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MRD:ED2618

2020-05-15T18:38:01Z

Ed2618: /* Molecular Reaction Dynamics */

= '''Molecular Reaction Dynamics''' =
In this lab, molecular dynamic trajectories are used to study the reactivity of triatomic systems; namely, when an atom collides with a diatomic molecule.

== EXERCISE 1: H + H2 system ==
For this exercise, the following reaction is analysed and discussed: HA + HBHC → HAHB + HC, where r1 = BC and r2 = AB.

=== Dynamics from the transition state region ===
'''Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?'''[[File:Saddle plot ED2618.png|thumb|440x440px|Figure 1: Plot of a saddle point, orthogonal directions from which have opposite signs as a result of their partial second derivative. ]]

'''Answer 1:''' On a potential energy surface diagram the transition state is defined, mathematically, as a saddle point. A transition state is the maximum energy present in the minimum energy pathway taken from reactants to products. A saddle point is of symmetric configuration at which the distance between three separate atoms is the same.

A saddle point is different to a local minima. For both, the first partial derivative is equal to zero:

∂V(ri)/ ∂ri = 0

However, the saddle point and local minima may be distinguished though inspection of their second partial derivatives. For a local minima, calculating partial second derivatives in orthogonal directions to the saddle point will give a result that is greater than 0, independent from which plane it has been measured. By contrast, for the saddle point, its second partial derivative will either be greater than zero or smaller than zero, dependent on which plane it has been measured. This explanation is summarised below:

For a local minima: ∂2V(ri)/ ∂ri2 > 0

For a saddle point: ∂2V(ri)/ ∂ri2 > 0 or ∂2V(ri)/ ∂ri2 < 0<ref>J. I. Steinfeld, J. S. Francisco, W. L. Hase, ''Chemical Kinetic and Dynamics'' ''2nd ed.'', Prentice-Hall, Upper Saddle River, 1998.</ref>

An example of a saddle point can be seen in Figure 1.
==== Trajectories from r1 = r2: locating the transition state ====
'''Question 2: Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'''

'''Answer 2: '''The transition state occurs at the saddle point, which is of symmetric configuration. Hence, r1 = r2 as the H + H2 surface is symmetric, where, r1 = BC and r2 = AB. At the transition state p1 = p2 = 0 g.mol-1.pm.fs-1. Through trial and error an estimate for the transition state position is when AB = BC = 90.775 pm. Using this distance produces the contour plot below. It can be seen that there is no trajectory of motion orthogonal to the point of the transition state (red cross). There are no forces acting along either AB or BC, thus confirming that this is a transition state.
[[File:Transition State Contour Plot ED2618.png|thumb|440x440px|Figure 2: A contour plot showing the transition state. There are no oscillations or directions of motion towards either reactants or products, thus the system is stationary|centre]]
The plot of 'Internuclear Distances vs Time' below shows the distances of A-B, B-C, and A-C constant with time, consolidating the fact that this is the transition state position. Running an animation with the r1 = r2 = 90.775 pm shows the species as stationary, as there are no fluctuations of internuclear distance with time.
[[File:ED2618 IND vs Time.png|thumb|489x489px|Figure 3: A plot of Internuclear Distances vs Time|centre]]

==== Trajectories from r1 = rts + δ, r2 = rts ====
'''Question 3: Comment on how the ''mep'' and the trajectory you just calculated differ.'''

'''Answer 3: '''The initial conditions are set so that the system is slightly displaced from the transition state, and so that the system has no momentum. r1 = rts + 1 pm, r2 = rts and the momenta p1 = p2 = 0 g.mol-1.pm.fs-1. This is shown in the figures below, where r1 (BC) = 91.775 pm and r2 (AB) = 90.775 pm. To obtain a complete MEP the step number was increased to 3000.
{| class="wikitable"
|[[File:ED2618 Q3 Dynamic.png|none|thumb|366x366px|Figure 4a: Contour Plot (Calculation Type: Dynamics)]]
|
|[[File:ED2618 Q3 MEP.png|none|thumb|366x366px|Figure 5a: Contour Plot (Calculation Type: MEP)]]
|-
|
|
|
|-
|[[File:ED261 Q3 Dynamic INDvsT.png|none|thumb|366x366px|Figure 4b: Internuclear Distance vs Time Plot (Calculation Type: Dynamics)]]
|
|[[File:ED2618 Q3 MEP INDvsT.png|none|thumb|366x366px|Figure 5b: Internuclear Distance vs Time Plot (Calculation Type: MEP)]]
|-
|
|
|
|-
|[[File:ED2618 Q3 Dynamic MvsT.png|none|thumb|366x366px|Figure 4c: Momenta vs Time Plot (Calculation Type: Dynamics)]]
|
|[[File:ED2618 Q3 MEP MvsT.png|none|thumb|366x366px|Figure 5c: Momenta vs Time Plot (Calculation Type: MEP)]]
|}
It can be deduced that the trajectory is greatly dependent on calculation type. Using the MEP calculation, the momentum is reset to zero after each step. This provides an unrealistic account of the motion of the atoms during the reaction thus, explaining the absence of any type of oscillation in the MEP calculation. The Dynamics calculation accounts for the vibrational energies possessed by the molecules, highlighting the presence of oscillations along the trajectory in Figure 4a.

Plots of "Internuclear Distances vs Time” and “Momenta vs Time” are also provided and more clearly depict the differences in these calculation types.

The Dynamics calculation accounts for the extra kinetic energy obtained after bond formation, which manifests itself into excess vibrational energy. This can be seen in the small oscillations of the A-B bond in Figure 4b. Using the MEP calculation, this A-B bond distance remains constant over time, indicating the excess vibrational energy has not been accounted for. The increased momentum of the B-C bond indicates HC moves with translational energy. In addition, the oscillating momentum of the A-B bond indicates the new HAHB bond moves with increased vibrational energy.

Figure 5c shows a straight line at 0 g.mol-1.pm.fs-1. The momentum is zero for the reaction pathway. This calculation type resets the momentum to zero after every step, hence this result is expected. Figure 4c shows an oscillation of momentum with time, as through this calculation type, momenta is not reset to zero after each step.

If the initial conditions were changed so that r1 = rts pm and r2 = rts + 1pm, the blue and orange lines would swap positions in Figures 4b, 4c, 5b, and 5c.

=== Reactive and unreactive trajectories ===
'''Question 4: Complete the table below by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?'''

With the initial positions r1 = 74 pm and r2 = 200 pm, reaction trajectories were run with the following momenta combination:
{| class="wikitable"
!Row Number
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot/ kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|1
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|The kinetic energy of the system is sufficient to over come the activation barrier and hence the reaction proceeds. Originally, the reactants have little vibrational energy, reflected by the lack of oscillatory behaviour. After the reaction occurs, the products have more oscillatory behaviour - they have greater vibrational energy. The BC bond is broken and the AB bond is formed.
|[[File:ED2618 Q4 C1.png|frameless|366x366px]]
|-
|2
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|HA approaches an oscillating HBHC molecule however, the activation energy required for the reaction to proceed exceeds the kinetic energy possessed by the reactants and thus, the reaction does not occur.
|[[File:ED2618 Q4 C2.png|frameless|366x366px]]
|-
|3
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|This trajectory is similar to that of row 1, the only difference being that the reactants possess more vibrational energy. Hence, there is presence of oscillatory behaviour as depicted in the figure on the right.
|[[File:ED2618 Q4 C3.png|frameless|366x366px]]
|-
|4
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|For this trajectory, the kinetic energy is high so the activation barrier is crossed and the products are formed. However, these products contain an excess of vibrational energy namely, the newly formed HAHB bond which oscillates with such great strength that the bond breaks and the reactants are reformed. The reactants then move away from one another with greater vibrational energy than when the reaction began.
|[[File:ED2618 Q4 C4.png|frameless|366x366px]]
|-
|5
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|This trajectory is similar to that of the row above, where the kinetic energy is sufficient to overcome the activation barrier however, the newly formed HAHB bond possess so much vibrational energy that it dissociates back to the products (HBHC and HA). The only difference is that the products are reformed as the system has a high vibrational energy. The reactant HBHC bond is broken for a second time.
|[[File:ED2618 Q4 C5.png|none|thumb|366x366px]]
|}
From this table it may be concluded that the reactants must have sufficient energy to overcome the activation barrier to reach and surpass the transition state. This, however, is not the only factor that has to be considered for reaction success. The products must not have too much vibrational energy or they could convert back to the reactants. This example is observed in row 4 of the above table.

==== Transition State Theory<ref>K. J. Laidler, ''Chemical Kinetics 3rd ed.'', Harper-Collins, Upper Saddle River, 1987.</ref> ====
'''Question 5: Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''

'''Answer 1: '''Transition State Theory is derived through the postulated quasi-equilibrium between a transition state species and the reactant molecules. There are a number of assumptions that are made upon derivation of the Transition State Theory rate expression:
# Electronic and nuclear motions are separated, similar to the Born-Oppenheimer approximation in quantum mechanics.
# Reactant molecule are distributed among their states in accordance with the Maxwell-Boltzmann distribution.
# Once a molecular system has crossed a transition state, in a direction towards the products, it is unable to turn around and re-form the reactants.
# In the transition state, any motion that occurs along the reaction coordinate can be treated independently of other motions. This motion is treated classically as a translation.
# Even in the absence of an equilibrium between reactant and product molecules, the transition states in the process of becoming products are distributed among their states in accordance with the Maxwell-Boltzmann laws.
In rows 4 and 5 of the [[MRD:ED2618#Reactive and unreactive trajectories|table above]], the transition state is crossed and re-crossed numerous times thus, contradicting the assumption made by the Transition State Theory that a molecular system cannot regress back to the reactants once it has passed the transition state towards the products. The recrossing of the transition state ultimately lowers the rate of reaction, and thus Transition State Theory predicts a reaction rate that is larger to what is experimentally observed. Transition State Theory fails to evaluate molecular systems using quantum mechanical phenomena such as tunnelling, hence highlighting the discrepancy between the observed and theoretical reaction rates.

== EXERCISE 2: F - H - H system ==
In this exercise A=F, B=H, and C=H. r1 = BC and r2 = AB.

=== PES inspection ===
'''Question 1: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''[[File:ED2618 Surface Plot FHH.png|thumb|320x320px|Figure 6: Surface plot of FHH system]]'''Answer 1: '''

In this discussion, atoms F, H, and H are consigned to A, B, and C respectively. A large BC distance means that A and B are bonded, thus the HF + H system is consigned to this axis. An increasing AB distance means that B and C are bonded, thus this axis is consigned to the F + H2 system. As shown in Figure 6, the HF + H system is lower in energy than the F + H2 system.

Thus the reactions:
* HF + H → H2 + F is endothermic. (Products higher in energy than reactants)

* F + H2 → HF + H is exothermic. (Reactants higher in energy than products)
The HF (565 kJ.mol-1) bond is much stronger than the HH bond (436 kJ.mol-1)<ref name=":0">P. Atkins, J. de Paula, J. Keeler, ''Physical Chemistry 11th ed''., OUP, Oxford, 2018.</ref>, thus the HF molecule is more stable than the H2 molecule. This is due to the large difference in electronegativity between H and F, resulting in a strongly ionic bond. This is a stronger bonding interaction than the purely covalent one that occurs between the two hydrogen atoms.

'''Question 2: Locate the approximate position of the transition state.'''

'''Answer 2: '''Hammond's postulate<ref>J. Clayden, ''N. Greeves, S. Warren, Organic Chemistry 2nd ed.'', OUP, New York, 2012.</ref> states that the transition state of a given reaction will resemble either the products or the reactants, depending on which one is higher in energy, as the transition state is the maximum energy present in the minimum energy pathway taken from reactants to products. Thus, for an endothermic reaction, the products are higher in energy than the reactants, and so the transition state will resemble the products. Conversely, for an exothermic reaction, the reactants are higher in energy than the products, and so the transition state will resemble the reactants.

The reaction of HF + H → H2 + F is endothermic and thus the transition state will resemble the products in structure. Thus the HH bond distance will be set to around 74.14 pm.<ref name=":0" />

F + H2 → HF + H is exothermic and thus the transition state will resemble the reactants in structure. Hence, the HH bond distance will be set around 74.14 pm, as above.

The value of the HF bond distance is expected to be much greater than its 91.68 pm<ref name=":0" /> bond distance as there is limited association between the two atoms.

The transition state was found when r1 (BC) = 74.485 and pm r2 (AB) = 181.200 pm. At this position there are no forces acting along either AB or BC.
{| class="wikitable"
|[[File:ED2618 Q2 TS Contour.png|none|thumb|320x320px|Figure 7: Contour plot of transition state of FHH system]]
|
|[[File:ED2618 Q2 TS INDvsT.png|none|thumb|320x320px|Figure 8: Internuclear Distance vs Time plot of transition state for FHH system]]
|}
Figure 7 above shows that there is no trajectory of motion at right angles to the point of the transition state (red cross). The plot of 'Internuclear Distances vs Time' in Figure 8 shows the distances of A-B, B-C, and A-C constant with time, consolidating that this is the transition state position. These lines are straight and have no oscillatory character, indicating no intermolecular vibrations are occurring.

'''Question 3: Report the activation energy for both reactions.'''

'''Answer 3: '''The kinetic energy of the transition state is -433.981 kJ.mol-1. The activation energy can be found through running an MEP calculation of a structure slightly offset from that of the transition state. The initial positions were set so that:

For the exothermic reaction, where F + H2 → HF + H:
* r1 (BC) = 74.485 pm = rts
* r2 (AB) = 182.200 pm = rts + 1 pm
* p1 = p2 = 0 g.mol-1.pm.fs-1

For the endothermic reaction, where HF + H → H2 + F:
* r1 (BC) = 74.485 pm = rts
* r2 (AB) = 180.200 pm = rts - 1 pm
* p1 = p2 = 0 g.mol-1.pm.fs-1
{| class="wikitable"
!Reaction
!Energy of Reactants / kJ.mol-1
!Transition State Energy / kJ.mol-1
!Activation Energy / kJ.mol-1
!Energy vs Time Plot of Reaction Trajectory
|-
|F + H2 → HF + H
|<nowiki>-435.5</nowiki>
|<nowiki>-434.0</nowiki>
|1.0
|[[File:ED2618 Q3 Exo.png|frameless|400x400px]]
|-
|HF + H → H2 + F
|<nowiki>-560.5</nowiki>
|<nowiki>-434.0</nowiki>
|126.5
|[[File:ED2618 Q3 Endo.png|frameless|400x400px]]
|}
=== Reaction dynamics ===
[[File:ED2618 Q4 Contour.png|thumb|320x320px|Figure 9: Contour plot of initial conditions that result in a reactive trajectory for F + H2 system]]
'''Question 4: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.'''

'''Answer 4: '''The set of initial conditions that results in a reactive trajectory for the reaction F + H2 → HF + H is given as follows:
* r1 (BC) = 74 pm
* r2 (AB) = 230 pm
* p1 (BC) = -0.75 g.mol-1.pm.fs-1
* p2 (AB) = -2.00 0 g.mol-1.pm.fs-1
* 2000 steps at 0.1 fs
This reactive trajectory is show in Figure 9:

The reaction F + H2 → HF + H is exothermic, meaning that the products possess a lower energy than the reactants. The energy discrepancy between reactants and products is released as heat to the surroundings. Figure 10 shows a plot of momenta vs time for this system. The H2 molecule approaches the F atom with low oscillatory momenta. The HH bond is broken and a new HF bond is formed. The energy released upon the HH bond breaking is greater than the energy required to form the HF bond. Upon collision of the molecules, the HF bond now oscillates with a much greater amplitude than possessed by the H2 molecule prior to collision. This is due to the release of excess energy. The unexpected drop in momenta at t = 110 fs is due to recrossing of the transition state, as can be seen in Figure 9. After collision, the H atom moves off with a much slower velocity, as is indicated by the straight line in Figure 10.
[[File:ED2618 Q4 Momenta.png|none|thumb|320x320px|Figure 10: Momenta vs Time plot of initial conditions that result in a reactive trajectory for F + H2 system]]
As energy must always be conserved, the increased kinetic energy of the system is compensated for by a decrease in potential energy, as indicated in Figure 11. This plot shows that the total energy of the system remains constant at - 433.372 kJ.mol-1. There is no energy loss but instead, it is converted from one form to another, in line with the law of conservation of energy.
[[File:ED2618 Q4 Energy.png|none|thumb|320x320px|Figure 11: Energy vs Time plot of initial conditions that result in a reactive trajectory for F + H2 system]]
As this reaction is exothermic, energy is released to the surroundings in the form of heat. To confirm experimentally that this reaction has occurred, calorimetry could be used. Calorimetry may be used to identify changes in enthalpy of a reaction. However, this method may underestimate the heat energy released by the reaction as heat may be lost to the surrounding environment.

Another method to confirm that the reaction has gone to completion is through analysis of IR data. The vibrational energy of the HF molecule after collision is greatly increased after collision of F with H2. This would produce overtone bands as the HF molecule is excited from a lower vibrational energy to a higher vibrational energy.

'''Question 5: Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.'''

'''Answer 5: '''Polanyi's empirical rules<ref>K. J. Laidler, ''Chemical Kinetics 3rd ed.'', Harper & Row, London, 1987.</ref> explain the dependence of a reaction's success on the way the energy of the reactants is distributed among different modes. For an exothermic reaction, the reactants are higher in energy than the products, thus the reactants resemble the transition state more than the products do. This system is said to have an early transition state. For a reaction to be successful, not only must the reactants have sufficient energy to overcome the activation barrier, they must also possess the right type of energy.

For an exothermic reaction, one possessing an early transition state, Polanyi's rules state that translational energy proves more useful than vibrational energy when overcoming the energy barrier from reactants to products. Conversely, for an endothermic reaction with a late transition state, vibrational energy is more critical than translational energy when overcoming this energy barrier.

It should be noted that overcoming the transition state energy barrier leads to a high reaction success rate, however sometimes the reaction does not go to completion as the energy barrier may be recrossed.

The reaction F + H2 → HF + H is exothermic. The reactants with more translational energy as opposed to vibrational energy should prove effective at overcoming the energy barrier of the transition state. A calculation was set up for this reaction using the initial conditions:
* r1 (BC) = 74 pm
* r2 (AB) = 200 pm
* p2 (AB) = -1.0 g.mol-1.pm.fs-1
* p1 (BC) varied from -6.1 to 6.1 g.mol-1.pm.fs-1
* 5000 steps at 0.1 fs
Observations of these parameters are produced in the table below:
{| class="wikitable"
!P1 (BC) / g.mol-1.pm.fs-1
!Contour Plot of Reaction Trajectory
!Reaction Success
|-
|<nowiki>-6.1</nowiki>
|[[File:ED2618 Q5 R1.png|frameless|400x400px]]
|No
|-
|<nowiki>-3.1</nowiki>
|[[File:ED2618 Q5 R2.png|frameless|400x400px]]
|No
|-
|<nowiki>-1.1</nowiki>
|[[File:ED2618 Q5 R3.png|frameless|400x400px]]
|Yes
|-
|<nowiki>-0.5</nowiki>
|[[File:ED2618 Q5 R4.png|frameless|400x400px]]
|Yes
|-
|0.0
|[[File:ED2618 Q5 R5.png|frameless|400x400px]]
|Yes
|-
|<nowiki>+0.5</nowiki>
|[[File:ED2618 Q5 R6.png|frameless|400x400px]]
|Yes
|-
|<nowiki>+1.1</nowiki>
|[[File:ED2618 Q5 R7.png|frameless|400x400px]]
|No
|-
|<nowiki>+3.1</nowiki>
|[[File:ED2618 Q5 R8.png|frameless|400x400px]]
|No
|-
|<nowiki>+6.1</nowiki>
|[[File:ED2618 Q5 R9.png|frameless|400x400px]]
|No
|}
Demonstrated by the conditions in the table above, a low absolute value (<1.1) for p1 is crucial for reaction success. Reactant molecules may pass through a transition state multiple times before either regressing back to being reactants or continuing onto becoming products. The result of these configurations may sometimes be due to dependence on more than just the type of energy possessed by the reactant molecules. To conclude, increasing the absolute value of p1 correlates to an increase in vibrational energy of the H2 molecule. In accordance with Polanyi's empirical rules, an exothermic reaction's success in crossing the energy barrier is more dependent on translational energy than it is on vibrational energy. This is demonstrated in the table above, where large absolute values of p1 do not invoke a successful reaction.
[[File:ED2618 Q5 p2.1.png|thumb|320x320px|Figure 12: Contour plot of successful endothermic reaction between HF and H.]]
The reverse reaction: HF + H → H2 + F is endothermic, which incurs a late transition state and a preference for the reactants to possess vibrational energy over translational energy when overcoming the energy barrier.

A calculation was set up for this reaction using the initial conditions:
* r1 (BC) = 200 pm
* r2 (AB) = 74 pm
* p2 (AB) = 6 g.mol-1.pm.fs-1
* p1 (BC) = -3 g.mol-1.pm.fs-1
* 5000 steps at 0.1 fs
A high value of p2 corresponds to a high vibrational energy of the HF molecule. These initial conditions yield the successful reaction trajectory shown in Figure 12 below. Changing the initial reaction conditions so that p2 is lowered to 4 g.mol-1.pm.fs-1 (HF molecule has a lower vibrational energy) yields an unsuccessful reaction. This complements Polanyi's empirical rules of the dependence of a reactant to possess high vibrational energy (and low translational energy) to overcome a late transitions state.

To conclude, Polanyi's rule as a set of rough guidelines that can be used to measure if a reaction will proceed or not. However, sometimes small tweaks to initial conditions incur a successful reaction that may not have been deemed possible by the rules.

== References ==
<references />

MRD:ED2618

2020-05-15T17:14:19Z

Ed2618: /* EXERCISE 1: H + H2 system */

= '''Molecular Reaction Dynamics''' =
In this lab we are using molecular dynamic trajectories to study the reactivity of triatomic systems, namely when an atom collides with a diatomic molecule.

== EXERCISE 1: H + H2 system ==
For this exercise, the following reaction is analysed and discussed: HA + HBHC → HAHB + HC, where r1 = BC and r2 = AB.

=== Dynamics from the transition state region ===
'''Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?'''[[File:Saddle plot ED2618.png|thumb|440x440px|Figure 1: Plot of a saddle point, orthogonal directions from which have opposite signs as a result of their partial second derivative. ]]

'''Answer 1:''' On a potential energy surface diagram the transition state is defined, mathematically, as a saddle point. A transition state is the maximum energy present in the minimum energy pathway taken from reactants to products. A saddle point is of symmetric configuration at which the distance between three separate atoms is the same.

A saddle point is different to a local minima. For both, the first partial derivative equates to zero:

∂V(ri)/ ∂ri = 0

However, the saddle point and local minima may be distinguished though inspection of their second partial derivatives. For a local minima, calculating partial second derivatives in orthogonal directions to the saddle point will give a result that is greater than 0, independent from which plane it has been measured. By contrast, for the saddle point, its second partial derivative will either be greater than zero or smaller than zero, dependent on which plane it has been measured. This explanation is summarised below:

For a local minima: ∂2V(ri)/ ∂ri2 > 0

For a saddle point: ∂2V(ri)/ ∂ri2 > 0 and ∂2V(ri)/ ∂ri2 < 0<ref>J. I. Steinfeld, J. S. Francisco, W. L. Hase, ''Chemical Kinetic and Dynamics'' ''2nd ed.'', Prentice-Hall, Upper Saddle River, 1998.</ref>

An example of a saddle point can be seen in Figure 1.
==== Trajectories from r1 = r2: locating the transition state ====
'''Question 2: Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'''

'''Answer 2: '''The transition state occurs at the saddle point, which is of symmetric configuration. Hence, r1 = r2 as the H + H2 surface is symmetric, where, r1 = BC and r2 = AB. At the transition state p1 = p2 = 0 g.mol-1.pm.fs-1. Through trial and error an estimate for the transition state position is when AB = BC = 90.775 pm. Using this distance produces the contour plot below. It can be seen that there is no gradient/trajectory of motion at right angles to the point of the transition state (red cross). There are no forces acting along either AB or BC.
[[File:Transition State Contour Plot ED2618.png|thumb|440x440px|Figure 2: A contour plot showing the transition state. There are no oscillations or directions of motion towards either reactants or products, thus the system is stationary|centre]]
The plot of 'Internuclear Distances vs Time' below shows the distances of A-B, B-C, and A-C constant with time, consolidating the fact that this position is the transition state position. Running an animation with the r1 = r2 = 90.775 pm shows the species as stationary, as there are no fluctuations of internuclear distance with time.
[[File:ED2618 IND vs Time.png|thumb|489x489px|Figure 3: A plot of Internuclear Distances vs Time|centre]]

==== Trajectories from r1 = rts + δ, r2 = rts ====
'''Question 3: Comment on how the ''mep'' and the trajectory you just calculated differ.'''

'''Answer 3: '''The initial conditions are set so that the system is slightly displaced from the transition state and so that there is no momenta, where r1 = rts + 1 pm, r2 = rts and the momenta p1 = p2 = 0 g.mol-1.pm.fs-1. This is shown in the figures below, where r1 (BC) = 91.775 pm and r2 (AB) = 90.775 pm. To obtain a complete MEP the step number was increased to 3000.
{| class="wikitable"
|[[File:ED2618 Q3 Dynamic.png|none|thumb|366x366px|Figure 4a: Contour Plot (Calculation Type: Dynamics)]]
|
|[[File:ED2618 Q3 MEP.png|none|thumb|366x366px|Figure 5a: Contour Plot (Calculation Type: MEP)]]
|-
|
|
|
|-
|[[File:ED261 Q3 Dynamic INDvsT.png|none|thumb|366x366px|Figure 4b: Internuclear Distance vs Time Plot (Calculation Type: Dynamics)]]
|
|[[File:ED2618 Q3 MEP INDvsT.png|none|thumb|366x366px|Figure 5b: Internuclear Distance vs Time Plot (Calculation Type: MEP)]]
|-
|
|
|
|-
|[[File:ED2618 Q3 Dynamic MvsT.png|none|thumb|366x366px|Figure 4c: Momenta vs Time Plot (Calculation Type: Dynamics)]]
|
|[[File:ED2618 Q3 MEP MvsT.png|none|thumb|366x366px|Figure 5c: Momenta vs Time Plot (Calculation Type: MEP)]]
|}
It can be deduced that the trajectory is greatly dependent on calculation type. Using the MEP calculation, the momenta are reset to zero after each step. This provides an unrealistic account of the motion of the atoms during the reaction. This explains the absence of any type of oscillation in the MEP calculation, Figure 5a, compared to the Dynamics calculation figure. The Dynamics calculation accounts for the vibrational energies possessed by the molecules, hence explaining the presence of oscillations along the trajectory in Figure 4a.

Plots of "Internuclear Distances vs Time” and “Momenta vs Time” are also provided and more clearly depict the differences in the calculation types.

The Dynamics calculation accounts for the extra kinetic energy obtained after bond formation, which manifests itself into excess vibrational energy. This can be seen in the small oscillations of the A-B bond in Figure 4b. Using the MEP calculation, this A-B bond distance remain constant over time, indicating this excess vibrational energy has not been accounted for. The increased momentum of the B-C bond is indicative of HC moving with translational energy and the oscillating momentum of the A-B bond is indicative of the new HAHB bond moving with vibrational energy.

Figure 5c shows a straight line at 0 g.mol-1.pm.fs-1. The momenta is zero for the reaction pathway. This calculation type resets the momenta to zero after every step, hence this result is expected. Figure 4c shows an oscillation of momentum with time, as through this calculation type, momenta is not reset to zero after each step.

If the initial conditions were changed so that r1 = rts pm and r2 = rts + 1pm, the blue and orange lines would swap positions in figures 4b, 4c, 5b, and 5c.

=== Reactive and unreactive trajectories ===
'''Question 4: Complete the table below by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?'''

With the initial positions r1 = 74 pm and r2 = 200 pm, reaction trajectories were run with the following momenta combination:
{| class="wikitable"
!Row Number
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot/ kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|1
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|The kinetic energy of the system is sufficient to over come the activation barrier and hence the reaction proceeds. Originally, the reactants have little vibrational energy, reflected by the lack of oscillatory behaviour. After the reaction occurs, the products have more oscillatory behaviour, meaning that they have greater vibrational energy. The BC bond is broken and the AB bond is formed.
|[[File:ED2618 Q4 C1.png|frameless|366x366px]]
|-
|2
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|HA approaches an oscillating HBHC molecule, however the activation energy for the reaction to proceed exceeds the kinetic energy possessed by the reactants and thus the reaction does not occur.
|[[File:ED2618 Q4 C2.png|frameless|366x366px]]
|-
|3
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|This trajectory is similar to that of row 1, with the only difference being that the reactants possess more vibrational energy, hence there is presence of oscillatory behaviour as depicted in the figure on the right.
|[[File:ED2618 Q4 C3.png|frameless|366x366px]]
|-
|4
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|For this trajectory, the kinetic energy is high and so the activation barrier is crossed and the products are formed. However, these products contain an excess of vibrational energy, namely the newly formed HAHB bond, which oscillates with such a great strength that the bond breaks and the reactants are reformed. The reactants then move away from one another with greater vibrational energy than when the reaction began.
|[[File:ED2618 Q4 C4.png|frameless|366x366px]]
|-
|5
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|This trajectory is similar to that of the row above, where the kinetic energy is sufficient to overcome the activation barrier, however the newly formed HAHB bond possess so much vibrational energy that it dissociates back to the products (HBHC and HA). The only difference this time is that the products are reformed as the system has a high vibrational energy. The reactant HBHC bond is broken for a second time.
|[[File:ED2618 Q4 C5.png|none|thumb|366x366px]]
|}
From this table it may be concluded that the reactants must have sufficient energy to overcome the activation barrier to reach and surpass the transition state. This, however, is not the only factor that has to be considered for reaction success. The products must not have too much vibrational energy or they will convert back to the reactants. This example is observed in row 4 of the above table.

==== Transition State Theory<ref>K. J. Laidler, ''Chemical Kinetics 3rd ed.'', Harper-Collins, Upper Saddle River, 1987.</ref> ====
'''Question 5: Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''

'''Answer 1: '''Transition State Theory is derived through the postulated quasi-equilibrium between a transition state species and the reactant molecules. There are a number of assumptions that are made upon derivation of the Transition State Theory rate expression:
# Electronic and nuclear motions are separated, similar to the Born-Oppenheimer approximation in quantum mechanics.
# Reactant molecule are distributed among their states in accordance with the Maxwell-Boltzmann distribution.
# Once a molecular system has crossed a transition state, in a direction towards the products, it is unable to turn around and re-form the reactants.
# In the transition state, any motion that occurs along the reaction coordinate can be treated independently of other motions. This motion is treated classically as a translation.
# Even in the absence of an equilibrium between reactant and product molecules, the transition states that are en route to becoming products are distributed among their states in accordance with the Maxwell-Boltzmann laws.
In rows 4 and 5 of the [[MRD:ED2618#Reactive and unreactive trajectories|table above]], the transition state is crossed and re-crossed numerous times, thus contradicting the assumption made by the Transition State Theory that a molecular system cannot regress back to the reactants once it has passed the transition state towards the products. The recrossing of the transition state ultimately lowers the rate of reaction, and thus Transition State Theory predicts a reaction rate that is larger to what is experimentally observed. Transition State Theory fails to evaluate molecular systems using quantum mechanical phenomena such as tunnelling, hence highlighting the discrepancy between the observed and theoretical reaction rates.

== EXERCISE 2: F - H - H system ==
In this exercise A=F, B=H, and C=H. r1 = BC and r2 = AB.

=== PES inspection ===
'''Question 1: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''[[File:ED2618 Surface Plot FHH.png|thumb|320x320px|Figure 6: Surface plot of FHH system]]'''Answer 1: '''

In this discussion, atoms F, H, and H are consigned to A, B, and C respectively. A large BC distance means that A and B are bonded, thus the HF + H system is consigned to this axis. An increasing AB distance means that B and C are bonded, thus this axis is consigned to the F + H2 system. As shown in Figure 6, the HF + H system is lower in energy than the F + H2 system.

Thus the reactions:
* HF + H → H2 + F is endothermic. (Products higher in energy than reactants)

* F + H2 → HF + H is exothermic. (Reactants higher in energy than products)
The HF (565 kJ.mol-1) bond is much stronger than the HH bond (436 kJ.mol-1)<ref name=":0">P. Atkins, J. de Paula, J. Keeler, ''Physical Chemistry 11th ed''., OUP, Oxford, 2018.</ref>, thus the HF molecule is more stable than the H2 molecule. This is due to the large difference in electronegativity between H and F, resulting in a strongly ionic bond. This is a stronger bonding interaction than a purely covalent one that occurs between the two hydrogen atoms.

'''Question 2: Locate the approximate position of the transition state.'''

'''Answer 2: '''Hammond's postulate<ref>J. Clayden, ''N. Greeves, S. Warren, Organic Chemistry 2nd ed.'', OUP, New York, 2012.</ref> states that the transition state of a given reaction will resemble either the products or the reactants, depending on which one is higher in energy, as the transition state is the maximum energy present in the minimum energy pathway taken from reactants to products. Thus, for an endothermic reaction, the products are higher in energy than the reactants, and so the transition state will resemble the products. Conversely, for an exothermic reaction, the reactants are higher in energy than the products, and so the transition state will resemble the reactants.

The reaction of HF + H → H2 + F is endothermic and thus the transition state will resemble the products in structure. Thus the HH bond distance will be set to around 74.14 pm.<ref name=":0" />

F + H2 → HF + H is exothermic and thus the transition state will resemble the reactants in structure. Hence, the HH bond distance will be set around 74.14 pm, as above.

The value of the HF bond distance is expected to be much greater than its 91.68 pm<ref name=":0" /> bond distance as there is limited association between the two atoms.

The transition state was found when r1 (BC) = 74.485 and pm r2 (AB) = 181.200 pm. At this position there are no forces acting along either AB or BC.
{| class="wikitable"
|[[File:ED2618 Q2 TS Contour.png|none|thumb|320x320px|Figure 7: Contour plot of transition state of FHH system]]
|
|[[File:ED2618 Q2 TS INDvsT.png|none|thumb|320x320px|Figure 8: Internuclear Distance vs Time plot of transition state for FHH system]]
|}
Figure 7 above shows that there is no trajectory of motion at right angles to the point of the transition state (red cross). The plot of 'Internuclear Distances vs Time' in Figure 8 shows the distances of A-B, B-C, and A-C constant with time, consolidating the fact that this position is the transition state position. These lines are straight and have no oscillatory character, indicating no intermolecular vibrations are occurring.

'''Question 3: Report the activation energy for both reactions.'''

'''Answer 3: '''The kinetic energy of the transition state is calculated to be -433.981 kJ.mol-1. The activation energy can be found through running an MEP calculation of a structure slightly offset from that of the transition state. The initial positions were set so that:

For the exothermic reaction, where F + H2 → HF + H:
* r1 (BC) = 74.485 pm = rts
* r2 (AB) = 182.200 pm = rts + 1 pm
* p1 = p2 = 0 g.mol-1.pm.fs-1

For the endothermic reaction, where HF + H → H2 + F:
* r1 (BC) = 74.485 pm = rts
* r2 (AB) = 180.200 pm = rts - 1 pm
* p1 = p2 = 0 g.mol-1.pm.fs-1
{| class="wikitable"
!Reaction
!Energy of Reactants / kJ.mol-1
!Transition State Energy / kJ.mol-1
!Activation Energy / kJ.mol-1
!Energy vs Time Plot of Reaction Trajectory
|-
|F + H2 → HF + H
|<nowiki>-435.5</nowiki>
|<nowiki>-434.0</nowiki>
|1.0
|[[File:ED2618 Q3 Exo.png|frameless|400x400px]]
|-
|HF + H → H2 + F
|<nowiki>-560.5</nowiki>
|<nowiki>-434.0</nowiki>
|126.5
|[[File:ED2618 Q3 Endo.png|frameless|400x400px]]
|}
=== Reaction dynamics ===
[[File:ED2618 Q4 Contour.png|thumb|320x320px|Figure 9: Contour plot of initial conditions that result in a reactive trajectory for F + H2 system]]
'''Question 4: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.'''

'''Answer 4: '''The set of initial conditions that results in a reactive trajectory for the reaction F + H2 → HF + H is given as follows:
* r1 (BC) = 74 pm
* r2 (AB) = 230 pm
* p1 (BC) = -0.75 g.mol-1.pm.fs-1
* p2 (AB) = -2.00 0 g.mol-1.pm.fs-1
* 2000 steps at 0.1 fs
This reactive trajectory is show in Figure 9:

The reaction F + H2 → HF + H is exothermic, meaning that the products possess a lower energy than the reactants. The energy discrepancy between reactants and products is released as heat to the surroundings. Figure 10 shows a plot of momenta vs time for this system. The H2 molecule approaches the F atom with low oscillatory momenta. The HH bond is broken and a new HF bond is formed. The energy released upon the HH bond breaking is greater than the energy required to form the HF bond. Upon collision of the molecules, the HF bond now oscillates with a much greater amplitude than possessed by the H2 molecule prior to collision. This is due to the release of excess energy. The unexpected drop in momenta at t = 110 fs is due to recrossing of the transition state, as can be seen in Figure 9. After collision, the H atom moves off with a much slower velocity, as is indicated by the straight line in Figure 10.
[[File:ED2618 Q4 Momenta.png|none|thumb|320x320px|Figure 10: Momenta vs Time plot of initial conditions that result in a reactive trajectory for F + H2 system]]
As energy must always be conserved, the increased kinetic energy of the system is compensated for by a decrease in potential energy, as indicated in Figure 11. This plot shows that the total energy of the system remains constant at - 433.372 kJ.mol-1. There is no loss of energy, but instead it is being converted from form to another, in line with the law of conservation of energy.
[[File:ED2618 Q4 Energy.png|none|thumb|320x320px|Figure 11: Energy vs Time plot of initial conditions that result in a reactive trajectory for F + H2 system]]
As this reaction is exothermic, energy is released to the surroundings in the form of heat. To confirm experimentally that this reaction has occurred, calorimetry could be used. Calorimetry may be used to identify changes in enthalpy of a reaction. However, this method may underestimate the heat energy released by the reaction as heat may be lost to the surrounding environment.

Another method to confirm that the reaction has gone to completion is through analysis of IR data. The vibrational energy of the HF molecule after collision is greatly increased after collision of F with H2. This would produce overtone bands as the HF molecule is excited from a lower vibrational energy to a higher vibrational energy.

'''Question 5: Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.'''

'''Answer 5: '''Polanyi's empirical rules<ref>K. J. Laidler, ''Chemical Kinetics 3rd ed.'', Harper & Row, London, 1987.</ref> explain the dependence of a reaction's success on the way the energy of the reactants is distributed among different modes. For an exothermic reaction, the reactants are higher in energy than the products, thus the reactants resemble the transition state more than the products do. This system is said to have an early transition state. For a reaction to be successful, not only must the reactants have sufficient energy to overcome the activation barrier, they must also possess the right type of energy.

For an exothermic reaction, one possessing an early transition state, Polanyi's rules state that translational energy proves more useful than vibrational energy when overcoming the energy barrier from reactants to products. Conversely, for an endothermic reaction with a late transition state, vibrational energy is more critical than translational energy when overcoming this energy barrier.

It should be noted that overcoming the transition state energy barrier leads to a high reaction success rate, however sometimes the reaction does not go to completion as the energy barrier may be recrossed.

The reaction F + H2 → HF + H is exothermic. The reactants with more translational energy as opposed to vibrational energy should prove effective at overcoming the energy barrier of the transition state. A calculation was set up for this reaction using the initial conditions:
* r1 (BC) = 74 pm
* r2 (AB) = 200 pm
* p2 (AB) = -1.0 g.mol-1.pm.fs-1
* p1 (BC) varied from -6.1 to 6.1 g.mol-1.pm.fs-1
* 5000 steps at 0.1 fs
Observations of these parameters are produced in the table below:
{| class="wikitable"
!P1 (BC) / g.mol-1.pm.fs-1
!Contour Plot of Reaction Trajectory
!Reaction Success
|-
|<nowiki>-6.1</nowiki>
|[[File:ED2618 Q5 R1.png|frameless|400x400px]]
|No
|-
|<nowiki>-3.1</nowiki>
|[[File:ED2618 Q5 R2.png|frameless|400x400px]]
|No
|-
|<nowiki>-1.1</nowiki>
|[[File:ED2618 Q5 R3.png|frameless|400x400px]]
|Yes
|-
|<nowiki>-0.5</nowiki>
|[[File:ED2618 Q5 R4.png|frameless|400x400px]]
|Yes
|-
|0.0
|[[File:ED2618 Q5 R5.png|frameless|400x400px]]
|Yes
|-
|<nowiki>+0.5</nowiki>
|[[File:ED2618 Q5 R6.png|frameless|400x400px]]
|Yes
|-
|<nowiki>+1.1</nowiki>
|[[File:ED2618 Q5 R7.png|frameless|400x400px]]
|No
|-
|<nowiki>+3.1</nowiki>
|[[File:ED2618 Q5 R8.png|frameless|400x400px]]
|No
|-
|<nowiki>+6.1</nowiki>
|[[File:ED2618 Q5 R9.png|frameless|400x400px]]
|No
|}
Demonstrated by the conditions in the table above, a low absolute value (<1.1) for p1 is crucial for reaction success. Reactant molecules may pass through a transition state multiple times before either regressing back to being reactants or continuing onto becoming products. The result of these configurations may sometimes be due to dependence on more than just the type of energy possessed by the reactant molecules. To conclude, increasing the absolute value of p1 correlates to an increase in vibrational energy of the H2 molecule. In accordance with Polanyi's empirical rules, an exothermic reaction's success in crossing the energy barrier is more dependent on translational energy than it is on vibrational energy. This is demonstrated in the table above, where large absolute values of p1 do not invoke a successful reaction.
[[File:ED2618 Q5 p2.1.png|thumb|320x320px|Figure 12: Contour plot of successful endothermic reaction between HF and H.]]
The reverse reaction: HF + H → H2 + F is endothermic, which incurs a late transition state and a preference for the reactants to possess vibrational energy over translational energy when overcoming the energy barrier.

A calculation was set up for this reaction using the initial conditions:
* r1 (BC) = 200 pm
* r2 (AB) = 74 pm
* p2 (AB) = 6 g.mol-1.pm.fs-1
* p1 (BC) = -3 g.mol-1.pm.fs-1
* 5000 steps at 0.1 fs
A high value of p2 corresponds to a high vibrational energy of the HF molecule. These initial conditions yield the successful reaction trajectory shown in Figure 12 below. Changing the initial reaction conditions so that p2 is lowered to 4 g.mol-1.pm.fs-1 (HF molecule has a lower vibrational energy) yields an unsuccessful reaction. This complements Polanyi's empirical rules of the dependence of a reactant to possess high vibrational energy (and low translational energy) to overcome a late transitions state.

To conclude, Polanyi's rule as a set of rough guidelines that can be used to measure if a reaction will proceed or not. However, sometimes small tweaks to initial conditions incur a successful reaction that may not have been deemed possible by the rules.

== References ==
<references />

MRD:ED2618

2020-05-15T16:07:39Z

Ed2618: /* Trajectories from r1 = r2: locating the transition state */

= '''Molecular Reaction Dynamics''' =
In this lab we are using molecular dynamic trajectories to study the reactivity of triatomic systems, namely when an atom collides with a diatomic molecule.

== EXERCISE 1: H + H2 system ==
For this exercise, the following reaction is analysed and discussed: HA + HBHC → HAHB + HC, where r1 = BC and r2 = AB.

=== Dynamics from the transition state region ===
'''Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?'''[[File:Saddle plot ED2618.png|thumb|440x440px|'''Figure 1: An example of a saddle point, orthogonal directions from which have opposite signs for their partial second derivative.''' ]]

'''Answer 1:''' On a potential energy surface diagram the transition state is defined, mathematically, as a saddle point. A transition state is the maximum energy present in the minimum energy pathway taken from reactants to products. A saddle point is of symmetric configuration at which the distance between three separate atoms is the same.

A saddle point is different to a local minima. For both, the gradient of the potential energy surface is zero, is given as the first partial derivative:

∂V(ri)/ ∂ri = 0

However, the saddle point and local minima may be distinguished though inspection of their second partial derivatives. For a local minima, calculating partial second derivatives in orthogonal directions to the saddle point will give a result that is greater than 0, independent from which plane it has been measured. By contrast, for the saddle point, is second partial derivative will either be greater than zero or smaller than zero dependent on which plane it has been measured. This explanation is summarised below:

For a local minima: ∂2V(ri)/ ∂ri2 > 0

For a saddle point: ∂2V(ri)/ ∂ri2 > 0 and ∂2V(ri)/ ∂ri2 < 0 (INSERT REFERENCE)

An example of a saddle point can be seen in Figure 1
==== Trajectories from r1 = r2: locating the transition state ====
'''Question 2: Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. (REDO IF TIME - ENSURE FORCES ON RHS = 0)'''

'''Answer 2: '''The transition state occurs at the saddle point, which is of symmetric configuration. Hence, r1 = r2 as the H + H2 surface is symmetric, where, r1 = BC and r2 = AB. At the transition state p1 = p2 = 0 g.mol-1.pm.fs-1. Through trial and error an estimate for the transition state position is when AB = BC = 90.75 pm. Using this distance produces the contour plot below. It can be seen that there is no gradient/trajectory of motion at right angles to the point of the transition state (red cross).
[[File:Transition State Contour Plot ED2618.png|thumb|440x440px|Figure 2: A contour plot showing the transition state. There are no oscillations or directions of motion towards either reactants or products, thus the system is stationary|centre]]
The plot of 'Internuclear Distances vs Time' below shows the distances of A-B, B-C, and A-C constant with time, consolidating the fact that this position is the transition state position. Running an animation with the r1 = r2 = 90.75 pm shows the species as stationary, as there are no fluctuations of interatomic distance with time.
[[File:ED2618 IND vs Time.png|thumb|489x489px|Figure 3: A plot of Internuclear Distances vs Time|centre]]

==== Trajectories from r1 = rts + δ, r2 = rts ====
'''Question 3: Comment on how the ''mep'' and the trajectory you just calculated differ.'''

'''Answer 3: '''The initial conditions are set so that the system is slightly displaced from the transition state and so that there is zero initial momenta, where r1 = rts + 1 pm, r2 = rts and the momenta p1 = p2 = 0 g.mol-1.pm.fs-1. This can be shown in the figures below, where r1 (BC) = 91.75 pm and r2 (AB) = 90.75 pm. To obtain a complete MEP the step number was increased to 3000.
{| class="wikitable"
|[[File:ED2618 Q3 Dynamic.png|none|thumb|366x366px|Figure 4a: Contour Plot (Calculation Type: Dynamics)]]
|
|[[File:ED2618 Q3 MEP.png|none|thumb|366x366px|Figure 5a: Contour Plot (Calculation Type: MEP)]]
|-
|
|
|
|-
|[[File:ED261 Q3 Dynamic INDvsT.png|none|thumb|366x366px|Figure 4b: Internuclear Distance vs Time Plot (Calculation Type: Dynamics)]]
|
|[[File:ED2618 Q3 MEP INDvsT.png|none|thumb|366x366px|Figure 5b: Internuclear Distance vs Time Plot (Calculation Type: MEP)]]
|-
|
|
|
|-
|[[File:ED2618 Q3 Dynamic MvsT.png|none|thumb|366x366px|Figure 4c: Momenta vs Time Plot (Calculation Type: Dynamics)]]
|
|[[File:ED2618 Q3 MEP MvsT.png|none|thumb|366x366px|Figure 5c: Momenta vs Time Plot (Calculation Type: MEP)]]
|}
It can be deduced that the trajectory is greatly dependent on calculation type. Using the MEP calculation, the momenta are reset to zero after each step. This provides an unrealistic account of the motion of the atoms during the reaction. This explains the absence of any type of oscillation in the MEP calculation, Figure 5a, compared to the Dynamics calculation figure. The Dynamics calculation accounts for the vibrational energies possessed by the molecules, hence explaining the presence of oscillations along the trajectory in Figure 4a.

Plots of "Internuclear Distances vs Time” and “Momenta vs Time” are also provided and more clearly depict the differences in the calculation types.

The Dynamics calculation accounts for the extra kinetic energy obtained after bond formation, which manifests itself into excess vibrational energy. This can be seen in the small oscillations of the A-B bond in figure 4b. Using the MEP calculation, this A-B bond distance remain constant over time, indicating this excess vibrational energy has not been accounted for. The increased momentum of the B-C bond is indicative of HC moving with translational energy and the oscillating momentum of the A-B bond is indicative of the new HAHB bond moving with vibrational energy.

Figure 5c shows a straight line at 0. The momenta is zero for the reaction pathway. This calculation type resets the momenta to zero after every step hence this result is expected. Figure 4c shows an oscillation of momentum with time, as through this calculation type, momenta is not reset to zero after each step.

If the initial conditions were changed so that r1 = rts pm and r2 = rts + 1pm, the blue and orange lines would swap positions in figures 4b, 4c, 5b, and 5c.

=== Reactive and unreactive trajectories ===
'''Question 4: Complete the table below by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?'''

With the initial positions r1 = 74 pm and r2 = 200 pm, trajectories were run with the following momenta combination:
{| class="wikitable"
!Row Number
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot/ kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|1
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|The kinetic energy of the system is sufficient to over come the activation barrier and hence the reaction proceeds. Originally, the reactants have little vibrational energy, reflected by the lack of oscillatory behaviour. After the reaction occurs, the products have more oscillatory behaviour, meaning that they have greater vibrational energy. The BC bond is broken and the AB bond is formed.
|[[File:ED2618 Q4 C1.png|frameless|366x366px]]
|-
|2
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|HA approaches an oscillating HBHC molecule, however the activation energy exceeds the kinetic energy and thus the reaction does not occur. Both reactants remain unchanged.
|[[File:ED2618 Q4 C2.png|frameless|366x366px]]
|-
|3
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|This trajectory is similar to that of row 1, with the only difference being that the reactants possess more vibrational energy, hence the presence of oscillatory behaviour as depicted in the figure.
|[[File:ED2618 Q4 C3.png|frameless|366x366px]]
|-
|4
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|For this trajectory, the kinetic energy is high and so the activation barrier is crossed and the products are formed. However, these products contain an excess of vibrational energy, namely the newly formed HAHB bond, which oscillates with such a great strength that the bond breaks and the reactants of the reaction are reformed. The reactants then move away from one another with greater vibrational energy than when the reaction began.
|[[File:ED2618 Q4 C4.png|frameless|366x366px]]
|-
|5
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|This trajectory is similar to that of the row above, where the kinetic energy is sufficient to overcome the activation barrier, however the newly formed HAHB bond possess so much vibrational energy that it dissociates back to the products (HBHC and HA). The only difference this time is that the products are reformed as the system has a high vibrational energy. The reactant HBHC bond is broken for a second time.
|[[File:ED2618 Q4 C5.png|none|thumb|366x366px]]
|}
From this table it may be concluded that the reactants must have sufficient energy to overcome the activation barrier to reach and surpass the transition state. This, however, is not the only factor that has to be considered for reaction success. The products may not have too much vibrational energy or the will convert back to the reactants. This example is observed in row 4 of the above table.

==== Transition State Theory (ADD REFERENCE) ====
'''Question 5: Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''

'''Answer 1: '''Transition State Theory is derived through the postulated quasi-equilibrium between a transition state species and the reactant molecules. There are a number of assumptions that are made upon derivation of the Transition State Theory rate expression:
# Electronic and nuclear motions are separated, similar to the Born-Oppenheimer approximation in quantum mechanics.
# Reactant molecule are distributed among their states in a way that is in accordance with the Maxwell-Boltzmann distribution.
# Once a molecular system has crossed a transition state, in a direction towards the products, it is unable to turn around and re-form the reactants.
# In the transition state, any motion that occurs along the reaction coordinate can be treated independently of other motions. This motion is treated classically as a translation.
# Even in the absence of an equilibrium between reactant and product molecules, the transition states that en route to becoming products are distributed among their states in accordance with the Maxwell-Boltzmann laws.
In rows 4 and 5 of the [[MRD:ED2618#Reactive and unreactive trajectories|table above]], the transition state is crossed and re-crossed numerous times, thus contradicting the assumption made by the Transition State Theory that a molecular system cannot regress back to the reactants once it has passed the transition state towards the products. The recrossing of the transition state ultimately lowers the rate of reaction, and thus Transition State Theory predicts a reaction rate that is larger than what is experimentally observed. Transition State Theory fails to evaluate molecular systems using quantum mechanical phenomena such as tunnelling, hence highlighting the discrepancy between the observed and theoretical reaction rates.

== EXERCISE 2: F - H - H system ==
In this exercise A=F, B=H, and C=H. r1 = BC and r2 = AB.

=== PES inspection ===
'''Question 1: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''[[File:ED2618 Surface Plot FHH.png|thumb|320x320px|''''''Figure 6: Surface plot of FHH system'''''']]'''Answer 1: '''

In this discussion, atoms F, H, and H are consigned to A, B, and C respectively. An increasing BC distance means that A and B are bonded, thus the HF + H system is consigned to this axis. An increasing AB distance means that B and C are bonded, thus this axis is consigned to the F + H2 system. As shown in the surface plot on the right, the HF + H system is lower in energy than the F + H2 system.

Thus the reaction:
* HF + H → H2 + F is endothermic. (Products higher in energy than reactants)

* F + H2 → HF + H is exothermic. (Reactants higher in energy than products)
The HF (565 kJ.mol-1) bond is much stronger than the HH bond (436 kJ.mol-1) (ADD REFERENCE - PHYSICAL CHEM TEXTBOOK pG 885), thus the HF molecule is more stable than the H2 molecules. This is due to the large difference in electronegativity between H and F, resulting in an ionic bond. This is a stronger interaction than a purely covalent one that occurs between the two hydrogen atoms.

'''Question 2: Locate the approximate position of the transition state.'''

'''Answer 2: '''Hammond's postulate states the the transition state of a given reaction will resemble either the products or the reactants. The transition state will resemble either the products or the reactants, depending on which on is higher in energy, as the transition state is the maximum energy present in the minimum energy pathway taken from reactants to products. Thus, for an endothermic reaction, the products are higher in energy than the reactants, and so the transition state will resemble the products. Conversely, for an exothermic reaction, the reactants are higher in energy than the products, and so the transition state will resemble the reactants. (REFERENCE HAMMOND'S POSTULATE?)

The reaction of HF + H → H2 + F is endothermic and thus the transition state will resemble the products in structure. Thus the HH bond distance will be set to around 74.14 pm (ADD REFERENCE - ATKINS PHYSICAL CHEM PAGE 362)

F + H2 → HF + H is exothermic and thus the transition state will resemble the reactants in structure. Hence, the HH bond distance will be set around 74.14 pm, as above.

The value of the HF bond distance is expected to be much greater than 91.68 (ADD REFERENCE) as there is limited association between the two atoms.

The transition state was found when r1 (BC) = 74.485 and pm r2 (AB) = 181.200 pm. At this position there are no forces acting along either AB or BC.
{| class="wikitable"
|[[File:ED2618 Q2 TS Contour.png|none|thumb|320x320px|Figure 7: Contour plot of transition state of FHH system]]
|
|[[File:ED2618 Q2 TS INDvsT.png|none|thumb|320x320px|Figure 8: Internuclear Distance vs Time plot of transition state for FHH system]]
|}
Figure 7 above shows that there is no gradient/trajectory of motion at right angles to the point of the transition state (red cross). The plot of 'Internuclear Distances vs Time' in Figure 8 shows the distances of A-B, B-C, and A-C constant with time, consolidating the fact that this position is the transition state position. These lines are straight and have no oscillatory character, indicating no intermolecular vibrations are occurring.

'''Question 3: Report the activation energy for both reactions.'''

'''Answer 3: '''The kinetic energy of the transition state was calculated to be -433.981 kJ.mol-1. The activation energy can be found through running an MEP calculation of a structure slightly offset from that of the transition state. The initial positions were set so that:

For the exothermic reaction where F + H2 → HF + H
* r1 (BC) = 74.485 pm = rts
* r2 (AB) = 182.200 pm = rts + 1 pm
* p1 = p2 = 0 g.mol-1.pm.fs-1

For the endothermic reaction where HF + H → H2 + F
* r1 (BC) = 74.485 pm = rts
* r2 (AB) = 180.200 pm = rts - 1 pm
* p1 = p2 = 0 g.mol-1.pm.fs-1
{| class="wikitable"
!Reaction
!Energy of Reactants / kJ.mol-1
!Transition State Energy / kJ.mol-1
!Activation Energy / kJ.mol-1
!Energy vs Time Plot of Reaction Trajectory
|-
|F + H2 → HF + H
|<nowiki>- 435.5</nowiki>
|<nowiki>- 434.0</nowiki>
|<nowiki>-1.0</nowiki>
|[[File:ED2618 Q3 Exo.png|frameless|400x400px]]
|-
|HF + H → H2 + F
|<nowiki>- 560.5</nowiki>
|<nowiki>- 434.0</nowiki>
|<nowiki>-126.5</nowiki>
|[[File:ED2618 Q3 Endo.png|frameless|400x400px]]
|}
EA POSITIVE OR NEGATIVE??

=== Reaction dynamics ===
[[File:ED2618 Q4 Contour.png|thumb|320x320px|'''Figure 9: Contour plot of initial conditions that result in a reactive trajectory for F + H2 system''']]
'''Question 4: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.'''

'''Answer 4: '''The set of initial conditions that results in a reactive trajectory for the reaction F + H2 → HF + H is given as follows:
* r1 (BC) = 74 pm
* r2 (AB) = 230 pm
* p1 (BC) = - 0.75 g.mol-1.pm.fs-1
* p2 (AB) = -2.00 0 g.mol-1.pm.fs-1
* 2000 steps at 0.1 fs
This reactive trajectory is show in Figure 9:

The reaction F + H2 → HF + H is exothermic, meaning that the products possess a lower energy than the reactants. The energy discrepancy between reactants and products is released as heat to the surroundings. Figure 10 shows a plot of momenta vs time for this system. The H2 molecule approaches the F atom with low oscillatory momenta. The HH bond is broken and a new HF bond is formed. The energy released upon an HH bond breaking is greater than the energy required to form the HF bond. Upon collision of the molecules, the HF bond now oscillates with a much greater aptitude than the H2 had done prior to collision, due to the release of excess energy. The unexpected drop in momenta at t = 110 fs is due to recrossing of the transition state, as can be seen in Figure 9. After collision, the H atom moves off with a much slower velocity, as is indicated by the straight line in Figure 10. (ENERGY TO TRANSLATION AND VIBRATIONAL AS PRODUCT MOVES FAST?)
[[File:ED2618 Q4 Momenta.png|none|thumb|320x320px|Figure 10: Momenta vs Time plot of initial conditions that result in a reactive trajectory for F + H2 system]]
As energy must always be conserved, the increased kinetic energy of the system is compensated for by a decrease in potential energy, as indicated in Figure 11. This plot shows that the total energy of the system remains constant at - 433.372 kJ.mol-1. There is no loss of energy, but instead it is being converted from form to another, in line with the law of conservation of energy.
[[File:ED2618 Q4 Energy.png|none|thumb|320x320px|Figure 11: Energy vs Time plot of initial conditions that result in a reactive trajectory for F + H2 system]]
As this reaction is exothermic, energy is released to the surroundings in the form of heat. To confirm experimentally that this reaction has occurred, calorimetry could be used. Calorimetry may be used to identify changes in enthalpy of the proceeding reaction. However, this method may underestimate the heat energy released by the reaction as heat is lost to the environment.

Another method to confirm that the reaction has gone to completion is through analysis of IR data. The vibrational energy of the HF molecule after collision is greatly increased after collision of F with H2. This would produce overtone bands as the HF molecule is excited from a lower vibrational energy to a higher vibrational energy.

'''Question 5: Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.'''

'''Answer 5: '''Polanyi's empirical rules explain the dependence of a reaction's success on the way energy of the reactants is distributed in different modes. For an exothermic reaction, the reactants are higher in energy than the products, thus the reactants resemble the transition state more than the products do. This system is said to have an early transition state. For a reaction to be successful, not only must the reactants have sufficient energy to overcome the activation barrier, they must also possess the right type of energy.

For an exothermic reaction, one possessing an early transition state, Polanyi's rules state that translational energy proves more useful than vibrational energy when overcoming the energy barrier from reactants to products. Conversely, for an endothermic reaction with a late transition state, vibrational energy is more critical than translational energy when overcoming this energy barrier. (ADD REFERENCE)

It should be noted that overcoming the transition state energy barrier leads to a high reaction success rate, however sometimes the reaction does not go to completion as the energy barrier may be recrossed.

The reaction F + H2 → HF + H is exothermic, thus reactants with more translational energy as opposed to vibrational energy should prove effective at overcoming the energy barrier of the transition state. A calculation was set up for this reaction using the initial conditions:
* r1 (BC) = 74 pm
* r2 (AB) = 200 pm
* p2 (AB) = -1.0 g.mol-1.pm.fs-1
* p1 (BC) varied from -6.1 to 6.1 g.mol-1.pm.fs-1
* 5000 steps at 0.1 fs
Observations of these parameters are produced in the table below:
{| class="wikitable"
!P1 (BC) / g.mol-1.pm.fs-1
!Contour Plot of Reaction Trajectory
!Reaction Success
|-
|<nowiki>-6.1</nowiki>
|[[File:ED2618 Q5 R1.png|frameless|400x400px]]
|No
|-
|<nowiki>-3.1</nowiki>
|[[File:ED2618 Q5 R2.png|frameless|400x400px]]
|No
|-
|<nowiki>-1.1</nowiki>
|[[File:ED2618 Q5 R3.png|frameless|400x400px]]
|Yes
|-
|<nowiki>-0.5</nowiki>
|[[File:ED2618 Q5 R4.png|frameless|400x400px]]
|Yes
|-
|0.0
|[[File:ED2618 Q5 R5.png|frameless|400x400px]]
|Yes
|-
|<nowiki>+0.5</nowiki>
|[[File:ED2618 Q5 R6.png|frameless|400x400px]]
|Yes
|-
|<nowiki>+1.1</nowiki>
|[[File:ED2618 Q5 R7.png|frameless|400x400px]]
|No
|-
|<nowiki>+3.1</nowiki>
|[[File:ED2618 Q5 R8.png|frameless|400x400px]]
|No
|-
|<nowiki>+6.1</nowiki>
|[[File:ED2618 Q5 R9.png|frameless|400x400px]]
|No
|}
Demonstrated by the conditions in the table above, a low absolute value (<1.1) for p1 is crucial for reaction success. Reactant molecules may pass through a transition state multiple times before either regressing back to reactants or continuing onto products. The result of these configurations may sometimes be dependent on more than just the type of energy possessed by the reactant molecules. To conclude, increasing the absolute value of p1 correlates to an increase in vibrational energy. In accordance with Polanyi's empirical rules, an exothermic reaction's success in crossing the energy barrier is more dependent on translational energy than it is on vibrational energy. This is proved in the table above, where large absolute values of p1 to do invoke a successful reaction.
[[File:ED2618 Q5 p2.1.png|thumb|320x320px|Figure 12: Contour plot of successful endothermic reaction between HF and H.]]
The reverse reaction: HF + H → H2 + F is endothermic, which incurs a late transition state and a preference for the reactants to possess vibrational energy over translational energy when overcoming the energy barrier.

A calculation was set up for this reaction using the initial conditions:
* r1 (BC) = 200 pm
* r2 (AB) = 74 pm
* p2 (AB) = 6 g.mol-1.pm.fs-1
* p1 (BC) = -3 g.mol-1.pm.fs-1
* 5000 steps at 0.1 fs
A high value of p2 corresponds to a high vibrational energy of the HF molecule. This initial conditions yield the successful reaction trajectory shown in Figure 12 below. Changing the initial reaction conditions so that p2 is lowered to 4 g.mol-1.pm.fs-1 (HF molecule has a lower vibrational energy) yields an unsuccessful reaction. This complements Polanyi's empirical rules of the dependence of a reactant to possess high vibrational energy (and low translational energy) to overcome an energy barrier.

CONCLUSION?

MRD:ED2618

2020-05-15T16:06:56Z

Ed2618: /* Reaction dynamics */

= '''Molecular Reaction Dynamics''' =
In this lab we are using molecular dynamic trajectories to study the reactivity of triatomic systems, namely when an atom collides with a diatomic molecule.

== EXERCISE 1: H + H2 system ==
For this exercise, the following reaction is analysed and discussed: HA + HBHC → HAHB + HC, where r1 = BC and r2 = AB.

=== Dynamics from the transition state region ===
'''Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?'''[[File:Saddle plot ED2618.png|thumb|440x440px|'''Figure 1: An example of a saddle point, orthogonal directions from which have opposite signs for their partial second derivative.''' ]]

'''Answer 1:''' On a potential energy surface diagram the transition state is defined, mathematically, as a saddle point. A transition state is the maximum energy present in the minimum energy pathway taken from reactants to products. A saddle point is of symmetric configuration at which the distance between three separate atoms is the same.

A saddle point is different to a local minima. For both, the gradient of the potential energy surface is zero, is given as the first partial derivative:

∂V(ri)/ ∂ri = 0

However, the saddle point and local minima may be distinguished though inspection of their second partial derivatives. For a local minima, calculating partial second derivatives in orthogonal directions to the saddle point will give a result that is greater than 0, independent from which plane it has been measured. By contrast, for the saddle point, is second partial derivative will either be greater than zero or smaller than zero dependent on which plane it has been measured. This explanation is summarised below:

For a local minima: ∂2V(ri)/ ∂ri2 > 0

For a saddle point: ∂2V(ri)/ ∂ri2 > 0 and ∂2V(ri)/ ∂ri2 < 0 (INSERT REFERENCE)

An example of a saddle point can be seen in Figure 1
==== Trajectories from r1 = r2: locating the transition state ====
'''Question 2: Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. (REDO IF TIME - ENSURE FORCES ON RHS = 0)'''

'''Answer 2: '''The transition state occurs at the saddle point, which is of symmetric configuration. Hence, r1 = r2 as the H + H2 surface is symmetric, where, r1 = BC and r2 = AB. At the transition state p1 = p2 = 0 g.mol-1.pm.fs-1. Through trial and error an estimate for the transition state position is when AB = BC = 90.75 pm. Using this distance produces the contour plot below. It can be seen that there is no gradient/trajectory of motion at right angles to the point of the transition state (red cross).
[[File:Transition State Contour Plot ED2618.png|none|thumb|440x440px|Figure 2: A contour plot showing the transition state. There are no oscillations or directions of motion towards either reactants or products, thus the system is stationary]]
The plot of 'Internuclear Distances vs Time' below shows the distances of A-B, B-C, and A-C constant with time, consolidating the fact that this position is the transition state position. Running an animation with the r1 = r2 = 90.75 pm shows the species as stationary, as there are no fluctuations of interatomic distance with time.
[[File:ED2618 IND vs Time.png|none|thumb|489x489px|Figure 3: A plot of Internuclear Distances vs Time]]

==== Trajectories from r1 = rts + δ, r2 = rts ====
'''Question 3: Comment on how the ''mep'' and the trajectory you just calculated differ.'''

'''Answer 3: '''The initial conditions are set so that the system is slightly displaced from the transition state and so that there is zero initial momenta, where r1 = rts + 1 pm, r2 = rts and the momenta p1 = p2 = 0 g.mol-1.pm.fs-1. This can be shown in the figures below, where r1 (BC) = 91.75 pm and r2 (AB) = 90.75 pm. To obtain a complete MEP the step number was increased to 3000.
{| class="wikitable"
|[[File:ED2618 Q3 Dynamic.png|none|thumb|366x366px|Figure 4a: Contour Plot (Calculation Type: Dynamics)]]
|
|[[File:ED2618 Q3 MEP.png|none|thumb|366x366px|Figure 5a: Contour Plot (Calculation Type: MEP)]]
|-
|
|
|
|-
|[[File:ED261 Q3 Dynamic INDvsT.png|none|thumb|366x366px|Figure 4b: Internuclear Distance vs Time Plot (Calculation Type: Dynamics)]]
|
|[[File:ED2618 Q3 MEP INDvsT.png|none|thumb|366x366px|Figure 5b: Internuclear Distance vs Time Plot (Calculation Type: MEP)]]
|-
|
|
|
|-
|[[File:ED2618 Q3 Dynamic MvsT.png|none|thumb|366x366px|Figure 4c: Momenta vs Time Plot (Calculation Type: Dynamics)]]
|
|[[File:ED2618 Q3 MEP MvsT.png|none|thumb|366x366px|Figure 5c: Momenta vs Time Plot (Calculation Type: MEP)]]
|}
It can be deduced that the trajectory is greatly dependent on calculation type. Using the MEP calculation, the momenta are reset to zero after each step. This provides an unrealistic account of the motion of the atoms during the reaction. This explains the absence of any type of oscillation in the MEP calculation, Figure 5a, compared to the Dynamics calculation figure. The Dynamics calculation accounts for the vibrational energies possessed by the molecules, hence explaining the presence of oscillations along the trajectory in Figure 4a.

Plots of "Internuclear Distances vs Time” and “Momenta vs Time” are also provided and more clearly depict the differences in the calculation types.

The Dynamics calculation accounts for the extra kinetic energy obtained after bond formation, which manifests itself into excess vibrational energy. This can be seen in the small oscillations of the A-B bond in figure 4b. Using the MEP calculation, this A-B bond distance remain constant over time, indicating this excess vibrational energy has not been accounted for. The increased momentum of the B-C bond is indicative of HC moving with translational energy and the oscillating momentum of the A-B bond is indicative of the new HAHB bond moving with vibrational energy.

Figure 5c shows a straight line at 0. The momenta is zero for the reaction pathway. This calculation type resets the momenta to zero after every step hence this result is expected. Figure 4c shows an oscillation of momentum with time, as through this calculation type, momenta is not reset to zero after each step.

If the initial conditions were changed so that r1 = rts pm and r2 = rts + 1pm, the blue and orange lines would swap positions in figures 4b, 4c, 5b, and 5c.

=== Reactive and unreactive trajectories ===
'''Question 4: Complete the table below by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?'''

With the initial positions r1 = 74 pm and r2 = 200 pm, trajectories were run with the following momenta combination:
{| class="wikitable"
!Row Number
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot/ kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|1
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|The kinetic energy of the system is sufficient to over come the activation barrier and hence the reaction proceeds. Originally, the reactants have little vibrational energy, reflected by the lack of oscillatory behaviour. After the reaction occurs, the products have more oscillatory behaviour, meaning that they have greater vibrational energy. The BC bond is broken and the AB bond is formed.
|[[File:ED2618 Q4 C1.png|frameless|366x366px]]
|-
|2
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|HA approaches an oscillating HBHC molecule, however the activation energy exceeds the kinetic energy and thus the reaction does not occur. Both reactants remain unchanged.
|[[File:ED2618 Q4 C2.png|frameless|366x366px]]
|-
|3
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|This trajectory is similar to that of row 1, with the only difference being that the reactants possess more vibrational energy, hence the presence of oscillatory behaviour as depicted in the figure.
|[[File:ED2618 Q4 C3.png|frameless|366x366px]]
|-
|4
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|For this trajectory, the kinetic energy is high and so the activation barrier is crossed and the products are formed. However, these products contain an excess of vibrational energy, namely the newly formed HAHB bond, which oscillates with such a great strength that the bond breaks and the reactants of the reaction are reformed. The reactants then move away from one another with greater vibrational energy than when the reaction began.
|[[File:ED2618 Q4 C4.png|frameless|366x366px]]
|-
|5
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|This trajectory is similar to that of the row above, where the kinetic energy is sufficient to overcome the activation barrier, however the newly formed HAHB bond possess so much vibrational energy that it dissociates back to the products (HBHC and HA). The only difference this time is that the products are reformed as the system has a high vibrational energy. The reactant HBHC bond is broken for a second time.
|[[File:ED2618 Q4 C5.png|none|thumb|366x366px]]
|}
From this table it may be concluded that the reactants must have sufficient energy to overcome the activation barrier to reach and surpass the transition state. This, however, is not the only factor that has to be considered for reaction success. The products may not have too much vibrational energy or the will convert back to the reactants. This example is observed in row 4 of the above table.

==== Transition State Theory (ADD REFERENCE) ====
'''Question 5: Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''

'''Answer 1: '''Transition State Theory is derived through the postulated quasi-equilibrium between a transition state species and the reactant molecules. There are a number of assumptions that are made upon derivation of the Transition State Theory rate expression:
# Electronic and nuclear motions are separated, similar to the Born-Oppenheimer approximation in quantum mechanics.
# Reactant molecule are distributed among their states in a way that is in accordance with the Maxwell-Boltzmann distribution.
# Once a molecular system has crossed a transition state, in a direction towards the products, it is unable to turn around and re-form the reactants.
# In the transition state, any motion that occurs along the reaction coordinate can be treated independently of other motions. This motion is treated classically as a translation.
# Even in the absence of an equilibrium between reactant and product molecules, the transition states that en route to becoming products are distributed among their states in accordance with the Maxwell-Boltzmann laws.
In rows 4 and 5 of the [[MRD:ED2618#Reactive and unreactive trajectories|table above]], the transition state is crossed and re-crossed numerous times, thus contradicting the assumption made by the Transition State Theory that a molecular system cannot regress back to the reactants once it has passed the transition state towards the products. The recrossing of the transition state ultimately lowers the rate of reaction, and thus Transition State Theory predicts a reaction rate that is larger than what is experimentally observed. Transition State Theory fails to evaluate molecular systems using quantum mechanical phenomena such as tunnelling, hence highlighting the discrepancy between the observed and theoretical reaction rates.

== EXERCISE 2: F - H - H system ==
In this exercise A=F, B=H, and C=H. r1 = BC and r2 = AB.

=== PES inspection ===
'''Question 1: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''[[File:ED2618 Surface Plot FHH.png|thumb|320x320px|''''''Figure 6: Surface plot of FHH system'''''']]'''Answer 1: '''

In this discussion, atoms F, H, and H are consigned to A, B, and C respectively. An increasing BC distance means that A and B are bonded, thus the HF + H system is consigned to this axis. An increasing AB distance means that B and C are bonded, thus this axis is consigned to the F + H2 system. As shown in the surface plot on the right, the HF + H system is lower in energy than the F + H2 system.

Thus the reaction:
* HF + H → H2 + F is endothermic. (Products higher in energy than reactants)

* F + H2 → HF + H is exothermic. (Reactants higher in energy than products)
The HF (565 kJ.mol-1) bond is much stronger than the HH bond (436 kJ.mol-1) (ADD REFERENCE - PHYSICAL CHEM TEXTBOOK pG 885), thus the HF molecule is more stable than the H2 molecules. This is due to the large difference in electronegativity between H and F, resulting in an ionic bond. This is a stronger interaction than a purely covalent one that occurs between the two hydrogen atoms.

'''Question 2: Locate the approximate position of the transition state.'''

'''Answer 2: '''Hammond's postulate states the the transition state of a given reaction will resemble either the products or the reactants. The transition state will resemble either the products or the reactants, depending on which on is higher in energy, as the transition state is the maximum energy present in the minimum energy pathway taken from reactants to products. Thus, for an endothermic reaction, the products are higher in energy than the reactants, and so the transition state will resemble the products. Conversely, for an exothermic reaction, the reactants are higher in energy than the products, and so the transition state will resemble the reactants. (REFERENCE HAMMOND'S POSTULATE?)

The reaction of HF + H → H2 + F is endothermic and thus the transition state will resemble the products in structure. Thus the HH bond distance will be set to around 74.14 pm (ADD REFERENCE - ATKINS PHYSICAL CHEM PAGE 362)

F + H2 → HF + H is exothermic and thus the transition state will resemble the reactants in structure. Hence, the HH bond distance will be set around 74.14 pm, as above.

The value of the HF bond distance is expected to be much greater than 91.68 (ADD REFERENCE) as there is limited association between the two atoms.

The transition state was found when r1 (BC) = 74.485 and pm r2 (AB) = 181.200 pm. At this position there are no forces acting along either AB or BC.
{| class="wikitable"
|[[File:ED2618 Q2 TS Contour.png|none|thumb|320x320px|Figure 7: Contour plot of transition state of FHH system]]
|
|[[File:ED2618 Q2 TS INDvsT.png|none|thumb|320x320px|Figure 8: Internuclear Distance vs Time plot of transition state for FHH system]]
|}
Figure 7 above shows that there is no gradient/trajectory of motion at right angles to the point of the transition state (red cross). The plot of 'Internuclear Distances vs Time' in Figure 8 shows the distances of A-B, B-C, and A-C constant with time, consolidating the fact that this position is the transition state position. These lines are straight and have no oscillatory character, indicating no intermolecular vibrations are occurring.

'''Question 3: Report the activation energy for both reactions.'''

'''Answer 3: '''The kinetic energy of the transition state was calculated to be -433.981 kJ.mol-1. The activation energy can be found through running an MEP calculation of a structure slightly offset from that of the transition state. The initial positions were set so that:

For the exothermic reaction where F + H2 → HF + H
* r1 (BC) = 74.485 pm = rts
* r2 (AB) = 182.200 pm = rts + 1 pm
* p1 = p2 = 0 g.mol-1.pm.fs-1

For the endothermic reaction where HF + H → H2 + F
* r1 (BC) = 74.485 pm = rts
* r2 (AB) = 180.200 pm = rts - 1 pm
* p1 = p2 = 0 g.mol-1.pm.fs-1
{| class="wikitable"
!Reaction
!Energy of Reactants / kJ.mol-1
!Transition State Energy / kJ.mol-1
!Activation Energy / kJ.mol-1
!Energy vs Time Plot of Reaction Trajectory
|-
|F + H2 → HF + H
|<nowiki>- 435.5</nowiki>
|<nowiki>- 434.0</nowiki>
|<nowiki>-1.0</nowiki>
|[[File:ED2618 Q3 Exo.png|frameless|400x400px]]
|-
|HF + H → H2 + F
|<nowiki>- 560.5</nowiki>
|<nowiki>- 434.0</nowiki>
|<nowiki>-126.5</nowiki>
|[[File:ED2618 Q3 Endo.png|frameless|400x400px]]
|}
EA POSITIVE OR NEGATIVE??

=== Reaction dynamics ===
[[File:ED2618 Q4 Contour.png|thumb|320x320px|'''Figure 9: Contour plot of initial conditions that result in a reactive trajectory for F + H2 system''']]
'''Question 4: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.'''

'''Answer 4: '''The set of initial conditions that results in a reactive trajectory for the reaction F + H2 → HF + H is given as follows:
* r1 (BC) = 74 pm
* r2 (AB) = 230 pm
* p1 (BC) = - 0.75 g.mol-1.pm.fs-1
* p2 (AB) = -2.00 0 g.mol-1.pm.fs-1
* 2000 steps at 0.1 fs
This reactive trajectory is show in Figure 9:

The reaction F + H2 → HF + H is exothermic, meaning that the products possess a lower energy than the reactants. The energy discrepancy between reactants and products is released as heat to the surroundings. Figure 10 shows a plot of momenta vs time for this system. The H2 molecule approaches the F atom with low oscillatory momenta. The HH bond is broken and a new HF bond is formed. The energy released upon an HH bond breaking is greater than the energy required to form the HF bond. Upon collision of the molecules, the HF bond now oscillates with a much greater aptitude than the H2 had done prior to collision, due to the release of excess energy. The unexpected drop in momenta at t = 110 fs is due to recrossing of the transition state, as can be seen in Figure 9. After collision, the H atom moves off with a much slower velocity, as is indicated by the straight line in Figure 10. (ENERGY TO TRANSLATION AND VIBRATIONAL AS PRODUCT MOVES FAST?)
[[File:ED2618 Q4 Momenta.png|none|thumb|320x320px|Figure 10: Momenta vs Time plot of initial conditions that result in a reactive trajectory for F + H2 system]]
As energy must always be conserved, the increased kinetic energy of the system is compensated for by a decrease in potential energy, as indicated in Figure 11. This plot shows that the total energy of the system remains constant at - 433.372 kJ.mol-1. There is no loss of energy, but instead it is being converted from form to another, in line with the law of conservation of energy.
[[File:ED2618 Q4 Energy.png|none|thumb|320x320px|Figure 11: Energy vs Time plot of initial conditions that result in a reactive trajectory for F + H2 system]]
As this reaction is exothermic, energy is released to the surroundings in the form of heat. To confirm experimentally that this reaction has occurred, calorimetry could be used. Calorimetry may be used to identify changes in enthalpy of the proceeding reaction. However, this method may underestimate the heat energy released by the reaction as heat is lost to the environment.

Another method to confirm that the reaction has gone to completion is through analysis of IR data. The vibrational energy of the HF molecule after collision is greatly increased after collision of F with H2. This would produce overtone bands as the HF molecule is excited from a lower vibrational energy to a higher vibrational energy.

'''Question 5: Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.'''

'''Answer 5: '''Polanyi's empirical rules explain the dependence of a reaction's success on the way energy of the reactants is distributed in different modes. For an exothermic reaction, the reactants are higher in energy than the products, thus the reactants resemble the transition state more than the products do. This system is said to have an early transition state. For a reaction to be successful, not only must the reactants have sufficient energy to overcome the activation barrier, they must also possess the right type of energy.

For an exothermic reaction, one possessing an early transition state, Polanyi's rules state that translational energy proves more useful than vibrational energy when overcoming the energy barrier from reactants to products. Conversely, for an endothermic reaction with a late transition state, vibrational energy is more critical than translational energy when overcoming this energy barrier. (ADD REFERENCE)

It should be noted that overcoming the transition state energy barrier leads to a high reaction success rate, however sometimes the reaction does not go to completion as the energy barrier may be recrossed.

The reaction F + H2 → HF + H is exothermic, thus reactants with more translational energy as opposed to vibrational energy should prove effective at overcoming the energy barrier of the transition state. A calculation was set up for this reaction using the initial conditions:
* r1 (BC) = 74 pm
* r2 (AB) = 200 pm
* p2 (AB) = -1.0 g.mol-1.pm.fs-1
* p1 (BC) varied from -6.1 to 6.1 g.mol-1.pm.fs-1
* 5000 steps at 0.1 fs
Observations of these parameters are produced in the table below:
{| class="wikitable"
!P1 (BC) / g.mol-1.pm.fs-1
!Contour Plot of Reaction Trajectory
!Reaction Success
|-
|<nowiki>-6.1</nowiki>
|[[File:ED2618 Q5 R1.png|frameless|400x400px]]
|No
|-
|<nowiki>-3.1</nowiki>
|[[File:ED2618 Q5 R2.png|frameless|400x400px]]
|No
|-
|<nowiki>-1.1</nowiki>
|[[File:ED2618 Q5 R3.png|frameless|400x400px]]
|Yes
|-
|<nowiki>-0.5</nowiki>
|[[File:ED2618 Q5 R4.png|frameless|400x400px]]
|Yes
|-
|0.0
|[[File:ED2618 Q5 R5.png|frameless|400x400px]]
|Yes
|-
|<nowiki>+0.5</nowiki>
|[[File:ED2618 Q5 R6.png|frameless|400x400px]]
|Yes
|-
|<nowiki>+1.1</nowiki>
|[[File:ED2618 Q5 R7.png|frameless|400x400px]]
|No
|-
|<nowiki>+3.1</nowiki>
|[[File:ED2618 Q5 R8.png|frameless|400x400px]]
|No
|-
|<nowiki>+6.1</nowiki>
|[[File:ED2618 Q5 R9.png|frameless|400x400px]]
|No
|}
Demonstrated by the conditions in the table above, a low absolute value (<1.1) for p1 is crucial for reaction success. Reactant molecules may pass through a transition state multiple times before either regressing back to reactants or continuing onto products. The result of these configurations may sometimes be dependent on more than just the type of energy possessed by the reactant molecules. To conclude, increasing the absolute value of p1 correlates to an increase in vibrational energy. In accordance with Polanyi's empirical rules, an exothermic reaction's success in crossing the energy barrier is more dependent on translational energy than it is on vibrational energy. This is proved in the table above, where large absolute values of p1 to do invoke a successful reaction.
[[File:ED2618 Q5 p2.1.png|thumb|320x320px|Figure 12: Contour plot of successful endothermic reaction between HF and H.]]
The reverse reaction: HF + H → H2 + F is endothermic, which incurs a late transition state and a preference for the reactants to possess vibrational energy over translational energy when overcoming the energy barrier.

A calculation was set up for this reaction using the initial conditions:
* r1 (BC) = 200 pm
* r2 (AB) = 74 pm
* p2 (AB) = 6 g.mol-1.pm.fs-1
* p1 (BC) = -3 g.mol-1.pm.fs-1
* 5000 steps at 0.1 fs
A high value of p2 corresponds to a high vibrational energy of the HF molecule. This initial conditions yield the successful reaction trajectory shown in Figure 12 below. Changing the initial reaction conditions so that p2 is lowered to 4 g.mol-1.pm.fs-1 (HF molecule has a lower vibrational energy) yields an unsuccessful reaction. This complements Polanyi's empirical rules of the dependence of a reactant to possess high vibrational energy (and low translational energy) to overcome an energy barrier.

CONCLUSION?

File:ED2618 Q5 p2.1.png

2020-05-15T16:01:21Z

Ed2618:

MRD:ED2618

2020-05-15T16:01:01Z

Ed2618: /* Reaction dynamics */

= '''Molecular Reaction Dynamics''' =
In this lab we are using molecular dynamic trajectories to study the reactivity of triatomic systems, namely when an atom collides with a diatomic molecule.

== EXERCISE 1: H + H2 system ==
For this exercise, the following reaction is analysed and discussed: HA + HBHC → HAHB + HC, where r1 = BC and r2 = AB.

=== Dynamics from the transition state region ===
'''Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?'''

'''Answer 1:''' On a potential energy surface diagram the transition state is defined, mathematically, as a saddle point. A transition state is the maximum energy present in the minimum energy pathway taken from reactants to products. A saddle point is of symmetric configuration at which the distance between three separate atoms is the same.

A saddle point is different to a local minima. For both, the gradient of the potential energy surface is zero, is given as the first partial derivative:

∂V(ri)/ ∂ri = 0

However, the saddle point and local minima may be distinguished though inspection of their second partial derivatives. For a local minima, calculating partial second derivatives in orthogonal directions to the saddle point will give a result that is greater than 0, independent from which plane it has been measured. By contrast, for the saddle point, is second partial derivative will either be greater than zero or smaller than zero dependent on which plane it has been measured. This explanation is summarised below:

For a local minima: ∂2V(ri)/ ∂ri2 > 0

For a saddle point: ∂2V(ri)/ ∂ri2 > 0 and ∂2V(ri)/ ∂ri2 < 0 (INSERT REFERENCE)

An example of a saddle point can be seen in the figure below:
[[File:Saddle plot ED2618.png|thumb|440x440px|Figure 1: An example of a saddle point, orthogonal directions from which have opposite signs for their partial second derivative. |none]]

==== Trajectories from r1 = r2: locating the transition state ====
'''Question 2: Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. (REDO IF TIME - ENSURE FORCES ON RHS = 0)'''

'''Answer 2: '''The transition state occurs at the saddle point, which is of symmetric configuration. Hence, r1 = r2 as the H + H2 surface is symmetric, where, r1 = BC and r2 = AB. At the transition state p1 = p2 = 0 g.mol-1.pm.fs-1. Through trial and error an estimate for the transition state position is when AB = BC = 90.75 pm. Using this distance produces the contour plot below. It can be seen that there is no gradient/trajectory of motion at right angles to the point of the transition state (red cross).
[[File:Transition State Contour Plot ED2618.png|none|thumb|440x440px|Figure 2: A contour plot showing the transition state. There are no oscillations or directions of motion towards either reactants or products, thus the system is stationary]]
The plot of 'Internuclear Distances vs Time' below shows the distances of A-B, B-C, and A-C constant with time, consolidating the fact that this position is the transition state position. Running an animation with the r1 = r2 = 90.75 pm shows the species as stationary, as there are no fluctuations of interatomic distance with time.
[[File:ED2618 IND vs Time.png|none|thumb|489x489px|Figure 3: A plot of Internuclear Distances vs Time]]

==== Trajectories from r1 = rts + δ, r2 = rts ====
'''Question 3: Comment on how the ''mep'' and the trajectory you just calculated differ.'''

'''Answer 3: '''The initial conditions are set so that the system is slightly displaced from the transition state and so that there is zero initial momenta, where r1 = rts + 1 pm, r2 = rts and the momenta p1 = p2 = 0 g.mol-1.pm.fs-1. This can be shown in the figures below, where r1 (BC) = 91.75 pm and r2 (AB) = 90.75 pm. To obtain a complete MEP the step number was increased to 3000.
{| class="wikitable"
|[[File:ED2618 Q3 Dynamic.png|none|thumb|366x366px|Figure 4a: Contour Plot (Calculation Type: Dynamics)]]
|
|[[File:ED2618 Q3 MEP.png|none|thumb|366x366px|Figure 5a: Contour Plot (Calculation Type: MEP)]]
|-
|
|
|
|-
|[[File:ED261 Q3 Dynamic INDvsT.png|none|thumb|366x366px|Figure 4b: Internuclear Distance vs Time Plot (Calculation Type: Dynamics)]]
|
|[[File:ED2618 Q3 MEP INDvsT.png|none|thumb|366x366px|Figure 5b: Internuclear Distance vs Time Plot (Calculation Type: MEP)]]
|-
|
|
|
|-
|[[File:ED2618 Q3 Dynamic MvsT.png|none|thumb|366x366px|Figure 4c: Momenta vs Time Plot (Calculation Type: Dynamics)]]
|
|[[File:ED2618 Q3 MEP MvsT.png|none|thumb|366x366px|Figure 5c: Momenta vs Time Plot (Calculation Type: MEP)]]
|}
It can be deduced that the trajectory is greatly dependent on calculation type. Using the MEP calculation, the momenta are reset to zero after each step. This provides an unrealistic account of the motion of the atoms during the reaction. This explains the absence of any type of oscillation in the MEP calculation, Figure 5a, compared to the Dynamics calculation figure. The Dynamics calculation accounts for the vibrational energies possessed by the molecules, hence explaining the presence of oscillations along the trajectory in Figure 4a.

Plots of "Internuclear Distances vs Time” and “Momenta vs Time” are also provided and more clearly depict the differences in the calculation types.

The Dynamics calculation accounts for the extra kinetic energy obtained after bond formation, which manifests itself into excess vibrational energy. This can be seen in the small oscillations of the A-B bond in figure 4b. Using the MEP calculation, this A-B bond distance remain constant over time, indicating this excess vibrational energy has not been accounted for. The increased momentum of the B-C bond is indicative of HC moving with translational energy and the oscillating momentum of the A-B bond is indicative of the new HAHB bond moving with vibrational energy.

Figure 5c shows a straight line at 0. The momenta is zero for the reaction pathway. This calculation type resets the momenta to zero after every step hence this result is expected. Figure 4c shows an oscillation of momentum with time, as through this calculation type, momenta is not reset to zero after each step.

If the initial conditions were changed so that r1 = rts pm and r2 = rts + 1pm, the blue and orange lines would swap positions in figures 4b, 4c, 5b, and 5c.

=== Reactive and unreactive trajectories ===
'''Question 4: Complete the table below by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?'''

With the initial positions r1 = 74 pm and r2 = 200 pm, trajectories were run with the following momenta combination:
{| class="wikitable"
!Row Number
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot/ kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|1
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|The kinetic energy of the system is sufficient to over come the activation barrier and hence the reaction proceeds. Originally, the reactants have little vibrational energy, reflected by the lack of oscillatory behaviour. After the reaction occurs, the products have more oscillatory behaviour, meaning that they have greater vibrational energy. The BC bond is broken and the AB bond is formed.
|[[File:ED2618 Q4 C1.png|frameless|366x366px]]
|-
|2
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|HA approaches an oscillating HBHC molecule, however the activation energy exceeds the kinetic energy and thus the reaction does not occur. Both reactants remain unchanged.
|[[File:ED2618 Q4 C2.png|frameless|366x366px]]
|-
|3
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|This trajectory is similar to that of row 1, with the only difference being that the reactants possess more vibrational energy, hence the presence of oscillatory behaviour as depicted in the figure.
|[[File:ED2618 Q4 C3.png|frameless|366x366px]]
|-
|4
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|For this trajectory, the kinetic energy is high and so the activation barrier is crossed and the products are formed. However, these products contain an excess of vibrational energy, namely the newly formed HAHB bond, which oscillates with such a great strength that the bond breaks and the reactants of the reaction are reformed. The reactants then move away from one another with greater vibrational energy than when the reaction began.
|[[File:ED2618 Q4 C4.png|frameless|366x366px]]
|-
|5
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|This trajectory is similar to that of the row above, where the kinetic energy is sufficient to overcome the activation barrier, however the newly formed HAHB bond possess so much vibrational energy that it dissociates back to the products (HBHC and HA). The only difference this time is that the products are reformed as the system has a high vibrational energy. The reactant HBHC bond is broken for a second time.
|[[File:ED2618 Q4 C5.png|none|thumb|366x366px]]
|}
From this table it may be concluded that the reactants must have sufficient energy to overcome the activation barrier to reach and surpass the transition state. This, however, is not the only factor that has to be considered for reaction success. The products may not have too much vibrational energy or the will convert back to the reactants. This example is observed in row 4 of the above table.

==== Transition State Theory (ADD REFERENCE) ====
'''Question 5: Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''

'''Answer 1: '''Transition State Theory is derived through the postulated quasi-equilibrium between a transition state species and the reactant molecules. There are a number of assumptions that are made upon derivation of the Transition State Theory rate expression:
# Electronic and nuclear motions are separated, similar to the Born-Oppenheimer approximation in quantum mechanics.
# Reactant molecule are distributed among their states in a way that is in accordance with the Maxwell-Boltzmann distribution.
# Once a molecular system has crossed a transition state, in a direction towards the products, it is unable to turn around and re-form the reactants.
# In the transition state, any motion that occurs along the reaction coordinate can be treated independently of other motions. This motion is treated classically as a translation.
# Even in the absence of an equilibrium between reactant and product molecules, the transition states that en route to becoming products are distributed among their states in accordance with the Maxwell-Boltzmann laws.
In rows 4 and 5 of the [[MRD:ED2618#Reactive and unreactive trajectories|table above]], the transition state is crossed and re-crossed numerous times, thus contradicting the assumption made by the Transition State Theory that a molecular system cannot regress back to the reactants once it has passed the transition state towards the products. The recrossing of the transition state ultimately lowers the rate of reaction, and thus Transition State Theory predicts a reaction rate that is larger than what is experimentally observed. Transition State Theory fails to evaluate molecular systems using quantum mechanical phenomena such as tunnelling, hence highlighting the discrepancy between the observed and theoretical reaction rates.

== EXERCISE 2: F - H - H system ==
In this exercise A=F, B=H, and C=H. r1 = BC and r2 = AB.

=== PES inspection ===
'''Question 1: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''

'''Answer 1: '''In this discussion, atoms F, H, and H are consigned to A, B, and C respectively. An increasing BC distance means that A and B are bonded, thus the HF + H system is consigned to this axis. An increasing AB distance means that B and C are bonded, thus this axis is consigned to the F + H2 system. As shown in the surface plot below, the HF + H system is lower in energy than the F + H2 system.

Thus the reaction:
* HF + H → H2 + F is endothermic. (Products higher in energy than reactants)

* F + H2 → HF + H is exothermic. (Reactants higher in energy than products)
[[File:ED2618 Surface Plot FHH.png|none|thumb|320x320px|Figure 6: Surface plot of FHH system]]
The HF (565 kJ.mol-1) bond is much stronger than the HH bond (436 kJ.mol-1) (ADD REFERENCE - PHYSICAL CHEM TEXTBOOK pG 885), thus the HF molecule is more stable than the H2 molecules. This is due to the large difference in electronegativity between H and F, resulting in an ionic bond. This is a stronger interaction than a purely covalent one that occurs between the two hydrogen atoms.

'''Question 2: Locate the approximate position of the transition state.'''

'''Answer 2: '''Hammond's postulate states the the transition state of a given reaction will resemble either the products or the reactants. The transition state will resemble either the products or the reactants, depending on which on is higher in energy, as the transition state is the maximum energy present in the minimum energy pathway taken from reactants to products. Thus, for an endothermic reaction, the products are higher in energy than the reactants, and so the transition state will resemble the products. Conversely, for an exothermic reaction, the reactants are higher in energy than the products, and so the transition state will resemble the reactants. (REFERENCE HAMMOND'S POSTULATE?)

The reaction of HF + H → H2 + F is endothermic and thus the transition state will resemble the products in structure. Thus the HH bond distance will be set to around 74.14 pm (ADD REFERENCE - ATKINS PHYSICAL CHEM PAGE 362)

F + H2 → HF + H is exothermic and thus the transition state will resemble the reactants in structure. Hence, the HH bond distance will be set around 74.14 pm, as above.

The value of the HF bond distance is expected to be much greater than 91.68 (ADD REFERENCE) as there is limited association between the two atoms.

The transition state was found when r1 (BC) = 74.485 and pm r2 (AB) = 181.200 pm. At this position there are no forces acting along either AB or BC.
{| class="wikitable"
|[[File:ED2618 Q2 TS Contour.png|none|thumb|320x320px|Figure 7: Contour plot of transition state of FHH system]]
|
|[[File:ED2618 Q2 TS INDvsT.png|none|thumb|320x320px|Figure 8: Internuclear Distance vs Time plot of transition state for FHH system]]
|}
Figure 7 above shows that there is no gradient/trajectory of motion at right angles to the point of the transition state (red cross). The plot of 'Internuclear Distances vs Time' in Figure 8 shows the distances of A-B, B-C, and A-C constant with time, consolidating the fact that this position is the transition state position. These lines are straight and have no oscillatory character, indicating no intermolecular vibrations are occurring.

'''Question 3: Report the activation energy for both reactions.'''

'''Answer 3: '''The kinetic energy of the transition state was calculated to be -433.981 kJ.mol-1. The activation energy can be found through running an MEP calculation of a structure slightly offset from that of the transition state. The initial positions were set so that:

For the exothermic reaction where F + H2 → HF + H
* r1 (BC) = 74.485 pm = rts
* r2 (AB) = 182.200 pm = rts + 1 pm
* p1 = p2 = 0 g.mol-1.pm.fs-1

For the endothermic reaction where HF + H → H2 + F
* r1 (BC) = 74.485 pm = rts
* r2 (AB) = 180.200 pm = rts - 1 pm
* p1 = p2 = 0 g.mol-1.pm.fs-1
{| class="wikitable"
!Reaction
!Energy of Reactants / kJ.mol-1
!Transition State Energy / kJ.mol-1
!Activation Energy / kJ.mol-1
!Energy vs Time Plot of Reaction Trajectory
|-
|F + H2 → HF + H
|<nowiki>- 435.5</nowiki>
|<nowiki>- 434.0</nowiki>
|<nowiki>-1.0</nowiki>
|[[File:ED2618 Q3 Exo.png|frameless|400x400px]]
|-
|HF + H → H2 + F
|<nowiki>- 560.5</nowiki>
|<nowiki>- 434.0</nowiki>
|<nowiki>-126.5</nowiki>
|[[File:ED2618 Q3 Endo.png|frameless|400x400px]]
|}
EA POSITIVE OR NEGATIVE??

=== Reaction dynamics ===
'''Question 4: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.'''

'''Answer 4: '''The set of initial conditions that results in a reactive trajectory for the reaction F + H2 → HF + H is given as follows:
* r1 (BC) = 74 pm
* r2 (AB) = 230 pm
* p1 (BC) = - 0.75 g.mol-1.pm.fs-1
* p2 (AB) = -2.00 0 g.mol-1.pm.fs-1
* 2000 steps at 0.1 fs
This reactive trajectory is show in the contour plot below:
[[File:ED2618 Q4 Contour.png|none|thumb|320x320px|Figure 9: Contour plot of initial conditions that result in a reactive trajectory for F + H2 system]]
The reaction F + H2 → HF + H is exothermic, meaning that the products possess a lower energy than the reactants. The energy discrepancy between reactants and products is released as heat to the surroundings. Figure 10 shows a plot of momenta vs time for this system. The H2 molecule approaches the F atom with low oscillatory momenta. The HH bond is broken and a new HF bond is formed. The energy released upon an HH bond breaking is greater than the energy required to form the HF bond. Upon collision of the molecules, the HF bond now oscillates with a much greater aptitude than the H2 had done prior to collision, due to the release of excess energy. The unexpected drop in momenta at t = 110 fs is due to recrossing of the transition state, as can be seen in Figure 9. After collision, the H atom moves off with a much slower velocity, as is indicated by the straight line in Figure 10. (ENERGY TO TRANSLATION AND VIBRATIONAL AS PRODUCT MOVES FAST?)
[[File:ED2618 Q4 Momenta.png|none|thumb|320x320px|Figure 10: Momenta vs Time plot of initial conditions that result in a reactive trajectory for F + H2 system]]
As energy must always be conserved, the increased kinetic energy of the system is compensated for by a decrease in potential energy, as indicated in Figure 11. This plot shows that the total energy of the system remains constant at - 433.372 kJ.mol-1. There is no loss of energy, but instead it is being converted from form to another, in line with the law of conservation of energy.
[[File:ED2618 Q4 Energy.png|none|thumb|320x320px|Figure 11: Energy vs Time plot of initial conditions that result in a reactive trajectory for F + H2 system]]
As this reaction is exothermic, energy is released to the surroundings in the form of heat. To confirm experimentally that this reaction has occurred, calorimetry could be used. Calorimetry may be used to identify changes in enthalpy of the proceeding reaction. However, this method may underestimate the heat energy released by the reaction as heat is lost to the environment.

Another method to confirm that the reaction has gone to completion is through analysis of IR data. The vibrational energy of the HF molecule after collision is greatly increased after collision of F with H2. This would produce overtone bands as the HF molecule is excited from a lower vibrational energy to a higher vibrational energy.

'''Question 5: Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.'''

'''Answer 5: '''Polanyi's empirical rules explain the dependence of a reaction's success on the way energy of the reactants is distributed in different modes. For an exothermic reaction, the reactants are higher in energy than the products, thus the reactants resemble the transition state more than the products do. This system is said to have an early transition state. For a reaction to be successful, not only must the reactants have sufficient energy to overcome the activation barrier, they must also possess the right type of energy.

For an exothermic reaction, one possessing an early transition state, Polanyi's rules state that translational energy proves more useful than vibrational energy when overcoming the energy barrier from reactants to products. Conversely, for an endothermic reaction with a late transition state, vibrational energy is more critical than translational energy when overcoming this energy barrier. (ADD REFERENCE)

It should be noted that overcoming the transition state energy barrier leads to a high reaction success rate, however sometimes the reaction does not go to completion as the energy barrier may be recrossed.

The reaction F + H2 → HF + H is exothermic, thus reactants with more translational energy as opposed to vibrational energy should prove effective at overcoming the energy barrier of the transition state. A calculation was set up for this reaction using the initial conditions:
* r1 (BC) = 74 pm
* r2 (AB) = 200 pm
* p2 (AB) = -1.0 g.mol-1.pm.fs-1
* p1 (BC) varied from -6.1 to 6.1 g.mol-1.pm.fs-1
* 5000 steps at 0.1 fs
Observations of these parameters are produced in the table below:
{| class="wikitable"
!P1 (BC) / g.mol-1.pm.fs-1
!Contour Plot of Reaction Trajectory
!Reaction Success
|-
|<nowiki>-6.1</nowiki>
|[[File:ED2618 Q5 R1.png|frameless|400x400px]]
|No
|-
|<nowiki>-3.1</nowiki>
|[[File:ED2618 Q5 R2.png|frameless|400x400px]]
|No
|-
|<nowiki>-1.1</nowiki>
|[[File:ED2618 Q5 R3.png|frameless|400x400px]]
|Yes
|-
|<nowiki>-0.5</nowiki>
|[[File:ED2618 Q5 R4.png|frameless|400x400px]]
|Yes
|-
|0.0
|[[File:ED2618 Q5 R5.png|frameless|400x400px]]
|Yes
|-
|<nowiki>+0.5</nowiki>
|[[File:ED2618 Q5 R6.png|frameless|400x400px]]
|Yes
|-
|<nowiki>+1.1</nowiki>
|[[File:ED2618 Q5 R7.png|frameless|400x400px]]
|No
|-
|<nowiki>+3.1</nowiki>
|[[File:ED2618 Q5 R8.png|frameless|400x400px]]
|No
|-
|<nowiki>+6.1</nowiki>
|[[File:ED2618 Q5 R9.png|frameless|400x400px]]
|No
|}
Demonstrated by the conditions in the table above, a low absolute value (<1.1) for p1 is crucial for reaction success. Reactant molecules may pass through a transition state multiple times before either regressing back to reactants or continuing onto products. The result of these configurations may sometimes be dependent on more than just the type of energy possessed by the reactant molecules. To conclude, increasing the absolute value of p1 correlates to an increase in vibrational energy. In accordance with Polanyi's empirical rules, an exothermic reaction's success in crossing the energy barrier is more dependent on translational energy than it is on vibrational energy. This is proved in the table above, where large absolute values of p1 to do invoke a successful reaction.

The reverse reaction: HF + H → H2 + F is endothermic, which incurs a late transition state and a preference for the reactants to possess vibrational energy over translational energy when overcoming the energy barrier.

A calculation was set up for this reaction using the initial conditions:
* r1 (BC) = 200 pm
* r2 (AB) = 74 pm
* p2 (AB) = 6 g.mol-1.pm.fs-1
* p1 (BC) = -3 g.mol-1.pm.fs-1
* 5000 steps at 0.1 fs
A high value of p2 corresponds to a high vibrational energy of the HF molecule. This initial conditions yield the successful reaction trajectory shown in Figure 12 below. Changing the initial reaction conditions so that p2 is lowered to 4 g.mol-1.pm.fs-1 (HF molecule has a lower vibrational energy) yields an unsuccessful reaction. This complements Polanyi's empirical rules of the dependence of a reactant to possess high vibrational energy (and low translational energy) to overcome an energy barrier.

MRD:ED2618

2020-05-15T15:35:00Z

2020-05-15T15:06:32Z

Ed2618: /* EXERCISE 1: H + H2 system */

= '''Molecular Reaction Dynamics''' =
In this lab we are using molecular dynamic trajectories to study the reactivity of triatomic systems, namely when an atom collides with a diatomic molecule.

== EXERCISE 1: H + H2 system ==
For this exercise, the following reaction is analysed and discussed: HA + HBHC → HAHB + HC, where r1 = BC and r2 = AB.

=== Dynamics from the transition state region ===
'''Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?'''

'''Answer 1:''' On a potential energy surface diagram the transition state is defined, mathematically, as a saddle point. A transition state is the maximum energy present in the minimum energy pathway taken from reactants to products. A saddle point is of symmetric configuration at which the distance between three separate atoms is the same.

A saddle point is different to a local minima. For both, the gradient of the potential energy surface is zero, is given as the first partial derivative:

∂V(ri)/ ∂ri = 0

However, the saddle point and local minima may be distinguished though inspection of their second partial derivatives. For a local minima, calculating partial second derivatives in orthogonal directions to the saddle point will give a result that is greater than 0, independent from which plane it has been measured. By contrast, for the saddle point, is second partial derivative will either be greater than zero or smaller than zero dependent on which plane it has been measured. This explanation is summarised below:

For a local minima: ∂2V(ri)/ ∂ri2 > 0

For a saddle point: ∂2V(ri)/ ∂ri2 > 0 and ∂2V(ri)/ ∂ri2 < 0 (INSERT REFERENCE)

An example of a saddle point can be seen in the figure below:
[[File:Saddle plot ED2618.png|thumb|440x440px|Figure 1: An example of a saddle point, orthogonal directions from which have opposite signs for their partial second derivative. |none]]

==== Trajectories from r1 = r2: locating the transition state ====
'''Question 2: Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. (REDO IF TIME - ENSURE FORCES ON RHS = 0)'''

'''Answer 2: '''The transition state occurs at the saddle point, which is of symmetric configuration. Hence, r1 = r2 as the H + H2 surface is symmetric, where, r1 = BC and r2 = AB. At the transition state p1 = p2 = 0 g.mol-1.pm.fs-1. Through trial and error an estimate for the transition state position is when AB = BC = 90.75 pm. Using this distance produces the contour plot below. It can be seen that there is no gradient/trajectory of motion at right angles to the point of the transition state (red cross).
[[File:Transition State Contour Plot ED2618.png|none|thumb|440x440px|Figure 2: A contour plot showing the transition state. There are no oscillations or directions of motion towards either reactants or products, thus the system is stationary]]
The plot of 'Internuclear Distances vs Time' below shows the distances of A-B, B-C, and A-C constant with time, consolidating the fact that this position is the transition state position. Running an animation with the r1 = r2 = 90.75 pm shows the species as stationary, as there are no fluctuations of interatomic distance with time.
[[File:ED2618 IND vs Time.png|none|thumb|489x489px|Figure 3: A plot of Internuclear Distances vs Time]]

==== Trajectories from r1 = rts + δ, r2 = rts ====
'''Question 3: Comment on how the ''mep'' and the trajectory you just calculated differ.'''

'''Answer 3: '''The initial conditions are set so that the system is slightly displaced from the transition state and so that there is zero initial momenta, where r1 = rts + 1 pm, r2 = rts and the momenta p1 = p2 = 0 g.mol-1.pm.fs-1. This can be shown in the figures below, where r1 (BC) = 91.75 pm and r2 (AB) = 90.75 pm. To obtain a complete MEP the step number was increased to 3000.
{| class="wikitable"
|[[File:ED2618 Q3 Dynamic.png|none|thumb|366x366px|Figure 4a: Contour Plot (Calculation Type: Dynamics)]]
|
|[[File:ED2618 Q3 MEP.png|none|thumb|366x366px|Figure 5a: Contour Plot (Calculation Type: MEP)]]
|-
|
|
|
|-
|[[File:ED261 Q3 Dynamic INDvsT.png|none|thumb|366x366px|Figure 4b: Internuclear Distance vs Time Plot (Calculation Type: Dynamics)]]
|
|[[File:ED2618 Q3 MEP INDvsT.png|none|thumb|366x366px|Figure 5b: Internuclear Distance vs Time Plot (Calculation Type: MEP)]]
|-
|
|
|
|-
|[[File:ED2618 Q3 Dynamic MvsT.png|none|thumb|366x366px|Figure 4c: Momenta vs Time Plot (Calculation Type: Dynamics)]]
|
|[[File:ED2618 Q3 MEP MvsT.png|none|thumb|366x366px|Figure 5c: Momenta vs Time Plot (Calculation Type: MEP)]]
|}
It can be deduced that the trajectory is greatly dependent on calculation type. Using the MEP calculation, the momenta are reset to zero after each step. This provides an unrealistic account of the motion of the atoms during the reaction. This explains the absence of any type of oscillation in the MEP calculation, Figure 5a, compared to the Dynamics calculation figure. The Dynamics calculation accounts for the vibrational energies possessed by the molecules, hence explaining the presence of oscillations along the trajectory in Figure 4a.

Plots of "Internuclear Distances vs Time” and “Momenta vs Time” are also provided and more clearly depict the differences in the calculation types.

The Dynamics calculation accounts for the extra kinetic energy obtained after bond formation, which manifests itself into excess vibrational energy. This can be seen in the small oscillations of the A-B bond in figure 4b. Using the MEP calculation, this A-B bond distance remain constant over time, indicating this excess vibrational energy has not been accounted for. The increased momentum of the B-C bond is indicative of HC moving with translational energy and the oscillating momentum of the A-B bond is indicative of the new HAHB bond moving with vibrational energy.

Figure 5c shows a straight line at 0. The momenta is zero for the reaction pathway. This calculation type resets the momenta to zero after every step hence this result is expected. Figure 4c shows an oscillation of momentum with time, as through this calculation type, momenta is not reset to zero after each step.

If the initial conditions were changed so that r1 = rts pm and r2 = rts + 1pm, the blue and orange lines would swap positions in figures 4b, 4c, 5b, and 5c.

=== Reactive and unreactive trajectories ===
'''Question 4: Complete the table below by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?'''

With the initial positions r1 = 74 pm and r2 = 200 pm, trajectories were run with the following momenta combination:
{| class="wikitable"
!Row Number
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot/ kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|1
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|The kinetic energy of the system is sufficient to over come the activation barrier and hence the reaction proceeds. Originally, the reactants have little vibrational energy, reflected by the lack of oscillatory behaviour. After the reaction occurs, the products have more oscillatory behaviour, meaning that they have greater vibrational energy. The BC bond is broken and the AB bond is formed.
|[[File:ED2618 Q4 C1.png|frameless|366x366px]]
|-
|2
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|HA approaches an oscillating HBHC molecule, however the activation energy exceeds the kinetic energy and thus the reaction does not occur. Both reactants remain unchanged.
|[[File:ED2618 Q4 C2.png|frameless|366x366px]]
|-
|3
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|This trajectory is similar to that of row 1, with the only difference being that the reactants possess more vibrational energy, hence the presence of oscillatory behaviour as depicted in the figure.
|[[File:ED2618 Q4 C3.png|frameless|366x366px]]
|-
|4
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|For this trajectory, the kinetic energy is high and so the activation barrier is crossed and the products are formed. However, these products contain an excess of vibrational energy, namely the newly formed HAHB bond, which oscillates with such a great strength that the bond breaks and the reactants of the reaction are reformed. The reactants then move away from one another with greater vibrational energy than when the reaction began.
|[[File:ED2618 Q4 C4.png|frameless|366x366px]]
|-
|5
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|This trajectory is similar to that of the row above, where the kinetic energy is sufficient to overcome the activation barrier, however the newly formed HAHB bond possess so much vibrational energy that it dissociates back to the products (HBHC and HA). The only difference this time is that the products are reformed as the system has a high vibrational energy. The reactant HBHC bond is broken for a second time.
|[[File:ED2618 Q4 C5.png|none|thumb|366x366px]]
|}
From this table it may be concluded that the reactants must have sufficient energy to overcome the activation barrier to reach and surpass the transition state. This, however, is not the only factor that has to be considered for reaction success. The products may not have too much vibrational energy or the will convert back to the reactants. This example is observed in row 4 of the above table.

==== Transition State Theory (ADD REFERENCE) ====
'''Question 5: Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''

'''Answer 1: '''Transition State Theory is derived through the postulated quasi-equilibrium between a transition state species and the reactant molecules. There are a number of assumptions that are made upon derivation of the Transition State Theory rate expression:
# Electronic and nuclear motions are separated, similar to the Born-Oppenheimer approximation in quantum mechanics.
# Reactant molecule are distributed among their states in a way that is in accordance with the Maxwell-Boltzmann distribution.
# Once a molecular system has crossed a transition state, in a direction towards the products, it is unable to turn around and re-form the reactants.
# In the transition state, any motion that occurs along the reaction coordinate can be treated independently of other motions. This motion is treated classically as a translation.
# Even in the absence of an equilibrium between reactant and product molecules, the transition states that en route to becoming products are distributed among their states in accordance with the Maxwell-Boltzmann laws.
In rows 4 and 5 of the [[MRD:ED2618#Reactive and unreactive trajectories|table above]], the transition state is crossed and re-crossed numerous times, thus contradicting the assumption made by the Transition State Theory that a molecular system cannot regress back to the reactants once it has passed the transition state towards the products. The recrossing of the transition state ultimately lowers the rate of reaction, and thus Transition State Theory predicts a reaction rate that is larger than what is experimentally observed. Transition State Theory fails to evaluate molecular systems using quantum mechanical phenomena such as tunnelling, hence highlighting the discrepancy between the observed and theoretical reaction rates.

== EXERCISE 2: F - H - H system ==
In this exercise A=F, B=H, and C=H. r1 = BC and r2 = AB.

=== PES inspection ===
'''Question 1: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''

'''Answer 1: '''In this discussion, atoms F, H, and H are consigned to A, B, and C respectively. An increasing BC distance means that A and B are bonded, thus the HF + H system is consigned to this axis. An increasing AB distance means that B and C are bonded, thus this axis is consigned to the F + H2 system. As shown in the surface plot below, the HF + H system is lower in energy than the F + H2 system.

Thus the reaction:
* HF + H → H2 + F is endothermic. (Products higher in energy than reactants)

* F + H2 → HF + H is exothermic. (Reactants higher in energy than products)
[[File:ED2618 Surface Plot FHH.png|none|thumb|320x320px|Figure 6: Surface plot of FHH system]]
The HF (565 kJ.mol-1) bond is much stronger than the HH bond (436 kJ.mol-1) (ADD REFERENCE - PHYSICAL CHEM TEXTBOOK pG 885), thus the HF molecule is more stable than the H2 molecules. This is due to the large difference in electronegativity between H and F, resulting in an ionic bond. This is a stronger interaction than a purely covalent one that occurs between the two hydrogen atoms.

'''Question 2: Locate the approximate position of the transition state.'''

'''Answer 2: '''Hammond's postulate states the the transition state of a given reaction will resemble either the products or the reactants. The transition state will resemble either the products or the reactants, depending on which on is higher in energy, as the transition state is the maximum energy present in the minimum energy pathway taken from reactants to products. Thus, for an endothermic reaction, the products are higher in energy than the reactants, and so the transition state will resemble the products. Conversely, for an exothermic reaction, the reactants are higher in energy than the products, and so the transition state will resemble the reactants. (REFERENCE HAMMOND'S POSTULATE?)

The reaction of HF + H → H2 + F is endothermic and thus the transition state will resemble the products in structure. Thus the HH bond distance will be set to around 74.14 pm (ADD REFERENCE - ATKINS PHYSICAL CHEM PAGE 362)

F + H2 → HF + H is exothermic and thus the transition state will resemble the reactants in structure. Hence, the HH bond distance will be set around 74.14 pm, as above.

The value of the HF bond distance is expected to be much greater than 91.68 (ADD REFERENCE) as there is limited association between the two atoms.

The transition state was found when r1 (BC) = 74.485 and pm r2 (AB) = 181.200 pm. At this position there are no forces acting along either AB or BC.
{| class="wikitable"
|[[File:ED2618 Q2 TS Contour.png|none|thumb|320x320px|Figure 7: Contour plot of transition state of FHH system]]
|
|[[File:ED2618 Q2 TS INDvsT.png|none|thumb|320x320px|Figure 8: Internuclear Distance vs Time plot of transition state for FHH system]]
|}
Figure 7 above shows that there is no gradient/trajectory of motion at right angles to the point of the transition state (red cross). The plot of 'Internuclear Distances vs Time' in Figure 8 shows the distances of A-B, B-C, and A-C constant with time, consolidating the fact that this position is the transition state position. These lines are straight and have no oscillatory character, indicating no intermolecular vibrations are occurring.

'''Question 3: Report the activation energy for both reactions.'''

'''Answer 3: '''The kinetic energy of the transition state was calculated to be -433.981 kJ.mol-1. The activation energy can be found through running an MEP calculation of a structure slightly offset from that of the transition state. The initial positions were set so that:

For the exothermic reaction where F + H2 → HF + H
* r1 (BC) = 74.485 pm = rts
* r2 (AB) = 182.200 pm = rts + 1 pm
* p1 = p2 = 0 g.mol-1.pm.fs-1

For the endothermic reaction where HF + H → H2 + F
* r1 (BC) = 74.485 pm = rts
* r2 (AB) = 180.200 pm = rts - 1 pm
* p1 = p2 = 0 g.mol-1.pm.fs-1
{| class="wikitable"
!Reaction
!Energy of Reactants / kJ.mol-1
!Transition State Energy / kJ.mol-1
!Activation Energy / kJ.mol-1
!Energy vs Time Plot of Reaction Trajectory
|-
|F + H2 → HF + H
|<nowiki>- 435.5</nowiki>
|<nowiki>- 434.0</nowiki>
|<nowiki>-1.0</nowiki>
|[[File:ED2618 Q3 Exo.png|frameless|400x400px]]
|-
|HF + H → H2 + F
|<nowiki>- 560.5</nowiki>
|<nowiki>- 434.0</nowiki>
|<nowiki>-126.5</nowiki>
|[[File:ED2618 Q3 Endo.png|frameless|400x400px]]
|}
EA POSITIVE OR NEGATIVE??

=== Reaction dynamics ===
'''Question 4: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.'''

'''Answer 4: '''The set of initial conditions that results in a reactive trajectory for the reaction F + H2 → HF + H is given as follows:
* r1 (BC) = 74 pm
* r2 (AB) = 230 pm
* p1 (BC) = - 0.75 g.mol-1.pm.fs-1
* p2 (AB) = -2.00 0 g.mol-1.pm.fs-1
* 2000 steps at 0.1 fs
This reactive trajectory is show in the contour plot below:
[[File:ED2618 Q4 Contour.png|none|thumb|320x320px|Figure 9: Contour plot of initial conditions that result in a reactive trajectory for F + H2 system]]
The reaction F + H2 → HF + H is exothermic, meaning that the products possess a lower energy than the reactants. The energy discrepancy between reactants and products is released as heat to the surroundings. Figure 10 shows a plot of momenta vs time for this system. The H2 molecule approaches the F atom with low oscillatory momenta. The HH bond is broken and a new HF bond is formed. The energy released upon an HH bond breaking is greater than the energy required to form the HF bond. Upon collision of the molecules, the HF bond now oscillates with a much greater aptitude than the H2 had done prior to collision, due to the release of excess energy. The unexpected drop in momenta at t = 110 fs is due to recrossing of the transition state, as can be seen in Figure 9. After collision, the H atom moves off with a much slower velocity, as is indicated by the straight line in Figure 10. (ENERGY TO TRANSLATION AND VIBRATIONAL AS PRODUCT MOVES FAST?)
[[File:ED2618 Q4 Momenta.png|none|thumb|320x320px|Figure 10: Momenta vs Time plot of initial conditions that result in a reactive trajectory for F + H2 system]]
As energy must always be conserved, the increased kinetic energy of the system is compensated for by a decrease in potential energy, as indicated in Figure 11. This plot shows that the total energy of the system remains constant at - 433.372 kJ.mol-1. There is no loss of energy, but instead it is being converted from form to another, in line with the law of conservation of energy.
[[File:ED2618 Q4 Energy.png|none|thumb|320x320px|Figure 11: Energy vs Time plot of initial conditions that result in a reactive trajectory for F + H2 system]]
As this reaction is exothermic, energy is released to the surroundings in the form of heat. To confirm experimentally that this reaction has occurred, calorimetry could be used. Calorimetry may be used to identify changes in enthalpy of the proceeding reaction. However, this method may underestimate the heat energy released by the reaction as heat is lost to the environment.

Another method to confirm that the reaction has gone to completion is through analysis of IR data. The vibrational energy of the HF molecule after collision is greatly increased after collision of F with H2. This would produce overtone bands as the HF molecule is excited from a lower vibrational energy to a higher vibrational energy.

'''Question 5: Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.'''

'''Answer 5: '''Polanyi's empirical rules explain the dependence of a reaction's success on the way energy of the reactants is distributed in different modes. For an exothermic reaction, the reactants are higher in energy than the products, thus the reactants resemble the transition state more than the products do. This system is said to have an early transition state. For a reaction to be successful, not only must the reactants have sufficient energy to overcome the activation barrier, they must also possess the right type of energy.

For an exothermic reaction, one possessing an early transition state, Polanyi's rules state that translational energy proves more useful than vibrational energy when overcoming the energy barrier from reactants to products. Conversely, for an endothermic reaction with a late transition state, vibrational energy is more critical than translational energy when overcoming this energy barrier. (ADD REFERENCE)

It should be noted that overcoming the transition state energy barrier leads to a high reaction success rate, however sometimes the reaction does not go to completion as the energy barrier may be recrossed.

The reaction reaction F + H2 → HF + H is exothermic, thus reactants with more translational energy as opposed to vibrational energy should prove effective at overcoming the energy barrier of the transition state. A calculation was set up for this reaction with r1 (BC) = 74 pm, r2 = (AB) = 200 pm, and p2 (AB) = -1.0 g.mol-1.pm.fs-1. p1 (BC) was varied with values ranging from -6.1 to 6.1 g.mol-1.pm.fs-1. Observations of these parameters are produced in the table below:
{| class="wikitable"
!P1 (BC) / g.mol-1.pm.fs-1
!Contour Plot of Reaction Trajectory
!Reaction Success
!Reaction Energy
|-
|<nowiki>-6.1</nowiki>
|
|
|
|-
|<nowiki>-3.1</nowiki>
|
|
|
|-
|<nowiki>-1.1</nowiki>
|
|
|
|-
|<nowiki>-0.5</nowiki>
|
|
|
|-
|0.0
|
|
|
|-
|<nowiki>+0.5</nowiki>
|
|
|
|-
|<nowiki>+1.1</nowiki>
|
|
|
|-
|<nowiki>+3.1</nowiki>
|
|
|
|-
|<nowiki>+6.1</nowiki>
|
|
|
|}

MRD:ED2618

2020-05-15T14:30:40Z

Ed2618: /* EXERCISE 1: H + H2 system */

= '''Molecular Reaction Dynamics''' =
In this lab we are using molecular dynamic trajectories to study the reactivity of triatomic systems, namely when an atom collides with a diatomic molecule.

== EXERCISE 1: H + H2 system ==
For this exercise, the following reaction is analysed and discussed: HA + HBHC → HAHB + HC, where r1 = BC and r2 = AB.

=== Dynamics from the transition state region ===
'''Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?'''

'''Answer 1:''' On a potential energy surface diagram the transition state is defined, mathematically, as a saddle point. A transition state is the maximum energy present in the minimum energy pathway taken from reactants to products. A saddle point is of symmetric configuration at which the distance between three separate atoms is the same.

A saddle point is different to a local minima. For both, the gradient of the potential energy surface is zero, is given as the first partial derivative:

∂V(ri)/ ∂ri = 0

However, the saddle point and local minima may be distinguished though inspection of their second partial derivatives. For a local minima, calculating partial second derivatives in orthogonal directions to the saddle point will give a result that is greater than 0, independent from which plane it has been measured. By contrast, for the saddle point, is second partial derivative will either be greater than zero or smaller than zero dependent on which plane it has been measured. This explanation is summarised below:

For a local minima: ∂2V(ri)/ ∂ri2 > 0

For a saddle point: ∂2V(ri)/ ∂ri2 > 0 and ∂2V(ri)/ ∂ri2 < 0 (INSERT REFERENCE)

An example of a saddle point can be seen in the figure below:
[[File:Saddle plot ED2618.png|thumb|440x440px|Figure 1: An example of a saddle point, orthogonal directions from which have opposite signs for their partial second derivative. |none]]

==== Trajectories from r1 = r2: locating the transition state ====
'''Question 2: Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. (REDO IF TIME - ENSURE FORCES ON RHS = 0)'''

'''Answer 2: '''The transition state occurs at the saddle point, which is of symmetric configuration. Hence, r1 = r2 as the H + H2 surface is symmetric, where, r1 = BC and r2 = AB. At the transition state p1 = p2 = 0 g.mol-1.pm.fs-1. Through trial and error an estimate for the transition state position is when AB = BC = 90.75 pm. Using this distance produces the contour plot below. It can be seen that there is no gradient/trajectory of motion at right angles to the point of the transition state (red cross).
[[File:Transition State Contour Plot ED2618.png|none|thumb|440x440px|Figure 2: A contour plot showing the transition state. There are no oscillations or directions of motion towards either reactants or products, thus the system is stationary]]
The plot of 'Internuclear Distances vs Time' below shows the distances of A-B, B-C, and A-C constant with time, consolidating the fact that this position is the transition state position. Running an animation with the r1 = r2 = 90.75 pm shows the species as stationary, as there are no fluctuations of interatomic distance with time.
[[File:ED2618 IND vs Time.png|none|thumb|489x489px|Figure 3: A plot of Internuclear Distances vs Time]]

==== Trajectories from r1 = rts + δ, r2 = rts ====
'''Question 3: Comment on how the ''mep'' and the trajectory you just calculated differ.'''

'''Answer 3: '''The initial conditions are set so that the system is slightly displaced from the transition state and so that there is zero initial momenta, where r1 = rts + 1 pm, r2 = rts and the momenta p1 = p2 = 0 g.mol-1.pm.fs-1. This can be shown in the figures below, where r1 (BC) = 91.75 pm and r2 (AB) = 90.75 pm. To obtain a complete MEP the step number was increased to 3000.
{| class="wikitable"
|[[File:ED2618 Q3 Dynamic.png|none|thumb|366x366px|Figure 4a: Contour Plot (Calculation Type: Dynamics)]]
|
|[[File:ED2618 Q3 MEP.png|none|thumb|366x366px|Figure 5a: Contour Plot (Calculation Type: MEP)]]
|-
|
|
|
|-
|[[File:ED261 Q3 Dynamic INDvsT.png|none|thumb|366x366px|Figure 4b: Internuclear Distance vs Time Plot (Calculation Type: Dynamics)]]
|
|[[File:ED2618 Q3 MEP INDvsT.png|none|thumb|366x366px|Figure 5b: Internuclear Distance vs Time Plot (Calculation Type: MEP)]]
|-
|
|
|
|-
|[[File:ED2618 Q3 Dynamic MvsT.png|none|thumb|366x366px|Figure 4c: Momenta vs Time Plot (Calculation Type: Dynamics)]]
|
|[[File:ED2618 Q3 MEP MvsT.png|none|thumb|366x366px|Figure 5c: Momenta vs Time Plot (Calculation Type: MEP)]]
|}
It can be deduced that the trajectory is greatly dependent on calculation type. Using the MEP calculation, the momenta are reset to zero after each step. This provides an unrealistic account of the motion of the atoms during the reaction. This explains the absence of any type of oscillation in the MEP calculation, Figure 5a, compared to the Dynamics calculation figure. The Dynamics calculation accounts for the vibrational energies possessed by the molecules, hence explaining the presence of oscillations along the trajectory in Figure 4a.

Plots of "Internuclear Distances vs Time” and “Momenta vs Time” are also provided and more clearly depict the differences in the calculation types.

The Dynamics calculation accounts for the extra kinetic energy obtained after bond formation, which manifests itself into excess vibrational energy. This can be seen in the small oscillations of the A-B bond in figure 4b. Using the MEP calculation, this A-B bond distance remain constant over time, indicating this excess vibrational energy has not been accounted for. The increased momentum of the B-C bond is indicative of HC moving with translational energy and the oscillating momentum of the A-B bond is indicative of the new HAHB bond moving with vibrational energy.

Figure 5c shows a straight line at 0. The momenta is zero for the reaction pathway. This calculation type resets the momenta to zero after every step hence this result is expected. Figure 4c shows an oscillation of momentum with time, as through this calculation type, momenta is not reset to zero after each step.

If the initial conditions were changed so that r1 = rts pm and r2 = rts + 1pm, the blue and orange lines would swap positions in figures 4b, 4c, 5b, and 5c.

=== Reactive and unreactive trajectories ===
'''Question 4: Complete the table below by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?'''

With the initial positions r1 = 74 pm and r2 = 200 pm, trajectories were run with the following momenta combination:
{| class="wikitable"
!Row Number
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot/ kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|1
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|The kinetic energy of the system is sufficient to over come the activation barrier and hence the reaction proceeds. Originally, the reactants have little vibrational energy, reflected by the lack of oscillatory behaviour. After the reaction occurs, the products have more oscillatory behaviour, meaning that they have greater vibrational energy. The BC bond is broken and the AB bond is formed.
|[[File:ED2618 Q4 C1.png|frameless|366x366px]]
|-
|2
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|HA approaches an oscillating HBHC molecule, however the activation energy exceeds the kinetic energy and thus the reaction does not occur. Both reactants remain unchanged.
|[[File:ED2618 Q4 C2.png|frameless|366x366px]]
|-
|3
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|This trajectory is similar to that of row 1, with the only difference being that the reactants possess more vibrational energy, hence the presence of oscillatory behaviour as depicted in the figure.
|[[File:ED2618 Q4 C3.png|frameless|366x366px]]
|-
|4
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|For this trajectory, the kinetic energy is high and so the activation barrier is crossed and the products are formed. However, these products contain an excess of vibrational energy, namely the newly formed HAHB bond, which oscillates with such a great strength that the bond breaks and the reactants of the reaction are reformed. The reactants then move away from one another with greater vibrational energy than when the reaction began.
|[[File:ED2618 Q4 C4.png|frameless|366x366px]]
|-
|5
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|This trajectory is similar to that of the row above, where the kinetic energy is sufficient to overcome the activation barrier, however the newly formed HAHB bond possess so much vibrational energy that it dissociates back to the products (HBHC and HA). The only difference this time is that the products are reformed as the system has a high vibrational energy. The reactant HBHC bond is broken for a second time.
|[[File:ED2618 Q4 C5.png|none|thumb|366x366px]]
|}
From this table it may be concluded that the reactants must have sufficient energy to overcome the activation barrier to reach and surpass the transition state. This, however, is not the only factor that has to be considered for reaction success. The products may not have too much vibrational energy or the will convert back to the reactants. This example is observed in row 4 of the above table.

==== Transition State Theory (ADD REFERENCE) ====
'''Question 5: Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''

'''Answer 1: '''Transition State Theory is derived through the postulated quasi-equilibrium between a transition state species and the reactant molecules. There are a number of assumptions that are made upon derivation of the Transition State Theory rate expression:
# Electronic and nuclear motions are separated, similar to the Born-Oppenheimer approximation in quantum mechanics.
# Reactant molecule are distributed among their states in a way that is in accordance with the Maxwell-Boltzmann distribution.
# Once a molecular system has crossed a transition state, in a direction towards the products, it is unable to turn around and re-form the reactants.
# In the transition state, any motion that occurs along the reaction coordinate can be treated independently of other motions. This motion is treated classically as a translation.
# Even in the absence of an equilibrium between reactant and product molecules, the transition states that en route to becoming products are distributed among their states in accordance with the Maxwell-Boltzmann laws.
In rows 4 and 5 of the [[MRD:ED2618#Reactive and unreactive trajectories|table above]], the transition state is crossed and re-crossed numerous times, thus contradicting the assumption made by the Transition State Theory that a molecular system cannot regress back to the reactants once it has passed the transition state towards the products. The recrossing of the transition state ultimately lowers the rate of reaction, and thus Transition State Theory predicts a reaction rate that is larger than what is experimentally observed. Transition State Theory fails to evaluate molecular systems using quantum mechanical phenomena such as tunnelling, hence highlighting the discrepancy between the observed and theoretical reaction rates.

== EXERCISE 2: F - H - H system ==
In this exercise A=F, B=H, and C=H. r1 = BC and r2 = AB.

=== PES inspection ===
'''Question 1: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''

'''Answer 1: '''In this discussion, atoms F, H, and H are consigned to A, B, and C respectively. An increasing BC distance means that A and B are bonded, thus the HF + H system is consigned to this axis. An increasing AB distance means that B and C are bonded, thus this axis is consigned to the F + H2 system. As shown in the surface plot below, the HF + H system is lower in energy than the F + H2 system.

Thus the reaction:
* HF + H → H2 + F is endothermic. (Products higher in energy than reactants)

* F + H2 → HF + H is exothermic. (Reactants higher in energy than products)
[[File:ED2618 Surface Plot FHH.png|none|thumb|320x320px|Figure 6: Surface plot of FHH system]]
The HF (565 kJ.mol-1) bond is much stronger than the HH bond (436 kJ.mol-1) (ADD REFERENCE - PHYSICAL CHEM TEXTBOOK pG 885), thus the HF molecule is more stable than the H2 molecules. This is due to the large difference in electronegativity between H and F, resulting in an ionic bond. This is a stronger interaction than a purely covalent one that occurs between the two hydrogen atoms.

'''Question 2: Locate the approximate position of the transition state.'''

'''Answer 2: '''Hammond's postulate states the the transition state of a given reaction will resemble either the products or the reactants. The transition state will resemble either the products or the reactants, depending on which on is higher in energy, as the transition state is the maximum energy present in the minimum energy pathway taken from reactants to products. Thus, for an endothermic reaction, the products are higher in energy than the reactants, and so the transition state will resemble the products. Conversely, for an exothermic reaction, the reactants are higher in energy than the products, and so the transition state will resemble the reactants. (REFERENCE HAMMOND'S POSTULATE?)

The reaction of HF + H → H2 + F is endothermic and thus the transition state will resemble the products in structure. Thus the HH bond distance will be set to around 74.14 pm (ADD REFERENCE - ATKINS PHYSICAL CHEM PAGE 362)

F + H2 → HF + H is exothermic and thus the transition state will resemble the reactants in structure. Hence, the HH bond distance will be set around 74.14 pm, as above.

The value of the HF bond distance is expected to be much greater than 91.68 (ADD REFERENCE) as there is limited association between the two atoms.

The transition state was found when r1 (BC) = 74.485 and pm r2 (AB) = 181.200 pm. At this position there are no forces acting along either AB or BC.
{| class="wikitable"
|[[File:ED2618 Q2 TS Contour.png|none|thumb|320x320px|Figure 7: Contour plot of transition state of FHH system]]
|
|[[File:ED2618 Q2 TS INDvsT.png|none|thumb|320x320px|Figure 8: Internuclear Distance vs Time plot of transition state for FHH system]]
|}
Figure 7 above shows that there is no gradient/trajectory of motion at right angles to the point of the transition state (red cross). The plot of 'Internuclear Distances vs Time' in Figure 8 shows the distances of A-B, B-C, and A-C constant with time, consolidating the fact that this position is the transition state position. These lines are straight and have no oscillatory character, indicating no intermolecular vibrations are occurring.

'''Question 3: Report the activation energy for both reactions.'''

'''Answer 3: '''The kinetic energy of the transition state was calculated to be -433.981 kJ.mol-1. The activation energy can be found through running an MEP calculation of a structure slightly offset from that of the transition state. The initial positions were set so that:

For the exothermic reaction where F + H2 → HF + H
* r1 (BC) = 74.485 pm = rts
* r2 (AB) = 182.200 pm = rts + 1 pm
* p1 = p2 = 0 g.mol-1.pm.fs-1

For the endothermic reaction where HF + H → H2 + F
* r1 (BC) = 74.485 pm = rts
* r2 (AB) = 180.200 pm = rts - 1 pm
* p1 = p2 = 0 g.mol-1.pm.fs-1
{| class="wikitable"
!Reaction
!Energy of Reactants / kJ.mol-1
!Transition State Energy / kJ.mol-1
!Activation Energy / kJ.mol-1
!Energy vs Time Plot of Reaction Trajectory
|-
|F + H2 → HF + H
|<nowiki>- 435.5</nowiki>
|<nowiki>- 434.0</nowiki>
|<nowiki>-1.0</nowiki>
|[[File:ED2618 Q3 Exo.png|frameless|400x400px]]
|-
|HF + H → H2 + F
|<nowiki>- 560.5</nowiki>
|<nowiki>- 434.0</nowiki>
|<nowiki>-126.5</nowiki>
|[[File:ED2618 Q3 Endo.png|frameless|400x400px]]
|}
EA POSITIVE OR NEGATIVE??

=== Reaction dynamics ===
'''Question 4: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.'''

'''Answer 4: '''The set of initial conditions that results in a reactive trajectory for the reaction F + H2 → HF + H is given as follows:
* r1 (BC) = 74 pm
* r2 (AB) = 230 pm
* p1 (BC) = - 0.75 g.mol-1.pm.fs-1
* p2 (AB) = -2.00 0 g.mol-1.pm.fs-1
* 2000 steps at 0.1 fs
This reactive trajectory is show in the contour plot below:
[[File:ED2618 Q4 Contour.png|none|thumb|320x320px|Figure 9: Contour plot of initial conditions that result in a reactive trajectory for F + H2 system]]
The reaction F + H2 → HF + H is exothermic, meaning that the products possess a lower energy than the reactants. The energy discrepancy between reactants and products is released as heat to the surroundings. Figure 10 shows a plot of momenta vs time for this system. The H2 molecule approaches the F atom with low oscillatory momenta. The HH bond is broken and a new HF bond is formed. The energy released upon an HH bond breaking is greater than the energy required to form the HF bond. Upon collision of the molecules, the HF bond now oscillates with a much greater aptitude than the H2 had done prior to collision, due to the release of excess energy. The unexpected drop in momenta at t = 110 fs is due to recrossing of the transition state, as can be seen in Figure 9. After collision, the H atom moves off with a much slower velocity, as is indicated by the straight line in Figure 10. (ENERGY TO TRANSLATION AND VIBRATIONAL AS PRODUCT MOVES FAST?)
[[File:ED2618 Q4 Momenta.png|none|thumb|320x320px|Figure 10: Momenta vs Time plot of initial conditions that result in a reactive trajectory for F + H2 system]]
As energy must always be conserved, the increased kinetic energy of the system is compensated for by a decrease in potential energy, as indicated in Figure 11. This plot shows that the total energy of the system remains constant at - 433.372 kJ.mol-1. There is no loss of energy, but instead it is being converted from form to another, in line with the law of conservation of energy.
[[File:ED2618 Q4 Energy.png|none|thumb|320x320px|Figure 11: Energy vs Time plot of initial conditions that result in a reactive trajectory for F + H2 system]]
As this reaction is exothermic, energy is released to the surroundings in the form of heat. To confirm experimentally that this reaction has occurred, calorimetry could be used. Calorimetry may be used to identify changes in enthalpy of the proceeding reaction. However, this method may underestimate the heat energy released by the reaction as heat is lost to the environment.

Another method to confirm that the reaction has gone to completion is through analysis of IR data. The vibrational energy of the HF molecule after collision is greatly increased after collision of F with H2. This would produce overtone bands as the HF molecule is excited from a lower vibrational energy to a higher vibrational energy.

'''Question 5: Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.'''

'''Answer 5: '''Polanyi's empirical rules explain the dependence of a reaction's success on the way energy of the reactants is distributed in different modes. For an exothermic reaction, the reactants are higher in energy than the products, thus the reactants resemble the transition state more than the products do. This system is said to have an early transition state. For a reaction to be successful, not only must the reactants have sufficient energy to overcome the activation barrier, they must also possess the right type of energy.

For an exothermic reaction, one possessing an early transition state, Polanyi's rules state that translational energy proves more useful than vibrational energy when overcoming the energy barrier from reactants to products. Conversely, for an endothermic reaction with a late transition state, vibrational energy is more critical than translational energy when overcoming this energy barrier. (ADD REFERENCE)

The reaction reaction F + H2 → HF + H is exothermic, thus reactants with more translational energy as opposed to vibrational energy should prove effective at overcoming the energy barrier of the transition state. A calculation is set up for this reaction with r1 (BC) = 74 pm

Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, with a momentum pFH = -1.0 g.mol-1.pm.fs-1, and explore several values of pHH in the range -6.1 to 6.1 g.mol-1.pm.fs-1 (explore values also close to these limits). What do you observe? Note that we are putting a significant amount of energy (much more than the activation energy) into the system on the H - H vibration

MRD:ED2618

2020-05-15T13:37:02Z

Ed2618: /* Reaction dynamicsQuestion 4: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally. */

= '''Molecular Reaction Dynamics''' =
In this lab we are using molecular dynamic trajectories to study the reactivity of triatomic systems, namely when an atom collides with a diatomic molecule.

== EXERCISE 1: H + H2 system ==
For this exercise, the following reaction is analysed and discussed: HA + HBHC → HAHB + HC, where r1 = BC and r2 = AB.

=== Dynamics from the transition state region ===
'''Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?'''

'''Answer 1:''' On a potential energy surface diagram the transition state is defined, mathematically, as a saddle point. A transition state is the maximum energy present in the minimum energy pathway taken from reactants to products. A saddle point is of symmetric configuration at which the distance between three separate atoms is the same.

A saddle point is different to a local minima. For both, the gradient of the potential energy surface is zero, is given as the first partial derivative:

∂V(ri)/ ∂ri = 0

However, the saddle point and local minima may be distinguished though inspection of their second partial derivatives. For a local minima, calculating partial second derivatives in orthogonal directions to the saddle point will give a result that is greater than 0, independent from which plane it has been measured. By contrast, for the saddle point, is second partial derivative will either be greater than zero or smaller than zero dependent on which plane it has been measured. This explanation is summarised below:

For a local minima: ∂2V(ri)/ ∂ri2 > 0

For a saddle point: ∂2V(ri)/ ∂ri2 > 0 and ∂2V(ri)/ ∂ri2 < 0 (INSERT REFERENCE)

An example of a saddle point can be seen in the figure below:
[[File:Saddle plot ED2618.png|thumb|440x440px|Figure 1: An example of a saddle point, orthogonal directions from which have opposite signs for their partial second derivative. |none]]

==== Trajectories from r1 = r2: locating the transition state ====
'''Question 2: Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. (REDO IF TIME - ENSURE FORCES ON RHS = 0)'''

'''Answer 2: '''The transition state occurs at the saddle point, which is of symmetric configuration. Hence, r1 = r2 as the H + H2 surface is symmetric, where, r1 = BC and r2 = AB. At the transition state p1 = p2 = 0 g.mol-1.pm.fs-1. Through trial and error an estimate for the transition state position is when AB = BC = 90.75 pm. Using this distance produces the contour plot below. It can be seen that there is no gradient/trajectory of motion at right angles to the point of the transition state (red cross).
[[File:Transition State Contour Plot ED2618.png|none|thumb|440x440px|Figure 2: A contour plot showing the transition state. There are no oscillations or directions of motion towards either reactants or products, thus the system is stationary]]
The plot of 'Internuclear Distances vs Time' below shows the distances of A-B, B-C, and A-C constant with time, consolidating the fact that this position is the transition state position. Running an animation with the r1 = r2 = 90.75 pm shows the species as stationary, as there are no fluctuations of interatomic distance with time.
[[File:ED2618 IND vs Time.png|none|thumb|489x489px|Figure 3: A plot of Internuclear Distances vs Time]]

==== Trajectories from r1 = rts + δ, r2 = rts ====
'''Question 3: Comment on how the ''mep'' and the trajectory you just calculated differ.'''

'''Answer 3: '''The initial conditions are set so that the system is slightly displaced from the transition state and so that there is zero initial momenta, where r1 = rts + 1 pm, r2 = rts and the momenta p1 = p2 = 0 g.mol-1.pm.fs-1. This can be shown in the figures below, where r1 (BC) = 91.75 pm and r2 (AB) = 90.75 pm. To obtain a complete MEP the step number was increased to 3000.
{| class="wikitable"
|[[File:ED2618 Q3 Dynamic.png|none|thumb|366x366px|Figure 4a: Contour Plot (Calculation Type: Dynamics)]]
|
|[[File:ED2618 Q3 MEP.png|none|thumb|366x366px|Figure 5a: Contour Plot (Calculation Type: MEP)]]
|-
|
|
|
|-
|[[File:ED261 Q3 Dynamic INDvsT.png|none|thumb|366x366px|Figure 4b: Internuclear Distance vs Time Plot (Calculation Type: Dynamics)]]
|
|[[File:ED2618 Q3 MEP INDvsT.png|none|thumb|366x366px|Figure 5b: Internuclear Distance vs Time Plot (Calculation Type: MEP)]]
|-
|
|
|
|-
|[[File:ED2618 Q3 Dynamic MvsT.png|none|thumb|366x366px|Figure 4c: Momenta vs Time Plot (Calculation Type: Dynamics)]]
|
|[[File:ED2618 Q3 MEP MvsT.png|none|thumb|366x366px|Figure 5c: Momenta vs Time Plot (Calculation Type: MEP)]]
|}
It can be deduced that the trajectory is greatly dependent on calculation type. Using the MEP calculation, the momenta are reset to zero after each step. This provides an unrealistic account of the motion of the atoms during the reaction. This explains the absence of any type of oscillation in the MEP calculation, Figure 5a, compared to the Dynamics calculation figure. The Dynamics calculation accounts for the vibrational energies possessed by the molecules, hence explaining the presence of oscillations along the trajectory in Figure 4a.

Plots of "Internuclear Distances vs Time” and “Momenta vs Time” are also provided and more clearly depict the differences in the calculation types.

The Dynamics calculation accounts for the extra kinetic energy obtained after bond formation, which manifests itself into excess vibrational energy. This can be seen in the small oscillations of the A-B bond in figure 4b. Using the MEP calculation, this A-B bond distance remain constant over time, indicating this excess vibrational energy has not been accounted for. The increased momentum of the B-C bond is indicative of HC moving with translational energy and the oscillating momentum of the A-B bond is indicative of the new HAHB bond moving with vibrational energy.

Figure 5c shows a straight line at 0. The momenta is zero for the reaction pathway. This calculation type resets the momenta to zero after every step hence this result is expected. Figure 4c shows an oscillation of momentum with time, as through this calculation type, momenta is not reset to zero after each step.

If the initial conditions were changed so that r1 = rts pm and r2 = rts + 1pm, the blue and orange lines would swap positions in figures 4b, 4c, 5b, and 5c.

=== Reactive and unreactive trajectories ===
'''Question 4: Complete the table below by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?'''

With the initial positions r1 = 74 pm and r2 = 200 pm, trajectories were run with the following momenta combination:
{| class="wikitable"
!Row Number
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot/ kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|1
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|The kinetic energy of the system is sufficient to over come the activation barrier and hence the reaction proceeds. Originally, the reactants have little vibrational energy, reflected by the lack of oscillatory behaviour. After the reaction occurs, the products have more oscillatory behaviour, meaning that they have greater vibrational energy. The BC bond is broken and the AB bond is formed.
|[[File:ED2618 Q4 C1.png|frameless|366x366px]]
|-
|2
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|HA approaches an oscillating HBHC molecule, however the activation energy exceeds the kinetic energy and thus the reaction does not occur. Both reactants remain unchanged.
|[[File:ED2618 Q4 C2.png|frameless|366x366px]]
|-
|3
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|This trajectory is similar to that of row 1, with the only difference being that the reactants possess more vibrational energy, hence the presence of oscillatory behaviour as depicted in the figure.
|[[File:ED2618 Q4 C3.png|frameless|366x366px]]
|-
|4
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|For this trajectory, the kinetic energy is high and so the activation barrier is crossed and the products are formed. However, these products contain an excess of vibrational energy, namely the newly formed HAHB bond, which oscillates with such a great strength that the bond breaks and the reactants of the reaction are reformed. The reactants then move away from one another with greater vibrational energy than when the reaction began.
|[[File:ED2618 Q4 C4.png|frameless|366x366px]]
|-
|5
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|This trajectory is similar to that of the row above, where the kinetic energy is sufficient to overcome the activation barrier, however the newly formed HAHB bond possess so much vibrational energy that it dissociates back to the products (HBHC and HA). The only difference this time is that the products are reformed as the system has a high vibrational energy. The reactant HBHC bond is broken for a second time.
|[[File:ED2618 Q4 C5.png|none|thumb|366x366px]]
|}
From this table it may be concluded that the reactants must have sufficient energy to overcome the activation barrier to reach and surpass the transition state. This, however, is not the only factor that has to be considered for reaction success. The products may not have too much vibrational energy or the will convert back to the reactants. This example is observed in row 4 of the above table.

==== Transition State Theory (ADD REFERENCE) ====
'''Question 5: Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''

'''Answer 1: '''Transition State Theory is derived through the postulated quasi-equilibrium between a transition state species and the reactant molecules. There are a number of assumptions that are made upon derivation of the Transition State Theory rate expression:
# Electronic and nuclear motions are separated, similar to the Born-Oppenheimer approximation in quantum mechanics.
# Reactant molecule are distributed among their states in a way that is in accordance with the Maxwell-Boltzmann distribution.
# Once a molecular system has crossed a transition state, in a direction towards the products, it is unable to turn around and re-form the reactants.
# In the transition state, any motion that occurs along the reaction coordinate can be treated independently of other motions. This motion is treated classically as a translation.
# Even in the absence of an equilibrium between reactant and product molecules, the transition states that en route to becoming products are distributed among their states in accordance with the Maxwell-Boltzmann laws.
In rows 4 and 5 of the [[MRD:ED2618#Reactive and unreactive trajectories|table above]], the transition state is crossed and re-crossed numerous times, thus contradicting the assumption made by the Transition State Theory that a molecular system cannot regress back to the reactants once it has passed the transition state towards the products. The recrossing of the transition state ultimately lowers the rate of reaction, and thus Transition State Theory predicts a reaction rate that is larger than what is experimentally observed. Transition State Theory fails to evaluate molecular systems using quantum mechanical phenomena such as tunnelling, hence highlighting the discrepancy between the observed and theoretical reaction rates.

== EXERCISE 2: F - H - H system ==
In this exercise A=F, B=H, and C=H. r1 = BC and r2 = AB.

=== PES inspection ===
'''Question 1: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''

'''Answer 1: '''In this discussion, atoms F, H, and H are consigned to A, B, and C respectively. An increasing BC distance means that A and B are bonded, thus the HF + H system is consigned to this axis. An increasing AB distance means that B and C are bonded, thus this axis is consigned to the F + H2 system. As shown in the surface plot below, the HF + H system is lower in energy than the F + H2 system.

Thus the reaction:
* HF + H → H2 + F is endothermic. (Products higher in energy than reactants)

* F + H2 → HF + H is exothermic. (Reactants higher in energy than products)
[[File:ED2618 Surface Plot FHH.png|none|thumb|320x320px|Figure 6: Surface plot of FHH system]]
The HF (565 kJ.mol-1) bond is much stronger than the HH bond (436 kJ.mol-1) (ADD REFERENCE - PHYSICAL CHEM TEXTBOOK pG 885), thus the HF molecule is more stable than the H2 molecules. This is due to the large difference in electronegativity between H and F, resulting in an ionic bond. This is a stronger interaction than a purely covalent one that occurs between the two hydrogen atoms.

'''Question 2: Locate the approximate position of the transition state.'''

'''Answer 2: '''Hammond's postulate states the the transition state of a given reaction will resemble either the products or the reactants. The transition state will resemble either the products or the reactants, depending on which on is higher in energy, as the transition state is the maximum energy present in the minimum energy pathway taken from reactants to products. Thus, for an endothermic reaction, the products are higher in energy than the reactants, and so the transition state will resemble the products. Conversely, for an exothermic reaction, the reactants are higher in energy than the products, and so the transition state will resemble the reactants. (REFERENCE HAMMOND'S POSTULATE?)

The reaction of HF + H → H2 + F is endothermic and thus the transition state will resemble the products in structure. Thus the HH bond distance will be set to around 74.14 pm (ADD REFERENCE - ATKINS PHYSICAL CHEM PAGE 362)

F + H2 → HF + H is exothermic and thus the transition state will resemble the reactants in structure. Hence, the HH bond distance will be set around 74.14 pm, as above.

The value of the HF bond distance is expected to be much greater than 91.68 (ADD REFERENCE) as there is limited association between the two atoms.

The transition state was found when r1 (BC) = 74.485 and pm r2 (AB) = 181.200 pm. At this position there are no forces acting along either AB or BC.
{| class="wikitable"
|[[File:ED2618 Q2 TS Contour.png|none|thumb|320x320px|Figure 7: Contour plot of transition state of FHH system]]
|
|[[File:ED2618 Q2 TS INDvsT.png|none|thumb|320x320px|Figure 8: Internuclear Distance vs Time plot of transition state for FHH system]]
|}
Figure 7 above shows that there is no gradient/trajectory of motion at right angles to the point of the transition state (red cross). The plot of 'Internuclear Distances vs Time' in Figure 8 shows the distances of A-B, B-C, and A-C constant with time, consolidating the fact that this position is the transition state position. These lines are straight and have no oscillatory character, indicating no intermolecular vibrations are occurring.

'''Question 3: Report the activation energy for both reactions.'''

'''Answer 3: '''The kinetic energy of the transition state was calculated to be -433.981 kJ.mol-1. The activation energy can be found through running an MEP calculation of a structure slightly offset from that of the transition state. The initial positions were set so that:

For the exothermic reaction where F + H2 → HF + H
* r1 (BC) = 74.485 pm = rts
* r2 (AB) = 182.200 pm = rts + 1 pm
* p1 = p2 = 0 g.mol-1.pm.fs-1

For the endothermic reaction where HF + H → H2 + F
* r1 (BC) = 74.485 pm = rts
* r2 (AB) = 180.200 pm = rts - 1 pm
* p1 = p2 = 0 g.mol-1.pm.fs-1
{| class="wikitable"
!Reaction
!Energy of Reactants / kJ.mol-1
!Transition State Energy / kJ.mol-1
!Activation Energy / kJ.mol-1
!Energy vs Time Plot of Reaction Trajectory
|-
|F + H2 → HF + H
|<nowiki>- 435.5</nowiki>
|<nowiki>- 434.0</nowiki>
|<nowiki>-1.0</nowiki>
|[[File:ED2618 Q3 Exo.png|frameless|400x400px]]
|-
|HF + H → H2 + F
|<nowiki>- 560.5</nowiki>
|<nowiki>- 434.0</nowiki>
|<nowiki>-126.5</nowiki>
|[[File:ED2618 Q3 Endo.png|frameless|400x400px]]
|}

=== Reaction dynamics ===
'''Question 4: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.'''

'''Answer 4: '''The set of initial conditions that results in a reactive trajectory for the reaction F + H2 → HF + H is given as follows:
* r1 (BC) = 74 pm
* r2 (AB) = 230 pm
* p1 (BC) = - 0.75 g.mol-1.pm.fs-1
* p2 (AB) = -2.00 0 g.mol-1.pm.fs-1
* 2000 steps at 0.1 fs
This reactive trajectory is show in the contour plot below:
[[File:ED2618 Q4 Contour.png|none|thumb|320x320px|Figure 9: Contour plot of initial conditions that result in a reactive trajectory for F + H2 system]]
The reaction F + H2 → HF + H is exothermic, meaning that the products possess a lower energy than the reactants. The energy discrepancy between reactants and products is released as heat to the surroundings. Figure 10 shows a plot of momenta vs time for this system. The H2 molecule approaches the F atom with low oscillatory momenta. The HH bond is broken and a new HF bond is formed. The energy released upon an HH bond breaking is greater than the energy required to form the HF bond. Upon collision of the molecules, the HF bond now oscillates with a much greater aptitude than the H2 had done prior to collision, due to the release of excess energy. The unexpected drop in momenta at t = 110 fs is due to recrossing of the transition state, as can be seen in Figure 9. After collision, the H atom moves off with a much slower velocity, as is indicated by the straight line in Figure 10. (ENERGY TO TRANSLATION AND VIBRATIONAL AS PRODUCT MOVES FAST?)
[[File:ED2618 Q4 Momenta.png|none|thumb|320x320px|Figure 10: Momenta vs Time plot of initial conditions that result in a reactive trajectory for F + H2 system]]
As energy must always be conserved, the increased kinetic energy of the system is compensated for by a decrease in potential energy, as indicated in Figure 11. This plot shows that the total energy of the system remains constant at - 433.372 kJ.mol-1. There is no loss of energy, but instead it is being converted from form to another, in line with the law of conservation of energy.
[[File:ED2618 Q4 Energy.png|none|thumb|320x320px|Figure 11: Energy vs Time plot of initial conditions that result in a reactive trajectory for F + H2 system]]
As this reaction is exothermic, energy is released to the surroundings in the form of heat. To confirm experimentally that this reaction has occurred, calorimetry could be used. Calorimetry may be used to identify changes in enthalpy in the surrounding in the immediate vicinity of the reaction. Another method to confirm that the reaction has gone to completion is through analysis of IR data. The vibrational energy of the HF molecule after collision is greatly increased after collision of F with H2. This would produce overtone bands as the HF molecule is excited to higher vibrational states.

'''Question 5: Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.'''

'''Answer 5:'''

MRD:ED2618

2020-05-15T13:33:29Z

Ed2618: /* Reaction dynamicsQuestion 4: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally. */

= '''Molecular Reaction Dynamics''' =
In this lab we are using molecular dynamic trajectories to study the reactivity of triatomic systems, namely when an atom collides with a diatomic molecule.

== EXERCISE 1: H + H2 system ==
For this exercise, the following reaction is analysed and discussed: HA + HBHC → HAHB + HC, where r1 = BC and r2 = AB.

=== Dynamics from the transition state region ===
'''Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?'''

'''Answer 1:''' On a potential energy surface diagram the transition state is defined, mathematically, as a saddle point. A transition state is the maximum energy present in the minimum energy pathway taken from reactants to products. A saddle point is of symmetric configuration at which the distance between three separate atoms is the same.

A saddle point is different to a local minima. For both, the gradient of the potential energy surface is zero, is given as the first partial derivative:

∂V(ri)/ ∂ri = 0

However, the saddle point and local minima may be distinguished though inspection of their second partial derivatives. For a local minima, calculating partial second derivatives in orthogonal directions to the saddle point will give a result that is greater than 0, independent from which plane it has been measured. By contrast, for the saddle point, is second partial derivative will either be greater than zero or smaller than zero dependent on which plane it has been measured. This explanation is summarised below:

For a local minima: ∂2V(ri)/ ∂ri2 > 0

For a saddle point: ∂2V(ri)/ ∂ri2 > 0 and ∂2V(ri)/ ∂ri2 < 0 (INSERT REFERENCE)

An example of a saddle point can be seen in the figure below:
[[File:Saddle plot ED2618.png|thumb|440x440px|Figure 1: An example of a saddle point, orthogonal directions from which have opposite signs for their partial second derivative. |none]]

==== Trajectories from r1 = r2: locating the transition state ====
'''Question 2: Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. (REDO IF TIME - ENSURE FORCES ON RHS = 0)'''

'''Answer 2: '''The transition state occurs at the saddle point, which is of symmetric configuration. Hence, r1 = r2 as the H + H2 surface is symmetric, where, r1 = BC and r2 = AB. At the transition state p1 = p2 = 0 g.mol-1.pm.fs-1. Through trial and error an estimate for the transition state position is when AB = BC = 90.75 pm. Using this distance produces the contour plot below. It can be seen that there is no gradient/trajectory of motion at right angles to the point of the transition state (red cross).
[[File:Transition State Contour Plot ED2618.png|none|thumb|440x440px|Figure 2: A contour plot showing the transition state. There are no oscillations or directions of motion towards either reactants or products, thus the system is stationary]]
The plot of 'Internuclear Distances vs Time' below shows the distances of A-B, B-C, and A-C constant with time, consolidating the fact that this position is the transition state position. Running an animation with the r1 = r2 = 90.75 pm shows the species as stationary, as there are no fluctuations of interatomic distance with time.
[[File:ED2618 IND vs Time.png|none|thumb|489x489px|Figure 3: A plot of Internuclear Distances vs Time]]

==== Trajectories from r1 = rts + δ, r2 = rts ====
'''Question 3: Comment on how the ''mep'' and the trajectory you just calculated differ.'''

'''Answer 3: '''The initial conditions are set so that the system is slightly displaced from the transition state and so that there is zero initial momenta, where r1 = rts + 1 pm, r2 = rts and the momenta p1 = p2 = 0 g.mol-1.pm.fs-1. This can be shown in the figures below, where r1 (BC) = 91.75 pm and r2 (AB) = 90.75 pm. To obtain a complete MEP the step number was increased to 3000.
{| class="wikitable"
|[[File:ED2618 Q3 Dynamic.png|none|thumb|366x366px|Figure 4a: Contour Plot (Calculation Type: Dynamics)]]
|
|[[File:ED2618 Q3 MEP.png|none|thumb|366x366px|Figure 5a: Contour Plot (Calculation Type: MEP)]]
|-
|
|
|
|-
|[[File:ED261 Q3 Dynamic INDvsT.png|none|thumb|366x366px|Figure 4b: Internuclear Distance vs Time Plot (Calculation Type: Dynamics)]]
|
|[[File:ED2618 Q3 MEP INDvsT.png|none|thumb|366x366px|Figure 5b: Internuclear Distance vs Time Plot (Calculation Type: MEP)]]
|-
|
|
|
|-
|[[File:ED2618 Q3 Dynamic MvsT.png|none|thumb|366x366px|Figure 4c: Momenta vs Time Plot (Calculation Type: Dynamics)]]
|
|[[File:ED2618 Q3 MEP MvsT.png|none|thumb|366x366px|Figure 5c: Momenta vs Time Plot (Calculation Type: MEP)]]
|}
It can be deduced that the trajectory is greatly dependent on calculation type. Using the MEP calculation, the momenta are reset to zero after each step. This provides an unrealistic account of the motion of the atoms during the reaction. This explains the absence of any type of oscillation in the MEP calculation, Figure 5a, compared to the Dynamics calculation figure. The Dynamics calculation accounts for the vibrational energies possessed by the molecules, hence explaining the presence of oscillations along the trajectory in Figure 4a.

Plots of "Internuclear Distances vs Time” and “Momenta vs Time” are also provided and more clearly depict the differences in the calculation types.

The Dynamics calculation accounts for the extra kinetic energy obtained after bond formation, which manifests itself into excess vibrational energy. This can be seen in the small oscillations of the A-B bond in figure 4b. Using the MEP calculation, this A-B bond distance remain constant over time, indicating this excess vibrational energy has not been accounted for. The increased momentum of the B-C bond is indicative of HC moving with translational energy and the oscillating momentum of the A-B bond is indicative of the new HAHB bond moving with vibrational energy.

Figure 5c shows a straight line at 0. The momenta is zero for the reaction pathway. This calculation type resets the momenta to zero after every step hence this result is expected. Figure 4c shows an oscillation of momentum with time, as through this calculation type, momenta is not reset to zero after each step.

If the initial conditions were changed so that r1 = rts pm and r2 = rts + 1pm, the blue and orange lines would swap positions in figures 4b, 4c, 5b, and 5c.

=== Reactive and unreactive trajectories ===
'''Question 4: Complete the table below by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?'''

With the initial positions r1 = 74 pm and r2 = 200 pm, trajectories were run with the following momenta combination:
{| class="wikitable"
!Row Number
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot/ kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|1
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|The kinetic energy of the system is sufficient to over come the activation barrier and hence the reaction proceeds. Originally, the reactants have little vibrational energy, reflected by the lack of oscillatory behaviour. After the reaction occurs, the products have more oscillatory behaviour, meaning that they have greater vibrational energy. The BC bond is broken and the AB bond is formed.
|[[File:ED2618 Q4 C1.png|frameless|366x366px]]
|-
|2
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|HA approaches an oscillating HBHC molecule, however the activation energy exceeds the kinetic energy and thus the reaction does not occur. Both reactants remain unchanged.
|[[File:ED2618 Q4 C2.png|frameless|366x366px]]
|-
|3
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|This trajectory is similar to that of row 1, with the only difference being that the reactants possess more vibrational energy, hence the presence of oscillatory behaviour as depicted in the figure.
|[[File:ED2618 Q4 C3.png|frameless|366x366px]]
|-
|4
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|For this trajectory, the kinetic energy is high and so the activation barrier is crossed and the products are formed. However, these products contain an excess of vibrational energy, namely the newly formed HAHB bond, which oscillates with such a great strength that the bond breaks and the reactants of the reaction are reformed. The reactants then move away from one another with greater vibrational energy than when the reaction began.
|[[File:ED2618 Q4 C4.png|frameless|366x366px]]
|-
|5
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|This trajectory is similar to that of the row above, where the kinetic energy is sufficient to overcome the activation barrier, however the newly formed HAHB bond possess so much vibrational energy that it dissociates back to the products (HBHC and HA). The only difference this time is that the products are reformed as the system has a high vibrational energy. The reactant HBHC bond is broken for a second time.
|[[File:ED2618 Q4 C5.png|none|thumb|366x366px]]
|}
From this table it may be concluded that the reactants must have sufficient energy to overcome the activation barrier to reach and surpass the transition state. This, however, is not the only factor that has to be considered for reaction success. The products may not have too much vibrational energy or the will convert back to the reactants. This example is observed in row 4 of the above table.

==== Transition State Theory (ADD REFERENCE) ====
'''Question 5: Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''

'''Answer 1: '''Transition State Theory is derived through the postulated quasi-equilibrium between a transition state species and the reactant molecules. There are a number of assumptions that are made upon derivation of the Transition State Theory rate expression:
# Electronic and nuclear motions are separated, similar to the Born-Oppenheimer approximation in quantum mechanics.
# Reactant molecule are distributed among their states in a way that is in accordance with the Maxwell-Boltzmann distribution.
# Once a molecular system has crossed a transition state, in a direction towards the products, it is unable to turn around and re-form the reactants.
# In the transition state, any motion that occurs along the reaction coordinate can be treated independently of other motions. This motion is treated classically as a translation.
# Even in the absence of an equilibrium between reactant and product molecules, the transition states that en route to becoming products are distributed among their states in accordance with the Maxwell-Boltzmann laws.
In rows 4 and 5 of the [[MRD:ED2618#Reactive and unreactive trajectories|table above]], the transition state is crossed and re-crossed numerous times, thus contradicting the assumption made by the Transition State Theory that a molecular system cannot regress back to the reactants once it has passed the transition state towards the products. The recrossing of the transition state ultimately lowers the rate of reaction, and thus Transition State Theory predicts a reaction rate that is larger than what is experimentally observed. Transition State Theory fails to evaluate molecular systems using quantum mechanical phenomena such as tunnelling, hence highlighting the discrepancy between the observed and theoretical reaction rates.

== EXERCISE 2: F - H - H system ==
In this exercise A=F, B=H, and C=H. r1 = BC and r2 = AB.

=== PES inspection ===
'''Question 1: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''

'''Answer 1: '''In this discussion, atoms F, H, and H are consigned to A, B, and C respectively. An increasing BC distance means that A and B are bonded, thus the HF + H system is consigned to this axis. An increasing AB distance means that B and C are bonded, thus this axis is consigned to the F + H2 system. As shown in the surface plot below, the HF + H system is lower in energy than the F + H2 system.

Thus the reaction:
* HF + H → H2 + F is endothermic. (Products higher in energy than reactants)

* F + H2 → HF + H is exothermic. (Reactants higher in energy than products)
[[File:ED2618 Surface Plot FHH.png|none|thumb|320x320px|Figure 6: Surface plot of FHH system]]
The HF (565 kJ.mol-1) bond is much stronger than the HH bond (436 kJ.mol-1) (ADD REFERENCE - PHYSICAL CHEM TEXTBOOK pG 885), thus the HF molecule is more stable than the H2 molecules. This is due to the large difference in electronegativity between H and F, resulting in an ionic bond. This is a stronger interaction than a purely covalent one that occurs between the two hydrogen atoms.

'''Question 2: Locate the approximate position of the transition state.'''

'''Answer 2: '''Hammond's postulate states the the transition state of a given reaction will resemble either the products or the reactants. The transition state will resemble either the products or the reactants, depending on which on is higher in energy, as the transition state is the maximum energy present in the minimum energy pathway taken from reactants to products. Thus, for an endothermic reaction, the products are higher in energy than the reactants, and so the transition state will resemble the products. Conversely, for an exothermic reaction, the reactants are higher in energy than the products, and so the transition state will resemble the reactants. (REFERENCE HAMMOND'S POSTULATE?)

The reaction of HF + H → H2 + F is endothermic and thus the transition state will resemble the products in structure. Thus the HH bond distance will be set to around 74.14 pm (ADD REFERENCE - ATKINS PHYSICAL CHEM PAGE 362)

F + H2 → HF + H is exothermic and thus the transition state will resemble the reactants in structure. Hence, the HH bond distance will be set around 74.14 pm, as above.

The value of the HF bond distance is expected to be much greater than 91.68 (ADD REFERENCE) as there is limited association between the two atoms.

The transition state was found when r1 (BC) = 74.485 and pm r2 (AB) = 181.200 pm. At this position there are no forces acting along either AB or BC.
{| class="wikitable"
|[[File:ED2618 Q2 TS Contour.png|none|thumb|320x320px|Figure 7: Contour plot of transition state of FHH system]]
|
|[[File:ED2618 Q2 TS INDvsT.png|none|thumb|320x320px|Figure 8: Internuclear Distance vs Time plot of transition state for FHH system]]
|}
Figure 7 above shows that there is no gradient/trajectory of motion at right angles to the point of the transition state (red cross). The plot of 'Internuclear Distances vs Time' in Figure 8 shows the distances of A-B, B-C, and A-C constant with time, consolidating the fact that this position is the transition state position. These lines are straight and have no oscillatory character, indicating no intermolecular vibrations are occurring.

'''Question 3: Report the activation energy for both reactions.'''

'''Answer 3: '''The kinetic energy of the transition state was calculated to be -433.981 kJ.mol-1. The activation energy can be found through running an MEP calculation of a structure slightly offset from that of the transition state. The initial positions were set so that:

For the exothermic reaction where F + H2 → HF + H
* r1 (BC) = 74.485 pm = rts
* r2 (AB) = 182.200 pm = rts + 1 pm
* p1 = p2 = 0 g.mol-1.pm.fs-1

For the endothermic reaction where HF + H → H2 + F
* r1 (BC) = 74.485 pm = rts
* r2 (AB) = 180.200 pm = rts - 1 pm
* p1 = p2 = 0 g.mol-1.pm.fs-1
{| class="wikitable"
!Reaction
!Energy of Reactants / kJ.mol-1
!Transition State Energy / kJ.mol-1
!Activation Energy / kJ.mol-1
!Energy vs Time Plot of Reaction Trajectory
|-
|F + H2 → HF + H
|<nowiki>- 435.5</nowiki>
|<nowiki>- 434.0</nowiki>
|<nowiki>-1.0</nowiki>
|[[File:ED2618 Q3 Exo.png|frameless|400x400px]]
|-
|HF + H → H2 + F
|<nowiki>- 560.5</nowiki>
|<nowiki>- 434.0</nowiki>
|<nowiki>-126.5</nowiki>
|[[File:ED2618 Q3 Endo.png|frameless|400x400px]]
|}

=== Reaction dynamics'''Question 4: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.''' ===
'''Answer 4: '''The set of initial conditions that results in a reactive trajectory for the reaction F + H2 → HF + H is given as follows:
* r1 (BC) = 74 pm
* r2 (AB) = 230 pm
* p1 (BC) = - 0.75 g.mol-1.pm.fs-1
* p2 (AB) = -2.00 0 g.mol-1.pm.fs-1
* 2000 steps at 0.1 fs
This reactive trajectory is show in the contour plot below:
[[File:ED2618 Q4 Contour.png|none|thumb|320x320px|Figure 9: Contour plot of initial conditions that result in a reactive trajectory for F + H2 system]]
The reaction F + H2 → HF + H is exothermic, meaning that the products possess a lower energy than the reactants. The energy discrepancy between reactants and products is released as heat to the surroundings. Figure 10 shows a plot of momenta vs time for this system. The H2 molecule approaches the F atom with low oscillatory momenta. The HH bond is broken and a new HF bond is formed. The energy released upon an HH bond breaking is greater than the energy required to form the HF bond. Upon collision of the molecules, the HF bond now oscillates with a much greater aptitude than the H2 had done prior to collision, due to the release of excess energy. The unexpected drop in momenta at t = 110 fs is due to recrossing of the transition state, as can be seen in Figure 9. After collision, the H atom moves off with a much slower velocity, as is indicated by the straight line in Figure 10. (ENERGY TO TRANSLATION AND VIBRATIONAL AS PRODUCT MOVES FAST?)
[[File:ED2618 Q4 Momenta.png|none|thumb|320x320px|Figure 10: Momenta vs Time plot of initial conditions that result in a reactive trajectory for F + H2 system]]
As energy must always be conserved, the increased kinetic energy of the system is compensated for by a decrease in potential energy, as indicated in Figure 11. This plot shows that the total energy of the system remains constant at - 433.372 kJ.mol-1. There is no loss of energy, but instead it is being converted from form to another, in line with the law of conservation of energy.
[[File:ED2618 Q4 Energy.png|none|thumb|320x320px|Figure 11: Energy vs Time plot of initial conditions that result in a reactive trajectory for F + H2 system]]
As this reaction is exothermic, energy is released to the surroundings in the form of heat. To confirm experimentally that this reaction has occurred, calorimetry could be used. Calorimetry may be used to identify changes in enthalpy in the surrounding in the immediate vicinity of the reaction. Another method to confirm that the reaction has gone to completion is through analysis of IR data. The vibrational energy of the HF molecule after collision is greatly increased after collision of F with H2. This would produce overtone bands as the HF molecule is excited to higher vibrational states.

File:ED2618 Q4 Energy.png

2020-05-15T12:57:25Z

Ed2618:

File:ED2618 Q4 Momenta.png

2020-05-15T12:57:13Z

Ed2618:

MRD:ED2618