that rAB = 0.917743 Å, rBC = 0.907743 Å and pA=pBC=0:
[[File:01409469 01409469 Reversed values dynamic mep.jpg|thumb|center|800px|'''Figure 8.''' A graph of internuclear distance against steps using minimum energy path '''Figure 9.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In this case, a hydrogen bond of length 0.77-0.79 Å is formed between HB and HC while HA is repelled away to beyond 3.45 Å.

Additionally, from the internuclear distance plot it can be seen that the distance between HB and HC increases in a rectangular hyperbola fashion when calculated using the minimum energy path. On the other hand, the internuclear distance between HB and HC increases in a linear manner when dynamic calculation is applied. A possible reason for this difference is that in the minimum energy path, the atoms are not vibrating and as such the reach the maximum distance and plateau at a faster rate. However, in the dynamic calculation, the atoms have momentum and as such move further away from each other.

=== Reactive and unreactive trajectories ===

With the initial positions r'''1''' = 0.74 and r'''2''' = 2.0, trajectories were run with the momenta combination shown below.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Contour plot !! Reactivity !! Description of the dynamics
|-
| -1.25 || -2.5 ||-99.018 ||[[File:01409469 trajectory1.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. There are no oscillations observed in HA-HB initially, indicating that the diatomic molecule is stable. A new bond is formed between HB and HC. Oscillatory behaviour is observed indicating that the bond is vibrating. On the other hand, HA is ejected away to beyond 10 Å.
|-
| -1.5 || -2.0 ||-100.456 ||[[File:01409469 trajectory2.jpg|thumb|center|400px]] ||Unreactive ||HC collides with the vibrating diatomic HA-HB. However, no bond is formed between HB and HC as the collision is unsuccessful. This is because the molecules lack sufficient kinetic energy to overcome the activation energy barrier and so no reaction occurs. HC is subsequently repelled to beyond 8 Å.
|-
| -1.5 || -2.5 ||-98.956 ||[[File:01409469 trajectory3.jpg|thumb|center|400px]] ||Reactive |||HC collides with HA-HB. The collision is successful and a bond is formed between HB and HC. This is because the atoms have sufficient kinetic energy to overcome the activation energy barrier and form a new HB-HC bond. HA is ejected away to beyond 10 Å. The new hydrogen bond formed is vibrating as seen by the oscillatory behaviour.
|-
| -2.5 || -5.0 ||-84.956 ||[[File:01409469 trajectory4.jpg|thumb|center|400px]] ||Unreactive ||HC collides with HA-HB. Initially the HA-HB bond is elongating, indicating that the hydrogen bond is weakening and going to break. However, after 0.5 s, the HA-HB bond length decreases and the HAHB is not broken. HC is repelled beyond 6 Å with very strong vibrations.
A reaction did not occur even though the atoms have sufficient kinetic energy to cross the activation energy barrier. This is because the vibrations are too strong. The system crosses the transition state region and a bond is formed in the product temporarily but then the system reverts back to the reactants. This is an example of barrier recrossing.
|-
| -2.5 || -5.2 ||-83.416 ||[[File:01409469 trajectory5.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. The collision is successful and a new hydrogen bond is formed between HB and HC with HA being ejected away. The products formed experience very strong vibrations as evidenced by the oscillatory motion which occur with a very high frequency.
|}

From the table, it can be seen that the chance of a reaction successfully occurring is not solely dependent on the amount of kinetic energy the atoms possess. Even if the atoms have sufficient kinetic energy to cross the activation energy barrier, barrier recrossing can occur if the atoms are too vibrationally excited.

=== Transition State Theory ===

Transition state theory provided the first theoretical attempt to determine absolute reaction rates. In this theory, a transition state separating the reactants from products is used to find an expression for the thermal rate constant.

The main assumptions of transition state theory are:

1. Electronic and nuclear motions are separated (equivalent to the Born-Oppenheimer approximation in quantum mechanics)

2. Reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution

3. Molecular systems that have crossed the transition state in the direction of products cannot turn around and reform reactants

4. In the transition state, motion along the reaction coordinate may be separated from the other motions and treated classically as a translation

5. Even in the absence of an equilibrium between reactant and product molecules, the transition states that are becoming products are distributed among their states according to the Maxwell-Boltzmann laws

The goal of transition state theory is to predict the rate constant, ''k'', of a reaction given a potential energy surface for the reaction.

However, transition state theory suffers from a set of limitations due to the assumptions stated above not holding true in reality.

Firstly, transition state theory assumes that barrier recrossing is negligible. However, as seen from the fourth trajectory in table 1, barrier recrossing occurred because the atoms were too vibrationally excited and the system reverted back to the reactants even though the system had already crossed the transition state region. In this situation, transition state theory would predict a faster rate of reaction compared to what was obtained experimentally.

Secondly, the transition state theory breaks down for systems with light atoms such as hydrogens. In such systems, quantum effects become crucial. Quantum effects tend to increase the rate of the reaction for two reasons. First, particles lie in energy levels which are higher then the bottom of the potential energy well (there is additional zero-point energy). Secondly, the particles can tunnel through the barrier. These two effects effectively lower the activation energy barrier. As a result, the rate of reaction predicted by transition state theory would be less than what is obtained experimentally.

Lastly, transition state theory does not hold true for systems which are far from equilibrium. In such cases, the assumption of a starting Boltzmann distribution breaks down.

== Exercise 2: F - H - H system ==

=== Potential energy surface inspection===

F + HA-HB → HA-F + HB

[[File:01409469 Surface_Plot F 3.jpg|thumb|center|400px|'''Figure 10.''' Potential energy surface for the reaction between F and H2]]

The reaction between F and H2 is an exothermic reaction. This means that the reactants are of a higher energy than the products. The overall reaction involves the breaking of a hydrogen bond and the formation of a hydrogen-fluorine bond. The HF bond formed is stronger than the hydrogen bond that was broken. This is because fluorine is more electronegative than hydrogen resulting in a polar HF bond. The negative enthalpy of H-F bond formation is greater than the positive enthalpy of hydrogen bond dissociation. Therefore, the reaction between F and H2 is an exothermic reaction.

HA + HB-F → HA-HB + F

[[File:01409469 Surface_Plot F 4.jpg|thumb|center|400px|'''Figure 11.''' Potential energy surface for the reaction between H and HF]]

The reaction between H and HF is an endothermic reaction. The products are at a higher energy level than the reactants. The overall reaction involves the breaking of a HF bond and the formation of a hydrogen bond. The formation of a homonuclear H2 molecule results in a non-polar hydrogen bond being formed. The HF bond broken is stronger than the hydrogen bond that was formed. The negative enthalpy of hydrogen bond formation is less than the positive enthalpy of hydrogen-fluorine dissociation. Therefore, the reaction between H and HF is endothermic.

=== Locating the transition state ===

[[File:01409469 Surface_Plot F TS4.jpg|thumb|center|800px|'''Figure 12.''' Determining the position of the transition state for the reaction between F and H2]]

As mentioned previously, when the molecules are not at the transition state, a reaction path will be plotted. However, if the initial conditions of the system are set such that the trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

An initial estimate for the transition state was made by setting rAB = 0.74 which is the length of a hydrogen bond. The position of the transition state was further refined by increasing the number of decimal places such that the final value of rAB was found to be 0.74478 and rBC = 1.81079. This set of distances gave a reaction path that was represented by a point indicating that no trajectory was plotted and that the molecules remained at the transition state since their initial momentum was 0.

Moreover, the kinetic energy as well as the forces acting on the atoms was calculated to be 0, which is consistent with the transition state.

Therefore, the position of the transition state was found to be at rAB = 0.74478 and rBC = 1.81079 when pA = pBC = 0.

=== Determining the activation energy ===

[[File:01409469 activation energy r1.1 combined.jpg|thumb|center|800px|'''Figure 13.''' Determining the activation energy for the reaction between F and H2]]

By reducing the value of rBC by 0.01, the atoms are no longer at the transition state and would move towards the formation of products. The activation energy was found by calculating the change in total energy after the reaction. The activation energy of F + H2 → HF + H was found to be +23.279 kcal/ mol.

[[File:01409469 activation energy r2.1 combined.jpg|thumb|center|800px|'''Figure 14.''' Determining the activation energy for the reaction between FH and H]]

The activation energy of FH + H → F + H2 was found by calculating the small dip in potential energy. The activation energy was found to be +0.280 kcal/ mol.

=== Reaction dynamics of F + HA-HB → HA-F + HB ===

The set of initial conditions that result in a reactive trajectory for F + H2 is as follows: rHH = 0.74 Å, rHF = 1.79 Å and pHH = pHF = 0.

[[File:01409469 momentum r1.jpg|thumb|center|400px|'''Figure 15.''' Graph of momentum against time for the reaction between F and H2]]

Initially, the H2 molecule is travelling in a relatively straight line with very little oscillations as evidenced by the low amplitude of momentum between 0-2 s. Following successful collision between F and H2, the amplitude of momentum of the product, HF, increases dramatically. This is because the reaction between F and H2 is exothermic and the excess energy released causes the molecule to become vibrationally excited and the amplitude of oscillations in the product to increase.

[[File:01409469 energy1 r1.jpg|thumb|center|400px|'''Figure 16.''' Graph of energy against time for the reaction between F and H2]]

The law of conservation of energy states that energy cannot be created or destroyed. It simply changes from one form to another. As can be seen from the graph in figure 16, total energy remains constant. In this reaction, potential energy is converted to kinetic energy. A decrease in potential energy of the atoms following a successful collision between F and H2 results in an equal increase in kinetic energy which manifests in the form of increased product (HF) oscillations.

This can be confirmed experimentally using infrared chemiluminescence. Observation of radiation from excited product molecules is a clear indication that these molecules are produced in the excited state. If the excitation is primarily vibrational, the radiation will appear in the infrared region of the spectrum (between 3-15 μm). This technique has been used extensively in hydrogen halogen reactions which produce HX molecules in the excited state.

One such experiment involves the use of the arrested relaxation method. In this technique, the reaction is carried out in a chamber whose walls are maintained at a very low temperature (20 K). This results in cryopumping of the chamber in that all the products and unreacted starting materials are condensed on the walls and only those product molecules that radiate before being pumped away are observed.

=== Polanyi's empirical rules ===

In a typical chemical reaction, there is an energetic barrier — a saddle point that the reactants must surmount to become products. Which form of energy is initially deposited in the reactants, translational or vibrational, is more efficacious in surmounting the activation energy barrier is one of the core topics in the area of reaction dynamics. The Polyani rules state that vibrational energy is more effective at promoting a late-barrier reaction (transition state resembles products). Conversely, translational energy is better at promoting an early-barrier reaction (transition state resembles the reactants).

The forward reaction between H2 and F is exothermic. Therefore, the transition state is closer to the reactants than to the products in energy. Since the transition state occurs early in an exothermic reaction, translational energy is more effective at promoting a chemical reaction than vibrational energy.

Conversely, the backward reaction between H and HF is endothermic. According to Hammond's postulate, the transition state resembles the products more than it resembles the reactants. Therefore, in an endothermic reaction, a late transition state is observed and vibrational energy is more effective at promoting a reaction than translational energy.

=== Promoting the forward reaction: F + HA-HB → HA-F + HB ===

1. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 2.95 and pHF = -0.5.

[[File:01409469 polyani 1+2.jpg|thumb|center|800px|'''Figure 17.''' Contour plot of the reaction between F and H2 when pHH = 2.95 '''Figure 18.''' Graph of momentum against time for the reaction between F and H2 when pHH = 2.95 (Left-right)]]

Vibrational energy is represented by pHH while translational energy is represented by pHF.

A molecule of H2 collides with F. However, the collision is unsuccessful and no reaction occurs. A reaction did not occur even though the molecules have sufficient kinetic energy to overcome the activation energy barrier. This is because most of the energy is deposited in the reactants as H-H vibrations as seen by the large oscillations in H2. Barrier recrossing has occured as a result of the large vibrational energy causing the system to revert back to the reactants even though the system has already crossed the transition state and a bond was formed between HB and F temporarily.

This appears to run counter to what one would expect to happen when more energy is put into the system. It is predicted that when more energy is put into the system, the reactants would possess more kinetic energy to overcome the activation energy barrier. However, as can be seen from the simulation above, an increase in the vibrational energy of the reactants is not effective at promoting the occurrence of an exothermic reaction with an early transition state. This is in accordance to Polyani's rules.

2. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 0.1 and pHF = -0.8.

[[File:01409469 polyani 3+4 2.jpg|thumb|center|800px|'''Figure 19.''' Contour plot of the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 '''Figure 20.''' Graph of momentum against time for the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 (Left-right)]]

The collision between a molecule of H2 and F is successful and a chemical reaction has occurred. A reaction has occurred even though the overall energy of the system has been significantly reduced by lowering the amount of vibrational energy in the reactants but slightly increasing the amount of translational energy deposited in the reactants. Therefore, H2 and F have more translational energy to move towards each other and form a H-F bond. The product formed is vibrationally excited as seen by the large amplitude of oscillations in the product HF which correspond to the strong H-F vibrations.

From the second simulation, it can be seen that an increase in translational energy in the reactants is more effective at promoting an exothermic reaction with an early transition state to occur, which again follows Polyani's rules.

=== Promoting the backward reaction: HA + HB-F → HA-HB + F ===

1. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 0.2 and pHH = -3.0.

[[File:01409469 polyani 5+6 .jpg|thumb|center|800px|'''Figure 21.''' Contour plot of the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 '''Figure 22.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 (Left-right)]]

An atom of HA collides with a molecule of HBF. However, the collision is unsuccessful and no chemical reaction occurs. This is because the reactants lack sufficient vibrational energy to surmount the activation energy barrier. Therefore, following an unsuccessful collision with HBF, HA simply bounces off and moves away at constant momentum. This is in agreement with Polyani's rules which state that vibrational energy is more effective than translational energy at promoting reactions with a late transition state, for example, endothermic chemical reactions.

2. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 7.1 and pHH = -0.5.

[[File:01409469 polyani 7+8 .jpg|thumb|center|800px|'''Figure 23.''' Contour plot of the reaction between HA + HB-F when pFH = 7.1 and pHH = -0.5 '''Figure 24.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 7.1 and pHH = -0.5 (Left-right)]]

Increasing the amount of vibrational energy in the reactants by increasing pFH resulted in a chemical reaction successfully occurring. This is because the reactants now have sufficient kinetic energy in the form of H-F vibrations to overcome the activation energy barrier. In addition, because the reaction between HA + HB-F is endothermic, energy is taken in from the surroundings during the reaction and the products are less vibrationally excited than the reactants. This can be seen in the lower amplitude of oscillations in the product formed.

In conclusion, the vibrational energy of the reactants should be increased to promote an endothermic reaction with a late transition state while the translational energy of the reactants should be increased to promote an exothermic reaction with an early transition state. All 4 simulations shown above obey Polyani's rules.

== References ==
<references>
<ref name="ref1">A. C. Vaucher, M. Reiher, ''J. Chem. Theory. Comput.'', 2018, '''14''', 3091-3099.</ref>
</references>

MRD:01409469

2019-05-10T14:58:44Z

Stc1917: /* Definition of a transition state */

== Exercise 1: H + H2 system ==

=== Definition of a transition state ===

A potential energy profile results from plotting the potential versus reaction path.

[[File:Surface_Plot01409469_pic1.jpg|thumb|center|400px|'''Figure 1.''' A potential energy surface plot]]

The transition state is defined as the maximum on the minimum energy path linking reactants and products. This means that the transition state is a stationary point on the potential energy surface. Thus, at the transition state, the derivative of the potential energy is zero with respect to each Cartesian coordinate.

<math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>

However, a local minimum on the potential energy surface corresponding to a well would have the same <math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>. Therefore, to distinguish between a local minimum on the potential energy surface and the transition state, the second derivative, <math>\tfrac{\partial^2 V(r_i)}{\partial r_i ^2}\ </math> is determined. This can be calculated by finding the determinant of the Hessian matrix shown below.

:<math>\mathbf H =
\begin{bmatrix}
\dfrac{\partial^2 \ V(r_i)}{\partial \ x^2} & \dfrac{\partial^2 \ V(r_i)}{\partial \ x \partial \mathbf y} \\
\dfrac{\partial^2 \ V(r_i)}{\partial \ y \partial \mathbf x} & \dfrac{\partial^2 \ V(r_i)}{\partial \mathbf y^2}
\end{bmatrix}</math>

The Hessian matrix of a convex function is positive semi-definite. Refining this property allows one to test if a critical point (x,y) is a local maximum, local minimum or saddle point.<ref name="ref1" />

If <math>f</math> < 0 (exactly one negative eigenvalue), then (x,y) is an isolated local maximum.

If <math>f</math> > 0 (exactly one positive eigenvalue), then (x,y) is an isolated local minimum.

If the Hessian has both positive and negative eigenvalues then (x,y) is a saddle point.

=== Locating the transition state ===

[[File:01409469 combined pic 2 and 3.jpg|thumb|center|800px|'''Figure 2.''' A surface and contour plot of the trajectory when rAB=rBC=0.907743 Å and pA=pBC=0]]

When the molecules are not at the transition state, a surface plot of the conditions shows a trajectory being plotted. This means that the trajectory is "rolling" towards the products or reactants. However, if a trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

[[File:01409469_internuclearpic3.png|thumb|center|400px|'''Figure 3.''' A plot of internuclear distance against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

An initial estimate for the position of the transition state was made by observing the potential energy surface and was found to be 0.90 Å. At 0.90 Å, the internuclear distance against time plot showed that the internuclear distance was constantly oscillating. Therefore, the position of the transition state was further refined by increasing the number of decimal places until a horizontal line in the internuclear distance againt time plot was obtained. This means that the internuclear distance is constant and that the molecules are simply undergoing periodic symmetric vibration at the transition state. Therefore, the position of the transition state was determined to be 0.907743 Å.

[[File:01409469 combined energy pic.png|thumb|center|400px|'''Figure 4.''' A plot of energy against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

When rAB=rBC=0.907743 Å and pA=pBC=0, the kinetic energy is 0. This means that HA and HBC are stationary, which is as predicted when the trajectory is started at the transition state with no initial momentum. Furthermore, the forces acting on HA and HBC are 0 as should be the case in the transition state.

=== Calculation type: Dynamic and minimum energy path ===

The minimum energy path, is the lowest energy path from the reactants to the products. The minimum energy path is determined by following the mode corresponding to the negative Hessian eigenvalue in both directions along the steepest descent direction. Following this path allows one to approach two valleys of stable structures. The resulting path generated is the minimum energy path. The reaction paths were chartered for a dynamic and a minimum energy path. The initial conditions used were that of the transition state except that the internuclear distance between HA and HB had increased to 0.917743 Å. The reaction path obtained using the two different calculation methods was different.

[[File:01409469 mep and dynamics.jpg|thumb|center|800px|'''Figure 5.''' A surface plot of the trajectory using minimum energy path '''Figure 6.''' A surface plot of the trajectory using dynamics (Left-right)]]

On the surface plot, the reaction path is depicted by the black line. Following the minimum energy path, the trajectory simply follows the valley floor to Ha+Hb-Hc. The reaction path appears to be a relatively straight line indicating that the atoms are not vibrating. In the minimum energy path, the reactants change into the products in an infinitely slow motion. This is done by having the momenta/velocities reset to zero in each time step. While the minimum energy path is useful in characterising a chemical reaction, it does not provide a realistic representation of the motion of the atoms during the reaction. This is because the mass and inertia of the atoms have been ignored.

Since the atoms simply follow the minimum energy path, they do not undergo slight displacements into surrounding valleys of higher potential energy. Furthermore, because the atoms are assumed to have no inertia, they do not undergo vibrations even if they were to be displaced to valleys of higher potential energy.

However, when dynamic calculation is performed, the reaction path is wavy indicating that the atoms are oscillating. This is because the atoms are displaced to regions of higher potential on the path from reactants to products. This leads to atom oscillation due to the inertia of the atoms.

'''Internuclear distance against time plot'''

Setting the initial system conditions such that rAB= 0.907743, rBC= 0.917743 and pA= pBC= 0:

[[File:01409469 Internuclear distance mep and dynamic.jpg|thumb|center|800px|'''Figure 6.''' A graph of internuclear distance against steps using minimum energy path '''Figure 7.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In the minimum energy path, the distance between HA and HB decreases from 0.90 Å to approximately 0.79 Å during the first 50 steps before plateauing off to a value of 0.74 Å which is the approximate length of a hydrogen bond. This means that a bond was formed between HA and HB. In addition, the plateau indicates that the bond is static. On the other hand, the distance between HB and HC has increased to 1.91 Å at large step values. This shows that HC has been ejected off following a successful collision between HA and HBC.

Similarly, in the dynamic trajectory, the distance between HA and HB decreases to approximately 0.75 Å after 0.3 s. This is indicative of hydrogen bond formation between HA and HB. In addition, the distance between HB and HC increases to 3.44 Å after HC is ejected following a successful collision.

One obvious distance between the two plots is that in the minimum energy path trajectory, the internuclear distance between HA and HB is constant indicating that the bond is static. However, in the dynamic trajectory the internuclear distance is oscillating showing the the bond is vibrating. This difference could be attributed to the fact that minimum energy path does not take oscillatory motion into account while dynamic trajectory does.

However, changing the initial conditions of the system such that rAB = 0.917743 Å, rBC = 0.907743 Å and pA=pBC=0:
[[File:01409469 01409469 Reversed values dynamic mep.jpg|thumb|center|800px|'''Figure 8.''' A graph of internuclear distance against steps using minimum energy path '''Figure 9.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In this case, a hydrogen bond of length 0.77-0.79 Å is formed between HB and HC while HA is repelled away to beyond 3.45 Å.

Additionally, from the internuclear distance plot it can be seen that the distance between HB and HC increases in a rectangular hyperbola fashion when calculated using the minimum energy path. On the other hand, the internuclear distance between HB and HC increases in a linear manner when dynamic calculation is applied. A possible reason for this difference is that in the minimum energy path, the atoms are not vibrating and as such the reach the maximum distance and plateau at a faster rate. However, in the dynamic calculation, the atoms have momentum and as such move further away from each other.

=== Reactive and unreactive trajectories ===

With the initial positions r'''1''' = 0.74 and r'''2''' = 2.0, trajectories were run with the momenta combination shown below.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Contour plot !! Reactivity !! Description of the dynamics
|-
| -1.25 || -2.5 ||-99.018 ||[[File:01409469 trajectory1.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. There are no oscillations observed in HA-HB initially, indicating that the diatomic molecule is stable. A new bond is formed between HB and HC. Oscillatory behaviour is observed indicating that the bond is vibrating. On the other hand, HA is ejected away to beyond 10 Å.
|-
| -1.5 || -2.0 ||-100.456 ||[[File:01409469 trajectory2.jpg|thumb|center|400px]] ||Unreactive ||HC collides with the vibrating diatomic HA-HB. However, no bond is formed between HB and HC as the collision is unsuccessful. This is because the molecules lack sufficient kinetic energy to overcome the activation energy barrier and so no reaction occurs. HC is subsequently repelled to beyond 8 Å.
|-
| -1.5 || -2.5 ||-98.956 ||[[File:01409469 trajectory3.jpg|thumb|center|400px]] ||Reactive |||HC collides with HA-HB. The collision is successful and a bond is formed between HB and HC. This is because the atoms have sufficient kinetic energy to overcome the activation energy barrier and form a new HB-HC bond. HA is ejected away to beyond 10 Å. The new hydrogen bond formed is vibrating as seen by the oscillatory behaviour.
|-
| -2.5 || -5.0 ||-84.956 ||[[File:01409469 trajectory4.jpg|thumb|center|400px]] ||Unreactive ||HC collides with HA-HB. Initially the HA-HB bond is elongating, indicating that the hydrogen bond is weakening and going to break. However, after 0.5 s, the HA-HB bond length decreases and the HAHB is not broken. HC is repelled beyond 6 Å with very strong vibrations.
A reaction did not occur even though the atoms have sufficient kinetic energy to cross the activation energy barrier. This is because the vibrations are too strong. The system crosses the transition state region and a bond is formed in the product temporarily but then the system reverts back to the reactants. This is an example of barrier recrossing.
|-
| -2.5 || -5.2 ||-83.416 ||[[File:01409469 trajectory5.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. The collision is successful and a new hydrogen bond is formed between HB and HC with HA being ejected away. The products formed experience very strong vibrations as evidenced by the oscillatory motion which occur with a very high frequency.
|}

From the table, it can be seen that the chance of a reaction successfully occurring is not solely dependent on the amount of kinetic energy the atoms possess. Even if the atoms have sufficient kinetic energy to cross the activation energy barrier, barrier recrossing can occur if the atoms are too vibrationally excited.

=== Transition State Theory ===

Transition state theory provided the first theoretical attempt to determine absolute reaction rates. In this theory, a transition state separating the reactants from products is used to find an expression for the thermal rate constant.

The main assumptions of transition state theory are:

1. Electronic and nuclear motions are separated (equivalent to the Born-Oppenheimer approximation in quantum mechanics)

2. Reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution

3. Molecular systems that have crossed the transition state in the direction of products cannot turn around and reform reactants

4. In the transition state, motion along the reaction coordinate may be separated from the other motions and treated classically as a translation

5. Even in the absence of an equilibrium between reactant and product molecules, the transition states that are becoming products are distributed among their states according to the Maxwell-Boltzmann laws

The goal of transition state theory is to predict the rate constant, ''k'', of a reaction given a potential energy surface for the reaction.

However, transition state theory suffers from a set of limitations due to the assumptions stated above not holding true in reality.

Firstly, transition state theory assumes that barrier recrossing is negligible. However, as seen from the fourth trajectory in table 1, barrier recrossing occurred because the atoms were too vibrationally excited and the system reverted back to the reactants even though the system had already crossed the transition state region. In this situation, transition state theory would predict a faster rate of reaction compared to what was obtained experimentally.

Secondly, the transition state theory breaks down for systems with light atoms such as hydrogens. In such systems, quantum effects become crucial. Quantum effects tend to increase the rate of the reaction for two reasons. First, particles lie in energy levels which are higher then the bottom of the potential energy well (there is additional zero-point energy). Secondly, the particles can tunnel through the barrier. These two effects effectively lower the activation energy barrier. As a result, the rate of reaction predicted by transition state theory would be less than what is obtained experimentally.

Lastly, transition state theory does not hold true for systems which are far from equilibrium. In such cases, the assumption of a starting Boltzmann distribution breaks down.

== Exercise 2: F - H - H system ==

=== Potential energy surface inspection===

F + HA-HB → HA-F + HB

[[File:01409469 Surface_Plot F 3.jpg|thumb|center|400px|'''Figure 10.''' Potential energy surface for the reaction between F and H2]]

The reaction between F and H2 is an exothermic reaction. This means that the reactants are of a higher energy than the products. The overall reaction involves the breaking of a hydrogen bond and the formation of a hydrogen-fluorine bond. The HF bond formed is stronger than the hydrogen bond that was broken. This is because fluorine is more electronegative than hydrogen resulting in a polar HF bond. The negative enthalpy of H-F bond formation is greater than the positive enthalpy of hydrogen bond dissociation. Therefore, the reaction between F and H2 is an exothermic reaction.

HA + HB-F → HA-HB + F

[[File:01409469 Surface_Plot F 4.jpg|thumb|center|400px|'''Figure 11.''' Potential energy surface for the reaction between H and HF]]

The reaction between H and HF is an endothermic reaction. The products are at a higher energy level than the reactants. The overall reaction involves the breaking of a HF bond and the formation of a hydrogen bond. The formation of a homonuclear H2 molecule results in a non-polar hydrogen bond being formed. The HF bond broken is stronger than the hydrogen bond that was formed. The negative enthalpy of hydrogen bond formation is less than the positive enthalpy of hydrogen-fluorine dissociation. Therefore, the reaction between H and HF is endothermic.

=== Locating the transition state ===

[[File:01409469 Surface_Plot F TS4.jpg|thumb|center|800px|'''Figure 12.''' Determining the position of the transition state for the reaction between F and H2]]

As mentioned previously, when the molecules are not at the transition state, a reaction path will be plotted. However, if the initial conditions of the system are set such that the trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

An initial estimate for the transition state was made by setting rAB = 0.74 which is the length of a hydrogen bond. The position of the transition state was further refined by increasing the number of decimal places such that the final value of rAB was found to be 0.74478 and rBC = 1.81079. This set of distances gave a reaction path that was represented by a point indicating that no trajectory was plotted and that the molecules remained at the transition state since their initial momentum was 0.

Moreover, the kinetic energy as well as the forces acting on the atoms was calculated to be 0, which is consistent with the transition state.

Therefore, the position of the transition state was found to be at rAB = 0.74478 and rBC = 1.81079 when pA = pBC = 0.

=== Determining the activation energy ===

[[File:01409469 activation energy r1.1 combined.jpg|thumb|center|800px|'''Figure 13.''' Determining the activation energy for the reaction between F and H2]]

By reducing the value of rBC by 0.01, the atoms are no longer at the transition state and would move towards the formation of products. The activation energy was found by calculating the change in total energy after the reaction. The activation energy of F + H2 → HF + H was found to be +23.279 kcal/ mol.

[[File:01409469 activation energy r2.1 combined.jpg|thumb|center|800px|'''Figure 14.''' Determining the activation energy for the reaction between FH and H]]

The activation energy of FH + H → F + H2 was found by calculating the small dip in potential energy. The activation energy was found to be +0.280 kcal/ mol.

=== Reaction dynamics of F + HA-HB → HA-F + HB ===

The set of initial conditions that result in a reactive trajectory for F + H2 is as follows: rHH = 0.74 Å, rHF = 1.79 Å and pHH = pHF = 0.

[[File:01409469 momentum r1.jpg|thumb|center|400px|'''Figure 15.''' Graph of momentum against time for the reaction between F and H2]]

Initially, the H2 molecule is travelling in a relatively straight line with very little oscillations as evidenced by the low amplitude of momentum between 0-2 s. Following successful collision between F and H2, the amplitude of momentum of the product, HF, increases dramatically. This is because the reaction between F and H2 is exothermic and the excess energy released causes the molecule to become vibrationally excited and the amplitude of oscillations in the product to increase.

[[File:01409469 energy1 r1.jpg|thumb|center|400px|'''Figure 16.''' Graph of energy against time for the reaction between F and H2]]

The law of conservation of energy states that energy cannot be created or destroyed. It simply changes from one form to another. As can be seen from the graph in figure 16, total energy remains constant. In this reaction, potential energy is converted to kinetic energy. A decrease in potential energy of the atoms following a successful collision between F and H2 results in an equal increase in kinetic energy which manifests in the form of increased product (HF) oscillations.

This can be confirmed experimentally using infrared chemiluminescence. Observation of radiation from excited product molecules is a clear indication that these molecules are produced in the excited state. If the excitation is primarily vibrational, the radiation will appear in the infrared region of the spectrum (between 3-15 μm). This technique has been used extensively in hydrogen halogen reactions which produce HX molecules in the excited state.

One such experiment involves the use of the arrested relaxation method. In this technique, the reaction is carried out in a chamber whose walls are maintained at a very low temperature (20 K). This results in cryopumping of the chamber in that all the products and unreacted starting materials are condensed on the walls and only those product molecules that radiate before being pumped away are observed.

=== Polanyi's empirical rules ===

In a typical chemical reaction, there is an energetic barrier — a saddle point that the reactants must surmount to become products. Which form of energy is initially deposited in the reactants, translational or vibrational, is more efficacious in surmounting the activation energy barrier is one of the core topics in the area of reaction dynamics. The Polyani rules state that vibrational energy is more effective at promoting a late-barrier reaction (transition state resembles products). Conversely, translational energy is better at promoting an early-barrier reaction (transition state resembles the reactants).

The forward reaction between H2 and F is exothermic. Therefore, the transition state is closer to the reactants than to the products in energy. Since the transition state occurs early in an exothermic reaction, translational energy is more effective at promoting a chemical reaction than vibrational energy.

Conversely, the backward reaction between H and HF is endothermic. According to Hammond's postulate, the transition state resembles the products more than it resembles the reactants. Therefore, in an endothermic reaction, a late transition state is observed and vibrational energy is more effective at promoting a reaction than translational energy.

=== Promoting the forward reaction: F + HA-HB → HA-F + HB ===

1. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 2.95 and pHF = -0.5.

[[File:01409469 polyani 1+2.jpg|thumb|center|800px|'''Figure 17.''' Contour plot of the reaction between F and H2 when pHH = 2.95 '''Figure 18.''' Graph of momentum against time for the reaction between F and H2 when pHH = 2.95 (Left-right)]]

Vibrational energy is represented by pHH while translational energy is represented by pHF.

A molecule of H2 collides with F. However, the collision is unsuccessful and no reaction occurs. A reaction did not occur even though the molecules have sufficient kinetic energy to overcome the activation energy barrier. This is because most of the energy is deposited in the reactants as H-H vibrations as seen by the large oscillations in H2. Barrier recrossing has occured as a result of the large vibrational energy causing the system to revert back to the reactants even though the system has already crossed the transition state and a bond was formed between HB and F temporarily.

This appears to run counter to what one would expect to happen when more energy is put into the system. It is predicted that when more energy is put into the system, the reactants would possess more kinetic energy to overcome the activation energy barrier. However, as can be seen from the simulation above, an increase in the vibrational energy of the reactants is not effective at promoting the occurrence of an exothermic reaction with an early transition state. This is in accordance to Polyani's rules.

2. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 0.1 and pHF = -0.8.

[[File:01409469 polyani 3+4 2.jpg|thumb|center|800px|'''Figure 19.''' Contour plot of the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 '''Figure 20.''' Graph of momentum against time for the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 (Left-right)]]

The collision between a molecule of H2 and F is successful and a chemical reaction has occurred. A reaction has occurred even though the overall energy of the system has been significantly reduced by lowering the amount of vibrational energy in the reactants but slightly increasing the amount of translational energy deposited in the reactants. Therefore, H2 and F have more translational energy to move towards each other and form a H-F bond. The product formed is vibrationally excited as seen by the large amplitude of oscillations in the product HF which correspond to the strong H-F vibrations.

From the second simulation, it can be seen that an increase in translational energy in the reactants is more effective at promoting an exothermic reaction with an early transition state to occur, which again follows Polyani's rules.

=== Promoting the backward reaction: HA + HB-F → HA-HB + F ===

1. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 0.2 and pHH = -3.0.

[[File:01409469 polyani 5+6 .jpg|thumb|center|800px|'''Figure 21.''' Contour plot of the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 '''Figure 22.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 (Left-right)]]

An atom of HA collides with a molecule of HBF. However, the collision is unsuccessful and no chemical reaction occurs. This is because the reactants lack sufficient vibrational energy to surmount the activation energy barrier. Therefore, following an unsuccessful collision with HBF, HA simply bounces off and moves away at constant momentum. This is in agreement with Polyani's rules which state that vibrational energy is more effective than translational energy at promoting reactions with a late transition state, for example, endothermic chemical reactions.

2. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 7.1 and pHH = -0.5.

[[File:01409469 polyani 7+8 .jpg|thumb|center|800px|'''Figure 23.''' Contour plot of the reaction between HA + HB-F when pFH = 7.1 and pHH = -0.5 '''Figure 24.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 7.1 and pHH = -0.5 (Left-right)]]

Increasing the amount of vibrational energy in the reactants by increasing pFH resulted in a chemical reaction successfully occurring. This is because the reactants now have sufficient kinetic energy in the form of H-F vibrations to overcome the activation energy barrier. In addition, because the reaction between HA + HB-F is endothermic, energy is taken in from the surroundings during the reaction and the products are less vibrationally excited than the reactants. This can be seen in the lower amplitude of oscillations in the product formed.

In conclusion, the vibrational energy of the reactants should be increased to promote an endothermic reaction with a late transition state while the translational energy of the reactants should be increased to promote an exothermic reaction with an early transition state. All 4 simulations shown above obey Polyani's rules.

== References ==
<references>
<ref name="ref1">A. C. Vaucher, M. Reiher, ''J. Chem. Theory. Comput.'', 2018, '''14''', 3091-3099.</ref>
</references>

MRD:01409469

2019-05-10T14:55:38Z

Stc1917: /* References */

== Exercise 1: H + H2 system ==

=== Definition of a transition state ===

A potential energy profile results from plotting the potential versus reaction path.

[[File:Surface_Plot01409469_pic1.jpg|thumb|center|400px|'''Figure 1.''' A potential energy surface plot]]

The transition state is defined as the maximum on the minimum energy path linking reactants and products. This means that the transition state is a stationary point on the potential energy surface. Thus, at the transition state, the derivative of the potential energy is zero with respect to each Cartesian coordinate.

<math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>

However, a local minimum on the potential energy surface corresponding to a well would have the same <math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>. Therefore, to distinguish between a local minimum on the potential energy surface and the transition state, the second derivative, <math>\tfrac{\partial^2 V(r_i)}{\partial r_i ^2}\ </math> is determined. This can be calculated by finding the determinant of the Hessian matrix shown below.

:<math>\mathbf H =
\begin{bmatrix}
\dfrac{\partial^2 \ V(r_i)}{\partial \ x^2} & \dfrac{\partial^2 \ V(r_i)}{\partial \ x \partial \mathbf y} \\
\dfrac{\partial^2 \ V(r_i)}{\partial \ y \partial \mathbf x} & \dfrac{\partial^2 \ V(r_i)}{\partial \mathbf y^2}
\end{bmatrix}</math>

The Hessian matrix of a convex function is positive semi-definite. Refining this property allows one to test if a critical point (x,y) is a local maximum, local minimum or saddle point.<ref name="ref1" />

If <math>f</math> < 0, then (x,y) is an isolated local maximum.

If <math>f</math> > 0, then (x,y) is an isolated local minimum.

If the Hessian has both positive and negative eigenvalues then (x,y) is a saddle point.

=== Locating the transition state ===

[[File:01409469 combined pic 2 and 3.jpg|thumb|center|800px|'''Figure 2.''' A surface and contour plot of the trajectory when rAB=rBC=0.907743 Å and pA=pBC=0]]

When the molecules are not at the transition state, a surface plot of the conditions shows a trajectory being plotted. This means that the trajectory is "rolling" towards the products or reactants. However, if a trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

[[File:01409469_internuclearpic3.png|thumb|center|400px|'''Figure 3.''' A plot of internuclear distance against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

An initial estimate for the position of the transition state was made by observing the potential energy surface and was found to be 0.90 Å. At 0.90 Å, the internuclear distance against time plot showed that the internuclear distance was constantly oscillating. Therefore, the position of the transition state was further refined by increasing the number of decimal places until a horizontal line in the internuclear distance againt time plot was obtained. This means that the internuclear distance is constant and that the molecules are simply undergoing periodic symmetric vibration at the transition state. Therefore, the position of the transition state was determined to be 0.907743 Å.

[[File:01409469 combined energy pic.png|thumb|center|400px|'''Figure 4.''' A plot of energy against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

When rAB=rBC=0.907743 Å and pA=pBC=0, the kinetic energy is 0. This means that HA and HBC are stationary, which is as predicted when the trajectory is started at the transition state with no initial momentum. Furthermore, the forces acting on HA and HBC are 0 as should be the case in the transition state.

=== Calculation type: Dynamic and minimum energy path ===

The minimum energy path, is the lowest energy path from the reactants to the products. The minimum energy path is determined by following the mode corresponding to the negative Hessian eigenvalue in both directions along the steepest descent direction. Following this path allows one to approach two valleys of stable structures. The resulting path generated is the minimum energy path. The reaction paths were chartered for a dynamic and a minimum energy path. The initial conditions used were that of the transition state except that the internuclear distance between HA and HB had increased to 0.917743 Å. The reaction path obtained using the two different calculation methods was different.

[[File:01409469 mep and dynamics.jpg|thumb|center|800px|'''Figure 5.''' A surface plot of the trajectory using minimum energy path '''Figure 6.''' A surface plot of the trajectory using dynamics (Left-right)]]

On the surface plot, the reaction path is depicted by the black line. Following the minimum energy path, the trajectory simply follows the valley floor to Ha+Hb-Hc. The reaction path appears to be a relatively straight line indicating that the atoms are not vibrating. In the minimum energy path, the reactants change into the products in an infinitely slow motion. This is done by having the momenta/velocities reset to zero in each time step. While the minimum energy path is useful in characterising a chemical reaction, it does not provide a realistic representation of the motion of the atoms during the reaction. This is because the mass and inertia of the atoms have been ignored.

Since the atoms simply follow the minimum energy path, they do not undergo slight displacements into surrounding valleys of higher potential energy. Furthermore, because the atoms are assumed to have no inertia, they do not undergo vibrations even if they were to be displaced to valleys of higher potential energy.

However, when dynamic calculation is performed, the reaction path is wavy indicating that the atoms are oscillating. This is because the atoms are displaced to regions of higher potential on the path from reactants to products. This leads to atom oscillation due to the inertia of the atoms.

'''Internuclear distance against time plot'''

Setting the initial system conditions such that rAB= 0.907743, rBC= 0.917743 and pA= pBC= 0:

[[File:01409469 Internuclear distance mep and dynamic.jpg|thumb|center|800px|'''Figure 6.''' A graph of internuclear distance against steps using minimum energy path '''Figure 7.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In the minimum energy path, the distance between HA and HB decreases from 0.90 Å to approximately 0.79 Å during the first 50 steps before plateauing off to a value of 0.74 Å which is the approximate length of a hydrogen bond. This means that a bond was formed between HA and HB. In addition, the plateau indicates that the bond is static. On the other hand, the distance between HB and HC has increased to 1.91 Å at large step values. This shows that HC has been ejected off following a successful collision between HA and HBC.

Similarly, in the dynamic trajectory, the distance between HA and HB decreases to approximately 0.75 Å after 0.3 s. This is indicative of hydrogen bond formation between HA and HB. In addition, the distance between HB and HC increases to 3.44 Å after HC is ejected following a successful collision.

One obvious distance between the two plots is that in the minimum energy path trajectory, the internuclear distance between HA and HB is constant indicating that the bond is static. However, in the dynamic trajectory the internuclear distance is oscillating showing the the bond is vibrating. This difference could be attributed to the fact that minimum energy path does not take oscillatory motion into account while dynamic trajectory does.

However, changing the initial conditions of the system such that rAB = 0.917743 Å, rBC = 0.907743 Å and pA=pBC=0:
[[File:01409469 01409469 Reversed values dynamic mep.jpg|thumb|center|800px|'''Figure 8.''' A graph of internuclear distance against steps using minimum energy path '''Figure 9.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In this case, a hydrogen bond of length 0.77-0.79 Å is formed between HB and HC while HA is repelled away to beyond 3.45 Å.

Additionally, from the internuclear distance plot it can be seen that the distance between HB and HC increases in a rectangular hyperbola fashion when calculated using the minimum energy path. On the other hand, the internuclear distance between HB and HC increases in a linear manner when dynamic calculation is applied. A possible reason for this difference is that in the minimum energy path, the atoms are not vibrating and as such the reach the maximum distance and plateau at a faster rate. However, in the dynamic calculation, the atoms have momentum and as such move further away from each other.

=== Reactive and unreactive trajectories ===

With the initial positions r'''1''' = 0.74 and r'''2''' = 2.0, trajectories were run with the momenta combination shown below.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Contour plot !! Reactivity !! Description of the dynamics
|-
| -1.25 || -2.5 ||-99.018 ||[[File:01409469 trajectory1.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. There are no oscillations observed in HA-HB initially, indicating that the diatomic molecule is stable. A new bond is formed between HB and HC. Oscillatory behaviour is observed indicating that the bond is vibrating. On the other hand, HA is ejected away to beyond 10 Å.
|-
| -1.5 || -2.0 ||-100.456 ||[[File:01409469 trajectory2.jpg|thumb|center|400px]] ||Unreactive ||HC collides with the vibrating diatomic HA-HB. However, no bond is formed between HB and HC as the collision is unsuccessful. This is because the molecules lack sufficient kinetic energy to overcome the activation energy barrier and so no reaction occurs. HC is subsequently repelled to beyond 8 Å.
|-
| -1.5 || -2.5 ||-98.956 ||[[File:01409469 trajectory3.jpg|thumb|center|400px]] ||Reactive |||HC collides with HA-HB. The collision is successful and a bond is formed between HB and HC. This is because the atoms have sufficient kinetic energy to overcome the activation energy barrier and form a new HB-HC bond. HA is ejected away to beyond 10 Å. The new hydrogen bond formed is vibrating as seen by the oscillatory behaviour.
|-
| -2.5 || -5.0 ||-84.956 ||[[File:01409469 trajectory4.jpg|thumb|center|400px]] ||Unreactive ||HC collides with HA-HB. Initially the HA-HB bond is elongating, indicating that the hydrogen bond is weakening and going to break. However, after 0.5 s, the HA-HB bond length decreases and the HAHB is not broken. HC is repelled beyond 6 Å with very strong vibrations.
A reaction did not occur even though the atoms have sufficient kinetic energy to cross the activation energy barrier. This is because the vibrations are too strong. The system crosses the transition state region and a bond is formed in the product temporarily but then the system reverts back to the reactants. This is an example of barrier recrossing.
|-
| -2.5 || -5.2 ||-83.416 ||[[File:01409469 trajectory5.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. The collision is successful and a new hydrogen bond is formed between HB and HC with HA being ejected away. The products formed experience very strong vibrations as evidenced by the oscillatory motion which occur with a very high frequency.
|}

From the table, it can be seen that the chance of a reaction successfully occurring is not solely dependent on the amount of kinetic energy the atoms possess. Even if the atoms have sufficient kinetic energy to cross the activation energy barrier, barrier recrossing can occur if the atoms are too vibrationally excited.

=== Transition State Theory ===

Transition state theory provided the first theoretical attempt to determine absolute reaction rates. In this theory, a transition state separating the reactants from products is used to find an expression for the thermal rate constant.

The main assumptions of transition state theory are:

1. Electronic and nuclear motions are separated (equivalent to the Born-Oppenheimer approximation in quantum mechanics)

2. Reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution

3. Molecular systems that have crossed the transition state in the direction of products cannot turn around and reform reactants

4. In the transition state, motion along the reaction coordinate may be separated from the other motions and treated classically as a translation

5. Even in the absence of an equilibrium between reactant and product molecules, the transition states that are becoming products are distributed among their states according to the Maxwell-Boltzmann laws

The goal of transition state theory is to predict the rate constant, ''k'', of a reaction given a potential energy surface for the reaction.

However, transition state theory suffers from a set of limitations due to the assumptions stated above not holding true in reality.

Firstly, transition state theory assumes that barrier recrossing is negligible. However, as seen from the fourth trajectory in table 1, barrier recrossing occurred because the atoms were too vibrationally excited and the system reverted back to the reactants even though the system had already crossed the transition state region. In this situation, transition state theory would predict a faster rate of reaction compared to what was obtained experimentally.

Secondly, the transition state theory breaks down for systems with light atoms such as hydrogens. In such systems, quantum effects become crucial. Quantum effects tend to increase the rate of the reaction for two reasons. First, particles lie in energy levels which are higher then the bottom of the potential energy well (there is additional zero-point energy). Secondly, the particles can tunnel through the barrier. These two effects effectively lower the activation energy barrier. As a result, the rate of reaction predicted by transition state theory would be less than what is obtained experimentally.

Lastly, transition state theory does not hold true for systems which are far from equilibrium. In such cases, the assumption of a starting Boltzmann distribution breaks down.

== Exercise 2: F - H - H system ==

=== Potential energy surface inspection===

F + HA-HB → HA-F + HB

[[File:01409469 Surface_Plot F 3.jpg|thumb|center|400px|'''Figure 10.''' Potential energy surface for the reaction between F and H2]]

The reaction between F and H2 is an exothermic reaction. This means that the reactants are of a higher energy than the products. The overall reaction involves the breaking of a hydrogen bond and the formation of a hydrogen-fluorine bond. The HF bond formed is stronger than the hydrogen bond that was broken. This is because fluorine is more electronegative than hydrogen resulting in a polar HF bond. The negative enthalpy of H-F bond formation is greater than the positive enthalpy of hydrogen bond dissociation. Therefore, the reaction between F and H2 is an exothermic reaction.

HA + HB-F → HA-HB + F

[[File:01409469 Surface_Plot F 4.jpg|thumb|center|400px|'''Figure 11.''' Potential energy surface for the reaction between H and HF]]

The reaction between H and HF is an endothermic reaction. The products are at a higher energy level than the reactants. The overall reaction involves the breaking of a HF bond and the formation of a hydrogen bond. The formation of a homonuclear H2 molecule results in a non-polar hydrogen bond being formed. The HF bond broken is stronger than the hydrogen bond that was formed. The negative enthalpy of hydrogen bond formation is less than the positive enthalpy of hydrogen-fluorine dissociation. Therefore, the reaction between H and HF is endothermic.

=== Locating the transition state ===

[[File:01409469 Surface_Plot F TS4.jpg|thumb|center|800px|'''Figure 12.''' Determining the position of the transition state for the reaction between F and H2]]

As mentioned previously, when the molecules are not at the transition state, a reaction path will be plotted. However, if the initial conditions of the system are set such that the trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

An initial estimate for the transition state was made by setting rAB = 0.74 which is the length of a hydrogen bond. The position of the transition state was further refined by increasing the number of decimal places such that the final value of rAB was found to be 0.74478 and rBC = 1.81079. This set of distances gave a reaction path that was represented by a point indicating that no trajectory was plotted and that the molecules remained at the transition state since their initial momentum was 0.

Moreover, the kinetic energy as well as the forces acting on the atoms was calculated to be 0, which is consistent with the transition state.

Therefore, the position of the transition state was found to be at rAB = 0.74478 and rBC = 1.81079 when pA = pBC = 0.

=== Determining the activation energy ===

[[File:01409469 activation energy r1.1 combined.jpg|thumb|center|800px|'''Figure 13.''' Determining the activation energy for the reaction between F and H2]]

By reducing the value of rBC by 0.01, the atoms are no longer at the transition state and would move towards the formation of products. The activation energy was found by calculating the change in total energy after the reaction. The activation energy of F + H2 → HF + H was found to be +23.279 kcal/ mol.

[[File:01409469 activation energy r2.1 combined.jpg|thumb|center|800px|'''Figure 14.''' Determining the activation energy for the reaction between FH and H]]

The activation energy of FH + H → F + H2 was found by calculating the small dip in potential energy. The activation energy was found to be +0.280 kcal/ mol.

=== Reaction dynamics of F + HA-HB → HA-F + HB ===

The set of initial conditions that result in a reactive trajectory for F + H2 is as follows: rHH = 0.74 Å, rHF = 1.79 Å and pHH = pHF = 0.

[[File:01409469 momentum r1.jpg|thumb|center|400px|'''Figure 15.''' Graph of momentum against time for the reaction between F and H2]]

Initially, the H2 molecule is travelling in a relatively straight line with very little oscillations as evidenced by the low amplitude of momentum between 0-2 s. Following successful collision between F and H2, the amplitude of momentum of the product, HF, increases dramatically. This is because the reaction between F and H2 is exothermic and the excess energy released causes the molecule to become vibrationally excited and the amplitude of oscillations in the product to increase.

[[File:01409469 energy1 r1.jpg|thumb|center|400px|'''Figure 16.''' Graph of energy against time for the reaction between F and H2]]

The law of conservation of energy states that energy cannot be created or destroyed. It simply changes from one form to another. As can be seen from the graph in figure 16, total energy remains constant. In this reaction, potential energy is converted to kinetic energy. A decrease in potential energy of the atoms following a successful collision between F and H2 results in an equal increase in kinetic energy which manifests in the form of increased product (HF) oscillations.

This can be confirmed experimentally using infrared chemiluminescence. Observation of radiation from excited product molecules is a clear indication that these molecules are produced in the excited state. If the excitation is primarily vibrational, the radiation will appear in the infrared region of the spectrum (between 3-15 μm). This technique has been used extensively in hydrogen halogen reactions which produce HX molecules in the excited state.

One such experiment involves the use of the arrested relaxation method. In this technique, the reaction is carried out in a chamber whose walls are maintained at a very low temperature (20 K). This results in cryopumping of the chamber in that all the products and unreacted starting materials are condensed on the walls and only those product molecules that radiate before being pumped away are observed.

=== Polanyi's empirical rules ===

In a typical chemical reaction, there is an energetic barrier — a saddle point that the reactants must surmount to become products. Which form of energy is initially deposited in the reactants, translational or vibrational, is more efficacious in surmounting the activation energy barrier is one of the core topics in the area of reaction dynamics. The Polyani rules state that vibrational energy is more effective at promoting a late-barrier reaction (transition state resembles products). Conversely, translational energy is better at promoting an early-barrier reaction (transition state resembles the reactants).

The forward reaction between H2 and F is exothermic. Therefore, the transition state is closer to the reactants than to the products in energy. Since the transition state occurs early in an exothermic reaction, translational energy is more effective at promoting a chemical reaction than vibrational energy.

Conversely, the backward reaction between H and HF is endothermic. According to Hammond's postulate, the transition state resembles the products more than it resembles the reactants. Therefore, in an endothermic reaction, a late transition state is observed and vibrational energy is more effective at promoting a reaction than translational energy.

=== Promoting the forward reaction: F + HA-HB → HA-F + HB ===

1. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 2.95 and pHF = -0.5.

[[File:01409469 polyani 1+2.jpg|thumb|center|800px|'''Figure 17.''' Contour plot of the reaction between F and H2 when pHH = 2.95 '''Figure 18.''' Graph of momentum against time for the reaction between F and H2 when pHH = 2.95 (Left-right)]]

Vibrational energy is represented by pHH while translational energy is represented by pHF.

A molecule of H2 collides with F. However, the collision is unsuccessful and no reaction occurs. A reaction did not occur even though the molecules have sufficient kinetic energy to overcome the activation energy barrier. This is because most of the energy is deposited in the reactants as H-H vibrations as seen by the large oscillations in H2. Barrier recrossing has occured as a result of the large vibrational energy causing the system to revert back to the reactants even though the system has already crossed the transition state and a bond was formed between HB and F temporarily.

This appears to run counter to what one would expect to happen when more energy is put into the system. It is predicted that when more energy is put into the system, the reactants would possess more kinetic energy to overcome the activation energy barrier. However, as can be seen from the simulation above, an increase in the vibrational energy of the reactants is not effective at promoting the occurrence of an exothermic reaction with an early transition state. This is in accordance to Polyani's rules.

2. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 0.1 and pHF = -0.8.

[[File:01409469 polyani 3+4 2.jpg|thumb|center|800px|'''Figure 19.''' Contour plot of the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 '''Figure 20.''' Graph of momentum against time for the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 (Left-right)]]

The collision between a molecule of H2 and F is successful and a chemical reaction has occurred. A reaction has occurred even though the overall energy of the system has been significantly reduced by lowering the amount of vibrational energy in the reactants but slightly increasing the amount of translational energy deposited in the reactants. Therefore, H2 and F have more translational energy to move towards each other and form a H-F bond. The product formed is vibrationally excited as seen by the large amplitude of oscillations in the product HF which correspond to the strong H-F vibrations.

From the second simulation, it can be seen that an increase in translational energy in the reactants is more effective at promoting an exothermic reaction with an early transition state to occur, which again follows Polyani's rules.

=== Promoting the backward reaction: HA + HB-F → HA-HB + F ===

1. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 0.2 and pHH = -3.0.

[[File:01409469 polyani 5+6 .jpg|thumb|center|800px|'''Figure 21.''' Contour plot of the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 '''Figure 22.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 (Left-right)]]

An atom of HA collides with a molecule of HBF. However, the collision is unsuccessful and no chemical reaction occurs. This is because the reactants lack sufficient vibrational energy to surmount the activation energy barrier. Therefore, following an unsuccessful collision with HBF, HA simply bounces off and moves away at constant momentum. This is in agreement with Polyani's rules which state that vibrational energy is more effective than translational energy at promoting reactions with a late transition state, for example, endothermic chemical reactions.

2. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 7.1 and pHH = -0.5.

[[File:01409469 polyani 7+8 .jpg|thumb|center|800px|'''Figure 23.''' Contour plot of the reaction between HA + HB-F when pFH = 7.1 and pHH = -0.5 '''Figure 24.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 7.1 and pHH = -0.5 (Left-right)]]

Increasing the amount of vibrational energy in the reactants by increasing pFH resulted in a chemical reaction successfully occurring. This is because the reactants now have sufficient kinetic energy in the form of H-F vibrations to overcome the activation energy barrier. In addition, because the reaction between HA + HB-F is endothermic, energy is taken in from the surroundings during the reaction and the products are less vibrationally excited than the reactants. This can be seen in the lower amplitude of oscillations in the product formed.

In conclusion, the vibrational energy of the reactants should be increased to promote an endothermic reaction with a late transition state while the translational energy of the reactants should be increased to promote an exothermic reaction with an early transition state. All 4 simulations shown above obey Polyani's rules.

== References ==
<references>
<ref name="ref1">A. C. Vaucher, M. Reiher, ''J. Chem. Theory. Comput.'', 2018, '''14''', 3091-3099.</ref>
</references>

MRD:01409469

2019-05-10T14:54:53Z

Stc1917: /* References */

== Exercise 1: H + H2 system ==

=== Definition of a transition state ===

A potential energy profile results from plotting the potential versus reaction path.

[[File:Surface_Plot01409469_pic1.jpg|thumb|center|400px|'''Figure 1.''' A potential energy surface plot]]

The transition state is defined as the maximum on the minimum energy path linking reactants and products. This means that the transition state is a stationary point on the potential energy surface. Thus, at the transition state, the derivative of the potential energy is zero with respect to each Cartesian coordinate.

<math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>

However, a local minimum on the potential energy surface corresponding to a well would have the same <math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>. Therefore, to distinguish between a local minimum on the potential energy surface and the transition state, the second derivative, <math>\tfrac{\partial^2 V(r_i)}{\partial r_i ^2}\ </math> is determined. This can be calculated by finding the determinant of the Hessian matrix shown below.

:<math>\mathbf H =
\begin{bmatrix}
\dfrac{\partial^2 \ V(r_i)}{\partial \ x^2} & \dfrac{\partial^2 \ V(r_i)}{\partial \ x \partial \mathbf y} \\
\dfrac{\partial^2 \ V(r_i)}{\partial \ y \partial \mathbf x} & \dfrac{\partial^2 \ V(r_i)}{\partial \mathbf y^2}
\end{bmatrix}</math>

The Hessian matrix of a convex function is positive semi-definite. Refining this property allows one to test if a critical point (x,y) is a local maximum, local minimum or saddle point.<ref name="ref1" />

If <math>f</math> < 0, then (x,y) is an isolated local maximum.

If <math>f</math> > 0, then (x,y) is an isolated local minimum.

If the Hessian has both positive and negative eigenvalues then (x,y) is a saddle point.

=== Locating the transition state ===

[[File:01409469 combined pic 2 and 3.jpg|thumb|center|800px|'''Figure 2.''' A surface and contour plot of the trajectory when rAB=rBC=0.907743 Å and pA=pBC=0]]

When the molecules are not at the transition state, a surface plot of the conditions shows a trajectory being plotted. This means that the trajectory is "rolling" towards the products or reactants. However, if a trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

[[File:01409469_internuclearpic3.png|thumb|center|400px|'''Figure 3.''' A plot of internuclear distance against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

An initial estimate for the position of the transition state was made by observing the potential energy surface and was found to be 0.90 Å. At 0.90 Å, the internuclear distance against time plot showed that the internuclear distance was constantly oscillating. Therefore, the position of the transition state was further refined by increasing the number of decimal places until a horizontal line in the internuclear distance againt time plot was obtained. This means that the internuclear distance is constant and that the molecules are simply undergoing periodic symmetric vibration at the transition state. Therefore, the position of the transition state was determined to be 0.907743 Å.

[[File:01409469 combined energy pic.png|thumb|center|400px|'''Figure 4.''' A plot of energy against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

When rAB=rBC=0.907743 Å and pA=pBC=0, the kinetic energy is 0. This means that HA and HBC are stationary, which is as predicted when the trajectory is started at the transition state with no initial momentum. Furthermore, the forces acting on HA and HBC are 0 as should be the case in the transition state.

=== Calculation type: Dynamic and minimum energy path ===

The minimum energy path, is the lowest energy path from the reactants to the products. The minimum energy path is determined by following the mode corresponding to the negative Hessian eigenvalue in both directions along the steepest descent direction. Following this path allows one to approach two valleys of stable structures. The resulting path generated is the minimum energy path. The reaction paths were chartered for a dynamic and a minimum energy path. The initial conditions used were that of the transition state except that the internuclear distance between HA and HB had increased to 0.917743 Å. The reaction path obtained using the two different calculation methods was different.

[[File:01409469 mep and dynamics.jpg|thumb|center|800px|'''Figure 5.''' A surface plot of the trajectory using minimum energy path '''Figure 6.''' A surface plot of the trajectory using dynamics (Left-right)]]

On the surface plot, the reaction path is depicted by the black line. Following the minimum energy path, the trajectory simply follows the valley floor to Ha+Hb-Hc. The reaction path appears to be a relatively straight line indicating that the atoms are not vibrating. In the minimum energy path, the reactants change into the products in an infinitely slow motion. This is done by having the momenta/velocities reset to zero in each time step. While the minimum energy path is useful in characterising a chemical reaction, it does not provide a realistic representation of the motion of the atoms during the reaction. This is because the mass and inertia of the atoms have been ignored.

Since the atoms simply follow the minimum energy path, they do not undergo slight displacements into surrounding valleys of higher potential energy. Furthermore, because the atoms are assumed to have no inertia, they do not undergo vibrations even if they were to be displaced to valleys of higher potential energy.

However, when dynamic calculation is performed, the reaction path is wavy indicating that the atoms are oscillating. This is because the atoms are displaced to regions of higher potential on the path from reactants to products. This leads to atom oscillation due to the inertia of the atoms.

'''Internuclear distance against time plot'''

Setting the initial system conditions such that rAB= 0.907743, rBC= 0.917743 and pA= pBC= 0:

[[File:01409469 Internuclear distance mep and dynamic.jpg|thumb|center|800px|'''Figure 6.''' A graph of internuclear distance against steps using minimum energy path '''Figure 7.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In the minimum energy path, the distance between HA and HB decreases from 0.90 Å to approximately 0.79 Å during the first 50 steps before plateauing off to a value of 0.74 Å which is the approximate length of a hydrogen bond. This means that a bond was formed between HA and HB. In addition, the plateau indicates that the bond is static. On the other hand, the distance between HB and HC has increased to 1.91 Å at large step values. This shows that HC has been ejected off following a successful collision between HA and HBC.

Similarly, in the dynamic trajectory, the distance between HA and HB decreases to approximately 0.75 Å after 0.3 s. This is indicative of hydrogen bond formation between HA and HB. In addition, the distance between HB and HC increases to 3.44 Å after HC is ejected following a successful collision.

One obvious distance between the two plots is that in the minimum energy path trajectory, the internuclear distance between HA and HB is constant indicating that the bond is static. However, in the dynamic trajectory the internuclear distance is oscillating showing the the bond is vibrating. This difference could be attributed to the fact that minimum energy path does not take oscillatory motion into account while dynamic trajectory does.

However, changing the initial conditions of the system such that rAB = 0.917743 Å, rBC = 0.907743 Å and pA=pBC=0:
[[File:01409469 01409469 Reversed values dynamic mep.jpg|thumb|center|800px|'''Figure 8.''' A graph of internuclear distance against steps using minimum energy path '''Figure 9.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In this case, a hydrogen bond of length 0.77-0.79 Å is formed between HB and HC while HA is repelled away to beyond 3.45 Å.

Additionally, from the internuclear distance plot it can be seen that the distance between HB and HC increases in a rectangular hyperbola fashion when calculated using the minimum energy path. On the other hand, the internuclear distance between HB and HC increases in a linear manner when dynamic calculation is applied. A possible reason for this difference is that in the minimum energy path, the atoms are not vibrating and as such the reach the maximum distance and plateau at a faster rate. However, in the dynamic calculation, the atoms have momentum and as such move further away from each other.

=== Reactive and unreactive trajectories ===

With the initial positions r'''1''' = 0.74 and r'''2''' = 2.0, trajectories were run with the momenta combination shown below.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Contour plot !! Reactivity !! Description of the dynamics
|-
| -1.25 || -2.5 ||-99.018 ||[[File:01409469 trajectory1.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. There are no oscillations observed in HA-HB initially, indicating that the diatomic molecule is stable. A new bond is formed between HB and HC. Oscillatory behaviour is observed indicating that the bond is vibrating. On the other hand, HA is ejected away to beyond 10 Å.
|-
| -1.5 || -2.0 ||-100.456 ||[[File:01409469 trajectory2.jpg|thumb|center|400px]] ||Unreactive ||HC collides with the vibrating diatomic HA-HB. However, no bond is formed between HB and HC as the collision is unsuccessful. This is because the molecules lack sufficient kinetic energy to overcome the activation energy barrier and so no reaction occurs. HC is subsequently repelled to beyond 8 Å.
|-
| -1.5 || -2.5 ||-98.956 ||[[File:01409469 trajectory3.jpg|thumb|center|400px]] ||Reactive |||HC collides with HA-HB. The collision is successful and a bond is formed between HB and HC. This is because the atoms have sufficient kinetic energy to overcome the activation energy barrier and form a new HB-HC bond. HA is ejected away to beyond 10 Å. The new hydrogen bond formed is vibrating as seen by the oscillatory behaviour.
|-
| -2.5 || -5.0 ||-84.956 ||[[File:01409469 trajectory4.jpg|thumb|center|400px]] ||Unreactive ||HC collides with HA-HB. Initially the HA-HB bond is elongating, indicating that the hydrogen bond is weakening and going to break. However, after 0.5 s, the HA-HB bond length decreases and the HAHB is not broken. HC is repelled beyond 6 Å with very strong vibrations.
A reaction did not occur even though the atoms have sufficient kinetic energy to cross the activation energy barrier. This is because the vibrations are too strong. The system crosses the transition state region and a bond is formed in the product temporarily but then the system reverts back to the reactants. This is an example of barrier recrossing.
|-
| -2.5 || -5.2 ||-83.416 ||[[File:01409469 trajectory5.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. The collision is successful and a new hydrogen bond is formed between HB and HC with HA being ejected away. The products formed experience very strong vibrations as evidenced by the oscillatory motion which occur with a very high frequency.
|}

From the table, it can be seen that the chance of a reaction successfully occurring is not solely dependent on the amount of kinetic energy the atoms possess. Even if the atoms have sufficient kinetic energy to cross the activation energy barrier, barrier recrossing can occur if the atoms are too vibrationally excited.

=== Transition State Theory ===

Transition state theory provided the first theoretical attempt to determine absolute reaction rates. In this theory, a transition state separating the reactants from products is used to find an expression for the thermal rate constant.

The main assumptions of transition state theory are:

1. Electronic and nuclear motions are separated (equivalent to the Born-Oppenheimer approximation in quantum mechanics)

2. Reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution

3. Molecular systems that have crossed the transition state in the direction of products cannot turn around and reform reactants

4. In the transition state, motion along the reaction coordinate may be separated from the other motions and treated classically as a translation

5. Even in the absence of an equilibrium between reactant and product molecules, the transition states that are becoming products are distributed among their states according to the Maxwell-Boltzmann laws

The goal of transition state theory is to predict the rate constant, ''k'', of a reaction given a potential energy surface for the reaction.

However, transition state theory suffers from a set of limitations due to the assumptions stated above not holding true in reality.

Firstly, transition state theory assumes that barrier recrossing is negligible. However, as seen from the fourth trajectory in table 1, barrier recrossing occurred because the atoms were too vibrationally excited and the system reverted back to the reactants even though the system had already crossed the transition state region. In this situation, transition state theory would predict a faster rate of reaction compared to what was obtained experimentally.

Secondly, the transition state theory breaks down for systems with light atoms such as hydrogens. In such systems, quantum effects become crucial. Quantum effects tend to increase the rate of the reaction for two reasons. First, particles lie in energy levels which are higher then the bottom of the potential energy well (there is additional zero-point energy). Secondly, the particles can tunnel through the barrier. These two effects effectively lower the activation energy barrier. As a result, the rate of reaction predicted by transition state theory would be less than what is obtained experimentally.

Lastly, transition state theory does not hold true for systems which are far from equilibrium. In such cases, the assumption of a starting Boltzmann distribution breaks down.

== Exercise 2: F - H - H system ==

=== Potential energy surface inspection===

F + HA-HB → HA-F + HB

[[File:01409469 Surface_Plot F 3.jpg|thumb|center|400px|'''Figure 10.''' Potential energy surface for the reaction between F and H2]]

The reaction between F and H2 is an exothermic reaction. This means that the reactants are of a higher energy than the products. The overall reaction involves the breaking of a hydrogen bond and the formation of a hydrogen-fluorine bond. The HF bond formed is stronger than the hydrogen bond that was broken. This is because fluorine is more electronegative than hydrogen resulting in a polar HF bond. The negative enthalpy of H-F bond formation is greater than the positive enthalpy of hydrogen bond dissociation. Therefore, the reaction between F and H2 is an exothermic reaction.

HA + HB-F → HA-HB + F

[[File:01409469 Surface_Plot F 4.jpg|thumb|center|400px|'''Figure 11.''' Potential energy surface for the reaction between H and HF]]

The reaction between H and HF is an endothermic reaction. The products are at a higher energy level than the reactants. The overall reaction involves the breaking of a HF bond and the formation of a hydrogen bond. The formation of a homonuclear H2 molecule results in a non-polar hydrogen bond being formed. The HF bond broken is stronger than the hydrogen bond that was formed. The negative enthalpy of hydrogen bond formation is less than the positive enthalpy of hydrogen-fluorine dissociation. Therefore, the reaction between H and HF is endothermic.

=== Locating the transition state ===

[[File:01409469 Surface_Plot F TS4.jpg|thumb|center|800px|'''Figure 12.''' Determining the position of the transition state for the reaction between F and H2]]

As mentioned previously, when the molecules are not at the transition state, a reaction path will be plotted. However, if the initial conditions of the system are set such that the trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

An initial estimate for the transition state was made by setting rAB = 0.74 which is the length of a hydrogen bond. The position of the transition state was further refined by increasing the number of decimal places such that the final value of rAB was found to be 0.74478 and rBC = 1.81079. This set of distances gave a reaction path that was represented by a point indicating that no trajectory was plotted and that the molecules remained at the transition state since their initial momentum was 0.

Moreover, the kinetic energy as well as the forces acting on the atoms was calculated to be 0, which is consistent with the transition state.

Therefore, the position of the transition state was found to be at rAB = 0.74478 and rBC = 1.81079 when pA = pBC = 0.

=== Determining the activation energy ===

[[File:01409469 activation energy r1.1 combined.jpg|thumb|center|800px|'''Figure 13.''' Determining the activation energy for the reaction between F and H2]]

By reducing the value of rBC by 0.01, the atoms are no longer at the transition state and would move towards the formation of products. The activation energy was found by calculating the change in total energy after the reaction. The activation energy of F + H2 → HF + H was found to be +23.279 kcal/ mol.

[[File:01409469 activation energy r2.1 combined.jpg|thumb|center|800px|'''Figure 14.''' Determining the activation energy for the reaction between FH and H]]

The activation energy of FH + H → F + H2 was found by calculating the small dip in potential energy. The activation energy was found to be +0.280 kcal/ mol.

=== Reaction dynamics of F + HA-HB → HA-F + HB ===

The set of initial conditions that result in a reactive trajectory for F + H2 is as follows: rHH = 0.74 Å, rHF = 1.79 Å and pHH = pHF = 0.

[[File:01409469 momentum r1.jpg|thumb|center|400px|'''Figure 15.''' Graph of momentum against time for the reaction between F and H2]]

Initially, the H2 molecule is travelling in a relatively straight line with very little oscillations as evidenced by the low amplitude of momentum between 0-2 s. Following successful collision between F and H2, the amplitude of momentum of the product, HF, increases dramatically. This is because the reaction between F and H2 is exothermic and the excess energy released causes the molecule to become vibrationally excited and the amplitude of oscillations in the product to increase.

[[File:01409469 energy1 r1.jpg|thumb|center|400px|'''Figure 16.''' Graph of energy against time for the reaction between F and H2]]

The law of conservation of energy states that energy cannot be created or destroyed. It simply changes from one form to another. As can be seen from the graph in figure 16, total energy remains constant. In this reaction, potential energy is converted to kinetic energy. A decrease in potential energy of the atoms following a successful collision between F and H2 results in an equal increase in kinetic energy which manifests in the form of increased product (HF) oscillations.

This can be confirmed experimentally using infrared chemiluminescence. Observation of radiation from excited product molecules is a clear indication that these molecules are produced in the excited state. If the excitation is primarily vibrational, the radiation will appear in the infrared region of the spectrum (between 3-15 μm). This technique has been used extensively in hydrogen halogen reactions which produce HX molecules in the excited state.

One such experiment involves the use of the arrested relaxation method. In this technique, the reaction is carried out in a chamber whose walls are maintained at a very low temperature (20 K). This results in cryopumping of the chamber in that all the products and unreacted starting materials are condensed on the walls and only those product molecules that radiate before being pumped away are observed.

=== Polanyi's empirical rules ===

In a typical chemical reaction, there is an energetic barrier — a saddle point that the reactants must surmount to become products. Which form of energy is initially deposited in the reactants, translational or vibrational, is more efficacious in surmounting the activation energy barrier is one of the core topics in the area of reaction dynamics. The Polyani rules state that vibrational energy is more effective at promoting a late-barrier reaction (transition state resembles products). Conversely, translational energy is better at promoting an early-barrier reaction (transition state resembles the reactants).

The forward reaction between H2 and F is exothermic. Therefore, the transition state is closer to the reactants than to the products in energy. Since the transition state occurs early in an exothermic reaction, translational energy is more effective at promoting a chemical reaction than vibrational energy.

Conversely, the backward reaction between H and HF is endothermic. According to Hammond's postulate, the transition state resembles the products more than it resembles the reactants. Therefore, in an endothermic reaction, a late transition state is observed and vibrational energy is more effective at promoting a reaction than translational energy.

=== Promoting the forward reaction: F + HA-HB → HA-F + HB ===

1. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 2.95 and pHF = -0.5.

[[File:01409469 polyani 1+2.jpg|thumb|center|800px|'''Figure 17.''' Contour plot of the reaction between F and H2 when pHH = 2.95 '''Figure 18.''' Graph of momentum against time for the reaction between F and H2 when pHH = 2.95 (Left-right)]]

Vibrational energy is represented by pHH while translational energy is represented by pHF.

A molecule of H2 collides with F. However, the collision is unsuccessful and no reaction occurs. A reaction did not occur even though the molecules have sufficient kinetic energy to overcome the activation energy barrier. This is because most of the energy is deposited in the reactants as H-H vibrations as seen by the large oscillations in H2. Barrier recrossing has occured as a result of the large vibrational energy causing the system to revert back to the reactants even though the system has already crossed the transition state and a bond was formed between HB and F temporarily.

This appears to run counter to what one would expect to happen when more energy is put into the system. It is predicted that when more energy is put into the system, the reactants would possess more kinetic energy to overcome the activation energy barrier. However, as can be seen from the simulation above, an increase in the vibrational energy of the reactants is not effective at promoting the occurrence of an exothermic reaction with an early transition state. This is in accordance to Polyani's rules.

2. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 0.1 and pHF = -0.8.

[[File:01409469 polyani 3+4 2.jpg|thumb|center|800px|'''Figure 19.''' Contour plot of the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 '''Figure 20.''' Graph of momentum against time for the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 (Left-right)]]

The collision between a molecule of H2 and F is successful and a chemical reaction has occurred. A reaction has occurred even though the overall energy of the system has been significantly reduced by lowering the amount of vibrational energy in the reactants but slightly increasing the amount of translational energy deposited in the reactants. Therefore, H2 and F have more translational energy to move towards each other and form a H-F bond. The product formed is vibrationally excited as seen by the large amplitude of oscillations in the product HF which correspond to the strong H-F vibrations.

From the second simulation, it can be seen that an increase in translational energy in the reactants is more effective at promoting an exothermic reaction with an early transition state to occur, which again follows Polyani's rules.

=== Promoting the backward reaction: HA + HB-F → HA-HB + F ===

1. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 0.2 and pHH = -3.0.

[[File:01409469 polyani 5+6 .jpg|thumb|center|800px|'''Figure 21.''' Contour plot of the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 '''Figure 22.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 (Left-right)]]

An atom of HA collides with a molecule of HBF. However, the collision is unsuccessful and no chemical reaction occurs. This is because the reactants lack sufficient vibrational energy to surmount the activation energy barrier. Therefore, following an unsuccessful collision with HBF, HA simply bounces off and moves away at constant momentum. This is in agreement with Polyani's rules which state that vibrational energy is more effective than translational energy at promoting reactions with a late transition state, for example, endothermic chemical reactions.

2. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 7.1 and pHH = -0.5.

[[File:01409469 polyani 7+8 .jpg|thumb|center|800px|'''Figure 23.''' Contour plot of the reaction between HA + HB-F when pFH = 7.1 and pHH = -0.5 '''Figure 24.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 7.1 and pHH = -0.5 (Left-right)]]

Increasing the amount of vibrational energy in the reactants by increasing pFH resulted in a chemical reaction successfully occurring. This is because the reactants now have sufficient kinetic energy in the form of H-F vibrations to overcome the activation energy barrier. In addition, because the reaction between HA + HB-F is endothermic, energy is taken in from the surroundings during the reaction and the products are less vibrationally excited than the reactants. This can be seen in the lower amplitude of oscillations in the product formed.

In conclusion, the vibrational energy of the reactants should be increased to promote an endothermic reaction with a late transition state while the translational energy of the reactants should be increased to promote an exothermic reaction with an early transition state. All 4 simulations shown above obey Polyani's rules.

== References ==
<references>
<ref name="ref1" />A. C. Vaucher, M. Reiher, ''J. Chem. Theory. Comput.'', 2018, '''14''', 3091-3099.</ref>
</references>

MRD:01409469

2019-05-10T14:51:22Z

Stc1917: /* Exercise 1: H + H2 system */

== Exercise 1: H + H2 system ==

=== Definition of a transition state ===

A potential energy profile results from plotting the potential versus reaction path.

[[File:Surface_Plot01409469_pic1.jpg|thumb|center|400px|'''Figure 1.''' A potential energy surface plot]]

The transition state is defined as the maximum on the minimum energy path linking reactants and products. This means that the transition state is a stationary point on the potential energy surface. Thus, at the transition state, the derivative of the potential energy is zero with respect to each Cartesian coordinate.

<math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>

However, a local minimum on the potential energy surface corresponding to a well would have the same <math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>. Therefore, to distinguish between a local minimum on the potential energy surface and the transition state, the second derivative, <math>\tfrac{\partial^2 V(r_i)}{\partial r_i ^2}\ </math> is determined. This can be calculated by finding the determinant of the Hessian matrix shown below.

:<math>\mathbf H =
\begin{bmatrix}
\dfrac{\partial^2 \ V(r_i)}{\partial \ x^2} & \dfrac{\partial^2 \ V(r_i)}{\partial \ x \partial \mathbf y} \\
\dfrac{\partial^2 \ V(r_i)}{\partial \ y \partial \mathbf x} & \dfrac{\partial^2 \ V(r_i)}{\partial \mathbf y^2}
\end{bmatrix}</math>

The Hessian matrix of a convex function is positive semi-definite. Refining this property allows one to test if a critical point (x,y) is a local maximum, local minimum or saddle point.<ref name="ref1" />

If <math>f</math> < 0, then (x,y) is an isolated local maximum.

If <math>f</math> > 0, then (x,y) is an isolated local minimum.

If the Hessian has both positive and negative eigenvalues then (x,y) is a saddle point.

=== Locating the transition state ===

[[File:01409469 combined pic 2 and 3.jpg|thumb|center|800px|'''Figure 2.''' A surface and contour plot of the trajectory when rAB=rBC=0.907743 Å and pA=pBC=0]]

When the molecules are not at the transition state, a surface plot of the conditions shows a trajectory being plotted. This means that the trajectory is "rolling" towards the products or reactants. However, if a trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

[[File:01409469_internuclearpic3.png|thumb|center|400px|'''Figure 3.''' A plot of internuclear distance against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

An initial estimate for the position of the transition state was made by observing the potential energy surface and was found to be 0.90 Å. At 0.90 Å, the internuclear distance against time plot showed that the internuclear distance was constantly oscillating. Therefore, the position of the transition state was further refined by increasing the number of decimal places until a horizontal line in the internuclear distance againt time plot was obtained. This means that the internuclear distance is constant and that the molecules are simply undergoing periodic symmetric vibration at the transition state. Therefore, the position of the transition state was determined to be 0.907743 Å.

[[File:01409469 combined energy pic.png|thumb|center|400px|'''Figure 4.''' A plot of energy against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

When rAB=rBC=0.907743 Å and pA=pBC=0, the kinetic energy is 0. This means that HA and HBC are stationary, which is as predicted when the trajectory is started at the transition state with no initial momentum. Furthermore, the forces acting on HA and HBC are 0 as should be the case in the transition state.

=== Calculation type: Dynamic and minimum energy path ===

The minimum energy path, is the lowest energy path from the reactants to the products. The minimum energy path is determined by following the mode corresponding to the negative Hessian eigenvalue in both directions along the steepest descent direction. Following this path allows one to approach two valleys of stable structures. The resulting path generated is the minimum energy path. The reaction paths were chartered for a dynamic and a minimum energy path. The initial conditions used were that of the transition state except that the internuclear distance between HA and HB had increased to 0.917743 Å. The reaction path obtained using the two different calculation methods was different.

[[File:01409469 mep and dynamics.jpg|thumb|center|800px|'''Figure 5.''' A surface plot of the trajectory using minimum energy path '''Figure 6.''' A surface plot of the trajectory using dynamics (Left-right)]]

On the surface plot, the reaction path is depicted by the black line. Following the minimum energy path, the trajectory simply follows the valley floor to Ha+Hb-Hc. The reaction path appears to be a relatively straight line indicating that the atoms are not vibrating. In the minimum energy path, the reactants change into the products in an infinitely slow motion. This is done by having the momenta/velocities reset to zero in each time step. While the minimum energy path is useful in characterising a chemical reaction, it does not provide a realistic representation of the motion of the atoms during the reaction. This is because the mass and inertia of the atoms have been ignored.

Since the atoms simply follow the minimum energy path, they do not undergo slight displacements into surrounding valleys of higher potential energy. Furthermore, because the atoms are assumed to have no inertia, they do not undergo vibrations even if they were to be displaced to valleys of higher potential energy.

However, when dynamic calculation is performed, the reaction path is wavy indicating that the atoms are oscillating. This is because the atoms are displaced to regions of higher potential on the path from reactants to products. This leads to atom oscillation due to the inertia of the atoms.

'''Internuclear distance against time plot'''

Setting the initial system conditions such that rAB= 0.907743, rBC= 0.917743 and pA= pBC= 0:

[[File:01409469 Internuclear distance mep and dynamic.jpg|thumb|center|800px|'''Figure 6.''' A graph of internuclear distance against steps using minimum energy path '''Figure 7.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In the minimum energy path, the distance between HA and HB decreases from 0.90 Å to approximately 0.79 Å during the first 50 steps before plateauing off to a value of 0.74 Å which is the approximate length of a hydrogen bond. This means that a bond was formed between HA and HB. In addition, the plateau indicates that the bond is static. On the other hand, the distance between HB and HC has increased to 1.91 Å at large step values. This shows that HC has been ejected off following a successful collision between HA and HBC.

Similarly, in the dynamic trajectory, the distance between HA and HB decreases to approximately 0.75 Å after 0.3 s. This is indicative of hydrogen bond formation between HA and HB. In addition, the distance between HB and HC increases to 3.44 Å after HC is ejected following a successful collision.

One obvious distance between the two plots is that in the minimum energy path trajectory, the internuclear distance between HA and HB is constant indicating that the bond is static. However, in the dynamic trajectory the internuclear distance is oscillating showing the the bond is vibrating. This difference could be attributed to the fact that minimum energy path does not take oscillatory motion into account while dynamic trajectory does.

However, changing the initial conditions of the system such that rAB = 0.917743 Å, rBC = 0.907743 Å and pA=pBC=0:
[[File:01409469 01409469 Reversed values dynamic mep.jpg|thumb|center|800px|'''Figure 8.''' A graph of internuclear distance against steps using minimum energy path '''Figure 9.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In this case, a hydrogen bond of length 0.77-0.79 Å is formed between HB and HC while HA is repelled away to beyond 3.45 Å.

Additionally, from the internuclear distance plot it can be seen that the distance between HB and HC increases in a rectangular hyperbola fashion when calculated using the minimum energy path. On the other hand, the internuclear distance between HB and HC increases in a linear manner when dynamic calculation is applied. A possible reason for this difference is that in the minimum energy path, the atoms are not vibrating and as such the reach the maximum distance and plateau at a faster rate. However, in the dynamic calculation, the atoms have momentum and as such move further away from each other.

=== Reactive and unreactive trajectories ===

With the initial positions r'''1''' = 0.74 and r'''2''' = 2.0, trajectories were run with the momenta combination shown below.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Contour plot !! Reactivity !! Description of the dynamics
|-
| -1.25 || -2.5 ||-99.018 ||[[File:01409469 trajectory1.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. There are no oscillations observed in HA-HB initially, indicating that the diatomic molecule is stable. A new bond is formed between HB and HC. Oscillatory behaviour is observed indicating that the bond is vibrating. On the other hand, HA is ejected away to beyond 10 Å.
|-
| -1.5 || -2.0 ||-100.456 ||[[File:01409469 trajectory2.jpg|thumb|center|400px]] ||Unreactive ||HC collides with the vibrating diatomic HA-HB. However, no bond is formed between HB and HC as the collision is unsuccessful. This is because the molecules lack sufficient kinetic energy to overcome the activation energy barrier and so no reaction occurs. HC is subsequently repelled to beyond 8 Å.
|-
| -1.5 || -2.5 ||-98.956 ||[[File:01409469 trajectory3.jpg|thumb|center|400px]] ||Reactive |||HC collides with HA-HB. The collision is successful and a bond is formed between HB and HC. This is because the atoms have sufficient kinetic energy to overcome the activation energy barrier and form a new HB-HC bond. HA is ejected away to beyond 10 Å. The new hydrogen bond formed is vibrating as seen by the oscillatory behaviour.
|-
| -2.5 || -5.0 ||-84.956 ||[[File:01409469 trajectory4.jpg|thumb|center|400px]] ||Unreactive ||HC collides with HA-HB. Initially the HA-HB bond is elongating, indicating that the hydrogen bond is weakening and going to break. However, after 0.5 s, the HA-HB bond length decreases and the HAHB is not broken. HC is repelled beyond 6 Å with very strong vibrations.
A reaction did not occur even though the atoms have sufficient kinetic energy to cross the activation energy barrier. This is because the vibrations are too strong. The system crosses the transition state region and a bond is formed in the product temporarily but then the system reverts back to the reactants. This is an example of barrier recrossing.
|-
| -2.5 || -5.2 ||-83.416 ||[[File:01409469 trajectory5.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. The collision is successful and a new hydrogen bond is formed between HB and HC with HA being ejected away. The products formed experience very strong vibrations as evidenced by the oscillatory motion which occur with a very high frequency.
|}

From the table, it can be seen that the chance of a reaction successfully occurring is not solely dependent on the amount of kinetic energy the atoms possess. Even if the atoms have sufficient kinetic energy to cross the activation energy barrier, barrier recrossing can occur if the atoms are too vibrationally excited.

=== Transition State Theory ===

Transition state theory provided the first theoretical attempt to determine absolute reaction rates. In this theory, a transition state separating the reactants from products is used to find an expression for the thermal rate constant.

The main assumptions of transition state theory are:

1. Electronic and nuclear motions are separated (equivalent to the Born-Oppenheimer approximation in quantum mechanics)

2. Reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution

3. Molecular systems that have crossed the transition state in the direction of products cannot turn around and reform reactants

4. In the transition state, motion along the reaction coordinate may be separated from the other motions and treated classically as a translation

5. Even in the absence of an equilibrium between reactant and product molecules, the transition states that are becoming products are distributed among their states according to the Maxwell-Boltzmann laws

The goal of transition state theory is to predict the rate constant, ''k'', of a reaction given a potential energy surface for the reaction.

However, transition state theory suffers from a set of limitations due to the assumptions stated above not holding true in reality.

Firstly, transition state theory assumes that barrier recrossing is negligible. However, as seen from the fourth trajectory in table 1, barrier recrossing occurred because the atoms were too vibrationally excited and the system reverted back to the reactants even though the system had already crossed the transition state region. In this situation, transition state theory would predict a faster rate of reaction compared to what was obtained experimentally.

Secondly, the transition state theory breaks down for systems with light atoms such as hydrogens. In such systems, quantum effects become crucial. Quantum effects tend to increase the rate of the reaction for two reasons. First, particles lie in energy levels which are higher then the bottom of the potential energy well (there is additional zero-point energy). Secondly, the particles can tunnel through the barrier. These two effects effectively lower the activation energy barrier. As a result, the rate of reaction predicted by transition state theory would be less than what is obtained experimentally.

Lastly, transition state theory does not hold true for systems which are far from equilibrium. In such cases, the assumption of a starting Boltzmann distribution breaks down.

== Exercise 2: F - H - H system ==

=== Potential energy surface inspection===

F + HA-HB → HA-F + HB

[[File:01409469 Surface_Plot F 3.jpg|thumb|center|400px|'''Figure 10.''' Potential energy surface for the reaction between F and H2]]

The reaction between F and H2 is an exothermic reaction. This means that the reactants are of a higher energy than the products. The overall reaction involves the breaking of a hydrogen bond and the formation of a hydrogen-fluorine bond. The HF bond formed is stronger than the hydrogen bond that was broken. This is because fluorine is more electronegative than hydrogen resulting in a polar HF bond. The negative enthalpy of H-F bond formation is greater than the positive enthalpy of hydrogen bond dissociation. Therefore, the reaction between F and H2 is an exothermic reaction.

HA + HB-F → HA-HB + F

[[File:01409469 Surface_Plot F 4.jpg|thumb|center|400px|'''Figure 11.''' Potential energy surface for the reaction between H and HF]]

The reaction between H and HF is an endothermic reaction. The products are at a higher energy level than the reactants. The overall reaction involves the breaking of a HF bond and the formation of a hydrogen bond. The formation of a homonuclear H2 molecule results in a non-polar hydrogen bond being formed. The HF bond broken is stronger than the hydrogen bond that was formed. The negative enthalpy of hydrogen bond formation is less than the positive enthalpy of hydrogen-fluorine dissociation. Therefore, the reaction between H and HF is endothermic.

=== Locating the transition state ===

[[File:01409469 Surface_Plot F TS4.jpg|thumb|center|800px|'''Figure 12.''' Determining the position of the transition state for the reaction between F and H2]]

As mentioned previously, when the molecules are not at the transition state, a reaction path will be plotted. However, if the initial conditions of the system are set such that the trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

An initial estimate for the transition state was made by setting rAB = 0.74 which is the length of a hydrogen bond. The position of the transition state was further refined by increasing the number of decimal places such that the final value of rAB was found to be 0.74478 and rBC = 1.81079. This set of distances gave a reaction path that was represented by a point indicating that no trajectory was plotted and that the molecules remained at the transition state since their initial momentum was 0.

Moreover, the kinetic energy as well as the forces acting on the atoms was calculated to be 0, which is consistent with the transition state.

Therefore, the position of the transition state was found to be at rAB = 0.74478 and rBC = 1.81079 when pA = pBC = 0.

=== Determining the activation energy ===

[[File:01409469 activation energy r1.1 combined.jpg|thumb|center|800px|'''Figure 13.''' Determining the activation energy for the reaction between F and H2]]

By reducing the value of rBC by 0.01, the atoms are no longer at the transition state and would move towards the formation of products. The activation energy was found by calculating the change in total energy after the reaction. The activation energy of F + H2 → HF + H was found to be +23.279 kcal/ mol.

[[File:01409469 activation energy r2.1 combined.jpg|thumb|center|800px|'''Figure 14.''' Determining the activation energy for the reaction between FH and H]]

The activation energy of FH + H → F + H2 was found by calculating the small dip in potential energy. The activation energy was found to be +0.280 kcal/ mol.

=== Reaction dynamics of F + HA-HB → HA-F + HB ===

The set of initial conditions that result in a reactive trajectory for F + H2 is as follows: rHH = 0.74 Å, rHF = 1.79 Å and pHH = pHF = 0.

[[File:01409469 momentum r1.jpg|thumb|center|400px|'''Figure 15.''' Graph of momentum against time for the reaction between F and H2]]

Initially, the H2 molecule is travelling in a relatively straight line with very little oscillations as evidenced by the low amplitude of momentum between 0-2 s. Following successful collision between F and H2, the amplitude of momentum of the product, HF, increases dramatically. This is because the reaction between F and H2 is exothermic and the excess energy released causes the molecule to become vibrationally excited and the amplitude of oscillations in the product to increase.

[[File:01409469 energy1 r1.jpg|thumb|center|400px|'''Figure 16.''' Graph of energy against time for the reaction between F and H2]]

The law of conservation of energy states that energy cannot be created or destroyed. It simply changes from one form to another. As can be seen from the graph in figure 16, total energy remains constant. In this reaction, potential energy is converted to kinetic energy. A decrease in potential energy of the atoms following a successful collision between F and H2 results in an equal increase in kinetic energy which manifests in the form of increased product (HF) oscillations.

This can be confirmed experimentally using infrared chemiluminescence. Observation of radiation from excited product molecules is a clear indication that these molecules are produced in the excited state. If the excitation is primarily vibrational, the radiation will appear in the infrared region of the spectrum (between 3-15 μm). This technique has been used extensively in hydrogen halogen reactions which produce HX molecules in the excited state.

One such experiment involves the use of the arrested relaxation method. In this technique, the reaction is carried out in a chamber whose walls are maintained at a very low temperature (20 K). This results in cryopumping of the chamber in that all the products and unreacted starting materials are condensed on the walls and only those product molecules that radiate before being pumped away are observed.

=== Polanyi's empirical rules ===

In a typical chemical reaction, there is an energetic barrier — a saddle point that the reactants must surmount to become products. Which form of energy is initially deposited in the reactants, translational or vibrational, is more efficacious in surmounting the activation energy barrier is one of the core topics in the area of reaction dynamics. The Polyani rules state that vibrational energy is more effective at promoting a late-barrier reaction (transition state resembles products). Conversely, translational energy is better at promoting an early-barrier reaction (transition state resembles the reactants).

The forward reaction between H2 and F is exothermic. Therefore, the transition state is closer to the reactants than to the products in energy. Since the transition state occurs early in an exothermic reaction, translational energy is more effective at promoting a chemical reaction than vibrational energy.

Conversely, the backward reaction between H and HF is endothermic. According to Hammond's postulate, the transition state resembles the products more than it resembles the reactants. Therefore, in an endothermic reaction, a late transition state is observed and vibrational energy is more effective at promoting a reaction than translational energy.

=== Promoting the forward reaction: F + HA-HB → HA-F + HB ===

1. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 2.95 and pHF = -0.5.

[[File:01409469 polyani 1+2.jpg|thumb|center|800px|'''Figure 17.''' Contour plot of the reaction between F and H2 when pHH = 2.95 '''Figure 18.''' Graph of momentum against time for the reaction between F and H2 when pHH = 2.95 (Left-right)]]

Vibrational energy is represented by pHH while translational energy is represented by pHF.

A molecule of H2 collides with F. However, the collision is unsuccessful and no reaction occurs. A reaction did not occur even though the molecules have sufficient kinetic energy to overcome the activation energy barrier. This is because most of the energy is deposited in the reactants as H-H vibrations as seen by the large oscillations in H2. Barrier recrossing has occured as a result of the large vibrational energy causing the system to revert back to the reactants even though the system has already crossed the transition state and a bond was formed between HB and F temporarily.

This appears to run counter to what one would expect to happen when more energy is put into the system. It is predicted that when more energy is put into the system, the reactants would possess more kinetic energy to overcome the activation energy barrier. However, as can be seen from the simulation above, an increase in the vibrational energy of the reactants is not effective at promoting the occurrence of an exothermic reaction with an early transition state. This is in accordance to Polyani's rules.

2. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 0.1 and pHF = -0.8.

[[File:01409469 polyani 3+4 2.jpg|thumb|center|800px|'''Figure 19.''' Contour plot of the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 '''Figure 20.''' Graph of momentum against time for the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 (Left-right)]]

The collision between a molecule of H2 and F is successful and a chemical reaction has occurred. A reaction has occurred even though the overall energy of the system has been significantly reduced by lowering the amount of vibrational energy in the reactants but slightly increasing the amount of translational energy deposited in the reactants. Therefore, H2 and F have more translational energy to move towards each other and form a H-F bond. The product formed is vibrationally excited as seen by the large amplitude of oscillations in the product HF which correspond to the strong H-F vibrations.

From the second simulation, it can be seen that an increase in translational energy in the reactants is more effective at promoting an exothermic reaction with an early transition state to occur, which again follows Polyani's rules.

=== Promoting the backward reaction: HA + HB-F → HA-HB + F ===

1. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 0.2 and pHH = -3.0.

[[File:01409469 polyani 5+6 .jpg|thumb|center|800px|'''Figure 21.''' Contour plot of the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 '''Figure 22.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 (Left-right)]]

An atom of HA collides with a molecule of HBF. However, the collision is unsuccessful and no chemical reaction occurs. This is because the reactants lack sufficient vibrational energy to surmount the activation energy barrier. Therefore, following an unsuccessful collision with HBF, HA simply bounces off and moves away at constant momentum. This is in agreement with Polyani's rules which state that vibrational energy is more effective than translational energy at promoting reactions with a late transition state, for example, endothermic chemical reactions.

2. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 7.1 and pHH = -0.5.

[[File:01409469 polyani 7+8 .jpg|thumb|center|800px|'''Figure 23.''' Contour plot of the reaction between HA + HB-F when pFH = 7.1 and pHH = -0.5 '''Figure 24.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 7.1 and pHH = -0.5 (Left-right)]]

Increasing the amount of vibrational energy in the reactants by increasing pFH resulted in a chemical reaction successfully occurring. This is because the reactants now have sufficient kinetic energy in the form of H-F vibrations to overcome the activation energy barrier. In addition, because the reaction between HA + HB-F is endothermic, energy is taken in from the surroundings during the reaction and the products are less vibrationally excited than the reactants. This can be seen in the lower amplitude of oscillations in the product formed.

In conclusion, the vibrational energy of the reactants should be increased to promote an endothermic reaction with a late transition state while the translational energy of the reactants should be increased to promote an exothermic reaction with an early transition state. All 4 simulations shown above obey Polyani's rules.

== References ==
<references>
<ref name="LazyDog">This is the lazy dog reference.</ref>
</references>

MRD:01409469

2019-05-10T14:51:06Z

Stc1917: /* Definition of a transition state */

== Exercise 1: H + H2 system ==

=== Definition of a transition state ===

A potential energy profile results from plotting the potential versus reaction path.<ref name="ref1" />

[[File:Surface_Plot01409469_pic1.jpg|thumb|center|400px|'''Figure 1.''' A potential energy surface plot]]

The transition state is defined as the maximum on the minimum energy path linking reactants and products. This means that the transition state is a stationary point on the potential energy surface. Thus, at the transition state, the derivative of the potential energy is zero with respect to each Cartesian coordinate.

<math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>

However, a local minimum on the potential energy surface corresponding to a well would have the same <math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>. Therefore, to distinguish between a local minimum on the potential energy surface and the transition state, the second derivative, <math>\tfrac{\partial^2 V(r_i)}{\partial r_i ^2}\ </math> is determined. This can be calculated by finding the determinant of the Hessian matrix shown below.

:<math>\mathbf H =
\begin{bmatrix}
\dfrac{\partial^2 \ V(r_i)}{\partial \ x^2} & \dfrac{\partial^2 \ V(r_i)}{\partial \ x \partial \mathbf y} \\
\dfrac{\partial^2 \ V(r_i)}{\partial \ y \partial \mathbf x} & \dfrac{\partial^2 \ V(r_i)}{\partial \mathbf y^2}
\end{bmatrix}</math>

The Hessian matrix of a convex function is positive semi-definite. Refining this property allows one to test if a critical point (x,y) is a local maximum, local minimum or saddle point.

If <math>f</math> < 0, then (x,y) is an isolated local maximum.

If <math>f</math> > 0, then (x,y) is an isolated local minimum.

If the Hessian has both positive and negative eigenvalues then (x,y) is a saddle point.

=== Locating the transition state ===

[[File:01409469 combined pic 2 and 3.jpg|thumb|center|800px|'''Figure 2.''' A surface and contour plot of the trajectory when rAB=rBC=0.907743 Å and pA=pBC=0]]

When the molecules are not at the transition state, a surface plot of the conditions shows a trajectory being plotted. This means that the trajectory is "rolling" towards the products or reactants. However, if a trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

[[File:01409469_internuclearpic3.png|thumb|center|400px|'''Figure 3.''' A plot of internuclear distance against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

An initial estimate for the position of the transition state was made by observing the potential energy surface and was found to be 0.90 Å. At 0.90 Å, the internuclear distance against time plot showed that the internuclear distance was constantly oscillating. Therefore, the position of the transition state was further refined by increasing the number of decimal places until a horizontal line in the internuclear distance againt time plot was obtained. This means that the internuclear distance is constant and that the molecules are simply undergoing periodic symmetric vibration at the transition state. Therefore, the position of the transition state was determined to be 0.907743 Å.

[[File:01409469 combined energy pic.png|thumb|center|400px|'''Figure 4.''' A plot of energy against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

When rAB=rBC=0.907743 Å and pA=pBC=0, the kinetic energy is 0. This means that HA and HBC are stationary, which is as predicted when the trajectory is started at the transition state with no initial momentum. Furthermore, the forces acting on HA and HBC are 0 as should be the case in the transition state.

=== Calculation type: Dynamic and minimum energy path ===

The minimum energy path, is the lowest energy path from the reactants to the products. The minimum energy path is determined by following the mode corresponding to the negative Hessian eigenvalue in both directions along the steepest descent direction. Following this path allows one to approach two valleys of stable structures. The resulting path generated is the minimum energy path. The reaction paths were chartered for a dynamic and a minimum energy path. The initial conditions used were that of the transition state except that the internuclear distance between HA and HB had increased to 0.917743 Å. The reaction path obtained using the two different calculation methods was different.

[[File:01409469 mep and dynamics.jpg|thumb|center|800px|'''Figure 5.''' A surface plot of the trajectory using minimum energy path '''Figure 6.''' A surface plot of the trajectory using dynamics (Left-right)]]

On the surface plot, the reaction path is depicted by the black line. Following the minimum energy path, the trajectory simply follows the valley floor to Ha+Hb-Hc. The reaction path appears to be a relatively straight line indicating that the atoms are not vibrating. In the minimum energy path, the reactants change into the products in an infinitely slow motion. This is done by having the momenta/velocities reset to zero in each time step. While the minimum energy path is useful in characterising a chemical reaction, it does not provide a realistic representation of the motion of the atoms during the reaction. This is because the mass and inertia of the atoms have been ignored.

Since the atoms simply follow the minimum energy path, they do not undergo slight displacements into surrounding valleys of higher potential energy. Furthermore, because the atoms are assumed to have no inertia, they do not undergo vibrations even if they were to be displaced to valleys of higher potential energy.

However, when dynamic calculation is performed, the reaction path is wavy indicating that the atoms are oscillating. This is because the atoms are displaced to regions of higher potential on the path from reactants to products. This leads to atom oscillation due to the inertia of the atoms.

'''Internuclear distance against time plot'''

Setting the initial system conditions such that rAB= 0.907743, rBC= 0.917743 and pA= pBC= 0:

[[File:01409469 Internuclear distance mep and dynamic.jpg|thumb|center|800px|'''Figure 6.''' A graph of internuclear distance against steps using minimum energy path '''Figure 7.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In the minimum energy path, the distance between HA and HB decreases from 0.90 Å to approximately 0.79 Å during the first 50 steps before plateauing off to a value of 0.74 Å which is the approximate length of a hydrogen bond. This means that a bond was formed between HA and HB. In addition, the plateau indicates that the bond is static. On the other hand, the distance between HB and HC has increased to 1.91 Å at large step values. This shows that HC has been ejected off following a successful collision between HA and HBC.

Similarly, in the dynamic trajectory, the distance between HA and HB decreases to approximately 0.75 Å after 0.3 s. This is indicative of hydrogen bond formation between HA and HB. In addition, the distance between HB and HC increases to 3.44 Å after HC is ejected following a successful collision.

One obvious distance between the two plots is that in the minimum energy path trajectory, the internuclear distance between HA and HB is constant indicating that the bond is static. However, in the dynamic trajectory the internuclear distance is oscillating showing the the bond is vibrating. This difference could be attributed to the fact that minimum energy path does not take oscillatory motion into account while dynamic trajectory does.

However, changing the initial conditions of the system such that rAB = 0.917743 Å, rBC = 0.907743 Å and pA=pBC=0:
[[File:01409469 01409469 Reversed values dynamic mep.jpg|thumb|center|800px|'''Figure 8.''' A graph of internuclear distance against steps using minimum energy path '''Figure 9.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In this case, a hydrogen bond of length 0.77-0.79 Å is formed between HB and HC while HA is repelled away to beyond 3.45 Å.

Additionally, from the internuclear distance plot it can be seen that the distance between HB and HC increases in a rectangular hyperbola fashion when calculated using the minimum energy path. On the other hand, the internuclear distance between HB and HC increases in a linear manner when dynamic calculation is applied. A possible reason for this difference is that in the minimum energy path, the atoms are not vibrating and as such the reach the maximum distance and plateau at a faster rate. However, in the dynamic calculation, the atoms have momentum and as such move further away from each other.

=== Reactive and unreactive trajectories ===

With the initial positions r'''1''' = 0.74 and r'''2''' = 2.0, trajectories were run with the momenta combination shown below.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Contour plot !! Reactivity !! Description of the dynamics
|-
| -1.25 || -2.5 ||-99.018 ||[[File:01409469 trajectory1.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. There are no oscillations observed in HA-HB initially, indicating that the diatomic molecule is stable. A new bond is formed between HB and HC. Oscillatory behaviour is observed indicating that the bond is vibrating. On the other hand, HA is ejected away to beyond 10 Å.
|-
| -1.5 || -2.0 ||-100.456 ||[[File:01409469 trajectory2.jpg|thumb|center|400px]] ||Unreactive ||HC collides with the vibrating diatomic HA-HB. However, no bond is formed between HB and HC as the collision is unsuccessful. This is because the molecules lack sufficient kinetic energy to overcome the activation energy barrier and so no reaction occurs. HC is subsequently repelled to beyond 8 Å.
|-
| -1.5 || -2.5 ||-98.956 ||[[File:01409469 trajectory3.jpg|thumb|center|400px]] ||Reactive |||HC collides with HA-HB. The collision is successful and a bond is formed between HB and HC. This is because the atoms have sufficient kinetic energy to overcome the activation energy barrier and form a new HB-HC bond. HA is ejected away to beyond 10 Å. The new hydrogen bond formed is vibrating as seen by the oscillatory behaviour.
|-
| -2.5 || -5.0 ||-84.956 ||[[File:01409469 trajectory4.jpg|thumb|center|400px]] ||Unreactive ||HC collides with HA-HB. Initially the HA-HB bond is elongating, indicating that the hydrogen bond is weakening and going to break. However, after 0.5 s, the HA-HB bond length decreases and the HAHB is not broken. HC is repelled beyond 6 Å with very strong vibrations.
A reaction did not occur even though the atoms have sufficient kinetic energy to cross the activation energy barrier. This is because the vibrations are too strong. The system crosses the transition state region and a bond is formed in the product temporarily but then the system reverts back to the reactants. This is an example of barrier recrossing.
|-
| -2.5 || -5.2 ||-83.416 ||[[File:01409469 trajectory5.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. The collision is successful and a new hydrogen bond is formed between HB and HC with HA being ejected away. The products formed experience very strong vibrations as evidenced by the oscillatory motion which occur with a very high frequency.
|}

From the table, it can be seen that the chance of a reaction successfully occurring is not solely dependent on the amount of kinetic energy the atoms possess. Even if the atoms have sufficient kinetic energy to cross the activation energy barrier, barrier recrossing can occur if the atoms are too vibrationally excited.

=== Transition State Theory ===

Transition state theory provided the first theoretical attempt to determine absolute reaction rates. In this theory, a transition state separating the reactants from products is used to find an expression for the thermal rate constant.

The main assumptions of transition state theory are:

1. Electronic and nuclear motions are separated (equivalent to the Born-Oppenheimer approximation in quantum mechanics)

2. Reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution

3. Molecular systems that have crossed the transition state in the direction of products cannot turn around and reform reactants

4. In the transition state, motion along the reaction coordinate may be separated from the other motions and treated classically as a translation

5. Even in the absence of an equilibrium between reactant and product molecules, the transition states that are becoming products are distributed among their states according to the Maxwell-Boltzmann laws

The goal of transition state theory is to predict the rate constant, ''k'', of a reaction given a potential energy surface for the reaction.

However, transition state theory suffers from a set of limitations due to the assumptions stated above not holding true in reality.

Firstly, transition state theory assumes that barrier recrossing is negligible. However, as seen from the fourth trajectory in table 1, barrier recrossing occurred because the atoms were too vibrationally excited and the system reverted back to the reactants even though the system had already crossed the transition state region. In this situation, transition state theory would predict a faster rate of reaction compared to what was obtained experimentally.

Secondly, the transition state theory breaks down for systems with light atoms such as hydrogens. In such systems, quantum effects become crucial. Quantum effects tend to increase the rate of the reaction for two reasons. First, particles lie in energy levels which are higher then the bottom of the potential energy well (there is additional zero-point energy). Secondly, the particles can tunnel through the barrier. These two effects effectively lower the activation energy barrier. As a result, the rate of reaction predicted by transition state theory would be less than what is obtained experimentally.

Lastly, transition state theory does not hold true for systems which are far from equilibrium. In such cases, the assumption of a starting Boltzmann distribution breaks down.

== Exercise 2: F - H - H system ==

=== Potential energy surface inspection===

F + HA-HB → HA-F + HB

[[File:01409469 Surface_Plot F 3.jpg|thumb|center|400px|'''Figure 10.''' Potential energy surface for the reaction between F and H2]]

The reaction between F and H2 is an exothermic reaction. This means that the reactants are of a higher energy than the products. The overall reaction involves the breaking of a hydrogen bond and the formation of a hydrogen-fluorine bond. The HF bond formed is stronger than the hydrogen bond that was broken. This is because fluorine is more electronegative than hydrogen resulting in a polar HF bond. The negative enthalpy of H-F bond formation is greater than the positive enthalpy of hydrogen bond dissociation. Therefore, the reaction between F and H2 is an exothermic reaction.

HA + HB-F → HA-HB + F

[[File:01409469 Surface_Plot F 4.jpg|thumb|center|400px|'''Figure 11.''' Potential energy surface for the reaction between H and HF]]

The reaction between H and HF is an endothermic reaction. The products are at a higher energy level than the reactants. The overall reaction involves the breaking of a HF bond and the formation of a hydrogen bond. The formation of a homonuclear H2 molecule results in a non-polar hydrogen bond being formed. The HF bond broken is stronger than the hydrogen bond that was formed. The negative enthalpy of hydrogen bond formation is less than the positive enthalpy of hydrogen-fluorine dissociation. Therefore, the reaction between H and HF is endothermic.

=== Locating the transition state ===

[[File:01409469 Surface_Plot F TS4.jpg|thumb|center|800px|'''Figure 12.''' Determining the position of the transition state for the reaction between F and H2]]

As mentioned previously, when the molecules are not at the transition state, a reaction path will be plotted. However, if the initial conditions of the system are set such that the trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

An initial estimate for the transition state was made by setting rAB = 0.74 which is the length of a hydrogen bond. The position of the transition state was further refined by increasing the number of decimal places such that the final value of rAB was found to be 0.74478 and rBC = 1.81079. This set of distances gave a reaction path that was represented by a point indicating that no trajectory was plotted and that the molecules remained at the transition state since their initial momentum was 0.

Moreover, the kinetic energy as well as the forces acting on the atoms was calculated to be 0, which is consistent with the transition state.

Therefore, the position of the transition state was found to be at rAB = 0.74478 and rBC = 1.81079 when pA = pBC = 0.

=== Determining the activation energy ===

[[File:01409469 activation energy r1.1 combined.jpg|thumb|center|800px|'''Figure 13.''' Determining the activation energy for the reaction between F and H2]]

By reducing the value of rBC by 0.01, the atoms are no longer at the transition state and would move towards the formation of products. The activation energy was found by calculating the change in total energy after the reaction. The activation energy of F + H2 → HF + H was found to be +23.279 kcal/ mol.

[[File:01409469 activation energy r2.1 combined.jpg|thumb|center|800px|'''Figure 14.''' Determining the activation energy for the reaction between FH and H]]

The activation energy of FH + H → F + H2 was found by calculating the small dip in potential energy. The activation energy was found to be +0.280 kcal/ mol.

=== Reaction dynamics of F + HA-HB → HA-F + HB ===

The set of initial conditions that result in a reactive trajectory for F + H2 is as follows: rHH = 0.74 Å, rHF = 1.79 Å and pHH = pHF = 0.

[[File:01409469 momentum r1.jpg|thumb|center|400px|'''Figure 15.''' Graph of momentum against time for the reaction between F and H2]]

Initially, the H2 molecule is travelling in a relatively straight line with very little oscillations as evidenced by the low amplitude of momentum between 0-2 s. Following successful collision between F and H2, the amplitude of momentum of the product, HF, increases dramatically. This is because the reaction between F and H2 is exothermic and the excess energy released causes the molecule to become vibrationally excited and the amplitude of oscillations in the product to increase.

[[File:01409469 energy1 r1.jpg|thumb|center|400px|'''Figure 16.''' Graph of energy against time for the reaction between F and H2]]

The law of conservation of energy states that energy cannot be created or destroyed. It simply changes from one form to another. As can be seen from the graph in figure 16, total energy remains constant. In this reaction, potential energy is converted to kinetic energy. A decrease in potential energy of the atoms following a successful collision between F and H2 results in an equal increase in kinetic energy which manifests in the form of increased product (HF) oscillations.

This can be confirmed experimentally using infrared chemiluminescence. Observation of radiation from excited product molecules is a clear indication that these molecules are produced in the excited state. If the excitation is primarily vibrational, the radiation will appear in the infrared region of the spectrum (between 3-15 μm). This technique has been used extensively in hydrogen halogen reactions which produce HX molecules in the excited state.

One such experiment involves the use of the arrested relaxation method. In this technique, the reaction is carried out in a chamber whose walls are maintained at a very low temperature (20 K). This results in cryopumping of the chamber in that all the products and unreacted starting materials are condensed on the walls and only those product molecules that radiate before being pumped away are observed.

=== Polanyi's empirical rules ===

In a typical chemical reaction, there is an energetic barrier — a saddle point that the reactants must surmount to become products. Which form of energy is initially deposited in the reactants, translational or vibrational, is more efficacious in surmounting the activation energy barrier is one of the core topics in the area of reaction dynamics. The Polyani rules state that vibrational energy is more effective at promoting a late-barrier reaction (transition state resembles products). Conversely, translational energy is better at promoting an early-barrier reaction (transition state resembles the reactants).

The forward reaction between H2 and F is exothermic. Therefore, the transition state is closer to the reactants than to the products in energy. Since the transition state occurs early in an exothermic reaction, translational energy is more effective at promoting a chemical reaction than vibrational energy.

Conversely, the backward reaction between H and HF is endothermic. According to Hammond's postulate, the transition state resembles the products more than it resembles the reactants. Therefore, in an endothermic reaction, a late transition state is observed and vibrational energy is more effective at promoting a reaction than translational energy.

=== Promoting the forward reaction: F + HA-HB → HA-F + HB ===

1. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 2.95 and pHF = -0.5.

[[File:01409469 polyani 1+2.jpg|thumb|center|800px|'''Figure 17.''' Contour plot of the reaction between F and H2 when pHH = 2.95 '''Figure 18.''' Graph of momentum against time for the reaction between F and H2 when pHH = 2.95 (Left-right)]]

Vibrational energy is represented by pHH while translational energy is represented by pHF.

A molecule of H2 collides with F. However, the collision is unsuccessful and no reaction occurs. A reaction did not occur even though the molecules have sufficient kinetic energy to overcome the activation energy barrier. This is because most of the energy is deposited in the reactants as H-H vibrations as seen by the large oscillations in H2. Barrier recrossing has occured as a result of the large vibrational energy causing the system to revert back to the reactants even though the system has already crossed the transition state and a bond was formed between HB and F temporarily.

This appears to run counter to what one would expect to happen when more energy is put into the system. It is predicted that when more energy is put into the system, the reactants would possess more kinetic energy to overcome the activation energy barrier. However, as can be seen from the simulation above, an increase in the vibrational energy of the reactants is not effective at promoting the occurrence of an exothermic reaction with an early transition state. This is in accordance to Polyani's rules.

2. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 0.1 and pHF = -0.8.

[[File:01409469 polyani 3+4 2.jpg|thumb|center|800px|'''Figure 19.''' Contour plot of the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 '''Figure 20.''' Graph of momentum against time for the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 (Left-right)]]

The collision between a molecule of H2 and F is successful and a chemical reaction has occurred. A reaction has occurred even though the overall energy of the system has been significantly reduced by lowering the amount of vibrational energy in the reactants but slightly increasing the amount of translational energy deposited in the reactants. Therefore, H2 and F have more translational energy to move towards each other and form a H-F bond. The product formed is vibrationally excited as seen by the large amplitude of oscillations in the product HF which correspond to the strong H-F vibrations.

From the second simulation, it can be seen that an increase in translational energy in the reactants is more effective at promoting an exothermic reaction with an early transition state to occur, which again follows Polyani's rules.

=== Promoting the backward reaction: HA + HB-F → HA-HB + F ===

1. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 0.2 and pHH = -3.0.

[[File:01409469 polyani 5+6 .jpg|thumb|center|800px|'''Figure 21.''' Contour plot of the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 '''Figure 22.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 (Left-right)]]

An atom of HA collides with a molecule of HBF. However, the collision is unsuccessful and no chemical reaction occurs. This is because the reactants lack sufficient vibrational energy to surmount the activation energy barrier. Therefore, following an unsuccessful collision with HBF, HA simply bounces off and moves away at constant momentum. This is in agreement with Polyani's rules which state that vibrational energy is more effective than translational energy at promoting reactions with a late transition state, for example, endothermic chemical reactions.

2. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 7.1 and pHH = -0.5.

[[File:01409469 polyani 7+8 .jpg|thumb|center|800px|'''Figure 23.''' Contour plot of the reaction between HA + HB-F when pFH = 7.1 and pHH = -0.5 '''Figure 24.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 7.1 and pHH = -0.5 (Left-right)]]

Increasing the amount of vibrational energy in the reactants by increasing pFH resulted in a chemical reaction successfully occurring. This is because the reactants now have sufficient kinetic energy in the form of H-F vibrations to overcome the activation energy barrier. In addition, because the reaction between HA + HB-F is endothermic, energy is taken in from the surroundings during the reaction and the products are less vibrationally excited than the reactants. This can be seen in the lower amplitude of oscillations in the product formed.

In conclusion, the vibrational energy of the reactants should be increased to promote an endothermic reaction with a late transition state while the translational energy of the reactants should be increased to promote an exothermic reaction with an early transition state. All 4 simulations shown above obey Polyani's rules.

== References ==
<references>
<ref name="LazyDog">This is the lazy dog reference.</ref>
</references>

MRD:01409469

2019-05-10T14:50:39Z

Stc1917: /* References */

== Exercise 1: H + H2 system ==

=== Definition of a transition state ===

A potential energy profile results from plotting the potential versus reaction path.

[[File:Surface_Plot01409469_pic1.jpg|thumb|center|400px|'''Figure 1.''' A potential energy surface plot]]

The transition state is defined as the maximum on the minimum energy path linking reactants and products. This means that the transition state is a stationary point on the potential energy surface. Thus, at the transition state, the derivative of the potential energy is zero with respect to each Cartesian coordinate.

<math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>

However, a local minimum on the potential energy surface corresponding to a well would have the same <math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>. Therefore, to distinguish between a local minimum on the potential energy surface and the transition state, the second derivative, <math>\tfrac{\partial^2 V(r_i)}{\partial r_i ^2}\ </math> is determined. This can be calculated by finding the determinant of the Hessian matrix shown below.

:<math>\mathbf H =
\begin{bmatrix}
\dfrac{\partial^2 \ V(r_i)}{\partial \ x^2} & \dfrac{\partial^2 \ V(r_i)}{\partial \ x \partial \mathbf y} \\
\dfrac{\partial^2 \ V(r_i)}{\partial \ y \partial \mathbf x} & \dfrac{\partial^2 \ V(r_i)}{\partial \mathbf y^2}
\end{bmatrix}</math>

The Hessian matrix of a convex function is positive semi-definite. Refining this property allows one to test if a critical point (x,y) is a local maximum, local minimum or saddle point.

If <math>f</math> < 0, then (x,y) is an isolated local maximum.

If <math>f</math> > 0, then (x,y) is an isolated local minimum.

If the Hessian has both positive and negative eigenvalues then (x,y) is a saddle point.

=== Locating the transition state ===

[[File:01409469 combined pic 2 and 3.jpg|thumb|center|800px|'''Figure 2.''' A surface and contour plot of the trajectory when rAB=rBC=0.907743 Å and pA=pBC=0]]

When the molecules are not at the transition state, a surface plot of the conditions shows a trajectory being plotted. This means that the trajectory is "rolling" towards the products or reactants. However, if a trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

[[File:01409469_internuclearpic3.png|thumb|center|400px|'''Figure 3.''' A plot of internuclear distance against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

An initial estimate for the position of the transition state was made by observing the potential energy surface and was found to be 0.90 Å. At 0.90 Å, the internuclear distance against time plot showed that the internuclear distance was constantly oscillating. Therefore, the position of the transition state was further refined by increasing the number of decimal places until a horizontal line in the internuclear distance againt time plot was obtained. This means that the internuclear distance is constant and that the molecules are simply undergoing periodic symmetric vibration at the transition state. Therefore, the position of the transition state was determined to be 0.907743 Å.

[[File:01409469 combined energy pic.png|thumb|center|400px|'''Figure 4.''' A plot of energy against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

When rAB=rBC=0.907743 Å and pA=pBC=0, the kinetic energy is 0. This means that HA and HBC are stationary, which is as predicted when the trajectory is started at the transition state with no initial momentum. Furthermore, the forces acting on HA and HBC are 0 as should be the case in the transition state.

=== Calculation type: Dynamic and minimum energy path ===

The minimum energy path, is the lowest energy path from the reactants to the products. The minimum energy path is determined by following the mode corresponding to the negative Hessian eigenvalue in both directions along the steepest descent direction. Following this path allows one to approach two valleys of stable structures. The resulting path generated is the minimum energy path. The reaction paths were chartered for a dynamic and a minimum energy path. The initial conditions used were that of the transition state except that the internuclear distance between HA and HB had increased to 0.917743 Å. The reaction path obtained using the two different calculation methods was different.

[[File:01409469 mep and dynamics.jpg|thumb|center|800px|'''Figure 5.''' A surface plot of the trajectory using minimum energy path '''Figure 6.''' A surface plot of the trajectory using dynamics (Left-right)]]

On the surface plot, the reaction path is depicted by the black line. Following the minimum energy path, the trajectory simply follows the valley floor to Ha+Hb-Hc. The reaction path appears to be a relatively straight line indicating that the atoms are not vibrating. In the minimum energy path, the reactants change into the products in an infinitely slow motion. This is done by having the momenta/velocities reset to zero in each time step. While the minimum energy path is useful in characterising a chemical reaction, it does not provide a realistic representation of the motion of the atoms during the reaction. This is because the mass and inertia of the atoms have been ignored.

Since the atoms simply follow the minimum energy path, they do not undergo slight displacements into surrounding valleys of higher potential energy. Furthermore, because the atoms are assumed to have no inertia, they do not undergo vibrations even if they were to be displaced to valleys of higher potential energy.

However, when dynamic calculation is performed, the reaction path is wavy indicating that the atoms are oscillating. This is because the atoms are displaced to regions of higher potential on the path from reactants to products. This leads to atom oscillation due to the inertia of the atoms.

'''Internuclear distance against time plot'''

Setting the initial system conditions such that rAB= 0.907743, rBC= 0.917743 and pA= pBC= 0:

[[File:01409469 Internuclear distance mep and dynamic.jpg|thumb|center|800px|'''Figure 6.''' A graph of internuclear distance against steps using minimum energy path '''Figure 7.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In the minimum energy path, the distance between HA and HB decreases from 0.90 Å to approximately 0.79 Å during the first 50 steps before plateauing off to a value of 0.74 Å which is the approximate length of a hydrogen bond. This means that a bond was formed between HA and HB. In addition, the plateau indicates that the bond is static. On the other hand, the distance between HB and HC has increased to 1.91 Å at large step values. This shows that HC has been ejected off following a successful collision between HA and HBC.

Similarly, in the dynamic trajectory, the distance between HA and HB decreases to approximately 0.75 Å after 0.3 s. This is indicative of hydrogen bond formation between HA and HB. In addition, the distance between HB and HC increases to 3.44 Å after HC is ejected following a successful collision.

One obvious distance between the two plots is that in the minimum energy path trajectory, the internuclear distance between HA and HB is constant indicating that the bond is static. However, in the dynamic trajectory the internuclear distance is oscillating showing the the bond is vibrating. This difference could be attributed to the fact that minimum energy path does not take oscillatory motion into account while dynamic trajectory does.

However, changing the initial conditions of the system such that rAB = 0.917743 Å, rBC = 0.907743 Å and pA=pBC=0:
[[File:01409469 01409469 Reversed values dynamic mep.jpg|thumb|center|800px|'''Figure 8.''' A graph of internuclear distance against steps using minimum energy path '''Figure 9.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In this case, a hydrogen bond of length 0.77-0.79 Å is formed between HB and HC while HA is repelled away to beyond 3.45 Å.

Additionally, from the internuclear distance plot it can be seen that the distance between HB and HC increases in a rectangular hyperbola fashion when calculated using the minimum energy path. On the other hand, the internuclear distance between HB and HC increases in a linear manner when dynamic calculation is applied. A possible reason for this difference is that in the minimum energy path, the atoms are not vibrating and as such the reach the maximum distance and plateau at a faster rate. However, in the dynamic calculation, the atoms have momentum and as such move further away from each other.

=== Reactive and unreactive trajectories ===

With the initial positions r'''1''' = 0.74 and r'''2''' = 2.0, trajectories were run with the momenta combination shown below.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Contour plot !! Reactivity !! Description of the dynamics
|-
| -1.25 || -2.5 ||-99.018 ||[[File:01409469 trajectory1.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. There are no oscillations observed in HA-HB initially, indicating that the diatomic molecule is stable. A new bond is formed between HB and HC. Oscillatory behaviour is observed indicating that the bond is vibrating. On the other hand, HA is ejected away to beyond 10 Å.
|-
| -1.5 || -2.0 ||-100.456 ||[[File:01409469 trajectory2.jpg|thumb|center|400px]] ||Unreactive ||HC collides with the vibrating diatomic HA-HB. However, no bond is formed between HB and HC as the collision is unsuccessful. This is because the molecules lack sufficient kinetic energy to overcome the activation energy barrier and so no reaction occurs. HC is subsequently repelled to beyond 8 Å.
|-
| -1.5 || -2.5 ||-98.956 ||[[File:01409469 trajectory3.jpg|thumb|center|400px]] ||Reactive |||HC collides with HA-HB. The collision is successful and a bond is formed between HB and HC. This is because the atoms have sufficient kinetic energy to overcome the activation energy barrier and form a new HB-HC bond. HA is ejected away to beyond 10 Å. The new hydrogen bond formed is vibrating as seen by the oscillatory behaviour.
|-
| -2.5 || -5.0 ||-84.956 ||[[File:01409469 trajectory4.jpg|thumb|center|400px]] ||Unreactive ||HC collides with HA-HB. Initially the HA-HB bond is elongating, indicating that the hydrogen bond is weakening and going to break. However, after 0.5 s, the HA-HB bond length decreases and the HAHB is not broken. HC is repelled beyond 6 Å with very strong vibrations.
A reaction did not occur even though the atoms have sufficient kinetic energy to cross the activation energy barrier. This is because the vibrations are too strong. The system crosses the transition state region and a bond is formed in the product temporarily but then the system reverts back to the reactants. This is an example of barrier recrossing.
|-
| -2.5 || -5.2 ||-83.416 ||[[File:01409469 trajectory5.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. The collision is successful and a new hydrogen bond is formed between HB and HC with HA being ejected away. The products formed experience very strong vibrations as evidenced by the oscillatory motion which occur with a very high frequency.
|}

From the table, it can be seen that the chance of a reaction successfully occurring is not solely dependent on the amount of kinetic energy the atoms possess. Even if the atoms have sufficient kinetic energy to cross the activation energy barrier, barrier recrossing can occur if the atoms are too vibrationally excited.

=== Transition State Theory ===

Transition state theory provided the first theoretical attempt to determine absolute reaction rates. In this theory, a transition state separating the reactants from products is used to find an expression for the thermal rate constant.

The main assumptions of transition state theory are:

1. Electronic and nuclear motions are separated (equivalent to the Born-Oppenheimer approximation in quantum mechanics)

2. Reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution

3. Molecular systems that have crossed the transition state in the direction of products cannot turn around and reform reactants

4. In the transition state, motion along the reaction coordinate may be separated from the other motions and treated classically as a translation

5. Even in the absence of an equilibrium between reactant and product molecules, the transition states that are becoming products are distributed among their states according to the Maxwell-Boltzmann laws

The goal of transition state theory is to predict the rate constant, ''k'', of a reaction given a potential energy surface for the reaction.

However, transition state theory suffers from a set of limitations due to the assumptions stated above not holding true in reality.

Firstly, transition state theory assumes that barrier recrossing is negligible. However, as seen from the fourth trajectory in table 1, barrier recrossing occurred because the atoms were too vibrationally excited and the system reverted back to the reactants even though the system had already crossed the transition state region. In this situation, transition state theory would predict a faster rate of reaction compared to what was obtained experimentally.

Secondly, the transition state theory breaks down for systems with light atoms such as hydrogens. In such systems, quantum effects become crucial. Quantum effects tend to increase the rate of the reaction for two reasons. First, particles lie in energy levels which are higher then the bottom of the potential energy well (there is additional zero-point energy). Secondly, the particles can tunnel through the barrier. These two effects effectively lower the activation energy barrier. As a result, the rate of reaction predicted by transition state theory would be less than what is obtained experimentally.

Lastly, transition state theory does not hold true for systems which are far from equilibrium. In such cases, the assumption of a starting Boltzmann distribution breaks down.

== Exercise 2: F - H - H system ==

=== Potential energy surface inspection===

F + HA-HB → HA-F + HB

[[File:01409469 Surface_Plot F 3.jpg|thumb|center|400px|'''Figure 10.''' Potential energy surface for the reaction between F and H2]]

The reaction between F and H2 is an exothermic reaction. This means that the reactants are of a higher energy than the products. The overall reaction involves the breaking of a hydrogen bond and the formation of a hydrogen-fluorine bond. The HF bond formed is stronger than the hydrogen bond that was broken. This is because fluorine is more electronegative than hydrogen resulting in a polar HF bond. The negative enthalpy of H-F bond formation is greater than the positive enthalpy of hydrogen bond dissociation. Therefore, the reaction between F and H2 is an exothermic reaction.

HA + HB-F → HA-HB + F

[[File:01409469 Surface_Plot F 4.jpg|thumb|center|400px|'''Figure 11.''' Potential energy surface for the reaction between H and HF]]

The reaction between H and HF is an endothermic reaction. The products are at a higher energy level than the reactants. The overall reaction involves the breaking of a HF bond and the formation of a hydrogen bond. The formation of a homonuclear H2 molecule results in a non-polar hydrogen bond being formed. The HF bond broken is stronger than the hydrogen bond that was formed. The negative enthalpy of hydrogen bond formation is less than the positive enthalpy of hydrogen-fluorine dissociation. Therefore, the reaction between H and HF is endothermic.

=== Locating the transition state ===

[[File:01409469 Surface_Plot F TS4.jpg|thumb|center|800px|'''Figure 12.''' Determining the position of the transition state for the reaction between F and H2]]

As mentioned previously, when the molecules are not at the transition state, a reaction path will be plotted. However, if the initial conditions of the system are set such that the trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

An initial estimate for the transition state was made by setting rAB = 0.74 which is the length of a hydrogen bond. The position of the transition state was further refined by increasing the number of decimal places such that the final value of rAB was found to be 0.74478 and rBC = 1.81079. This set of distances gave a reaction path that was represented by a point indicating that no trajectory was plotted and that the molecules remained at the transition state since their initial momentum was 0.

Moreover, the kinetic energy as well as the forces acting on the atoms was calculated to be 0, which is consistent with the transition state.

Therefore, the position of the transition state was found to be at rAB = 0.74478 and rBC = 1.81079 when pA = pBC = 0.

=== Determining the activation energy ===

[[File:01409469 activation energy r1.1 combined.jpg|thumb|center|800px|'''Figure 13.''' Determining the activation energy for the reaction between F and H2]]

By reducing the value of rBC by 0.01, the atoms are no longer at the transition state and would move towards the formation of products. The activation energy was found by calculating the change in total energy after the reaction. The activation energy of F + H2 → HF + H was found to be +23.279 kcal/ mol.

[[File:01409469 activation energy r2.1 combined.jpg|thumb|center|800px|'''Figure 14.''' Determining the activation energy for the reaction between FH and H]]

The activation energy of FH + H → F + H2 was found by calculating the small dip in potential energy. The activation energy was found to be +0.280 kcal/ mol.

=== Reaction dynamics of F + HA-HB → HA-F + HB ===

The set of initial conditions that result in a reactive trajectory for F + H2 is as follows: rHH = 0.74 Å, rHF = 1.79 Å and pHH = pHF = 0.

[[File:01409469 momentum r1.jpg|thumb|center|400px|'''Figure 15.''' Graph of momentum against time for the reaction between F and H2]]

Initially, the H2 molecule is travelling in a relatively straight line with very little oscillations as evidenced by the low amplitude of momentum between 0-2 s. Following successful collision between F and H2, the amplitude of momentum of the product, HF, increases dramatically. This is because the reaction between F and H2 is exothermic and the excess energy released causes the molecule to become vibrationally excited and the amplitude of oscillations in the product to increase.

[[File:01409469 energy1 r1.jpg|thumb|center|400px|'''Figure 16.''' Graph of energy against time for the reaction between F and H2]]

The law of conservation of energy states that energy cannot be created or destroyed. It simply changes from one form to another. As can be seen from the graph in figure 16, total energy remains constant. In this reaction, potential energy is converted to kinetic energy. A decrease in potential energy of the atoms following a successful collision between F and H2 results in an equal increase in kinetic energy which manifests in the form of increased product (HF) oscillations.

This can be confirmed experimentally using infrared chemiluminescence. Observation of radiation from excited product molecules is a clear indication that these molecules are produced in the excited state. If the excitation is primarily vibrational, the radiation will appear in the infrared region of the spectrum (between 3-15 μm). This technique has been used extensively in hydrogen halogen reactions which produce HX molecules in the excited state.

One such experiment involves the use of the arrested relaxation method. In this technique, the reaction is carried out in a chamber whose walls are maintained at a very low temperature (20 K). This results in cryopumping of the chamber in that all the products and unreacted starting materials are condensed on the walls and only those product molecules that radiate before being pumped away are observed.

=== Polanyi's empirical rules ===

In a typical chemical reaction, there is an energetic barrier — a saddle point that the reactants must surmount to become products. Which form of energy is initially deposited in the reactants, translational or vibrational, is more efficacious in surmounting the activation energy barrier is one of the core topics in the area of reaction dynamics. The Polyani rules state that vibrational energy is more effective at promoting a late-barrier reaction (transition state resembles products). Conversely, translational energy is better at promoting an early-barrier reaction (transition state resembles the reactants).

The forward reaction between H2 and F is exothermic. Therefore, the transition state is closer to the reactants than to the products in energy. Since the transition state occurs early in an exothermic reaction, translational energy is more effective at promoting a chemical reaction than vibrational energy.

Conversely, the backward reaction between H and HF is endothermic. According to Hammond's postulate, the transition state resembles the products more than it resembles the reactants. Therefore, in an endothermic reaction, a late transition state is observed and vibrational energy is more effective at promoting a reaction than translational energy.

=== Promoting the forward reaction: F + HA-HB → HA-F + HB ===

1. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 2.95 and pHF = -0.5.

[[File:01409469 polyani 1+2.jpg|thumb|center|800px|'''Figure 17.''' Contour plot of the reaction between F and H2 when pHH = 2.95 '''Figure 18.''' Graph of momentum against time for the reaction between F and H2 when pHH = 2.95 (Left-right)]]

Vibrational energy is represented by pHH while translational energy is represented by pHF.

A molecule of H2 collides with F. However, the collision is unsuccessful and no reaction occurs. A reaction did not occur even though the molecules have sufficient kinetic energy to overcome the activation energy barrier. This is because most of the energy is deposited in the reactants as H-H vibrations as seen by the large oscillations in H2. Barrier recrossing has occured as a result of the large vibrational energy causing the system to revert back to the reactants even though the system has already crossed the transition state and a bond was formed between HB and F temporarily.

This appears to run counter to what one would expect to happen when more energy is put into the system. It is predicted that when more energy is put into the system, the reactants would possess more kinetic energy to overcome the activation energy barrier. However, as can be seen from the simulation above, an increase in the vibrational energy of the reactants is not effective at promoting the occurrence of an exothermic reaction with an early transition state. This is in accordance to Polyani's rules.

2. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 0.1 and pHF = -0.8.

[[File:01409469 polyani 3+4 2.jpg|thumb|center|800px|'''Figure 19.''' Contour plot of the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 '''Figure 20.''' Graph of momentum against time for the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 (Left-right)]]

The collision between a molecule of H2 and F is successful and a chemical reaction has occurred. A reaction has occurred even though the overall energy of the system has been significantly reduced by lowering the amount of vibrational energy in the reactants but slightly increasing the amount of translational energy deposited in the reactants. Therefore, H2 and F have more translational energy to move towards each other and form a H-F bond. The product formed is vibrationally excited as seen by the large amplitude of oscillations in the product HF which correspond to the strong H-F vibrations.

From the second simulation, it can be seen that an increase in translational energy in the reactants is more effective at promoting an exothermic reaction with an early transition state to occur, which again follows Polyani's rules.

=== Promoting the backward reaction: HA + HB-F → HA-HB + F ===

1. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 0.2 and pHH = -3.0.

[[File:01409469 polyani 5+6 .jpg|thumb|center|800px|'''Figure 21.''' Contour plot of the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 '''Figure 22.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 (Left-right)]]

An atom of HA collides with a molecule of HBF. However, the collision is unsuccessful and no chemical reaction occurs. This is because the reactants lack sufficient vibrational energy to surmount the activation energy barrier. Therefore, following an unsuccessful collision with HBF, HA simply bounces off and moves away at constant momentum. This is in agreement with Polyani's rules which state that vibrational energy is more effective than translational energy at promoting reactions with a late transition state, for example, endothermic chemical reactions.

2. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 7.1 and pHH = -0.5.

[[File:01409469 polyani 7+8 .jpg|thumb|center|800px|'''Figure 23.''' Contour plot of the reaction between HA + HB-F when pFH = 7.1 and pHH = -0.5 '''Figure 24.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 7.1 and pHH = -0.5 (Left-right)]]

Increasing the amount of vibrational energy in the reactants by increasing pFH resulted in a chemical reaction successfully occurring. This is because the reactants now have sufficient kinetic energy in the form of H-F vibrations to overcome the activation energy barrier. In addition, because the reaction between HA + HB-F is endothermic, energy is taken in from the surroundings during the reaction and the products are less vibrationally excited than the reactants. This can be seen in the lower amplitude of oscillations in the product formed.

In conclusion, the vibrational energy of the reactants should be increased to promote an endothermic reaction with a late transition state while the translational energy of the reactants should be increased to promote an exothermic reaction with an early transition state. All 4 simulations shown above obey Polyani's rules.

== References ==
<references>
<ref name="LazyDog">This is the lazy dog reference.</ref>
</references>

MRD:01409469

2019-05-10T14:49:47Z

Stc1917: /* References */

== Exercise 1: H + H2 system ==

=== Definition of a transition state ===

A potential energy profile results from plotting the potential versus reaction path.

[[File:Surface_Plot01409469_pic1.jpg|thumb|center|400px|'''Figure 1.''' A potential energy surface plot]]

The transition state is defined as the maximum on the minimum energy path linking reactants and products. This means that the transition state is a stationary point on the potential energy surface. Thus, at the transition state, the derivative of the potential energy is zero with respect to each Cartesian coordinate.

<math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>

However, a local minimum on the potential energy surface corresponding to a well would have the same <math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>. Therefore, to distinguish between a local minimum on the potential energy surface and the transition state, the second derivative, <math>\tfrac{\partial^2 V(r_i)}{\partial r_i ^2}\ </math> is determined. This can be calculated by finding the determinant of the Hessian matrix shown below.

:<math>\mathbf H =
\begin{bmatrix}
\dfrac{\partial^2 \ V(r_i)}{\partial \ x^2} & \dfrac{\partial^2 \ V(r_i)}{\partial \ x \partial \mathbf y} \\
\dfrac{\partial^2 \ V(r_i)}{\partial \ y \partial \mathbf x} & \dfrac{\partial^2 \ V(r_i)}{\partial \mathbf y^2}
\end{bmatrix}</math>

The Hessian matrix of a convex function is positive semi-definite. Refining this property allows one to test if a critical point (x,y) is a local maximum, local minimum or saddle point.

If <math>f</math> < 0, then (x,y) is an isolated local maximum.

If <math>f</math> > 0, then (x,y) is an isolated local minimum.

If the Hessian has both positive and negative eigenvalues then (x,y) is a saddle point.

=== Locating the transition state ===

[[File:01409469 combined pic 2 and 3.jpg|thumb|center|800px|'''Figure 2.''' A surface and contour plot of the trajectory when rAB=rBC=0.907743 Å and pA=pBC=0]]

When the molecules are not at the transition state, a surface plot of the conditions shows a trajectory being plotted. This means that the trajectory is "rolling" towards the products or reactants. However, if a trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

[[File:01409469_internuclearpic3.png|thumb|center|400px|'''Figure 3.''' A plot of internuclear distance against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

An initial estimate for the position of the transition state was made by observing the potential energy surface and was found to be 0.90 Å. At 0.90 Å, the internuclear distance against time plot showed that the internuclear distance was constantly oscillating. Therefore, the position of the transition state was further refined by increasing the number of decimal places until a horizontal line in the internuclear distance againt time plot was obtained. This means that the internuclear distance is constant and that the molecules are simply undergoing periodic symmetric vibration at the transition state. Therefore, the position of the transition state was determined to be 0.907743 Å.

[[File:01409469 combined energy pic.png|thumb|center|400px|'''Figure 4.''' A plot of energy against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

When rAB=rBC=0.907743 Å and pA=pBC=0, the kinetic energy is 0. This means that HA and HBC are stationary, which is as predicted when the trajectory is started at the transition state with no initial momentum. Furthermore, the forces acting on HA and HBC are 0 as should be the case in the transition state.

=== Calculation type: Dynamic and minimum energy path ===

The minimum energy path, is the lowest energy path from the reactants to the products. The minimum energy path is determined by following the mode corresponding to the negative Hessian eigenvalue in both directions along the steepest descent direction. Following this path allows one to approach two valleys of stable structures. The resulting path generated is the minimum energy path. The reaction paths were chartered for a dynamic and a minimum energy path. The initial conditions used were that of the transition state except that the internuclear distance between HA and HB had increased to 0.917743 Å. The reaction path obtained using the two different calculation methods was different.

[[File:01409469 mep and dynamics.jpg|thumb|center|800px|'''Figure 5.''' A surface plot of the trajectory using minimum energy path '''Figure 6.''' A surface plot of the trajectory using dynamics (Left-right)]]

On the surface plot, the reaction path is depicted by the black line. Following the minimum energy path, the trajectory simply follows the valley floor to Ha+Hb-Hc. The reaction path appears to be a relatively straight line indicating that the atoms are not vibrating. In the minimum energy path, the reactants change into the products in an infinitely slow motion. This is done by having the momenta/velocities reset to zero in each time step. While the minimum energy path is useful in characterising a chemical reaction, it does not provide a realistic representation of the motion of the atoms during the reaction. This is because the mass and inertia of the atoms have been ignored.

Since the atoms simply follow the minimum energy path, they do not undergo slight displacements into surrounding valleys of higher potential energy. Furthermore, because the atoms are assumed to have no inertia, they do not undergo vibrations even if they were to be displaced to valleys of higher potential energy.

However, when dynamic calculation is performed, the reaction path is wavy indicating that the atoms are oscillating. This is because the atoms are displaced to regions of higher potential on the path from reactants to products. This leads to atom oscillation due to the inertia of the atoms.

'''Internuclear distance against time plot'''

Setting the initial system conditions such that rAB= 0.907743, rBC= 0.917743 and pA= pBC= 0:

[[File:01409469 Internuclear distance mep and dynamic.jpg|thumb|center|800px|'''Figure 6.''' A graph of internuclear distance against steps using minimum energy path '''Figure 7.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In the minimum energy path, the distance between HA and HB decreases from 0.90 Å to approximately 0.79 Å during the first 50 steps before plateauing off to a value of 0.74 Å which is the approximate length of a hydrogen bond. This means that a bond was formed between HA and HB. In addition, the plateau indicates that the bond is static. On the other hand, the distance between HB and HC has increased to 1.91 Å at large step values. This shows that HC has been ejected off following a successful collision between HA and HBC.

Similarly, in the dynamic trajectory, the distance between HA and HB decreases to approximately 0.75 Å after 0.3 s. This is indicative of hydrogen bond formation between HA and HB. In addition, the distance between HB and HC increases to 3.44 Å after HC is ejected following a successful collision.

One obvious distance between the two plots is that in the minimum energy path trajectory, the internuclear distance between HA and HB is constant indicating that the bond is static. However, in the dynamic trajectory the internuclear distance is oscillating showing the the bond is vibrating. This difference could be attributed to the fact that minimum energy path does not take oscillatory motion into account while dynamic trajectory does.

However, changing the initial conditions of the system such that rAB = 0.917743 Å, rBC = 0.907743 Å and pA=pBC=0:
[[File:01409469 01409469 Reversed values dynamic mep.jpg|thumb|center|800px|'''Figure 8.''' A graph of internuclear distance against steps using minimum energy path '''Figure 9.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In this case, a hydrogen bond of length 0.77-0.79 Å is formed between HB and HC while HA is repelled away to beyond 3.45 Å.

Additionally, from the internuclear distance plot it can be seen that the distance between HB and HC increases in a rectangular hyperbola fashion when calculated using the minimum energy path. On the other hand, the internuclear distance between HB and HC increases in a linear manner when dynamic calculation is applied. A possible reason for this difference is that in the minimum energy path, the atoms are not vibrating and as such the reach the maximum distance and plateau at a faster rate. However, in the dynamic calculation, the atoms have momentum and as such move further away from each other.

=== Reactive and unreactive trajectories ===

With the initial positions r'''1''' = 0.74 and r'''2''' = 2.0, trajectories were run with the momenta combination shown below.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Contour plot !! Reactivity !! Description of the dynamics
|-
| -1.25 || -2.5 ||-99.018 ||[[File:01409469 trajectory1.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. There are no oscillations observed in HA-HB initially, indicating that the diatomic molecule is stable. A new bond is formed between HB and HC. Oscillatory behaviour is observed indicating that the bond is vibrating. On the other hand, HA is ejected away to beyond 10 Å.
|-
| -1.5 || -2.0 ||-100.456 ||[[File:01409469 trajectory2.jpg|thumb|center|400px]] ||Unreactive ||HC collides with the vibrating diatomic HA-HB. However, no bond is formed between HB and HC as the collision is unsuccessful. This is because the molecules lack sufficient kinetic energy to overcome the activation energy barrier and so no reaction occurs. HC is subsequently repelled to beyond 8 Å.
|-
| -1.5 || -2.5 ||-98.956 ||[[File:01409469 trajectory3.jpg|thumb|center|400px]] ||Reactive |||HC collides with HA-HB. The collision is successful and a bond is formed between HB and HC. This is because the atoms have sufficient kinetic energy to overcome the activation energy barrier and form a new HB-HC bond. HA is ejected away to beyond 10 Å. The new hydrogen bond formed is vibrating as seen by the oscillatory behaviour.
|-
| -2.5 || -5.0 ||-84.956 ||[[File:01409469 trajectory4.jpg|thumb|center|400px]] ||Unreactive ||HC collides with HA-HB. Initially the HA-HB bond is elongating, indicating that the hydrogen bond is weakening and going to break. However, after 0.5 s, the HA-HB bond length decreases and the HAHB is not broken. HC is repelled beyond 6 Å with very strong vibrations.
A reaction did not occur even though the atoms have sufficient kinetic energy to cross the activation energy barrier. This is because the vibrations are too strong. The system crosses the transition state region and a bond is formed in the product temporarily but then the system reverts back to the reactants. This is an example of barrier recrossing.
|-
| -2.5 || -5.2 ||-83.416 ||[[File:01409469 trajectory5.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. The collision is successful and a new hydrogen bond is formed between HB and HC with HA being ejected away. The products formed experience very strong vibrations as evidenced by the oscillatory motion which occur with a very high frequency.
|}

From the table, it can be seen that the chance of a reaction successfully occurring is not solely dependent on the amount of kinetic energy the atoms possess. Even if the atoms have sufficient kinetic energy to cross the activation energy barrier, barrier recrossing can occur if the atoms are too vibrationally excited.

=== Transition State Theory ===

Transition state theory provided the first theoretical attempt to determine absolute reaction rates. In this theory, a transition state separating the reactants from products is used to find an expression for the thermal rate constant.

The main assumptions of transition state theory are:

1. Electronic and nuclear motions are separated (equivalent to the Born-Oppenheimer approximation in quantum mechanics)

2. Reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution

3. Molecular systems that have crossed the transition state in the direction of products cannot turn around and reform reactants

4. In the transition state, motion along the reaction coordinate may be separated from the other motions and treated classically as a translation

5. Even in the absence of an equilibrium between reactant and product molecules, the transition states that are becoming products are distributed among their states according to the Maxwell-Boltzmann laws

The goal of transition state theory is to predict the rate constant, ''k'', of a reaction given a potential energy surface for the reaction.

However, transition state theory suffers from a set of limitations due to the assumptions stated above not holding true in reality.

Firstly, transition state theory assumes that barrier recrossing is negligible. However, as seen from the fourth trajectory in table 1, barrier recrossing occurred because the atoms were too vibrationally excited and the system reverted back to the reactants even though the system had already crossed the transition state region. In this situation, transition state theory would predict a faster rate of reaction compared to what was obtained experimentally.

Secondly, the transition state theory breaks down for systems with light atoms such as hydrogens. In such systems, quantum effects become crucial. Quantum effects tend to increase the rate of the reaction for two reasons. First, particles lie in energy levels which are higher then the bottom of the potential energy well (there is additional zero-point energy). Secondly, the particles can tunnel through the barrier. These two effects effectively lower the activation energy barrier. As a result, the rate of reaction predicted by transition state theory would be less than what is obtained experimentally.

Lastly, transition state theory does not hold true for systems which are far from equilibrium. In such cases, the assumption of a starting Boltzmann distribution breaks down.

== Exercise 2: F - H - H system ==

=== Potential energy surface inspection===

F + HA-HB → HA-F + HB

[[File:01409469 Surface_Plot F 3.jpg|thumb|center|400px|'''Figure 10.''' Potential energy surface for the reaction between F and H2]]

The reaction between F and H2 is an exothermic reaction. This means that the reactants are of a higher energy than the products. The overall reaction involves the breaking of a hydrogen bond and the formation of a hydrogen-fluorine bond. The HF bond formed is stronger than the hydrogen bond that was broken. This is because fluorine is more electronegative than hydrogen resulting in a polar HF bond. The negative enthalpy of H-F bond formation is greater than the positive enthalpy of hydrogen bond dissociation. Therefore, the reaction between F and H2 is an exothermic reaction.

HA + HB-F → HA-HB + F

[[File:01409469 Surface_Plot F 4.jpg|thumb|center|400px|'''Figure 11.''' Potential energy surface for the reaction between H and HF]]

The reaction between H and HF is an endothermic reaction. The products are at a higher energy level than the reactants. The overall reaction involves the breaking of a HF bond and the formation of a hydrogen bond. The formation of a homonuclear H2 molecule results in a non-polar hydrogen bond being formed. The HF bond broken is stronger than the hydrogen bond that was formed. The negative enthalpy of hydrogen bond formation is less than the positive enthalpy of hydrogen-fluorine dissociation. Therefore, the reaction between H and HF is endothermic.

=== Locating the transition state ===

[[File:01409469 Surface_Plot F TS4.jpg|thumb|center|800px|'''Figure 12.''' Determining the position of the transition state for the reaction between F and H2]]

As mentioned previously, when the molecules are not at the transition state, a reaction path will be plotted. However, if the initial conditions of the system are set such that the trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

An initial estimate for the transition state was made by setting rAB = 0.74 which is the length of a hydrogen bond. The position of the transition state was further refined by increasing the number of decimal places such that the final value of rAB was found to be 0.74478 and rBC = 1.81079. This set of distances gave a reaction path that was represented by a point indicating that no trajectory was plotted and that the molecules remained at the transition state since their initial momentum was 0.

Moreover, the kinetic energy as well as the forces acting on the atoms was calculated to be 0, which is consistent with the transition state.

Therefore, the position of the transition state was found to be at rAB = 0.74478 and rBC = 1.81079 when pA = pBC = 0.

=== Determining the activation energy ===

[[File:01409469 activation energy r1.1 combined.jpg|thumb|center|800px|'''Figure 13.''' Determining the activation energy for the reaction between F and H2]]

By reducing the value of rBC by 0.01, the atoms are no longer at the transition state and would move towards the formation of products. The activation energy was found by calculating the change in total energy after the reaction. The activation energy of F + H2 → HF + H was found to be +23.279 kcal/ mol.

[[File:01409469 activation energy r2.1 combined.jpg|thumb|center|800px|'''Figure 14.''' Determining the activation energy for the reaction between FH and H]]

The activation energy of FH + H → F + H2 was found by calculating the small dip in potential energy. The activation energy was found to be +0.280 kcal/ mol.

=== Reaction dynamics of F + HA-HB → HA-F + HB ===

The set of initial conditions that result in a reactive trajectory for F + H2 is as follows: rHH = 0.74 Å, rHF = 1.79 Å and pHH = pHF = 0.

[[File:01409469 momentum r1.jpg|thumb|center|400px|'''Figure 15.''' Graph of momentum against time for the reaction between F and H2]]

Initially, the H2 molecule is travelling in a relatively straight line with very little oscillations as evidenced by the low amplitude of momentum between 0-2 s. Following successful collision between F and H2, the amplitude of momentum of the product, HF, increases dramatically. This is because the reaction between F and H2 is exothermic and the excess energy released causes the molecule to become vibrationally excited and the amplitude of oscillations in the product to increase.

[[File:01409469 energy1 r1.jpg|thumb|center|400px|'''Figure 16.''' Graph of energy against time for the reaction between F and H2]]

The law of conservation of energy states that energy cannot be created or destroyed. It simply changes from one form to another. As can be seen from the graph in figure 16, total energy remains constant. In this reaction, potential energy is converted to kinetic energy. A decrease in potential energy of the atoms following a successful collision between F and H2 results in an equal increase in kinetic energy which manifests in the form of increased product (HF) oscillations.

This can be confirmed experimentally using infrared chemiluminescence. Observation of radiation from excited product molecules is a clear indication that these molecules are produced in the excited state. If the excitation is primarily vibrational, the radiation will appear in the infrared region of the spectrum (between 3-15 μm). This technique has been used extensively in hydrogen halogen reactions which produce HX molecules in the excited state.

One such experiment involves the use of the arrested relaxation method. In this technique, the reaction is carried out in a chamber whose walls are maintained at a very low temperature (20 K). This results in cryopumping of the chamber in that all the products and unreacted starting materials are condensed on the walls and only those product molecules that radiate before being pumped away are observed.

=== Polanyi's empirical rules ===

In a typical chemical reaction, there is an energetic barrier — a saddle point that the reactants must surmount to become products. Which form of energy is initially deposited in the reactants, translational or vibrational, is more efficacious in surmounting the activation energy barrier is one of the core topics in the area of reaction dynamics. The Polyani rules state that vibrational energy is more effective at promoting a late-barrier reaction (transition state resembles products). Conversely, translational energy is better at promoting an early-barrier reaction (transition state resembles the reactants).

The forward reaction between H2 and F is exothermic. Therefore, the transition state is closer to the reactants than to the products in energy. Since the transition state occurs early in an exothermic reaction, translational energy is more effective at promoting a chemical reaction than vibrational energy.

Conversely, the backward reaction between H and HF is endothermic. According to Hammond's postulate, the transition state resembles the products more than it resembles the reactants. Therefore, in an endothermic reaction, a late transition state is observed and vibrational energy is more effective at promoting a reaction than translational energy.

=== Promoting the forward reaction: F + HA-HB → HA-F + HB ===

1. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 2.95 and pHF = -0.5.

[[File:01409469 polyani 1+2.jpg|thumb|center|800px|'''Figure 17.''' Contour plot of the reaction between F and H2 when pHH = 2.95 '''Figure 18.''' Graph of momentum against time for the reaction between F and H2 when pHH = 2.95 (Left-right)]]

Vibrational energy is represented by pHH while translational energy is represented by pHF.

A molecule of H2 collides with F. However, the collision is unsuccessful and no reaction occurs. A reaction did not occur even though the molecules have sufficient kinetic energy to overcome the activation energy barrier. This is because most of the energy is deposited in the reactants as H-H vibrations as seen by the large oscillations in H2. Barrier recrossing has occured as a result of the large vibrational energy causing the system to revert back to the reactants even though the system has already crossed the transition state and a bond was formed between HB and F temporarily.

This appears to run counter to what one would expect to happen when more energy is put into the system. It is predicted that when more energy is put into the system, the reactants would possess more kinetic energy to overcome the activation energy barrier. However, as can be seen from the simulation above, an increase in the vibrational energy of the reactants is not effective at promoting the occurrence of an exothermic reaction with an early transition state. This is in accordance to Polyani's rules.

2. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 0.1 and pHF = -0.8.

[[File:01409469 polyani 3+4 2.jpg|thumb|center|800px|'''Figure 19.''' Contour plot of the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 '''Figure 20.''' Graph of momentum against time for the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 (Left-right)]]

The collision between a molecule of H2 and F is successful and a chemical reaction has occurred. A reaction has occurred even though the overall energy of the system has been significantly reduced by lowering the amount of vibrational energy in the reactants but slightly increasing the amount of translational energy deposited in the reactants. Therefore, H2 and F have more translational energy to move towards each other and form a H-F bond. The product formed is vibrationally excited as seen by the large amplitude of oscillations in the product HF which correspond to the strong H-F vibrations.

From the second simulation, it can be seen that an increase in translational energy in the reactants is more effective at promoting an exothermic reaction with an early transition state to occur, which again follows Polyani's rules.

=== Promoting the backward reaction: HA + HB-F → HA-HB + F ===

1. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 0.2 and pHH = -3.0.

[[File:01409469 polyani 5+6 .jpg|thumb|center|800px|'''Figure 21.''' Contour plot of the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 '''Figure 22.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 (Left-right)]]

An atom of HA collides with a molecule of HBF. However, the collision is unsuccessful and no chemical reaction occurs. This is because the reactants lack sufficient vibrational energy to surmount the activation energy barrier. Therefore, following an unsuccessful collision with HBF, HA simply bounces off and moves away at constant momentum. This is in agreement with Polyani's rules which state that vibrational energy is more effective than translational energy at promoting reactions with a late transition state, for example, endothermic chemical reactions.

2. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 7.1 and pHH = -0.5.

[[File:01409469 polyani 7+8 .jpg|thumb|center|800px|'''Figure 23.''' Contour plot of the reaction between HA + HB-F when pFH = 7.1 and pHH = -0.5 '''Figure 24.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 7.1 and pHH = -0.5 (Left-right)]]

Increasing the amount of vibrational energy in the reactants by increasing pFH resulted in a chemical reaction successfully occurring. This is because the reactants now have sufficient kinetic energy in the form of H-F vibrations to overcome the activation energy barrier. In addition, because the reaction between HA + HB-F is endothermic, energy is taken in from the surroundings during the reaction and the products are less vibrationally excited than the reactants. This can be seen in the lower amplitude of oscillations in the product formed.

In conclusion, the vibrational energy of the reactants should be increased to promote an endothermic reaction with a late transition state while the translational energy of the reactants should be increased to promote an exothermic reaction with an early transition state. All 4 simulations shown above obey Polyani's rules.

== References ==

The quick brown fox jumps over the lazy dog.<ref name="LazyDog" />
<references>
<ref name="LazyDog">This is the lazy dog reference.</ref>
</references>

MRD:01409469

2019-05-10T14:32:56Z

Stc1917: /* Promoting the backward reaction: HA + HB-F → HA-HB + F */

== Exercise 1: H + H2 system ==

=== Definition of a transition state ===

A potential energy profile results from plotting the potential versus reaction path.

[[File:Surface_Plot01409469_pic1.jpg|thumb|center|400px|'''Figure 1.''' A potential energy surface plot]]

The transition state is defined as the maximum on the minimum energy path linking reactants and products. This means that the transition state is a stationary point on the potential energy surface. Thus, at the transition state, the derivative of the potential energy is zero with respect to each Cartesian coordinate.

<math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>

However, a local minimum on the potential energy surface corresponding to a well would have the same <math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>. Therefore, to distinguish between a local minimum on the potential energy surface and the transition state, the second derivative, <math>\tfrac{\partial^2 V(r_i)}{\partial r_i ^2}\ </math> is determined. This can be calculated by finding the determinant of the Hessian matrix shown below.

:<math>\mathbf H =
\begin{bmatrix}
\dfrac{\partial^2 \ V(r_i)}{\partial \ x^2} & \dfrac{\partial^2 \ V(r_i)}{\partial \ x \partial \mathbf y} \\
\dfrac{\partial^2 \ V(r_i)}{\partial \ y \partial \mathbf x} & \dfrac{\partial^2 \ V(r_i)}{\partial \mathbf y^2}
\end{bmatrix}</math>

The Hessian matrix of a convex function is positive semi-definite. Refining this property allows one to test if a critical point (x,y) is a local maximum, local minimum or saddle point.

If <math>f</math> < 0, then (x,y) is an isolated local maximum.

If <math>f</math> > 0, then (x,y) is an isolated local minimum.

If the Hessian has both positive and negative eigenvalues then (x,y) is a saddle point.

=== Locating the transition state ===

[[File:01409469 combined pic 2 and 3.jpg|thumb|center|800px|'''Figure 2.''' A surface and contour plot of the trajectory when rAB=rBC=0.907743 Å and pA=pBC=0]]

When the molecules are not at the transition state, a surface plot of the conditions shows a trajectory being plotted. This means that the trajectory is "rolling" towards the products or reactants. However, if a trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

[[File:01409469_internuclearpic3.png|thumb|center|400px|'''Figure 3.''' A plot of internuclear distance against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

An initial estimate for the position of the transition state was made by observing the potential energy surface and was found to be 0.90 Å. At 0.90 Å, the internuclear distance against time plot showed that the internuclear distance was constantly oscillating. Therefore, the position of the transition state was further refined by increasing the number of decimal places until a horizontal line in the internuclear distance againt time plot was obtained. This means that the internuclear distance is constant and that the molecules are simply undergoing periodic symmetric vibration at the transition state. Therefore, the position of the transition state was determined to be 0.907743 Å.

[[File:01409469 combined energy pic.png|thumb|center|400px|'''Figure 4.''' A plot of energy against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

When rAB=rBC=0.907743 Å and pA=pBC=0, the kinetic energy is 0. This means that HA and HBC are stationary, which is as predicted when the trajectory is started at the transition state with no initial momentum. Furthermore, the forces acting on HA and HBC are 0 as should be the case in the transition state.

=== Calculation type: Dynamic and minimum energy path ===

The minimum energy path, is the lowest energy path from the reactants to the products. The minimum energy path is determined by following the mode corresponding to the negative Hessian eigenvalue in both directions along the steepest descent direction. Following this path allows one to approach two valleys of stable structures. The resulting path generated is the minimum energy path. The reaction paths were chartered for a dynamic and a minimum energy path. The initial conditions used were that of the transition state except that the internuclear distance between HA and HB had increased to 0.917743 Å. The reaction path obtained using the two different calculation methods was different.

[[File:01409469 mep and dynamics.jpg|thumb|center|800px|'''Figure 5.''' A surface plot of the trajectory using minimum energy path '''Figure 6.''' A surface plot of the trajectory using dynamics (Left-right)]]

On the surface plot, the reaction path is depicted by the black line. Following the minimum energy path, the trajectory simply follows the valley floor to Ha+Hb-Hc. The reaction path appears to be a relatively straight line indicating that the atoms are not vibrating. In the minimum energy path, the reactants change into the products in an infinitely slow motion. This is done by having the momenta/velocities reset to zero in each time step. While the minimum energy path is useful in characterising a chemical reaction, it does not provide a realistic representation of the motion of the atoms during the reaction. This is because the mass and inertia of the atoms have been ignored.

Since the atoms simply follow the minimum energy path, they do not undergo slight displacements into surrounding valleys of higher potential energy. Furthermore, because the atoms are assumed to have no inertia, they do not undergo vibrations even if they were to be displaced to valleys of higher potential energy.

However, when dynamic calculation is performed, the reaction path is wavy indicating that the atoms are oscillating. This is because the atoms are displaced to regions of higher potential on the path from reactants to products. This leads to atom oscillation due to the inertia of the atoms.

'''Internuclear distance against time plot'''

Setting the initial system conditions such that rAB= 0.907743, rBC= 0.917743 and pA= pBC= 0:

[[File:01409469 Internuclear distance mep and dynamic.jpg|thumb|center|800px|'''Figure 6.''' A graph of internuclear distance against steps using minimum energy path '''Figure 7.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In the minimum energy path, the distance between HA and HB decreases from 0.90 Å to approximately 0.79 Å during the first 50 steps before plateauing off to a value of 0.74 Å which is the approximate length of a hydrogen bond. This means that a bond was formed between HA and HB. In addition, the plateau indicates that the bond is static. On the other hand, the distance between HB and HC has increased to 1.91 Å at large step values. This shows that HC has been ejected off following a successful collision between HA and HBC.

Similarly, in the dynamic trajectory, the distance between HA and HB decreases to approximately 0.75 Å after 0.3 s. This is indicative of hydrogen bond formation between HA and HB. In addition, the distance between HB and HC increases to 3.44 Å after HC is ejected following a successful collision.

One obvious distance between the two plots is that in the minimum energy path trajectory, the internuclear distance between HA and HB is constant indicating that the bond is static. However, in the dynamic trajectory the internuclear distance is oscillating showing the the bond is vibrating. This difference could be attributed to the fact that minimum energy path does not take oscillatory motion into account while dynamic trajectory does.

However, changing the initial conditions of the system such that rAB = 0.917743 Å, rBC = 0.907743 Å and pA=pBC=0:
[[File:01409469 01409469 Reversed values dynamic mep.jpg|thumb|center|800px|'''Figure 8.''' A graph of internuclear distance against steps using minimum energy path '''Figure 9.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In this case, a hydrogen bond of length 0.77-0.79 Å is formed between HB and HC while HA is repelled away to beyond 3.45 Å.

Additionally, from the internuclear distance plot it can be seen that the distance between HB and HC increases in a rectangular hyperbola fashion when calculated using the minimum energy path. On the other hand, the internuclear distance between HB and HC increases in a linear manner when dynamic calculation is applied. A possible reason for this difference is that in the minimum energy path, the atoms are not vibrating and as such the reach the maximum distance and plateau at a faster rate. However, in the dynamic calculation, the atoms have momentum and as such move further away from each other.

=== Reactive and unreactive trajectories ===

With the initial positions r'''1''' = 0.74 and r'''2''' = 2.0, trajectories were run with the momenta combination shown below.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Contour plot !! Reactivity !! Description of the dynamics
|-
| -1.25 || -2.5 ||-99.018 ||[[File:01409469 trajectory1.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. There are no oscillations observed in HA-HB initially, indicating that the diatomic molecule is stable. A new bond is formed between HB and HC. Oscillatory behaviour is observed indicating that the bond is vibrating. On the other hand, HA is ejected away to beyond 10 Å.
|-
| -1.5 || -2.0 ||-100.456 ||[[File:01409469 trajectory2.jpg|thumb|center|400px]] ||Unreactive ||HC collides with the vibrating diatomic HA-HB. However, no bond is formed between HB and HC as the collision is unsuccessful. This is because the molecules lack sufficient kinetic energy to overcome the activation energy barrier and so no reaction occurs. HC is subsequently repelled to beyond 8 Å.
|-
| -1.5 || -2.5 ||-98.956 ||[[File:01409469 trajectory3.jpg|thumb|center|400px]] ||Reactive |||HC collides with HA-HB. The collision is successful and a bond is formed between HB and HC. This is because the atoms have sufficient kinetic energy to overcome the activation energy barrier and form a new HB-HC bond. HA is ejected away to beyond 10 Å. The new hydrogen bond formed is vibrating as seen by the oscillatory behaviour.
|-
| -2.5 || -5.0 ||-84.956 ||[[File:01409469 trajectory4.jpg|thumb|center|400px]] ||Unreactive ||HC collides with HA-HB. Initially the HA-HB bond is elongating, indicating that the hydrogen bond is weakening and going to break. However, after 0.5 s, the HA-HB bond length decreases and the HAHB is not broken. HC is repelled beyond 6 Å with very strong vibrations.
A reaction did not occur even though the atoms have sufficient kinetic energy to cross the activation energy barrier. This is because the vibrations are too strong. The system crosses the transition state region and a bond is formed in the product temporarily but then the system reverts back to the reactants. This is an example of barrier recrossing.
|-
| -2.5 || -5.2 ||-83.416 ||[[File:01409469 trajectory5.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. The collision is successful and a new hydrogen bond is formed between HB and HC with HA being ejected away. The products formed experience very strong vibrations as evidenced by the oscillatory motion which occur with a very high frequency.
|}

From the table, it can be seen that the chance of a reaction successfully occurring is not solely dependent on the amount of kinetic energy the atoms possess. Even if the atoms have sufficient kinetic energy to cross the activation energy barrier, barrier recrossing can occur if the atoms are too vibrationally excited.

=== Transition State Theory ===

Transition state theory provided the first theoretical attempt to determine absolute reaction rates. In this theory, a transition state separating the reactants from products is used to find an expression for the thermal rate constant.

The main assumptions of transition state theory are:

1. Electronic and nuclear motions are separated (equivalent to the Born-Oppenheimer approximation in quantum mechanics)

2. Reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution

3. Molecular systems that have crossed the transition state in the direction of products cannot turn around and reform reactants

4. In the transition state, motion along the reaction coordinate may be separated from the other motions and treated classically as a translation

5. Even in the absence of an equilibrium between reactant and product molecules, the transition states that are becoming products are distributed among their states according to the Maxwell-Boltzmann laws

The goal of transition state theory is to predict the rate constant, ''k'', of a reaction given a potential energy surface for the reaction.

However, transition state theory suffers from a set of limitations due to the assumptions stated above not holding true in reality.

Firstly, transition state theory assumes that barrier recrossing is negligible. However, as seen from the fourth trajectory in table 1, barrier recrossing occurred because the atoms were too vibrationally excited and the system reverted back to the reactants even though the system had already crossed the transition state region. In this situation, transition state theory would predict a faster rate of reaction compared to what was obtained experimentally.

Secondly, the transition state theory breaks down for systems with light atoms such as hydrogens. In such systems, quantum effects become crucial. Quantum effects tend to increase the rate of the reaction for two reasons. First, particles lie in energy levels which are higher then the bottom of the potential energy well (there is additional zero-point energy). Secondly, the particles can tunnel through the barrier. These two effects effectively lower the activation energy barrier. As a result, the rate of reaction predicted by transition state theory would be less than what is obtained experimentally.

Lastly, transition state theory does not hold true for systems which are far from equilibrium. In such cases, the assumption of a starting Boltzmann distribution breaks down.

== Exercise 2: F - H - H system ==

=== Potential energy surface inspection===

F + HA-HB → HA-F + HB

[[File:01409469 Surface_Plot F 3.jpg|thumb|center|400px|'''Figure 10.''' Potential energy surface for the reaction between F and H2]]

The reaction between F and H2 is an exothermic reaction. This means that the reactants are of a higher energy than the products. The overall reaction involves the breaking of a hydrogen bond and the formation of a hydrogen-fluorine bond. The HF bond formed is stronger than the hydrogen bond that was broken. This is because fluorine is more electronegative than hydrogen resulting in a polar HF bond. The negative enthalpy of H-F bond formation is greater than the positive enthalpy of hydrogen bond dissociation. Therefore, the reaction between F and H2 is an exothermic reaction.

HA + HB-F → HA-HB + F

[[File:01409469 Surface_Plot F 4.jpg|thumb|center|400px|'''Figure 11.''' Potential energy surface for the reaction between H and HF]]

The reaction between H and HF is an endothermic reaction. The products are at a higher energy level than the reactants. The overall reaction involves the breaking of a HF bond and the formation of a hydrogen bond. The formation of a homonuclear H2 molecule results in a non-polar hydrogen bond being formed. The HF bond broken is stronger than the hydrogen bond that was formed. The negative enthalpy of hydrogen bond formation is less than the positive enthalpy of hydrogen-fluorine dissociation. Therefore, the reaction between H and HF is endothermic.

=== Locating the transition state ===

[[File:01409469 Surface_Plot F TS4.jpg|thumb|center|800px|'''Figure 12.''' Determining the position of the transition state for the reaction between F and H2]]

As mentioned previously, when the molecules are not at the transition state, a reaction path will be plotted. However, if the initial conditions of the system are set such that the trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

An initial estimate for the transition state was made by setting rAB = 0.74 which is the length of a hydrogen bond. The position of the transition state was further refined by increasing the number of decimal places such that the final value of rAB was found to be 0.74478 and rBC = 1.81079. This set of distances gave a reaction path that was represented by a point indicating that no trajectory was plotted and that the molecules remained at the transition state since their initial momentum was 0.

Moreover, the kinetic energy as well as the forces acting on the atoms was calculated to be 0, which is consistent with the transition state.

Therefore, the position of the transition state was found to be at rAB = 0.74478 and rBC = 1.81079 when pA = pBC = 0.

=== Determining the activation energy ===

[[File:01409469 activation energy r1.1 combined.jpg|thumb|center|800px|'''Figure 13.''' Determining the activation energy for the reaction between F and H2]]

By reducing the value of rBC by 0.01, the atoms are no longer at the transition state and would move towards the formation of products. The activation energy was found by calculating the change in total energy after the reaction. The activation energy of F + H2 → HF + H was found to be +23.279 kcal/ mol.

[[File:01409469 activation energy r2.1 combined.jpg|thumb|center|800px|'''Figure 14.''' Determining the activation energy for the reaction between FH and H]]

The activation energy of FH + H → F + H2 was found by calculating the small dip in potential energy. The activation energy was found to be +0.280 kcal/ mol.

=== Reaction dynamics of F + HA-HB → HA-F + HB ===

The set of initial conditions that result in a reactive trajectory for F + H2 is as follows: rHH = 0.74 Å, rHF = 1.79 Å and pHH = pHF = 0.

[[File:01409469 momentum r1.jpg|thumb|center|400px|'''Figure 15.''' Graph of momentum against time for the reaction between F and H2]]

Initially, the H2 molecule is travelling in a relatively straight line with very little oscillations as evidenced by the low amplitude of momentum between 0-2 s. Following successful collision between F and H2, the amplitude of momentum of the product, HF, increases dramatically. This is because the reaction between F and H2 is exothermic and the excess energy released causes the molecule to become vibrationally excited and the amplitude of oscillations in the product to increase.

[[File:01409469 energy1 r1.jpg|thumb|center|400px|'''Figure 16.''' Graph of energy against time for the reaction between F and H2]]

The law of conservation of energy states that energy cannot be created or destroyed. It simply changes from one form to another. As can be seen from the graph in figure 16, total energy remains constant. In this reaction, potential energy is converted to kinetic energy. A decrease in potential energy of the atoms following a successful collision between F and H2 results in an equal increase in kinetic energy which manifests in the form of increased product (HF) oscillations.

This can be confirmed experimentally using infrared chemiluminescence. Observation of radiation from excited product molecules is a clear indication that these molecules are produced in the excited state. If the excitation is primarily vibrational, the radiation will appear in the infrared region of the spectrum (between 3-15 μm). This technique has been used extensively in hydrogen halogen reactions which produce HX molecules in the excited state.

One such experiment involves the use of the arrested relaxation method. In this technique, the reaction is carried out in a chamber whose walls are maintained at a very low temperature (20 K). This results in cryopumping of the chamber in that all the products and unreacted starting materials are condensed on the walls and only those product molecules that radiate before being pumped away are observed.

=== Polanyi's empirical rules ===

In a typical chemical reaction, there is an energetic barrier — a saddle point that the reactants must surmount to become products. Which form of energy is initially deposited in the reactants, translational or vibrational, is more efficacious in surmounting the activation energy barrier is one of the core topics in the area of reaction dynamics. The Polyani rules state that vibrational energy is more effective at promoting a late-barrier reaction (transition state resembles products). Conversely, translational energy is better at promoting an early-barrier reaction (transition state resembles the reactants).

The forward reaction between H2 and F is exothermic. Therefore, the transition state is closer to the reactants than to the products in energy. Since the transition state occurs early in an exothermic reaction, translational energy is more effective at promoting a chemical reaction than vibrational energy.

Conversely, the backward reaction between H and HF is endothermic. According to Hammond's postulate, the transition state resembles the products more than it resembles the reactants. Therefore, in an endothermic reaction, a late transition state is observed and vibrational energy is more effective at promoting a reaction than translational energy.

=== Promoting the forward reaction: F + HA-HB → HA-F + HB ===

1. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 2.95 and pHF = -0.5.

[[File:01409469 polyani 1+2.jpg|thumb|center|800px|'''Figure 17.''' Contour plot of the reaction between F and H2 when pHH = 2.95 '''Figure 18.''' Graph of momentum against time for the reaction between F and H2 when pHH = 2.95 (Left-right)]]

Vibrational energy is represented by pHH while translational energy is represented by pHF.

A molecule of H2 collides with F. However, the collision is unsuccessful and no reaction occurs. A reaction did not occur even though the molecules have sufficient kinetic energy to overcome the activation energy barrier. This is because most of the energy is deposited in the reactants as H-H vibrations as seen by the large oscillations in H2. Barrier recrossing has occured as a result of the large vibrational energy causing the system to revert back to the reactants even though the system has already crossed the transition state and a bond was formed between HB and F temporarily.

This appears to run counter to what one would expect to happen when more energy is put into the system. It is predicted that when more energy is put into the system, the reactants would possess more kinetic energy to overcome the activation energy barrier. However, as can be seen from the simulation above, an increase in the vibrational energy of the reactants is not effective at promoting the occurrence of an exothermic reaction with an early transition state. This is in accordance to Polyani's rules.

2. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 0.1 and pHF = -0.8.

[[File:01409469 polyani 3+4 2.jpg|thumb|center|800px|'''Figure 19.''' Contour plot of the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 '''Figure 20.''' Graph of momentum against time for the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 (Left-right)]]

The collision between a molecule of H2 and F is successful and a chemical reaction has occurred. A reaction has occurred even though the overall energy of the system has been significantly reduced by lowering the amount of vibrational energy in the reactants but slightly increasing the amount of translational energy deposited in the reactants. Therefore, H2 and F have more translational energy to move towards each other and form a H-F bond. The product formed is vibrationally excited as seen by the large amplitude of oscillations in the product HF which correspond to the strong H-F vibrations.

From the second simulation, it can be seen that an increase in translational energy in the reactants is more effective at promoting an exothermic reaction with an early transition state to occur, which again follows Polyani's rules.

=== Promoting the backward reaction: HA + HB-F → HA-HB + F ===

1. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 0.2 and pHH = -3.0.

[[File:01409469 polyani 5+6 .jpg|thumb|center|800px|'''Figure 21.''' Contour plot of the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 '''Figure 22.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 (Left-right)]]

An atom of HA collides with a molecule of HBF. However, the collision is unsuccessful and no chemical reaction occurs. This is because the reactants lack sufficient vibrational energy to surmount the activation energy barrier. Therefore, following an unsuccessful collision with HBF, HA simply bounces off and moves away at constant momentum. This is in agreement with Polyani's rules which state that vibrational energy is more effective than translational energy at promoting reactions with a late transition state, for example, endothermic chemical reactions.

2. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 7.1 and pHH = -0.5.

[[File:01409469 polyani 7+8 .jpg|thumb|center|800px|'''Figure 23.''' Contour plot of the reaction between HA + HB-F when pFH = 7.1 and pHH = -0.5 '''Figure 24.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 7.1 and pHH = -0.5 (Left-right)]]

Increasing the amount of vibrational energy in the reactants by increasing pFH resulted in a chemical reaction successfully occurring. This is because the reactants now have sufficient kinetic energy in the form of H-F vibrations to overcome the activation energy barrier. In addition, because the reaction between HA + HB-F is endothermic, energy is taken in from the surroundings during the reaction and the products are less vibrationally excited than the reactants. This can be seen in the lower amplitude of oscillations in the product formed.

In conclusion, the vibrational energy of the reactants should be increased to promote an endothermic reaction with a late transition state while the translational energy of the reactants should be increased to promote an exothermic reaction with an early transition state. All 4 simulations shown above obey Polyani's rules.

== References ==

MRD:01409469

2019-05-10T14:31:27Z

Stc1917: /* Promoting the backward reaction: HA + HB-F → HA-HB + F */

== Exercise 1: H + H2 system ==

=== Definition of a transition state ===

A potential energy profile results from plotting the potential versus reaction path.

[[File:Surface_Plot01409469_pic1.jpg|thumb|center|400px|'''Figure 1.''' A potential energy surface plot]]

The transition state is defined as the maximum on the minimum energy path linking reactants and products. This means that the transition state is a stationary point on the potential energy surface. Thus, at the transition state, the derivative of the potential energy is zero with respect to each Cartesian coordinate.

<math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>

However, a local minimum on the potential energy surface corresponding to a well would have the same <math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>. Therefore, to distinguish between a local minimum on the potential energy surface and the transition state, the second derivative, <math>\tfrac{\partial^2 V(r_i)}{\partial r_i ^2}\ </math> is determined. This can be calculated by finding the determinant of the Hessian matrix shown below.

:<math>\mathbf H =
\begin{bmatrix}
\dfrac{\partial^2 \ V(r_i)}{\partial \ x^2} & \dfrac{\partial^2 \ V(r_i)}{\partial \ x \partial \mathbf y} \\
\dfrac{\partial^2 \ V(r_i)}{\partial \ y \partial \mathbf x} & \dfrac{\partial^2 \ V(r_i)}{\partial \mathbf y^2}
\end{bmatrix}</math>

The Hessian matrix of a convex function is positive semi-definite. Refining this property allows one to test if a critical point (x,y) is a local maximum, local minimum or saddle point.

If <math>f</math> < 0, then (x,y) is an isolated local maximum.

If <math>f</math> > 0, then (x,y) is an isolated local minimum.

If the Hessian has both positive and negative eigenvalues then (x,y) is a saddle point.

=== Locating the transition state ===

[[File:01409469 combined pic 2 and 3.jpg|thumb|center|800px|'''Figure 2.''' A surface and contour plot of the trajectory when rAB=rBC=0.907743 Å and pA=pBC=0]]

When the molecules are not at the transition state, a surface plot of the conditions shows a trajectory being plotted. This means that the trajectory is "rolling" towards the products or reactants. However, if a trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

[[File:01409469_internuclearpic3.png|thumb|center|400px|'''Figure 3.''' A plot of internuclear distance against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

An initial estimate for the position of the transition state was made by observing the potential energy surface and was found to be 0.90 Å. At 0.90 Å, the internuclear distance against time plot showed that the internuclear distance was constantly oscillating. Therefore, the position of the transition state was further refined by increasing the number of decimal places until a horizontal line in the internuclear distance againt time plot was obtained. This means that the internuclear distance is constant and that the molecules are simply undergoing periodic symmetric vibration at the transition state. Therefore, the position of the transition state was determined to be 0.907743 Å.

[[File:01409469 combined energy pic.png|thumb|center|400px|'''Figure 4.''' A plot of energy against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

When rAB=rBC=0.907743 Å and pA=pBC=0, the kinetic energy is 0. This means that HA and HBC are stationary, which is as predicted when the trajectory is started at the transition state with no initial momentum. Furthermore, the forces acting on HA and HBC are 0 as should be the case in the transition state.

=== Calculation type: Dynamic and minimum energy path ===

The minimum energy path, is the lowest energy path from the reactants to the products. The minimum energy path is determined by following the mode corresponding to the negative Hessian eigenvalue in both directions along the steepest descent direction. Following this path allows one to approach two valleys of stable structures. The resulting path generated is the minimum energy path. The reaction paths were chartered for a dynamic and a minimum energy path. The initial conditions used were that of the transition state except that the internuclear distance between HA and HB had increased to 0.917743 Å. The reaction path obtained using the two different calculation methods was different.

[[File:01409469 mep and dynamics.jpg|thumb|center|800px|'''Figure 5.''' A surface plot of the trajectory using minimum energy path '''Figure 6.''' A surface plot of the trajectory using dynamics (Left-right)]]

On the surface plot, the reaction path is depicted by the black line. Following the minimum energy path, the trajectory simply follows the valley floor to Ha+Hb-Hc. The reaction path appears to be a relatively straight line indicating that the atoms are not vibrating. In the minimum energy path, the reactants change into the products in an infinitely slow motion. This is done by having the momenta/velocities reset to zero in each time step. While the minimum energy path is useful in characterising a chemical reaction, it does not provide a realistic representation of the motion of the atoms during the reaction. This is because the mass and inertia of the atoms have been ignored.

Since the atoms simply follow the minimum energy path, they do not undergo slight displacements into surrounding valleys of higher potential energy. Furthermore, because the atoms are assumed to have no inertia, they do not undergo vibrations even if they were to be displaced to valleys of higher potential energy.

However, when dynamic calculation is performed, the reaction path is wavy indicating that the atoms are oscillating. This is because the atoms are displaced to regions of higher potential on the path from reactants to products. This leads to atom oscillation due to the inertia of the atoms.

'''Internuclear distance against time plot'''

Setting the initial system conditions such that rAB= 0.907743, rBC= 0.917743 and pA= pBC= 0:

[[File:01409469 Internuclear distance mep and dynamic.jpg|thumb|center|800px|'''Figure 6.''' A graph of internuclear distance against steps using minimum energy path '''Figure 7.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In the minimum energy path, the distance between HA and HB decreases from 0.90 Å to approximately 0.79 Å during the first 50 steps before plateauing off to a value of 0.74 Å which is the approximate length of a hydrogen bond. This means that a bond was formed between HA and HB. In addition, the plateau indicates that the bond is static. On the other hand, the distance between HB and HC has increased to 1.91 Å at large step values. This shows that HC has been ejected off following a successful collision between HA and HBC.

Similarly, in the dynamic trajectory, the distance between HA and HB decreases to approximately 0.75 Å after 0.3 s. This is indicative of hydrogen bond formation between HA and HB. In addition, the distance between HB and HC increases to 3.44 Å after HC is ejected following a successful collision.

One obvious distance between the two plots is that in the minimum energy path trajectory, the internuclear distance between HA and HB is constant indicating that the bond is static. However, in the dynamic trajectory the internuclear distance is oscillating showing the the bond is vibrating. This difference could be attributed to the fact that minimum energy path does not take oscillatory motion into account while dynamic trajectory does.

However, changing the initial conditions of the system such that rAB = 0.917743 Å, rBC = 0.907743 Å and pA=pBC=0:
[[File:01409469 01409469 Reversed values dynamic mep.jpg|thumb|center|800px|'''Figure 8.''' A graph of internuclear distance against steps using minimum energy path '''Figure 9.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In this case, a hydrogen bond of length 0.77-0.79 Å is formed between HB and HC while HA is repelled away to beyond 3.45 Å.

Additionally, from the internuclear distance plot it can be seen that the distance between HB and HC increases in a rectangular hyperbola fashion when calculated using the minimum energy path. On the other hand, the internuclear distance between HB and HC increases in a linear manner when dynamic calculation is applied. A possible reason for this difference is that in the minimum energy path, the atoms are not vibrating and as such the reach the maximum distance and plateau at a faster rate. However, in the dynamic calculation, the atoms have momentum and as such move further away from each other.

=== Reactive and unreactive trajectories ===

With the initial positions r'''1''' = 0.74 and r'''2''' = 2.0, trajectories were run with the momenta combination shown below.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Contour plot !! Reactivity !! Description of the dynamics
|-
| -1.25 || -2.5 ||-99.018 ||[[File:01409469 trajectory1.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. There are no oscillations observed in HA-HB initially, indicating that the diatomic molecule is stable. A new bond is formed between HB and HC. Oscillatory behaviour is observed indicating that the bond is vibrating. On the other hand, HA is ejected away to beyond 10 Å.
|-
| -1.5 || -2.0 ||-100.456 ||[[File:01409469 trajectory2.jpg|thumb|center|400px]] ||Unreactive ||HC collides with the vibrating diatomic HA-HB. However, no bond is formed between HB and HC as the collision is unsuccessful. This is because the molecules lack sufficient kinetic energy to overcome the activation energy barrier and so no reaction occurs. HC is subsequently repelled to beyond 8 Å.
|-
| -1.5 || -2.5 ||-98.956 ||[[File:01409469 trajectory3.jpg|thumb|center|400px]] ||Reactive |||HC collides with HA-HB. The collision is successful and a bond is formed between HB and HC. This is because the atoms have sufficient kinetic energy to overcome the activation energy barrier and form a new HB-HC bond. HA is ejected away to beyond 10 Å. The new hydrogen bond formed is vibrating as seen by the oscillatory behaviour.
|-
| -2.5 || -5.0 ||-84.956 ||[[File:01409469 trajectory4.jpg|thumb|center|400px]] ||Unreactive ||HC collides with HA-HB. Initially the HA-HB bond is elongating, indicating that the hydrogen bond is weakening and going to break. However, after 0.5 s, the HA-HB bond length decreases and the HAHB is not broken. HC is repelled beyond 6 Å with very strong vibrations.
A reaction did not occur even though the atoms have sufficient kinetic energy to cross the activation energy barrier. This is because the vibrations are too strong. The system crosses the transition state region and a bond is formed in the product temporarily but then the system reverts back to the reactants. This is an example of barrier recrossing.
|-
| -2.5 || -5.2 ||-83.416 ||[[File:01409469 trajectory5.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. The collision is successful and a new hydrogen bond is formed between HB and HC with HA being ejected away. The products formed experience very strong vibrations as evidenced by the oscillatory motion which occur with a very high frequency.
|}

From the table, it can be seen that the chance of a reaction successfully occurring is not solely dependent on the amount of kinetic energy the atoms possess. Even if the atoms have sufficient kinetic energy to cross the activation energy barrier, barrier recrossing can occur if the atoms are too vibrationally excited.

=== Transition State Theory ===

Transition state theory provided the first theoretical attempt to determine absolute reaction rates. In this theory, a transition state separating the reactants from products is used to find an expression for the thermal rate constant.

The main assumptions of transition state theory are:

1. Electronic and nuclear motions are separated (equivalent to the Born-Oppenheimer approximation in quantum mechanics)

2. Reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution

3. Molecular systems that have crossed the transition state in the direction of products cannot turn around and reform reactants

4. In the transition state, motion along the reaction coordinate may be separated from the other motions and treated classically as a translation

5. Even in the absence of an equilibrium between reactant and product molecules, the transition states that are becoming products are distributed among their states according to the Maxwell-Boltzmann laws

The goal of transition state theory is to predict the rate constant, ''k'', of a reaction given a potential energy surface for the reaction.

However, transition state theory suffers from a set of limitations due to the assumptions stated above not holding true in reality.

Firstly, transition state theory assumes that barrier recrossing is negligible. However, as seen from the fourth trajectory in table 1, barrier recrossing occurred because the atoms were too vibrationally excited and the system reverted back to the reactants even though the system had already crossed the transition state region. In this situation, transition state theory would predict a faster rate of reaction compared to what was obtained experimentally.

Secondly, the transition state theory breaks down for systems with light atoms such as hydrogens. In such systems, quantum effects become crucial. Quantum effects tend to increase the rate of the reaction for two reasons. First, particles lie in energy levels which are higher then the bottom of the potential energy well (there is additional zero-point energy). Secondly, the particles can tunnel through the barrier. These two effects effectively lower the activation energy barrier. As a result, the rate of reaction predicted by transition state theory would be less than what is obtained experimentally.

Lastly, transition state theory does not hold true for systems which are far from equilibrium. In such cases, the assumption of a starting Boltzmann distribution breaks down.

== Exercise 2: F - H - H system ==

=== Potential energy surface inspection===

F + HA-HB → HA-F + HB

[[File:01409469 Surface_Plot F 3.jpg|thumb|center|400px|'''Figure 10.''' Potential energy surface for the reaction between F and H2]]

The reaction between F and H2 is an exothermic reaction. This means that the reactants are of a higher energy than the products. The overall reaction involves the breaking of a hydrogen bond and the formation of a hydrogen-fluorine bond. The HF bond formed is stronger than the hydrogen bond that was broken. This is because fluorine is more electronegative than hydrogen resulting in a polar HF bond. The negative enthalpy of H-F bond formation is greater than the positive enthalpy of hydrogen bond dissociation. Therefore, the reaction between F and H2 is an exothermic reaction.

HA + HB-F → HA-HB + F

[[File:01409469 Surface_Plot F 4.jpg|thumb|center|400px|'''Figure 11.''' Potential energy surface for the reaction between H and HF]]

The reaction between H and HF is an endothermic reaction. The products are at a higher energy level than the reactants. The overall reaction involves the breaking of a HF bond and the formation of a hydrogen bond. The formation of a homonuclear H2 molecule results in a non-polar hydrogen bond being formed. The HF bond broken is stronger than the hydrogen bond that was formed. The negative enthalpy of hydrogen bond formation is less than the positive enthalpy of hydrogen-fluorine dissociation. Therefore, the reaction between H and HF is endothermic.

=== Locating the transition state ===

[[File:01409469 Surface_Plot F TS4.jpg|thumb|center|800px|'''Figure 12.''' Determining the position of the transition state for the reaction between F and H2]]

As mentioned previously, when the molecules are not at the transition state, a reaction path will be plotted. However, if the initial conditions of the system are set such that the trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

An initial estimate for the transition state was made by setting rAB = 0.74 which is the length of a hydrogen bond. The position of the transition state was further refined by increasing the number of decimal places such that the final value of rAB was found to be 0.74478 and rBC = 1.81079. This set of distances gave a reaction path that was represented by a point indicating that no trajectory was plotted and that the molecules remained at the transition state since their initial momentum was 0.

Moreover, the kinetic energy as well as the forces acting on the atoms was calculated to be 0, which is consistent with the transition state.

Therefore, the position of the transition state was found to be at rAB = 0.74478 and rBC = 1.81079 when pA = pBC = 0.

=== Determining the activation energy ===

[[File:01409469 activation energy r1.1 combined.jpg|thumb|center|800px|'''Figure 13.''' Determining the activation energy for the reaction between F and H2]]

By reducing the value of rBC by 0.01, the atoms are no longer at the transition state and would move towards the formation of products. The activation energy was found by calculating the change in total energy after the reaction. The activation energy of F + H2 → HF + H was found to be +23.279 kcal/ mol.

[[File:01409469 activation energy r2.1 combined.jpg|thumb|center|800px|'''Figure 14.''' Determining the activation energy for the reaction between FH and H]]

The activation energy of FH + H → F + H2 was found by calculating the small dip in potential energy. The activation energy was found to be +0.280 kcal/ mol.

=== Reaction dynamics of F + HA-HB → HA-F + HB ===

The set of initial conditions that result in a reactive trajectory for F + H2 is as follows: rHH = 0.74 Å, rHF = 1.79 Å and pHH = pHF = 0.

[[File:01409469 momentum r1.jpg|thumb|center|400px|'''Figure 15.''' Graph of momentum against time for the reaction between F and H2]]

Initially, the H2 molecule is travelling in a relatively straight line with very little oscillations as evidenced by the low amplitude of momentum between 0-2 s. Following successful collision between F and H2, the amplitude of momentum of the product, HF, increases dramatically. This is because the reaction between F and H2 is exothermic and the excess energy released causes the molecule to become vibrationally excited and the amplitude of oscillations in the product to increase.

[[File:01409469 energy1 r1.jpg|thumb|center|400px|'''Figure 16.''' Graph of energy against time for the reaction between F and H2]]

The law of conservation of energy states that energy cannot be created or destroyed. It simply changes from one form to another. As can be seen from the graph in figure 16, total energy remains constant. In this reaction, potential energy is converted to kinetic energy. A decrease in potential energy of the atoms following a successful collision between F and H2 results in an equal increase in kinetic energy which manifests in the form of increased product (HF) oscillations.

This can be confirmed experimentally using infrared chemiluminescence. Observation of radiation from excited product molecules is a clear indication that these molecules are produced in the excited state. If the excitation is primarily vibrational, the radiation will appear in the infrared region of the spectrum (between 3-15 μm). This technique has been used extensively in hydrogen halogen reactions which produce HX molecules in the excited state.

One such experiment involves the use of the arrested relaxation method. In this technique, the reaction is carried out in a chamber whose walls are maintained at a very low temperature (20 K). This results in cryopumping of the chamber in that all the products and unreacted starting materials are condensed on the walls and only those product molecules that radiate before being pumped away are observed.

=== Polanyi's empirical rules ===

In a typical chemical reaction, there is an energetic barrier — a saddle point that the reactants must surmount to become products. Which form of energy is initially deposited in the reactants, translational or vibrational, is more efficacious in surmounting the activation energy barrier is one of the core topics in the area of reaction dynamics. The Polyani rules state that vibrational energy is more effective at promoting a late-barrier reaction (transition state resembles products). Conversely, translational energy is better at promoting an early-barrier reaction (transition state resembles the reactants).

The forward reaction between H2 and F is exothermic. Therefore, the transition state is closer to the reactants than to the products in energy. Since the transition state occurs early in an exothermic reaction, translational energy is more effective at promoting a chemical reaction than vibrational energy.

Conversely, the backward reaction between H and HF is endothermic. According to Hammond's postulate, the transition state resembles the products more than it resembles the reactants. Therefore, in an endothermic reaction, a late transition state is observed and vibrational energy is more effective at promoting a reaction than translational energy.

=== Promoting the forward reaction: F + HA-HB → HA-F + HB ===

1. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 2.95 and pHF = -0.5.

[[File:01409469 polyani 1+2.jpg|thumb|center|800px|'''Figure 17.''' Contour plot of the reaction between F and H2 when pHH = 2.95 '''Figure 18.''' Graph of momentum against time for the reaction between F and H2 when pHH = 2.95 (Left-right)]]

Vibrational energy is represented by pHH while translational energy is represented by pHF.

A molecule of H2 collides with F. However, the collision is unsuccessful and no reaction occurs. A reaction did not occur even though the molecules have sufficient kinetic energy to overcome the activation energy barrier. This is because most of the energy is deposited in the reactants as H-H vibrations as seen by the large oscillations in H2. Barrier recrossing has occured as a result of the large vibrational energy causing the system to revert back to the reactants even though the system has already crossed the transition state and a bond was formed between HB and F temporarily.

This appears to run counter to what one would expect to happen when more energy is put into the system. It is predicted that when more energy is put into the system, the reactants would possess more kinetic energy to overcome the activation energy barrier. However, as can be seen from the simulation above, an increase in the vibrational energy of the reactants is not effective at promoting the occurrence of an exothermic reaction with an early transition state. This is in accordance to Polyani's rules.

2. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 0.1 and pHF = -0.8.

[[File:01409469 polyani 3+4 2.jpg|thumb|center|800px|'''Figure 19.''' Contour plot of the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 '''Figure 20.''' Graph of momentum against time for the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 (Left-right)]]

The collision between a molecule of H2 and F is successful and a chemical reaction has occurred. A reaction has occurred even though the overall energy of the system has been significantly reduced by lowering the amount of vibrational energy in the reactants but slightly increasing the amount of translational energy deposited in the reactants. Therefore, H2 and F have more translational energy to move towards each other and form a H-F bond. The product formed is vibrationally excited as seen by the large amplitude of oscillations in the product HF which correspond to the strong H-F vibrations.

From the second simulation, it can be seen that an increase in translational energy in the reactants is more effective at promoting an exothermic reaction with an early transition state to occur, which again follows Polyani's rules.

=== Promoting the backward reaction: HA + HB-F → HA-HB + F ===

1. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 0.2 and pHH = -3.0.

[[File:01409469 polyani 5+6 .jpg|thumb|center|800px|'''Figure 21.''' Contour plot of the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 '''Figure 22.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 (Left-right)]]

An atom of HA collides with a molecule of HBF. However, the collision is unsuccessful and no chemical reaction occurs. This is because the reactants lack sufficient vibrational energy to surmount the activation energy barrier. Therefore, following an unsuccessful collision with HBF, HA simply bounces off and moves away at constant momentum. This is in agreement with Polyani's rules which state that vibrational energy is more effective than translational energy at promoting reactions with a late transition state, for example, endothermic chemical reactions.

2. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 7.1 and pHH = -0.5.

[[File:01409469 polyani 7+8 .jpg|thumb|center|800px|'''Figure 23.''' Contour plot of the reaction between HA + HB-F when pFH = 7.1 and pHH = -0.5 '''Figure 24.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 7.1 and pHH = -0.5 (Left-right)]]

Increasing the amount of vibrational energy in the reactants by increasing pFH resulted in a chemical reaction successfully occurring. This is because the reactants now have sufficient kinetic energy in the form of H-F vibrations to overcome the activation energy barrier. In addition, because the reaction between HA + HB-F is endothermic, energy is taken in from the surroundings during the reaction and the products are less vibrationally excited than the reactants. This can be seen in the lower amplitude of oscillations in the product formed.

In conclusion, the vibrational energy of the reactants should be increased to promote an endothermic reaction with a late transition state while the translational energy of the reactants should be increased to promote an exothermic reaction with an early transition state. All 4 simulations shown above obey Polyani's rules.

File:01409469 polyani 7+8 .jpg

2019-05-10T14:26:42Z

Stc1917:

MRD:01409469

2019-05-10T14:23:14Z

Stc1917: /* Promoting the backward reaction: HA + HB-F → HA-HB + F */

== Exercise 1: H + H2 system ==

=== Definition of a transition state ===

A potential energy profile results from plotting the potential versus reaction path.

[[File:Surface_Plot01409469_pic1.jpg|thumb|center|400px|'''Figure 1.''' A potential energy surface plot]]

The transition state is defined as the maximum on the minimum energy path linking reactants and products. This means that the transition state is a stationary point on the potential energy surface. Thus, at the transition state, the derivative of the potential energy is zero with respect to each Cartesian coordinate.

<math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>

However, a local minimum on the potential energy surface corresponding to a well would have the same <math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>. Therefore, to distinguish between a local minimum on the potential energy surface and the transition state, the second derivative, <math>\tfrac{\partial^2 V(r_i)}{\partial r_i ^2}\ </math> is determined. This can be calculated by finding the determinant of the Hessian matrix shown below.

:<math>\mathbf H =
\begin{bmatrix}
\dfrac{\partial^2 \ V(r_i)}{\partial \ x^2} & \dfrac{\partial^2 \ V(r_i)}{\partial \ x \partial \mathbf y} \\
\dfrac{\partial^2 \ V(r_i)}{\partial \ y \partial \mathbf x} & \dfrac{\partial^2 \ V(r_i)}{\partial \mathbf y^2}
\end{bmatrix}</math>

The Hessian matrix of a convex function is positive semi-definite. Refining this property allows one to test if a critical point (x,y) is a local maximum, local minimum or saddle point.

If <math>f</math> < 0, then (x,y) is an isolated local maximum.

If <math>f</math> > 0, then (x,y) is an isolated local minimum.

If the Hessian has both positive and negative eigenvalues then (x,y) is a saddle point.

=== Locating the transition state ===

[[File:01409469 combined pic 2 and 3.jpg|thumb|center|800px|'''Figure 2.''' A surface and contour plot of the trajectory when rAB=rBC=0.907743 Å and pA=pBC=0]]

When the molecules are not at the transition state, a surface plot of the conditions shows a trajectory being plotted. This means that the trajectory is "rolling" towards the products or reactants. However, if a trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

[[File:01409469_internuclearpic3.png|thumb|center|400px|'''Figure 3.''' A plot of internuclear distance against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

An initial estimate for the position of the transition state was made by observing the potential energy surface and was found to be 0.90 Å. At 0.90 Å, the internuclear distance against time plot showed that the internuclear distance was constantly oscillating. Therefore, the position of the transition state was further refined by increasing the number of decimal places until a horizontal line in the internuclear distance againt time plot was obtained. This means that the internuclear distance is constant and that the molecules are simply undergoing periodic symmetric vibration at the transition state. Therefore, the position of the transition state was determined to be 0.907743 Å.

[[File:01409469 combined energy pic.png|thumb|center|400px|'''Figure 4.''' A plot of energy against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

When rAB=rBC=0.907743 Å and pA=pBC=0, the kinetic energy is 0. This means that HA and HBC are stationary, which is as predicted when the trajectory is started at the transition state with no initial momentum. Furthermore, the forces acting on HA and HBC are 0 as should be the case in the transition state.

=== Calculation type: Dynamic and minimum energy path ===

The minimum energy path, is the lowest energy path from the reactants to the products. The minimum energy path is determined by following the mode corresponding to the negative Hessian eigenvalue in both directions along the steepest descent direction. Following this path allows one to approach two valleys of stable structures. The resulting path generated is the minimum energy path. The reaction paths were chartered for a dynamic and a minimum energy path. The initial conditions used were that of the transition state except that the internuclear distance between HA and HB had increased to 0.917743 Å. The reaction path obtained using the two different calculation methods was different.

[[File:01409469 mep and dynamics.jpg|thumb|center|800px|'''Figure 5.''' A surface plot of the trajectory using minimum energy path '''Figure 6.''' A surface plot of the trajectory using dynamics (Left-right)]]

On the surface plot, the reaction path is depicted by the black line. Following the minimum energy path, the trajectory simply follows the valley floor to Ha+Hb-Hc. The reaction path appears to be a relatively straight line indicating that the atoms are not vibrating. In the minimum energy path, the reactants change into the products in an infinitely slow motion. This is done by having the momenta/velocities reset to zero in each time step. While the minimum energy path is useful in characterising a chemical reaction, it does not provide a realistic representation of the motion of the atoms during the reaction. This is because the mass and inertia of the atoms have been ignored.

Since the atoms simply follow the minimum energy path, they do not undergo slight displacements into surrounding valleys of higher potential energy. Furthermore, because the atoms are assumed to have no inertia, they do not undergo vibrations even if they were to be displaced to valleys of higher potential energy.

However, when dynamic calculation is performed, the reaction path is wavy indicating that the atoms are oscillating. This is because the atoms are displaced to regions of higher potential on the path from reactants to products. This leads to atom oscillation due to the inertia of the atoms.

'''Internuclear distance against time plot'''

Setting the initial system conditions such that rAB= 0.907743, rBC= 0.917743 and pA= pBC= 0:

[[File:01409469 Internuclear distance mep and dynamic.jpg|thumb|center|800px|'''Figure 6.''' A graph of internuclear distance against steps using minimum energy path '''Figure 7.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In the minimum energy path, the distance between HA and HB decreases from 0.90 Å to approximately 0.79 Å during the first 50 steps before plateauing off to a value of 0.74 Å which is the approximate length of a hydrogen bond. This means that a bond was formed between HA and HB. In addition, the plateau indicates that the bond is static. On the other hand, the distance between HB and HC has increased to 1.91 Å at large step values. This shows that HC has been ejected off following a successful collision between HA and HBC.

Similarly, in the dynamic trajectory, the distance between HA and HB decreases to approximately 0.75 Å after 0.3 s. This is indicative of hydrogen bond formation between HA and HB. In addition, the distance between HB and HC increases to 3.44 Å after HC is ejected following a successful collision.

One obvious distance between the two plots is that in the minimum energy path trajectory, the internuclear distance between HA and HB is constant indicating that the bond is static. However, in the dynamic trajectory the internuclear distance is oscillating showing the the bond is vibrating. This difference could be attributed to the fact that minimum energy path does not take oscillatory motion into account while dynamic trajectory does.

However, changing the initial conditions of the system such that rAB = 0.917743 Å, rBC = 0.907743 Å and pA=pBC=0:
[[File:01409469 01409469 Reversed values dynamic mep.jpg|thumb|center|800px|'''Figure 8.''' A graph of internuclear distance against steps using minimum energy path '''Figure 9.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In this case, a hydrogen bond of length 0.77-0.79 Å is formed between HB and HC while HA is repelled away to beyond 3.45 Å.

Additionally, from the internuclear distance plot it can be seen that the distance between HB and HC increases in a rectangular hyperbola fashion when calculated using the minimum energy path. On the other hand, the internuclear distance between HB and HC increases in a linear manner when dynamic calculation is applied. A possible reason for this difference is that in the minimum energy path, the atoms are not vibrating and as such the reach the maximum distance and plateau at a faster rate. However, in the dynamic calculation, the atoms have momentum and as such move further away from each other.

=== Reactive and unreactive trajectories ===

With the initial positions r'''1''' = 0.74 and r'''2''' = 2.0, trajectories were run with the momenta combination shown below.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Contour plot !! Reactivity !! Description of the dynamics
|-
| -1.25 || -2.5 ||-99.018 ||[[File:01409469 trajectory1.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. There are no oscillations observed in HA-HB initially, indicating that the diatomic molecule is stable. A new bond is formed between HB and HC. Oscillatory behaviour is observed indicating that the bond is vibrating. On the other hand, HA is ejected away to beyond 10 Å.
|-
| -1.5 || -2.0 ||-100.456 ||[[File:01409469 trajectory2.jpg|thumb|center|400px]] ||Unreactive ||HC collides with the vibrating diatomic HA-HB. However, no bond is formed between HB and HC as the collision is unsuccessful. This is because the molecules lack sufficient kinetic energy to overcome the activation energy barrier and so no reaction occurs. HC is subsequently repelled to beyond 8 Å.
|-
| -1.5 || -2.5 ||-98.956 ||[[File:01409469 trajectory3.jpg|thumb|center|400px]] ||Reactive |||HC collides with HA-HB. The collision is successful and a bond is formed between HB and HC. This is because the atoms have sufficient kinetic energy to overcome the activation energy barrier and form a new HB-HC bond. HA is ejected away to beyond 10 Å. The new hydrogen bond formed is vibrating as seen by the oscillatory behaviour.
|-
| -2.5 || -5.0 ||-84.956 ||[[File:01409469 trajectory4.jpg|thumb|center|400px]] ||Unreactive ||HC collides with HA-HB. Initially the HA-HB bond is elongating, indicating that the hydrogen bond is weakening and going to break. However, after 0.5 s, the HA-HB bond length decreases and the HAHB is not broken. HC is repelled beyond 6 Å with very strong vibrations.
A reaction did not occur even though the atoms have sufficient kinetic energy to cross the activation energy barrier. This is because the vibrations are too strong. The system crosses the transition state region and a bond is formed in the product temporarily but then the system reverts back to the reactants. This is an example of barrier recrossing.
|-
| -2.5 || -5.2 ||-83.416 ||[[File:01409469 trajectory5.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. The collision is successful and a new hydrogen bond is formed between HB and HC with HA being ejected away. The products formed experience very strong vibrations as evidenced by the oscillatory motion which occur with a very high frequency.
|}

From the table, it can be seen that the chance of a reaction successfully occurring is not solely dependent on the amount of kinetic energy the atoms possess. Even if the atoms have sufficient kinetic energy to cross the activation energy barrier, barrier recrossing can occur if the atoms are too vibrationally excited.

=== Transition State Theory ===

Transition state theory provided the first theoretical attempt to determine absolute reaction rates. In this theory, a transition state separating the reactants from products is used to find an expression for the thermal rate constant.

The main assumptions of transition state theory are:

1. Electronic and nuclear motions are separated (equivalent to the Born-Oppenheimer approximation in quantum mechanics)

2. Reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution

3. Molecular systems that have crossed the transition state in the direction of products cannot turn around and reform reactants

4. In the transition state, motion along the reaction coordinate may be separated from the other motions and treated classically as a translation

5. Even in the absence of an equilibrium between reactant and product molecules, the transition states that are becoming products are distributed among their states according to the Maxwell-Boltzmann laws

The goal of transition state theory is to predict the rate constant, ''k'', of a reaction given a potential energy surface for the reaction.

However, transition state theory suffers from a set of limitations due to the assumptions stated above not holding true in reality.

Firstly, transition state theory assumes that barrier recrossing is negligible. However, as seen from the fourth trajectory in table 1, barrier recrossing occurred because the atoms were too vibrationally excited and the system reverted back to the reactants even though the system had already crossed the transition state region. In this situation, transition state theory would predict a faster rate of reaction compared to what was obtained experimentally.

Secondly, the transition state theory breaks down for systems with light atoms such as hydrogens. In such systems, quantum effects become crucial. Quantum effects tend to increase the rate of the reaction for two reasons. First, particles lie in energy levels which are higher then the bottom of the potential energy well (there is additional zero-point energy). Secondly, the particles can tunnel through the barrier. These two effects effectively lower the activation energy barrier. As a result, the rate of reaction predicted by transition state theory would be less than what is obtained experimentally.

Lastly, transition state theory does not hold true for systems which are far from equilibrium. In such cases, the assumption of a starting Boltzmann distribution breaks down.

== Exercise 2: F - H - H system ==

=== Potential energy surface inspection===

F + HA-HB → HA-F + HB

[[File:01409469 Surface_Plot F 3.jpg|thumb|center|400px|'''Figure 10.''' Potential energy surface for the reaction between F and H2]]

The reaction between F and H2 is an exothermic reaction. This means that the reactants are of a higher energy than the products. The overall reaction involves the breaking of a hydrogen bond and the formation of a hydrogen-fluorine bond. The HF bond formed is stronger than the hydrogen bond that was broken. This is because fluorine is more electronegative than hydrogen resulting in a polar HF bond. The negative enthalpy of H-F bond formation is greater than the positive enthalpy of hydrogen bond dissociation. Therefore, the reaction between F and H2 is an exothermic reaction.

HA + HB-F → HA-HB + F

[[File:01409469 Surface_Plot F 4.jpg|thumb|center|400px|'''Figure 11.''' Potential energy surface for the reaction between H and HF]]

The reaction between H and HF is an endothermic reaction. The products are at a higher energy level than the reactants. The overall reaction involves the breaking of a HF bond and the formation of a hydrogen bond. The formation of a homonuclear H2 molecule results in a non-polar hydrogen bond being formed. The HF bond broken is stronger than the hydrogen bond that was formed. The negative enthalpy of hydrogen bond formation is less than the positive enthalpy of hydrogen-fluorine dissociation. Therefore, the reaction between H and HF is endothermic.

=== Locating the transition state ===

[[File:01409469 Surface_Plot F TS4.jpg|thumb|center|800px|'''Figure 12.''' Determining the position of the transition state for the reaction between F and H2]]

As mentioned previously, when the molecules are not at the transition state, a reaction path will be plotted. However, if the initial conditions of the system are set such that the trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

An initial estimate for the transition state was made by setting rAB = 0.74 which is the length of a hydrogen bond. The position of the transition state was further refined by increasing the number of decimal places such that the final value of rAB was found to be 0.74478 and rBC = 1.81079. This set of distances gave a reaction path that was represented by a point indicating that no trajectory was plotted and that the molecules remained at the transition state since their initial momentum was 0.

Moreover, the kinetic energy as well as the forces acting on the atoms was calculated to be 0, which is consistent with the transition state.

Therefore, the position of the transition state was found to be at rAB = 0.74478 and rBC = 1.81079 when pA = pBC = 0.

=== Determining the activation energy ===

[[File:01409469 activation energy r1.1 combined.jpg|thumb|center|800px|'''Figure 13.''' Determining the activation energy for the reaction between F and H2]]

By reducing the value of rBC by 0.01, the atoms are no longer at the transition state and would move towards the formation of products. The activation energy was found by calculating the change in total energy after the reaction. The activation energy of F + H2 → HF + H was found to be +23.279 kcal/ mol.

[[File:01409469 activation energy r2.1 combined.jpg|thumb|center|800px|'''Figure 14.''' Determining the activation energy for the reaction between FH and H]]

The activation energy of FH + H → F + H2 was found by calculating the small dip in potential energy. The activation energy was found to be +0.280 kcal/ mol.

=== Reaction dynamics of F + HA-HB → HA-F + HB ===

The set of initial conditions that result in a reactive trajectory for F + H2 is as follows: rHH = 0.74 Å, rHF = 1.79 Å and pHH = pHF = 0.

[[File:01409469 momentum r1.jpg|thumb|center|400px|'''Figure 15.''' Graph of momentum against time for the reaction between F and H2]]

Initially, the H2 molecule is travelling in a relatively straight line with very little oscillations as evidenced by the low amplitude of momentum between 0-2 s. Following successful collision between F and H2, the amplitude of momentum of the product, HF, increases dramatically. This is because the reaction between F and H2 is exothermic and the excess energy released causes the molecule to become vibrationally excited and the amplitude of oscillations in the product to increase.

[[File:01409469 energy1 r1.jpg|thumb|center|400px|'''Figure 16.''' Graph of energy against time for the reaction between F and H2]]

The law of conservation of energy states that energy cannot be created or destroyed. It simply changes from one form to another. As can be seen from the graph in figure 16, total energy remains constant. In this reaction, potential energy is converted to kinetic energy. A decrease in potential energy of the atoms following a successful collision between F and H2 results in an equal increase in kinetic energy which manifests in the form of increased product (HF) oscillations.

This can be confirmed experimentally using infrared chemiluminescence. Observation of radiation from excited product molecules is a clear indication that these molecules are produced in the excited state. If the excitation is primarily vibrational, the radiation will appear in the infrared region of the spectrum (between 3-15 μm). This technique has been used extensively in hydrogen halogen reactions which produce HX molecules in the excited state.

One such experiment involves the use of the arrested relaxation method. In this technique, the reaction is carried out in a chamber whose walls are maintained at a very low temperature (20 K). This results in cryopumping of the chamber in that all the products and unreacted starting materials are condensed on the walls and only those product molecules that radiate before being pumped away are observed.

=== Polanyi's empirical rules ===

In a typical chemical reaction, there is an energetic barrier — a saddle point that the reactants must surmount to become products. Which form of energy is initially deposited in the reactants, translational or vibrational, is more efficacious in surmounting the activation energy barrier is one of the core topics in the area of reaction dynamics. The Polyani rules state that vibrational energy is more effective at promoting a late-barrier reaction (transition state resembles products). Conversely, translational energy is better at promoting an early-barrier reaction (transition state resembles the reactants).

The forward reaction between H2 and F is exothermic. Therefore, the transition state is closer to the reactants than to the products in energy. Since the transition state occurs early in an exothermic reaction, translational energy is more effective at promoting a chemical reaction than vibrational energy.

Conversely, the backward reaction between H and HF is endothermic. According to Hammond's postulate, the transition state resembles the products more than it resembles the reactants. Therefore, in an endothermic reaction, a late transition state is observed and vibrational energy is more effective at promoting a reaction than translational energy.

=== Promoting the forward reaction: F + HA-HB → HA-F + HB ===

1. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 2.95 and pHF = -0.5.

[[File:01409469 polyani 1+2.jpg|thumb|center|800px|'''Figure 17.''' Contour plot of the reaction between F and H2 when pHH = 2.95 '''Figure 18.''' Graph of momentum against time for the reaction between F and H2 when pHH = 2.95 (Left-right)]]

Vibrational energy is represented by pHH while translational energy is represented by pHF.

A molecule of H2 collides with F. However, the collision is unsuccessful and no reaction occurs. A reaction did not occur even though the molecules have sufficient kinetic energy to overcome the activation energy barrier. This is because most of the energy is deposited in the reactants as H-H vibrations as seen by the large oscillations in H2. Barrier recrossing has occured as a result of the large vibrational energy causing the system to revert back to the reactants even though the system has already crossed the transition state and a bond was formed between HB and F temporarily.

This appears to run counter to what one would expect to happen when more energy is put into the system. It is predicted that when more energy is put into the system, the reactants would possess more kinetic energy to overcome the activation energy barrier. However, as can be seen from the simulation above, an increase in the vibrational energy of the reactants is not effective at promoting the occurrence of an exothermic reaction with an early transition state. This is in accordance to Polyani's rules.

2. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 0.1 and pHF = -0.8.

[[File:01409469 polyani 3+4 2.jpg|thumb|center|800px|'''Figure 19.''' Contour plot of the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 '''Figure 20.''' Graph of momentum against time for the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 (Left-right)]]

The collision between a molecule of H2 and F is successful and a chemical reaction has occurred. A reaction has occurred even though the overall energy of the system has been significantly reduced by lowering the amount of vibrational energy in the reactants but slightly increasing the amount of translational energy deposited in the reactants. Therefore, H2 and F have more translational energy to move towards each other and form a H-F bond. The product formed is vibrationally excited as seen by the large amplitude of oscillations in the product HF which correspond to the strong H-F vibrations.

From the second simulation, it can be seen that an increase in translational energy in the reactants is more effective at promoting an exothermic reaction with an early transition state to occur, which again follows Polyani's rules.

=== Promoting the backward reaction: HA + HB-F → HA-HB + F ===

1. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 0.2 and pHH = -3.0.

[[File:01409469 polyani 5+6 .jpg|thumb|center|800px|'''Figure 21.''' Contour plot of the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 '''Figure 22.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 (Left-right)]]

An atom of HA collides with a molecule of HBF. However, the collision is unsuccessful and no chemical reaction occurs. This is because the reactants lack sufficient vibrational energy to surmount the activation energy barrier. Therefore, following an unsuccessful collision with HBF, HA simply bounces off and moves away at constant momentum. This is in agreement with Polyani's rules which state that vibrational energy is more effective than translational energy at promoting reactions with a late transition state, for example, endothermic chemical reactions.

2. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 7.1 and pHH = -0.5.

[[File:01409469 polyani 7+8 .jpg|thumb|center|800px|'''Figure 23.''' Contour plot of the reaction between HA + HB-F when pFH = 7.1 and pHH = -0.5 '''Figure 24.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 7.1 and pHH = -0.5 (Left-right)]]

In addition, because the reaction between HA + HB-F is endothermic, energy is taken in from the surroundings during the reaction and the products are less vibrationally excited than the reactants.

MRD:01409469

2019-05-10T14:22:53Z

Stc1917: /* Promoting the backward reaction: HA + HB-F → HA-HB + F */

== Exercise 1: H + H2 system ==

=== Definition of a transition state ===

A potential energy profile results from plotting the potential versus reaction path.

[[File:Surface_Plot01409469_pic1.jpg|thumb|center|400px|'''Figure 1.''' A potential energy surface plot]]

The transition state is defined as the maximum on the minimum energy path linking reactants and products. This means that the transition state is a stationary point on the potential energy surface. Thus, at the transition state, the derivative of the potential energy is zero with respect to each Cartesian coordinate.

<math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>

However, a local minimum on the potential energy surface corresponding to a well would have the same <math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>. Therefore, to distinguish between a local minimum on the potential energy surface and the transition state, the second derivative, <math>\tfrac{\partial^2 V(r_i)}{\partial r_i ^2}\ </math> is determined. This can be calculated by finding the determinant of the Hessian matrix shown below.

:<math>\mathbf H =
\begin{bmatrix}
\dfrac{\partial^2 \ V(r_i)}{\partial \ x^2} & \dfrac{\partial^2 \ V(r_i)}{\partial \ x \partial \mathbf y} \\
\dfrac{\partial^2 \ V(r_i)}{\partial \ y \partial \mathbf x} & \dfrac{\partial^2 \ V(r_i)}{\partial \mathbf y^2}
\end{bmatrix}</math>

The Hessian matrix of a convex function is positive semi-definite. Refining this property allows one to test if a critical point (x,y) is a local maximum, local minimum or saddle point.

If <math>f</math> < 0, then (x,y) is an isolated local maximum.

If <math>f</math> > 0, then (x,y) is an isolated local minimum.

If the Hessian has both positive and negative eigenvalues then (x,y) is a saddle point.

=== Locating the transition state ===

[[File:01409469 combined pic 2 and 3.jpg|thumb|center|800px|'''Figure 2.''' A surface and contour plot of the trajectory when rAB=rBC=0.907743 Å and pA=pBC=0]]

When the molecules are not at the transition state, a surface plot of the conditions shows a trajectory being plotted. This means that the trajectory is "rolling" towards the products or reactants. However, if a trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

[[File:01409469_internuclearpic3.png|thumb|center|400px|'''Figure 3.''' A plot of internuclear distance against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

An initial estimate for the position of the transition state was made by observing the potential energy surface and was found to be 0.90 Å. At 0.90 Å, the internuclear distance against time plot showed that the internuclear distance was constantly oscillating. Therefore, the position of the transition state was further refined by increasing the number of decimal places until a horizontal line in the internuclear distance againt time plot was obtained. This means that the internuclear distance is constant and that the molecules are simply undergoing periodic symmetric vibration at the transition state. Therefore, the position of the transition state was determined to be 0.907743 Å.

[[File:01409469 combined energy pic.png|thumb|center|400px|'''Figure 4.''' A plot of energy against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

When rAB=rBC=0.907743 Å and pA=pBC=0, the kinetic energy is 0. This means that HA and HBC are stationary, which is as predicted when the trajectory is started at the transition state with no initial momentum. Furthermore, the forces acting on HA and HBC are 0 as should be the case in the transition state.

=== Calculation type: Dynamic and minimum energy path ===

The minimum energy path, is the lowest energy path from the reactants to the products. The minimum energy path is determined by following the mode corresponding to the negative Hessian eigenvalue in both directions along the steepest descent direction. Following this path allows one to approach two valleys of stable structures. The resulting path generated is the minimum energy path. The reaction paths were chartered for a dynamic and a minimum energy path. The initial conditions used were that of the transition state except that the internuclear distance between HA and HB had increased to 0.917743 Å. The reaction path obtained using the two different calculation methods was different.

[[File:01409469 mep and dynamics.jpg|thumb|center|800px|'''Figure 5.''' A surface plot of the trajectory using minimum energy path '''Figure 6.''' A surface plot of the trajectory using dynamics (Left-right)]]

On the surface plot, the reaction path is depicted by the black line. Following the minimum energy path, the trajectory simply follows the valley floor to Ha+Hb-Hc. The reaction path appears to be a relatively straight line indicating that the atoms are not vibrating. In the minimum energy path, the reactants change into the products in an infinitely slow motion. This is done by having the momenta/velocities reset to zero in each time step. While the minimum energy path is useful in characterising a chemical reaction, it does not provide a realistic representation of the motion of the atoms during the reaction. This is because the mass and inertia of the atoms have been ignored.

Since the atoms simply follow the minimum energy path, they do not undergo slight displacements into surrounding valleys of higher potential energy. Furthermore, because the atoms are assumed to have no inertia, they do not undergo vibrations even if they were to be displaced to valleys of higher potential energy.

However, when dynamic calculation is performed, the reaction path is wavy indicating that the atoms are oscillating. This is because the atoms are displaced to regions of higher potential on the path from reactants to products. This leads to atom oscillation due to the inertia of the atoms.

'''Internuclear distance against time plot'''

Setting the initial system conditions such that rAB= 0.907743, rBC= 0.917743 and pA= pBC= 0:

[[File:01409469 Internuclear distance mep and dynamic.jpg|thumb|center|800px|'''Figure 6.''' A graph of internuclear distance against steps using minimum energy path '''Figure 7.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In the minimum energy path, the distance between HA and HB decreases from 0.90 Å to approximately 0.79 Å during the first 50 steps before plateauing off to a value of 0.74 Å which is the approximate length of a hydrogen bond. This means that a bond was formed between HA and HB. In addition, the plateau indicates that the bond is static. On the other hand, the distance between HB and HC has increased to 1.91 Å at large step values. This shows that HC has been ejected off following a successful collision between HA and HBC.

Similarly, in the dynamic trajectory, the distance between HA and HB decreases to approximately 0.75 Å after 0.3 s. This is indicative of hydrogen bond formation between HA and HB. In addition, the distance between HB and HC increases to 3.44 Å after HC is ejected following a successful collision.

One obvious distance between the two plots is that in the minimum energy path trajectory, the internuclear distance between HA and HB is constant indicating that the bond is static. However, in the dynamic trajectory the internuclear distance is oscillating showing the the bond is vibrating. This difference could be attributed to the fact that minimum energy path does not take oscillatory motion into account while dynamic trajectory does.

However, changing the initial conditions of the system such that rAB = 0.917743 Å, rBC = 0.907743 Å and pA=pBC=0:
[[File:01409469 01409469 Reversed values dynamic mep.jpg|thumb|center|800px|'''Figure 8.''' A graph of internuclear distance against steps using minimum energy path '''Figure 9.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In this case, a hydrogen bond of length 0.77-0.79 Å is formed between HB and HC while HA is repelled away to beyond 3.45 Å.

Additionally, from the internuclear distance plot it can be seen that the distance between HB and HC increases in a rectangular hyperbola fashion when calculated using the minimum energy path. On the other hand, the internuclear distance between HB and HC increases in a linear manner when dynamic calculation is applied. A possible reason for this difference is that in the minimum energy path, the atoms are not vibrating and as such the reach the maximum distance and plateau at a faster rate. However, in the dynamic calculation, the atoms have momentum and as such move further away from each other.

=== Reactive and unreactive trajectories ===

With the initial positions r'''1''' = 0.74 and r'''2''' = 2.0, trajectories were run with the momenta combination shown below.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Contour plot !! Reactivity !! Description of the dynamics
|-
| -1.25 || -2.5 ||-99.018 ||[[File:01409469 trajectory1.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. There are no oscillations observed in HA-HB initially, indicating that the diatomic molecule is stable. A new bond is formed between HB and HC. Oscillatory behaviour is observed indicating that the bond is vibrating. On the other hand, HA is ejected away to beyond 10 Å.
|-
| -1.5 || -2.0 ||-100.456 ||[[File:01409469 trajectory2.jpg|thumb|center|400px]] ||Unreactive ||HC collides with the vibrating diatomic HA-HB. However, no bond is formed between HB and HC as the collision is unsuccessful. This is because the molecules lack sufficient kinetic energy to overcome the activation energy barrier and so no reaction occurs. HC is subsequently repelled to beyond 8 Å.
|-
| -1.5 || -2.5 ||-98.956 ||[[File:01409469 trajectory3.jpg|thumb|center|400px]] ||Reactive |||HC collides with HA-HB. The collision is successful and a bond is formed between HB and HC. This is because the atoms have sufficient kinetic energy to overcome the activation energy barrier and form a new HB-HC bond. HA is ejected away to beyond 10 Å. The new hydrogen bond formed is vibrating as seen by the oscillatory behaviour.
|-
| -2.5 || -5.0 ||-84.956 ||[[File:01409469 trajectory4.jpg|thumb|center|400px]] ||Unreactive ||HC collides with HA-HB. Initially the HA-HB bond is elongating, indicating that the hydrogen bond is weakening and going to break. However, after 0.5 s, the HA-HB bond length decreases and the HAHB is not broken. HC is repelled beyond 6 Å with very strong vibrations.
A reaction did not occur even though the atoms have sufficient kinetic energy to cross the activation energy barrier. This is because the vibrations are too strong. The system crosses the transition state region and a bond is formed in the product temporarily but then the system reverts back to the reactants. This is an example of barrier recrossing.
|-
| -2.5 || -5.2 ||-83.416 ||[[File:01409469 trajectory5.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. The collision is successful and a new hydrogen bond is formed between HB and HC with HA being ejected away. The products formed experience very strong vibrations as evidenced by the oscillatory motion which occur with a very high frequency.
|}

From the table, it can be seen that the chance of a reaction successfully occurring is not solely dependent on the amount of kinetic energy the atoms possess. Even if the atoms have sufficient kinetic energy to cross the activation energy barrier, barrier recrossing can occur if the atoms are too vibrationally excited.

=== Transition State Theory ===

Transition state theory provided the first theoretical attempt to determine absolute reaction rates. In this theory, a transition state separating the reactants from products is used to find an expression for the thermal rate constant.

The main assumptions of transition state theory are:

1. Electronic and nuclear motions are separated (equivalent to the Born-Oppenheimer approximation in quantum mechanics)

2. Reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution

3. Molecular systems that have crossed the transition state in the direction of products cannot turn around and reform reactants

4. In the transition state, motion along the reaction coordinate may be separated from the other motions and treated classically as a translation

5. Even in the absence of an equilibrium between reactant and product molecules, the transition states that are becoming products are distributed among their states according to the Maxwell-Boltzmann laws

The goal of transition state theory is to predict the rate constant, ''k'', of a reaction given a potential energy surface for the reaction.

However, transition state theory suffers from a set of limitations due to the assumptions stated above not holding true in reality.

Firstly, transition state theory assumes that barrier recrossing is negligible. However, as seen from the fourth trajectory in table 1, barrier recrossing occurred because the atoms were too vibrationally excited and the system reverted back to the reactants even though the system had already crossed the transition state region. In this situation, transition state theory would predict a faster rate of reaction compared to what was obtained experimentally.

Secondly, the transition state theory breaks down for systems with light atoms such as hydrogens. In such systems, quantum effects become crucial. Quantum effects tend to increase the rate of the reaction for two reasons. First, particles lie in energy levels which are higher then the bottom of the potential energy well (there is additional zero-point energy). Secondly, the particles can tunnel through the barrier. These two effects effectively lower the activation energy barrier. As a result, the rate of reaction predicted by transition state theory would be less than what is obtained experimentally.

Lastly, transition state theory does not hold true for systems which are far from equilibrium. In such cases, the assumption of a starting Boltzmann distribution breaks down.

== Exercise 2: F - H - H system ==

=== Potential energy surface inspection===

F + HA-HB → HA-F + HB

[[File:01409469 Surface_Plot F 3.jpg|thumb|center|400px|'''Figure 10.''' Potential energy surface for the reaction between F and H2]]

The reaction between F and H2 is an exothermic reaction. This means that the reactants are of a higher energy than the products. The overall reaction involves the breaking of a hydrogen bond and the formation of a hydrogen-fluorine bond. The HF bond formed is stronger than the hydrogen bond that was broken. This is because fluorine is more electronegative than hydrogen resulting in a polar HF bond. The negative enthalpy of H-F bond formation is greater than the positive enthalpy of hydrogen bond dissociation. Therefore, the reaction between F and H2 is an exothermic reaction.

HA + HB-F → HA-HB + F

[[File:01409469 Surface_Plot F 4.jpg|thumb|center|400px|'''Figure 11.''' Potential energy surface for the reaction between H and HF]]

The reaction between H and HF is an endothermic reaction. The products are at a higher energy level than the reactants. The overall reaction involves the breaking of a HF bond and the formation of a hydrogen bond. The formation of a homonuclear H2 molecule results in a non-polar hydrogen bond being formed. The HF bond broken is stronger than the hydrogen bond that was formed. The negative enthalpy of hydrogen bond formation is less than the positive enthalpy of hydrogen-fluorine dissociation. Therefore, the reaction between H and HF is endothermic.

=== Locating the transition state ===

[[File:01409469 Surface_Plot F TS4.jpg|thumb|center|800px|'''Figure 12.''' Determining the position of the transition state for the reaction between F and H2]]

As mentioned previously, when the molecules are not at the transition state, a reaction path will be plotted. However, if the initial conditions of the system are set such that the trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

An initial estimate for the transition state was made by setting rAB = 0.74 which is the length of a hydrogen bond. The position of the transition state was further refined by increasing the number of decimal places such that the final value of rAB was found to be 0.74478 and rBC = 1.81079. This set of distances gave a reaction path that was represented by a point indicating that no trajectory was plotted and that the molecules remained at the transition state since their initial momentum was 0.

Moreover, the kinetic energy as well as the forces acting on the atoms was calculated to be 0, which is consistent with the transition state.

Therefore, the position of the transition state was found to be at rAB = 0.74478 and rBC = 1.81079 when pA = pBC = 0.

=== Determining the activation energy ===

[[File:01409469 activation energy r1.1 combined.jpg|thumb|center|800px|'''Figure 13.''' Determining the activation energy for the reaction between F and H2]]

By reducing the value of rBC by 0.01, the atoms are no longer at the transition state and would move towards the formation of products. The activation energy was found by calculating the change in total energy after the reaction. The activation energy of F + H2 → HF + H was found to be +23.279 kcal/ mol.

[[File:01409469 activation energy r2.1 combined.jpg|thumb|center|800px|'''Figure 14.''' Determining the activation energy for the reaction between FH and H]]

The activation energy of FH + H → F + H2 was found by calculating the small dip in potential energy. The activation energy was found to be +0.280 kcal/ mol.

=== Reaction dynamics of F + HA-HB → HA-F + HB ===

The set of initial conditions that result in a reactive trajectory for F + H2 is as follows: rHH = 0.74 Å, rHF = 1.79 Å and pHH = pHF = 0.

[[File:01409469 momentum r1.jpg|thumb|center|400px|'''Figure 15.''' Graph of momentum against time for the reaction between F and H2]]

Initially, the H2 molecule is travelling in a relatively straight line with very little oscillations as evidenced by the low amplitude of momentum between 0-2 s. Following successful collision between F and H2, the amplitude of momentum of the product, HF, increases dramatically. This is because the reaction between F and H2 is exothermic and the excess energy released causes the molecule to become vibrationally excited and the amplitude of oscillations in the product to increase.

[[File:01409469 energy1 r1.jpg|thumb|center|400px|'''Figure 16.''' Graph of energy against time for the reaction between F and H2]]

The law of conservation of energy states that energy cannot be created or destroyed. It simply changes from one form to another. As can be seen from the graph in figure 16, total energy remains constant. In this reaction, potential energy is converted to kinetic energy. A decrease in potential energy of the atoms following a successful collision between F and H2 results in an equal increase in kinetic energy which manifests in the form of increased product (HF) oscillations.

This can be confirmed experimentally using infrared chemiluminescence. Observation of radiation from excited product molecules is a clear indication that these molecules are produced in the excited state. If the excitation is primarily vibrational, the radiation will appear in the infrared region of the spectrum (between 3-15 μm). This technique has been used extensively in hydrogen halogen reactions which produce HX molecules in the excited state.

One such experiment involves the use of the arrested relaxation method. In this technique, the reaction is carried out in a chamber whose walls are maintained at a very low temperature (20 K). This results in cryopumping of the chamber in that all the products and unreacted starting materials are condensed on the walls and only those product molecules that radiate before being pumped away are observed.

=== Polanyi's empirical rules ===

In a typical chemical reaction, there is an energetic barrier — a saddle point that the reactants must surmount to become products. Which form of energy is initially deposited in the reactants, translational or vibrational, is more efficacious in surmounting the activation energy barrier is one of the core topics in the area of reaction dynamics. The Polyani rules state that vibrational energy is more effective at promoting a late-barrier reaction (transition state resembles products). Conversely, translational energy is better at promoting an early-barrier reaction (transition state resembles the reactants).

The forward reaction between H2 and F is exothermic. Therefore, the transition state is closer to the reactants than to the products in energy. Since the transition state occurs early in an exothermic reaction, translational energy is more effective at promoting a chemical reaction than vibrational energy.

Conversely, the backward reaction between H and HF is endothermic. According to Hammond's postulate, the transition state resembles the products more than it resembles the reactants. Therefore, in an endothermic reaction, a late transition state is observed and vibrational energy is more effective at promoting a reaction than translational energy.

=== Promoting the forward reaction: F + HA-HB → HA-F + HB ===

1. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 2.95 and pHF = -0.5.

[[File:01409469 polyani 1+2.jpg|thumb|center|800px|'''Figure 17.''' Contour plot of the reaction between F and H2 when pHH = 2.95 '''Figure 18.''' Graph of momentum against time for the reaction between F and H2 when pHH = 2.95 (Left-right)]]

Vibrational energy is represented by pHH while translational energy is represented by pHF.

A molecule of H2 collides with F. However, the collision is unsuccessful and no reaction occurs. A reaction did not occur even though the molecules have sufficient kinetic energy to overcome the activation energy barrier. This is because most of the energy is deposited in the reactants as H-H vibrations as seen by the large oscillations in H2. Barrier recrossing has occured as a result of the large vibrational energy causing the system to revert back to the reactants even though the system has already crossed the transition state and a bond was formed between HB and F temporarily.

This appears to run counter to what one would expect to happen when more energy is put into the system. It is predicted that when more energy is put into the system, the reactants would possess more kinetic energy to overcome the activation energy barrier. However, as can be seen from the simulation above, an increase in the vibrational energy of the reactants is not effective at promoting the occurrence of an exothermic reaction with an early transition state. This is in accordance to Polyani's rules.

2. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 0.1 and pHF = -0.8.

[[File:01409469 polyani 3+4 2.jpg|thumb|center|800px|'''Figure 19.''' Contour plot of the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 '''Figure 20.''' Graph of momentum against time for the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 (Left-right)]]

The collision between a molecule of H2 and F is successful and a chemical reaction has occurred. A reaction has occurred even though the overall energy of the system has been significantly reduced by lowering the amount of vibrational energy in the reactants but slightly increasing the amount of translational energy deposited in the reactants. Therefore, H2 and F have more translational energy to move towards each other and form a H-F bond. The product formed is vibrationally excited as seen by the large amplitude of oscillations in the product HF which correspond to the strong H-F vibrations.

From the second simulation, it can be seen that an increase in translational energy in the reactants is more effective at promoting an exothermic reaction with an early transition state to occur, which again follows Polyani's rules.

=== Promoting the backward reaction: HA + HB-F → HA-HB + F ===

1. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 0.2 and pHH = -3.0.

[[File:01409469 polyani 5+6 .jpg|thumb|center|1000px|'''Figure 21.''' Contour plot of the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 '''Figure 22.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 (Left-right)]]

An atom of HA collides with a molecule of HBF. However, the collision is unsuccessful and no chemical reaction occurs. This is because the reactants lack sufficient vibrational energy to surmount the activation energy barrier. Therefore, following an unsuccessful collision with HBF, HA simply bounces off and moves away at constant momentum. This is in agreement with Polyani's rules which state that vibrational energy is more effective than translational energy at promoting reactions with a late transition state, for example, endothermic chemical reactions.

2. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 7.1 and pHH = -0.5.

[[File:01409469 polyani 7+8 .jpg|thumb|center|1000px|'''Figure 23.''' Contour plot of the reaction between HA + HB-F when pFH = 7.1 and pHH = -0.5 '''Figure 24.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 7.1 and pHH = -0.5 (Left-right)]]

In addition, because the reaction between HA + HB-F is endothermic, energy is taken in from the surroundings during the reaction and the products are less vibrationally excited than the reactants.

MRD:01409469

2019-05-10T14:21:11Z

Stc1917: /* Promoting the backward reaction: HA + HB-F → HA-HB + F */

== Exercise 1: H + H2 system ==

=== Definition of a transition state ===

A potential energy profile results from plotting the potential versus reaction path.

[[File:Surface_Plot01409469_pic1.jpg|thumb|center|400px|'''Figure 1.''' A potential energy surface plot]]

The transition state is defined as the maximum on the minimum energy path linking reactants and products. This means that the transition state is a stationary point on the potential energy surface. Thus, at the transition state, the derivative of the potential energy is zero with respect to each Cartesian coordinate.

<math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>

However, a local minimum on the potential energy surface corresponding to a well would have the same <math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>. Therefore, to distinguish between a local minimum on the potential energy surface and the transition state, the second derivative, <math>\tfrac{\partial^2 V(r_i)}{\partial r_i ^2}\ </math> is determined. This can be calculated by finding the determinant of the Hessian matrix shown below.

:<math>\mathbf H =
\begin{bmatrix}
\dfrac{\partial^2 \ V(r_i)}{\partial \ x^2} & \dfrac{\partial^2 \ V(r_i)}{\partial \ x \partial \mathbf y} \\
\dfrac{\partial^2 \ V(r_i)}{\partial \ y \partial \mathbf x} & \dfrac{\partial^2 \ V(r_i)}{\partial \mathbf y^2}
\end{bmatrix}</math>

The Hessian matrix of a convex function is positive semi-definite. Refining this property allows one to test if a critical point (x,y) is a local maximum, local minimum or saddle point.

If <math>f</math> < 0, then (x,y) is an isolated local maximum.

If <math>f</math> > 0, then (x,y) is an isolated local minimum.

If the Hessian has both positive and negative eigenvalues then (x,y) is a saddle point.

=== Locating the transition state ===

[[File:01409469 combined pic 2 and 3.jpg|thumb|center|800px|'''Figure 2.''' A surface and contour plot of the trajectory when rAB=rBC=0.907743 Å and pA=pBC=0]]

When the molecules are not at the transition state, a surface plot of the conditions shows a trajectory being plotted. This means that the trajectory is "rolling" towards the products or reactants. However, if a trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

[[File:01409469_internuclearpic3.png|thumb|center|400px|'''Figure 3.''' A plot of internuclear distance against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

An initial estimate for the position of the transition state was made by observing the potential energy surface and was found to be 0.90 Å. At 0.90 Å, the internuclear distance against time plot showed that the internuclear distance was constantly oscillating. Therefore, the position of the transition state was further refined by increasing the number of decimal places until a horizontal line in the internuclear distance againt time plot was obtained. This means that the internuclear distance is constant and that the molecules are simply undergoing periodic symmetric vibration at the transition state. Therefore, the position of the transition state was determined to be 0.907743 Å.

[[File:01409469 combined energy pic.png|thumb|center|400px|'''Figure 4.''' A plot of energy against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

When rAB=rBC=0.907743 Å and pA=pBC=0, the kinetic energy is 0. This means that HA and HBC are stationary, which is as predicted when the trajectory is started at the transition state with no initial momentum. Furthermore, the forces acting on HA and HBC are 0 as should be the case in the transition state.

=== Calculation type: Dynamic and minimum energy path ===

The minimum energy path, is the lowest energy path from the reactants to the products. The minimum energy path is determined by following the mode corresponding to the negative Hessian eigenvalue in both directions along the steepest descent direction. Following this path allows one to approach two valleys of stable structures. The resulting path generated is the minimum energy path. The reaction paths were chartered for a dynamic and a minimum energy path. The initial conditions used were that of the transition state except that the internuclear distance between HA and HB had increased to 0.917743 Å. The reaction path obtained using the two different calculation methods was different.

[[File:01409469 mep and dynamics.jpg|thumb|center|800px|'''Figure 5.''' A surface plot of the trajectory using minimum energy path '''Figure 6.''' A surface plot of the trajectory using dynamics (Left-right)]]

On the surface plot, the reaction path is depicted by the black line. Following the minimum energy path, the trajectory simply follows the valley floor to Ha+Hb-Hc. The reaction path appears to be a relatively straight line indicating that the atoms are not vibrating. In the minimum energy path, the reactants change into the products in an infinitely slow motion. This is done by having the momenta/velocities reset to zero in each time step. While the minimum energy path is useful in characterising a chemical reaction, it does not provide a realistic representation of the motion of the atoms during the reaction. This is because the mass and inertia of the atoms have been ignored.

Since the atoms simply follow the minimum energy path, they do not undergo slight displacements into surrounding valleys of higher potential energy. Furthermore, because the atoms are assumed to have no inertia, they do not undergo vibrations even if they were to be displaced to valleys of higher potential energy.

However, when dynamic calculation is performed, the reaction path is wavy indicating that the atoms are oscillating. This is because the atoms are displaced to regions of higher potential on the path from reactants to products. This leads to atom oscillation due to the inertia of the atoms.

'''Internuclear distance against time plot'''

Setting the initial system conditions such that rAB= 0.907743, rBC= 0.917743 and pA= pBC= 0:

[[File:01409469 Internuclear distance mep and dynamic.jpg|thumb|center|800px|'''Figure 6.''' A graph of internuclear distance against steps using minimum energy path '''Figure 7.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In the minimum energy path, the distance between HA and HB decreases from 0.90 Å to approximately 0.79 Å during the first 50 steps before plateauing off to a value of 0.74 Å which is the approximate length of a hydrogen bond. This means that a bond was formed between HA and HB. In addition, the plateau indicates that the bond is static. On the other hand, the distance between HB and HC has increased to 1.91 Å at large step values. This shows that HC has been ejected off following a successful collision between HA and HBC.

Similarly, in the dynamic trajectory, the distance between HA and HB decreases to approximately 0.75 Å after 0.3 s. This is indicative of hydrogen bond formation between HA and HB. In addition, the distance between HB and HC increases to 3.44 Å after HC is ejected following a successful collision.

One obvious distance between the two plots is that in the minimum energy path trajectory, the internuclear distance between HA and HB is constant indicating that the bond is static. However, in the dynamic trajectory the internuclear distance is oscillating showing the the bond is vibrating. This difference could be attributed to the fact that minimum energy path does not take oscillatory motion into account while dynamic trajectory does.

However, changing the initial conditions of the system such that rAB = 0.917743 Å, rBC = 0.907743 Å and pA=pBC=0:
[[File:01409469 01409469 Reversed values dynamic mep.jpg|thumb|center|800px|'''Figure 8.''' A graph of internuclear distance against steps using minimum energy path '''Figure 9.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In this case, a hydrogen bond of length 0.77-0.79 Å is formed between HB and HC while HA is repelled away to beyond 3.45 Å.

Additionally, from the internuclear distance plot it can be seen that the distance between HB and HC increases in a rectangular hyperbola fashion when calculated using the minimum energy path. On the other hand, the internuclear distance between HB and HC increases in a linear manner when dynamic calculation is applied. A possible reason for this difference is that in the minimum energy path, the atoms are not vibrating and as such the reach the maximum distance and plateau at a faster rate. However, in the dynamic calculation, the atoms have momentum and as such move further away from each other.

=== Reactive and unreactive trajectories ===

With the initial positions r'''1''' = 0.74 and r'''2''' = 2.0, trajectories were run with the momenta combination shown below.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Contour plot !! Reactivity !! Description of the dynamics
|-
| -1.25 || -2.5 ||-99.018 ||[[File:01409469 trajectory1.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. There are no oscillations observed in HA-HB initially, indicating that the diatomic molecule is stable. A new bond is formed between HB and HC. Oscillatory behaviour is observed indicating that the bond is vibrating. On the other hand, HA is ejected away to beyond 10 Å.
|-
| -1.5 || -2.0 ||-100.456 ||[[File:01409469 trajectory2.jpg|thumb|center|400px]] ||Unreactive ||HC collides with the vibrating diatomic HA-HB. However, no bond is formed between HB and HC as the collision is unsuccessful. This is because the molecules lack sufficient kinetic energy to overcome the activation energy barrier and so no reaction occurs. HC is subsequently repelled to beyond 8 Å.
|-
| -1.5 || -2.5 ||-98.956 ||[[File:01409469 trajectory3.jpg|thumb|center|400px]] ||Reactive |||HC collides with HA-HB. The collision is successful and a bond is formed between HB and HC. This is because the atoms have sufficient kinetic energy to overcome the activation energy barrier and form a new HB-HC bond. HA is ejected away to beyond 10 Å. The new hydrogen bond formed is vibrating as seen by the oscillatory behaviour.
|-
| -2.5 || -5.0 ||-84.956 ||[[File:01409469 trajectory4.jpg|thumb|center|400px]] ||Unreactive ||HC collides with HA-HB. Initially the HA-HB bond is elongating, indicating that the hydrogen bond is weakening and going to break. However, after 0.5 s, the HA-HB bond length decreases and the HAHB is not broken. HC is repelled beyond 6 Å with very strong vibrations.
A reaction did not occur even though the atoms have sufficient kinetic energy to cross the activation energy barrier. This is because the vibrations are too strong. The system crosses the transition state region and a bond is formed in the product temporarily but then the system reverts back to the reactants. This is an example of barrier recrossing.
|-
| -2.5 || -5.2 ||-83.416 ||[[File:01409469 trajectory5.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. The collision is successful and a new hydrogen bond is formed between HB and HC with HA being ejected away. The products formed experience very strong vibrations as evidenced by the oscillatory motion which occur with a very high frequency.
|}

From the table, it can be seen that the chance of a reaction successfully occurring is not solely dependent on the amount of kinetic energy the atoms possess. Even if the atoms have sufficient kinetic energy to cross the activation energy barrier, barrier recrossing can occur if the atoms are too vibrationally excited.

=== Transition State Theory ===

Transition state theory provided the first theoretical attempt to determine absolute reaction rates. In this theory, a transition state separating the reactants from products is used to find an expression for the thermal rate constant.

The main assumptions of transition state theory are:

1. Electronic and nuclear motions are separated (equivalent to the Born-Oppenheimer approximation in quantum mechanics)

2. Reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution

3. Molecular systems that have crossed the transition state in the direction of products cannot turn around and reform reactants

4. In the transition state, motion along the reaction coordinate may be separated from the other motions and treated classically as a translation

5. Even in the absence of an equilibrium between reactant and product molecules, the transition states that are becoming products are distributed among their states according to the Maxwell-Boltzmann laws

The goal of transition state theory is to predict the rate constant, ''k'', of a reaction given a potential energy surface for the reaction.

However, transition state theory suffers from a set of limitations due to the assumptions stated above not holding true in reality.

Firstly, transition state theory assumes that barrier recrossing is negligible. However, as seen from the fourth trajectory in table 1, barrier recrossing occurred because the atoms were too vibrationally excited and the system reverted back to the reactants even though the system had already crossed the transition state region. In this situation, transition state theory would predict a faster rate of reaction compared to what was obtained experimentally.

Secondly, the transition state theory breaks down for systems with light atoms such as hydrogens. In such systems, quantum effects become crucial. Quantum effects tend to increase the rate of the reaction for two reasons. First, particles lie in energy levels which are higher then the bottom of the potential energy well (there is additional zero-point energy). Secondly, the particles can tunnel through the barrier. These two effects effectively lower the activation energy barrier. As a result, the rate of reaction predicted by transition state theory would be less than what is obtained experimentally.

Lastly, transition state theory does not hold true for systems which are far from equilibrium. In such cases, the assumption of a starting Boltzmann distribution breaks down.

== Exercise 2: F - H - H system ==

=== Potential energy surface inspection===

F + HA-HB → HA-F + HB

[[File:01409469 Surface_Plot F 3.jpg|thumb|center|400px|'''Figure 10.''' Potential energy surface for the reaction between F and H2]]

The reaction between F and H2 is an exothermic reaction. This means that the reactants are of a higher energy than the products. The overall reaction involves the breaking of a hydrogen bond and the formation of a hydrogen-fluorine bond. The HF bond formed is stronger than the hydrogen bond that was broken. This is because fluorine is more electronegative than hydrogen resulting in a polar HF bond. The negative enthalpy of H-F bond formation is greater than the positive enthalpy of hydrogen bond dissociation. Therefore, the reaction between F and H2 is an exothermic reaction.

HA + HB-F → HA-HB + F

[[File:01409469 Surface_Plot F 4.jpg|thumb|center|400px|'''Figure 11.''' Potential energy surface for the reaction between H and HF]]

The reaction between H and HF is an endothermic reaction. The products are at a higher energy level than the reactants. The overall reaction involves the breaking of a HF bond and the formation of a hydrogen bond. The formation of a homonuclear H2 molecule results in a non-polar hydrogen bond being formed. The HF bond broken is stronger than the hydrogen bond that was formed. The negative enthalpy of hydrogen bond formation is less than the positive enthalpy of hydrogen-fluorine dissociation. Therefore, the reaction between H and HF is endothermic.

=== Locating the transition state ===

[[File:01409469 Surface_Plot F TS4.jpg|thumb|center|800px|'''Figure 12.''' Determining the position of the transition state for the reaction between F and H2]]

As mentioned previously, when the molecules are not at the transition state, a reaction path will be plotted. However, if the initial conditions of the system are set such that the trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

An initial estimate for the transition state was made by setting rAB = 0.74 which is the length of a hydrogen bond. The position of the transition state was further refined by increasing the number of decimal places such that the final value of rAB was found to be 0.74478 and rBC = 1.81079. This set of distances gave a reaction path that was represented by a point indicating that no trajectory was plotted and that the molecules remained at the transition state since their initial momentum was 0.

Moreover, the kinetic energy as well as the forces acting on the atoms was calculated to be 0, which is consistent with the transition state.

Therefore, the position of the transition state was found to be at rAB = 0.74478 and rBC = 1.81079 when pA = pBC = 0.

=== Determining the activation energy ===

[[File:01409469 activation energy r1.1 combined.jpg|thumb|center|800px|'''Figure 13.''' Determining the activation energy for the reaction between F and H2]]

By reducing the value of rBC by 0.01, the atoms are no longer at the transition state and would move towards the formation of products. The activation energy was found by calculating the change in total energy after the reaction. The activation energy of F + H2 → HF + H was found to be +23.279 kcal/ mol.

[[File:01409469 activation energy r2.1 combined.jpg|thumb|center|800px|'''Figure 14.''' Determining the activation energy for the reaction between FH and H]]

The activation energy of FH + H → F + H2 was found by calculating the small dip in potential energy. The activation energy was found to be +0.280 kcal/ mol.

=== Reaction dynamics of F + HA-HB → HA-F + HB ===

The set of initial conditions that result in a reactive trajectory for F + H2 is as follows: rHH = 0.74 Å, rHF = 1.79 Å and pHH = pHF = 0.

[[File:01409469 momentum r1.jpg|thumb|center|400px|'''Figure 15.''' Graph of momentum against time for the reaction between F and H2]]

Initially, the H2 molecule is travelling in a relatively straight line with very little oscillations as evidenced by the low amplitude of momentum between 0-2 s. Following successful collision between F and H2, the amplitude of momentum of the product, HF, increases dramatically. This is because the reaction between F and H2 is exothermic and the excess energy released causes the molecule to become vibrationally excited and the amplitude of oscillations in the product to increase.

[[File:01409469 energy1 r1.jpg|thumb|center|400px|'''Figure 16.''' Graph of energy against time for the reaction between F and H2]]

The law of conservation of energy states that energy cannot be created or destroyed. It simply changes from one form to another. As can be seen from the graph in figure 16, total energy remains constant. In this reaction, potential energy is converted to kinetic energy. A decrease in potential energy of the atoms following a successful collision between F and H2 results in an equal increase in kinetic energy which manifests in the form of increased product (HF) oscillations.

This can be confirmed experimentally using infrared chemiluminescence. Observation of radiation from excited product molecules is a clear indication that these molecules are produced in the excited state. If the excitation is primarily vibrational, the radiation will appear in the infrared region of the spectrum (between 3-15 μm). This technique has been used extensively in hydrogen halogen reactions which produce HX molecules in the excited state.

One such experiment involves the use of the arrested relaxation method. In this technique, the reaction is carried out in a chamber whose walls are maintained at a very low temperature (20 K). This results in cryopumping of the chamber in that all the products and unreacted starting materials are condensed on the walls and only those product molecules that radiate before being pumped away are observed.

=== Polanyi's empirical rules ===

In a typical chemical reaction, there is an energetic barrier — a saddle point that the reactants must surmount to become products. Which form of energy is initially deposited in the reactants, translational or vibrational, is more efficacious in surmounting the activation energy barrier is one of the core topics in the area of reaction dynamics. The Polyani rules state that vibrational energy is more effective at promoting a late-barrier reaction (transition state resembles products). Conversely, translational energy is better at promoting an early-barrier reaction (transition state resembles the reactants).

The forward reaction between H2 and F is exothermic. Therefore, the transition state is closer to the reactants than to the products in energy. Since the transition state occurs early in an exothermic reaction, translational energy is more effective at promoting a chemical reaction than vibrational energy.

Conversely, the backward reaction between H and HF is endothermic. According to Hammond's postulate, the transition state resembles the products more than it resembles the reactants. Therefore, in an endothermic reaction, a late transition state is observed and vibrational energy is more effective at promoting a reaction than translational energy.

=== Promoting the forward reaction: F + HA-HB → HA-F + HB ===

1. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 2.95 and pHF = -0.5.

[[File:01409469 polyani 1+2.jpg|thumb|center|800px|'''Figure 17.''' Contour plot of the reaction between F and H2 when pHH = 2.95 '''Figure 18.''' Graph of momentum against time for the reaction between F and H2 when pHH = 2.95 (Left-right)]]

Vibrational energy is represented by pHH while translational energy is represented by pHF.

A molecule of H2 collides with F. However, the collision is unsuccessful and no reaction occurs. A reaction did not occur even though the molecules have sufficient kinetic energy to overcome the activation energy barrier. This is because most of the energy is deposited in the reactants as H-H vibrations as seen by the large oscillations in H2. Barrier recrossing has occured as a result of the large vibrational energy causing the system to revert back to the reactants even though the system has already crossed the transition state and a bond was formed between HB and F temporarily.

This appears to run counter to what one would expect to happen when more energy is put into the system. It is predicted that when more energy is put into the system, the reactants would possess more kinetic energy to overcome the activation energy barrier. However, as can be seen from the simulation above, an increase in the vibrational energy of the reactants is not effective at promoting the occurrence of an exothermic reaction with an early transition state. This is in accordance to Polyani's rules.

2. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 0.1 and pHF = -0.8.

[[File:01409469 polyani 3+4 2.jpg|thumb|center|800px|'''Figure 19.''' Contour plot of the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 '''Figure 20.''' Graph of momentum against time for the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 (Left-right)]]

The collision between a molecule of H2 and F is successful and a chemical reaction has occurred. A reaction has occurred even though the overall energy of the system has been significantly reduced by lowering the amount of vibrational energy in the reactants but slightly increasing the amount of translational energy deposited in the reactants. Therefore, H2 and F have more translational energy to move towards each other and form a H-F bond. The product formed is vibrationally excited as seen by the large amplitude of oscillations in the product HF which correspond to the strong H-F vibrations.

From the second simulation, it can be seen that an increase in translational energy in the reactants is more effective at promoting an exothermic reaction with an early transition state to occur, which again follows Polyani's rules.

=== Promoting the backward reaction: HA + HB-F → HA-HB + F ===

1. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 0.2 and pHH = -3.0.

[[File:01409469 polyani 5+6 .jpg|thumb|center|800px|'''Figure 21.''' Contour plot of the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 '''Figure 22.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 (Left-right)]]

An atom of HA collides with a molecule of HBF. However, the collision is unsuccessful and no chemical reaction occurs. This is because the reactants lack sufficient vibrational energy to surmount the activation energy barrier. Therefore, following an unsuccessful collision with HBF, HA simply bounces off and moves away at constant momentum. This is in agreement with Polyani's rules which state that vibrational energy is more effective than translational energy at promoting reactions with a late transition state, for example, endothermic chemical reactions.

2. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 7.1 and pHH = -0.5.

[[File:01409469 polyani 7+8 .jpg|thumb|center|800px|'''Figure 23.''' Contour plot of the reaction between HA + HB-F when pFH = 7.1 and pHH = -0.5 '''Figure 24.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 7.1 and pHH = -0.5 (Left-right)]]

In addition, because the reaction between HA + HB-F is endothermic, energy is taken in from the surroundings during the reaction and the products are less vibrationally excited than the reactants.

MRD:01409469

2019-05-10T14:06:12Z

Stc1917: /* Promoting the backward reaction: HA + HB-F → HA-HB + F */

== Exercise 1: H + H2 system ==

=== Definition of a transition state ===

A potential energy profile results from plotting the potential versus reaction path.

[[File:Surface_Plot01409469_pic1.jpg|thumb|center|400px|'''Figure 1.''' A potential energy surface plot]]

The transition state is defined as the maximum on the minimum energy path linking reactants and products. This means that the transition state is a stationary point on the potential energy surface. Thus, at the transition state, the derivative of the potential energy is zero with respect to each Cartesian coordinate.

<math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>

However, a local minimum on the potential energy surface corresponding to a well would have the same <math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>. Therefore, to distinguish between a local minimum on the potential energy surface and the transition state, the second derivative, <math>\tfrac{\partial^2 V(r_i)}{\partial r_i ^2}\ </math> is determined. This can be calculated by finding the determinant of the Hessian matrix shown below.

:<math>\mathbf H =
\begin{bmatrix}
\dfrac{\partial^2 \ V(r_i)}{\partial \ x^2} & \dfrac{\partial^2 \ V(r_i)}{\partial \ x \partial \mathbf y} \\
\dfrac{\partial^2 \ V(r_i)}{\partial \ y \partial \mathbf x} & \dfrac{\partial^2 \ V(r_i)}{\partial \mathbf y^2}
\end{bmatrix}</math>

The Hessian matrix of a convex function is positive semi-definite. Refining this property allows one to test if a critical point (x,y) is a local maximum, local minimum or saddle point.

If <math>f</math> < 0, then (x,y) is an isolated local maximum.

If <math>f</math> > 0, then (x,y) is an isolated local minimum.

If the Hessian has both positive and negative eigenvalues then (x,y) is a saddle point.

=== Locating the transition state ===

[[File:01409469 combined pic 2 and 3.jpg|thumb|center|800px|'''Figure 2.''' A surface and contour plot of the trajectory when rAB=rBC=0.907743 Å and pA=pBC=0]]

When the molecules are not at the transition state, a surface plot of the conditions shows a trajectory being plotted. This means that the trajectory is "rolling" towards the products or reactants. However, if a trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

[[File:01409469_internuclearpic3.png|thumb|center|400px|'''Figure 3.''' A plot of internuclear distance against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

An initial estimate for the position of the transition state was made by observing the potential energy surface and was found to be 0.90 Å. At 0.90 Å, the internuclear distance against time plot showed that the internuclear distance was constantly oscillating. Therefore, the position of the transition state was further refined by increasing the number of decimal places until a horizontal line in the internuclear distance againt time plot was obtained. This means that the internuclear distance is constant and that the molecules are simply undergoing periodic symmetric vibration at the transition state. Therefore, the position of the transition state was determined to be 0.907743 Å.

[[File:01409469 combined energy pic.png|thumb|center|400px|'''Figure 4.''' A plot of energy against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

When rAB=rBC=0.907743 Å and pA=pBC=0, the kinetic energy is 0. This means that HA and HBC are stationary, which is as predicted when the trajectory is started at the transition state with no initial momentum. Furthermore, the forces acting on HA and HBC are 0 as should be the case in the transition state.

=== Calculation type: Dynamic and minimum energy path ===

The minimum energy path, is the lowest energy path from the reactants to the products. The minimum energy path is determined by following the mode corresponding to the negative Hessian eigenvalue in both directions along the steepest descent direction. Following this path allows one to approach two valleys of stable structures. The resulting path generated is the minimum energy path. The reaction paths were chartered for a dynamic and a minimum energy path. The initial conditions used were that of the transition state except that the internuclear distance between HA and HB had increased to 0.917743 Å. The reaction path obtained using the two different calculation methods was different.

[[File:01409469 mep and dynamics.jpg|thumb|center|800px|'''Figure 5.''' A surface plot of the trajectory using minimum energy path '''Figure 6.''' A surface plot of the trajectory using dynamics (Left-right)]]

On the surface plot, the reaction path is depicted by the black line. Following the minimum energy path, the trajectory simply follows the valley floor to Ha+Hb-Hc. The reaction path appears to be a relatively straight line indicating that the atoms are not vibrating. In the minimum energy path, the reactants change into the products in an infinitely slow motion. This is done by having the momenta/velocities reset to zero in each time step. While the minimum energy path is useful in characterising a chemical reaction, it does not provide a realistic representation of the motion of the atoms during the reaction. This is because the mass and inertia of the atoms have been ignored.

Since the atoms simply follow the minimum energy path, they do not undergo slight displacements into surrounding valleys of higher potential energy. Furthermore, because the atoms are assumed to have no inertia, they do not undergo vibrations even if they were to be displaced to valleys of higher potential energy.

However, when dynamic calculation is performed, the reaction path is wavy indicating that the atoms are oscillating. This is because the atoms are displaced to regions of higher potential on the path from reactants to products. This leads to atom oscillation due to the inertia of the atoms.

'''Internuclear distance against time plot'''

Setting the initial system conditions such that rAB= 0.907743, rBC= 0.917743 and pA= pBC= 0:

[[File:01409469 Internuclear distance mep and dynamic.jpg|thumb|center|800px|'''Figure 6.''' A graph of internuclear distance against steps using minimum energy path '''Figure 7.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In the minimum energy path, the distance between HA and HB decreases from 0.90 Å to approximately 0.79 Å during the first 50 steps before plateauing off to a value of 0.74 Å which is the approximate length of a hydrogen bond. This means that a bond was formed between HA and HB. In addition, the plateau indicates that the bond is static. On the other hand, the distance between HB and HC has increased to 1.91 Å at large step values. This shows that HC has been ejected off following a successful collision between HA and HBC.

Similarly, in the dynamic trajectory, the distance between HA and HB decreases to approximately 0.75 Å after 0.3 s. This is indicative of hydrogen bond formation between HA and HB. In addition, the distance between HB and HC increases to 3.44 Å after HC is ejected following a successful collision.

One obvious distance between the two plots is that in the minimum energy path trajectory, the internuclear distance between HA and HB is constant indicating that the bond is static. However, in the dynamic trajectory the internuclear distance is oscillating showing the the bond is vibrating. This difference could be attributed to the fact that minimum energy path does not take oscillatory motion into account while dynamic trajectory does.

However, changing the initial conditions of the system such that rAB = 0.917743 Å, rBC = 0.907743 Å and pA=pBC=0:
[[File:01409469 01409469 Reversed values dynamic mep.jpg|thumb|center|800px|'''Figure 8.''' A graph of internuclear distance against steps using minimum energy path '''Figure 9.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In this case, a hydrogen bond of length 0.77-0.79 Å is formed between HB and HC while HA is repelled away to beyond 3.45 Å.

Additionally, from the internuclear distance plot it can be seen that the distance between HB and HC increases in a rectangular hyperbola fashion when calculated using the minimum energy path. On the other hand, the internuclear distance between HB and HC increases in a linear manner when dynamic calculation is applied. A possible reason for this difference is that in the minimum energy path, the atoms are not vibrating and as such the reach the maximum distance and plateau at a faster rate. However, in the dynamic calculation, the atoms have momentum and as such move further away from each other.

=== Reactive and unreactive trajectories ===

With the initial positions r'''1''' = 0.74 and r'''2''' = 2.0, trajectories were run with the momenta combination shown below.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Contour plot !! Reactivity !! Description of the dynamics
|-
| -1.25 || -2.5 ||-99.018 ||[[File:01409469 trajectory1.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. There are no oscillations observed in HA-HB initially, indicating that the diatomic molecule is stable. A new bond is formed between HB and HC. Oscillatory behaviour is observed indicating that the bond is vibrating. On the other hand, HA is ejected away to beyond 10 Å.
|-
| -1.5 || -2.0 ||-100.456 ||[[File:01409469 trajectory2.jpg|thumb|center|400px]] ||Unreactive ||HC collides with the vibrating diatomic HA-HB. However, no bond is formed between HB and HC as the collision is unsuccessful. This is because the molecules lack sufficient kinetic energy to overcome the activation energy barrier and so no reaction occurs. HC is subsequently repelled to beyond 8 Å.
|-
| -1.5 || -2.5 ||-98.956 ||[[File:01409469 trajectory3.jpg|thumb|center|400px]] ||Reactive |||HC collides with HA-HB. The collision is successful and a bond is formed between HB and HC. This is because the atoms have sufficient kinetic energy to overcome the activation energy barrier and form a new HB-HC bond. HA is ejected away to beyond 10 Å. The new hydrogen bond formed is vibrating as seen by the oscillatory behaviour.
|-
| -2.5 || -5.0 ||-84.956 ||[[File:01409469 trajectory4.jpg|thumb|center|400px]] ||Unreactive ||HC collides with HA-HB. Initially the HA-HB bond is elongating, indicating that the hydrogen bond is weakening and going to break. However, after 0.5 s, the HA-HB bond length decreases and the HAHB is not broken. HC is repelled beyond 6 Å with very strong vibrations.
A reaction did not occur even though the atoms have sufficient kinetic energy to cross the activation energy barrier. This is because the vibrations are too strong. The system crosses the transition state region and a bond is formed in the product temporarily but then the system reverts back to the reactants. This is an example of barrier recrossing.
|-
| -2.5 || -5.2 ||-83.416 ||[[File:01409469 trajectory5.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. The collision is successful and a new hydrogen bond is formed between HB and HC with HA being ejected away. The products formed experience very strong vibrations as evidenced by the oscillatory motion which occur with a very high frequency.
|}

From the table, it can be seen that the chance of a reaction successfully occurring is not solely dependent on the amount of kinetic energy the atoms possess. Even if the atoms have sufficient kinetic energy to cross the activation energy barrier, barrier recrossing can occur if the atoms are too vibrationally excited.

=== Transition State Theory ===

Transition state theory provided the first theoretical attempt to determine absolute reaction rates. In this theory, a transition state separating the reactants from products is used to find an expression for the thermal rate constant.

The main assumptions of transition state theory are:

1. Electronic and nuclear motions are separated (equivalent to the Born-Oppenheimer approximation in quantum mechanics)

2. Reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution

3. Molecular systems that have crossed the transition state in the direction of products cannot turn around and reform reactants

4. In the transition state, motion along the reaction coordinate may be separated from the other motions and treated classically as a translation

5. Even in the absence of an equilibrium between reactant and product molecules, the transition states that are becoming products are distributed among their states according to the Maxwell-Boltzmann laws

The goal of transition state theory is to predict the rate constant, ''k'', of a reaction given a potential energy surface for the reaction.

However, transition state theory suffers from a set of limitations due to the assumptions stated above not holding true in reality.

Firstly, transition state theory assumes that barrier recrossing is negligible. However, as seen from the fourth trajectory in table 1, barrier recrossing occurred because the atoms were too vibrationally excited and the system reverted back to the reactants even though the system had already crossed the transition state region. In this situation, transition state theory would predict a faster rate of reaction compared to what was obtained experimentally.

Secondly, the transition state theory breaks down for systems with light atoms such as hydrogens. In such systems, quantum effects become crucial. Quantum effects tend to increase the rate of the reaction for two reasons. First, particles lie in energy levels which are higher then the bottom of the potential energy well (there is additional zero-point energy). Secondly, the particles can tunnel through the barrier. These two effects effectively lower the activation energy barrier. As a result, the rate of reaction predicted by transition state theory would be less than what is obtained experimentally.

Lastly, transition state theory does not hold true for systems which are far from equilibrium. In such cases, the assumption of a starting Boltzmann distribution breaks down.

== Exercise 2: F - H - H system ==

=== Potential energy surface inspection===

F + HA-HB → HA-F + HB

[[File:01409469 Surface_Plot F 3.jpg|thumb|center|400px|'''Figure 10.''' Potential energy surface for the reaction between F and H2]]

The reaction between F and H2 is an exothermic reaction. This means that the reactants are of a higher energy than the products. The overall reaction involves the breaking of a hydrogen bond and the formation of a hydrogen-fluorine bond. The HF bond formed is stronger than the hydrogen bond that was broken. This is because fluorine is more electronegative than hydrogen resulting in a polar HF bond. The negative enthalpy of H-F bond formation is greater than the positive enthalpy of hydrogen bond dissociation. Therefore, the reaction between F and H2 is an exothermic reaction.

HA + HB-F → HA-HB + F

[[File:01409469 Surface_Plot F 4.jpg|thumb|center|400px|'''Figure 11.''' Potential energy surface for the reaction between H and HF]]

The reaction between H and HF is an endothermic reaction. The products are at a higher energy level than the reactants. The overall reaction involves the breaking of a HF bond and the formation of a hydrogen bond. The formation of a homonuclear H2 molecule results in a non-polar hydrogen bond being formed. The HF bond broken is stronger than the hydrogen bond that was formed. The negative enthalpy of hydrogen bond formation is less than the positive enthalpy of hydrogen-fluorine dissociation. Therefore, the reaction between H and HF is endothermic.

=== Locating the transition state ===

[[File:01409469 Surface_Plot F TS4.jpg|thumb|center|800px|'''Figure 12.''' Determining the position of the transition state for the reaction between F and H2]]

As mentioned previously, when the molecules are not at the transition state, a reaction path will be plotted. However, if the initial conditions of the system are set such that the trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

An initial estimate for the transition state was made by setting rAB = 0.74 which is the length of a hydrogen bond. The position of the transition state was further refined by increasing the number of decimal places such that the final value of rAB was found to be 0.74478 and rBC = 1.81079. This set of distances gave a reaction path that was represented by a point indicating that no trajectory was plotted and that the molecules remained at the transition state since their initial momentum was 0.

Moreover, the kinetic energy as well as the forces acting on the atoms was calculated to be 0, which is consistent with the transition state.

Therefore, the position of the transition state was found to be at rAB = 0.74478 and rBC = 1.81079 when pA = pBC = 0.

=== Determining the activation energy ===

[[File:01409469 activation energy r1.1 combined.jpg|thumb|center|800px|'''Figure 13.''' Determining the activation energy for the reaction between F and H2]]

By reducing the value of rBC by 0.01, the atoms are no longer at the transition state and would move towards the formation of products. The activation energy was found by calculating the change in total energy after the reaction. The activation energy of F + H2 → HF + H was found to be +23.279 kcal/ mol.

[[File:01409469 activation energy r2.1 combined.jpg|thumb|center|800px|'''Figure 14.''' Determining the activation energy for the reaction between FH and H]]

The activation energy of FH + H → F + H2 was found by calculating the small dip in potential energy. The activation energy was found to be +0.280 kcal/ mol.

=== Reaction dynamics of F + HA-HB → HA-F + HB ===

The set of initial conditions that result in a reactive trajectory for F + H2 is as follows: rHH = 0.74 Å, rHF = 1.79 Å and pHH = pHF = 0.

[[File:01409469 momentum r1.jpg|thumb|center|400px|'''Figure 15.''' Graph of momentum against time for the reaction between F and H2]]

Initially, the H2 molecule is travelling in a relatively straight line with very little oscillations as evidenced by the low amplitude of momentum between 0-2 s. Following successful collision between F and H2, the amplitude of momentum of the product, HF, increases dramatically. This is because the reaction between F and H2 is exothermic and the excess energy released causes the molecule to become vibrationally excited and the amplitude of oscillations in the product to increase.

[[File:01409469 energy1 r1.jpg|thumb|center|400px|'''Figure 16.''' Graph of energy against time for the reaction between F and H2]]

The law of conservation of energy states that energy cannot be created or destroyed. It simply changes from one form to another. As can be seen from the graph in figure 16, total energy remains constant. In this reaction, potential energy is converted to kinetic energy. A decrease in potential energy of the atoms following a successful collision between F and H2 results in an equal increase in kinetic energy which manifests in the form of increased product (HF) oscillations.

This can be confirmed experimentally using infrared chemiluminescence. Observation of radiation from excited product molecules is a clear indication that these molecules are produced in the excited state. If the excitation is primarily vibrational, the radiation will appear in the infrared region of the spectrum (between 3-15 μm). This technique has been used extensively in hydrogen halogen reactions which produce HX molecules in the excited state.

One such experiment involves the use of the arrested relaxation method. In this technique, the reaction is carried out in a chamber whose walls are maintained at a very low temperature (20 K). This results in cryopumping of the chamber in that all the products and unreacted starting materials are condensed on the walls and only those product molecules that radiate before being pumped away are observed.

=== Polanyi's empirical rules ===

In a typical chemical reaction, there is an energetic barrier — a saddle point that the reactants must surmount to become products. Which form of energy is initially deposited in the reactants, translational or vibrational, is more efficacious in surmounting the activation energy barrier is one of the core topics in the area of reaction dynamics. The Polyani rules state that vibrational energy is more effective at promoting a late-barrier reaction (transition state resembles products). Conversely, translational energy is better at promoting an early-barrier reaction (transition state resembles the reactants).

The forward reaction between H2 and F is exothermic. Therefore, the transition state is closer to the reactants than to the products in energy. Since the transition state occurs early in an exothermic reaction, translational energy is more effective at promoting a chemical reaction than vibrational energy.

Conversely, the backward reaction between H and HF is endothermic. According to Hammond's postulate, the transition state resembles the products more than it resembles the reactants. Therefore, in an endothermic reaction, a late transition state is observed and vibrational energy is more effective at promoting a reaction than translational energy.

=== Promoting the forward reaction: F + HA-HB → HA-F + HB ===

1. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 2.95 and pHF = -0.5.

[[File:01409469 polyani 1+2.jpg|thumb|center|800px|'''Figure 17.''' Contour plot of the reaction between F and H2 when pHH = 2.95 '''Figure 18.''' Graph of momentum against time for the reaction between F and H2 when pHH = 2.95 (Left-right)]]

Vibrational energy is represented by pHH while translational energy is represented by pHF.

A molecule of H2 collides with F. However, the collision is unsuccessful and no reaction occurs. A reaction did not occur even though the molecules have sufficient kinetic energy to overcome the activation energy barrier. This is because most of the energy is deposited in the reactants as H-H vibrations as seen by the large oscillations in H2. Barrier recrossing has occured as a result of the large vibrational energy causing the system to revert back to the reactants even though the system has already crossed the transition state and a bond was formed between HB and F temporarily.

This appears to run counter to what one would expect to happen when more energy is put into the system. It is predicted that when more energy is put into the system, the reactants would possess more kinetic energy to overcome the activation energy barrier. However, as can be seen from the simulation above, an increase in the vibrational energy of the reactants is not effective at promoting the occurrence of an exothermic reaction with an early transition state. This is in accordance to Polyani's rules.

2. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 0.1 and pHF = -0.8.

[[File:01409469 polyani 3+4 2.jpg|thumb|center|800px|'''Figure 19.''' Contour plot of the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 '''Figure 20.''' Graph of momentum against time for the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 (Left-right)]]

The collision between a molecule of H2 and F is successful and a chemical reaction has occurred. A reaction has occurred even though the overall energy of the system has been significantly reduced by lowering the amount of vibrational energy in the reactants but slightly increasing the amount of translational energy deposited in the reactants. Therefore, H2 and F have more translational energy to move towards each other and form a H-F bond. The product formed is vibrationally excited as seen by the large amplitude of oscillations in the product HF which correspond to the strong H-F vibrations.

From the second simulation, it can be seen that an increase in translational energy in the reactants is more effective at promoting an exothermic reaction with an early transition state to occur, which again follows Polyani's rules.

=== Promoting the backward reaction: HA + HB-F → HA-HB + F ===

1. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 0.2 and pHH = -3.0.

[[File:01409469 polyani 5+6 .jpg|thumb|center|800px|'''Figure 21.''' Contour plot of the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 '''Figure 22.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 (Left-right)]]

An atom of HA collides with a molecule of HBF. However, the collision is unsuccessful and no chemical reaction occurs. This is because the reactants lack sufficient vibrational energy to surmount the activation energy barrier. Therefore, following an unsuccessful collision with HBF, HA simply bounces off and moves away at constant momentum. This is in agreement with Polyani's rules which state that vibrational energy is more effective than translational energy at promoting reactions with a late transition state, for example, endothermic chemical reactions.

In addition, because the reaction between HA + HB-F is endothermic, energy is taken in from the surroundings during the reaction and the products are less vibrationally excited than the reactants.

File:01409469 polyani 5+6 .jpg

2019-05-10T13:58:33Z

Stc1917:

MRD:01409469

2019-05-10T13:58:19Z

Stc1917: /* Promoting the backward reaction: HA + HB-F → HA-HB + F */

== Exercise 1: H + H2 system ==

=== Definition of a transition state ===

A potential energy profile results from plotting the potential versus reaction path.

[[File:Surface_Plot01409469_pic1.jpg|thumb|center|400px|'''Figure 1.''' A potential energy surface plot]]

The transition state is defined as the maximum on the minimum energy path linking reactants and products. This means that the transition state is a stationary point on the potential energy surface. Thus, at the transition state, the derivative of the potential energy is zero with respect to each Cartesian coordinate.

<math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>

However, a local minimum on the potential energy surface corresponding to a well would have the same <math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>. Therefore, to distinguish between a local minimum on the potential energy surface and the transition state, the second derivative, <math>\tfrac{\partial^2 V(r_i)}{\partial r_i ^2}\ </math> is determined. This can be calculated by finding the determinant of the Hessian matrix shown below.

:<math>\mathbf H =
\begin{bmatrix}
\dfrac{\partial^2 \ V(r_i)}{\partial \ x^2} & \dfrac{\partial^2 \ V(r_i)}{\partial \ x \partial \mathbf y} \\
\dfrac{\partial^2 \ V(r_i)}{\partial \ y \partial \mathbf x} & \dfrac{\partial^2 \ V(r_i)}{\partial \mathbf y^2}
\end{bmatrix}</math>

The Hessian matrix of a convex function is positive semi-definite. Refining this property allows one to test if a critical point (x,y) is a local maximum, local minimum or saddle point.

If <math>f</math> < 0, then (x,y) is an isolated local maximum.

If <math>f</math> > 0, then (x,y) is an isolated local minimum.

If the Hessian has both positive and negative eigenvalues then (x,y) is a saddle point.

=== Locating the transition state ===

[[File:01409469 combined pic 2 and 3.jpg|thumb|center|800px|'''Figure 2.''' A surface and contour plot of the trajectory when rAB=rBC=0.907743 Å and pA=pBC=0]]

When the molecules are not at the transition state, a surface plot of the conditions shows a trajectory being plotted. This means that the trajectory is "rolling" towards the products or reactants. However, if a trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

[[File:01409469_internuclearpic3.png|thumb|center|400px|'''Figure 3.''' A plot of internuclear distance against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

An initial estimate for the position of the transition state was made by observing the potential energy surface and was found to be 0.90 Å. At 0.90 Å, the internuclear distance against time plot showed that the internuclear distance was constantly oscillating. Therefore, the position of the transition state was further refined by increasing the number of decimal places until a horizontal line in the internuclear distance againt time plot was obtained. This means that the internuclear distance is constant and that the molecules are simply undergoing periodic symmetric vibration at the transition state. Therefore, the position of the transition state was determined to be 0.907743 Å.

[[File:01409469 combined energy pic.png|thumb|center|400px|'''Figure 4.''' A plot of energy against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

When rAB=rBC=0.907743 Å and pA=pBC=0, the kinetic energy is 0. This means that HA and HBC are stationary, which is as predicted when the trajectory is started at the transition state with no initial momentum. Furthermore, the forces acting on HA and HBC are 0 as should be the case in the transition state.

=== Calculation type: Dynamic and minimum energy path ===

The minimum energy path, is the lowest energy path from the reactants to the products. The minimum energy path is determined by following the mode corresponding to the negative Hessian eigenvalue in both directions along the steepest descent direction. Following this path allows one to approach two valleys of stable structures. The resulting path generated is the minimum energy path. The reaction paths were chartered for a dynamic and a minimum energy path. The initial conditions used were that of the transition state except that the internuclear distance between HA and HB had increased to 0.917743 Å. The reaction path obtained using the two different calculation methods was different.

[[File:01409469 mep and dynamics.jpg|thumb|center|800px|'''Figure 5.''' A surface plot of the trajectory using minimum energy path '''Figure 6.''' A surface plot of the trajectory using dynamics (Left-right)]]

On the surface plot, the reaction path is depicted by the black line. Following the minimum energy path, the trajectory simply follows the valley floor to Ha+Hb-Hc. The reaction path appears to be a relatively straight line indicating that the atoms are not vibrating. In the minimum energy path, the reactants change into the products in an infinitely slow motion. This is done by having the momenta/velocities reset to zero in each time step. While the minimum energy path is useful in characterising a chemical reaction, it does not provide a realistic representation of the motion of the atoms during the reaction. This is because the mass and inertia of the atoms have been ignored.

Since the atoms simply follow the minimum energy path, they do not undergo slight displacements into surrounding valleys of higher potential energy. Furthermore, because the atoms are assumed to have no inertia, they do not undergo vibrations even if they were to be displaced to valleys of higher potential energy.

However, when dynamic calculation is performed, the reaction path is wavy indicating that the atoms are oscillating. This is because the atoms are displaced to regions of higher potential on the path from reactants to products. This leads to atom oscillation due to the inertia of the atoms.

'''Internuclear distance against time plot'''

Setting the initial system conditions such that rAB= 0.907743, rBC= 0.917743 and pA= pBC= 0:

[[File:01409469 Internuclear distance mep and dynamic.jpg|thumb|center|800px|'''Figure 6.''' A graph of internuclear distance against steps using minimum energy path '''Figure 7.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In the minimum energy path, the distance between HA and HB decreases from 0.90 Å to approximately 0.79 Å during the first 50 steps before plateauing off to a value of 0.74 Å which is the approximate length of a hydrogen bond. This means that a bond was formed between HA and HB. In addition, the plateau indicates that the bond is static. On the other hand, the distance between HB and HC has increased to 1.91 Å at large step values. This shows that HC has been ejected off following a successful collision between HA and HBC.

Similarly, in the dynamic trajectory, the distance between HA and HB decreases to approximately 0.75 Å after 0.3 s. This is indicative of hydrogen bond formation between HA and HB. In addition, the distance between HB and HC increases to 3.44 Å after HC is ejected following a successful collision.

One obvious distance between the two plots is that in the minimum energy path trajectory, the internuclear distance between HA and HB is constant indicating that the bond is static. However, in the dynamic trajectory the internuclear distance is oscillating showing the the bond is vibrating. This difference could be attributed to the fact that minimum energy path does not take oscillatory motion into account while dynamic trajectory does.

However, changing the initial conditions of the system such that rAB = 0.917743 Å, rBC = 0.907743 Å and pA=pBC=0:
[[File:01409469 01409469 Reversed values dynamic mep.jpg|thumb|center|800px|'''Figure 8.''' A graph of internuclear distance against steps using minimum energy path '''Figure 9.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In this case, a hydrogen bond of length 0.77-0.79 Å is formed between HB and HC while HA is repelled away to beyond 3.45 Å.

Additionally, from the internuclear distance plot it can be seen that the distance between HB and HC increases in a rectangular hyperbola fashion when calculated using the minimum energy path. On the other hand, the internuclear distance between HB and HC increases in a linear manner when dynamic calculation is applied. A possible reason for this difference is that in the minimum energy path, the atoms are not vibrating and as such the reach the maximum distance and plateau at a faster rate. However, in the dynamic calculation, the atoms have momentum and as such move further away from each other.

=== Reactive and unreactive trajectories ===

With the initial positions r'''1''' = 0.74 and r'''2''' = 2.0, trajectories were run with the momenta combination shown below.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Contour plot !! Reactivity !! Description of the dynamics
|-
| -1.25 || -2.5 ||-99.018 ||[[File:01409469 trajectory1.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. There are no oscillations observed in HA-HB initially, indicating that the diatomic molecule is stable. A new bond is formed between HB and HC. Oscillatory behaviour is observed indicating that the bond is vibrating. On the other hand, HA is ejected away to beyond 10 Å.
|-
| -1.5 || -2.0 ||-100.456 ||[[File:01409469 trajectory2.jpg|thumb|center|400px]] ||Unreactive ||HC collides with the vibrating diatomic HA-HB. However, no bond is formed between HB and HC as the collision is unsuccessful. This is because the molecules lack sufficient kinetic energy to overcome the activation energy barrier and so no reaction occurs. HC is subsequently repelled to beyond 8 Å.
|-
| -1.5 || -2.5 ||-98.956 ||[[File:01409469 trajectory3.jpg|thumb|center|400px]] ||Reactive |||HC collides with HA-HB. The collision is successful and a bond is formed between HB and HC. This is because the atoms have sufficient kinetic energy to overcome the activation energy barrier and form a new HB-HC bond. HA is ejected away to beyond 10 Å. The new hydrogen bond formed is vibrating as seen by the oscillatory behaviour.
|-
| -2.5 || -5.0 ||-84.956 ||[[File:01409469 trajectory4.jpg|thumb|center|400px]] ||Unreactive ||HC collides with HA-HB. Initially the HA-HB bond is elongating, indicating that the hydrogen bond is weakening and going to break. However, after 0.5 s, the HA-HB bond length decreases and the HAHB is not broken. HC is repelled beyond 6 Å with very strong vibrations.
A reaction did not occur even though the atoms have sufficient kinetic energy to cross the activation energy barrier. This is because the vibrations are too strong. The system crosses the transition state region and a bond is formed in the product temporarily but then the system reverts back to the reactants. This is an example of barrier recrossing.
|-
| -2.5 || -5.2 ||-83.416 ||[[File:01409469 trajectory5.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. The collision is successful and a new hydrogen bond is formed between HB and HC with HA being ejected away. The products formed experience very strong vibrations as evidenced by the oscillatory motion which occur with a very high frequency.
|}

From the table, it can be seen that the chance of a reaction successfully occurring is not solely dependent on the amount of kinetic energy the atoms possess. Even if the atoms have sufficient kinetic energy to cross the activation energy barrier, barrier recrossing can occur if the atoms are too vibrationally excited.

=== Transition State Theory ===

Transition state theory provided the first theoretical attempt to determine absolute reaction rates. In this theory, a transition state separating the reactants from products is used to find an expression for the thermal rate constant.

The main assumptions of transition state theory are:

1. Electronic and nuclear motions are separated (equivalent to the Born-Oppenheimer approximation in quantum mechanics)

2. Reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution

3. Molecular systems that have crossed the transition state in the direction of products cannot turn around and reform reactants

4. In the transition state, motion along the reaction coordinate may be separated from the other motions and treated classically as a translation

5. Even in the absence of an equilibrium between reactant and product molecules, the transition states that are becoming products are distributed among their states according to the Maxwell-Boltzmann laws

The goal of transition state theory is to predict the rate constant, ''k'', of a reaction given a potential energy surface for the reaction.

However, transition state theory suffers from a set of limitations due to the assumptions stated above not holding true in reality.

Firstly, transition state theory assumes that barrier recrossing is negligible. However, as seen from the fourth trajectory in table 1, barrier recrossing occurred because the atoms were too vibrationally excited and the system reverted back to the reactants even though the system had already crossed the transition state region. In this situation, transition state theory would predict a faster rate of reaction compared to what was obtained experimentally.

Secondly, the transition state theory breaks down for systems with light atoms such as hydrogens. In such systems, quantum effects become crucial. Quantum effects tend to increase the rate of the reaction for two reasons. First, particles lie in energy levels which are higher then the bottom of the potential energy well (there is additional zero-point energy). Secondly, the particles can tunnel through the barrier. These two effects effectively lower the activation energy barrier. As a result, the rate of reaction predicted by transition state theory would be less than what is obtained experimentally.

Lastly, transition state theory does not hold true for systems which are far from equilibrium. In such cases, the assumption of a starting Boltzmann distribution breaks down.

== Exercise 2: F - H - H system ==

=== Potential energy surface inspection===

F + HA-HB → HA-F + HB

[[File:01409469 Surface_Plot F 3.jpg|thumb|center|400px|'''Figure 10.''' Potential energy surface for the reaction between F and H2]]

The reaction between F and H2 is an exothermic reaction. This means that the reactants are of a higher energy than the products. The overall reaction involves the breaking of a hydrogen bond and the formation of a hydrogen-fluorine bond. The HF bond formed is stronger than the hydrogen bond that was broken. This is because fluorine is more electronegative than hydrogen resulting in a polar HF bond. The negative enthalpy of H-F bond formation is greater than the positive enthalpy of hydrogen bond dissociation. Therefore, the reaction between F and H2 is an exothermic reaction.

HA + HB-F → HA-HB + F

[[File:01409469 Surface_Plot F 4.jpg|thumb|center|400px|'''Figure 11.''' Potential energy surface for the reaction between H and HF]]

The reaction between H and HF is an endothermic reaction. The products are at a higher energy level than the reactants. The overall reaction involves the breaking of a HF bond and the formation of a hydrogen bond. The formation of a homonuclear H2 molecule results in a non-polar hydrogen bond being formed. The HF bond broken is stronger than the hydrogen bond that was formed. The negative enthalpy of hydrogen bond formation is less than the positive enthalpy of hydrogen-fluorine dissociation. Therefore, the reaction between H and HF is endothermic.

=== Locating the transition state ===

[[File:01409469 Surface_Plot F TS4.jpg|thumb|center|800px|'''Figure 12.''' Determining the position of the transition state for the reaction between F and H2]]

As mentioned previously, when the molecules are not at the transition state, a reaction path will be plotted. However, if the initial conditions of the system are set such that the trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

An initial estimate for the transition state was made by setting rAB = 0.74 which is the length of a hydrogen bond. The position of the transition state was further refined by increasing the number of decimal places such that the final value of rAB was found to be 0.74478 and rBC = 1.81079. This set of distances gave a reaction path that was represented by a point indicating that no trajectory was plotted and that the molecules remained at the transition state since their initial momentum was 0.

Moreover, the kinetic energy as well as the forces acting on the atoms was calculated to be 0, which is consistent with the transition state.

Therefore, the position of the transition state was found to be at rAB = 0.74478 and rBC = 1.81079 when pA = pBC = 0.

=== Determining the activation energy ===

[[File:01409469 activation energy r1.1 combined.jpg|thumb|center|800px|'''Figure 13.''' Determining the activation energy for the reaction between F and H2]]

By reducing the value of rBC by 0.01, the atoms are no longer at the transition state and would move towards the formation of products. The activation energy was found by calculating the change in total energy after the reaction. The activation energy of F + H2 → HF + H was found to be +23.279 kcal/ mol.

[[File:01409469 activation energy r2.1 combined.jpg|thumb|center|800px|'''Figure 14.''' Determining the activation energy for the reaction between FH and H]]

The activation energy of FH + H → F + H2 was found by calculating the small dip in potential energy. The activation energy was found to be +0.280 kcal/ mol.

=== Reaction dynamics of F + HA-HB → HA-F + HB ===

The set of initial conditions that result in a reactive trajectory for F + H2 is as follows: rHH = 0.74 Å, rHF = 1.79 Å and pHH = pHF = 0.

[[File:01409469 momentum r1.jpg|thumb|center|400px|'''Figure 15.''' Graph of momentum against time for the reaction between F and H2]]

Initially, the H2 molecule is travelling in a relatively straight line with very little oscillations as evidenced by the low amplitude of momentum between 0-2 s. Following successful collision between F and H2, the amplitude of momentum of the product, HF, increases dramatically. This is because the reaction between F and H2 is exothermic and the excess energy released causes the molecule to become vibrationally excited and the amplitude of oscillations in the product to increase.

[[File:01409469 energy1 r1.jpg|thumb|center|400px|'''Figure 16.''' Graph of energy against time for the reaction between F and H2]]

The law of conservation of energy states that energy cannot be created or destroyed. It simply changes from one form to another. As can be seen from the graph in figure 16, total energy remains constant. In this reaction, potential energy is converted to kinetic energy. A decrease in potential energy of the atoms following a successful collision between F and H2 results in an equal increase in kinetic energy which manifests in the form of increased product (HF) oscillations.

This can be confirmed experimentally using infrared chemiluminescence. Observation of radiation from excited product molecules is a clear indication that these molecules are produced in the excited state. If the excitation is primarily vibrational, the radiation will appear in the infrared region of the spectrum (between 3-15 μm). This technique has been used extensively in hydrogen halogen reactions which produce HX molecules in the excited state.

One such experiment involves the use of the arrested relaxation method. In this technique, the reaction is carried out in a chamber whose walls are maintained at a very low temperature (20 K). This results in cryopumping of the chamber in that all the products and unreacted starting materials are condensed on the walls and only those product molecules that radiate before being pumped away are observed.

=== Polanyi's empirical rules ===

In a typical chemical reaction, there is an energetic barrier — a saddle point that the reactants must surmount to become products. Which form of energy is initially deposited in the reactants, translational or vibrational, is more efficacious in surmounting the activation energy barrier is one of the core topics in the area of reaction dynamics. The Polyani rules state that vibrational energy is more effective at promoting a late-barrier reaction (transition state resembles products). Conversely, translational energy is better at promoting an early-barrier reaction (transition state resembles the reactants).

The forward reaction between H2 and F is exothermic. Therefore, the transition state is closer to the reactants than to the products in energy. Since the transition state occurs early in an exothermic reaction, translational energy is more effective at promoting a chemical reaction than vibrational energy.

Conversely, the backward reaction between H and HF is endothermic. According to Hammond's postulate, the transition state resembles the products more than it resembles the reactants. Therefore, in an endothermic reaction, a late transition state is observed and vibrational energy is more effective at promoting a reaction than translational energy.

=== Promoting the forward reaction: F + HA-HB → HA-F + HB ===

1. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 2.95 and pHF = -0.5.

[[File:01409469 polyani 1+2.jpg|thumb|center|800px|'''Figure 17.''' Contour plot of the reaction between F and H2 when pHH = 2.95 '''Figure 18.''' Graph of momentum against time for the reaction between F and H2 when pHH = 2.95 (Left-right)]]

Vibrational energy is represented by pHH while translational energy is represented by pHF.

A molecule of H2 collides with F. However, the collision is unsuccessful and no reaction occurs. A reaction did not occur even though the molecules have sufficient kinetic energy to overcome the activation energy barrier. This is because most of the energy is deposited in the reactants as H-H vibrations as seen by the large oscillations in H2. Barrier recrossing has occured as a result of the large vibrational energy causing the system to revert back to the reactants even though the system has already crossed the transition state and a bond was formed between HB and F temporarily.

This appears to run counter to what one would expect to happen when more energy is put into the system. It is predicted that when more energy is put into the system, the reactants would possess more kinetic energy to overcome the activation energy barrier. However, as can be seen from the simulation above, an increase in the vibrational energy of the reactants is not effective at promoting the occurrence of an exothermic reaction with an early transition state. This is in accordance to Polyani's rules.

2. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 0.1 and pHF = -0.8.

[[File:01409469 polyani 3+4 2.jpg|thumb|center|800px|'''Figure 19.''' Contour plot of the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 '''Figure 20.''' Graph of momentum against time for the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 (Left-right)]]

The collision between a molecule of H2 and F is successful and a chemical reaction has occurred. A reaction has occurred even though the overall energy of the system has been significantly reduced by lowering the amount of vibrational energy in the reactants but slightly increasing the amount of translational energy deposited in the reactants. Therefore, H2 and F have more translational energy to move towards each other and form a H-F bond. The product formed is vibrationally excited as seen by the large amplitude of oscillations in the product HF which correspond to the strong H-F vibrations.

From the second simulation, it can be seen that an increase in translational energy in the reactants is more effective at promoting an exothermic reaction with an early transition state to occur, which again follows Polyani's rules.

=== Promoting the backward reaction: HA + HB-F → HA-HB + F ===

1. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 0.2 and pHH = -3.0.

[[File:01409469 polyani 5+6 .jpg|thumb|center|800px|'''Figure 21.''' Contour plot of the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 '''Figure 22.''' Graph of momentum against time for the reaction between HA + HB-F when pFH = 0.2 and pHH = -3.0 (Left-right)]]

MRD:01409469

2019-05-10T13:55:45Z

Stc1917: /* Promoting the backward reaction: HA + HB-F → HA-HB + F */

== Exercise 1: H + H2 system ==

=== Definition of a transition state ===

A potential energy profile results from plotting the potential versus reaction path.

[[File:Surface_Plot01409469_pic1.jpg|thumb|center|400px|'''Figure 1.''' A potential energy surface plot]]

The transition state is defined as the maximum on the minimum energy path linking reactants and products. This means that the transition state is a stationary point on the potential energy surface. Thus, at the transition state, the derivative of the potential energy is zero with respect to each Cartesian coordinate.

<math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>

However, a local minimum on the potential energy surface corresponding to a well would have the same <math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>. Therefore, to distinguish between a local minimum on the potential energy surface and the transition state, the second derivative, <math>\tfrac{\partial^2 V(r_i)}{\partial r_i ^2}\ </math> is determined. This can be calculated by finding the determinant of the Hessian matrix shown below.

:<math>\mathbf H =
\begin{bmatrix}
\dfrac{\partial^2 \ V(r_i)}{\partial \ x^2} & \dfrac{\partial^2 \ V(r_i)}{\partial \ x \partial \mathbf y} \\
\dfrac{\partial^2 \ V(r_i)}{\partial \ y \partial \mathbf x} & \dfrac{\partial^2 \ V(r_i)}{\partial \mathbf y^2}
\end{bmatrix}</math>

The Hessian matrix of a convex function is positive semi-definite. Refining this property allows one to test if a critical point (x,y) is a local maximum, local minimum or saddle point.

If <math>f</math> < 0, then (x,y) is an isolated local maximum.

If <math>f</math> > 0, then (x,y) is an isolated local minimum.

If the Hessian has both positive and negative eigenvalues then (x,y) is a saddle point.

=== Locating the transition state ===

[[File:01409469 combined pic 2 and 3.jpg|thumb|center|800px|'''Figure 2.''' A surface and contour plot of the trajectory when rAB=rBC=0.907743 Å and pA=pBC=0]]

When the molecules are not at the transition state, a surface plot of the conditions shows a trajectory being plotted. This means that the trajectory is "rolling" towards the products or reactants. However, if a trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

[[File:01409469_internuclearpic3.png|thumb|center|400px|'''Figure 3.''' A plot of internuclear distance against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

An initial estimate for the position of the transition state was made by observing the potential energy surface and was found to be 0.90 Å. At 0.90 Å, the internuclear distance against time plot showed that the internuclear distance was constantly oscillating. Therefore, the position of the transition state was further refined by increasing the number of decimal places until a horizontal line in the internuclear distance againt time plot was obtained. This means that the internuclear distance is constant and that the molecules are simply undergoing periodic symmetric vibration at the transition state. Therefore, the position of the transition state was determined to be 0.907743 Å.

[[File:01409469 combined energy pic.png|thumb|center|400px|'''Figure 4.''' A plot of energy against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

When rAB=rBC=0.907743 Å and pA=pBC=0, the kinetic energy is 0. This means that HA and HBC are stationary, which is as predicted when the trajectory is started at the transition state with no initial momentum. Furthermore, the forces acting on HA and HBC are 0 as should be the case in the transition state.

=== Calculation type: Dynamic and minimum energy path ===

The minimum energy path, is the lowest energy path from the reactants to the products. The minimum energy path is determined by following the mode corresponding to the negative Hessian eigenvalue in both directions along the steepest descent direction. Following this path allows one to approach two valleys of stable structures. The resulting path generated is the minimum energy path. The reaction paths were chartered for a dynamic and a minimum energy path. The initial conditions used were that of the transition state except that the internuclear distance between HA and HB had increased to 0.917743 Å. The reaction path obtained using the two different calculation methods was different.

[[File:01409469 mep and dynamics.jpg|thumb|center|800px|'''Figure 5.''' A surface plot of the trajectory using minimum energy path '''Figure 6.''' A surface plot of the trajectory using dynamics (Left-right)]]

On the surface plot, the reaction path is depicted by the black line. Following the minimum energy path, the trajectory simply follows the valley floor to Ha+Hb-Hc. The reaction path appears to be a relatively straight line indicating that the atoms are not vibrating. In the minimum energy path, the reactants change into the products in an infinitely slow motion. This is done by having the momenta/velocities reset to zero in each time step. While the minimum energy path is useful in characterising a chemical reaction, it does not provide a realistic representation of the motion of the atoms during the reaction. This is because the mass and inertia of the atoms have been ignored.

Since the atoms simply follow the minimum energy path, they do not undergo slight displacements into surrounding valleys of higher potential energy. Furthermore, because the atoms are assumed to have no inertia, they do not undergo vibrations even if they were to be displaced to valleys of higher potential energy.

However, when dynamic calculation is performed, the reaction path is wavy indicating that the atoms are oscillating. This is because the atoms are displaced to regions of higher potential on the path from reactants to products. This leads to atom oscillation due to the inertia of the atoms.

'''Internuclear distance against time plot'''

Setting the initial system conditions such that rAB= 0.907743, rBC= 0.917743 and pA= pBC= 0:

[[File:01409469 Internuclear distance mep and dynamic.jpg|thumb|center|800px|'''Figure 6.''' A graph of internuclear distance against steps using minimum energy path '''Figure 7.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In the minimum energy path, the distance between HA and HB decreases from 0.90 Å to approximately 0.79 Å during the first 50 steps before plateauing off to a value of 0.74 Å which is the approximate length of a hydrogen bond. This means that a bond was formed between HA and HB. In addition, the plateau indicates that the bond is static. On the other hand, the distance between HB and HC has increased to 1.91 Å at large step values. This shows that HC has been ejected off following a successful collision between HA and HBC.

Similarly, in the dynamic trajectory, the distance between HA and HB decreases to approximately 0.75 Å after 0.3 s. This is indicative of hydrogen bond formation between HA and HB. In addition, the distance between HB and HC increases to 3.44 Å after HC is ejected following a successful collision.

One obvious distance between the two plots is that in the minimum energy path trajectory, the internuclear distance between HA and HB is constant indicating that the bond is static. However, in the dynamic trajectory the internuclear distance is oscillating showing the the bond is vibrating. This difference could be attributed to the fact that minimum energy path does not take oscillatory motion into account while dynamic trajectory does.

However, changing the initial conditions of the system such that rAB = 0.917743 Å, rBC = 0.907743 Å and pA=pBC=0:
[[File:01409469 01409469 Reversed values dynamic mep.jpg|thumb|center|800px|'''Figure 8.''' A graph of internuclear distance against steps using minimum energy path '''Figure 9.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In this case, a hydrogen bond of length 0.77-0.79 Å is formed between HB and HC while HA is repelled away to beyond 3.45 Å.

Additionally, from the internuclear distance plot it can be seen that the distance between HB and HC increases in a rectangular hyperbola fashion when calculated using the minimum energy path. On the other hand, the internuclear distance between HB and HC increases in a linear manner when dynamic calculation is applied. A possible reason for this difference is that in the minimum energy path, the atoms are not vibrating and as such the reach the maximum distance and plateau at a faster rate. However, in the dynamic calculation, the atoms have momentum and as such move further away from each other.

=== Reactive and unreactive trajectories ===

With the initial positions r'''1''' = 0.74 and r'''2''' = 2.0, trajectories were run with the momenta combination shown below.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Contour plot !! Reactivity !! Description of the dynamics
|-
| -1.25 || -2.5 ||-99.018 ||[[File:01409469 trajectory1.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. There are no oscillations observed in HA-HB initially, indicating that the diatomic molecule is stable. A new bond is formed between HB and HC. Oscillatory behaviour is observed indicating that the bond is vibrating. On the other hand, HA is ejected away to beyond 10 Å.
|-
| -1.5 || -2.0 ||-100.456 ||[[File:01409469 trajectory2.jpg|thumb|center|400px]] ||Unreactive ||HC collides with the vibrating diatomic HA-HB. However, no bond is formed between HB and HC as the collision is unsuccessful. This is because the molecules lack sufficient kinetic energy to overcome the activation energy barrier and so no reaction occurs. HC is subsequently repelled to beyond 8 Å.
|-
| -1.5 || -2.5 ||-98.956 ||[[File:01409469 trajectory3.jpg|thumb|center|400px]] ||Reactive |||HC collides with HA-HB. The collision is successful and a bond is formed between HB and HC. This is because the atoms have sufficient kinetic energy to overcome the activation energy barrier and form a new HB-HC bond. HA is ejected away to beyond 10 Å. The new hydrogen bond formed is vibrating as seen by the oscillatory behaviour.
|-
| -2.5 || -5.0 ||-84.956 ||[[File:01409469 trajectory4.jpg|thumb|center|400px]] ||Unreactive ||HC collides with HA-HB. Initially the HA-HB bond is elongating, indicating that the hydrogen bond is weakening and going to break. However, after 0.5 s, the HA-HB bond length decreases and the HAHB is not broken. HC is repelled beyond 6 Å with very strong vibrations.
A reaction did not occur even though the atoms have sufficient kinetic energy to cross the activation energy barrier. This is because the vibrations are too strong. The system crosses the transition state region and a bond is formed in the product temporarily but then the system reverts back to the reactants. This is an example of barrier recrossing.
|-
| -2.5 || -5.2 ||-83.416 ||[[File:01409469 trajectory5.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. The collision is successful and a new hydrogen bond is formed between HB and HC with HA being ejected away. The products formed experience very strong vibrations as evidenced by the oscillatory motion which occur with a very high frequency.
|}

From the table, it can be seen that the chance of a reaction successfully occurring is not solely dependent on the amount of kinetic energy the atoms possess. Even if the atoms have sufficient kinetic energy to cross the activation energy barrier, barrier recrossing can occur if the atoms are too vibrationally excited.

=== Transition State Theory ===

Transition state theory provided the first theoretical attempt to determine absolute reaction rates. In this theory, a transition state separating the reactants from products is used to find an expression for the thermal rate constant.

The main assumptions of transition state theory are:

1. Electronic and nuclear motions are separated (equivalent to the Born-Oppenheimer approximation in quantum mechanics)

2. Reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution

3. Molecular systems that have crossed the transition state in the direction of products cannot turn around and reform reactants

4. In the transition state, motion along the reaction coordinate may be separated from the other motions and treated classically as a translation

5. Even in the absence of an equilibrium between reactant and product molecules, the transition states that are becoming products are distributed among their states according to the Maxwell-Boltzmann laws

The goal of transition state theory is to predict the rate constant, ''k'', of a reaction given a potential energy surface for the reaction.

However, transition state theory suffers from a set of limitations due to the assumptions stated above not holding true in reality.

Firstly, transition state theory assumes that barrier recrossing is negligible. However, as seen from the fourth trajectory in table 1, barrier recrossing occurred because the atoms were too vibrationally excited and the system reverted back to the reactants even though the system had already crossed the transition state region. In this situation, transition state theory would predict a faster rate of reaction compared to what was obtained experimentally.

Secondly, the transition state theory breaks down for systems with light atoms such as hydrogens. In such systems, quantum effects become crucial. Quantum effects tend to increase the rate of the reaction for two reasons. First, particles lie in energy levels which are higher then the bottom of the potential energy well (there is additional zero-point energy). Secondly, the particles can tunnel through the barrier. These two effects effectively lower the activation energy barrier. As a result, the rate of reaction predicted by transition state theory would be less than what is obtained experimentally.

Lastly, transition state theory does not hold true for systems which are far from equilibrium. In such cases, the assumption of a starting Boltzmann distribution breaks down.

== Exercise 2: F - H - H system ==

=== Potential energy surface inspection===

F + HA-HB → HA-F + HB

[[File:01409469 Surface_Plot F 3.jpg|thumb|center|400px|'''Figure 10.''' Potential energy surface for the reaction between F and H2]]

The reaction between F and H2 is an exothermic reaction. This means that the reactants are of a higher energy than the products. The overall reaction involves the breaking of a hydrogen bond and the formation of a hydrogen-fluorine bond. The HF bond formed is stronger than the hydrogen bond that was broken. This is because fluorine is more electronegative than hydrogen resulting in a polar HF bond. The negative enthalpy of H-F bond formation is greater than the positive enthalpy of hydrogen bond dissociation. Therefore, the reaction between F and H2 is an exothermic reaction.

HA + HB-F → HA-HB + F

[[File:01409469 Surface_Plot F 4.jpg|thumb|center|400px|'''Figure 11.''' Potential energy surface for the reaction between H and HF]]

The reaction between H and HF is an endothermic reaction. The products are at a higher energy level than the reactants. The overall reaction involves the breaking of a HF bond and the formation of a hydrogen bond. The formation of a homonuclear H2 molecule results in a non-polar hydrogen bond being formed. The HF bond broken is stronger than the hydrogen bond that was formed. The negative enthalpy of hydrogen bond formation is less than the positive enthalpy of hydrogen-fluorine dissociation. Therefore, the reaction between H and HF is endothermic.

=== Locating the transition state ===

[[File:01409469 Surface_Plot F TS4.jpg|thumb|center|800px|'''Figure 12.''' Determining the position of the transition state for the reaction between F and H2]]

As mentioned previously, when the molecules are not at the transition state, a reaction path will be plotted. However, if the initial conditions of the system are set such that the trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

An initial estimate for the transition state was made by setting rAB = 0.74 which is the length of a hydrogen bond. The position of the transition state was further refined by increasing the number of decimal places such that the final value of rAB was found to be 0.74478 and rBC = 1.81079. This set of distances gave a reaction path that was represented by a point indicating that no trajectory was plotted and that the molecules remained at the transition state since their initial momentum was 0.

Moreover, the kinetic energy as well as the forces acting on the atoms was calculated to be 0, which is consistent with the transition state.

Therefore, the position of the transition state was found to be at rAB = 0.74478 and rBC = 1.81079 when pA = pBC = 0.

=== Determining the activation energy ===

[[File:01409469 activation energy r1.1 combined.jpg|thumb|center|800px|'''Figure 13.''' Determining the activation energy for the reaction between F and H2]]

By reducing the value of rBC by 0.01, the atoms are no longer at the transition state and would move towards the formation of products. The activation energy was found by calculating the change in total energy after the reaction. The activation energy of F + H2 → HF + H was found to be +23.279 kcal/ mol.

[[File:01409469 activation energy r2.1 combined.jpg|thumb|center|800px|'''Figure 14.''' Determining the activation energy for the reaction between FH and H]]

The activation energy of FH + H → F + H2 was found by calculating the small dip in potential energy. The activation energy was found to be +0.280 kcal/ mol.

=== Reaction dynamics of F + HA-HB → HA-F + HB ===

The set of initial conditions that result in a reactive trajectory for F + H2 is as follows: rHH = 0.74 Å, rHF = 1.79 Å and pHH = pHF = 0.

[[File:01409469 momentum r1.jpg|thumb|center|400px|'''Figure 15.''' Graph of momentum against time for the reaction between F and H2]]

Initially, the H2 molecule is travelling in a relatively straight line with very little oscillations as evidenced by the low amplitude of momentum between 0-2 s. Following successful collision between F and H2, the amplitude of momentum of the product, HF, increases dramatically. This is because the reaction between F and H2 is exothermic and the excess energy released causes the molecule to become vibrationally excited and the amplitude of oscillations in the product to increase.

[[File:01409469 energy1 r1.jpg|thumb|center|400px|'''Figure 16.''' Graph of energy against time for the reaction between F and H2]]

The law of conservation of energy states that energy cannot be created or destroyed. It simply changes from one form to another. As can be seen from the graph in figure 16, total energy remains constant. In this reaction, potential energy is converted to kinetic energy. A decrease in potential energy of the atoms following a successful collision between F and H2 results in an equal increase in kinetic energy which manifests in the form of increased product (HF) oscillations.

This can be confirmed experimentally using infrared chemiluminescence. Observation of radiation from excited product molecules is a clear indication that these molecules are produced in the excited state. If the excitation is primarily vibrational, the radiation will appear in the infrared region of the spectrum (between 3-15 μm). This technique has been used extensively in hydrogen halogen reactions which produce HX molecules in the excited state.

One such experiment involves the use of the arrested relaxation method. In this technique, the reaction is carried out in a chamber whose walls are maintained at a very low temperature (20 K). This results in cryopumping of the chamber in that all the products and unreacted starting materials are condensed on the walls and only those product molecules that radiate before being pumped away are observed.

=== Polanyi's empirical rules ===

In a typical chemical reaction, there is an energetic barrier — a saddle point that the reactants must surmount to become products. Which form of energy is initially deposited in the reactants, translational or vibrational, is more efficacious in surmounting the activation energy barrier is one of the core topics in the area of reaction dynamics. The Polyani rules state that vibrational energy is more effective at promoting a late-barrier reaction (transition state resembles products). Conversely, translational energy is better at promoting an early-barrier reaction (transition state resembles the reactants).

The forward reaction between H2 and F is exothermic. Therefore, the transition state is closer to the reactants than to the products in energy. Since the transition state occurs early in an exothermic reaction, translational energy is more effective at promoting a chemical reaction than vibrational energy.

Conversely, the backward reaction between H and HF is endothermic. According to Hammond's postulate, the transition state resembles the products more than it resembles the reactants. Therefore, in an endothermic reaction, a late transition state is observed and vibrational energy is more effective at promoting a reaction than translational energy.

=== Promoting the forward reaction: F + HA-HB → HA-F + HB ===

1. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 2.95 and pHF = -0.5.

[[File:01409469 polyani 1+2.jpg|thumb|center|800px|'''Figure 17.''' Contour plot of the reaction between F and H2 when pHH = 2.95 '''Figure 18.''' Graph of momentum against time for the reaction between F and H2 when pHH = 2.95 (Left-right)]]

Vibrational energy is represented by pHH while translational energy is represented by pHF.

A molecule of H2 collides with F. However, the collision is unsuccessful and no reaction occurs. A reaction did not occur even though the molecules have sufficient kinetic energy to overcome the activation energy barrier. This is because most of the energy is deposited in the reactants as H-H vibrations as seen by the large oscillations in H2. Barrier recrossing has occured as a result of the large vibrational energy causing the system to revert back to the reactants even though the system has already crossed the transition state and a bond was formed between HB and F temporarily.

This appears to run counter to what one would expect to happen when more energy is put into the system. It is predicted that when more energy is put into the system, the reactants would possess more kinetic energy to overcome the activation energy barrier. However, as can be seen from the simulation above, an increase in the vibrational energy of the reactants is not effective at promoting the occurrence of an exothermic reaction with an early transition state. This is in accordance to Polyani's rules.

2. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 0.1 and pHF = -0.8.

[[File:01409469 polyani 3+4 2.jpg|thumb|center|800px|'''Figure 19.''' Contour plot of the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 '''Figure 20.''' Graph of momentum against time for the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 (Left-right)]]

The collision between a molecule of H2 and F is successful and a chemical reaction has occurred. A reaction has occurred even though the overall energy of the system has been significantly reduced by lowering the amount of vibrational energy in the reactants but slightly increasing the amount of translational energy deposited in the reactants. Therefore, H2 and F have more translational energy to move towards each other and form a H-F bond. The product formed is vibrationally excited as seen by the large amplitude of oscillations in the product HF which correspond to the strong H-F vibrations.

From the second simulation, it can be seen that an increase in translational energy in the reactants is more effective at promoting an exothermic reaction with an early transition state to occur, which again follows Polyani's rules.

=== Promoting the backward reaction: HA + HB-F → HA-HB + F ===

1. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 0.2 and pHH = -3.0.

MRD:01409469

2019-05-10T13:55:20Z

Stc1917: /* Promoting the backward reaction: HA + HB-F → HA-HB + F */

== Exercise 1: H + H2 system ==

=== Definition of a transition state ===

A potential energy profile results from plotting the potential versus reaction path.

[[File:Surface_Plot01409469_pic1.jpg|thumb|center|400px|'''Figure 1.''' A potential energy surface plot]]

The transition state is defined as the maximum on the minimum energy path linking reactants and products. This means that the transition state is a stationary point on the potential energy surface. Thus, at the transition state, the derivative of the potential energy is zero with respect to each Cartesian coordinate.

<math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>

However, a local minimum on the potential energy surface corresponding to a well would have the same <math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>. Therefore, to distinguish between a local minimum on the potential energy surface and the transition state, the second derivative, <math>\tfrac{\partial^2 V(r_i)}{\partial r_i ^2}\ </math> is determined. This can be calculated by finding the determinant of the Hessian matrix shown below.

:<math>\mathbf H =
\begin{bmatrix}
\dfrac{\partial^2 \ V(r_i)}{\partial \ x^2} & \dfrac{\partial^2 \ V(r_i)}{\partial \ x \partial \mathbf y} \\
\dfrac{\partial^2 \ V(r_i)}{\partial \ y \partial \mathbf x} & \dfrac{\partial^2 \ V(r_i)}{\partial \mathbf y^2}
\end{bmatrix}</math>

The Hessian matrix of a convex function is positive semi-definite. Refining this property allows one to test if a critical point (x,y) is a local maximum, local minimum or saddle point.

If <math>f</math> < 0, then (x,y) is an isolated local maximum.

If <math>f</math> > 0, then (x,y) is an isolated local minimum.

If the Hessian has both positive and negative eigenvalues then (x,y) is a saddle point.

=== Locating the transition state ===

[[File:01409469 combined pic 2 and 3.jpg|thumb|center|800px|'''Figure 2.''' A surface and contour plot of the trajectory when rAB=rBC=0.907743 Å and pA=pBC=0]]

When the molecules are not at the transition state, a surface plot of the conditions shows a trajectory being plotted. This means that the trajectory is "rolling" towards the products or reactants. However, if a trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

[[File:01409469_internuclearpic3.png|thumb|center|400px|'''Figure 3.''' A plot of internuclear distance against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

An initial estimate for the position of the transition state was made by observing the potential energy surface and was found to be 0.90 Å. At 0.90 Å, the internuclear distance against time plot showed that the internuclear distance was constantly oscillating. Therefore, the position of the transition state was further refined by increasing the number of decimal places until a horizontal line in the internuclear distance againt time plot was obtained. This means that the internuclear distance is constant and that the molecules are simply undergoing periodic symmetric vibration at the transition state. Therefore, the position of the transition state was determined to be 0.907743 Å.

[[File:01409469 combined energy pic.png|thumb|center|400px|'''Figure 4.''' A plot of energy against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

When rAB=rBC=0.907743 Å and pA=pBC=0, the kinetic energy is 0. This means that HA and HBC are stationary, which is as predicted when the trajectory is started at the transition state with no initial momentum. Furthermore, the forces acting on HA and HBC are 0 as should be the case in the transition state.

=== Calculation type: Dynamic and minimum energy path ===

The minimum energy path, is the lowest energy path from the reactants to the products. The minimum energy path is determined by following the mode corresponding to the negative Hessian eigenvalue in both directions along the steepest descent direction. Following this path allows one to approach two valleys of stable structures. The resulting path generated is the minimum energy path. The reaction paths were chartered for a dynamic and a minimum energy path. The initial conditions used were that of the transition state except that the internuclear distance between HA and HB had increased to 0.917743 Å. The reaction path obtained using the two different calculation methods was different.

[[File:01409469 mep and dynamics.jpg|thumb|center|800px|'''Figure 5.''' A surface plot of the trajectory using minimum energy path '''Figure 6.''' A surface plot of the trajectory using dynamics (Left-right)]]

On the surface plot, the reaction path is depicted by the black line. Following the minimum energy path, the trajectory simply follows the valley floor to Ha+Hb-Hc. The reaction path appears to be a relatively straight line indicating that the atoms are not vibrating. In the minimum energy path, the reactants change into the products in an infinitely slow motion. This is done by having the momenta/velocities reset to zero in each time step. While the minimum energy path is useful in characterising a chemical reaction, it does not provide a realistic representation of the motion of the atoms during the reaction. This is because the mass and inertia of the atoms have been ignored.

Since the atoms simply follow the minimum energy path, they do not undergo slight displacements into surrounding valleys of higher potential energy. Furthermore, because the atoms are assumed to have no inertia, they do not undergo vibrations even if they were to be displaced to valleys of higher potential energy.

However, when dynamic calculation is performed, the reaction path is wavy indicating that the atoms are oscillating. This is because the atoms are displaced to regions of higher potential on the path from reactants to products. This leads to atom oscillation due to the inertia of the atoms.

'''Internuclear distance against time plot'''

Setting the initial system conditions such that rAB= 0.907743, rBC= 0.917743 and pA= pBC= 0:

[[File:01409469 Internuclear distance mep and dynamic.jpg|thumb|center|800px|'''Figure 6.''' A graph of internuclear distance against steps using minimum energy path '''Figure 7.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In the minimum energy path, the distance between HA and HB decreases from 0.90 Å to approximately 0.79 Å during the first 50 steps before plateauing off to a value of 0.74 Å which is the approximate length of a hydrogen bond. This means that a bond was formed between HA and HB. In addition, the plateau indicates that the bond is static. On the other hand, the distance between HB and HC has increased to 1.91 Å at large step values. This shows that HC has been ejected off following a successful collision between HA and HBC.

Similarly, in the dynamic trajectory, the distance between HA and HB decreases to approximately 0.75 Å after 0.3 s. This is indicative of hydrogen bond formation between HA and HB. In addition, the distance between HB and HC increases to 3.44 Å after HC is ejected following a successful collision.

One obvious distance between the two plots is that in the minimum energy path trajectory, the internuclear distance between HA and HB is constant indicating that the bond is static. However, in the dynamic trajectory the internuclear distance is oscillating showing the the bond is vibrating. This difference could be attributed to the fact that minimum energy path does not take oscillatory motion into account while dynamic trajectory does.

However, changing the initial conditions of the system such that rAB = 0.917743 Å, rBC = 0.907743 Å and pA=pBC=0:
[[File:01409469 01409469 Reversed values dynamic mep.jpg|thumb|center|800px|'''Figure 8.''' A graph of internuclear distance against steps using minimum energy path '''Figure 9.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In this case, a hydrogen bond of length 0.77-0.79 Å is formed between HB and HC while HA is repelled away to beyond 3.45 Å.

Additionally, from the internuclear distance plot it can be seen that the distance between HB and HC increases in a rectangular hyperbola fashion when calculated using the minimum energy path. On the other hand, the internuclear distance between HB and HC increases in a linear manner when dynamic calculation is applied. A possible reason for this difference is that in the minimum energy path, the atoms are not vibrating and as such the reach the maximum distance and plateau at a faster rate. However, in the dynamic calculation, the atoms have momentum and as such move further away from each other.

=== Reactive and unreactive trajectories ===

With the initial positions r'''1''' = 0.74 and r'''2''' = 2.0, trajectories were run with the momenta combination shown below.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Contour plot !! Reactivity !! Description of the dynamics
|-
| -1.25 || -2.5 ||-99.018 ||[[File:01409469 trajectory1.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. There are no oscillations observed in HA-HB initially, indicating that the diatomic molecule is stable. A new bond is formed between HB and HC. Oscillatory behaviour is observed indicating that the bond is vibrating. On the other hand, HA is ejected away to beyond 10 Å.
|-
| -1.5 || -2.0 ||-100.456 ||[[File:01409469 trajectory2.jpg|thumb|center|400px]] ||Unreactive ||HC collides with the vibrating diatomic HA-HB. However, no bond is formed between HB and HC as the collision is unsuccessful. This is because the molecules lack sufficient kinetic energy to overcome the activation energy barrier and so no reaction occurs. HC is subsequently repelled to beyond 8 Å.
|-
| -1.5 || -2.5 ||-98.956 ||[[File:01409469 trajectory3.jpg|thumb|center|400px]] ||Reactive |||HC collides with HA-HB. The collision is successful and a bond is formed between HB and HC. This is because the atoms have sufficient kinetic energy to overcome the activation energy barrier and form a new HB-HC bond. HA is ejected away to beyond 10 Å. The new hydrogen bond formed is vibrating as seen by the oscillatory behaviour.
|-
| -2.5 || -5.0 ||-84.956 ||[[File:01409469 trajectory4.jpg|thumb|center|400px]] ||Unreactive ||HC collides with HA-HB. Initially the HA-HB bond is elongating, indicating that the hydrogen bond is weakening and going to break. However, after 0.5 s, the HA-HB bond length decreases and the HAHB is not broken. HC is repelled beyond 6 Å with very strong vibrations.
A reaction did not occur even though the atoms have sufficient kinetic energy to cross the activation energy barrier. This is because the vibrations are too strong. The system crosses the transition state region and a bond is formed in the product temporarily but then the system reverts back to the reactants. This is an example of barrier recrossing.
|-
| -2.5 || -5.2 ||-83.416 ||[[File:01409469 trajectory5.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. The collision is successful and a new hydrogen bond is formed between HB and HC with HA being ejected away. The products formed experience very strong vibrations as evidenced by the oscillatory motion which occur with a very high frequency.
|}

From the table, it can be seen that the chance of a reaction successfully occurring is not solely dependent on the amount of kinetic energy the atoms possess. Even if the atoms have sufficient kinetic energy to cross the activation energy barrier, barrier recrossing can occur if the atoms are too vibrationally excited.

=== Transition State Theory ===

Transition state theory provided the first theoretical attempt to determine absolute reaction rates. In this theory, a transition state separating the reactants from products is used to find an expression for the thermal rate constant.

The main assumptions of transition state theory are:

1. Electronic and nuclear motions are separated (equivalent to the Born-Oppenheimer approximation in quantum mechanics)

2. Reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution

3. Molecular systems that have crossed the transition state in the direction of products cannot turn around and reform reactants

4. In the transition state, motion along the reaction coordinate may be separated from the other motions and treated classically as a translation

5. Even in the absence of an equilibrium between reactant and product molecules, the transition states that are becoming products are distributed among their states according to the Maxwell-Boltzmann laws

The goal of transition state theory is to predict the rate constant, ''k'', of a reaction given a potential energy surface for the reaction.

However, transition state theory suffers from a set of limitations due to the assumptions stated above not holding true in reality.

Firstly, transition state theory assumes that barrier recrossing is negligible. However, as seen from the fourth trajectory in table 1, barrier recrossing occurred because the atoms were too vibrationally excited and the system reverted back to the reactants even though the system had already crossed the transition state region. In this situation, transition state theory would predict a faster rate of reaction compared to what was obtained experimentally.

Secondly, the transition state theory breaks down for systems with light atoms such as hydrogens. In such systems, quantum effects become crucial. Quantum effects tend to increase the rate of the reaction for two reasons. First, particles lie in energy levels which are higher then the bottom of the potential energy well (there is additional zero-point energy). Secondly, the particles can tunnel through the barrier. These two effects effectively lower the activation energy barrier. As a result, the rate of reaction predicted by transition state theory would be less than what is obtained experimentally.

Lastly, transition state theory does not hold true for systems which are far from equilibrium. In such cases, the assumption of a starting Boltzmann distribution breaks down.

== Exercise 2: F - H - H system ==

=== Potential energy surface inspection===

F + HA-HB → HA-F + HB

[[File:01409469 Surface_Plot F 3.jpg|thumb|center|400px|'''Figure 10.''' Potential energy surface for the reaction between F and H2]]

The reaction between F and H2 is an exothermic reaction. This means that the reactants are of a higher energy than the products. The overall reaction involves the breaking of a hydrogen bond and the formation of a hydrogen-fluorine bond. The HF bond formed is stronger than the hydrogen bond that was broken. This is because fluorine is more electronegative than hydrogen resulting in a polar HF bond. The negative enthalpy of H-F bond formation is greater than the positive enthalpy of hydrogen bond dissociation. Therefore, the reaction between F and H2 is an exothermic reaction.

HA + HB-F → HA-HB + F

[[File:01409469 Surface_Plot F 4.jpg|thumb|center|400px|'''Figure 11.''' Potential energy surface for the reaction between H and HF]]

The reaction between H and HF is an endothermic reaction. The products are at a higher energy level than the reactants. The overall reaction involves the breaking of a HF bond and the formation of a hydrogen bond. The formation of a homonuclear H2 molecule results in a non-polar hydrogen bond being formed. The HF bond broken is stronger than the hydrogen bond that was formed. The negative enthalpy of hydrogen bond formation is less than the positive enthalpy of hydrogen-fluorine dissociation. Therefore, the reaction between H and HF is endothermic.

=== Locating the transition state ===

[[File:01409469 Surface_Plot F TS4.jpg|thumb|center|800px|'''Figure 12.''' Determining the position of the transition state for the reaction between F and H2]]

As mentioned previously, when the molecules are not at the transition state, a reaction path will be plotted. However, if the initial conditions of the system are set such that the trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

An initial estimate for the transition state was made by setting rAB = 0.74 which is the length of a hydrogen bond. The position of the transition state was further refined by increasing the number of decimal places such that the final value of rAB was found to be 0.74478 and rBC = 1.81079. This set of distances gave a reaction path that was represented by a point indicating that no trajectory was plotted and that the molecules remained at the transition state since their initial momentum was 0.

Moreover, the kinetic energy as well as the forces acting on the atoms was calculated to be 0, which is consistent with the transition state.

Therefore, the position of the transition state was found to be at rAB = 0.74478 and rBC = 1.81079 when pA = pBC = 0.

=== Determining the activation energy ===

[[File:01409469 activation energy r1.1 combined.jpg|thumb|center|800px|'''Figure 13.''' Determining the activation energy for the reaction between F and H2]]

By reducing the value of rBC by 0.01, the atoms are no longer at the transition state and would move towards the formation of products. The activation energy was found by calculating the change in total energy after the reaction. The activation energy of F + H2 → HF + H was found to be +23.279 kcal/ mol.

[[File:01409469 activation energy r2.1 combined.jpg|thumb|center|800px|'''Figure 14.''' Determining the activation energy for the reaction between FH and H]]

The activation energy of FH + H → F + H2 was found by calculating the small dip in potential energy. The activation energy was found to be +0.280 kcal/ mol.

=== Reaction dynamics of F + HA-HB → HA-F + HB ===

The set of initial conditions that result in a reactive trajectory for F + H2 is as follows: rHH = 0.74 Å, rHF = 1.79 Å and pHH = pHF = 0.

[[File:01409469 momentum r1.jpg|thumb|center|400px|'''Figure 15.''' Graph of momentum against time for the reaction between F and H2]]

Initially, the H2 molecule is travelling in a relatively straight line with very little oscillations as evidenced by the low amplitude of momentum between 0-2 s. Following successful collision between F and H2, the amplitude of momentum of the product, HF, increases dramatically. This is because the reaction between F and H2 is exothermic and the excess energy released causes the molecule to become vibrationally excited and the amplitude of oscillations in the product to increase.

[[File:01409469 energy1 r1.jpg|thumb|center|400px|'''Figure 16.''' Graph of energy against time for the reaction between F and H2]]

The law of conservation of energy states that energy cannot be created or destroyed. It simply changes from one form to another. As can be seen from the graph in figure 16, total energy remains constant. In this reaction, potential energy is converted to kinetic energy. A decrease in potential energy of the atoms following a successful collision between F and H2 results in an equal increase in kinetic energy which manifests in the form of increased product (HF) oscillations.

This can be confirmed experimentally using infrared chemiluminescence. Observation of radiation from excited product molecules is a clear indication that these molecules are produced in the excited state. If the excitation is primarily vibrational, the radiation will appear in the infrared region of the spectrum (between 3-15 μm). This technique has been used extensively in hydrogen halogen reactions which produce HX molecules in the excited state.

One such experiment involves the use of the arrested relaxation method. In this technique, the reaction is carried out in a chamber whose walls are maintained at a very low temperature (20 K). This results in cryopumping of the chamber in that all the products and unreacted starting materials are condensed on the walls and only those product molecules that radiate before being pumped away are observed.

=== Polanyi's empirical rules ===

In a typical chemical reaction, there is an energetic barrier — a saddle point that the reactants must surmount to become products. Which form of energy is initially deposited in the reactants, translational or vibrational, is more efficacious in surmounting the activation energy barrier is one of the core topics in the area of reaction dynamics. The Polyani rules state that vibrational energy is more effective at promoting a late-barrier reaction (transition state resembles products). Conversely, translational energy is better at promoting an early-barrier reaction (transition state resembles the reactants).

The forward reaction between H2 and F is exothermic. Therefore, the transition state is closer to the reactants than to the products in energy. Since the transition state occurs early in an exothermic reaction, translational energy is more effective at promoting a chemical reaction than vibrational energy.

Conversely, the backward reaction between H and HF is endothermic. According to Hammond's postulate, the transition state resembles the products more than it resembles the reactants. Therefore, in an endothermic reaction, a late transition state is observed and vibrational energy is more effective at promoting a reaction than translational energy.

=== Promoting the forward reaction: F + HA-HB → HA-F + HB ===

1. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 2.95 and pHF = -0.5.

[[File:01409469 polyani 1+2.jpg|thumb|center|800px|'''Figure 17.''' Contour plot of the reaction between F and H2 when pHH = 2.95 '''Figure 18.''' Graph of momentum against time for the reaction between F and H2 when pHH = 2.95 (Left-right)]]

Vibrational energy is represented by pHH while translational energy is represented by pHF.

A molecule of H2 collides with F. However, the collision is unsuccessful and no reaction occurs. A reaction did not occur even though the molecules have sufficient kinetic energy to overcome the activation energy barrier. This is because most of the energy is deposited in the reactants as H-H vibrations as seen by the large oscillations in H2. Barrier recrossing has occured as a result of the large vibrational energy causing the system to revert back to the reactants even though the system has already crossed the transition state and a bond was formed between HB and F temporarily.

This appears to run counter to what one would expect to happen when more energy is put into the system. It is predicted that when more energy is put into the system, the reactants would possess more kinetic energy to overcome the activation energy barrier. However, as can be seen from the simulation above, an increase in the vibrational energy of the reactants is not effective at promoting the occurrence of an exothermic reaction with an early transition state. This is in accordance to Polyani's rules.

2. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 0.1 and pHF = -0.8.

[[File:01409469 polyani 3+4 2.jpg|thumb|center|800px|'''Figure 19.''' Contour plot of the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 '''Figure 20.''' Graph of momentum against time for the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 (Left-right)]]

The collision between a molecule of H2 and F is successful and a chemical reaction has occurred. A reaction has occurred even though the overall energy of the system has been significantly reduced by lowering the amount of vibrational energy in the reactants but slightly increasing the amount of translational energy deposited in the reactants. Therefore, H2 and F have more translational energy to move towards each other and form a H-F bond. The product formed is vibrationally excited as seen by the large amplitude of oscillations in the product HF which correspond to the strong H-F vibrations.

From the second simulation, it can be seen that an increase in translational energy in the reactants is more effective at promoting an exothermic reaction with an early transition state to occur, which again follows Polyani's rules.

=== Promoting the backward reaction: HA + HB-F → HA-HB + F ===

1. The initial system conditions used are: rFH = 0.92 Å, rHH = 2.0 Å, pFH = 0.2 and pHF = -3.0.

MRD:01409469

2019-05-10T13:34:00Z

Stc1917: /* Promoting the backward reaction: HA + HB-FB → HA-HB + F */

== Exercise 1: H + H2 system ==

=== Definition of a transition state ===

A potential energy profile results from plotting the potential versus reaction path.

[[File:Surface_Plot01409469_pic1.jpg|thumb|center|400px|'''Figure 1.''' A potential energy surface plot]]

The transition state is defined as the maximum on the minimum energy path linking reactants and products. This means that the transition state is a stationary point on the potential energy surface. Thus, at the transition state, the derivative of the potential energy is zero with respect to each Cartesian coordinate.

<math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>

However, a local minimum on the potential energy surface corresponding to a well would have the same <math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>. Therefore, to distinguish between a local minimum on the potential energy surface and the transition state, the second derivative, <math>\tfrac{\partial^2 V(r_i)}{\partial r_i ^2}\ </math> is determined. This can be calculated by finding the determinant of the Hessian matrix shown below.

:<math>\mathbf H =
\begin{bmatrix}
\dfrac{\partial^2 \ V(r_i)}{\partial \ x^2} & \dfrac{\partial^2 \ V(r_i)}{\partial \ x \partial \mathbf y} \\
\dfrac{\partial^2 \ V(r_i)}{\partial \ y \partial \mathbf x} & \dfrac{\partial^2 \ V(r_i)}{\partial \mathbf y^2}
\end{bmatrix}</math>

The Hessian matrix of a convex function is positive semi-definite. Refining this property allows one to test if a critical point (x,y) is a local maximum, local minimum or saddle point.

If <math>f</math> < 0, then (x,y) is an isolated local maximum.

If <math>f</math> > 0, then (x,y) is an isolated local minimum.

If the Hessian has both positive and negative eigenvalues then (x,y) is a saddle point.

=== Locating the transition state ===

[[File:01409469 combined pic 2 and 3.jpg|thumb|center|800px|'''Figure 2.''' A surface and contour plot of the trajectory when rAB=rBC=0.907743 Å and pA=pBC=0]]

When the molecules are not at the transition state, a surface plot of the conditions shows a trajectory being plotted. This means that the trajectory is "rolling" towards the products or reactants. However, if a trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

[[File:01409469_internuclearpic3.png|thumb|center|400px|'''Figure 3.''' A plot of internuclear distance against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

An initial estimate for the position of the transition state was made by observing the potential energy surface and was found to be 0.90 Å. At 0.90 Å, the internuclear distance against time plot showed that the internuclear distance was constantly oscillating. Therefore, the position of the transition state was further refined by increasing the number of decimal places until a horizontal line in the internuclear distance againt time plot was obtained. This means that the internuclear distance is constant and that the molecules are simply undergoing periodic symmetric vibration at the transition state. Therefore, the position of the transition state was determined to be 0.907743 Å.

[[File:01409469 combined energy pic.png|thumb|center|400px|'''Figure 4.''' A plot of energy against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

When rAB=rBC=0.907743 Å and pA=pBC=0, the kinetic energy is 0. This means that HA and HBC are stationary, which is as predicted when the trajectory is started at the transition state with no initial momentum. Furthermore, the forces acting on HA and HBC are 0 as should be the case in the transition state.

=== Calculation type: Dynamic and minimum energy path ===

The minimum energy path, is the lowest energy path from the reactants to the products. The minimum energy path is determined by following the mode corresponding to the negative Hessian eigenvalue in both directions along the steepest descent direction. Following this path allows one to approach two valleys of stable structures. The resulting path generated is the minimum energy path. The reaction paths were chartered for a dynamic and a minimum energy path. The initial conditions used were that of the transition state except that the internuclear distance between HA and HB had increased to 0.917743 Å. The reaction path obtained using the two different calculation methods was different.

[[File:01409469 mep and dynamics.jpg|thumb|center|800px|'''Figure 5.''' A surface plot of the trajectory using minimum energy path '''Figure 6.''' A surface plot of the trajectory using dynamics (Left-right)]]

On the surface plot, the reaction path is depicted by the black line. Following the minimum energy path, the trajectory simply follows the valley floor to Ha+Hb-Hc. The reaction path appears to be a relatively straight line indicating that the atoms are not vibrating. In the minimum energy path, the reactants change into the products in an infinitely slow motion. This is done by having the momenta/velocities reset to zero in each time step. While the minimum energy path is useful in characterising a chemical reaction, it does not provide a realistic representation of the motion of the atoms during the reaction. This is because the mass and inertia of the atoms have been ignored.

Since the atoms simply follow the minimum energy path, they do not undergo slight displacements into surrounding valleys of higher potential energy. Furthermore, because the atoms are assumed to have no inertia, they do not undergo vibrations even if they were to be displaced to valleys of higher potential energy.

However, when dynamic calculation is performed, the reaction path is wavy indicating that the atoms are oscillating. This is because the atoms are displaced to regions of higher potential on the path from reactants to products. This leads to atom oscillation due to the inertia of the atoms.

'''Internuclear distance against time plot'''

Setting the initial system conditions such that rAB= 0.907743, rBC= 0.917743 and pA= pBC= 0:

[[File:01409469 Internuclear distance mep and dynamic.jpg|thumb|center|800px|'''Figure 6.''' A graph of internuclear distance against steps using minimum energy path '''Figure 7.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In the minimum energy path, the distance between HA and HB decreases from 0.90 Å to approximately 0.79 Å during the first 50 steps before plateauing off to a value of 0.74 Å which is the approximate length of a hydrogen bond. This means that a bond was formed between HA and HB. In addition, the plateau indicates that the bond is static. On the other hand, the distance between HB and HC has increased to 1.91 Å at large step values. This shows that HC has been ejected off following a successful collision between HA and HBC.

Similarly, in the dynamic trajectory, the distance between HA and HB decreases to approximately 0.75 Å after 0.3 s. This is indicative of hydrogen bond formation between HA and HB. In addition, the distance between HB and HC increases to 3.44 Å after HC is ejected following a successful collision.

One obvious distance between the two plots is that in the minimum energy path trajectory, the internuclear distance between HA and HB is constant indicating that the bond is static. However, in the dynamic trajectory the internuclear distance is oscillating showing the the bond is vibrating. This difference could be attributed to the fact that minimum energy path does not take oscillatory motion into account while dynamic trajectory does.

However, changing the initial conditions of the system such that rAB = 0.917743 Å, rBC = 0.907743 Å and pA=pBC=0:
[[File:01409469 01409469 Reversed values dynamic mep.jpg|thumb|center|800px|'''Figure 8.''' A graph of internuclear distance against steps using minimum energy path '''Figure 9.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In this case, a hydrogen bond of length 0.77-0.79 Å is formed between HB and HC while HA is repelled away to beyond 3.45 Å.

Additionally, from the internuclear distance plot it can be seen that the distance between HB and HC increases in a rectangular hyperbola fashion when calculated using the minimum energy path. On the other hand, the internuclear distance between HB and HC increases in a linear manner when dynamic calculation is applied. A possible reason for this difference is that in the minimum energy path, the atoms are not vibrating and as such the reach the maximum distance and plateau at a faster rate. However, in the dynamic calculation, the atoms have momentum and as such move further away from each other.

=== Reactive and unreactive trajectories ===

With the initial positions r'''1''' = 0.74 and r'''2''' = 2.0, trajectories were run with the momenta combination shown below.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Contour plot !! Reactivity !! Description of the dynamics
|-
| -1.25 || -2.5 ||-99.018 ||[[File:01409469 trajectory1.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. There are no oscillations observed in HA-HB initially, indicating that the diatomic molecule is stable. A new bond is formed between HB and HC. Oscillatory behaviour is observed indicating that the bond is vibrating. On the other hand, HA is ejected away to beyond 10 Å.
|-
| -1.5 || -2.0 ||-100.456 ||[[File:01409469 trajectory2.jpg|thumb|center|400px]] ||Unreactive ||HC collides with the vibrating diatomic HA-HB. However, no bond is formed between HB and HC as the collision is unsuccessful. This is because the molecules lack sufficient kinetic energy to overcome the activation energy barrier and so no reaction occurs. HC is subsequently repelled to beyond 8 Å.
|-
| -1.5 || -2.5 ||-98.956 ||[[File:01409469 trajectory3.jpg|thumb|center|400px]] ||Reactive |||HC collides with HA-HB. The collision is successful and a bond is formed between HB and HC. This is because the atoms have sufficient kinetic energy to overcome the activation energy barrier and form a new HB-HC bond. HA is ejected away to beyond 10 Å. The new hydrogen bond formed is vibrating as seen by the oscillatory behaviour.
|-
| -2.5 || -5.0 ||-84.956 ||[[File:01409469 trajectory4.jpg|thumb|center|400px]] ||Unreactive ||HC collides with HA-HB. Initially the HA-HB bond is elongating, indicating that the hydrogen bond is weakening and going to break. However, after 0.5 s, the HA-HB bond length decreases and the HAHB is not broken. HC is repelled beyond 6 Å with very strong vibrations.
A reaction did not occur even though the atoms have sufficient kinetic energy to cross the activation energy barrier. This is because the vibrations are too strong. The system crosses the transition state region and a bond is formed in the product temporarily but then the system reverts back to the reactants. This is an example of barrier recrossing.
|-
| -2.5 || -5.2 ||-83.416 ||[[File:01409469 trajectory5.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. The collision is successful and a new hydrogen bond is formed between HB and HC with HA being ejected away. The products formed experience very strong vibrations as evidenced by the oscillatory motion which occur with a very high frequency.
|}

From the table, it can be seen that the chance of a reaction successfully occurring is not solely dependent on the amount of kinetic energy the atoms possess. Even if the atoms have sufficient kinetic energy to cross the activation energy barrier, barrier recrossing can occur if the atoms are too vibrationally excited.

=== Transition State Theory ===

Transition state theory provided the first theoretical attempt to determine absolute reaction rates. In this theory, a transition state separating the reactants from products is used to find an expression for the thermal rate constant.

The main assumptions of transition state theory are:

1. Electronic and nuclear motions are separated (equivalent to the Born-Oppenheimer approximation in quantum mechanics)

2. Reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution

3. Molecular systems that have crossed the transition state in the direction of products cannot turn around and reform reactants

4. In the transition state, motion along the reaction coordinate may be separated from the other motions and treated classically as a translation

5. Even in the absence of an equilibrium between reactant and product molecules, the transition states that are becoming products are distributed among their states according to the Maxwell-Boltzmann laws

The goal of transition state theory is to predict the rate constant, ''k'', of a reaction given a potential energy surface for the reaction.

However, transition state theory suffers from a set of limitations due to the assumptions stated above not holding true in reality.

Firstly, transition state theory assumes that barrier recrossing is negligible. However, as seen from the fourth trajectory in table 1, barrier recrossing occurred because the atoms were too vibrationally excited and the system reverted back to the reactants even though the system had already crossed the transition state region. In this situation, transition state theory would predict a faster rate of reaction compared to what was obtained experimentally.

Secondly, the transition state theory breaks down for systems with light atoms such as hydrogens. In such systems, quantum effects become crucial. Quantum effects tend to increase the rate of the reaction for two reasons. First, particles lie in energy levels which are higher then the bottom of the potential energy well (there is additional zero-point energy). Secondly, the particles can tunnel through the barrier. These two effects effectively lower the activation energy barrier. As a result, the rate of reaction predicted by transition state theory would be less than what is obtained experimentally.

Lastly, transition state theory does not hold true for systems which are far from equilibrium. In such cases, the assumption of a starting Boltzmann distribution breaks down.

== Exercise 2: F - H - H system ==

=== Potential energy surface inspection===

F + HA-HB → HA-F + HB

[[File:01409469 Surface_Plot F 3.jpg|thumb|center|400px|'''Figure 10.''' Potential energy surface for the reaction between F and H2]]

The reaction between F and H2 is an exothermic reaction. This means that the reactants are of a higher energy than the products. The overall reaction involves the breaking of a hydrogen bond and the formation of a hydrogen-fluorine bond. The HF bond formed is stronger than the hydrogen bond that was broken. This is because fluorine is more electronegative than hydrogen resulting in a polar HF bond. The negative enthalpy of H-F bond formation is greater than the positive enthalpy of hydrogen bond dissociation. Therefore, the reaction between F and H2 is an exothermic reaction.

HA + HB-F → HA-HB + F

[[File:01409469 Surface_Plot F 4.jpg|thumb|center|400px|'''Figure 11.''' Potential energy surface for the reaction between H and HF]]

The reaction between H and HF is an endothermic reaction. The products are at a higher energy level than the reactants. The overall reaction involves the breaking of a HF bond and the formation of a hydrogen bond. The formation of a homonuclear H2 molecule results in a non-polar hydrogen bond being formed. The HF bond broken is stronger than the hydrogen bond that was formed. The negative enthalpy of hydrogen bond formation is less than the positive enthalpy of hydrogen-fluorine dissociation. Therefore, the reaction between H and HF is endothermic.

=== Locating the transition state ===

[[File:01409469 Surface_Plot F TS4.jpg|thumb|center|800px|'''Figure 12.''' Determining the position of the transition state for the reaction between F and H2]]

As mentioned previously, when the molecules are not at the transition state, a reaction path will be plotted. However, if the initial conditions of the system are set such that the trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

An initial estimate for the transition state was made by setting rAB = 0.74 which is the length of a hydrogen bond. The position of the transition state was further refined by increasing the number of decimal places such that the final value of rAB was found to be 0.74478 and rBC = 1.81079. This set of distances gave a reaction path that was represented by a point indicating that no trajectory was plotted and that the molecules remained at the transition state since their initial momentum was 0.

Moreover, the kinetic energy as well as the forces acting on the atoms was calculated to be 0, which is consistent with the transition state.

Therefore, the position of the transition state was found to be at rAB = 0.74478 and rBC = 1.81079 when pA = pBC = 0.

=== Determining the activation energy ===

[[File:01409469 activation energy r1.1 combined.jpg|thumb|center|800px|'''Figure 13.''' Determining the activation energy for the reaction between F and H2]]

By reducing the value of rBC by 0.01, the atoms are no longer at the transition state and would move towards the formation of products. The activation energy was found by calculating the change in total energy after the reaction. The activation energy of F + H2 → HF + H was found to be +23.279 kcal/ mol.

[[File:01409469 activation energy r2.1 combined.jpg|thumb|center|800px|'''Figure 14.''' Determining the activation energy for the reaction between FH and H]]

The activation energy of FH + H → F + H2 was found by calculating the small dip in potential energy. The activation energy was found to be +0.280 kcal/ mol.

=== Reaction dynamics of F + HA-HB → HA-F + HB ===

The set of initial conditions that result in a reactive trajectory for F + H2 is as follows: rHH = 0.74 Å, rHF = 1.79 Å and pHH = pHF = 0.

[[File:01409469 momentum r1.jpg|thumb|center|400px|'''Figure 15.''' Graph of momentum against time for the reaction between F and H2]]

Initially, the H2 molecule is travelling in a relatively straight line with very little oscillations as evidenced by the low amplitude of momentum between 0-2 s. Following successful collision between F and H2, the amplitude of momentum of the product, HF, increases dramatically. This is because the reaction between F and H2 is exothermic and the excess energy released causes the molecule to become vibrationally excited and the amplitude of oscillations in the product to increase.

[[File:01409469 energy1 r1.jpg|thumb|center|400px|'''Figure 16.''' Graph of energy against time for the reaction between F and H2]]

The law of conservation of energy states that energy cannot be created or destroyed. It simply changes from one form to another. As can be seen from the graph in figure 16, total energy remains constant. In this reaction, potential energy is converted to kinetic energy. A decrease in potential energy of the atoms following a successful collision between F and H2 results in an equal increase in kinetic energy which manifests in the form of increased product (HF) oscillations.

This can be confirmed experimentally using infrared chemiluminescence. Observation of radiation from excited product molecules is a clear indication that these molecules are produced in the excited state. If the excitation is primarily vibrational, the radiation will appear in the infrared region of the spectrum (between 3-15 μm). This technique has been used extensively in hydrogen halogen reactions which produce HX molecules in the excited state.

One such experiment involves the use of the arrested relaxation method. In this technique, the reaction is carried out in a chamber whose walls are maintained at a very low temperature (20 K). This results in cryopumping of the chamber in that all the products and unreacted starting materials are condensed on the walls and only those product molecules that radiate before being pumped away are observed.

=== Polanyi's empirical rules ===

In a typical chemical reaction, there is an energetic barrier — a saddle point that the reactants must surmount to become products. Which form of energy is initially deposited in the reactants, translational or vibrational, is more efficacious in surmounting the activation energy barrier is one of the core topics in the area of reaction dynamics. The Polyani rules state that vibrational energy is more effective at promoting a late-barrier reaction (transition state resembles products). Conversely, translational energy is better at promoting an early-barrier reaction (transition state resembles the reactants).

The forward reaction between H2 and F is exothermic. Therefore, the transition state is closer to the reactants than to the products in energy. Since the transition state occurs early in an exothermic reaction, translational energy is more effective at promoting a chemical reaction than vibrational energy.

Conversely, the backward reaction between H and HF is endothermic. According to Hammond's postulate, the transition state resembles the products more than it resembles the reactants. Therefore, in an endothermic reaction, a late transition state is observed and vibrational energy is more effective at promoting a reaction than translational energy.

=== Promoting the forward reaction: F + HA-HB → HA-F + HB ===

1. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 2.95 and pHF = -0.5.

[[File:01409469 polyani 1+2.jpg|thumb|center|800px|'''Figure 17.''' Contour plot of the reaction between F and H2 when pHH = 2.95 '''Figure 18.''' Graph of momentum against time for the reaction between F and H2 when pHH = 2.95 (Left-right)]]

Vibrational energy is represented by pHH while translational energy is represented by pHF.

A molecule of H2 collides with F. However, the collision is unsuccessful and no reaction occurs. A reaction did not occur even though the molecules have sufficient kinetic energy to overcome the activation energy barrier. This is because most of the energy is deposited in the reactants as H-H vibrations as seen by the large oscillations in H2. Barrier recrossing has occured as a result of the large vibrational energy causing the system to revert back to the reactants even though the system has already crossed the transition state and a bond was formed between HB and F temporarily.

This appears to run counter to what one would expect to happen when more energy is put into the system. It is predicted that when more energy is put into the system, the reactants would possess more kinetic energy to overcome the activation energy barrier. However, as can be seen from the simulation above, an increase in the vibrational energy of the reactants is not effective at promoting the occurrence of an exothermic reaction with an early transition state. This is in accordance to Polyani's rules.

2. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 0.1 and pHF = -0.8.

[[File:01409469 polyani 3+4 2.jpg|thumb|center|800px|'''Figure 19.''' Contour plot of the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 '''Figure 20.''' Graph of momentum against time for the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 (Left-right)]]

The collision between a molecule of H2 and F is successful and a chemical reaction has occurred. A reaction has occurred even though the overall energy of the system has been significantly reduced by lowering the amount of vibrational energy in the reactants but slightly increasing the amount of translational energy deposited in the reactants. Therefore, H2 and F have more translational energy to move towards each other and form a H-F bond. The product formed is vibrationally excited as seen by the large amplitude of oscillations in the product HF which correspond to the strong H-F vibrations.

From the second simulation, it can be seen that an increase in translational energy in the reactants is more effective at promoting an exothermic reaction with an early transition state to occur, which again follows Polyani's rules.

=== Promoting the backward reaction: HA + HB-F → HA-HB + F ===

1. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 2.95 and pHF = -0.5.

MRD:01409469

2019-05-10T13:33:47Z

Stc1917: /* Promoting the forward reaction: F + HA-HB → HA-F + HB */

== Exercise 1: H + H2 system ==

=== Definition of a transition state ===

A potential energy profile results from plotting the potential versus reaction path.

[[File:Surface_Plot01409469_pic1.jpg|thumb|center|400px|'''Figure 1.''' A potential energy surface plot]]

The transition state is defined as the maximum on the minimum energy path linking reactants and products. This means that the transition state is a stationary point on the potential energy surface. Thus, at the transition state, the derivative of the potential energy is zero with respect to each Cartesian coordinate.

<math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>

However, a local minimum on the potential energy surface corresponding to a well would have the same <math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>. Therefore, to distinguish between a local minimum on the potential energy surface and the transition state, the second derivative, <math>\tfrac{\partial^2 V(r_i)}{\partial r_i ^2}\ </math> is determined. This can be calculated by finding the determinant of the Hessian matrix shown below.

:<math>\mathbf H =
\begin{bmatrix}
\dfrac{\partial^2 \ V(r_i)}{\partial \ x^2} & \dfrac{\partial^2 \ V(r_i)}{\partial \ x \partial \mathbf y} \\
\dfrac{\partial^2 \ V(r_i)}{\partial \ y \partial \mathbf x} & \dfrac{\partial^2 \ V(r_i)}{\partial \mathbf y^2}
\end{bmatrix}</math>

The Hessian matrix of a convex function is positive semi-definite. Refining this property allows one to test if a critical point (x,y) is a local maximum, local minimum or saddle point.

If <math>f</math> < 0, then (x,y) is an isolated local maximum.

If <math>f</math> > 0, then (x,y) is an isolated local minimum.

If the Hessian has both positive and negative eigenvalues then (x,y) is a saddle point.

=== Locating the transition state ===

[[File:01409469 combined pic 2 and 3.jpg|thumb|center|800px|'''Figure 2.''' A surface and contour plot of the trajectory when rAB=rBC=0.907743 Å and pA=pBC=0]]

When the molecules are not at the transition state, a surface plot of the conditions shows a trajectory being plotted. This means that the trajectory is "rolling" towards the products or reactants. However, if a trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

[[File:01409469_internuclearpic3.png|thumb|center|400px|'''Figure 3.''' A plot of internuclear distance against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

An initial estimate for the position of the transition state was made by observing the potential energy surface and was found to be 0.90 Å. At 0.90 Å, the internuclear distance against time plot showed that the internuclear distance was constantly oscillating. Therefore, the position of the transition state was further refined by increasing the number of decimal places until a horizontal line in the internuclear distance againt time plot was obtained. This means that the internuclear distance is constant and that the molecules are simply undergoing periodic symmetric vibration at the transition state. Therefore, the position of the transition state was determined to be 0.907743 Å.

[[File:01409469 combined energy pic.png|thumb|center|400px|'''Figure 4.''' A plot of energy against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

When rAB=rBC=0.907743 Å and pA=pBC=0, the kinetic energy is 0. This means that HA and HBC are stationary, which is as predicted when the trajectory is started at the transition state with no initial momentum. Furthermore, the forces acting on HA and HBC are 0 as should be the case in the transition state.

=== Calculation type: Dynamic and minimum energy path ===

The minimum energy path, is the lowest energy path from the reactants to the products. The minimum energy path is determined by following the mode corresponding to the negative Hessian eigenvalue in both directions along the steepest descent direction. Following this path allows one to approach two valleys of stable structures. The resulting path generated is the minimum energy path. The reaction paths were chartered for a dynamic and a minimum energy path. The initial conditions used were that of the transition state except that the internuclear distance between HA and HB had increased to 0.917743 Å. The reaction path obtained using the two different calculation methods was different.

[[File:01409469 mep and dynamics.jpg|thumb|center|800px|'''Figure 5.''' A surface plot of the trajectory using minimum energy path '''Figure 6.''' A surface plot of the trajectory using dynamics (Left-right)]]

On the surface plot, the reaction path is depicted by the black line. Following the minimum energy path, the trajectory simply follows the valley floor to Ha+Hb-Hc. The reaction path appears to be a relatively straight line indicating that the atoms are not vibrating. In the minimum energy path, the reactants change into the products in an infinitely slow motion. This is done by having the momenta/velocities reset to zero in each time step. While the minimum energy path is useful in characterising a chemical reaction, it does not provide a realistic representation of the motion of the atoms during the reaction. This is because the mass and inertia of the atoms have been ignored.

Since the atoms simply follow the minimum energy path, they do not undergo slight displacements into surrounding valleys of higher potential energy. Furthermore, because the atoms are assumed to have no inertia, they do not undergo vibrations even if they were to be displaced to valleys of higher potential energy.

However, when dynamic calculation is performed, the reaction path is wavy indicating that the atoms are oscillating. This is because the atoms are displaced to regions of higher potential on the path from reactants to products. This leads to atom oscillation due to the inertia of the atoms.

'''Internuclear distance against time plot'''

Setting the initial system conditions such that rAB= 0.907743, rBC= 0.917743 and pA= pBC= 0:

[[File:01409469 Internuclear distance mep and dynamic.jpg|thumb|center|800px|'''Figure 6.''' A graph of internuclear distance against steps using minimum energy path '''Figure 7.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In the minimum energy path, the distance between HA and HB decreases from 0.90 Å to approximately 0.79 Å during the first 50 steps before plateauing off to a value of 0.74 Å which is the approximate length of a hydrogen bond. This means that a bond was formed between HA and HB. In addition, the plateau indicates that the bond is static. On the other hand, the distance between HB and HC has increased to 1.91 Å at large step values. This shows that HC has been ejected off following a successful collision between HA and HBC.

Similarly, in the dynamic trajectory, the distance between HA and HB decreases to approximately 0.75 Å after 0.3 s. This is indicative of hydrogen bond formation between HA and HB. In addition, the distance between HB and HC increases to 3.44 Å after HC is ejected following a successful collision.

One obvious distance between the two plots is that in the minimum energy path trajectory, the internuclear distance between HA and HB is constant indicating that the bond is static. However, in the dynamic trajectory the internuclear distance is oscillating showing the the bond is vibrating. This difference could be attributed to the fact that minimum energy path does not take oscillatory motion into account while dynamic trajectory does.

However, changing the initial conditions of the system such that rAB = 0.917743 Å, rBC = 0.907743 Å and pA=pBC=0:
[[File:01409469 01409469 Reversed values dynamic mep.jpg|thumb|center|800px|'''Figure 8.''' A graph of internuclear distance against steps using minimum energy path '''Figure 9.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In this case, a hydrogen bond of length 0.77-0.79 Å is formed between HB and HC while HA is repelled away to beyond 3.45 Å.

Additionally, from the internuclear distance plot it can be seen that the distance between HB and HC increases in a rectangular hyperbola fashion when calculated using the minimum energy path. On the other hand, the internuclear distance between HB and HC increases in a linear manner when dynamic calculation is applied. A possible reason for this difference is that in the minimum energy path, the atoms are not vibrating and as such the reach the maximum distance and plateau at a faster rate. However, in the dynamic calculation, the atoms have momentum and as such move further away from each other.

=== Reactive and unreactive trajectories ===

With the initial positions r'''1''' = 0.74 and r'''2''' = 2.0, trajectories were run with the momenta combination shown below.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Contour plot !! Reactivity !! Description of the dynamics
|-
| -1.25 || -2.5 ||-99.018 ||[[File:01409469 trajectory1.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. There are no oscillations observed in HA-HB initially, indicating that the diatomic molecule is stable. A new bond is formed between HB and HC. Oscillatory behaviour is observed indicating that the bond is vibrating. On the other hand, HA is ejected away to beyond 10 Å.
|-
| -1.5 || -2.0 ||-100.456 ||[[File:01409469 trajectory2.jpg|thumb|center|400px]] ||Unreactive ||HC collides with the vibrating diatomic HA-HB. However, no bond is formed between HB and HC as the collision is unsuccessful. This is because the molecules lack sufficient kinetic energy to overcome the activation energy barrier and so no reaction occurs. HC is subsequently repelled to beyond 8 Å.
|-
| -1.5 || -2.5 ||-98.956 ||[[File:01409469 trajectory3.jpg|thumb|center|400px]] ||Reactive |||HC collides with HA-HB. The collision is successful and a bond is formed between HB and HC. This is because the atoms have sufficient kinetic energy to overcome the activation energy barrier and form a new HB-HC bond. HA is ejected away to beyond 10 Å. The new hydrogen bond formed is vibrating as seen by the oscillatory behaviour.
|-
| -2.5 || -5.0 ||-84.956 ||[[File:01409469 trajectory4.jpg|thumb|center|400px]] ||Unreactive ||HC collides with HA-HB. Initially the HA-HB bond is elongating, indicating that the hydrogen bond is weakening and going to break. However, after 0.5 s, the HA-HB bond length decreases and the HAHB is not broken. HC is repelled beyond 6 Å with very strong vibrations.
A reaction did not occur even though the atoms have sufficient kinetic energy to cross the activation energy barrier. This is because the vibrations are too strong. The system crosses the transition state region and a bond is formed in the product temporarily but then the system reverts back to the reactants. This is an example of barrier recrossing.
|-
| -2.5 || -5.2 ||-83.416 ||[[File:01409469 trajectory5.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. The collision is successful and a new hydrogen bond is formed between HB and HC with HA being ejected away. The products formed experience very strong vibrations as evidenced by the oscillatory motion which occur with a very high frequency.
|}

From the table, it can be seen that the chance of a reaction successfully occurring is not solely dependent on the amount of kinetic energy the atoms possess. Even if the atoms have sufficient kinetic energy to cross the activation energy barrier, barrier recrossing can occur if the atoms are too vibrationally excited.

=== Transition State Theory ===

Transition state theory provided the first theoretical attempt to determine absolute reaction rates. In this theory, a transition state separating the reactants from products is used to find an expression for the thermal rate constant.

The main assumptions of transition state theory are:

1. Electronic and nuclear motions are separated (equivalent to the Born-Oppenheimer approximation in quantum mechanics)

2. Reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution

3. Molecular systems that have crossed the transition state in the direction of products cannot turn around and reform reactants

4. In the transition state, motion along the reaction coordinate may be separated from the other motions and treated classically as a translation

5. Even in the absence of an equilibrium between reactant and product molecules, the transition states that are becoming products are distributed among their states according to the Maxwell-Boltzmann laws

The goal of transition state theory is to predict the rate constant, ''k'', of a reaction given a potential energy surface for the reaction.

However, transition state theory suffers from a set of limitations due to the assumptions stated above not holding true in reality.

Firstly, transition state theory assumes that barrier recrossing is negligible. However, as seen from the fourth trajectory in table 1, barrier recrossing occurred because the atoms were too vibrationally excited and the system reverted back to the reactants even though the system had already crossed the transition state region. In this situation, transition state theory would predict a faster rate of reaction compared to what was obtained experimentally.

Secondly, the transition state theory breaks down for systems with light atoms such as hydrogens. In such systems, quantum effects become crucial. Quantum effects tend to increase the rate of the reaction for two reasons. First, particles lie in energy levels which are higher then the bottom of the potential energy well (there is additional zero-point energy). Secondly, the particles can tunnel through the barrier. These two effects effectively lower the activation energy barrier. As a result, the rate of reaction predicted by transition state theory would be less than what is obtained experimentally.

Lastly, transition state theory does not hold true for systems which are far from equilibrium. In such cases, the assumption of a starting Boltzmann distribution breaks down.

== Exercise 2: F - H - H system ==

=== Potential energy surface inspection===

F + HA-HB → HA-F + HB

[[File:01409469 Surface_Plot F 3.jpg|thumb|center|400px|'''Figure 10.''' Potential energy surface for the reaction between F and H2]]

The reaction between F and H2 is an exothermic reaction. This means that the reactants are of a higher energy than the products. The overall reaction involves the breaking of a hydrogen bond and the formation of a hydrogen-fluorine bond. The HF bond formed is stronger than the hydrogen bond that was broken. This is because fluorine is more electronegative than hydrogen resulting in a polar HF bond. The negative enthalpy of H-F bond formation is greater than the positive enthalpy of hydrogen bond dissociation. Therefore, the reaction between F and H2 is an exothermic reaction.

HA + HB-F → HA-HB + F

[[File:01409469 Surface_Plot F 4.jpg|thumb|center|400px|'''Figure 11.''' Potential energy surface for the reaction between H and HF]]

The reaction between H and HF is an endothermic reaction. The products are at a higher energy level than the reactants. The overall reaction involves the breaking of a HF bond and the formation of a hydrogen bond. The formation of a homonuclear H2 molecule results in a non-polar hydrogen bond being formed. The HF bond broken is stronger than the hydrogen bond that was formed. The negative enthalpy of hydrogen bond formation is less than the positive enthalpy of hydrogen-fluorine dissociation. Therefore, the reaction between H and HF is endothermic.

=== Locating the transition state ===

[[File:01409469 Surface_Plot F TS4.jpg|thumb|center|800px|'''Figure 12.''' Determining the position of the transition state for the reaction between F and H2]]

As mentioned previously, when the molecules are not at the transition state, a reaction path will be plotted. However, if the initial conditions of the system are set such that the trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

An initial estimate for the transition state was made by setting rAB = 0.74 which is the length of a hydrogen bond. The position of the transition state was further refined by increasing the number of decimal places such that the final value of rAB was found to be 0.74478 and rBC = 1.81079. This set of distances gave a reaction path that was represented by a point indicating that no trajectory was plotted and that the molecules remained at the transition state since their initial momentum was 0.

Moreover, the kinetic energy as well as the forces acting on the atoms was calculated to be 0, which is consistent with the transition state.

Therefore, the position of the transition state was found to be at rAB = 0.74478 and rBC = 1.81079 when pA = pBC = 0.

=== Determining the activation energy ===

[[File:01409469 activation energy r1.1 combined.jpg|thumb|center|800px|'''Figure 13.''' Determining the activation energy for the reaction between F and H2]]

By reducing the value of rBC by 0.01, the atoms are no longer at the transition state and would move towards the formation of products. The activation energy was found by calculating the change in total energy after the reaction. The activation energy of F + H2 → HF + H was found to be +23.279 kcal/ mol.

[[File:01409469 activation energy r2.1 combined.jpg|thumb|center|800px|'''Figure 14.''' Determining the activation energy for the reaction between FH and H]]

The activation energy of FH + H → F + H2 was found by calculating the small dip in potential energy. The activation energy was found to be +0.280 kcal/ mol.

=== Reaction dynamics of F + HA-HB → HA-F + HB ===

The set of initial conditions that result in a reactive trajectory for F + H2 is as follows: rHH = 0.74 Å, rHF = 1.79 Å and pHH = pHF = 0.

[[File:01409469 momentum r1.jpg|thumb|center|400px|'''Figure 15.''' Graph of momentum against time for the reaction between F and H2]]

Initially, the H2 molecule is travelling in a relatively straight line with very little oscillations as evidenced by the low amplitude of momentum between 0-2 s. Following successful collision between F and H2, the amplitude of momentum of the product, HF, increases dramatically. This is because the reaction between F and H2 is exothermic and the excess energy released causes the molecule to become vibrationally excited and the amplitude of oscillations in the product to increase.

[[File:01409469 energy1 r1.jpg|thumb|center|400px|'''Figure 16.''' Graph of energy against time for the reaction between F and H2]]

The law of conservation of energy states that energy cannot be created or destroyed. It simply changes from one form to another. As can be seen from the graph in figure 16, total energy remains constant. In this reaction, potential energy is converted to kinetic energy. A decrease in potential energy of the atoms following a successful collision between F and H2 results in an equal increase in kinetic energy which manifests in the form of increased product (HF) oscillations.

This can be confirmed experimentally using infrared chemiluminescence. Observation of radiation from excited product molecules is a clear indication that these molecules are produced in the excited state. If the excitation is primarily vibrational, the radiation will appear in the infrared region of the spectrum (between 3-15 μm). This technique has been used extensively in hydrogen halogen reactions which produce HX molecules in the excited state.

One such experiment involves the use of the arrested relaxation method. In this technique, the reaction is carried out in a chamber whose walls are maintained at a very low temperature (20 K). This results in cryopumping of the chamber in that all the products and unreacted starting materials are condensed on the walls and only those product molecules that radiate before being pumped away are observed.

=== Polanyi's empirical rules ===

In a typical chemical reaction, there is an energetic barrier — a saddle point that the reactants must surmount to become products. Which form of energy is initially deposited in the reactants, translational or vibrational, is more efficacious in surmounting the activation energy barrier is one of the core topics in the area of reaction dynamics. The Polyani rules state that vibrational energy is more effective at promoting a late-barrier reaction (transition state resembles products). Conversely, translational energy is better at promoting an early-barrier reaction (transition state resembles the reactants).

The forward reaction between H2 and F is exothermic. Therefore, the transition state is closer to the reactants than to the products in energy. Since the transition state occurs early in an exothermic reaction, translational energy is more effective at promoting a chemical reaction than vibrational energy.

Conversely, the backward reaction between H and HF is endothermic. According to Hammond's postulate, the transition state resembles the products more than it resembles the reactants. Therefore, in an endothermic reaction, a late transition state is observed and vibrational energy is more effective at promoting a reaction than translational energy.

=== Promoting the forward reaction: F + HA-HB → HA-F + HB ===

1. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 2.95 and pHF = -0.5.

[[File:01409469 polyani 1+2.jpg|thumb|center|800px|'''Figure 17.''' Contour plot of the reaction between F and H2 when pHH = 2.95 '''Figure 18.''' Graph of momentum against time for the reaction between F and H2 when pHH = 2.95 (Left-right)]]

Vibrational energy is represented by pHH while translational energy is represented by pHF.

A molecule of H2 collides with F. However, the collision is unsuccessful and no reaction occurs. A reaction did not occur even though the molecules have sufficient kinetic energy to overcome the activation energy barrier. This is because most of the energy is deposited in the reactants as H-H vibrations as seen by the large oscillations in H2. Barrier recrossing has occured as a result of the large vibrational energy causing the system to revert back to the reactants even though the system has already crossed the transition state and a bond was formed between HB and F temporarily.

This appears to run counter to what one would expect to happen when more energy is put into the system. It is predicted that when more energy is put into the system, the reactants would possess more kinetic energy to overcome the activation energy barrier. However, as can be seen from the simulation above, an increase in the vibrational energy of the reactants is not effective at promoting the occurrence of an exothermic reaction with an early transition state. This is in accordance to Polyani's rules.

2. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 0.1 and pHF = -0.8.

[[File:01409469 polyani 3+4 2.jpg|thumb|center|800px|'''Figure 19.''' Contour plot of the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 '''Figure 20.''' Graph of momentum against time for the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 (Left-right)]]

The collision between a molecule of H2 and F is successful and a chemical reaction has occurred. A reaction has occurred even though the overall energy of the system has been significantly reduced by lowering the amount of vibrational energy in the reactants but slightly increasing the amount of translational energy deposited in the reactants. Therefore, H2 and F have more translational energy to move towards each other and form a H-F bond. The product formed is vibrationally excited as seen by the large amplitude of oscillations in the product HF which correspond to the strong H-F vibrations.

From the second simulation, it can be seen that an increase in translational energy in the reactants is more effective at promoting an exothermic reaction with an early transition state to occur, which again follows Polyani's rules.

=== Promoting the backward reaction: HA + HB-FB → HA-HB + F ===

1. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 2.95 and pHF = -0.5.

MRD:01409469

2019-05-10T13:32:03Z

Stc1917: /* Promoting the forward reaction: F + HA-HB → HA-F + HB */

== Exercise 1: H + H2 system ==

=== Definition of a transition state ===

A potential energy profile results from plotting the potential versus reaction path.

[[File:Surface_Plot01409469_pic1.jpg|thumb|center|400px|'''Figure 1.''' A potential energy surface plot]]

The transition state is defined as the maximum on the minimum energy path linking reactants and products. This means that the transition state is a stationary point on the potential energy surface. Thus, at the transition state, the derivative of the potential energy is zero with respect to each Cartesian coordinate.

<math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>

However, a local minimum on the potential energy surface corresponding to a well would have the same <math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>. Therefore, to distinguish between a local minimum on the potential energy surface and the transition state, the second derivative, <math>\tfrac{\partial^2 V(r_i)}{\partial r_i ^2}\ </math> is determined. This can be calculated by finding the determinant of the Hessian matrix shown below.

:<math>\mathbf H =
\begin{bmatrix}
\dfrac{\partial^2 \ V(r_i)}{\partial \ x^2} & \dfrac{\partial^2 \ V(r_i)}{\partial \ x \partial \mathbf y} \\
\dfrac{\partial^2 \ V(r_i)}{\partial \ y \partial \mathbf x} & \dfrac{\partial^2 \ V(r_i)}{\partial \mathbf y^2}
\end{bmatrix}</math>

The Hessian matrix of a convex function is positive semi-definite. Refining this property allows one to test if a critical point (x,y) is a local maximum, local minimum or saddle point.

If <math>f</math> < 0, then (x,y) is an isolated local maximum.

If <math>f</math> > 0, then (x,y) is an isolated local minimum.

If the Hessian has both positive and negative eigenvalues then (x,y) is a saddle point.

=== Locating the transition state ===

[[File:01409469 combined pic 2 and 3.jpg|thumb|center|800px|'''Figure 2.''' A surface and contour plot of the trajectory when rAB=rBC=0.907743 Å and pA=pBC=0]]

When the molecules are not at the transition state, a surface plot of the conditions shows a trajectory being plotted. This means that the trajectory is "rolling" towards the products or reactants. However, if a trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

[[File:01409469_internuclearpic3.png|thumb|center|400px|'''Figure 3.''' A plot of internuclear distance against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

An initial estimate for the position of the transition state was made by observing the potential energy surface and was found to be 0.90 Å. At 0.90 Å, the internuclear distance against time plot showed that the internuclear distance was constantly oscillating. Therefore, the position of the transition state was further refined by increasing the number of decimal places until a horizontal line in the internuclear distance againt time plot was obtained. This means that the internuclear distance is constant and that the molecules are simply undergoing periodic symmetric vibration at the transition state. Therefore, the position of the transition state was determined to be 0.907743 Å.

[[File:01409469 combined energy pic.png|thumb|center|400px|'''Figure 4.''' A plot of energy against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

When rAB=rBC=0.907743 Å and pA=pBC=0, the kinetic energy is 0. This means that HA and HBC are stationary, which is as predicted when the trajectory is started at the transition state with no initial momentum. Furthermore, the forces acting on HA and HBC are 0 as should be the case in the transition state.

=== Calculation type: Dynamic and minimum energy path ===

The minimum energy path, is the lowest energy path from the reactants to the products. The minimum energy path is determined by following the mode corresponding to the negative Hessian eigenvalue in both directions along the steepest descent direction. Following this path allows one to approach two valleys of stable structures. The resulting path generated is the minimum energy path. The reaction paths were chartered for a dynamic and a minimum energy path. The initial conditions used were that of the transition state except that the internuclear distance between HA and HB had increased to 0.917743 Å. The reaction path obtained using the two different calculation methods was different.

[[File:01409469 mep and dynamics.jpg|thumb|center|800px|'''Figure 5.''' A surface plot of the trajectory using minimum energy path '''Figure 6.''' A surface plot of the trajectory using dynamics (Left-right)]]

On the surface plot, the reaction path is depicted by the black line. Following the minimum energy path, the trajectory simply follows the valley floor to Ha+Hb-Hc. The reaction path appears to be a relatively straight line indicating that the atoms are not vibrating. In the minimum energy path, the reactants change into the products in an infinitely slow motion. This is done by having the momenta/velocities reset to zero in each time step. While the minimum energy path is useful in characterising a chemical reaction, it does not provide a realistic representation of the motion of the atoms during the reaction. This is because the mass and inertia of the atoms have been ignored.

Since the atoms simply follow the minimum energy path, they do not undergo slight displacements into surrounding valleys of higher potential energy. Furthermore, because the atoms are assumed to have no inertia, they do not undergo vibrations even if they were to be displaced to valleys of higher potential energy.

However, when dynamic calculation is performed, the reaction path is wavy indicating that the atoms are oscillating. This is because the atoms are displaced to regions of higher potential on the path from reactants to products. This leads to atom oscillation due to the inertia of the atoms.

'''Internuclear distance against time plot'''

Setting the initial system conditions such that rAB= 0.907743, rBC= 0.917743 and pA= pBC= 0:

[[File:01409469 Internuclear distance mep and dynamic.jpg|thumb|center|800px|'''Figure 6.''' A graph of internuclear distance against steps using minimum energy path '''Figure 7.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In the minimum energy path, the distance between HA and HB decreases from 0.90 Å to approximately 0.79 Å during the first 50 steps before plateauing off to a value of 0.74 Å which is the approximate length of a hydrogen bond. This means that a bond was formed between HA and HB. In addition, the plateau indicates that the bond is static. On the other hand, the distance between HB and HC has increased to 1.91 Å at large step values. This shows that HC has been ejected off following a successful collision between HA and HBC.

Similarly, in the dynamic trajectory, the distance between HA and HB decreases to approximately 0.75 Å after 0.3 s. This is indicative of hydrogen bond formation between HA and HB. In addition, the distance between HB and HC increases to 3.44 Å after HC is ejected following a successful collision.

One obvious distance between the two plots is that in the minimum energy path trajectory, the internuclear distance between HA and HB is constant indicating that the bond is static. However, in the dynamic trajectory the internuclear distance is oscillating showing the the bond is vibrating. This difference could be attributed to the fact that minimum energy path does not take oscillatory motion into account while dynamic trajectory does.

However, changing the initial conditions of the system such that rAB = 0.917743 Å, rBC = 0.907743 Å and pA=pBC=0:
[[File:01409469 01409469 Reversed values dynamic mep.jpg|thumb|center|800px|'''Figure 8.''' A graph of internuclear distance against steps using minimum energy path '''Figure 9.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In this case, a hydrogen bond of length 0.77-0.79 Å is formed between HB and HC while HA is repelled away to beyond 3.45 Å.

Additionally, from the internuclear distance plot it can be seen that the distance between HB and HC increases in a rectangular hyperbola fashion when calculated using the minimum energy path. On the other hand, the internuclear distance between HB and HC increases in a linear manner when dynamic calculation is applied. A possible reason for this difference is that in the minimum energy path, the atoms are not vibrating and as such the reach the maximum distance and plateau at a faster rate. However, in the dynamic calculation, the atoms have momentum and as such move further away from each other.

=== Reactive and unreactive trajectories ===

With the initial positions r'''1''' = 0.74 and r'''2''' = 2.0, trajectories were run with the momenta combination shown below.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Contour plot !! Reactivity !! Description of the dynamics
|-
| -1.25 || -2.5 ||-99.018 ||[[File:01409469 trajectory1.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. There are no oscillations observed in HA-HB initially, indicating that the diatomic molecule is stable. A new bond is formed between HB and HC. Oscillatory behaviour is observed indicating that the bond is vibrating. On the other hand, HA is ejected away to beyond 10 Å.
|-
| -1.5 || -2.0 ||-100.456 ||[[File:01409469 trajectory2.jpg|thumb|center|400px]] ||Unreactive ||HC collides with the vibrating diatomic HA-HB. However, no bond is formed between HB and HC as the collision is unsuccessful. This is because the molecules lack sufficient kinetic energy to overcome the activation energy barrier and so no reaction occurs. HC is subsequently repelled to beyond 8 Å.
|-
| -1.5 || -2.5 ||-98.956 ||[[File:01409469 trajectory3.jpg|thumb|center|400px]] ||Reactive |||HC collides with HA-HB. The collision is successful and a bond is formed between HB and HC. This is because the atoms have sufficient kinetic energy to overcome the activation energy barrier and form a new HB-HC bond. HA is ejected away to beyond 10 Å. The new hydrogen bond formed is vibrating as seen by the oscillatory behaviour.
|-
| -2.5 || -5.0 ||-84.956 ||[[File:01409469 trajectory4.jpg|thumb|center|400px]] ||Unreactive ||HC collides with HA-HB. Initially the HA-HB bond is elongating, indicating that the hydrogen bond is weakening and going to break. However, after 0.5 s, the HA-HB bond length decreases and the HAHB is not broken. HC is repelled beyond 6 Å with very strong vibrations.
A reaction did not occur even though the atoms have sufficient kinetic energy to cross the activation energy barrier. This is because the vibrations are too strong. The system crosses the transition state region and a bond is formed in the product temporarily but then the system reverts back to the reactants. This is an example of barrier recrossing.
|-
| -2.5 || -5.2 ||-83.416 ||[[File:01409469 trajectory5.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. The collision is successful and a new hydrogen bond is formed between HB and HC with HA being ejected away. The products formed experience very strong vibrations as evidenced by the oscillatory motion which occur with a very high frequency.
|}

From the table, it can be seen that the chance of a reaction successfully occurring is not solely dependent on the amount of kinetic energy the atoms possess. Even if the atoms have sufficient kinetic energy to cross the activation energy barrier, barrier recrossing can occur if the atoms are too vibrationally excited.

=== Transition State Theory ===

Transition state theory provided the first theoretical attempt to determine absolute reaction rates. In this theory, a transition state separating the reactants from products is used to find an expression for the thermal rate constant.

The main assumptions of transition state theory are:

1. Electronic and nuclear motions are separated (equivalent to the Born-Oppenheimer approximation in quantum mechanics)

2. Reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution

3. Molecular systems that have crossed the transition state in the direction of products cannot turn around and reform reactants

4. In the transition state, motion along the reaction coordinate may be separated from the other motions and treated classically as a translation

5. Even in the absence of an equilibrium between reactant and product molecules, the transition states that are becoming products are distributed among their states according to the Maxwell-Boltzmann laws

The goal of transition state theory is to predict the rate constant, ''k'', of a reaction given a potential energy surface for the reaction.

However, transition state theory suffers from a set of limitations due to the assumptions stated above not holding true in reality.

Firstly, transition state theory assumes that barrier recrossing is negligible. However, as seen from the fourth trajectory in table 1, barrier recrossing occurred because the atoms were too vibrationally excited and the system reverted back to the reactants even though the system had already crossed the transition state region. In this situation, transition state theory would predict a faster rate of reaction compared to what was obtained experimentally.

Secondly, the transition state theory breaks down for systems with light atoms such as hydrogens. In such systems, quantum effects become crucial. Quantum effects tend to increase the rate of the reaction for two reasons. First, particles lie in energy levels which are higher then the bottom of the potential energy well (there is additional zero-point energy). Secondly, the particles can tunnel through the barrier. These two effects effectively lower the activation energy barrier. As a result, the rate of reaction predicted by transition state theory would be less than what is obtained experimentally.

Lastly, transition state theory does not hold true for systems which are far from equilibrium. In such cases, the assumption of a starting Boltzmann distribution breaks down.

== Exercise 2: F - H - H system ==

=== Potential energy surface inspection===

F + HA-HB → HA-F + HB

[[File:01409469 Surface_Plot F 3.jpg|thumb|center|400px|'''Figure 10.''' Potential energy surface for the reaction between F and H2]]

The reaction between F and H2 is an exothermic reaction. This means that the reactants are of a higher energy than the products. The overall reaction involves the breaking of a hydrogen bond and the formation of a hydrogen-fluorine bond. The HF bond formed is stronger than the hydrogen bond that was broken. This is because fluorine is more electronegative than hydrogen resulting in a polar HF bond. The negative enthalpy of H-F bond formation is greater than the positive enthalpy of hydrogen bond dissociation. Therefore, the reaction between F and H2 is an exothermic reaction.

HA + HB-F → HA-HB + F

[[File:01409469 Surface_Plot F 4.jpg|thumb|center|400px|'''Figure 11.''' Potential energy surface for the reaction between H and HF]]

The reaction between H and HF is an endothermic reaction. The products are at a higher energy level than the reactants. The overall reaction involves the breaking of a HF bond and the formation of a hydrogen bond. The formation of a homonuclear H2 molecule results in a non-polar hydrogen bond being formed. The HF bond broken is stronger than the hydrogen bond that was formed. The negative enthalpy of hydrogen bond formation is less than the positive enthalpy of hydrogen-fluorine dissociation. Therefore, the reaction between H and HF is endothermic.

=== Locating the transition state ===

[[File:01409469 Surface_Plot F TS4.jpg|thumb|center|800px|'''Figure 12.''' Determining the position of the transition state for the reaction between F and H2]]

As mentioned previously, when the molecules are not at the transition state, a reaction path will be plotted. However, if the initial conditions of the system are set such that the trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

An initial estimate for the transition state was made by setting rAB = 0.74 which is the length of a hydrogen bond. The position of the transition state was further refined by increasing the number of decimal places such that the final value of rAB was found to be 0.74478 and rBC = 1.81079. This set of distances gave a reaction path that was represented by a point indicating that no trajectory was plotted and that the molecules remained at the transition state since their initial momentum was 0.

Moreover, the kinetic energy as well as the forces acting on the atoms was calculated to be 0, which is consistent with the transition state.

Therefore, the position of the transition state was found to be at rAB = 0.74478 and rBC = 1.81079 when pA = pBC = 0.

=== Determining the activation energy ===

[[File:01409469 activation energy r1.1 combined.jpg|thumb|center|800px|'''Figure 13.''' Determining the activation energy for the reaction between F and H2]]

By reducing the value of rBC by 0.01, the atoms are no longer at the transition state and would move towards the formation of products. The activation energy was found by calculating the change in total energy after the reaction. The activation energy of F + H2 → HF + H was found to be +23.279 kcal/ mol.

[[File:01409469 activation energy r2.1 combined.jpg|thumb|center|800px|'''Figure 14.''' Determining the activation energy for the reaction between FH and H]]

The activation energy of FH + H → F + H2 was found by calculating the small dip in potential energy. The activation energy was found to be +0.280 kcal/ mol.

=== Reaction dynamics of F + HA-HB → HA-F + HB ===

The set of initial conditions that result in a reactive trajectory for F + H2 is as follows: rHH = 0.74 Å, rHF = 1.79 Å and pHH = pHF = 0.

[[File:01409469 momentum r1.jpg|thumb|center|400px|'''Figure 15.''' Graph of momentum against time for the reaction between F and H2]]

Initially, the H2 molecule is travelling in a relatively straight line with very little oscillations as evidenced by the low amplitude of momentum between 0-2 s. Following successful collision between F and H2, the amplitude of momentum of the product, HF, increases dramatically. This is because the reaction between F and H2 is exothermic and the excess energy released causes the molecule to become vibrationally excited and the amplitude of oscillations in the product to increase.

[[File:01409469 energy1 r1.jpg|thumb|center|400px|'''Figure 16.''' Graph of energy against time for the reaction between F and H2]]

The law of conservation of energy states that energy cannot be created or destroyed. It simply changes from one form to another. As can be seen from the graph in figure 16, total energy remains constant. In this reaction, potential energy is converted to kinetic energy. A decrease in potential energy of the atoms following a successful collision between F and H2 results in an equal increase in kinetic energy which manifests in the form of increased product (HF) oscillations.

This can be confirmed experimentally using infrared chemiluminescence. Observation of radiation from excited product molecules is a clear indication that these molecules are produced in the excited state. If the excitation is primarily vibrational, the radiation will appear in the infrared region of the spectrum (between 3-15 μm). This technique has been used extensively in hydrogen halogen reactions which produce HX molecules in the excited state.

One such experiment involves the use of the arrested relaxation method. In this technique, the reaction is carried out in a chamber whose walls are maintained at a very low temperature (20 K). This results in cryopumping of the chamber in that all the products and unreacted starting materials are condensed on the walls and only those product molecules that radiate before being pumped away are observed.

=== Polanyi's empirical rules ===

In a typical chemical reaction, there is an energetic barrier — a saddle point that the reactants must surmount to become products. Which form of energy is initially deposited in the reactants, translational or vibrational, is more efficacious in surmounting the activation energy barrier is one of the core topics in the area of reaction dynamics. The Polyani rules state that vibrational energy is more effective at promoting a late-barrier reaction (transition state resembles products). Conversely, translational energy is better at promoting an early-barrier reaction (transition state resembles the reactants).

The forward reaction between H2 and F is exothermic. Therefore, the transition state is closer to the reactants than to the products in energy. Since the transition state occurs early in an exothermic reaction, translational energy is more effective at promoting a chemical reaction than vibrational energy.

Conversely, the backward reaction between H and HF is endothermic. According to Hammond's postulate, the transition state resembles the products more than it resembles the reactants. Therefore, in an endothermic reaction, a late transition state is observed and vibrational energy is more effective at promoting a reaction than translational energy.

=== Promoting the forward reaction: F + HA-HB → HA-F + HB ===

1. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 2.95 and pHF = -0.5.

[[File:01409469 polyani 1+2.jpg|thumb|center|800px|'''Figure 17.''' Contour plot of the reaction between F and H2 when pHH = 2.95 '''Figure 18.''' Graph of momentum against time for the reaction between F and H2 when pHH = 2.95 (Left-right)]]

Vibrational energy is represented by pHH while translational energy is represented by pHF.

A molecule of H2 collides with F. However, the collision is unsuccessful and no reaction occurs. A reaction did not occur even though the molecules have sufficient kinetic energy to overcome the activation energy barrier. This is because most of the energy is deposited in the reactants as H-H vibrations as seen by the large oscillations in H2. Barrier recrossing has occured as a result of the large vibrational energy causing the system to revert back to the reactants even though the system has already crossed the transition state and a bond was formed between HB and F temporarily.

This appears to run counter to what one would expect to happen when more energy is put into the system. It is predicted that when more energy is put into the system, the reactants would possess more kinetic energy to overcome the activation energy barrier. However, as can be seen from the simulation above, an increase in the vibrational energy of the reactants is not effective at promoting the occurrence of an exothermic reaction with an early transition state. This is in accordance to Polyani's rules.

2. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 0.1 and pHF = -0.8.

[[File:01409469 polyani 3+4 2.jpg|thumb|center|800px|'''Figure 19.''' Contour plot of the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 '''Figure 20.''' Graph of momentum against time for the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 (Left-right)]]

The collision between a molecule of H2 and F is successful and a chemical reaction has occurred. A reaction has occurred even though the overall energy of the system has been significantly reduced by lowering the amount of vibrational energy in the reactants but slightly increasing the amount of translational energy deposited in the reactants. Therefore, H2 and F have more translational energy to move towards each other and form a H-F bond. The product formed is vibrationally excited as seen by the large amplitude of oscillations in the product HF which correspond to the strong H-F vibrations.

From the second simulation, it can be seen that an increase in translational energy in the reactants is more effective at promoting an exothermic reaction with an early transition state to occur, which again follows Polyani's rules.

File:01409469 polyani 3+4 2.jpg

2019-05-10T13:31:36Z

Stc1917:

MRD:01409469

2019-05-10T13:31:20Z

Stc1917: /* Promoting the forward reaction: F + HA-HB → HA-F + HB */

== Exercise 1: H + H2 system ==

=== Definition of a transition state ===

A potential energy profile results from plotting the potential versus reaction path.

[[File:Surface_Plot01409469_pic1.jpg|thumb|center|400px|'''Figure 1.''' A potential energy surface plot]]

The transition state is defined as the maximum on the minimum energy path linking reactants and products. This means that the transition state is a stationary point on the potential energy surface. Thus, at the transition state, the derivative of the potential energy is zero with respect to each Cartesian coordinate.

<math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>

However, a local minimum on the potential energy surface corresponding to a well would have the same <math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>. Therefore, to distinguish between a local minimum on the potential energy surface and the transition state, the second derivative, <math>\tfrac{\partial^2 V(r_i)}{\partial r_i ^2}\ </math> is determined. This can be calculated by finding the determinant of the Hessian matrix shown below.

:<math>\mathbf H =
\begin{bmatrix}
\dfrac{\partial^2 \ V(r_i)}{\partial \ x^2} & \dfrac{\partial^2 \ V(r_i)}{\partial \ x \partial \mathbf y} \\
\dfrac{\partial^2 \ V(r_i)}{\partial \ y \partial \mathbf x} & \dfrac{\partial^2 \ V(r_i)}{\partial \mathbf y^2}
\end{bmatrix}</math>

The Hessian matrix of a convex function is positive semi-definite. Refining this property allows one to test if a critical point (x,y) is a local maximum, local minimum or saddle point.

If <math>f</math> < 0, then (x,y) is an isolated local maximum.

If <math>f</math> > 0, then (x,y) is an isolated local minimum.

If the Hessian has both positive and negative eigenvalues then (x,y) is a saddle point.

=== Locating the transition state ===

[[File:01409469 combined pic 2 and 3.jpg|thumb|center|800px|'''Figure 2.''' A surface and contour plot of the trajectory when rAB=rBC=0.907743 Å and pA=pBC=0]]

When the molecules are not at the transition state, a surface plot of the conditions shows a trajectory being plotted. This means that the trajectory is "rolling" towards the products or reactants. However, if a trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

[[File:01409469_internuclearpic3.png|thumb|center|400px|'''Figure 3.''' A plot of internuclear distance against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

An initial estimate for the position of the transition state was made by observing the potential energy surface and was found to be 0.90 Å. At 0.90 Å, the internuclear distance against time plot showed that the internuclear distance was constantly oscillating. Therefore, the position of the transition state was further refined by increasing the number of decimal places until a horizontal line in the internuclear distance againt time plot was obtained. This means that the internuclear distance is constant and that the molecules are simply undergoing periodic symmetric vibration at the transition state. Therefore, the position of the transition state was determined to be 0.907743 Å.

[[File:01409469 combined energy pic.png|thumb|center|400px|'''Figure 4.''' A plot of energy against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

When rAB=rBC=0.907743 Å and pA=pBC=0, the kinetic energy is 0. This means that HA and HBC are stationary, which is as predicted when the trajectory is started at the transition state with no initial momentum. Furthermore, the forces acting on HA and HBC are 0 as should be the case in the transition state.

=== Calculation type: Dynamic and minimum energy path ===

The minimum energy path, is the lowest energy path from the reactants to the products. The minimum energy path is determined by following the mode corresponding to the negative Hessian eigenvalue in both directions along the steepest descent direction. Following this path allows one to approach two valleys of stable structures. The resulting path generated is the minimum energy path. The reaction paths were chartered for a dynamic and a minimum energy path. The initial conditions used were that of the transition state except that the internuclear distance between HA and HB had increased to 0.917743 Å. The reaction path obtained using the two different calculation methods was different.

[[File:01409469 mep and dynamics.jpg|thumb|center|800px|'''Figure 5.''' A surface plot of the trajectory using minimum energy path '''Figure 6.''' A surface plot of the trajectory using dynamics (Left-right)]]

On the surface plot, the reaction path is depicted by the black line. Following the minimum energy path, the trajectory simply follows the valley floor to Ha+Hb-Hc. The reaction path appears to be a relatively straight line indicating that the atoms are not vibrating. In the minimum energy path, the reactants change into the products in an infinitely slow motion. This is done by having the momenta/velocities reset to zero in each time step. While the minimum energy path is useful in characterising a chemical reaction, it does not provide a realistic representation of the motion of the atoms during the reaction. This is because the mass and inertia of the atoms have been ignored.

Since the atoms simply follow the minimum energy path, they do not undergo slight displacements into surrounding valleys of higher potential energy. Furthermore, because the atoms are assumed to have no inertia, they do not undergo vibrations even if they were to be displaced to valleys of higher potential energy.

However, when dynamic calculation is performed, the reaction path is wavy indicating that the atoms are oscillating. This is because the atoms are displaced to regions of higher potential on the path from reactants to products. This leads to atom oscillation due to the inertia of the atoms.

'''Internuclear distance against time plot'''

Setting the initial system conditions such that rAB= 0.907743, rBC= 0.917743 and pA= pBC= 0:

[[File:01409469 Internuclear distance mep and dynamic.jpg|thumb|center|800px|'''Figure 6.''' A graph of internuclear distance against steps using minimum energy path '''Figure 7.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In the minimum energy path, the distance between HA and HB decreases from 0.90 Å to approximately 0.79 Å during the first 50 steps before plateauing off to a value of 0.74 Å which is the approximate length of a hydrogen bond. This means that a bond was formed between HA and HB. In addition, the plateau indicates that the bond is static. On the other hand, the distance between HB and HC has increased to 1.91 Å at large step values. This shows that HC has been ejected off following a successful collision between HA and HBC.

Similarly, in the dynamic trajectory, the distance between HA and HB decreases to approximately 0.75 Å after 0.3 s. This is indicative of hydrogen bond formation between HA and HB. In addition, the distance between HB and HC increases to 3.44 Å after HC is ejected following a successful collision.

One obvious distance between the two plots is that in the minimum energy path trajectory, the internuclear distance between HA and HB is constant indicating that the bond is static. However, in the dynamic trajectory the internuclear distance is oscillating showing the the bond is vibrating. This difference could be attributed to the fact that minimum energy path does not take oscillatory motion into account while dynamic trajectory does.

However, changing the initial conditions of the system such that rAB = 0.917743 Å, rBC = 0.907743 Å and pA=pBC=0:
[[File:01409469 01409469 Reversed values dynamic mep.jpg|thumb|center|800px|'''Figure 8.''' A graph of internuclear distance against steps using minimum energy path '''Figure 9.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In this case, a hydrogen bond of length 0.77-0.79 Å is formed between HB and HC while HA is repelled away to beyond 3.45 Å.

Additionally, from the internuclear distance plot it can be seen that the distance between HB and HC increases in a rectangular hyperbola fashion when calculated using the minimum energy path. On the other hand, the internuclear distance between HB and HC increases in a linear manner when dynamic calculation is applied. A possible reason for this difference is that in the minimum energy path, the atoms are not vibrating and as such the reach the maximum distance and plateau at a faster rate. However, in the dynamic calculation, the atoms have momentum and as such move further away from each other.

=== Reactive and unreactive trajectories ===

With the initial positions r'''1''' = 0.74 and r'''2''' = 2.0, trajectories were run with the momenta combination shown below.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Contour plot !! Reactivity !! Description of the dynamics
|-
| -1.25 || -2.5 ||-99.018 ||[[File:01409469 trajectory1.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. There are no oscillations observed in HA-HB initially, indicating that the diatomic molecule is stable. A new bond is formed between HB and HC. Oscillatory behaviour is observed indicating that the bond is vibrating. On the other hand, HA is ejected away to beyond 10 Å.
|-
| -1.5 || -2.0 ||-100.456 ||[[File:01409469 trajectory2.jpg|thumb|center|400px]] ||Unreactive ||HC collides with the vibrating diatomic HA-HB. However, no bond is formed between HB and HC as the collision is unsuccessful. This is because the molecules lack sufficient kinetic energy to overcome the activation energy barrier and so no reaction occurs. HC is subsequently repelled to beyond 8 Å.
|-
| -1.5 || -2.5 ||-98.956 ||[[File:01409469 trajectory3.jpg|thumb|center|400px]] ||Reactive |||HC collides with HA-HB. The collision is successful and a bond is formed between HB and HC. This is because the atoms have sufficient kinetic energy to overcome the activation energy barrier and form a new HB-HC bond. HA is ejected away to beyond 10 Å. The new hydrogen bond formed is vibrating as seen by the oscillatory behaviour.
|-
| -2.5 || -5.0 ||-84.956 ||[[File:01409469 trajectory4.jpg|thumb|center|400px]] ||Unreactive ||HC collides with HA-HB. Initially the HA-HB bond is elongating, indicating that the hydrogen bond is weakening and going to break. However, after 0.5 s, the HA-HB bond length decreases and the HAHB is not broken. HC is repelled beyond 6 Å with very strong vibrations.
A reaction did not occur even though the atoms have sufficient kinetic energy to cross the activation energy barrier. This is because the vibrations are too strong. The system crosses the transition state region and a bond is formed in the product temporarily but then the system reverts back to the reactants. This is an example of barrier recrossing.
|-
| -2.5 || -5.2 ||-83.416 ||[[File:01409469 trajectory5.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. The collision is successful and a new hydrogen bond is formed between HB and HC with HA being ejected away. The products formed experience very strong vibrations as evidenced by the oscillatory motion which occur with a very high frequency.
|}

From the table, it can be seen that the chance of a reaction successfully occurring is not solely dependent on the amount of kinetic energy the atoms possess. Even if the atoms have sufficient kinetic energy to cross the activation energy barrier, barrier recrossing can occur if the atoms are too vibrationally excited.

=== Transition State Theory ===

Transition state theory provided the first theoretical attempt to determine absolute reaction rates. In this theory, a transition state separating the reactants from products is used to find an expression for the thermal rate constant.

The main assumptions of transition state theory are:

1. Electronic and nuclear motions are separated (equivalent to the Born-Oppenheimer approximation in quantum mechanics)

2. Reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution

3. Molecular systems that have crossed the transition state in the direction of products cannot turn around and reform reactants

4. In the transition state, motion along the reaction coordinate may be separated from the other motions and treated classically as a translation

5. Even in the absence of an equilibrium between reactant and product molecules, the transition states that are becoming products are distributed among their states according to the Maxwell-Boltzmann laws

The goal of transition state theory is to predict the rate constant, ''k'', of a reaction given a potential energy surface for the reaction.

However, transition state theory suffers from a set of limitations due to the assumptions stated above not holding true in reality.

Firstly, transition state theory assumes that barrier recrossing is negligible. However, as seen from the fourth trajectory in table 1, barrier recrossing occurred because the atoms were too vibrationally excited and the system reverted back to the reactants even though the system had already crossed the transition state region. In this situation, transition state theory would predict a faster rate of reaction compared to what was obtained experimentally.

Secondly, the transition state theory breaks down for systems with light atoms such as hydrogens. In such systems, quantum effects become crucial. Quantum effects tend to increase the rate of the reaction for two reasons. First, particles lie in energy levels which are higher then the bottom of the potential energy well (there is additional zero-point energy). Secondly, the particles can tunnel through the barrier. These two effects effectively lower the activation energy barrier. As a result, the rate of reaction predicted by transition state theory would be less than what is obtained experimentally.

Lastly, transition state theory does not hold true for systems which are far from equilibrium. In such cases, the assumption of a starting Boltzmann distribution breaks down.

== Exercise 2: F - H - H system ==

=== Potential energy surface inspection===

F + HA-HB → HA-F + HB

[[File:01409469 Surface_Plot F 3.jpg|thumb|center|400px|'''Figure 10.''' Potential energy surface for the reaction between F and H2]]

The reaction between F and H2 is an exothermic reaction. This means that the reactants are of a higher energy than the products. The overall reaction involves the breaking of a hydrogen bond and the formation of a hydrogen-fluorine bond. The HF bond formed is stronger than the hydrogen bond that was broken. This is because fluorine is more electronegative than hydrogen resulting in a polar HF bond. The negative enthalpy of H-F bond formation is greater than the positive enthalpy of hydrogen bond dissociation. Therefore, the reaction between F and H2 is an exothermic reaction.

HA + HB-F → HA-HB + F

[[File:01409469 Surface_Plot F 4.jpg|thumb|center|400px|'''Figure 11.''' Potential energy surface for the reaction between H and HF]]

The reaction between H and HF is an endothermic reaction. The products are at a higher energy level than the reactants. The overall reaction involves the breaking of a HF bond and the formation of a hydrogen bond. The formation of a homonuclear H2 molecule results in a non-polar hydrogen bond being formed. The HF bond broken is stronger than the hydrogen bond that was formed. The negative enthalpy of hydrogen bond formation is less than the positive enthalpy of hydrogen-fluorine dissociation. Therefore, the reaction between H and HF is endothermic.

=== Locating the transition state ===

[[File:01409469 Surface_Plot F TS4.jpg|thumb|center|800px|'''Figure 12.''' Determining the position of the transition state for the reaction between F and H2]]

As mentioned previously, when the molecules are not at the transition state, a reaction path will be plotted. However, if the initial conditions of the system are set such that the trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

An initial estimate for the transition state was made by setting rAB = 0.74 which is the length of a hydrogen bond. The position of the transition state was further refined by increasing the number of decimal places such that the final value of rAB was found to be 0.74478 and rBC = 1.81079. This set of distances gave a reaction path that was represented by a point indicating that no trajectory was plotted and that the molecules remained at the transition state since their initial momentum was 0.

Moreover, the kinetic energy as well as the forces acting on the atoms was calculated to be 0, which is consistent with the transition state.

Therefore, the position of the transition state was found to be at rAB = 0.74478 and rBC = 1.81079 when pA = pBC = 0.

=== Determining the activation energy ===

[[File:01409469 activation energy r1.1 combined.jpg|thumb|center|800px|'''Figure 13.''' Determining the activation energy for the reaction between F and H2]]

By reducing the value of rBC by 0.01, the atoms are no longer at the transition state and would move towards the formation of products. The activation energy was found by calculating the change in total energy after the reaction. The activation energy of F + H2 → HF + H was found to be +23.279 kcal/ mol.

[[File:01409469 activation energy r2.1 combined.jpg|thumb|center|800px|'''Figure 14.''' Determining the activation energy for the reaction between FH and H]]

The activation energy of FH + H → F + H2 was found by calculating the small dip in potential energy. The activation energy was found to be +0.280 kcal/ mol.

=== Reaction dynamics of F + HA-HB → HA-F + HB ===

The set of initial conditions that result in a reactive trajectory for F + H2 is as follows: rHH = 0.74 Å, rHF = 1.79 Å and pHH = pHF = 0.

[[File:01409469 momentum r1.jpg|thumb|center|400px|'''Figure 15.''' Graph of momentum against time for the reaction between F and H2]]

Initially, the H2 molecule is travelling in a relatively straight line with very little oscillations as evidenced by the low amplitude of momentum between 0-2 s. Following successful collision between F and H2, the amplitude of momentum of the product, HF, increases dramatically. This is because the reaction between F and H2 is exothermic and the excess energy released causes the molecule to become vibrationally excited and the amplitude of oscillations in the product to increase.

[[File:01409469 energy1 r1.jpg|thumb|center|400px|'''Figure 16.''' Graph of energy against time for the reaction between F and H2]]

The law of conservation of energy states that energy cannot be created or destroyed. It simply changes from one form to another. As can be seen from the graph in figure 16, total energy remains constant. In this reaction, potential energy is converted to kinetic energy. A decrease in potential energy of the atoms following a successful collision between F and H2 results in an equal increase in kinetic energy which manifests in the form of increased product (HF) oscillations.

This can be confirmed experimentally using infrared chemiluminescence. Observation of radiation from excited product molecules is a clear indication that these molecules are produced in the excited state. If the excitation is primarily vibrational, the radiation will appear in the infrared region of the spectrum (between 3-15 μm). This technique has been used extensively in hydrogen halogen reactions which produce HX molecules in the excited state.

One such experiment involves the use of the arrested relaxation method. In this technique, the reaction is carried out in a chamber whose walls are maintained at a very low temperature (20 K). This results in cryopumping of the chamber in that all the products and unreacted starting materials are condensed on the walls and only those product molecules that radiate before being pumped away are observed.

=== Polanyi's empirical rules ===

In a typical chemical reaction, there is an energetic barrier — a saddle point that the reactants must surmount to become products. Which form of energy is initially deposited in the reactants, translational or vibrational, is more efficacious in surmounting the activation energy barrier is one of the core topics in the area of reaction dynamics. The Polyani rules state that vibrational energy is more effective at promoting a late-barrier reaction (transition state resembles products). Conversely, translational energy is better at promoting an early-barrier reaction (transition state resembles the reactants).

The forward reaction between H2 and F is exothermic. Therefore, the transition state is closer to the reactants than to the products in energy. Since the transition state occurs early in an exothermic reaction, translational energy is more effective at promoting a chemical reaction than vibrational energy.

Conversely, the backward reaction between H and HF is endothermic. According to Hammond's postulate, the transition state resembles the products more than it resembles the reactants. Therefore, in an endothermic reaction, a late transition state is observed and vibrational energy is more effective at promoting a reaction than translational energy.

=== Promoting the forward reaction: F + HA-HB → HA-F + HB ===

1. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 2.95 and pHF = -0.5.

[[File:01409469 polyani 1+2.jpg|thumb|center|800px|'''Figure 17.''' Contour plot of the reaction between F and H2 when pHH = 2.95 '''Figure 18.''' Graph of momentum against time for the reaction between F and H2 when pHH = 2.95 (Left-right)]]

Vibrational energy is represented by pHH while translational energy is represented by pHF.

A molecule of H2 collides with F. However, the collision is unsuccessful and no reaction occurs. A reaction did not occur even though the molecules have sufficient kinetic energy to overcome the activation energy barrier. This is because most of the energy is deposited in the reactants as H-H vibrations as seen by the large oscillations in H2. Barrier recrossing has occured as a result of the large vibrational energy causing the system to revert back to the reactants even though the system has already crossed the transition state and a bond was formed between HB and F temporarily.

This appears to run counter to what one would expect to happen when more energy is put into the system. It is predicted that when more energy is put into the system, the reactants would possess more kinetic energy to overcome the activation energy barrier. However, as can be seen from the simulation above, an increase in the vibrational energy of the reactants is not effective at promoting the occurrence of an exothermic reaction with an early transition state. This is in accordance to Polyani's rules.

2. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 0.1 and pHF = -0.8.

[[File:01409469 polyani 3+4 2.jpg|thumb|center|800px|'''Figure 19.''' Contour plot of the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 '''Figure 20.''' Graph of momentum against time for the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 (Left-right)]]

The collision between a molecule of H2 and F is successful and a chemical reaction has occurred. A reaction has occurred even though the overall energy of the system has been significantly reduced by lowering the amount of vibrational energy in the reactants but slightly increasing the amount of translational energy deposited in the reactants. Therefore, H2 and F have more translational energy to move towards each other and form a H-F bond. The product is vibrationally excited as seen by the large amplitude of oscillations in the product HF which correspond to the strong H-F vibrations.

From the second simulation, it can be seen that an increase in translational energy in the reactants is more effective at promoting an exothermic reaction with an early transition state to occur, which again follows Polyani's rules.

MRD:01409469

2019-05-10T13:28:48Z

Stc1917: /* Promoting the forward reaction: F + HA-HB → HA-F + HB */

== Exercise 1: H + H2 system ==

=== Definition of a transition state ===

A potential energy profile results from plotting the potential versus reaction path.

[[File:Surface_Plot01409469_pic1.jpg|thumb|center|400px|'''Figure 1.''' A potential energy surface plot]]

The transition state is defined as the maximum on the minimum energy path linking reactants and products. This means that the transition state is a stationary point on the potential energy surface. Thus, at the transition state, the derivative of the potential energy is zero with respect to each Cartesian coordinate.

<math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>

However, a local minimum on the potential energy surface corresponding to a well would have the same <math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>. Therefore, to distinguish between a local minimum on the potential energy surface and the transition state, the second derivative, <math>\tfrac{\partial^2 V(r_i)}{\partial r_i ^2}\ </math> is determined. This can be calculated by finding the determinant of the Hessian matrix shown below.

:<math>\mathbf H =
\begin{bmatrix}
\dfrac{\partial^2 \ V(r_i)}{\partial \ x^2} & \dfrac{\partial^2 \ V(r_i)}{\partial \ x \partial \mathbf y} \\
\dfrac{\partial^2 \ V(r_i)}{\partial \ y \partial \mathbf x} & \dfrac{\partial^2 \ V(r_i)}{\partial \mathbf y^2}
\end{bmatrix}</math>

The Hessian matrix of a convex function is positive semi-definite. Refining this property allows one to test if a critical point (x,y) is a local maximum, local minimum or saddle point.

If <math>f</math> < 0, then (x,y) is an isolated local maximum.

If <math>f</math> > 0, then (x,y) is an isolated local minimum.

If the Hessian has both positive and negative eigenvalues then (x,y) is a saddle point.

=== Locating the transition state ===

[[File:01409469 combined pic 2 and 3.jpg|thumb|center|800px|'''Figure 2.''' A surface and contour plot of the trajectory when rAB=rBC=0.907743 Å and pA=pBC=0]]

When the molecules are not at the transition state, a surface plot of the conditions shows a trajectory being plotted. This means that the trajectory is "rolling" towards the products or reactants. However, if a trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

[[File:01409469_internuclearpic3.png|thumb|center|400px|'''Figure 3.''' A plot of internuclear distance against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

An initial estimate for the position of the transition state was made by observing the potential energy surface and was found to be 0.90 Å. At 0.90 Å, the internuclear distance against time plot showed that the internuclear distance was constantly oscillating. Therefore, the position of the transition state was further refined by increasing the number of decimal places until a horizontal line in the internuclear distance againt time plot was obtained. This means that the internuclear distance is constant and that the molecules are simply undergoing periodic symmetric vibration at the transition state. Therefore, the position of the transition state was determined to be 0.907743 Å.

[[File:01409469 combined energy pic.png|thumb|center|400px|'''Figure 4.''' A plot of energy against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

When rAB=rBC=0.907743 Å and pA=pBC=0, the kinetic energy is 0. This means that HA and HBC are stationary, which is as predicted when the trajectory is started at the transition state with no initial momentum. Furthermore, the forces acting on HA and HBC are 0 as should be the case in the transition state.

=== Calculation type: Dynamic and minimum energy path ===

The minimum energy path, is the lowest energy path from the reactants to the products. The minimum energy path is determined by following the mode corresponding to the negative Hessian eigenvalue in both directions along the steepest descent direction. Following this path allows one to approach two valleys of stable structures. The resulting path generated is the minimum energy path. The reaction paths were chartered for a dynamic and a minimum energy path. The initial conditions used were that of the transition state except that the internuclear distance between HA and HB had increased to 0.917743 Å. The reaction path obtained using the two different calculation methods was different.

[[File:01409469 mep and dynamics.jpg|thumb|center|800px|'''Figure 5.''' A surface plot of the trajectory using minimum energy path '''Figure 6.''' A surface plot of the trajectory using dynamics (Left-right)]]

On the surface plot, the reaction path is depicted by the black line. Following the minimum energy path, the trajectory simply follows the valley floor to Ha+Hb-Hc. The reaction path appears to be a relatively straight line indicating that the atoms are not vibrating. In the minimum energy path, the reactants change into the products in an infinitely slow motion. This is done by having the momenta/velocities reset to zero in each time step. While the minimum energy path is useful in characterising a chemical reaction, it does not provide a realistic representation of the motion of the atoms during the reaction. This is because the mass and inertia of the atoms have been ignored.

Since the atoms simply follow the minimum energy path, they do not undergo slight displacements into surrounding valleys of higher potential energy. Furthermore, because the atoms are assumed to have no inertia, they do not undergo vibrations even if they were to be displaced to valleys of higher potential energy.

However, when dynamic calculation is performed, the reaction path is wavy indicating that the atoms are oscillating. This is because the atoms are displaced to regions of higher potential on the path from reactants to products. This leads to atom oscillation due to the inertia of the atoms.

'''Internuclear distance against time plot'''

Setting the initial system conditions such that rAB= 0.907743, rBC= 0.917743 and pA= pBC= 0:

[[File:01409469 Internuclear distance mep and dynamic.jpg|thumb|center|800px|'''Figure 6.''' A graph of internuclear distance against steps using minimum energy path '''Figure 7.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In the minimum energy path, the distance between HA and HB decreases from 0.90 Å to approximately 0.79 Å during the first 50 steps before plateauing off to a value of 0.74 Å which is the approximate length of a hydrogen bond. This means that a bond was formed between HA and HB. In addition, the plateau indicates that the bond is static. On the other hand, the distance between HB and HC has increased to 1.91 Å at large step values. This shows that HC has been ejected off following a successful collision between HA and HBC.

Similarly, in the dynamic trajectory, the distance between HA and HB decreases to approximately 0.75 Å after 0.3 s. This is indicative of hydrogen bond formation between HA and HB. In addition, the distance between HB and HC increases to 3.44 Å after HC is ejected following a successful collision.

One obvious distance between the two plots is that in the minimum energy path trajectory, the internuclear distance between HA and HB is constant indicating that the bond is static. However, in the dynamic trajectory the internuclear distance is oscillating showing the the bond is vibrating. This difference could be attributed to the fact that minimum energy path does not take oscillatory motion into account while dynamic trajectory does.

However, changing the initial conditions of the system such that rAB = 0.917743 Å, rBC = 0.907743 Å and pA=pBC=0:
[[File:01409469 01409469 Reversed values dynamic mep.jpg|thumb|center|800px|'''Figure 8.''' A graph of internuclear distance against steps using minimum energy path '''Figure 9.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In this case, a hydrogen bond of length 0.77-0.79 Å is formed between HB and HC while HA is repelled away to beyond 3.45 Å.

Additionally, from the internuclear distance plot it can be seen that the distance between HB and HC increases in a rectangular hyperbola fashion when calculated using the minimum energy path. On the other hand, the internuclear distance between HB and HC increases in a linear manner when dynamic calculation is applied. A possible reason for this difference is that in the minimum energy path, the atoms are not vibrating and as such the reach the maximum distance and plateau at a faster rate. However, in the dynamic calculation, the atoms have momentum and as such move further away from each other.

=== Reactive and unreactive trajectories ===

With the initial positions r'''1''' = 0.74 and r'''2''' = 2.0, trajectories were run with the momenta combination shown below.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Contour plot !! Reactivity !! Description of the dynamics
|-
| -1.25 || -2.5 ||-99.018 ||[[File:01409469 trajectory1.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. There are no oscillations observed in HA-HB initially, indicating that the diatomic molecule is stable. A new bond is formed between HB and HC. Oscillatory behaviour is observed indicating that the bond is vibrating. On the other hand, HA is ejected away to beyond 10 Å.
|-
| -1.5 || -2.0 ||-100.456 ||[[File:01409469 trajectory2.jpg|thumb|center|400px]] ||Unreactive ||HC collides with the vibrating diatomic HA-HB. However, no bond is formed between HB and HC as the collision is unsuccessful. This is because the molecules lack sufficient kinetic energy to overcome the activation energy barrier and so no reaction occurs. HC is subsequently repelled to beyond 8 Å.
|-
| -1.5 || -2.5 ||-98.956 ||[[File:01409469 trajectory3.jpg|thumb|center|400px]] ||Reactive |||HC collides with HA-HB. The collision is successful and a bond is formed between HB and HC. This is because the atoms have sufficient kinetic energy to overcome the activation energy barrier and form a new HB-HC bond. HA is ejected away to beyond 10 Å. The new hydrogen bond formed is vibrating as seen by the oscillatory behaviour.
|-
| -2.5 || -5.0 ||-84.956 ||[[File:01409469 trajectory4.jpg|thumb|center|400px]] ||Unreactive ||HC collides with HA-HB. Initially the HA-HB bond is elongating, indicating that the hydrogen bond is weakening and going to break. However, after 0.5 s, the HA-HB bond length decreases and the HAHB is not broken. HC is repelled beyond 6 Å with very strong vibrations.
A reaction did not occur even though the atoms have sufficient kinetic energy to cross the activation energy barrier. This is because the vibrations are too strong. The system crosses the transition state region and a bond is formed in the product temporarily but then the system reverts back to the reactants. This is an example of barrier recrossing.
|-
| -2.5 || -5.2 ||-83.416 ||[[File:01409469 trajectory5.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. The collision is successful and a new hydrogen bond is formed between HB and HC with HA being ejected away. The products formed experience very strong vibrations as evidenced by the oscillatory motion which occur with a very high frequency.
|}

From the table, it can be seen that the chance of a reaction successfully occurring is not solely dependent on the amount of kinetic energy the atoms possess. Even if the atoms have sufficient kinetic energy to cross the activation energy barrier, barrier recrossing can occur if the atoms are too vibrationally excited.

=== Transition State Theory ===

Transition state theory provided the first theoretical attempt to determine absolute reaction rates. In this theory, a transition state separating the reactants from products is used to find an expression for the thermal rate constant.

The main assumptions of transition state theory are:

1. Electronic and nuclear motions are separated (equivalent to the Born-Oppenheimer approximation in quantum mechanics)

2. Reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution

3. Molecular systems that have crossed the transition state in the direction of products cannot turn around and reform reactants

4. In the transition state, motion along the reaction coordinate may be separated from the other motions and treated classically as a translation

5. Even in the absence of an equilibrium between reactant and product molecules, the transition states that are becoming products are distributed among their states according to the Maxwell-Boltzmann laws

The goal of transition state theory is to predict the rate constant, ''k'', of a reaction given a potential energy surface for the reaction.

However, transition state theory suffers from a set of limitations due to the assumptions stated above not holding true in reality.

Firstly, transition state theory assumes that barrier recrossing is negligible. However, as seen from the fourth trajectory in table 1, barrier recrossing occurred because the atoms were too vibrationally excited and the system reverted back to the reactants even though the system had already crossed the transition state region. In this situation, transition state theory would predict a faster rate of reaction compared to what was obtained experimentally.

Secondly, the transition state theory breaks down for systems with light atoms such as hydrogens. In such systems, quantum effects become crucial. Quantum effects tend to increase the rate of the reaction for two reasons. First, particles lie in energy levels which are higher then the bottom of the potential energy well (there is additional zero-point energy). Secondly, the particles can tunnel through the barrier. These two effects effectively lower the activation energy barrier. As a result, the rate of reaction predicted by transition state theory would be less than what is obtained experimentally.

Lastly, transition state theory does not hold true for systems which are far from equilibrium. In such cases, the assumption of a starting Boltzmann distribution breaks down.

== Exercise 2: F - H - H system ==

=== Potential energy surface inspection===

F + HA-HB → HA-F + HB

[[File:01409469 Surface_Plot F 3.jpg|thumb|center|400px|'''Figure 10.''' Potential energy surface for the reaction between F and H2]]

The reaction between F and H2 is an exothermic reaction. This means that the reactants are of a higher energy than the products. The overall reaction involves the breaking of a hydrogen bond and the formation of a hydrogen-fluorine bond. The HF bond formed is stronger than the hydrogen bond that was broken. This is because fluorine is more electronegative than hydrogen resulting in a polar HF bond. The negative enthalpy of H-F bond formation is greater than the positive enthalpy of hydrogen bond dissociation. Therefore, the reaction between F and H2 is an exothermic reaction.

HA + HB-F → HA-HB + F

[[File:01409469 Surface_Plot F 4.jpg|thumb|center|400px|'''Figure 11.''' Potential energy surface for the reaction between H and HF]]

The reaction between H and HF is an endothermic reaction. The products are at a higher energy level than the reactants. The overall reaction involves the breaking of a HF bond and the formation of a hydrogen bond. The formation of a homonuclear H2 molecule results in a non-polar hydrogen bond being formed. The HF bond broken is stronger than the hydrogen bond that was formed. The negative enthalpy of hydrogen bond formation is less than the positive enthalpy of hydrogen-fluorine dissociation. Therefore, the reaction between H and HF is endothermic.

=== Locating the transition state ===

[[File:01409469 Surface_Plot F TS4.jpg|thumb|center|800px|'''Figure 12.''' Determining the position of the transition state for the reaction between F and H2]]

As mentioned previously, when the molecules are not at the transition state, a reaction path will be plotted. However, if the initial conditions of the system are set such that the trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

An initial estimate for the transition state was made by setting rAB = 0.74 which is the length of a hydrogen bond. The position of the transition state was further refined by increasing the number of decimal places such that the final value of rAB was found to be 0.74478 and rBC = 1.81079. This set of distances gave a reaction path that was represented by a point indicating that no trajectory was plotted and that the molecules remained at the transition state since their initial momentum was 0.

Moreover, the kinetic energy as well as the forces acting on the atoms was calculated to be 0, which is consistent with the transition state.

Therefore, the position of the transition state was found to be at rAB = 0.74478 and rBC = 1.81079 when pA = pBC = 0.

=== Determining the activation energy ===

[[File:01409469 activation energy r1.1 combined.jpg|thumb|center|800px|'''Figure 13.''' Determining the activation energy for the reaction between F and H2]]

By reducing the value of rBC by 0.01, the atoms are no longer at the transition state and would move towards the formation of products. The activation energy was found by calculating the change in total energy after the reaction. The activation energy of F + H2 → HF + H was found to be +23.279 kcal/ mol.

[[File:01409469 activation energy r2.1 combined.jpg|thumb|center|800px|'''Figure 14.''' Determining the activation energy for the reaction between FH and H]]

The activation energy of FH + H → F + H2 was found by calculating the small dip in potential energy. The activation energy was found to be +0.280 kcal/ mol.

=== Reaction dynamics of F + HA-HB → HA-F + HB ===

The set of initial conditions that result in a reactive trajectory for F + H2 is as follows: rHH = 0.74 Å, rHF = 1.79 Å and pHH = pHF = 0.

[[File:01409469 momentum r1.jpg|thumb|center|400px|'''Figure 15.''' Graph of momentum against time for the reaction between F and H2]]

Initially, the H2 molecule is travelling in a relatively straight line with very little oscillations as evidenced by the low amplitude of momentum between 0-2 s. Following successful collision between F and H2, the amplitude of momentum of the product, HF, increases dramatically. This is because the reaction between F and H2 is exothermic and the excess energy released causes the molecule to become vibrationally excited and the amplitude of oscillations in the product to increase.

[[File:01409469 energy1 r1.jpg|thumb|center|400px|'''Figure 16.''' Graph of energy against time for the reaction between F and H2]]

The law of conservation of energy states that energy cannot be created or destroyed. It simply changes from one form to another. As can be seen from the graph in figure 16, total energy remains constant. In this reaction, potential energy is converted to kinetic energy. A decrease in potential energy of the atoms following a successful collision between F and H2 results in an equal increase in kinetic energy which manifests in the form of increased product (HF) oscillations.

This can be confirmed experimentally using infrared chemiluminescence. Observation of radiation from excited product molecules is a clear indication that these molecules are produced in the excited state. If the excitation is primarily vibrational, the radiation will appear in the infrared region of the spectrum (between 3-15 μm). This technique has been used extensively in hydrogen halogen reactions which produce HX molecules in the excited state.

One such experiment involves the use of the arrested relaxation method. In this technique, the reaction is carried out in a chamber whose walls are maintained at a very low temperature (20 K). This results in cryopumping of the chamber in that all the products and unreacted starting materials are condensed on the walls and only those product molecules that radiate before being pumped away are observed.

=== Polanyi's empirical rules ===

In a typical chemical reaction, there is an energetic barrier — a saddle point that the reactants must surmount to become products. Which form of energy is initially deposited in the reactants, translational or vibrational, is more efficacious in surmounting the activation energy barrier is one of the core topics in the area of reaction dynamics. The Polyani rules state that vibrational energy is more effective at promoting a late-barrier reaction (transition state resembles products). Conversely, translational energy is better at promoting an early-barrier reaction (transition state resembles the reactants).

The forward reaction between H2 and F is exothermic. Therefore, the transition state is closer to the reactants than to the products in energy. Since the transition state occurs early in an exothermic reaction, translational energy is more effective at promoting a chemical reaction than vibrational energy.

Conversely, the backward reaction between H and HF is endothermic. According to Hammond's postulate, the transition state resembles the products more than it resembles the reactants. Therefore, in an endothermic reaction, a late transition state is observed and vibrational energy is more effective at promoting a reaction than translational energy.

=== Promoting the forward reaction: F + HA-HB → HA-F + HB ===

1. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 2.95 and pHF = -0.5.

[[File:01409469 polyani 1+2.jpg|thumb|center|800px|'''Figure 17.''' Contour plot of the reaction between F and H2 when pHH = 2.95 '''Figure 18.''' Graph of momentum against time for the reaction between F and H2 when pHH = 2.95 (Left-right)]]

Vibrational energy is represented by pHH while translational energy is represented by pHF.

A molecule of H2 collides with F. However, the collision is unsuccessful and no reaction occurs. A reaction did not occur even though the molecules have sufficient kinetic energy to overcome the activation energy barrier. This is because most of the energy is deposited in the reactants as H-H vibrations as seen by the large oscillations in H2. Barrier recrossing has occured as a result of the large vibrational energy causing the system to revert back to the reactants even though the system has already crossed the transition state and a bond was formed between HB and F temporarily.

This appears to run counter to what one would expect to happen when more energy is put into the system. It is predicted that when more energy is put into the system, the reactants would possess more kinetic energy to overcome the activation energy barrier. However, as can be seen from the simulation above, an increase in the vibrational energy of the reactants is not effective at promoting the occurrence of an exothermic reaction with an early transition state. This is in accordance to Polyani's rules.

2. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 0.1 and pHF = -0.8.

[[File:01409469 polyani 3+4.jpg|thumb|center|800px|'''Figure 19.''' Contour plot of the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 '''Figure 20.''' Graph of momentum against time for the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 (Left-right)]]

The collision between a molecule of H2 and F is successful and a chemical reaction has occurred. A reaction has occurred even though the overall energy of the system has been significantly reduced by lowering the amount of vibrational energy in the reactants but slightly increasing the amount of translational energy deposited in the reactants. Therefore, H2 and F have more translational energy to move towards each other and form a H-F bond. The product is vibrationally excited as seen by the large amplitude of oscillations in the product HF which correspond to the strong H-F vibrations.

From the second simulation, it can be seen that an increase in translational energy in the reactants is more effective at promoting an exothermic reaction with an early transition state to occur, which again follows Polyani's rules.

File:01409469 polyani 3+4.jpg

2019-05-10T13:21:08Z

Stc1917:

MRD:01409469

2019-05-10T13:20:45Z

Stc1917: /* Promoting the forward reaction: F + HA-HB → HA-F + HB */

== Exercise 1: H + H2 system ==

=== Definition of a transition state ===

A potential energy profile results from plotting the potential versus reaction path.

[[File:Surface_Plot01409469_pic1.jpg|thumb|center|400px|'''Figure 1.''' A potential energy surface plot]]

The transition state is defined as the maximum on the minimum energy path linking reactants and products. This means that the transition state is a stationary point on the potential energy surface. Thus, at the transition state, the derivative of the potential energy is zero with respect to each Cartesian coordinate.

<math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>

However, a local minimum on the potential energy surface corresponding to a well would have the same <math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>. Therefore, to distinguish between a local minimum on the potential energy surface and the transition state, the second derivative, <math>\tfrac{\partial^2 V(r_i)}{\partial r_i ^2}\ </math> is determined. This can be calculated by finding the determinant of the Hessian matrix shown below.

:<math>\mathbf H =
\begin{bmatrix}
\dfrac{\partial^2 \ V(r_i)}{\partial \ x^2} & \dfrac{\partial^2 \ V(r_i)}{\partial \ x \partial \mathbf y} \\
\dfrac{\partial^2 \ V(r_i)}{\partial \ y \partial \mathbf x} & \dfrac{\partial^2 \ V(r_i)}{\partial \mathbf y^2}
\end{bmatrix}</math>

The Hessian matrix of a convex function is positive semi-definite. Refining this property allows one to test if a critical point (x,y) is a local maximum, local minimum or saddle point.

If <math>f</math> < 0, then (x,y) is an isolated local maximum.

If <math>f</math> > 0, then (x,y) is an isolated local minimum.

If the Hessian has both positive and negative eigenvalues then (x,y) is a saddle point.

=== Locating the transition state ===

[[File:01409469 combined pic 2 and 3.jpg|thumb|center|800px|'''Figure 2.''' A surface and contour plot of the trajectory when rAB=rBC=0.907743 Å and pA=pBC=0]]

When the molecules are not at the transition state, a surface plot of the conditions shows a trajectory being plotted. This means that the trajectory is "rolling" towards the products or reactants. However, if a trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

[[File:01409469_internuclearpic3.png|thumb|center|400px|'''Figure 3.''' A plot of internuclear distance against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

An initial estimate for the position of the transition state was made by observing the potential energy surface and was found to be 0.90 Å. At 0.90 Å, the internuclear distance against time plot showed that the internuclear distance was constantly oscillating. Therefore, the position of the transition state was further refined by increasing the number of decimal places until a horizontal line in the internuclear distance againt time plot was obtained. This means that the internuclear distance is constant and that the molecules are simply undergoing periodic symmetric vibration at the transition state. Therefore, the position of the transition state was determined to be 0.907743 Å.

[[File:01409469 combined energy pic.png|thumb|center|400px|'''Figure 4.''' A plot of energy against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

When rAB=rBC=0.907743 Å and pA=pBC=0, the kinetic energy is 0. This means that HA and HBC are stationary, which is as predicted when the trajectory is started at the transition state with no initial momentum. Furthermore, the forces acting on HA and HBC are 0 as should be the case in the transition state.

=== Calculation type: Dynamic and minimum energy path ===

The minimum energy path, is the lowest energy path from the reactants to the products. The minimum energy path is determined by following the mode corresponding to the negative Hessian eigenvalue in both directions along the steepest descent direction. Following this path allows one to approach two valleys of stable structures. The resulting path generated is the minimum energy path. The reaction paths were chartered for a dynamic and a minimum energy path. The initial conditions used were that of the transition state except that the internuclear distance between HA and HB had increased to 0.917743 Å. The reaction path obtained using the two different calculation methods was different.

[[File:01409469 mep and dynamics.jpg|thumb|center|800px|'''Figure 5.''' A surface plot of the trajectory using minimum energy path '''Figure 6.''' A surface plot of the trajectory using dynamics (Left-right)]]

On the surface plot, the reaction path is depicted by the black line. Following the minimum energy path, the trajectory simply follows the valley floor to Ha+Hb-Hc. The reaction path appears to be a relatively straight line indicating that the atoms are not vibrating. In the minimum energy path, the reactants change into the products in an infinitely slow motion. This is done by having the momenta/velocities reset to zero in each time step. While the minimum energy path is useful in characterising a chemical reaction, it does not provide a realistic representation of the motion of the atoms during the reaction. This is because the mass and inertia of the atoms have been ignored.

Since the atoms simply follow the minimum energy path, they do not undergo slight displacements into surrounding valleys of higher potential energy. Furthermore, because the atoms are assumed to have no inertia, they do not undergo vibrations even if they were to be displaced to valleys of higher potential energy.

However, when dynamic calculation is performed, the reaction path is wavy indicating that the atoms are oscillating. This is because the atoms are displaced to regions of higher potential on the path from reactants to products. This leads to atom oscillation due to the inertia of the atoms.

'''Internuclear distance against time plot'''

Setting the initial system conditions such that rAB= 0.907743, rBC= 0.917743 and pA= pBC= 0:

[[File:01409469 Internuclear distance mep and dynamic.jpg|thumb|center|800px|'''Figure 6.''' A graph of internuclear distance against steps using minimum energy path '''Figure 7.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In the minimum energy path, the distance between HA and HB decreases from 0.90 Å to approximately 0.79 Å during the first 50 steps before plateauing off to a value of 0.74 Å which is the approximate length of a hydrogen bond. This means that a bond was formed between HA and HB. In addition, the plateau indicates that the bond is static. On the other hand, the distance between HB and HC has increased to 1.91 Å at large step values. This shows that HC has been ejected off following a successful collision between HA and HBC.

Similarly, in the dynamic trajectory, the distance between HA and HB decreases to approximately 0.75 Å after 0.3 s. This is indicative of hydrogen bond formation between HA and HB. In addition, the distance between HB and HC increases to 3.44 Å after HC is ejected following a successful collision.

One obvious distance between the two plots is that in the minimum energy path trajectory, the internuclear distance between HA and HB is constant indicating that the bond is static. However, in the dynamic trajectory the internuclear distance is oscillating showing the the bond is vibrating. This difference could be attributed to the fact that minimum energy path does not take oscillatory motion into account while dynamic trajectory does.

However, changing the initial conditions of the system such that rAB = 0.917743 Å, rBC = 0.907743 Å and pA=pBC=0:
[[File:01409469 01409469 Reversed values dynamic mep.jpg|thumb|center|800px|'''Figure 8.''' A graph of internuclear distance against steps using minimum energy path '''Figure 9.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In this case, a hydrogen bond of length 0.77-0.79 Å is formed between HB and HC while HA is repelled away to beyond 3.45 Å.

Additionally, from the internuclear distance plot it can be seen that the distance between HB and HC increases in a rectangular hyperbola fashion when calculated using the minimum energy path. On the other hand, the internuclear distance between HB and HC increases in a linear manner when dynamic calculation is applied. A possible reason for this difference is that in the minimum energy path, the atoms are not vibrating and as such the reach the maximum distance and plateau at a faster rate. However, in the dynamic calculation, the atoms have momentum and as such move further away from each other.

=== Reactive and unreactive trajectories ===

With the initial positions r'''1''' = 0.74 and r'''2''' = 2.0, trajectories were run with the momenta combination shown below.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Contour plot !! Reactivity !! Description of the dynamics
|-
| -1.25 || -2.5 ||-99.018 ||[[File:01409469 trajectory1.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. There are no oscillations observed in HA-HB initially, indicating that the diatomic molecule is stable. A new bond is formed between HB and HC. Oscillatory behaviour is observed indicating that the bond is vibrating. On the other hand, HA is ejected away to beyond 10 Å.
|-
| -1.5 || -2.0 ||-100.456 ||[[File:01409469 trajectory2.jpg|thumb|center|400px]] ||Unreactive ||HC collides with the vibrating diatomic HA-HB. However, no bond is formed between HB and HC as the collision is unsuccessful. This is because the molecules lack sufficient kinetic energy to overcome the activation energy barrier and so no reaction occurs. HC is subsequently repelled to beyond 8 Å.
|-
| -1.5 || -2.5 ||-98.956 ||[[File:01409469 trajectory3.jpg|thumb|center|400px]] ||Reactive |||HC collides with HA-HB. The collision is successful and a bond is formed between HB and HC. This is because the atoms have sufficient kinetic energy to overcome the activation energy barrier and form a new HB-HC bond. HA is ejected away to beyond 10 Å. The new hydrogen bond formed is vibrating as seen by the oscillatory behaviour.
|-
| -2.5 || -5.0 ||-84.956 ||[[File:01409469 trajectory4.jpg|thumb|center|400px]] ||Unreactive ||HC collides with HA-HB. Initially the HA-HB bond is elongating, indicating that the hydrogen bond is weakening and going to break. However, after 0.5 s, the HA-HB bond length decreases and the HAHB is not broken. HC is repelled beyond 6 Å with very strong vibrations.
A reaction did not occur even though the atoms have sufficient kinetic energy to cross the activation energy barrier. This is because the vibrations are too strong. The system crosses the transition state region and a bond is formed in the product temporarily but then the system reverts back to the reactants. This is an example of barrier recrossing.
|-
| -2.5 || -5.2 ||-83.416 ||[[File:01409469 trajectory5.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. The collision is successful and a new hydrogen bond is formed between HB and HC with HA being ejected away. The products formed experience very strong vibrations as evidenced by the oscillatory motion which occur with a very high frequency.
|}

From the table, it can be seen that the chance of a reaction successfully occurring is not solely dependent on the amount of kinetic energy the atoms possess. Even if the atoms have sufficient kinetic energy to cross the activation energy barrier, barrier recrossing can occur if the atoms are too vibrationally excited.

=== Transition State Theory ===

Transition state theory provided the first theoretical attempt to determine absolute reaction rates. In this theory, a transition state separating the reactants from products is used to find an expression for the thermal rate constant.

The main assumptions of transition state theory are:

1. Electronic and nuclear motions are separated (equivalent to the Born-Oppenheimer approximation in quantum mechanics)

2. Reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution

3. Molecular systems that have crossed the transition state in the direction of products cannot turn around and reform reactants

4. In the transition state, motion along the reaction coordinate may be separated from the other motions and treated classically as a translation

5. Even in the absence of an equilibrium between reactant and product molecules, the transition states that are becoming products are distributed among their states according to the Maxwell-Boltzmann laws

The goal of transition state theory is to predict the rate constant, ''k'', of a reaction given a potential energy surface for the reaction.

However, transition state theory suffers from a set of limitations due to the assumptions stated above not holding true in reality.

Firstly, transition state theory assumes that barrier recrossing is negligible. However, as seen from the fourth trajectory in table 1, barrier recrossing occurred because the atoms were too vibrationally excited and the system reverted back to the reactants even though the system had already crossed the transition state region. In this situation, transition state theory would predict a faster rate of reaction compared to what was obtained experimentally.

Secondly, the transition state theory breaks down for systems with light atoms such as hydrogens. In such systems, quantum effects become crucial. Quantum effects tend to increase the rate of the reaction for two reasons. First, particles lie in energy levels which are higher then the bottom of the potential energy well (there is additional zero-point energy). Secondly, the particles can tunnel through the barrier. These two effects effectively lower the activation energy barrier. As a result, the rate of reaction predicted by transition state theory would be less than what is obtained experimentally.

Lastly, transition state theory does not hold true for systems which are far from equilibrium. In such cases, the assumption of a starting Boltzmann distribution breaks down.

== Exercise 2: F - H - H system ==

=== Potential energy surface inspection===

F + HA-HB → HA-F + HB

[[File:01409469 Surface_Plot F 3.jpg|thumb|center|400px|'''Figure 10.''' Potential energy surface for the reaction between F and H2]]

The reaction between F and H2 is an exothermic reaction. This means that the reactants are of a higher energy than the products. The overall reaction involves the breaking of a hydrogen bond and the formation of a hydrogen-fluorine bond. The HF bond formed is stronger than the hydrogen bond that was broken. This is because fluorine is more electronegative than hydrogen resulting in a polar HF bond. The negative enthalpy of H-F bond formation is greater than the positive enthalpy of hydrogen bond dissociation. Therefore, the reaction between F and H2 is an exothermic reaction.

HA + HB-F → HA-HB + F

[[File:01409469 Surface_Plot F 4.jpg|thumb|center|400px|'''Figure 11.''' Potential energy surface for the reaction between H and HF]]

The reaction between H and HF is an endothermic reaction. The products are at a higher energy level than the reactants. The overall reaction involves the breaking of a HF bond and the formation of a hydrogen bond. The formation of a homonuclear H2 molecule results in a non-polar hydrogen bond being formed. The HF bond broken is stronger than the hydrogen bond that was formed. The negative enthalpy of hydrogen bond formation is less than the positive enthalpy of hydrogen-fluorine dissociation. Therefore, the reaction between H and HF is endothermic.

=== Locating the transition state ===

[[File:01409469 Surface_Plot F TS4.jpg|thumb|center|800px|'''Figure 12.''' Determining the position of the transition state for the reaction between F and H2]]

As mentioned previously, when the molecules are not at the transition state, a reaction path will be plotted. However, if the initial conditions of the system are set such that the trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

An initial estimate for the transition state was made by setting rAB = 0.74 which is the length of a hydrogen bond. The position of the transition state was further refined by increasing the number of decimal places such that the final value of rAB was found to be 0.74478 and rBC = 1.81079. This set of distances gave a reaction path that was represented by a point indicating that no trajectory was plotted and that the molecules remained at the transition state since their initial momentum was 0.

Moreover, the kinetic energy as well as the forces acting on the atoms was calculated to be 0, which is consistent with the transition state.

Therefore, the position of the transition state was found to be at rAB = 0.74478 and rBC = 1.81079 when pA = pBC = 0.

=== Determining the activation energy ===

[[File:01409469 activation energy r1.1 combined.jpg|thumb|center|800px|'''Figure 13.''' Determining the activation energy for the reaction between F and H2]]

By reducing the value of rBC by 0.01, the atoms are no longer at the transition state and would move towards the formation of products. The activation energy was found by calculating the change in total energy after the reaction. The activation energy of F + H2 → HF + H was found to be +23.279 kcal/ mol.

[[File:01409469 activation energy r2.1 combined.jpg|thumb|center|800px|'''Figure 14.''' Determining the activation energy for the reaction between FH and H]]

The activation energy of FH + H → F + H2 was found by calculating the small dip in potential energy. The activation energy was found to be +0.280 kcal/ mol.

=== Reaction dynamics of F + HA-HB → HA-F + HB ===

The set of initial conditions that result in a reactive trajectory for F + H2 is as follows: rHH = 0.74 Å, rHF = 1.79 Å and pHH = pHF = 0.

[[File:01409469 momentum r1.jpg|thumb|center|400px|'''Figure 15.''' Graph of momentum against time for the reaction between F and H2]]

Initially, the H2 molecule is travelling in a relatively straight line with very little oscillations as evidenced by the low amplitude of momentum between 0-2 s. Following successful collision between F and H2, the amplitude of momentum of the product, HF, increases dramatically. This is because the reaction between F and H2 is exothermic and the excess energy released causes the molecule to become vibrationally excited and the amplitude of oscillations in the product to increase.

[[File:01409469 energy1 r1.jpg|thumb|center|400px|'''Figure 16.''' Graph of energy against time for the reaction between F and H2]]

The law of conservation of energy states that energy cannot be created or destroyed. It simply changes from one form to another. As can be seen from the graph in figure 16, total energy remains constant. In this reaction, potential energy is converted to kinetic energy. A decrease in potential energy of the atoms following a successful collision between F and H2 results in an equal increase in kinetic energy which manifests in the form of increased product (HF) oscillations.

This can be confirmed experimentally using infrared chemiluminescence. Observation of radiation from excited product molecules is a clear indication that these molecules are produced in the excited state. If the excitation is primarily vibrational, the radiation will appear in the infrared region of the spectrum (between 3-15 μm). This technique has been used extensively in hydrogen halogen reactions which produce HX molecules in the excited state.

One such experiment involves the use of the arrested relaxation method. In this technique, the reaction is carried out in a chamber whose walls are maintained at a very low temperature (20 K). This results in cryopumping of the chamber in that all the products and unreacted starting materials are condensed on the walls and only those product molecules that radiate before being pumped away are observed.

=== Polanyi's empirical rules ===

In a typical chemical reaction, there is an energetic barrier — a saddle point that the reactants must surmount to become products. Which form of energy is initially deposited in the reactants, translational or vibrational, is more efficacious in surmounting the activation energy barrier is one of the core topics in the area of reaction dynamics. The Polyani rules state that vibrational energy is more effective at promoting a late-barrier reaction (transition state resembles products). Conversely, translational energy is better at promoting an early-barrier reaction (transition state resembles the reactants).

The forward reaction between H2 and F is exothermic. Therefore, the transition state is closer to the reactants than to the products in energy. Since the transition state occurs early in an exothermic reaction, translational energy is more effective at promoting a chemical reaction than vibrational energy.

Conversely, the backward reaction between H and HF is endothermic. According to Hammond's postulate, the transition state resembles the products more than it resembles the reactants. Therefore, in an endothermic reaction, a late transition state is observed and vibrational energy is more effective at promoting a reaction than translational energy.

=== Promoting the forward reaction: F + HA-HB → HA-F + HB ===

1. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 2.95 and pHF = -0.5.

[[File:01409469 polyani 1+2.jpg|thumb|center|800px|'''Figure 17.''' Contour plot of the reaction between F and H2 when pHH = 2.95 '''Figure 18.''' Graph of momentum against time for the reaction between F and H2 when pHH = 2.95 (Left-right)]]

Vibrational energy is represented by pHH while translational energy is represented by pHF.

A molecule of H2 collides with F. However, the collision is unsuccessful and no reaction occurs. A reaction did not occur even though the molecules have sufficient kinetic energy to overcome the activation energy barrier. This is because most of the energy is deposited in the reactants as H-H vibrations as seen by the large oscillations in H2. Barrier recrossing has occured as a result of the large vibrational energy causing the system to revert back to the reactants even though the system has already crossed the transition state and a bond was formed between HB and F temporarily.

This appears to run counter to what one would expect to happen when more energy is put into the system. It is predicted that when more energy is put into the system, the reactants would possess more kinetic energy to overcome the activation energy barrier. However, as can be seen from the simulation above, an increase in the vibrational energy of the reactants is not effective at promoting the occurrence of an exothermic reaction with an early transition state. This is in accordance to Polyani's rules.

2. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 0.1 and pHF = -0.8.

[[File:01409469 polyani 3+4.jpg|thumb|center|800px|'''Figure 19.''' Contour plot of the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 '''Figure 20.''' Graph of momentum against time for the reaction between F and H2 when pHH = 0.1 and pHF = -0.8 (Left-right)]]

MRD:01409469

2019-05-10T13:17:32Z

Stc1917: /* Promoting the forward reaction: F + HA-HB → HA-F + HB */

== Exercise 1: H + H2 system ==

=== Definition of a transition state ===

A potential energy profile results from plotting the potential versus reaction path.

[[File:Surface_Plot01409469_pic1.jpg|thumb|center|400px|'''Figure 1.''' A potential energy surface plot]]

The transition state is defined as the maximum on the minimum energy path linking reactants and products. This means that the transition state is a stationary point on the potential energy surface. Thus, at the transition state, the derivative of the potential energy is zero with respect to each Cartesian coordinate.

<math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>

However, a local minimum on the potential energy surface corresponding to a well would have the same <math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>. Therefore, to distinguish between a local minimum on the potential energy surface and the transition state, the second derivative, <math>\tfrac{\partial^2 V(r_i)}{\partial r_i ^2}\ </math> is determined. This can be calculated by finding the determinant of the Hessian matrix shown below.

:<math>\mathbf H =
\begin{bmatrix}
\dfrac{\partial^2 \ V(r_i)}{\partial \ x^2} & \dfrac{\partial^2 \ V(r_i)}{\partial \ x \partial \mathbf y} \\
\dfrac{\partial^2 \ V(r_i)}{\partial \ y \partial \mathbf x} & \dfrac{\partial^2 \ V(r_i)}{\partial \mathbf y^2}
\end{bmatrix}</math>

The Hessian matrix of a convex function is positive semi-definite. Refining this property allows one to test if a critical point (x,y) is a local maximum, local minimum or saddle point.

If <math>f</math> < 0, then (x,y) is an isolated local maximum.

If <math>f</math> > 0, then (x,y) is an isolated local minimum.

If the Hessian has both positive and negative eigenvalues then (x,y) is a saddle point.

=== Locating the transition state ===

[[File:01409469 combined pic 2 and 3.jpg|thumb|center|800px|'''Figure 2.''' A surface and contour plot of the trajectory when rAB=rBC=0.907743 Å and pA=pBC=0]]

When the molecules are not at the transition state, a surface plot of the conditions shows a trajectory being plotted. This means that the trajectory is "rolling" towards the products or reactants. However, if a trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

[[File:01409469_internuclearpic3.png|thumb|center|400px|'''Figure 3.''' A plot of internuclear distance against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

An initial estimate for the position of the transition state was made by observing the potential energy surface and was found to be 0.90 Å. At 0.90 Å, the internuclear distance against time plot showed that the internuclear distance was constantly oscillating. Therefore, the position of the transition state was further refined by increasing the number of decimal places until a horizontal line in the internuclear distance againt time plot was obtained. This means that the internuclear distance is constant and that the molecules are simply undergoing periodic symmetric vibration at the transition state. Therefore, the position of the transition state was determined to be 0.907743 Å.

[[File:01409469 combined energy pic.png|thumb|center|400px|'''Figure 4.''' A plot of energy against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

When rAB=rBC=0.907743 Å and pA=pBC=0, the kinetic energy is 0. This means that HA and HBC are stationary, which is as predicted when the trajectory is started at the transition state with no initial momentum. Furthermore, the forces acting on HA and HBC are 0 as should be the case in the transition state.

=== Calculation type: Dynamic and minimum energy path ===

The minimum energy path, is the lowest energy path from the reactants to the products. The minimum energy path is determined by following the mode corresponding to the negative Hessian eigenvalue in both directions along the steepest descent direction. Following this path allows one to approach two valleys of stable structures. The resulting path generated is the minimum energy path. The reaction paths were chartered for a dynamic and a minimum energy path. The initial conditions used were that of the transition state except that the internuclear distance between HA and HB had increased to 0.917743 Å. The reaction path obtained using the two different calculation methods was different.

[[File:01409469 mep and dynamics.jpg|thumb|center|800px|'''Figure 5.''' A surface plot of the trajectory using minimum energy path '''Figure 6.''' A surface plot of the trajectory using dynamics (Left-right)]]

On the surface plot, the reaction path is depicted by the black line. Following the minimum energy path, the trajectory simply follows the valley floor to Ha+Hb-Hc. The reaction path appears to be a relatively straight line indicating that the atoms are not vibrating. In the minimum energy path, the reactants change into the products in an infinitely slow motion. This is done by having the momenta/velocities reset to zero in each time step. While the minimum energy path is useful in characterising a chemical reaction, it does not provide a realistic representation of the motion of the atoms during the reaction. This is because the mass and inertia of the atoms have been ignored.

Since the atoms simply follow the minimum energy path, they do not undergo slight displacements into surrounding valleys of higher potential energy. Furthermore, because the atoms are assumed to have no inertia, they do not undergo vibrations even if they were to be displaced to valleys of higher potential energy.

However, when dynamic calculation is performed, the reaction path is wavy indicating that the atoms are oscillating. This is because the atoms are displaced to regions of higher potential on the path from reactants to products. This leads to atom oscillation due to the inertia of the atoms.

'''Internuclear distance against time plot'''

Setting the initial system conditions such that rAB= 0.907743, rBC= 0.917743 and pA= pBC= 0:

[[File:01409469 Internuclear distance mep and dynamic.jpg|thumb|center|800px|'''Figure 6.''' A graph of internuclear distance against steps using minimum energy path '''Figure 7.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In the minimum energy path, the distance between HA and HB decreases from 0.90 Å to approximately 0.79 Å during the first 50 steps before plateauing off to a value of 0.74 Å which is the approximate length of a hydrogen bond. This means that a bond was formed between HA and HB. In addition, the plateau indicates that the bond is static. On the other hand, the distance between HB and HC has increased to 1.91 Å at large step values. This shows that HC has been ejected off following a successful collision between HA and HBC.

Similarly, in the dynamic trajectory, the distance between HA and HB decreases to approximately 0.75 Å after 0.3 s. This is indicative of hydrogen bond formation between HA and HB. In addition, the distance between HB and HC increases to 3.44 Å after HC is ejected following a successful collision.

One obvious distance between the two plots is that in the minimum energy path trajectory, the internuclear distance between HA and HB is constant indicating that the bond is static. However, in the dynamic trajectory the internuclear distance is oscillating showing the the bond is vibrating. This difference could be attributed to the fact that minimum energy path does not take oscillatory motion into account while dynamic trajectory does.

However, changing the initial conditions of the system such that rAB = 0.917743 Å, rBC = 0.907743 Å and pA=pBC=0:
[[File:01409469 01409469 Reversed values dynamic mep.jpg|thumb|center|800px|'''Figure 8.''' A graph of internuclear distance against steps using minimum energy path '''Figure 9.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In this case, a hydrogen bond of length 0.77-0.79 Å is formed between HB and HC while HA is repelled away to beyond 3.45 Å.

Additionally, from the internuclear distance plot it can be seen that the distance between HB and HC increases in a rectangular hyperbola fashion when calculated using the minimum energy path. On the other hand, the internuclear distance between HB and HC increases in a linear manner when dynamic calculation is applied. A possible reason for this difference is that in the minimum energy path, the atoms are not vibrating and as such the reach the maximum distance and plateau at a faster rate. However, in the dynamic calculation, the atoms have momentum and as such move further away from each other.

=== Reactive and unreactive trajectories ===

With the initial positions r'''1''' = 0.74 and r'''2''' = 2.0, trajectories were run with the momenta combination shown below.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Contour plot !! Reactivity !! Description of the dynamics
|-
| -1.25 || -2.5 ||-99.018 ||[[File:01409469 trajectory1.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. There are no oscillations observed in HA-HB initially, indicating that the diatomic molecule is stable. A new bond is formed between HB and HC. Oscillatory behaviour is observed indicating that the bond is vibrating. On the other hand, HA is ejected away to beyond 10 Å.
|-
| -1.5 || -2.0 ||-100.456 ||[[File:01409469 trajectory2.jpg|thumb|center|400px]] ||Unreactive ||HC collides with the vibrating diatomic HA-HB. However, no bond is formed between HB and HC as the collision is unsuccessful. This is because the molecules lack sufficient kinetic energy to overcome the activation energy barrier and so no reaction occurs. HC is subsequently repelled to beyond 8 Å.
|-
| -1.5 || -2.5 ||-98.956 ||[[File:01409469 trajectory3.jpg|thumb|center|400px]] ||Reactive |||HC collides with HA-HB. The collision is successful and a bond is formed between HB and HC. This is because the atoms have sufficient kinetic energy to overcome the activation energy barrier and form a new HB-HC bond. HA is ejected away to beyond 10 Å. The new hydrogen bond formed is vibrating as seen by the oscillatory behaviour.
|-
| -2.5 || -5.0 ||-84.956 ||[[File:01409469 trajectory4.jpg|thumb|center|400px]] ||Unreactive ||HC collides with HA-HB. Initially the HA-HB bond is elongating, indicating that the hydrogen bond is weakening and going to break. However, after 0.5 s, the HA-HB bond length decreases and the HAHB is not broken. HC is repelled beyond 6 Å with very strong vibrations.
A reaction did not occur even though the atoms have sufficient kinetic energy to cross the activation energy barrier. This is because the vibrations are too strong. The system crosses the transition state region and a bond is formed in the product temporarily but then the system reverts back to the reactants. This is an example of barrier recrossing.
|-
| -2.5 || -5.2 ||-83.416 ||[[File:01409469 trajectory5.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. The collision is successful and a new hydrogen bond is formed between HB and HC with HA being ejected away. The products formed experience very strong vibrations as evidenced by the oscillatory motion which occur with a very high frequency.
|}

From the table, it can be seen that the chance of a reaction successfully occurring is not solely dependent on the amount of kinetic energy the atoms possess. Even if the atoms have sufficient kinetic energy to cross the activation energy barrier, barrier recrossing can occur if the atoms are too vibrationally excited.

=== Transition State Theory ===

Transition state theory provided the first theoretical attempt to determine absolute reaction rates. In this theory, a transition state separating the reactants from products is used to find an expression for the thermal rate constant.

The main assumptions of transition state theory are:

1. Electronic and nuclear motions are separated (equivalent to the Born-Oppenheimer approximation in quantum mechanics)

2. Reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution

3. Molecular systems that have crossed the transition state in the direction of products cannot turn around and reform reactants

4. In the transition state, motion along the reaction coordinate may be separated from the other motions and treated classically as a translation

5. Even in the absence of an equilibrium between reactant and product molecules, the transition states that are becoming products are distributed among their states according to the Maxwell-Boltzmann laws

The goal of transition state theory is to predict the rate constant, ''k'', of a reaction given a potential energy surface for the reaction.

However, transition state theory suffers from a set of limitations due to the assumptions stated above not holding true in reality.

Firstly, transition state theory assumes that barrier recrossing is negligible. However, as seen from the fourth trajectory in table 1, barrier recrossing occurred because the atoms were too vibrationally excited and the system reverted back to the reactants even though the system had already crossed the transition state region. In this situation, transition state theory would predict a faster rate of reaction compared to what was obtained experimentally.

Secondly, the transition state theory breaks down for systems with light atoms such as hydrogens. In such systems, quantum effects become crucial. Quantum effects tend to increase the rate of the reaction for two reasons. First, particles lie in energy levels which are higher then the bottom of the potential energy well (there is additional zero-point energy). Secondly, the particles can tunnel through the barrier. These two effects effectively lower the activation energy barrier. As a result, the rate of reaction predicted by transition state theory would be less than what is obtained experimentally.

Lastly, transition state theory does not hold true for systems which are far from equilibrium. In such cases, the assumption of a starting Boltzmann distribution breaks down.

== Exercise 2: F - H - H system ==

=== Potential energy surface inspection===

F + HA-HB → HA-F + HB

[[File:01409469 Surface_Plot F 3.jpg|thumb|center|400px|'''Figure 10.''' Potential energy surface for the reaction between F and H2]]

The reaction between F and H2 is an exothermic reaction. This means that the reactants are of a higher energy than the products. The overall reaction involves the breaking of a hydrogen bond and the formation of a hydrogen-fluorine bond. The HF bond formed is stronger than the hydrogen bond that was broken. This is because fluorine is more electronegative than hydrogen resulting in a polar HF bond. The negative enthalpy of H-F bond formation is greater than the positive enthalpy of hydrogen bond dissociation. Therefore, the reaction between F and H2 is an exothermic reaction.

HA + HB-F → HA-HB + F

[[File:01409469 Surface_Plot F 4.jpg|thumb|center|400px|'''Figure 11.''' Potential energy surface for the reaction between H and HF]]

The reaction between H and HF is an endothermic reaction. The products are at a higher energy level than the reactants. The overall reaction involves the breaking of a HF bond and the formation of a hydrogen bond. The formation of a homonuclear H2 molecule results in a non-polar hydrogen bond being formed. The HF bond broken is stronger than the hydrogen bond that was formed. The negative enthalpy of hydrogen bond formation is less than the positive enthalpy of hydrogen-fluorine dissociation. Therefore, the reaction between H and HF is endothermic.

=== Locating the transition state ===

[[File:01409469 Surface_Plot F TS4.jpg|thumb|center|800px|'''Figure 12.''' Determining the position of the transition state for the reaction between F and H2]]

As mentioned previously, when the molecules are not at the transition state, a reaction path will be plotted. However, if the initial conditions of the system are set such that the trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

An initial estimate for the transition state was made by setting rAB = 0.74 which is the length of a hydrogen bond. The position of the transition state was further refined by increasing the number of decimal places such that the final value of rAB was found to be 0.74478 and rBC = 1.81079. This set of distances gave a reaction path that was represented by a point indicating that no trajectory was plotted and that the molecules remained at the transition state since their initial momentum was 0.

Moreover, the kinetic energy as well as the forces acting on the atoms was calculated to be 0, which is consistent with the transition state.

Therefore, the position of the transition state was found to be at rAB = 0.74478 and rBC = 1.81079 when pA = pBC = 0.

=== Determining the activation energy ===

[[File:01409469 activation energy r1.1 combined.jpg|thumb|center|800px|'''Figure 13.''' Determining the activation energy for the reaction between F and H2]]

By reducing the value of rBC by 0.01, the atoms are no longer at the transition state and would move towards the formation of products. The activation energy was found by calculating the change in total energy after the reaction. The activation energy of F + H2 → HF + H was found to be +23.279 kcal/ mol.

[[File:01409469 activation energy r2.1 combined.jpg|thumb|center|800px|'''Figure 14.''' Determining the activation energy for the reaction between FH and H]]

The activation energy of FH + H → F + H2 was found by calculating the small dip in potential energy. The activation energy was found to be +0.280 kcal/ mol.

=== Reaction dynamics of F + HA-HB → HA-F + HB ===

The set of initial conditions that result in a reactive trajectory for F + H2 is as follows: rHH = 0.74 Å, rHF = 1.79 Å and pHH = pHF = 0.

[[File:01409469 momentum r1.jpg|thumb|center|400px|'''Figure 15.''' Graph of momentum against time for the reaction between F and H2]]

Initially, the H2 molecule is travelling in a relatively straight line with very little oscillations as evidenced by the low amplitude of momentum between 0-2 s. Following successful collision between F and H2, the amplitude of momentum of the product, HF, increases dramatically. This is because the reaction between F and H2 is exothermic and the excess energy released causes the molecule to become vibrationally excited and the amplitude of oscillations in the product to increase.

[[File:01409469 energy1 r1.jpg|thumb|center|400px|'''Figure 16.''' Graph of energy against time for the reaction between F and H2]]

The law of conservation of energy states that energy cannot be created or destroyed. It simply changes from one form to another. As can be seen from the graph in figure 16, total energy remains constant. In this reaction, potential energy is converted to kinetic energy. A decrease in potential energy of the atoms following a successful collision between F and H2 results in an equal increase in kinetic energy which manifests in the form of increased product (HF) oscillations.

This can be confirmed experimentally using infrared chemiluminescence. Observation of radiation from excited product molecules is a clear indication that these molecules are produced in the excited state. If the excitation is primarily vibrational, the radiation will appear in the infrared region of the spectrum (between 3-15 μm). This technique has been used extensively in hydrogen halogen reactions which produce HX molecules in the excited state.

One such experiment involves the use of the arrested relaxation method. In this technique, the reaction is carried out in a chamber whose walls are maintained at a very low temperature (20 K). This results in cryopumping of the chamber in that all the products and unreacted starting materials are condensed on the walls and only those product molecules that radiate before being pumped away are observed.

=== Polanyi's empirical rules ===

In a typical chemical reaction, there is an energetic barrier — a saddle point that the reactants must surmount to become products. Which form of energy is initially deposited in the reactants, translational or vibrational, is more efficacious in surmounting the activation energy barrier is one of the core topics in the area of reaction dynamics. The Polyani rules state that vibrational energy is more effective at promoting a late-barrier reaction (transition state resembles products). Conversely, translational energy is better at promoting an early-barrier reaction (transition state resembles the reactants).

The forward reaction between H2 and F is exothermic. Therefore, the transition state is closer to the reactants than to the products in energy. Since the transition state occurs early in an exothermic reaction, translational energy is more effective at promoting a chemical reaction than vibrational energy.

Conversely, the backward reaction between H and HF is endothermic. According to Hammond's postulate, the transition state resembles the products more than it resembles the reactants. Therefore, in an endothermic reaction, a late transition state is observed and vibrational energy is more effective at promoting a reaction than translational energy.

=== Promoting the forward reaction: F + HA-HB → HA-F + HB ===

1. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 2.95 and pHF = -0.5.

[[File:01409469 polyani 1+2.jpg|thumb|center|800px|'''Figure 17.''' Contour plot of the reaction between F and H2 when pHH = 2.95 '''Figure 18.''' Graph of momentum against time for the reaction between F and H2 when pHH = 2.95 (Left-right)]]

Vibrational energy is represented by pHH while translational energy is represented by pHF.

A molecule of H2 collides with F. However, the collision is unsuccessful and no reaction occurs. A reaction did not occur even though the molecules have sufficient kinetic energy to overcome the activation energy barrier. This is because most of the energy is deposited in the reactants as H-H vibrations as seen by the large oscillations in H2. Barrier recrossing has occured as a result of the large vibrational energy causing the system to revert back to the reactants even though the system has already crossed the transition state and a bond was formed between HB and F temporarily.

This appears to run counter to what one would expect to happen when more energy is put into the system. It is predicted that when more energy is put into the system, the reactants would possess more kinetic energy to overcome the activation energy barrier. However, as can be seen from the simulation above, an increase in the vibrational energy of the reactants is not effective at promoting the occurrence of an exothermic reaction with an early transition state. This is in accordance to Polyani's rules.

2. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 0.1 and pHF = -0.8.

MRD:01409469

2019-05-10T13:11:06Z

Stc1917: /* Promoting the forward reaction: F + HA-HB → HA-F + HB */

== Exercise 1: H + H2 system ==

=== Definition of a transition state ===

A potential energy profile results from plotting the potential versus reaction path.

[[File:Surface_Plot01409469_pic1.jpg|thumb|center|400px|'''Figure 1.''' A potential energy surface plot]]

The transition state is defined as the maximum on the minimum energy path linking reactants and products. This means that the transition state is a stationary point on the potential energy surface. Thus, at the transition state, the derivative of the potential energy is zero with respect to each Cartesian coordinate.

<math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>

However, a local minimum on the potential energy surface corresponding to a well would have the same <math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>. Therefore, to distinguish between a local minimum on the potential energy surface and the transition state, the second derivative, <math>\tfrac{\partial^2 V(r_i)}{\partial r_i ^2}\ </math> is determined. This can be calculated by finding the determinant of the Hessian matrix shown below.

:<math>\mathbf H =
\begin{bmatrix}
\dfrac{\partial^2 \ V(r_i)}{\partial \ x^2} & \dfrac{\partial^2 \ V(r_i)}{\partial \ x \partial \mathbf y} \\
\dfrac{\partial^2 \ V(r_i)}{\partial \ y \partial \mathbf x} & \dfrac{\partial^2 \ V(r_i)}{\partial \mathbf y^2}
\end{bmatrix}</math>

The Hessian matrix of a convex function is positive semi-definite. Refining this property allows one to test if a critical point (x,y) is a local maximum, local minimum or saddle point.

If <math>f</math> < 0, then (x,y) is an isolated local maximum.

If <math>f</math> > 0, then (x,y) is an isolated local minimum.

If the Hessian has both positive and negative eigenvalues then (x,y) is a saddle point.

=== Locating the transition state ===

[[File:01409469 combined pic 2 and 3.jpg|thumb|center|800px|'''Figure 2.''' A surface and contour plot of the trajectory when rAB=rBC=0.907743 Å and pA=pBC=0]]

When the molecules are not at the transition state, a surface plot of the conditions shows a trajectory being plotted. This means that the trajectory is "rolling" towards the products or reactants. However, if a trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

[[File:01409469_internuclearpic3.png|thumb|center|400px|'''Figure 3.''' A plot of internuclear distance against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

An initial estimate for the position of the transition state was made by observing the potential energy surface and was found to be 0.90 Å. At 0.90 Å, the internuclear distance against time plot showed that the internuclear distance was constantly oscillating. Therefore, the position of the transition state was further refined by increasing the number of decimal places until a horizontal line in the internuclear distance againt time plot was obtained. This means that the internuclear distance is constant and that the molecules are simply undergoing periodic symmetric vibration at the transition state. Therefore, the position of the transition state was determined to be 0.907743 Å.

[[File:01409469 combined energy pic.png|thumb|center|400px|'''Figure 4.''' A plot of energy against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

When rAB=rBC=0.907743 Å and pA=pBC=0, the kinetic energy is 0. This means that HA and HBC are stationary, which is as predicted when the trajectory is started at the transition state with no initial momentum. Furthermore, the forces acting on HA and HBC are 0 as should be the case in the transition state.

=== Calculation type: Dynamic and minimum energy path ===

The minimum energy path, is the lowest energy path from the reactants to the products. The minimum energy path is determined by following the mode corresponding to the negative Hessian eigenvalue in both directions along the steepest descent direction. Following this path allows one to approach two valleys of stable structures. The resulting path generated is the minimum energy path. The reaction paths were chartered for a dynamic and a minimum energy path. The initial conditions used were that of the transition state except that the internuclear distance between HA and HB had increased to 0.917743 Å. The reaction path obtained using the two different calculation methods was different.

[[File:01409469 mep and dynamics.jpg|thumb|center|800px|'''Figure 5.''' A surface plot of the trajectory using minimum energy path '''Figure 6.''' A surface plot of the trajectory using dynamics (Left-right)]]

On the surface plot, the reaction path is depicted by the black line. Following the minimum energy path, the trajectory simply follows the valley floor to Ha+Hb-Hc. The reaction path appears to be a relatively straight line indicating that the atoms are not vibrating. In the minimum energy path, the reactants change into the products in an infinitely slow motion. This is done by having the momenta/velocities reset to zero in each time step. While the minimum energy path is useful in characterising a chemical reaction, it does not provide a realistic representation of the motion of the atoms during the reaction. This is because the mass and inertia of the atoms have been ignored.

Since the atoms simply follow the minimum energy path, they do not undergo slight displacements into surrounding valleys of higher potential energy. Furthermore, because the atoms are assumed to have no inertia, they do not undergo vibrations even if they were to be displaced to valleys of higher potential energy.

However, when dynamic calculation is performed, the reaction path is wavy indicating that the atoms are oscillating. This is because the atoms are displaced to regions of higher potential on the path from reactants to products. This leads to atom oscillation due to the inertia of the atoms.

'''Internuclear distance against time plot'''

Setting the initial system conditions such that rAB= 0.907743, rBC= 0.917743 and pA= pBC= 0:

[[File:01409469 Internuclear distance mep and dynamic.jpg|thumb|center|800px|'''Figure 6.''' A graph of internuclear distance against steps using minimum energy path '''Figure 7.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In the minimum energy path, the distance between HA and HB decreases from 0.90 Å to approximately 0.79 Å during the first 50 steps before plateauing off to a value of 0.74 Å which is the approximate length of a hydrogen bond. This means that a bond was formed between HA and HB. In addition, the plateau indicates that the bond is static. On the other hand, the distance between HB and HC has increased to 1.91 Å at large step values. This shows that HC has been ejected off following a successful collision between HA and HBC.

Similarly, in the dynamic trajectory, the distance between HA and HB decreases to approximately 0.75 Å after 0.3 s. This is indicative of hydrogen bond formation between HA and HB. In addition, the distance between HB and HC increases to 3.44 Å after HC is ejected following a successful collision.

One obvious distance between the two plots is that in the minimum energy path trajectory, the internuclear distance between HA and HB is constant indicating that the bond is static. However, in the dynamic trajectory the internuclear distance is oscillating showing the the bond is vibrating. This difference could be attributed to the fact that minimum energy path does not take oscillatory motion into account while dynamic trajectory does.

However, changing the initial conditions of the system such that rAB = 0.917743 Å, rBC = 0.907743 Å and pA=pBC=0:
[[File:01409469 01409469 Reversed values dynamic mep.jpg|thumb|center|800px|'''Figure 8.''' A graph of internuclear distance against steps using minimum energy path '''Figure 9.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In this case, a hydrogen bond of length 0.77-0.79 Å is formed between HB and HC while HA is repelled away to beyond 3.45 Å.

Additionally, from the internuclear distance plot it can be seen that the distance between HB and HC increases in a rectangular hyperbola fashion when calculated using the minimum energy path. On the other hand, the internuclear distance between HB and HC increases in a linear manner when dynamic calculation is applied. A possible reason for this difference is that in the minimum energy path, the atoms are not vibrating and as such the reach the maximum distance and plateau at a faster rate. However, in the dynamic calculation, the atoms have momentum and as such move further away from each other.

=== Reactive and unreactive trajectories ===

With the initial positions r'''1''' = 0.74 and r'''2''' = 2.0, trajectories were run with the momenta combination shown below.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Contour plot !! Reactivity !! Description of the dynamics
|-
| -1.25 || -2.5 ||-99.018 ||[[File:01409469 trajectory1.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. There are no oscillations observed in HA-HB initially, indicating that the diatomic molecule is stable. A new bond is formed between HB and HC. Oscillatory behaviour is observed indicating that the bond is vibrating. On the other hand, HA is ejected away to beyond 10 Å.
|-
| -1.5 || -2.0 ||-100.456 ||[[File:01409469 trajectory2.jpg|thumb|center|400px]] ||Unreactive ||HC collides with the vibrating diatomic HA-HB. However, no bond is formed between HB and HC as the collision is unsuccessful. This is because the molecules lack sufficient kinetic energy to overcome the activation energy barrier and so no reaction occurs. HC is subsequently repelled to beyond 8 Å.
|-
| -1.5 || -2.5 ||-98.956 ||[[File:01409469 trajectory3.jpg|thumb|center|400px]] ||Reactive |||HC collides with HA-HB. The collision is successful and a bond is formed between HB and HC. This is because the atoms have sufficient kinetic energy to overcome the activation energy barrier and form a new HB-HC bond. HA is ejected away to beyond 10 Å. The new hydrogen bond formed is vibrating as seen by the oscillatory behaviour.
|-
| -2.5 || -5.0 ||-84.956 ||[[File:01409469 trajectory4.jpg|thumb|center|400px]] ||Unreactive ||HC collides with HA-HB. Initially the HA-HB bond is elongating, indicating that the hydrogen bond is weakening and going to break. However, after 0.5 s, the HA-HB bond length decreases and the HAHB is not broken. HC is repelled beyond 6 Å with very strong vibrations.
A reaction did not occur even though the atoms have sufficient kinetic energy to cross the activation energy barrier. This is because the vibrations are too strong. The system crosses the transition state region and a bond is formed in the product temporarily but then the system reverts back to the reactants. This is an example of barrier recrossing.
|-
| -2.5 || -5.2 ||-83.416 ||[[File:01409469 trajectory5.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. The collision is successful and a new hydrogen bond is formed between HB and HC with HA being ejected away. The products formed experience very strong vibrations as evidenced by the oscillatory motion which occur with a very high frequency.
|}

From the table, it can be seen that the chance of a reaction successfully occurring is not solely dependent on the amount of kinetic energy the atoms possess. Even if the atoms have sufficient kinetic energy to cross the activation energy barrier, barrier recrossing can occur if the atoms are too vibrationally excited.

=== Transition State Theory ===

Transition state theory provided the first theoretical attempt to determine absolute reaction rates. In this theory, a transition state separating the reactants from products is used to find an expression for the thermal rate constant.

The main assumptions of transition state theory are:

1. Electronic and nuclear motions are separated (equivalent to the Born-Oppenheimer approximation in quantum mechanics)

2. Reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution

3. Molecular systems that have crossed the transition state in the direction of products cannot turn around and reform reactants

4. In the transition state, motion along the reaction coordinate may be separated from the other motions and treated classically as a translation

5. Even in the absence of an equilibrium between reactant and product molecules, the transition states that are becoming products are distributed among their states according to the Maxwell-Boltzmann laws

The goal of transition state theory is to predict the rate constant, ''k'', of a reaction given a potential energy surface for the reaction.

However, transition state theory suffers from a set of limitations due to the assumptions stated above not holding true in reality.

Firstly, transition state theory assumes that barrier recrossing is negligible. However, as seen from the fourth trajectory in table 1, barrier recrossing occurred because the atoms were too vibrationally excited and the system reverted back to the reactants even though the system had already crossed the transition state region. In this situation, transition state theory would predict a faster rate of reaction compared to what was obtained experimentally.

Secondly, the transition state theory breaks down for systems with light atoms such as hydrogens. In such systems, quantum effects become crucial. Quantum effects tend to increase the rate of the reaction for two reasons. First, particles lie in energy levels which are higher then the bottom of the potential energy well (there is additional zero-point energy). Secondly, the particles can tunnel through the barrier. These two effects effectively lower the activation energy barrier. As a result, the rate of reaction predicted by transition state theory would be less than what is obtained experimentally.

Lastly, transition state theory does not hold true for systems which are far from equilibrium. In such cases, the assumption of a starting Boltzmann distribution breaks down.

== Exercise 2: F - H - H system ==

=== Potential energy surface inspection===

F + HA-HB → HA-F + HB

[[File:01409469 Surface_Plot F 3.jpg|thumb|center|400px|'''Figure 10.''' Potential energy surface for the reaction between F and H2]]

The reaction between F and H2 is an exothermic reaction. This means that the reactants are of a higher energy than the products. The overall reaction involves the breaking of a hydrogen bond and the formation of a hydrogen-fluorine bond. The HF bond formed is stronger than the hydrogen bond that was broken. This is because fluorine is more electronegative than hydrogen resulting in a polar HF bond. The negative enthalpy of H-F bond formation is greater than the positive enthalpy of hydrogen bond dissociation. Therefore, the reaction between F and H2 is an exothermic reaction.

HA + HB-F → HA-HB + F

[[File:01409469 Surface_Plot F 4.jpg|thumb|center|400px|'''Figure 11.''' Potential energy surface for the reaction between H and HF]]

The reaction between H and HF is an endothermic reaction. The products are at a higher energy level than the reactants. The overall reaction involves the breaking of a HF bond and the formation of a hydrogen bond. The formation of a homonuclear H2 molecule results in a non-polar hydrogen bond being formed. The HF bond broken is stronger than the hydrogen bond that was formed. The negative enthalpy of hydrogen bond formation is less than the positive enthalpy of hydrogen-fluorine dissociation. Therefore, the reaction between H and HF is endothermic.

=== Locating the transition state ===

[[File:01409469 Surface_Plot F TS4.jpg|thumb|center|800px|'''Figure 12.''' Determining the position of the transition state for the reaction between F and H2]]

As mentioned previously, when the molecules are not at the transition state, a reaction path will be plotted. However, if the initial conditions of the system are set such that the trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

An initial estimate for the transition state was made by setting rAB = 0.74 which is the length of a hydrogen bond. The position of the transition state was further refined by increasing the number of decimal places such that the final value of rAB was found to be 0.74478 and rBC = 1.81079. This set of distances gave a reaction path that was represented by a point indicating that no trajectory was plotted and that the molecules remained at the transition state since their initial momentum was 0.

Moreover, the kinetic energy as well as the forces acting on the atoms was calculated to be 0, which is consistent with the transition state.

Therefore, the position of the transition state was found to be at rAB = 0.74478 and rBC = 1.81079 when pA = pBC = 0.

=== Determining the activation energy ===

[[File:01409469 activation energy r1.1 combined.jpg|thumb|center|800px|'''Figure 13.''' Determining the activation energy for the reaction between F and H2]]

By reducing the value of rBC by 0.01, the atoms are no longer at the transition state and would move towards the formation of products. The activation energy was found by calculating the change in total energy after the reaction. The activation energy of F + H2 → HF + H was found to be +23.279 kcal/ mol.

[[File:01409469 activation energy r2.1 combined.jpg|thumb|center|800px|'''Figure 14.''' Determining the activation energy for the reaction between FH and H]]

The activation energy of FH + H → F + H2 was found by calculating the small dip in potential energy. The activation energy was found to be +0.280 kcal/ mol.

=== Reaction dynamics of F + HA-HB → HA-F + HB ===

The set of initial conditions that result in a reactive trajectory for F + H2 is as follows: rHH = 0.74 Å, rHF = 1.79 Å and pHH = pHF = 0.

[[File:01409469 momentum r1.jpg|thumb|center|400px|'''Figure 15.''' Graph of momentum against time for the reaction between F and H2]]

Initially, the H2 molecule is travelling in a relatively straight line with very little oscillations as evidenced by the low amplitude of momentum between 0-2 s. Following successful collision between F and H2, the amplitude of momentum of the product, HF, increases dramatically. This is because the reaction between F and H2 is exothermic and the excess energy released causes the molecule to become vibrationally excited and the amplitude of oscillations in the product to increase.

[[File:01409469 energy1 r1.jpg|thumb|center|400px|'''Figure 16.''' Graph of energy against time for the reaction between F and H2]]

The law of conservation of energy states that energy cannot be created or destroyed. It simply changes from one form to another. As can be seen from the graph in figure 16, total energy remains constant. In this reaction, potential energy is converted to kinetic energy. A decrease in potential energy of the atoms following a successful collision between F and H2 results in an equal increase in kinetic energy which manifests in the form of increased product (HF) oscillations.

This can be confirmed experimentally using infrared chemiluminescence. Observation of radiation from excited product molecules is a clear indication that these molecules are produced in the excited state. If the excitation is primarily vibrational, the radiation will appear in the infrared region of the spectrum (between 3-15 μm). This technique has been used extensively in hydrogen halogen reactions which produce HX molecules in the excited state.

One such experiment involves the use of the arrested relaxation method. In this technique, the reaction is carried out in a chamber whose walls are maintained at a very low temperature (20 K). This results in cryopumping of the chamber in that all the products and unreacted starting materials are condensed on the walls and only those product molecules that radiate before being pumped away are observed.

=== Polanyi's empirical rules ===

In a typical chemical reaction, there is an energetic barrier — a saddle point that the reactants must surmount to become products. Which form of energy is initially deposited in the reactants, translational or vibrational, is more efficacious in surmounting the activation energy barrier is one of the core topics in the area of reaction dynamics. The Polyani rules state that vibrational energy is more effective at promoting a late-barrier reaction (transition state resembles products). Conversely, translational energy is better at promoting an early-barrier reaction (transition state resembles the reactants).

The forward reaction between H2 and F is exothermic. Therefore, the transition state is closer to the reactants than to the products in energy. Since the transition state occurs early in an exothermic reaction, translational energy is more effective at promoting a chemical reaction than vibrational energy.

Conversely, the backward reaction between H and HF is endothermic. According to Hammond's postulate, the transition state resembles the products more than it resembles the reactants. Therefore, in an endothermic reaction, a late transition state is observed and vibrational energy is more effective at promoting a reaction than translational energy.

=== Promoting the forward reaction: F + HA-HB → HA-F + HB ===

1. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 2.95 and pHF = -0.5.

[[File:01409469 polyani 1+2.jpg|thumb|center|800px|'''Figure 17.''' Contour plot of the reaction between F and H2 when pHH = 2.95 '''Figure 18.''' Graph of momentum against time for the reaction between F and H2 when pHH = 2.95 (Left-right)]]

Vibrational energy is represented by pHH while translational energy is represented by pHF.

A molecule of H2 collides with F. However, the collision is unsuccessful and no reaction occurs. A reaction did not occur even though the molecules have sufficient kinetic energy to overcome the activation energy barrier. This is because most of the energy is deposited in the reactants as H-H vibrations as seen by the large oscillations in H2. Barrier recrossing has occured as a result of the large vibrational energy causing the system to revert back to the reactants even though the system has already crossed the transition state and a bond was formed between HB and F temporarily.

This appears to run counter to what one would expect to happen when more energy is put into the system. It is predicted that when more energy is put into the system, the reactants would possess more kinetic energy to overcome the activation energy barrier. However, as can be seen from the simulation above, an increase in the vibrational energy of the reactants is not effective at promoting the occurrence of an exothermic reaction with an early transition state. This is in accordance to Polyani's rules.

MRD:01409469

2019-05-09T21:28:05Z

Stc1917: /* Promoting the forward reaction: F + HA-HB → HA-F + HB */

== Exercise 1: H + H2 system ==

=== Definition of a transition state ===

A potential energy profile results from plotting the potential versus reaction path.

[[File:Surface_Plot01409469_pic1.jpg|thumb|center|400px|'''Figure 1.''' A potential energy surface plot]]

The transition state is defined as the maximum on the minimum energy path linking reactants and products. This means that the transition state is a stationary point on the potential energy surface. Thus, at the transition state, the derivative of the potential energy is zero with respect to each Cartesian coordinate.

<math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>

However, a local minimum on the potential energy surface corresponding to a well would have the same <math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>. Therefore, to distinguish between a local minimum on the potential energy surface and the transition state, the second derivative, <math>\tfrac{\partial^2 V(r_i)}{\partial r_i ^2}\ </math> is determined. This can be calculated by finding the determinant of the Hessian matrix shown below.

:<math>\mathbf H =
\begin{bmatrix}
\dfrac{\partial^2 \ V(r_i)}{\partial \ x^2} & \dfrac{\partial^2 \ V(r_i)}{\partial \ x \partial \mathbf y} \\
\dfrac{\partial^2 \ V(r_i)}{\partial \ y \partial \mathbf x} & \dfrac{\partial^2 \ V(r_i)}{\partial \mathbf y^2}
\end{bmatrix}</math>

The Hessian matrix of a convex function is positive semi-definite. Refining this property allows one to test if a critical point (x,y) is a local maximum, local minimum or saddle point.

If <math>f</math> < 0, then (x,y) is an isolated local maximum.

If <math>f</math> > 0, then (x,y) is an isolated local minimum.

If the Hessian has both positive and negative eigenvalues then (x,y) is a saddle point.

=== Locating the transition state ===

[[File:01409469 combined pic 2 and 3.jpg|thumb|center|800px|'''Figure 2.''' A surface and contour plot of the trajectory when rAB=rBC=0.907743 Å and pA=pBC=0]]

When the molecules are not at the transition state, a surface plot of the conditions shows a trajectory being plotted. This means that the trajectory is "rolling" towards the products or reactants. However, if a trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

[[File:01409469_internuclearpic3.png|thumb|center|400px|'''Figure 3.''' A plot of internuclear distance against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

An initial estimate for the position of the transition state was made by observing the potential energy surface and was found to be 0.90 Å. At 0.90 Å, the internuclear distance against time plot showed that the internuclear distance was constantly oscillating. Therefore, the position of the transition state was further refined by increasing the number of decimal places until a horizontal line in the internuclear distance againt time plot was obtained. This means that the internuclear distance is constant and that the molecules are simply undergoing periodic symmetric vibration at the transition state. Therefore, the position of the transition state was determined to be 0.907743 Å.

[[File:01409469 combined energy pic.png|thumb|center|400px|'''Figure 4.''' A plot of energy against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

When rAB=rBC=0.907743 Å and pA=pBC=0, the kinetic energy is 0. This means that HA and HBC are stationary, which is as predicted when the trajectory is started at the transition state with no initial momentum. Furthermore, the forces acting on HA and HBC are 0 as should be the case in the transition state.

=== Calculation type: Dynamic and minimum energy path ===

The minimum energy path, is the lowest energy path from the reactants to the products. The minimum energy path is determined by following the mode corresponding to the negative Hessian eigenvalue in both directions along the steepest descent direction. Following this path allows one to approach two valleys of stable structures. The resulting path generated is the minimum energy path. The reaction paths were chartered for a dynamic and a minimum energy path. The initial conditions used were that of the transition state except that the internuclear distance between HA and HB had increased to 0.917743 Å. The reaction path obtained using the two different calculation methods was different.

[[File:01409469 mep and dynamics.jpg|thumb|center|800px|'''Figure 5.''' A surface plot of the trajectory using minimum energy path '''Figure 6.''' A surface plot of the trajectory using dynamics (Left-right)]]

On the surface plot, the reaction path is depicted by the black line. Following the minimum energy path, the trajectory simply follows the valley floor to Ha+Hb-Hc. The reaction path appears to be a relatively straight line indicating that the atoms are not vibrating. In the minimum energy path, the reactants change into the products in an infinitely slow motion. This is done by having the momenta/velocities reset to zero in each time step. While the minimum energy path is useful in characterising a chemical reaction, it does not provide a realistic representation of the motion of the atoms during the reaction. This is because the mass and inertia of the atoms have been ignored.

Since the atoms simply follow the minimum energy path, they do not undergo slight displacements into surrounding valleys of higher potential energy. Furthermore, because the atoms are assumed to have no inertia, they do not undergo vibrations even if they were to be displaced to valleys of higher potential energy.

However, when dynamic calculation is performed, the reaction path is wavy indicating that the atoms are oscillating. This is because the atoms are displaced to regions of higher potential on the path from reactants to products. This leads to atom oscillation due to the inertia of the atoms.

'''Internuclear distance against time plot'''

Setting the initial system conditions such that rAB= 0.907743, rBC= 0.917743 and pA= pBC= 0:

[[File:01409469 Internuclear distance mep and dynamic.jpg|thumb|center|800px|'''Figure 6.''' A graph of internuclear distance against steps using minimum energy path '''Figure 7.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In the minimum energy path, the distance between HA and HB decreases from 0.90 Å to approximately 0.79 Å during the first 50 steps before plateauing off to a value of 0.74 Å which is the approximate length of a hydrogen bond. This means that a bond was formed between HA and HB. In addition, the plateau indicates that the bond is static. On the other hand, the distance between HB and HC has increased to 1.91 Å at large step values. This shows that HC has been ejected off following a successful collision between HA and HBC.

Similarly, in the dynamic trajectory, the distance between HA and HB decreases to approximately 0.75 Å after 0.3 s. This is indicative of hydrogen bond formation between HA and HB. In addition, the distance between HB and HC increases to 3.44 Å after HC is ejected following a successful collision.

One obvious distance between the two plots is that in the minimum energy path trajectory, the internuclear distance between HA and HB is constant indicating that the bond is static. However, in the dynamic trajectory the internuclear distance is oscillating showing the the bond is vibrating. This difference could be attributed to the fact that minimum energy path does not take oscillatory motion into account while dynamic trajectory does.

However, changing the initial conditions of the system such that rAB = 0.917743 Å, rBC = 0.907743 Å and pA=pBC=0:
[[File:01409469 01409469 Reversed values dynamic mep.jpg|thumb|center|800px|'''Figure 8.''' A graph of internuclear distance against steps using minimum energy path '''Figure 9.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In this case, a hydrogen bond of length 0.77-0.79 Å is formed between HB and HC while HA is repelled away to beyond 3.45 Å.

Additionally, from the internuclear distance plot it can be seen that the distance between HB and HC increases in a rectangular hyperbola fashion when calculated using the minimum energy path. On the other hand, the internuclear distance between HB and HC increases in a linear manner when dynamic calculation is applied. A possible reason for this difference is that in the minimum energy path, the atoms are not vibrating and as such the reach the maximum distance and plateau at a faster rate. However, in the dynamic calculation, the atoms have momentum and as such move further away from each other.

=== Reactive and unreactive trajectories ===

With the initial positions r'''1''' = 0.74 and r'''2''' = 2.0, trajectories were run with the momenta combination shown below.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Contour plot !! Reactivity !! Description of the dynamics
|-
| -1.25 || -2.5 ||-99.018 ||[[File:01409469 trajectory1.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. There are no oscillations observed in HA-HB initially, indicating that the diatomic molecule is stable. A new bond is formed between HB and HC. Oscillatory behaviour is observed indicating that the bond is vibrating. On the other hand, HA is ejected away to beyond 10 Å.
|-
| -1.5 || -2.0 ||-100.456 ||[[File:01409469 trajectory2.jpg|thumb|center|400px]] ||Unreactive ||HC collides with the vibrating diatomic HA-HB. However, no bond is formed between HB and HC as the collision is unsuccessful. This is because the molecules lack sufficient kinetic energy to overcome the activation energy barrier and so no reaction occurs. HC is subsequently repelled to beyond 8 Å.
|-
| -1.5 || -2.5 ||-98.956 ||[[File:01409469 trajectory3.jpg|thumb|center|400px]] ||Reactive |||HC collides with HA-HB. The collision is successful and a bond is formed between HB and HC. This is because the atoms have sufficient kinetic energy to overcome the activation energy barrier and form a new HB-HC bond. HA is ejected away to beyond 10 Å. The new hydrogen bond formed is vibrating as seen by the oscillatory behaviour.
|-
| -2.5 || -5.0 ||-84.956 ||[[File:01409469 trajectory4.jpg|thumb|center|400px]] ||Unreactive ||HC collides with HA-HB. Initially the HA-HB bond is elongating, indicating that the hydrogen bond is weakening and going to break. However, after 0.5 s, the HA-HB bond length decreases and the HAHB is not broken. HC is repelled beyond 6 Å with very strong vibrations.
A reaction did not occur even though the atoms have sufficient kinetic energy to cross the activation energy barrier. This is because the vibrations are too strong. The system crosses the transition state region and a bond is formed in the product temporarily but then the system reverts back to the reactants. This is an example of barrier recrossing.
|-
| -2.5 || -5.2 ||-83.416 ||[[File:01409469 trajectory5.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. The collision is successful and a new hydrogen bond is formed between HB and HC with HA being ejected away. The products formed experience very strong vibrations as evidenced by the oscillatory motion which occur with a very high frequency.
|}

From the table, it can be seen that the chance of a reaction successfully occurring is not solely dependent on the amount of kinetic energy the atoms possess. Even if the atoms have sufficient kinetic energy to cross the activation energy barrier, barrier recrossing can occur if the atoms are too vibrationally excited.

=== Transition State Theory ===

Transition state theory provided the first theoretical attempt to determine absolute reaction rates. In this theory, a transition state separating the reactants from products is used to find an expression for the thermal rate constant.

The main assumptions of transition state theory are:

1. Electronic and nuclear motions are separated (equivalent to the Born-Oppenheimer approximation in quantum mechanics)

2. Reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution

3. Molecular systems that have crossed the transition state in the direction of products cannot turn around and reform reactants

4. In the transition state, motion along the reaction coordinate may be separated from the other motions and treated classically as a translation

5. Even in the absence of an equilibrium between reactant and product molecules, the transition states that are becoming products are distributed among their states according to the Maxwell-Boltzmann laws

The goal of transition state theory is to predict the rate constant, ''k'', of a reaction given a potential energy surface for the reaction.

However, transition state theory suffers from a set of limitations due to the assumptions stated above not holding true in reality.

Firstly, transition state theory assumes that barrier recrossing is negligible. However, as seen from the fourth trajectory in table 1, barrier recrossing occurred because the atoms were too vibrationally excited and the system reverted back to the reactants even though the system had already crossed the transition state region. In this situation, transition state theory would predict a faster rate of reaction compared to what was obtained experimentally.

Secondly, the transition state theory breaks down for systems with light atoms such as hydrogens. In such systems, quantum effects become crucial. Quantum effects tend to increase the rate of the reaction for two reasons. First, particles lie in energy levels which are higher then the bottom of the potential energy well (there is additional zero-point energy). Secondly, the particles can tunnel through the barrier. These two effects effectively lower the activation energy barrier. As a result, the rate of reaction predicted by transition state theory would be less than what is obtained experimentally.

Lastly, transition state theory does not hold true for systems which are far from equilibrium. In such cases, the assumption of a starting Boltzmann distribution breaks down.

== Exercise 2: F - H - H system ==

=== Potential energy surface inspection===

F + HA-HB → HA-F + HB

[[File:01409469 Surface_Plot F 3.jpg|thumb|center|400px|'''Figure 10.''' Potential energy surface for the reaction between F and H2]]

The reaction between F and H2 is an exothermic reaction. This means that the reactants are of a higher energy than the products. The overall reaction involves the breaking of a hydrogen bond and the formation of a hydrogen-fluorine bond. The HF bond formed is stronger than the hydrogen bond that was broken. This is because fluorine is more electronegative than hydrogen resulting in a polar HF bond. The negative enthalpy of H-F bond formation is greater than the positive enthalpy of hydrogen bond dissociation. Therefore, the reaction between F and H2 is an exothermic reaction.

HA + HB-F → HA-HB + F

[[File:01409469 Surface_Plot F 4.jpg|thumb|center|400px|'''Figure 11.''' Potential energy surface for the reaction between H and HF]]

The reaction between H and HF is an endothermic reaction. The products are at a higher energy level than the reactants. The overall reaction involves the breaking of a HF bond and the formation of a hydrogen bond. The formation of a homonuclear H2 molecule results in a non-polar hydrogen bond being formed. The HF bond broken is stronger than the hydrogen bond that was formed. The negative enthalpy of hydrogen bond formation is less than the positive enthalpy of hydrogen-fluorine dissociation. Therefore, the reaction between H and HF is endothermic.

=== Locating the transition state ===

[[File:01409469 Surface_Plot F TS4.jpg|thumb|center|800px|'''Figure 12.''' Determining the position of the transition state for the reaction between F and H2]]

As mentioned previously, when the molecules are not at the transition state, a reaction path will be plotted. However, if the initial conditions of the system are set such that the trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

An initial estimate for the transition state was made by setting rAB = 0.74 which is the length of a hydrogen bond. The position of the transition state was further refined by increasing the number of decimal places such that the final value of rAB was found to be 0.74478 and rBC = 1.81079. This set of distances gave a reaction path that was represented by a point indicating that no trajectory was plotted and that the molecules remained at the transition state since their initial momentum was 0.

Moreover, the kinetic energy as well as the forces acting on the atoms was calculated to be 0, which is consistent with the transition state.

Therefore, the position of the transition state was found to be at rAB = 0.74478 and rBC = 1.81079 when pA = pBC = 0.

=== Determining the activation energy ===

[[File:01409469 activation energy r1.1 combined.jpg|thumb|center|800px|'''Figure 13.''' Determining the activation energy for the reaction between F and H2]]

By reducing the value of rBC by 0.01, the atoms are no longer at the transition state and would move towards the formation of products. The activation energy was found by calculating the change in total energy after the reaction. The activation energy of F + H2 → HF + H was found to be +23.279 kcal/ mol.

[[File:01409469 activation energy r2.1 combined.jpg|thumb|center|800px|'''Figure 14.''' Determining the activation energy for the reaction between FH and H]]

The activation energy of FH + H → F + H2 was found by calculating the small dip in potential energy. The activation energy was found to be +0.280 kcal/ mol.

=== Reaction dynamics of F + HA-HB → HA-F + HB ===

The set of initial conditions that result in a reactive trajectory for F + H2 is as follows: rHH = 0.74 Å, rHF = 1.79 Å and pHH = pHF = 0.

[[File:01409469 momentum r1.jpg|thumb|center|400px|'''Figure 15.''' Graph of momentum against time for the reaction between F and H2]]

Initially, the H2 molecule is travelling in a relatively straight line with very little oscillations as evidenced by the low amplitude of momentum between 0-2 s. Following successful collision between F and H2, the amplitude of momentum of the product, HF, increases dramatically. This is because the reaction between F and H2 is exothermic and the excess energy released causes the molecule to become vibrationally excited and the amplitude of oscillations in the product to increase.

[[File:01409469 energy1 r1.jpg|thumb|center|400px|'''Figure 16.''' Graph of energy against time for the reaction between F and H2]]

The law of conservation of energy states that energy cannot be created or destroyed. It simply changes from one form to another. As can be seen from the graph in figure 16, total energy remains constant. In this reaction, potential energy is converted to kinetic energy. A decrease in potential energy of the atoms following a successful collision between F and H2 results in an equal increase in kinetic energy which manifests in the form of increased product (HF) oscillations.

This can be confirmed experimentally using infrared chemiluminescence. Observation of radiation from excited product molecules is a clear indication that these molecules are produced in the excited state. If the excitation is primarily vibrational, the radiation will appear in the infrared region of the spectrum (between 3-15 μm). This technique has been used extensively in hydrogen halogen reactions which produce HX molecules in the excited state.

One such experiment involves the use of the arrested relaxation method. In this technique, the reaction is carried out in a chamber whose walls are maintained at a very low temperature (20 K). This results in cryopumping of the chamber in that all the products and unreacted starting materials are condensed on the walls and only those product molecules that radiate before being pumped away are observed.

=== Polanyi's empirical rules ===

In a typical chemical reaction, there is an energetic barrier — a saddle point that the reactants must surmount to become products. Which form of energy is initially deposited in the reactants, translational or vibrational, is more efficacious in surmounting the activation energy barrier is one of the core topics in the area of reaction dynamics. The Polyani rules state that vibrational energy is more effective at promoting a late-barrier reaction (transition state resembles products). Conversely, translational energy is better at promoting an early-barrier reaction (transition state resembles the reactants).

The forward reaction between H2 and F is exothermic. Therefore, the transition state is closer to the reactants than to the products in energy. Since the transition state occurs early in an exothermic reaction, translational energy is more effective at promoting a chemical reaction than vibrational energy.

Conversely, the backward reaction between H and HF is endothermic. According to Hammond's postulate, the transition state resembles the products more than it resembles the reactants. Therefore, in an endothermic reaction, a late transition state is observed and vibrational energy is more effective at promoting a reaction than translational energy.

=== Promoting the forward reaction: F + HA-HB → HA-F + HB ===

1. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 2.95 and pHF = -0.5.

[[File:01409469 polyani 1+2.jpg|thumb|center|800px|'''Figure 17.''' Contour plot of the reaction between F and H2 when pHH = 2.95 '''Figure 18.''' Graph of momentum against time for the reaction between F and H2 when pHH = 2.95 (Left-right)]]

Vibrational energy is represented by pHH while translational energy is represented by pHF.

File:01409469 polyani 1+2.jpg

2019-05-09T21:26:37Z

Stc1917:

MRD:01409469

2019-05-09T21:26:24Z

Stc1917: /* Promoting the forward reaction: F + HA-HB → HA-F + HB */

== Exercise 1: H + H2 system ==

=== Definition of a transition state ===

A potential energy profile results from plotting the potential versus reaction path.

[[File:Surface_Plot01409469_pic1.jpg|thumb|center|400px|'''Figure 1.''' A potential energy surface plot]]

The transition state is defined as the maximum on the minimum energy path linking reactants and products. This means that the transition state is a stationary point on the potential energy surface. Thus, at the transition state, the derivative of the potential energy is zero with respect to each Cartesian coordinate.

<math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>

However, a local minimum on the potential energy surface corresponding to a well would have the same <math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>. Therefore, to distinguish between a local minimum on the potential energy surface and the transition state, the second derivative, <math>\tfrac{\partial^2 V(r_i)}{\partial r_i ^2}\ </math> is determined. This can be calculated by finding the determinant of the Hessian matrix shown below.

:<math>\mathbf H =
\begin{bmatrix}
\dfrac{\partial^2 \ V(r_i)}{\partial \ x^2} & \dfrac{\partial^2 \ V(r_i)}{\partial \ x \partial \mathbf y} \\
\dfrac{\partial^2 \ V(r_i)}{\partial \ y \partial \mathbf x} & \dfrac{\partial^2 \ V(r_i)}{\partial \mathbf y^2}
\end{bmatrix}</math>

The Hessian matrix of a convex function is positive semi-definite. Refining this property allows one to test if a critical point (x,y) is a local maximum, local minimum or saddle point.

If <math>f</math> < 0, then (x,y) is an isolated local maximum.

If <math>f</math> > 0, then (x,y) is an isolated local minimum.

If the Hessian has both positive and negative eigenvalues then (x,y) is a saddle point.

=== Locating the transition state ===

[[File:01409469 combined pic 2 and 3.jpg|thumb|center|800px|'''Figure 2.''' A surface and contour plot of the trajectory when rAB=rBC=0.907743 Å and pA=pBC=0]]

When the molecules are not at the transition state, a surface plot of the conditions shows a trajectory being plotted. This means that the trajectory is "rolling" towards the products or reactants. However, if a trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

[[File:01409469_internuclearpic3.png|thumb|center|400px|'''Figure 3.''' A plot of internuclear distance against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

An initial estimate for the position of the transition state was made by observing the potential energy surface and was found to be 0.90 Å. At 0.90 Å, the internuclear distance against time plot showed that the internuclear distance was constantly oscillating. Therefore, the position of the transition state was further refined by increasing the number of decimal places until a horizontal line in the internuclear distance againt time plot was obtained. This means that the internuclear distance is constant and that the molecules are simply undergoing periodic symmetric vibration at the transition state. Therefore, the position of the transition state was determined to be 0.907743 Å.

[[File:01409469 combined energy pic.png|thumb|center|400px|'''Figure 4.''' A plot of energy against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

When rAB=rBC=0.907743 Å and pA=pBC=0, the kinetic energy is 0. This means that HA and HBC are stationary, which is as predicted when the trajectory is started at the transition state with no initial momentum. Furthermore, the forces acting on HA and HBC are 0 as should be the case in the transition state.

=== Calculation type: Dynamic and minimum energy path ===

The minimum energy path, is the lowest energy path from the reactants to the products. The minimum energy path is determined by following the mode corresponding to the negative Hessian eigenvalue in both directions along the steepest descent direction. Following this path allows one to approach two valleys of stable structures. The resulting path generated is the minimum energy path. The reaction paths were chartered for a dynamic and a minimum energy path. The initial conditions used were that of the transition state except that the internuclear distance between HA and HB had increased to 0.917743 Å. The reaction path obtained using the two different calculation methods was different.

[[File:01409469 mep and dynamics.jpg|thumb|center|800px|'''Figure 5.''' A surface plot of the trajectory using minimum energy path '''Figure 6.''' A surface plot of the trajectory using dynamics (Left-right)]]

On the surface plot, the reaction path is depicted by the black line. Following the minimum energy path, the trajectory simply follows the valley floor to Ha+Hb-Hc. The reaction path appears to be a relatively straight line indicating that the atoms are not vibrating. In the minimum energy path, the reactants change into the products in an infinitely slow motion. This is done by having the momenta/velocities reset to zero in each time step. While the minimum energy path is useful in characterising a chemical reaction, it does not provide a realistic representation of the motion of the atoms during the reaction. This is because the mass and inertia of the atoms have been ignored.

Since the atoms simply follow the minimum energy path, they do not undergo slight displacements into surrounding valleys of higher potential energy. Furthermore, because the atoms are assumed to have no inertia, they do not undergo vibrations even if they were to be displaced to valleys of higher potential energy.

However, when dynamic calculation is performed, the reaction path is wavy indicating that the atoms are oscillating. This is because the atoms are displaced to regions of higher potential on the path from reactants to products. This leads to atom oscillation due to the inertia of the atoms.

'''Internuclear distance against time plot'''

Setting the initial system conditions such that rAB= 0.907743, rBC= 0.917743 and pA= pBC= 0:

[[File:01409469 Internuclear distance mep and dynamic.jpg|thumb|center|800px|'''Figure 6.''' A graph of internuclear distance against steps using minimum energy path '''Figure 7.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In the minimum energy path, the distance between HA and HB decreases from 0.90 Å to approximately 0.79 Å during the first 50 steps before plateauing off to a value of 0.74 Å which is the approximate length of a hydrogen bond. This means that a bond was formed between HA and HB. In addition, the plateau indicates that the bond is static. On the other hand, the distance between HB and HC has increased to 1.91 Å at large step values. This shows that HC has been ejected off following a successful collision between HA and HBC.

Similarly, in the dynamic trajectory, the distance between HA and HB decreases to approximately 0.75 Å after 0.3 s. This is indicative of hydrogen bond formation between HA and HB. In addition, the distance between HB and HC increases to 3.44 Å after HC is ejected following a successful collision.

One obvious distance between the two plots is that in the minimum energy path trajectory, the internuclear distance between HA and HB is constant indicating that the bond is static. However, in the dynamic trajectory the internuclear distance is oscillating showing the the bond is vibrating. This difference could be attributed to the fact that minimum energy path does not take oscillatory motion into account while dynamic trajectory does.

However, changing the initial conditions of the system such that rAB = 0.917743 Å, rBC = 0.907743 Å and pA=pBC=0:
[[File:01409469 01409469 Reversed values dynamic mep.jpg|thumb|center|800px|'''Figure 8.''' A graph of internuclear distance against steps using minimum energy path '''Figure 9.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In this case, a hydrogen bond of length 0.77-0.79 Å is formed between HB and HC while HA is repelled away to beyond 3.45 Å.

Additionally, from the internuclear distance plot it can be seen that the distance between HB and HC increases in a rectangular hyperbola fashion when calculated using the minimum energy path. On the other hand, the internuclear distance between HB and HC increases in a linear manner when dynamic calculation is applied. A possible reason for this difference is that in the minimum energy path, the atoms are not vibrating and as such the reach the maximum distance and plateau at a faster rate. However, in the dynamic calculation, the atoms have momentum and as such move further away from each other.

=== Reactive and unreactive trajectories ===

With the initial positions r'''1''' = 0.74 and r'''2''' = 2.0, trajectories were run with the momenta combination shown below.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Contour plot !! Reactivity !! Description of the dynamics
|-
| -1.25 || -2.5 ||-99.018 ||[[File:01409469 trajectory1.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. There are no oscillations observed in HA-HB initially, indicating that the diatomic molecule is stable. A new bond is formed between HB and HC. Oscillatory behaviour is observed indicating that the bond is vibrating. On the other hand, HA is ejected away to beyond 10 Å.
|-
| -1.5 || -2.0 ||-100.456 ||[[File:01409469 trajectory2.jpg|thumb|center|400px]] ||Unreactive ||HC collides with the vibrating diatomic HA-HB. However, no bond is formed between HB and HC as the collision is unsuccessful. This is because the molecules lack sufficient kinetic energy to overcome the activation energy barrier and so no reaction occurs. HC is subsequently repelled to beyond 8 Å.
|-
| -1.5 || -2.5 ||-98.956 ||[[File:01409469 trajectory3.jpg|thumb|center|400px]] ||Reactive |||HC collides with HA-HB. The collision is successful and a bond is formed between HB and HC. This is because the atoms have sufficient kinetic energy to overcome the activation energy barrier and form a new HB-HC bond. HA is ejected away to beyond 10 Å. The new hydrogen bond formed is vibrating as seen by the oscillatory behaviour.
|-
| -2.5 || -5.0 ||-84.956 ||[[File:01409469 trajectory4.jpg|thumb|center|400px]] ||Unreactive ||HC collides with HA-HB. Initially the HA-HB bond is elongating, indicating that the hydrogen bond is weakening and going to break. However, after 0.5 s, the HA-HB bond length decreases and the HAHB is not broken. HC is repelled beyond 6 Å with very strong vibrations.
A reaction did not occur even though the atoms have sufficient kinetic energy to cross the activation energy barrier. This is because the vibrations are too strong. The system crosses the transition state region and a bond is formed in the product temporarily but then the system reverts back to the reactants. This is an example of barrier recrossing.
|-
| -2.5 || -5.2 ||-83.416 ||[[File:01409469 trajectory5.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. The collision is successful and a new hydrogen bond is formed between HB and HC with HA being ejected away. The products formed experience very strong vibrations as evidenced by the oscillatory motion which occur with a very high frequency.
|}

From the table, it can be seen that the chance of a reaction successfully occurring is not solely dependent on the amount of kinetic energy the atoms possess. Even if the atoms have sufficient kinetic energy to cross the activation energy barrier, barrier recrossing can occur if the atoms are too vibrationally excited.

=== Transition State Theory ===

Transition state theory provided the first theoretical attempt to determine absolute reaction rates. In this theory, a transition state separating the reactants from products is used to find an expression for the thermal rate constant.

The main assumptions of transition state theory are:

1. Electronic and nuclear motions are separated (equivalent to the Born-Oppenheimer approximation in quantum mechanics)

2. Reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution

3. Molecular systems that have crossed the transition state in the direction of products cannot turn around and reform reactants

4. In the transition state, motion along the reaction coordinate may be separated from the other motions and treated classically as a translation

5. Even in the absence of an equilibrium between reactant and product molecules, the transition states that are becoming products are distributed among their states according to the Maxwell-Boltzmann laws

The goal of transition state theory is to predict the rate constant, ''k'', of a reaction given a potential energy surface for the reaction.

However, transition state theory suffers from a set of limitations due to the assumptions stated above not holding true in reality.

Firstly, transition state theory assumes that barrier recrossing is negligible. However, as seen from the fourth trajectory in table 1, barrier recrossing occurred because the atoms were too vibrationally excited and the system reverted back to the reactants even though the system had already crossed the transition state region. In this situation, transition state theory would predict a faster rate of reaction compared to what was obtained experimentally.

Secondly, the transition state theory breaks down for systems with light atoms such as hydrogens. In such systems, quantum effects become crucial. Quantum effects tend to increase the rate of the reaction for two reasons. First, particles lie in energy levels which are higher then the bottom of the potential energy well (there is additional zero-point energy). Secondly, the particles can tunnel through the barrier. These two effects effectively lower the activation energy barrier. As a result, the rate of reaction predicted by transition state theory would be less than what is obtained experimentally.

Lastly, transition state theory does not hold true for systems which are far from equilibrium. In such cases, the assumption of a starting Boltzmann distribution breaks down.

== Exercise 2: F - H - H system ==

=== Potential energy surface inspection===

F + HA-HB → HA-F + HB

[[File:01409469 Surface_Plot F 3.jpg|thumb|center|400px|'''Figure 10.''' Potential energy surface for the reaction between F and H2]]

The reaction between F and H2 is an exothermic reaction. This means that the reactants are of a higher energy than the products. The overall reaction involves the breaking of a hydrogen bond and the formation of a hydrogen-fluorine bond. The HF bond formed is stronger than the hydrogen bond that was broken. This is because fluorine is more electronegative than hydrogen resulting in a polar HF bond. The negative enthalpy of H-F bond formation is greater than the positive enthalpy of hydrogen bond dissociation. Therefore, the reaction between F and H2 is an exothermic reaction.

HA + HB-F → HA-HB + F

[[File:01409469 Surface_Plot F 4.jpg|thumb|center|400px|'''Figure 11.''' Potential energy surface for the reaction between H and HF]]

The reaction between H and HF is an endothermic reaction. The products are at a higher energy level than the reactants. The overall reaction involves the breaking of a HF bond and the formation of a hydrogen bond. The formation of a homonuclear H2 molecule results in a non-polar hydrogen bond being formed. The HF bond broken is stronger than the hydrogen bond that was formed. The negative enthalpy of hydrogen bond formation is less than the positive enthalpy of hydrogen-fluorine dissociation. Therefore, the reaction between H and HF is endothermic.

=== Locating the transition state ===

[[File:01409469 Surface_Plot F TS4.jpg|thumb|center|800px|'''Figure 12.''' Determining the position of the transition state for the reaction between F and H2]]

As mentioned previously, when the molecules are not at the transition state, a reaction path will be plotted. However, if the initial conditions of the system are set such that the trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

An initial estimate for the transition state was made by setting rAB = 0.74 which is the length of a hydrogen bond. The position of the transition state was further refined by increasing the number of decimal places such that the final value of rAB was found to be 0.74478 and rBC = 1.81079. This set of distances gave a reaction path that was represented by a point indicating that no trajectory was plotted and that the molecules remained at the transition state since their initial momentum was 0.

Moreover, the kinetic energy as well as the forces acting on the atoms was calculated to be 0, which is consistent with the transition state.

Therefore, the position of the transition state was found to be at rAB = 0.74478 and rBC = 1.81079 when pA = pBC = 0.

=== Determining the activation energy ===

[[File:01409469 activation energy r1.1 combined.jpg|thumb|center|800px|'''Figure 13.''' Determining the activation energy for the reaction between F and H2]]

By reducing the value of rBC by 0.01, the atoms are no longer at the transition state and would move towards the formation of products. The activation energy was found by calculating the change in total energy after the reaction. The activation energy of F + H2 → HF + H was found to be +23.279 kcal/ mol.

[[File:01409469 activation energy r2.1 combined.jpg|thumb|center|800px|'''Figure 14.''' Determining the activation energy for the reaction between FH and H]]

The activation energy of FH + H → F + H2 was found by calculating the small dip in potential energy. The activation energy was found to be +0.280 kcal/ mol.

=== Reaction dynamics of F + HA-HB → HA-F + HB ===

The set of initial conditions that result in a reactive trajectory for F + H2 is as follows: rHH = 0.74 Å, rHF = 1.79 Å and pHH = pHF = 0.

[[File:01409469 momentum r1.jpg|thumb|center|400px|'''Figure 15.''' Graph of momentum against time for the reaction between F and H2]]

Initially, the H2 molecule is travelling in a relatively straight line with very little oscillations as evidenced by the low amplitude of momentum between 0-2 s. Following successful collision between F and H2, the amplitude of momentum of the product, HF, increases dramatically. This is because the reaction between F and H2 is exothermic and the excess energy released causes the molecule to become vibrationally excited and the amplitude of oscillations in the product to increase.

[[File:01409469 energy1 r1.jpg|thumb|center|400px|'''Figure 16.''' Graph of energy against time for the reaction between F and H2]]

The law of conservation of energy states that energy cannot be created or destroyed. It simply changes from one form to another. As can be seen from the graph in figure 16, total energy remains constant. In this reaction, potential energy is converted to kinetic energy. A decrease in potential energy of the atoms following a successful collision between F and H2 results in an equal increase in kinetic energy which manifests in the form of increased product (HF) oscillations.

This can be confirmed experimentally using infrared chemiluminescence. Observation of radiation from excited product molecules is a clear indication that these molecules are produced in the excited state. If the excitation is primarily vibrational, the radiation will appear in the infrared region of the spectrum (between 3-15 μm). This technique has been used extensively in hydrogen halogen reactions which produce HX molecules in the excited state.

One such experiment involves the use of the arrested relaxation method. In this technique, the reaction is carried out in a chamber whose walls are maintained at a very low temperature (20 K). This results in cryopumping of the chamber in that all the products and unreacted starting materials are condensed on the walls and only those product molecules that radiate before being pumped away are observed.

=== Polanyi's empirical rules ===

In a typical chemical reaction, there is an energetic barrier — a saddle point that the reactants must surmount to become products. Which form of energy is initially deposited in the reactants, translational or vibrational, is more efficacious in surmounting the activation energy barrier is one of the core topics in the area of reaction dynamics. The Polyani rules state that vibrational energy is more effective at promoting a late-barrier reaction (transition state resembles products). Conversely, translational energy is better at promoting an early-barrier reaction (transition state resembles the reactants).

The forward reaction between H2 and F is exothermic. Therefore, the transition state is closer to the reactants than to the products in energy. Since the transition state occurs early in an exothermic reaction, translational energy is more effective at promoting a chemical reaction than vibrational energy.

Conversely, the backward reaction between H and HF is endothermic. According to Hammond's postulate, the transition state resembles the products more than it resembles the reactants. Therefore, in an endothermic reaction, a late transition state is observed and vibrational energy is more effective at promoting a reaction than translational energy.

=== Promoting the forward reaction: F + HA-HB → HA-F + HB ===

1. The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 2.95 and pHF = -0.5.

[[File:01409469 polyani 1+2.jpg|thumb|center|800px|'''Figure 17.''' Contour plot of the reaction between F and H2 when pHH = 2.95 '''Figure 18.''' Graph of momentum against time for the reaction between F and H2 when pHH = 2.95 (Left-right)]]

MRD:01409469

2019-05-09T21:25:20Z

Stc1917: /* Promoting the forward reaction: F + HA-HB → HA-F + HB */

== Exercise 1: H + H2 system ==

=== Definition of a transition state ===

A potential energy profile results from plotting the potential versus reaction path.

[[File:Surface_Plot01409469_pic1.jpg|thumb|center|400px|'''Figure 1.''' A potential energy surface plot]]

The transition state is defined as the maximum on the minimum energy path linking reactants and products. This means that the transition state is a stationary point on the potential energy surface. Thus, at the transition state, the derivative of the potential energy is zero with respect to each Cartesian coordinate.

<math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>

However, a local minimum on the potential energy surface corresponding to a well would have the same <math>\tfrac{\partial V(r_i)}{\partial r_i}\ = 0 </math>. Therefore, to distinguish between a local minimum on the potential energy surface and the transition state, the second derivative, <math>\tfrac{\partial^2 V(r_i)}{\partial r_i ^2}\ </math> is determined. This can be calculated by finding the determinant of the Hessian matrix shown below.

:<math>\mathbf H =
\begin{bmatrix}
\dfrac{\partial^2 \ V(r_i)}{\partial \ x^2} & \dfrac{\partial^2 \ V(r_i)}{\partial \ x \partial \mathbf y} \\
\dfrac{\partial^2 \ V(r_i)}{\partial \ y \partial \mathbf x} & \dfrac{\partial^2 \ V(r_i)}{\partial \mathbf y^2}
\end{bmatrix}</math>

The Hessian matrix of a convex function is positive semi-definite. Refining this property allows one to test if a critical point (x,y) is a local maximum, local minimum or saddle point.

If <math>f</math> < 0, then (x,y) is an isolated local maximum.

If <math>f</math> > 0, then (x,y) is an isolated local minimum.

If the Hessian has both positive and negative eigenvalues then (x,y) is a saddle point.

=== Locating the transition state ===

[[File:01409469 combined pic 2 and 3.jpg|thumb|center|800px|'''Figure 2.''' A surface and contour plot of the trajectory when rAB=rBC=0.907743 Å and pA=pBC=0]]

When the molecules are not at the transition state, a surface plot of the conditions shows a trajectory being plotted. This means that the trajectory is "rolling" towards the products or reactants. However, if a trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

[[File:01409469_internuclearpic3.png|thumb|center|400px|'''Figure 3.''' A plot of internuclear distance against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

An initial estimate for the position of the transition state was made by observing the potential energy surface and was found to be 0.90 Å. At 0.90 Å, the internuclear distance against time plot showed that the internuclear distance was constantly oscillating. Therefore, the position of the transition state was further refined by increasing the number of decimal places until a horizontal line in the internuclear distance againt time plot was obtained. This means that the internuclear distance is constant and that the molecules are simply undergoing periodic symmetric vibration at the transition state. Therefore, the position of the transition state was determined to be 0.907743 Å.

[[File:01409469 combined energy pic.png|thumb|center|400px|'''Figure 4.''' A plot of energy against time when rAB=rBC=0.907743 Å and pA=pBC=0]]

When rAB=rBC=0.907743 Å and pA=pBC=0, the kinetic energy is 0. This means that HA and HBC are stationary, which is as predicted when the trajectory is started at the transition state with no initial momentum. Furthermore, the forces acting on HA and HBC are 0 as should be the case in the transition state.

=== Calculation type: Dynamic and minimum energy path ===

The minimum energy path, is the lowest energy path from the reactants to the products. The minimum energy path is determined by following the mode corresponding to the negative Hessian eigenvalue in both directions along the steepest descent direction. Following this path allows one to approach two valleys of stable structures. The resulting path generated is the minimum energy path. The reaction paths were chartered for a dynamic and a minimum energy path. The initial conditions used were that of the transition state except that the internuclear distance between HA and HB had increased to 0.917743 Å. The reaction path obtained using the two different calculation methods was different.

[[File:01409469 mep and dynamics.jpg|thumb|center|800px|'''Figure 5.''' A surface plot of the trajectory using minimum energy path '''Figure 6.''' A surface plot of the trajectory using dynamics (Left-right)]]

On the surface plot, the reaction path is depicted by the black line. Following the minimum energy path, the trajectory simply follows the valley floor to Ha+Hb-Hc. The reaction path appears to be a relatively straight line indicating that the atoms are not vibrating. In the minimum energy path, the reactants change into the products in an infinitely slow motion. This is done by having the momenta/velocities reset to zero in each time step. While the minimum energy path is useful in characterising a chemical reaction, it does not provide a realistic representation of the motion of the atoms during the reaction. This is because the mass and inertia of the atoms have been ignored.

Since the atoms simply follow the minimum energy path, they do not undergo slight displacements into surrounding valleys of higher potential energy. Furthermore, because the atoms are assumed to have no inertia, they do not undergo vibrations even if they were to be displaced to valleys of higher potential energy.

However, when dynamic calculation is performed, the reaction path is wavy indicating that the atoms are oscillating. This is because the atoms are displaced to regions of higher potential on the path from reactants to products. This leads to atom oscillation due to the inertia of the atoms.

'''Internuclear distance against time plot'''

Setting the initial system conditions such that rAB= 0.907743, rBC= 0.917743 and pA= pBC= 0:

[[File:01409469 Internuclear distance mep and dynamic.jpg|thumb|center|800px|'''Figure 6.''' A graph of internuclear distance against steps using minimum energy path '''Figure 7.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In the minimum energy path, the distance between HA and HB decreases from 0.90 Å to approximately 0.79 Å during the first 50 steps before plateauing off to a value of 0.74 Å which is the approximate length of a hydrogen bond. This means that a bond was formed between HA and HB. In addition, the plateau indicates that the bond is static. On the other hand, the distance between HB and HC has increased to 1.91 Å at large step values. This shows that HC has been ejected off following a successful collision between HA and HBC.

Similarly, in the dynamic trajectory, the distance between HA and HB decreases to approximately 0.75 Å after 0.3 s. This is indicative of hydrogen bond formation between HA and HB. In addition, the distance between HB and HC increases to 3.44 Å after HC is ejected following a successful collision.

One obvious distance between the two plots is that in the minimum energy path trajectory, the internuclear distance between HA and HB is constant indicating that the bond is static. However, in the dynamic trajectory the internuclear distance is oscillating showing the the bond is vibrating. This difference could be attributed to the fact that minimum energy path does not take oscillatory motion into account while dynamic trajectory does.

However, changing the initial conditions of the system such that rAB = 0.917743 Å, rBC = 0.907743 Å and pA=pBC=0:
[[File:01409469 01409469 Reversed values dynamic mep.jpg|thumb|center|800px|'''Figure 8.''' A graph of internuclear distance against steps using minimum energy path '''Figure 9.''' A graph of internuclear distance against time using dynamic trajectory (Left-right)]]

In this case, a hydrogen bond of length 0.77-0.79 Å is formed between HB and HC while HA is repelled away to beyond 3.45 Å.

Additionally, from the internuclear distance plot it can be seen that the distance between HB and HC increases in a rectangular hyperbola fashion when calculated using the minimum energy path. On the other hand, the internuclear distance between HB and HC increases in a linear manner when dynamic calculation is applied. A possible reason for this difference is that in the minimum energy path, the atoms are not vibrating and as such the reach the maximum distance and plateau at a faster rate. However, in the dynamic calculation, the atoms have momentum and as such move further away from each other.

=== Reactive and unreactive trajectories ===

With the initial positions r'''1''' = 0.74 and r'''2''' = 2.0, trajectories were run with the momenta combination shown below.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Contour plot !! Reactivity !! Description of the dynamics
|-
| -1.25 || -2.5 ||-99.018 ||[[File:01409469 trajectory1.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. There are no oscillations observed in HA-HB initially, indicating that the diatomic molecule is stable. A new bond is formed between HB and HC. Oscillatory behaviour is observed indicating that the bond is vibrating. On the other hand, HA is ejected away to beyond 10 Å.
|-
| -1.5 || -2.0 ||-100.456 ||[[File:01409469 trajectory2.jpg|thumb|center|400px]] ||Unreactive ||HC collides with the vibrating diatomic HA-HB. However, no bond is formed between HB and HC as the collision is unsuccessful. This is because the molecules lack sufficient kinetic energy to overcome the activation energy barrier and so no reaction occurs. HC is subsequently repelled to beyond 8 Å.
|-
| -1.5 || -2.5 ||-98.956 ||[[File:01409469 trajectory3.jpg|thumb|center|400px]] ||Reactive |||HC collides with HA-HB. The collision is successful and a bond is formed between HB and HC. This is because the atoms have sufficient kinetic energy to overcome the activation energy barrier and form a new HB-HC bond. HA is ejected away to beyond 10 Å. The new hydrogen bond formed is vibrating as seen by the oscillatory behaviour.
|-
| -2.5 || -5.0 ||-84.956 ||[[File:01409469 trajectory4.jpg|thumb|center|400px]] ||Unreactive ||HC collides with HA-HB. Initially the HA-HB bond is elongating, indicating that the hydrogen bond is weakening and going to break. However, after 0.5 s, the HA-HB bond length decreases and the HAHB is not broken. HC is repelled beyond 6 Å with very strong vibrations.
A reaction did not occur even though the atoms have sufficient kinetic energy to cross the activation energy barrier. This is because the vibrations are too strong. The system crosses the transition state region and a bond is formed in the product temporarily but then the system reverts back to the reactants. This is an example of barrier recrossing.
|-
| -2.5 || -5.2 ||-83.416 ||[[File:01409469 trajectory5.jpg|thumb|center|400px]] ||Reactive ||HC collides with HA-HB. The collision is successful and a new hydrogen bond is formed between HB and HC with HA being ejected away. The products formed experience very strong vibrations as evidenced by the oscillatory motion which occur with a very high frequency.
|}

From the table, it can be seen that the chance of a reaction successfully occurring is not solely dependent on the amount of kinetic energy the atoms possess. Even if the atoms have sufficient kinetic energy to cross the activation energy barrier, barrier recrossing can occur if the atoms are too vibrationally excited.

=== Transition State Theory ===

Transition state theory provided the first theoretical attempt to determine absolute reaction rates. In this theory, a transition state separating the reactants from products is used to find an expression for the thermal rate constant.

The main assumptions of transition state theory are:

1. Electronic and nuclear motions are separated (equivalent to the Born-Oppenheimer approximation in quantum mechanics)

2. Reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution

3. Molecular systems that have crossed the transition state in the direction of products cannot turn around and reform reactants

4. In the transition state, motion along the reaction coordinate may be separated from the other motions and treated classically as a translation

5. Even in the absence of an equilibrium between reactant and product molecules, the transition states that are becoming products are distributed among their states according to the Maxwell-Boltzmann laws

The goal of transition state theory is to predict the rate constant, ''k'', of a reaction given a potential energy surface for the reaction.

However, transition state theory suffers from a set of limitations due to the assumptions stated above not holding true in reality.

Firstly, transition state theory assumes that barrier recrossing is negligible. However, as seen from the fourth trajectory in table 1, barrier recrossing occurred because the atoms were too vibrationally excited and the system reverted back to the reactants even though the system had already crossed the transition state region. In this situation, transition state theory would predict a faster rate of reaction compared to what was obtained experimentally.

Secondly, the transition state theory breaks down for systems with light atoms such as hydrogens. In such systems, quantum effects become crucial. Quantum effects tend to increase the rate of the reaction for two reasons. First, particles lie in energy levels which are higher then the bottom of the potential energy well (there is additional zero-point energy). Secondly, the particles can tunnel through the barrier. These two effects effectively lower the activation energy barrier. As a result, the rate of reaction predicted by transition state theory would be less than what is obtained experimentally.

Lastly, transition state theory does not hold true for systems which are far from equilibrium. In such cases, the assumption of a starting Boltzmann distribution breaks down.

== Exercise 2: F - H - H system ==

=== Potential energy surface inspection===

F + HA-HB → HA-F + HB

[[File:01409469 Surface_Plot F 3.jpg|thumb|center|400px|'''Figure 10.''' Potential energy surface for the reaction between F and H2]]

The reaction between F and H2 is an exothermic reaction. This means that the reactants are of a higher energy than the products. The overall reaction involves the breaking of a hydrogen bond and the formation of a hydrogen-fluorine bond. The HF bond formed is stronger than the hydrogen bond that was broken. This is because fluorine is more electronegative than hydrogen resulting in a polar HF bond. The negative enthalpy of H-F bond formation is greater than the positive enthalpy of hydrogen bond dissociation. Therefore, the reaction between F and H2 is an exothermic reaction.

HA + HB-F → HA-HB + F

[[File:01409469 Surface_Plot F 4.jpg|thumb|center|400px|'''Figure 11.''' Potential energy surface for the reaction between H and HF]]

The reaction between H and HF is an endothermic reaction. The products are at a higher energy level than the reactants. The overall reaction involves the breaking of a HF bond and the formation of a hydrogen bond. The formation of a homonuclear H2 molecule results in a non-polar hydrogen bond being formed. The HF bond broken is stronger than the hydrogen bond that was formed. The negative enthalpy of hydrogen bond formation is less than the positive enthalpy of hydrogen-fluorine dissociation. Therefore, the reaction between H and HF is endothermic.

=== Locating the transition state ===

[[File:01409469 Surface_Plot F TS4.jpg|thumb|center|800px|'''Figure 12.''' Determining the position of the transition state for the reaction between F and H2]]

As mentioned previously, when the molecules are not at the transition state, a reaction path will be plotted. However, if the initial conditions of the system are set such that the trajectory is started exactly at the transition state with no initial momentum, the molecules will remain at the transition state forever.

An initial estimate for the transition state was made by setting rAB = 0.74 which is the length of a hydrogen bond. The position of the transition state was further refined by increasing the number of decimal places such that the final value of rAB was found to be 0.74478 and rBC = 1.81079. This set of distances gave a reaction path that was represented by a point indicating that no trajectory was plotted and that the molecules remained at the transition state since their initial momentum was 0.

Moreover, the kinetic energy as well as the forces acting on the atoms was calculated to be 0, which is consistent with the transition state.

Therefore, the position of the transition state was found to be at rAB = 0.74478 and rBC = 1.81079 when pA = pBC = 0.

=== Determining the activation energy ===

[[File:01409469 activation energy r1.1 combined.jpg|thumb|center|800px|'''Figure 13.''' Determining the activation energy for the reaction between F and H2]]

By reducing the value of rBC by 0.01, the atoms are no longer at the transition state and would move towards the formation of products. The activation energy was found by calculating the change in total energy after the reaction. The activation energy of F + H2 → HF + H was found to be +23.279 kcal/ mol.

[[File:01409469 activation energy r2.1 combined.jpg|thumb|center|800px|'''Figure 14.''' Determining the activation energy for the reaction between FH and H]]

The activation energy of FH + H → F + H2 was found by calculating the small dip in potential energy. The activation energy was found to be +0.280 kcal/ mol.

=== Reaction dynamics of F + HA-HB → HA-F + HB ===

The set of initial conditions that result in a reactive trajectory for F + H2 is as follows: rHH = 0.74 Å, rHF = 1.79 Å and pHH = pHF = 0.

[[File:01409469 momentum r1.jpg|thumb|center|400px|'''Figure 15.''' Graph of momentum against time for the reaction between F and H2]]

Initially, the H2 molecule is travelling in a relatively straight line with very little oscillations as evidenced by the low amplitude of momentum between 0-2 s. Following successful collision between F and H2, the amplitude of momentum of the product, HF, increases dramatically. This is because the reaction between F and H2 is exothermic and the excess energy released causes the molecule to become vibrationally excited and the amplitude of oscillations in the product to increase.

[[File:01409469 energy1 r1.jpg|thumb|center|400px|'''Figure 16.''' Graph of energy against time for the reaction between F and H2]]

The law of conservation of energy states that energy cannot be created or destroyed. It simply changes from one form to another. As can be seen from the graph in figure 16, total energy remains constant. In this reaction, potential energy is converted to kinetic energy. A decrease in potential energy of the atoms following a successful collision between F and H2 results in an equal increase in kinetic energy which manifests in the form of increased product (HF) oscillations.

This can be confirmed experimentally using infrared chemiluminescence. Observation of radiation from excited product molecules is a clear indication that these molecules are produced in the excited state. If the excitation is primarily vibrational, the radiation will appear in the infrared region of the spectrum (between 3-15 μm). This technique has been used extensively in hydrogen halogen reactions which produce HX molecules in the excited state.

One such experiment involves the use of the arrested relaxation method. In this technique, the reaction is carried out in a chamber whose walls are maintained at a very low temperature (20 K). This results in cryopumping of the chamber in that all the products and unreacted starting materials are condensed on the walls and only those product molecules that radiate before being pumped away are observed.

=== Polanyi's empirical rules ===

In a typical chemical reaction, there is an energetic barrier — a saddle point that the reactants must surmount to become products. Which form of energy is initially deposited in the reactants, translational or vibrational, is more efficacious in surmounting the activation energy barrier is one of the core topics in the area of reaction dynamics. The Polyani rules state that vibrational energy is more effective at promoting a late-barrier reaction (transition state resembles products). Conversely, translational energy is better at promoting an early-barrier reaction (transition state resembles the reactants).

The forward reaction between H2 and F is exothermic. Therefore, the transition state is closer to the reactants than to the products in energy. Since the transition state occurs early in an exothermic reaction, translational energy is more effective at promoting a chemical reaction than vibrational energy.

Conversely, the backward reaction between H and HF is endothermic. According to Hammond's postulate, the transition state resembles the products more than it resembles the reactants. Therefore, in an endothermic reaction, a late transition state is observed and vibrational energy is more effective at promoting a reaction than translational energy.

=== Promoting the forward reaction: F + HA-HB → HA-F + HB ===

The initial system conditions used are: rHH = 0.74 Å, rHF = 1.8 Å, pHH = 2.95 and pHF = -0.5.

[[File:.jpg|thumb|center|800px|'''Figure 17.''' Graph of momentum against time for the reaction between F and H2 when pHH = -3 '''Figure 18.''' Graph of momentum against time for the reaction between F and H2 when pHH = -2.75 (Left-right)]]