ChemWiki - User contributions [en]

Talk:MRD:tts16

2018-06-07T17:40:36Z

Mm10114: Created page with "~~~~ Overall a good report. Some in-line comments to help it make even better - sometimes you need to be careful to explain more clearly. Otherwise, g..."

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 18:40, 7 June 2018 (BST) Overall a good report. Some in-line comments to help it make even better - sometimes you need to be careful to explain more clearly. Otherwise, good job.

MRD:tts16

2018-06-07T17:29:42Z

Mm10114: /* Exercise 1ː H + H2 system */

== Introduction to Molecular Reaction Dynamics ==

For a chemical reaction to occur, reactants much have sufficient energy to overcome its activation barrier as well as be in the right vibrational modes at the right time. Consequently, it is possible for a reactant trajectories crossing its energy barrier to form the product, yet fall back and result back to the starting reactants. During this process, energy is exchanged between translation and vibration energies.

== Exercise 1ː H + H2 system ==

=== Dynamics from the Transition State Region ===

What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

Given that the two axes represent the distance of r1 (distance of H1—H2) and r2 (distance of H2—H3), the gradient (first derivative) of the potential energy with respect to the atomic distances would be zero at either a maxima or minima of the surface as these are stationary points. Maxima in energy would indicate a transition state, while minima indicates equilibrium bond distances. However, to distinguish between maxima and minima, the second derivative (∂2V/∂r12 and ∂2V/∂r22) must be considered: when ∂2V/∂r1 and ∂2V/∂r22 > 0 a minima is obtained, while when ∂2V/∂r12 > 0, and ∂2V/∂r22 < 0, a maximum (transition state) will be obtained.

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 18:26, 7 June 2018 (BST) Are you sure the maxima in energy indicate a transition state? (Maxima is plural, maximum is singular). I think I know what you mean by that, a maximum along the minimum energy pathway for the reaction, however, you have to be very careful how you explain things. This is not clear enough.

=== Trajectories from r1 = r2ː locating the transition state ===

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.

By manipulating and evaluating different distance values for the transition state, my best estimate of the transition state position resulted to have a distance of 0.90780 Å. Looking at the inter-nuclear distances versus time graph, there is virtually no change in atomic distance for approximately 11 time periods before a steep change to form AB. This distance represented a transition state that had a considerably long time period (2654 steps) without oscillations, and was obtained through trial and error between high and low values, which eventually converged to this distance.

[[File:Tts16_e1q2_distance_time.PNG]]

=== Trajectories from r1 = rts +∂, r2=rts ===

Comment on how the mep and the trajectory you just calculated differ. 

It is evident that in the MEP figure, there is a smooth curve while the dynamic calculation shows an oscillating signal. This is due to the nature of the MEP calculation, where at all points (infinitesimal changes of distance) the velocity is set to zero and is absent of vibrations. By contrast, dynamic calculations do not manipulate the velocity, thus giving a more realistic portrayal of bond formation where oscillations are still present due to the element of momenta.

{| class="wikitable" border="1"
|+
! MEP Contour !! Dynamics Contour
|-
| [[File:Tts16_e1q3_contour_mep.PNG|400px]] || [[File:Tts16_e1q3_contour_dynamics.PNG|400px]]
|}

=== Reactive and Nonreactive Trajectories ===

Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory. 

{| class="wikitable" border="1"
|+ caption
! '''p1''' !!'''p2''' || Total Energy || Reactive? || Contour Plot || Description
|-
| -1.20 || -2.50 || -99.119 || Yes || [[File:Tts16_e1q4_contour1.PNG|400px]] || Atom C approaches AB which leads to the formation of BC, which is then oscillating with respect to its equilibrium position. As AB is dissociating, it has low vibration.
|-
| -1.50 || -2.00 || -99.119 || No || [[File:Tts16 e1q4 contour2.PNG|400px]] || Atom C approaches AB and BC distance begins to close in. However, there is not enough energy to exceed the energy barrier and atom C moves back away from BC. There is no significant change in AB distance and AB molecule is just oscillating with respect to its equilibrium.
|-
| -1.50 || -2.50 || -99.119 || Yes || [[File:Tts16_e1q4_contour3.PNG|400px]] || These values of momenta display a similar system to that of system 1. C approaches the vibrating AB bond which leads to the formation of BC. The AB bond is seen to be vibrating more than the AB bond in system 1.
|-
| -2.50 || -5.00 || -99.119|| No || [[File:Tts16_e1q4_contour4.PNG|400px]] || Atom C approaches B closing up the distance to around 0.70 Å (enough to cross the transition state barrier) and causes bond AB to elongate until 1.20 Å. There is high oscillation that causes AB to elongate to 1.20 Å, but atom C quickly bounces back away from B and the high momentum causes AB to oscillate with very high energy with respect to its equilibrium.
|-
| -2.50 || -5.20 || -99.119|| Yes || [[File:Tts16_e1q4_contour5a.PNG|400px]] || Atom C approaches AB with high momentum, closing the distance to around 0.70 Å crossing the transition state equilibrium distance. However, there is enough energy to cause full distortion of the bond which lead to the dissociation of AB and formation of BC.
|}

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 18:23, 7 June 2018 (BST) Are you sure the total energy should be the same in every case? (there was a special 'update' button in the software, that would have to be clicked each time to update the energy values) Also, what is the unit of energy here?

State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? [1][2]

'''Below are the main assumptions of transition state theory.'''

1. There exists a divided in potential energy surface separating reactant and product regions, and the transition state is defined as the highest value on the lowest energy path connecting reactants and products.

2. Any trajectory that passes through the transition state will form the product, and will not re-cross to give the reactant.

3. Born-Oppenheimer approximation is invoked, separating the electrons and nuclei: motion can be treated classically.

4. The reactants are distributed amount their states according to the Maxwell-Boltzmann distribution

Transition state theory provides a good model for reactions with high energy barrier where once the reactant has enough activation energy to overcome the transition state, it will not re-cross the barrier. However, in the examples that were speculated, situation 4 and 5 both cross the transition state, re-cross it, then ends up with one stabilizing back to the reactant while one crosses the transition state barrier again and ends up as product BC. Consequently, for systems with high energy collisions and low energy barriers, transition state theory may not yield an accurate depiction of reality. Moreover, as transition state theory treats the system classically, it ignores the possibility of quantum tunneling which would allow product formation that do not conform to transition state theory assumptions.

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 18:29, 7 June 2018 (BST) Well done here. You relate to the examples from previous point, I like that.

== Exercise 2ː F - H - H system ==

=== PES Inspection ===

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?

Looking at the figure below of F + H2 --> H + HF, it is seen from the potential energy surface that the reactants are of higher energy relative to the products hence it is an exothermic reaction. By tabulating energy changes through energetics, the H-H bond (436 kJ/mol) is broken to form a stronger HF bond (565 kJ/mol), thus forming bonds release more energy than making the bond, agreeing with the computational outcome as an exothermic reaction.[3]
On the other hand, HF + H --> H2 + F is an endothermic reaction as the opposite is demonstrated by the potential energy surfaces.

{| class="wikitable" border="1"
|+
! H2 + F Contour !! H + HF Contour
|-
|[[File:Tts16_e2q1_H2F_contour.PNG|400px]] || [[File:Tts16_e2q1_HHF_contour.PNG|400px]]
|}

Locate the approximate position of the transition state.

By manipulating and evaluating different distances for the transition state, my best estimate of the transition state positions resulted to have AB = 1.8900 Å, BC = 0.7445 Å distances and momenta set to zero for both. Looking at the inter-nuclear distance versus time graph, there is virtually no change in atomic distance for approximately 17 time periods before any deviation. This distance represented a transition state that had a considerably long time period (3411 steps) without oscillations and was obtained through trial and error between high and low values, which eventually converged to this distance.

[[File:Tts16_e2q2_ditsance_time_transitionstate_H2F.PNG|400px]]

Report the activation energy for both reactions.

[[File:Tts16_e2q3_energy_time_H2F.PNG|400px]]

To find the reactant energies, the distance of A-B was manipulated slightly, such that the F-H-H transition state would not be in its saddle point and fall to its reactants to determine the reactant energy. A energy vs. time graph was then plotted, indicating a reactant energy of -103.496 kcal/mol, and a product energy of -132.963 kcal/mol. The activation energy of a reaction can be calculated by finding the difference between the energy at its starting state and the transition state. From above, it was found that the energy of the transition state was -103.869 kcal/mol. Consequently, the activation energy for F + H2 is 0.127 kcal/mol, while the reverse reaction H + HF has an activation energy of 29.094 kcal/mol.

'''EA = ETS - Ereactants'''

'''HF + H'''

EA = -103.869 - (-103.996)

EA = 0.127 kcal/mol

'''H2 + F'''

EA = -103.869 - (-132.963)

EA = 29.094 kcal/mol

=== Reaction Dynamics ===
In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?

{| class="wikitable" border="1"
|+
! Dynamic Contour || Energy vs Time
|-
| [[File:Tts16_e2q4_contour1.PNG|400px]] || [[File:Tts16_e2q4_energies1.PNG|400px]]
|}

From the contour plot, it is seen that the reaction hovers over the transition state multiple times going from reactants to products back to reactants and finally results in the products. This is also shown through the energy versus time graph, where kinetic energy dramatically increases then decreases as the system goes through its transition state, while the potential energy decreases and increases as it mirrors the change in kinetic energy. In this exothermic reaction of F + H2 (discussed earlier), potential energy is converted to (kinetic) vibrational energy as it goes through its transition state. Consequently, to experimentally measure this, calorimetry techniques can determine the amount of energy released from this exothermic reaction.

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.

Polanyi’s empirical rules state that vibrational energy is more efficient in driving a system with a late transition state than translational energy, and vice versa for a system with an early transition state.[4] Transition states are a maximum in a reaction profile and cannot be captured or isolated. Consequently, Hammond’s postulate is used to determine what the transition state resembles, by stating that the transition state of a reaction is similar to either the reactants or products — whichever that is closer in energy. Consequently, an endothermic reaction has a later transition state, while an exothermic reaction has an early transition state. Below are plots that show systems with varying PHH between 3 and -3.

{| class="wikitable" border="1"
|+
! Momentum Values || Dynamic Contour !! Momentum vs Time
|-
| '''pHH''' = 0 || [[File:Tts16_e2q4_contour1aa0.PNG|350px]] || [[File:Tts16_e2q4_momentum_time1aa0.PNG|350px]]
|-
| '''pHH''' = 3 || [[File:Tts16_e2q4_contour1b3.PNG|350px]] || [[File:Tts16_e2q4_momentum_time1b3.PNG|350px]]
|-
| '''pHH''' = 1 || [[File:Tts16_e2q4_contour1e1.PNG|350px]] || [[File:Tts16_e2q4_momentum_timee1.PNG|350px]]
|-
| '''pHH''' = -2.5 || [[File:Tts16_e2q4_contour1dneg25.PNG|350px]] || [[File:Tts16_e2q4_momentum_time1dneg25.PNG|350px]]
|-
| '''pHH''' = 3 || [[File:Tts16_e2q4_contour1b3.PNG|350px]] || [[File:Tts16_e2q4_momentum_time1b3.PNG|350px]]
|}

According to Polanyi’s empirical rules, exothermic reactions are more likely to be successful if there is high translational energy relative to vibrational energy. In the plot below, PHF is set to -0.8, while PHH is set to 0.1. A successful reaction carries out, reflecting Polanyi’s rules. In the fourth row of the table above with PHH at -2.5, the reaction was still successful, meaning even though Polanyi's empirical rules lay out a general way of predicting reaction success, it is possible that high enough ranges of vibrational energy can drive the reaction over the transition state to its product.

[[File:Tts16_e2q4_contour3.PNG|400px]]

To consider the reverse reaction H + HF, HF was set to 0.74 Å, , and F H was set to 2.1 Å. As the reaction is an endothermic reaction, a high vibrational energy should lead to a successful reaction. Below are plots of a system with high and a system with low vibrational energy, and it is seen that in the system with high vibrational energy, the reaction is successful following Polanyi’s empirical rules.

{| class="wikitable" border="1"
|+
! '''pHH''' || '''pHF''' || Dynamic Contour || Energy vs Time
|-
| '''pHH''' = -0.5 || '''pHF''' = -1.5 || [[File:Tts16_e2q5_contour1.PNG|350px]] || [[File:Tts16_e2q5_momentum_time1.PNG|350px]]
|-
| '''pHH''' = -1.5 || '''pHF''' = -0.5 || [[File:Tts16_e2q5_contour3.PNG|350px]] || [[File:Tts16_e2q5_momentum_time3.PNG|350px]]
|}

== References ==

1. Felipe, M., Xiao, Y. & Kubicki, J. D. Molecular Orbital Modeling and Transition State Theory in Geochemistry. Rev. Mineral. Geochemistry 2001

2. Gonzalez_lafont, A., Villa, J., Lluch, J. M. & Juan, B. Variational Transition State Theory and Tunneling Calculations with Reorientation of the Generalized Transition States for Methyl Cation Transfer. J. Phys. Chem. A 3420–3428 (1998).

3. Texas, U. of. Bond Enthalpies. (2014). Available at: https://ch301.cm.utexas.edu/section2.php?target=thermo/thermochemistry/enthalpy-bonds.html.

4. Zhang, Z., Zhou, Y. & Zhang, D. Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction. J. Phys. Chem. A 3, 3416–3419 (2012).

MRD:tts16

2018-06-07T17:26:09Z

Mm10114: /* Dynamics from the Transition State Region */

== Introduction to Molecular Reaction Dynamics ==

For a chemical reaction to occur, reactants much have sufficient energy to overcome its activation barrier as well as be in the right vibrational modes at the right time. Consequently, it is possible for a reactant trajectories crossing its energy barrier to form the product, yet fall back and result back to the starting reactants. During this process, energy is exchanged between translation and vibration energies.

== Exercise 1ː H + H2 system ==

=== Dynamics from the Transition State Region ===

What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

Given that the two axes represent the distance of r1 (distance of H1—H2) and r2 (distance of H2—H3), the gradient (first derivative) of the potential energy with respect to the atomic distances would be zero at either a maxima or minima of the surface as these are stationary points. Maxima in energy would indicate a transition state, while minima indicates equilibrium bond distances. However, to distinguish between maxima and minima, the second derivative (∂2V/∂r12 and ∂2V/∂r22) must be considered: when ∂2V/∂r1 and ∂2V/∂r22 > 0 a minima is obtained, while when ∂2V/∂r12 > 0, and ∂2V/∂r22 < 0, a maximum (transition state) will be obtained.

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 18:26, 7 June 2018 (BST) Are you sure the maxima in energy indicate a transition state? (Maxima is plural, maximum is singular). I think I know what you mean by that, a maximum along the minimum energy pathway for the reaction, however, you have to be very careful how you explain things. This is not clear enough.

=== Trajectories from r1 = r2ː locating the transition state ===

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.

By manipulating and evaluating different distance values for the transition state, my best estimate of the transition state position resulted to have a distance of 0.90780 Å. Looking at the inter-nuclear distances versus time graph, there is virtually no change in atomic distance for approximately 11 time periods before a steep change to form AB. This distance represented a transition state that had a considerably long time period (2654 steps) without oscillations, and was obtained through trial and error between high and low values, which eventually converged to this distance.

[[File:Tts16_e1q2_distance_time.PNG]]

=== Trajectories from r1 = rts +∂, r2=rts ===

Comment on how the mep and the trajectory you just calculated differ. 

It is evident that in the MEP figure, there is a smooth curve while the dynamic calculation shows an oscillating signal. This is due to the nature of the MEP calculation, where at all points (infinitesimal changes of distance) the velocity is set to zero and is absent of vibrations. By contrast, dynamic calculations do not manipulate the velocity, thus giving a more realistic portrayal of bond formation where oscillations are still present due to the element of momenta.

{| class="wikitable" border="1"
|+
! MEP Contour !! Dynamics Contour
|-
| [[File:Tts16_e1q3_contour_mep.PNG|400px]] || [[File:Tts16_e1q3_contour_dynamics.PNG|400px]]
|}

=== Reactive and Nonreactive Trajectories ===

Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory. 

{| class="wikitable" border="1"
|+ caption
! '''p1''' !!'''p2''' || Total Energy || Reactive? || Contour Plot || Description
|-
| -1.20 || -2.50 || -99.119 || Yes || [[File:Tts16_e1q4_contour1.PNG|400px]] || Atom C approaches AB which leads to the formation of BC, which is then oscillating with respect to its equilibrium position. As AB is dissociating, it has low vibration.
|-
| -1.50 || -2.00 || -99.119 || No || [[File:Tts16 e1q4 contour2.PNG|400px]] || Atom C approaches AB and BC distance begins to close in. However, there is not enough energy to exceed the energy barrier and atom C moves back away from BC. There is no significant change in AB distance and AB molecule is just oscillating with respect to its equilibrium.
|-
| -1.50 || -2.50 || -99.119 || Yes || [[File:Tts16_e1q4_contour3.PNG|400px]] || These values of momenta display a similar system to that of system 1. C approaches the vibrating AB bond which leads to the formation of BC. The AB bond is seen to be vibrating more than the AB bond in system 1.
|-
| -2.50 || -5.00 || -99.119|| No || [[File:Tts16_e1q4_contour4.PNG|400px]] || Atom C approaches B closing up the distance to around 0.70 Å (enough to cross the transition state barrier) and causes bond AB to elongate until 1.20 Å. There is high oscillation that causes AB to elongate to 1.20 Å, but atom C quickly bounces back away from B and the high momentum causes AB to oscillate with very high energy with respect to its equilibrium.
|-
| -2.50 || -5.20 || -99.119|| Yes || [[File:Tts16_e1q4_contour5a.PNG|400px]] || Atom C approaches AB with high momentum, closing the distance to around 0.70 Å crossing the transition state equilibrium distance. However, there is enough energy to cause full distortion of the bond which lead to the dissociation of AB and formation of BC.
|}

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 18:23, 7 June 2018 (BST) Are you sure the total energy should be the same in every case? (there was a special 'update' button in the software, that would have to be clicked each time to update the energy values) Also, what is the unit of energy here?

State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? [1][2]

'''Below are the main assumptions of transition state theory.'''

1. There exists a divided in potential energy surface separating reactant and product regions, and the transition state is defined as the highest value on the lowest energy path connecting reactants and products.

2. Any trajectory that passes through the transition state will form the product, and will not re-cross to give the reactant.

3. Born-Oppenheimer approximation is invoked, separating the electrons and nuclei: motion can be treated classically.

4. The reactants are distributed amount their states according to the Maxwell-Boltzmann distribution

Transition state theory provides a good model for reactions with high energy barrier where once the reactant has enough activation energy to overcome the transition state, it will not re-cross the barrier. However, in the examples that were speculated, situation 4 and 5 both cross the transition state, re-cross it, then ends up with one stabilizing back to the reactant while one crosses the transition state barrier again and ends up as product BC. Consequently, for systems with high energy collisions and low energy barriers, transition state theory may not yield an accurate depiction of reality. Moreover, as transition state theory treats the system classically, it ignores the possibility of quantum tunneling which would allow product formation that do not conform to transition state theory assumptions.

== Exercise 2ː F - H - H system ==

=== PES Inspection ===

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?

Looking at the figure below of F + H2 --> H + HF, it is seen from the potential energy surface that the reactants are of higher energy relative to the products hence it is an exothermic reaction. By tabulating energy changes through energetics, the H-H bond (436 kJ/mol) is broken to form a stronger HF bond (565 kJ/mol), thus forming bonds release more energy than making the bond, agreeing with the computational outcome as an exothermic reaction.[3]
On the other hand, HF + H --> H2 + F is an endothermic reaction as the opposite is demonstrated by the potential energy surfaces.

{| class="wikitable" border="1"
|+
! H2 + F Contour !! H + HF Contour
|-
|[[File:Tts16_e2q1_H2F_contour.PNG|400px]] || [[File:Tts16_e2q1_HHF_contour.PNG|400px]]
|}

Locate the approximate position of the transition state.

By manipulating and evaluating different distances for the transition state, my best estimate of the transition state positions resulted to have AB = 1.8900 Å, BC = 0.7445 Å distances and momenta set to zero for both. Looking at the inter-nuclear distance versus time graph, there is virtually no change in atomic distance for approximately 17 time periods before any deviation. This distance represented a transition state that had a considerably long time period (3411 steps) without oscillations and was obtained through trial and error between high and low values, which eventually converged to this distance.

[[File:Tts16_e2q2_ditsance_time_transitionstate_H2F.PNG|400px]]

Report the activation energy for both reactions.

[[File:Tts16_e2q3_energy_time_H2F.PNG|400px]]

To find the reactant energies, the distance of A-B was manipulated slightly, such that the F-H-H transition state would not be in its saddle point and fall to its reactants to determine the reactant energy. A energy vs. time graph was then plotted, indicating a reactant energy of -103.496 kcal/mol, and a product energy of -132.963 kcal/mol. The activation energy of a reaction can be calculated by finding the difference between the energy at its starting state and the transition state. From above, it was found that the energy of the transition state was -103.869 kcal/mol. Consequently, the activation energy for F + H2 is 0.127 kcal/mol, while the reverse reaction H + HF has an activation energy of 29.094 kcal/mol.

'''EA = ETS - Ereactants'''

'''HF + H'''

EA = -103.869 - (-103.996)

EA = 0.127 kcal/mol

'''H2 + F'''

EA = -103.869 - (-132.963)

EA = 29.094 kcal/mol

=== Reaction Dynamics ===
In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?

{| class="wikitable" border="1"
|+
! Dynamic Contour || Energy vs Time
|-
| [[File:Tts16_e2q4_contour1.PNG|400px]] || [[File:Tts16_e2q4_energies1.PNG|400px]]
|}

From the contour plot, it is seen that the reaction hovers over the transition state multiple times going from reactants to products back to reactants and finally results in the products. This is also shown through the energy versus time graph, where kinetic energy dramatically increases then decreases as the system goes through its transition state, while the potential energy decreases and increases as it mirrors the change in kinetic energy. In this exothermic reaction of F + H2 (discussed earlier), potential energy is converted to (kinetic) vibrational energy as it goes through its transition state. Consequently, to experimentally measure this, calorimetry techniques can determine the amount of energy released from this exothermic reaction.

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.

Polanyi’s empirical rules state that vibrational energy is more efficient in driving a system with a late transition state than translational energy, and vice versa for a system with an early transition state.[4] Transition states are a maximum in a reaction profile and cannot be captured or isolated. Consequently, Hammond’s postulate is used to determine what the transition state resembles, by stating that the transition state of a reaction is similar to either the reactants or products — whichever that is closer in energy. Consequently, an endothermic reaction has a later transition state, while an exothermic reaction has an early transition state. Below are plots that show systems with varying PHH between 3 and -3.

{| class="wikitable" border="1"
|+
! Momentum Values || Dynamic Contour !! Momentum vs Time
|-
| '''pHH''' = 0 || [[File:Tts16_e2q4_contour1aa0.PNG|350px]] || [[File:Tts16_e2q4_momentum_time1aa0.PNG|350px]]
|-
| '''pHH''' = 3 || [[File:Tts16_e2q4_contour1b3.PNG|350px]] || [[File:Tts16_e2q4_momentum_time1b3.PNG|350px]]
|-
| '''pHH''' = 1 || [[File:Tts16_e2q4_contour1e1.PNG|350px]] || [[File:Tts16_e2q4_momentum_timee1.PNG|350px]]
|-
| '''pHH''' = -2.5 || [[File:Tts16_e2q4_contour1dneg25.PNG|350px]] || [[File:Tts16_e2q4_momentum_time1dneg25.PNG|350px]]
|-
| '''pHH''' = 3 || [[File:Tts16_e2q4_contour1b3.PNG|350px]] || [[File:Tts16_e2q4_momentum_time1b3.PNG|350px]]
|}

According to Polanyi’s empirical rules, exothermic reactions are more likely to be successful if there is high translational energy relative to vibrational energy. In the plot below, PHF is set to -0.8, while PHH is set to 0.1. A successful reaction carries out, reflecting Polanyi’s rules. In the fourth row of the table above with PHH at -2.5, the reaction was still successful, meaning even though Polanyi's empirical rules lay out a general way of predicting reaction success, it is possible that high enough ranges of vibrational energy can drive the reaction over the transition state to its product.

[[File:Tts16_e2q4_contour3.PNG|400px]]

To consider the reverse reaction H + HF, HF was set to 0.74 Å, , and F H was set to 2.1 Å. As the reaction is an endothermic reaction, a high vibrational energy should lead to a successful reaction. Below are plots of a system with high and a system with low vibrational energy, and it is seen that in the system with high vibrational energy, the reaction is successful following Polanyi’s empirical rules.

{| class="wikitable" border="1"
|+
! '''pHH''' || '''pHF''' || Dynamic Contour || Energy vs Time
|-
| '''pHH''' = -0.5 || '''pHF''' = -1.5 || [[File:Tts16_e2q5_contour1.PNG|350px]] || [[File:Tts16_e2q5_momentum_time1.PNG|350px]]
|-
| '''pHH''' = -1.5 || '''pHF''' = -0.5 || [[File:Tts16_e2q5_contour3.PNG|350px]] || [[File:Tts16_e2q5_momentum_time3.PNG|350px]]
|}

== References ==

1. Felipe, M., Xiao, Y. & Kubicki, J. D. Molecular Orbital Modeling and Transition State Theory in Geochemistry. Rev. Mineral. Geochemistry 2001

2. Gonzalez_lafont, A., Villa, J., Lluch, J. M. & Juan, B. Variational Transition State Theory and Tunneling Calculations with Reorientation of the Generalized Transition States for Methyl Cation Transfer. J. Phys. Chem. A 3420–3428 (1998).

3. Texas, U. of. Bond Enthalpies. (2014). Available at: https://ch301.cm.utexas.edu/section2.php?target=thermo/thermochemistry/enthalpy-bonds.html.

4. Zhang, Z., Zhou, Y. & Zhang, D. Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction. J. Phys. Chem. A 3, 3416–3419 (2012).

MRD:tts16

2018-06-07T17:23:54Z

Mm10114: /* Reactive and Nonreactive Trajectories */

== Introduction to Molecular Reaction Dynamics ==

For a chemical reaction to occur, reactants much have sufficient energy to overcome its activation barrier as well as be in the right vibrational modes at the right time. Consequently, it is possible for a reactant trajectories crossing its energy barrier to form the product, yet fall back and result back to the starting reactants. During this process, energy is exchanged between translation and vibration energies.

== Exercise 1ː H + H2 system ==

=== Dynamics from the Transition State Region ===

What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

Given that the two axes represent the distance of r1 (distance of H1—H2) and r2 (distance of H2—H3), the gradient (first derivative) of the potential energy with respect to the atomic distances would be zero at either a maxima or minima of the surface as these are stationary points. Maxima in energy would indicate a transition state, while minima indicates equilibrium bond distances. However, to distinguish between maxima and minima, the second derivative (∂2V/∂r12 and ∂2V/∂r22) must be considered: when ∂2V/∂r1 and ∂2V/∂r22 > 0 a minima is obtained, while when ∂2V/∂r12 > 0, and ∂2V/∂r22 < 0, a maximum (transition state) will be obtained.

=== Trajectories from r1 = r2ː locating the transition state ===

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.

By manipulating and evaluating different distance values for the transition state, my best estimate of the transition state position resulted to have a distance of 0.90780 Å. Looking at the inter-nuclear distances versus time graph, there is virtually no change in atomic distance for approximately 11 time periods before a steep change to form AB. This distance represented a transition state that had a considerably long time period (2654 steps) without oscillations, and was obtained through trial and error between high and low values, which eventually converged to this distance.

[[File:Tts16_e1q2_distance_time.PNG]]

=== Trajectories from r1 = rts +∂, r2=rts ===

Comment on how the mep and the trajectory you just calculated differ. 

It is evident that in the MEP figure, there is a smooth curve while the dynamic calculation shows an oscillating signal. This is due to the nature of the MEP calculation, where at all points (infinitesimal changes of distance) the velocity is set to zero and is absent of vibrations. By contrast, dynamic calculations do not manipulate the velocity, thus giving a more realistic portrayal of bond formation where oscillations are still present due to the element of momenta.

{| class="wikitable" border="1"
|+
! MEP Contour !! Dynamics Contour
|-
| [[File:Tts16_e1q3_contour_mep.PNG|400px]] || [[File:Tts16_e1q3_contour_dynamics.PNG|400px]]
|}

=== Reactive and Nonreactive Trajectories ===

Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory. 

{| class="wikitable" border="1"
|+ caption
! '''p1''' !!'''p2''' || Total Energy || Reactive? || Contour Plot || Description
|-
| -1.20 || -2.50 || -99.119 || Yes || [[File:Tts16_e1q4_contour1.PNG|400px]] || Atom C approaches AB which leads to the formation of BC, which is then oscillating with respect to its equilibrium position. As AB is dissociating, it has low vibration.
|-
| -1.50 || -2.00 || -99.119 || No || [[File:Tts16 e1q4 contour2.PNG|400px]] || Atom C approaches AB and BC distance begins to close in. However, there is not enough energy to exceed the energy barrier and atom C moves back away from BC. There is no significant change in AB distance and AB molecule is just oscillating with respect to its equilibrium.
|-
| -1.50 || -2.50 || -99.119 || Yes || [[File:Tts16_e1q4_contour3.PNG|400px]] || These values of momenta display a similar system to that of system 1. C approaches the vibrating AB bond which leads to the formation of BC. The AB bond is seen to be vibrating more than the AB bond in system 1.
|-
| -2.50 || -5.00 || -99.119|| No || [[File:Tts16_e1q4_contour4.PNG|400px]] || Atom C approaches B closing up the distance to around 0.70 Å (enough to cross the transition state barrier) and causes bond AB to elongate until 1.20 Å. There is high oscillation that causes AB to elongate to 1.20 Å, but atom C quickly bounces back away from B and the high momentum causes AB to oscillate with very high energy with respect to its equilibrium.
|-
| -2.50 || -5.20 || -99.119|| Yes || [[File:Tts16_e1q4_contour5a.PNG|400px]] || Atom C approaches AB with high momentum, closing the distance to around 0.70 Å crossing the transition state equilibrium distance. However, there is enough energy to cause full distortion of the bond which lead to the dissociation of AB and formation of BC.
|}

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 18:23, 7 June 2018 (BST) Are you sure the total energy should be the same in every case? (there was a special 'update' button in the software, that would have to be clicked each time to update the energy values) Also, what is the unit of energy here?

State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? [1][2]

'''Below are the main assumptions of transition state theory.'''

1. There exists a divided in potential energy surface separating reactant and product regions, and the transition state is defined as the highest value on the lowest energy path connecting reactants and products.

2. Any trajectory that passes through the transition state will form the product, and will not re-cross to give the reactant.

3. Born-Oppenheimer approximation is invoked, separating the electrons and nuclei: motion can be treated classically.

4. The reactants are distributed amount their states according to the Maxwell-Boltzmann distribution

Transition state theory provides a good model for reactions with high energy barrier where once the reactant has enough activation energy to overcome the transition state, it will not re-cross the barrier. However, in the examples that were speculated, situation 4 and 5 both cross the transition state, re-cross it, then ends up with one stabilizing back to the reactant while one crosses the transition state barrier again and ends up as product BC. Consequently, for systems with high energy collisions and low energy barriers, transition state theory may not yield an accurate depiction of reality. Moreover, as transition state theory treats the system classically, it ignores the possibility of quantum tunneling which would allow product formation that do not conform to transition state theory assumptions.

== Exercise 2ː F - H - H system ==

=== PES Inspection ===

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?

Looking at the figure below of F + H2 --> H + HF, it is seen from the potential energy surface that the reactants are of higher energy relative to the products hence it is an exothermic reaction. By tabulating energy changes through energetics, the H-H bond (436 kJ/mol) is broken to form a stronger HF bond (565 kJ/mol), thus forming bonds release more energy than making the bond, agreeing with the computational outcome as an exothermic reaction.[3]
On the other hand, HF + H --> H2 + F is an endothermic reaction as the opposite is demonstrated by the potential energy surfaces.

{| class="wikitable" border="1"
|+
! H2 + F Contour !! H + HF Contour
|-
|[[File:Tts16_e2q1_H2F_contour.PNG|400px]] || [[File:Tts16_e2q1_HHF_contour.PNG|400px]]
|}

Locate the approximate position of the transition state.

By manipulating and evaluating different distances for the transition state, my best estimate of the transition state positions resulted to have AB = 1.8900 Å, BC = 0.7445 Å distances and momenta set to zero for both. Looking at the inter-nuclear distance versus time graph, there is virtually no change in atomic distance for approximately 17 time periods before any deviation. This distance represented a transition state that had a considerably long time period (3411 steps) without oscillations and was obtained through trial and error between high and low values, which eventually converged to this distance.

[[File:Tts16_e2q2_ditsance_time_transitionstate_H2F.PNG|400px]]

Report the activation energy for both reactions.

[[File:Tts16_e2q3_energy_time_H2F.PNG|400px]]

To find the reactant energies, the distance of A-B was manipulated slightly, such that the F-H-H transition state would not be in its saddle point and fall to its reactants to determine the reactant energy. A energy vs. time graph was then plotted, indicating a reactant energy of -103.496 kcal/mol, and a product energy of -132.963 kcal/mol. The activation energy of a reaction can be calculated by finding the difference between the energy at its starting state and the transition state. From above, it was found that the energy of the transition state was -103.869 kcal/mol. Consequently, the activation energy for F + H2 is 0.127 kcal/mol, while the reverse reaction H + HF has an activation energy of 29.094 kcal/mol.

'''EA = ETS - Ereactants'''

'''HF + H'''

EA = -103.869 - (-103.996)

EA = 0.127 kcal/mol

'''H2 + F'''

EA = -103.869 - (-132.963)

EA = 29.094 kcal/mol

=== Reaction Dynamics ===
In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?

{| class="wikitable" border="1"
|+
! Dynamic Contour || Energy vs Time
|-
| [[File:Tts16_e2q4_contour1.PNG|400px]] || [[File:Tts16_e2q4_energies1.PNG|400px]]
|}

From the contour plot, it is seen that the reaction hovers over the transition state multiple times going from reactants to products back to reactants and finally results in the products. This is also shown through the energy versus time graph, where kinetic energy dramatically increases then decreases as the system goes through its transition state, while the potential energy decreases and increases as it mirrors the change in kinetic energy. In this exothermic reaction of F + H2 (discussed earlier), potential energy is converted to (kinetic) vibrational energy as it goes through its transition state. Consequently, to experimentally measure this, calorimetry techniques can determine the amount of energy released from this exothermic reaction.

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.

Polanyi’s empirical rules state that vibrational energy is more efficient in driving a system with a late transition state than translational energy, and vice versa for a system with an early transition state.[4] Transition states are a maximum in a reaction profile and cannot be captured or isolated. Consequently, Hammond’s postulate is used to determine what the transition state resembles, by stating that the transition state of a reaction is similar to either the reactants or products — whichever that is closer in energy. Consequently, an endothermic reaction has a later transition state, while an exothermic reaction has an early transition state. Below are plots that show systems with varying PHH between 3 and -3.

{| class="wikitable" border="1"
|+
! Momentum Values || Dynamic Contour !! Momentum vs Time
|-
| '''pHH''' = 0 || [[File:Tts16_e2q4_contour1aa0.PNG|350px]] || [[File:Tts16_e2q4_momentum_time1aa0.PNG|350px]]
|-
| '''pHH''' = 3 || [[File:Tts16_e2q4_contour1b3.PNG|350px]] || [[File:Tts16_e2q4_momentum_time1b3.PNG|350px]]
|-
| '''pHH''' = 1 || [[File:Tts16_e2q4_contour1e1.PNG|350px]] || [[File:Tts16_e2q4_momentum_timee1.PNG|350px]]
|-
| '''pHH''' = -2.5 || [[File:Tts16_e2q4_contour1dneg25.PNG|350px]] || [[File:Tts16_e2q4_momentum_time1dneg25.PNG|350px]]
|-
| '''pHH''' = 3 || [[File:Tts16_e2q4_contour1b3.PNG|350px]] || [[File:Tts16_e2q4_momentum_time1b3.PNG|350px]]
|}

According to Polanyi’s empirical rules, exothermic reactions are more likely to be successful if there is high translational energy relative to vibrational energy. In the plot below, PHF is set to -0.8, while PHH is set to 0.1. A successful reaction carries out, reflecting Polanyi’s rules. In the fourth row of the table above with PHH at -2.5, the reaction was still successful, meaning even though Polanyi's empirical rules lay out a general way of predicting reaction success, it is possible that high enough ranges of vibrational energy can drive the reaction over the transition state to its product.

[[File:Tts16_e2q4_contour3.PNG|400px]]

To consider the reverse reaction H + HF, HF was set to 0.74 Å, , and F H was set to 2.1 Å. As the reaction is an endothermic reaction, a high vibrational energy should lead to a successful reaction. Below are plots of a system with high and a system with low vibrational energy, and it is seen that in the system with high vibrational energy, the reaction is successful following Polanyi’s empirical rules.

{| class="wikitable" border="1"
|+
! '''pHH''' || '''pHF''' || Dynamic Contour || Energy vs Time
|-
| '''pHH''' = -0.5 || '''pHF''' = -1.5 || [[File:Tts16_e2q5_contour1.PNG|350px]] || [[File:Tts16_e2q5_momentum_time1.PNG|350px]]
|-
| '''pHH''' = -1.5 || '''pHF''' = -0.5 || [[File:Tts16_e2q5_contour3.PNG|350px]] || [[File:Tts16_e2q5_momentum_time3.PNG|350px]]
|}

== References ==

1. Felipe, M., Xiao, Y. & Kubicki, J. D. Molecular Orbital Modeling and Transition State Theory in Geochemistry. Rev. Mineral. Geochemistry 2001

2. Gonzalez_lafont, A., Villa, J., Lluch, J. M. & Juan, B. Variational Transition State Theory and Tunneling Calculations with Reorientation of the Generalized Transition States for Methyl Cation Transfer. J. Phys. Chem. A 3420–3428 (1998).

3. Texas, U. of. Bond Enthalpies. (2014). Available at: https://ch301.cm.utexas.edu/section2.php?target=thermo/thermochemistry/enthalpy-bonds.html.

4. Zhang, Z., Zhou, Y. & Zhang, D. Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction. J. Phys. Chem. A 3, 3416–3419 (2012).

Talk:MRD:bd316

2018-06-07T17:08:06Z

Mm10114: Created page with "~~~~ Overall your report is very good. The few minor things are regarding the second question, where you were supposed to locate the transition state..."

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 18:08, 7 June 2018 (BST) Overall your report is very good. The few minor things are regarding the second question, where you were supposed to locate the transition state more precisely and the energy unit (if you were not sure about it, don't assume, confirm with a demonstrator, this is what they are there for). Please, be more careful in the future, as these mistakes are probably more a source of misunderstanding (the expected answer for a question/ software internal units).

MRD:bd316

2018-06-07T16:59:54Z

Mm10114: /* Calculating the activation energies */

= Molecular reaction dynamics of an H2 + H reacting system =

In this section the reaction dynamics of the following system, whereby a free hydrogen atom collides with a H2 molecule to form a new H2 molecule and a new free H atom, are considered:

[[File: bd316scheme1.png|center|500px]]

The three atoms HC, HB and HA are constrained to a 180° angle such that a potential energy surface for the reaction may be calculated.2

=== The dynamics of the potential energy surface plot - the energy minima and transition state region ===

[[File:bd316figure1.png|thumb|Figure 1: the three-dimensional potential energy surface of the reacting system|500px]]

The trajectory of the reaction follows the path of energy minima. The transition state region is defined as the energy maximum on the path of minimum energy between the reactants and products.

Hydrogen atom A approaches hydrogen molecule BC (starting from the left hand side of figure 1) following the path of the potential energy surface along which ∂(V(rBC))/∂(rBC) = 0. Along this path the value of the gradient ∂(V(rBA))/∂(rBA) gradually increases until it reaches a maximum along the path of minimum energy - this is the transition state region. From the transition state region, the AB bond forms and the hydrogen atom C moves away from H2 molecule AB along the path along the potential energy surface on which ∂(V(rBA))/∂(rBA) = 0. The reaction coordinate continues along this path until it reaches the minimum of the potential energy surface on both axes, at which ∂(V(rBA))/∂(rBA) = ∂(V(rBC))/∂(rBC) = 0, at which point the reaction is complete.

In the transition state region, the potential energy surface has reached a saddle point, which has the minimum potential energy whereby rCB = rBA. In the transition state region, we can consider there to be a third 'axis', running diagonally from the rBC axis to the rBA axis through the saddle point (s), which defines the reaction coordinate (the view along this axis is given in figure 1). At the saddle point ∂(V(s))/∂(s)= 0; this is an energy maximum along the reaction pathway - ∂2(V(s))/∂(s)2 < 0. The 'axis' running orthogonal to the axis s, through the saddle point from the origin (o) also has ∂(V(o))/∂(o)= 0 and is an energy minimum with respect to this axis - ∂2(V(o))/∂(o)2 > 0.1 Taking the second derivative distinguishes the minima and maxima along a given axis.

[[File:bd316figure2.png|thumb|Figure 2: the contour potential energy surface of the reacting system|400px]]

=== Determining the geometry of the transition state ===

Computationally we can use the fact that, in the transition state, rBC = rBA (if this were not the case the potential energy surface in figure 1 would not be symmetrical) to determine the value of rTS. The results of this calculation are illustrated by the plot of rTS vs time in figure 3. When the transition state has zero momentum and the reaction does not proceed, its potential energy fluctuates in the transition state region and the value of rTS oscillates between approximately 0.8 to 1.0 Å.

An enlarged section of the potential energy surface plot (shown in full in figure 1) in figure 4 shows these inter-atomic vibrations. The value of rTS fluctuates because the transition state is able to interconvert between kinetic energy and potential energy, as is illustrated in figure 5. There are three vibrational degrees freedom at the saddle point - one of these corresponds to infinitesimal motion along the reaction path, and the other two are vibrational motion orthogonal to the reaction path (through the saddle point and along the energy minima).1,2

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 17:50, 7 June 2018 (BST) You correctly observe that the TS is somewhere between the values of 0.8 and 1.0 Å. However, there is only one TS at this potential energy surface. And it was the purpose of this exercise to find it. You were supposed to sample different r1=r2 until you find your best estimate. You would be able to tell this when, for example in Figure 5, you would have no oscillation for the distances and there would be no interconversion of the potential-kinetic energies. So, in the end you were supposed to report a single value for r1=r2, with your best estimate for the TS. In your current calculation, if you would increase number of steps significantly, the aggregation of numerical errors would probably result in your trajectory skewing towards reactants/products. With a good estimate for TS that would not happen.

<gallery widths=300px heights=300px>
File:bd316figure3.png|Figure 3: plot of internuclear distance vs time for the transition state
File:bd316figure4.png|Figure 4: the transition state region of the potential energy surface showing vibrational excitation of the transition state
File:bd316figure5.png|Figure 5: kinetic energy and potential energy vs time for the transition state
</gallery>

=== Calculating the reaction path ===

The minimum energy path, or reaction path, is the path that follows the energy minima from the saddle point to the products along a series of infinitesimally small steps - the vibrational degree of freedom mentioned previously (figure 7). At each of these steps, the kinetic energy is removed, such that the reacting system always has the minimum possible potential energy.1 This distinguishes the minimum energy path from the dynamic reaction path (figure 8), which gives a realistic picture of the trajectory that the reacting system will follow, vibrating orthogonally to the reaction path as the reaction coordinate progresses.1

<gallery heights=250px widths=250px>
File:bd316figure7.png|Figure 7: the minimum energy path
File:bd316figure8.png|Figure 8: the dynamic reaction path
</gallery>

Figure 9 shows the progression of the values rCB and rBA as the reaction progresses from the transition state region to the products, following the dynamic reaction trajectory. At time zero rCB = rBA = rTS. As HC leaves and the HB-HA bond forms, the value of rCB increases to infinity, and the value of rBA reaches the value of the equilibrium H-H bond length - this fluctuates between values of approximately 0.715 - 0.748 Å, with an average value of approximately 0.726 Å.

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 17:54, 7 June 2018 (BST) Very good observation for the MEP and dynamics calculations differences. However, as you have not given the value for rCB = rBA = rTS in the previous exercise, which value have you used here? Although, it does look on the plots like you're at the transition state, but without a value reported I can't be sure.

The progression of the values of interatomic momentum are shown in figure 10. As the reaction coordinate initially progresses from the transition state the values of both momenta initially increase (figure 11). The value of pCB increases sharply and plateaus at 2.5 when C and B are fully dissociated. The value of pBA first decreases as the HB-HA bond contracts on forming, then reaches the equilibrium value of approximately 1.2, oscillating between 0.95 and 1.53 as the HB-HA bond vibrates.

<gallery widths=250px heights=250px>
File:bd316figure9.png|Figure 9: plot of internuclear distance vs time for the dynamic reacting system
File:bd316figure10.png|Figure 10: plot of internuclear momentum vs time for the dynamic reacting system
File:bd316figure11.png|Figure 11: expanded region of the plot of internuclear momentum vs time for the dynamic reacting system
</gallery>

=== Reactive and unreactive trajectories ===

A trajectory may be reactive or unreactive depending on the values of pCB and pBA, which determine the kinetic energy with which the constituent species collide; they must have sufficient total energy to overcome the activation energy barrier and pass through the transition state region to the products. The reactivities of a series of trajectories of varying initial values of the momenta are compared below.

{| class="wikitable" border=1
! Trajectory !! pCB !! pBA !! Total energy (kJ mol-1) !! Reactivity!! Plot !! Comment
|-
| 1 || -1.25 || -2.5 || -99.1 || Reactive || [[File: bd316table1.png|200px]] || The reacting system has sufficient energy to pass over the transition state region. The products HC + HBHA are formed and the resultant HBHA molecule shows a greater degree of vibrational excitation relative to the HCHB molecule.
|-
| 2 ||-1.5 || -2.0 || -100.4 || Unreactive || [[File: bd316table2.png|200px]] || The reacting system has insufficient energy to pass over the transition state region.
|-
| 3 ||-1.5 || -2.5 || -99.0 || Reactive || [[File: bd316table3.png|200px]] || The reacting system has sufficient energy to pass over the transition state region. The resultant HBHC molecule shows a greater degree of vibrational excitation than that resulting from trajectory 1 because of the greater initial value of pCB.
|-
| 4 || -2.5 || -5.0 || -85.0 || Unreactive || [[File: bd316table4.png|200px]] || The species collide with a great deal of kinetic energy - the activation barrier is passed and re-passed, because the initially formed products HC + HB-HA have such a high degree of vibrational excitation that they recollide with sufficient kinetic energy to re-form the reactants.
|-
| 5 || -2.5 || -5.2 || -83.4 || Reactive || [[File: bd316table5.png|200px]] || The reacting species collide with even greater kinetic energy than in trajectory 4 - the activation barrier is crossed, re-crossed and crossed again, until the product species are formed with internuclear momenta such they do not re-collide.
|}

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 17:55, 7 June 2018 (BST) Good observations. For the reported total energies, are these the values taken directly from the LepsGUI? If yes, are you sure the unit is kJ mol-1?

== Using transition state theory to predict rates of reaction ==

Transition state theory provides a means of predicting the rate constant for a given reaction that passes through a transition state, based on the concepts of statistical thermodynamics.3 It is assumed that the reactants are in rapid equilibrium with the transition state complex; the products are formed when this complex decays in to the products. Upon further calculation, the rate constant of reaction is given by the Eyring equation:3

[[File: bd316eyring.png|center]]

In which kr is the reaction rate, κ is a proportionality constant between the rate of passage of the reacting system through the transition state and the vibration frequency of the reacting system along the reaction coordinate, k is the Botzmann constant, h is Planck's constant, T is the temperature and K-barcǂ is the rate of formation of the transition state.3 K-barcǂ may be determined from the partition functions of the reactants and products, which are themselves easily calculated spectroscopically. The subscript c denotes that the the motion along the reaction coordinate has been separated from the other modes of motion of the species (equivalent to the Born-Oppenheimer approximation) and treated entirely classically.1,3

In calculating the rate constant using transition state theory, three main assumptions are made:4

1. That the reacting species are in equilibrium with the transition state complex, as mentioned previously - this allows the rate constant to be determined using statistical thermodynamics expressions in which partition functions appear (is is assumed that the 'phase space' of the reactants is populated according to the Boltzmann distribution), which are derived with the assumption of equilibrium.

2. That the system can be described adequately with classical mechanics.

3. That systems which pass the transition state barrier always become products.

Each of these assumptions may cause deviations of the calculated rate constants from the experimentally determined rate constants to different extents, depending on the reality of the system considered. For our reacting system in which the molecules may repass the transition state region when they collide with sufficient kinetic energy, as in trajectory 4 in the table above, assumption three may lead to marked deviations between the experimental and calculated rate constants. Assumption two neglects the possibility of quantum tunneling over the activation barrier, underestimating the rate of reaction - as yet there is no transition state theory model that successfully takes quantum mechanical effects in to account.1,3

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 17:57, 7 June 2018 (BST) Well explained.

= Molecular reaction dynamics of an F - H - H reacting system =

In this section the dynamics of the following, slightly more complex, reaction between a free flourine atom and an H2 molecule are considered:

[[File: bd316scheme2.png|center|500px]]

=== Inspection of the potential energy surface ===
[[File: bd316figure13.png|thumb| Figure 12 a: the potential energy surface of the F-H-H reacting system showing a reactive trajectory in the exothermic direction|300px]]
[[File: bd316figure13b.png|thumb| Figure 12 b: the potential energy surface of the F-H-H reacting system from above|300px]]
[[File: bd316figure14.png|thumb| Figure 13: the contour plot of the F-H-H reacting system showing the location of the transition state|300px]]
[[File: bd316figure15.png|thumb| Figure 14: plot of internuclear momenta vs. time for the F + H2 reacting system following the trajectory illustrated in figure 13|300px]]

The reaction is exothermic in the F + H2 --> FH + H direction, because the products are lower in energy than the reactants, and endothermic in the reverse direction (figure 12a). This indicates the greater bond strength of the H-F bond relative to the H-H bond - this is confirmed by literature values.5

The activation energy of the F + H2 reaction is small - this makes the transition state difficult to locate by eye alone. Hammond's postulate states that the transition state will have the geometry closest to the species (reactants or products) that are closest in energy.6 The location of the transition state for the exothermic F + H2 reaction will therefore most closely resemble the geometry of the reactants. The location of the transition state can be found by starting from the F-H-H geometry with the H-H bond length and optimising until the potential energy surface no longer shows a trajectory with zero internuclear momentum - the location of the transition state is given in figure 13. The transition state complex obtained by this method has an F-H bond length of 1.813 Å and an H-H bond length of 0.741 Å.

=== Calculating the activation energies ===

The activation energy is defined as the difference in energy between the reactants and the transition state, i.e. the minimum energy that the reactants must have in order to react. Comparison of the energy of the transition state (-103.5 kJ mol-1) with the minimum energy of the reacting species (-133.3 kJ mol-1) gives a value of 29.8 kJ mol-1 for the activation energy of the endothermic H + HF --> F + H2 reaction. The minimum energy of H2 was found to be -104.3 kJ mol-1, hence the activation energy in the exothermic direction is 0.8 kJ mol -1.

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 17:59, 7 June 2018 (BST) This is good. However, the unit of energy you report here is wrong. It is kcal/mol. This might be a very honest mistake. However, in the future, if you're not certain please try to confirm with any of the demonstrators.

=== Energetics of the reacting system ===

According to the first law of thermodynamics, it is not possible for energy to be created or destroyed.3 Hence, when reactants collide with insufficient energy to reach the transition state they may not react (quantum mechanical possibilities aside), and when they collide with more than sufficient energy to react, the excess energy of reaction must 'go somewhere'. This excess energy may be dissipated as heat (random motion, which results in an increase in temperature), or stored in one of the molecules translational, vibrational, rotational or electronic modes of energy (excited electronic modes are typically thermally inaccessible). The amount of energy which is dissipated as heat relative to the amount which is stored in a given species' energetic modes is determined by how many energetic modes the species has available - this is the 'heat capacity'.

This is seen in figure 12 and in many of the reactive trajectories in the previous section, where the reactive trajectory oscillates to a greater extent orthogonal to the reaction coordinate after passing the transition state region, because some of the excess energy is converted to vibrational energy of the products.3 This is also clearly illustrated in figure 14, in which it can be seen that the internuculear momentum in the product H-F molecule oscillates to a significantly greater extent than in the reactant H2 molecule because there is more energy in the H-F vibrational modes.

Calorimetry may be used to experimentally measure the flow of heat resulting from an exothermic reaction, and from the difference in this value, the activation energy and the energy of the reactants the amount of energy stored in the reacting species' energetic modes may be deduced.3

=== Polanyi's empirical rules ===

The energy that contributes to a reacting system's ability to pass an activation barrier on collision may be stored in the colliding species' vibrational, translational, rotational or electronic modes, as mentioned previously. Polanyi's rules describe the relative efficacy of energy stored in translational and vibrational modes in allowing a reacting system to pass the energy barrier on collision. These state that vibrational energy is likely to be more efficient in supporting an endothermic reaction (i.e. one with a late-stage transition state resembling the products, according to Hammond's postulate), an example of which is the H + HF reaction, and that translational energy is likely to be more efficient in supporting an exothermic reaction, an example of which is the F + H2 reaction.7

An example of a successful reactive H + HF trajectory with a high initial value of H-F vibrational energy is given in figures 15 and 16 below; figures 17 and 18 show an unreactive trajectory with a lower initial value of H-F vibrational energy and a higher value for the translational energy of the colliding hydrogen atom. This demonstrates the effect described by Polanyi's rules. However, these relative values are approximate, and it is not possible to predict with certainty the reactivity of a given trajectory based on the relative amounts of vibrational and translational energy.7

<gallery heights=200px widths=200px>
File: bd316figure16.png|Figure 15: potential energy surface of an H + HF reactive trajectory
File: bd316figure17.png|Figure 16: contour plot of an H + HF reactive trajectory
File: bd316figure18.png|Figure 17: potential energy surface of an unreactive H + HF trajectory
File: bd316figure19.png|Figure 18: contour plot of an unreactive H + HF trajectory
</gallery>

= References =

1. Steinfeld, J.I., Francisco, J.S., Hase, W.L., Chemical Kinetics and Dynamics, Prentice Hall, 1989

2. Pilling, M. J., Seakins, P. W., Reaction Kinetics, Oxford University Press, 1995

3. Atkins, P. W., Atkins' Physical Chemistry, Oxford University Press, 2014

4. Mahan, B. H., J. Chem. Educ., Activated Complex Theory of Bimolecular Reactions, University of California, 1974

5. Polanyi, J. C., Tardy, D. C., J. Chem. Phys., Energy Distribution in the Exothermic Reaction F+ H2 and the Endothermic Reaction HF + H, 1969

6. IUPAC, Compendium of Chemical Terminology, 2nd ed. (the "Gold Book"), compiled by A. D. McNaught and A. Wilkinson, Blackwell Scientific Publications, Oxford, 1997

7. Zhang, Z., Zhou, Y., Zhang, D.H, Czakó, G., Bowman, J., J. Phys. Chem. Lett., Theoretical Study of the Validity of the Polanyi Rules for the Late Barrier Cl + CHD3 Reaction, 2012

MRD:bd316

2018-06-07T16:57:01Z

Mm10114: /* Using transition state theory to predict rates of reaction */

= Molecular reaction dynamics of an H2 + H reacting system =

In this section the reaction dynamics of the following system, whereby a free hydrogen atom collides with a H2 molecule to form a new H2 molecule and a new free H atom, are considered:

[[File: bd316scheme1.png|center|500px]]

The three atoms HC, HB and HA are constrained to a 180° angle such that a potential energy surface for the reaction may be calculated.2

=== The dynamics of the potential energy surface plot - the energy minima and transition state region ===

[[File:bd316figure1.png|thumb|Figure 1: the three-dimensional potential energy surface of the reacting system|500px]]

The trajectory of the reaction follows the path of energy minima. The transition state region is defined as the energy maximum on the path of minimum energy between the reactants and products.

Hydrogen atom A approaches hydrogen molecule BC (starting from the left hand side of figure 1) following the path of the potential energy surface along which ∂(V(rBC))/∂(rBC) = 0. Along this path the value of the gradient ∂(V(rBA))/∂(rBA) gradually increases until it reaches a maximum along the path of minimum energy - this is the transition state region. From the transition state region, the AB bond forms and the hydrogen atom C moves away from H2 molecule AB along the path along the potential energy surface on which ∂(V(rBA))/∂(rBA) = 0. The reaction coordinate continues along this path until it reaches the minimum of the potential energy surface on both axes, at which ∂(V(rBA))/∂(rBA) = ∂(V(rBC))/∂(rBC) = 0, at which point the reaction is complete.

In the transition state region, the potential energy surface has reached a saddle point, which has the minimum potential energy whereby rCB = rBA. In the transition state region, we can consider there to be a third 'axis', running diagonally from the rBC axis to the rBA axis through the saddle point (s), which defines the reaction coordinate (the view along this axis is given in figure 1). At the saddle point ∂(V(s))/∂(s)= 0; this is an energy maximum along the reaction pathway - ∂2(V(s))/∂(s)2 < 0. The 'axis' running orthogonal to the axis s, through the saddle point from the origin (o) also has ∂(V(o))/∂(o)= 0 and is an energy minimum with respect to this axis - ∂2(V(o))/∂(o)2 > 0.1 Taking the second derivative distinguishes the minima and maxima along a given axis.

[[File:bd316figure2.png|thumb|Figure 2: the contour potential energy surface of the reacting system|400px]]

=== Determining the geometry of the transition state ===

Computationally we can use the fact that, in the transition state, rBC = rBA (if this were not the case the potential energy surface in figure 1 would not be symmetrical) to determine the value of rTS. The results of this calculation are illustrated by the plot of rTS vs time in figure 3. When the transition state has zero momentum and the reaction does not proceed, its potential energy fluctuates in the transition state region and the value of rTS oscillates between approximately 0.8 to 1.0 Å.

An enlarged section of the potential energy surface plot (shown in full in figure 1) in figure 4 shows these inter-atomic vibrations. The value of rTS fluctuates because the transition state is able to interconvert between kinetic energy and potential energy, as is illustrated in figure 5. There are three vibrational degrees freedom at the saddle point - one of these corresponds to infinitesimal motion along the reaction path, and the other two are vibrational motion orthogonal to the reaction path (through the saddle point and along the energy minima).1,2

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 17:50, 7 June 2018 (BST) You correctly observe that the TS is somewhere between the values of 0.8 and 1.0 Å. However, there is only one TS at this potential energy surface. And it was the purpose of this exercise to find it. You were supposed to sample different r1=r2 until you find your best estimate. You would be able to tell this when, for example in Figure 5, you would have no oscillation for the distances and there would be no interconversion of the potential-kinetic energies. So, in the end you were supposed to report a single value for r1=r2, with your best estimate for the TS. In your current calculation, if you would increase number of steps significantly, the aggregation of numerical errors would probably result in your trajectory skewing towards reactants/products. With a good estimate for TS that would not happen.

<gallery widths=300px heights=300px>
File:bd316figure3.png|Figure 3: plot of internuclear distance vs time for the transition state
File:bd316figure4.png|Figure 4: the transition state region of the potential energy surface showing vibrational excitation of the transition state
File:bd316figure5.png|Figure 5: kinetic energy and potential energy vs time for the transition state
</gallery>

=== Calculating the reaction path ===

The minimum energy path, or reaction path, is the path that follows the energy minima from the saddle point to the products along a series of infinitesimally small steps - the vibrational degree of freedom mentioned previously (figure 7). At each of these steps, the kinetic energy is removed, such that the reacting system always has the minimum possible potential energy.1 This distinguishes the minimum energy path from the dynamic reaction path (figure 8), which gives a realistic picture of the trajectory that the reacting system will follow, vibrating orthogonally to the reaction path as the reaction coordinate progresses.1

<gallery heights=250px widths=250px>
File:bd316figure7.png|Figure 7: the minimum energy path
File:bd316figure8.png|Figure 8: the dynamic reaction path
</gallery>

Figure 9 shows the progression of the values rCB and rBA as the reaction progresses from the transition state region to the products, following the dynamic reaction trajectory. At time zero rCB = rBA = rTS. As HC leaves and the HB-HA bond forms, the value of rCB increases to infinity, and the value of rBA reaches the value of the equilibrium H-H bond length - this fluctuates between values of approximately 0.715 - 0.748 Å, with an average value of approximately 0.726 Å.

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 17:54, 7 June 2018 (BST) Very good observation for the MEP and dynamics calculations differences. However, as you have not given the value for rCB = rBA = rTS in the previous exercise, which value have you used here? Although, it does look on the plots like you're at the transition state, but without a value reported I can't be sure.

The progression of the values of interatomic momentum are shown in figure 10. As the reaction coordinate initially progresses from the transition state the values of both momenta initially increase (figure 11). The value of pCB increases sharply and plateaus at 2.5 when C and B are fully dissociated. The value of pBA first decreases as the HB-HA bond contracts on forming, then reaches the equilibrium value of approximately 1.2, oscillating between 0.95 and 1.53 as the HB-HA bond vibrates.

<gallery widths=250px heights=250px>
File:bd316figure9.png|Figure 9: plot of internuclear distance vs time for the dynamic reacting system
File:bd316figure10.png|Figure 10: plot of internuclear momentum vs time for the dynamic reacting system
File:bd316figure11.png|Figure 11: expanded region of the plot of internuclear momentum vs time for the dynamic reacting system
</gallery>

=== Reactive and unreactive trajectories ===

A trajectory may be reactive or unreactive depending on the values of pCB and pBA, which determine the kinetic energy with which the constituent species collide; they must have sufficient total energy to overcome the activation energy barrier and pass through the transition state region to the products. The reactivities of a series of trajectories of varying initial values of the momenta are compared below.

{| class="wikitable" border=1
! Trajectory !! pCB !! pBA !! Total energy (kJ mol-1) !! Reactivity!! Plot !! Comment
|-
| 1 || -1.25 || -2.5 || -99.1 || Reactive || [[File: bd316table1.png|200px]] || The reacting system has sufficient energy to pass over the transition state region. The products HC + HBHA are formed and the resultant HBHA molecule shows a greater degree of vibrational excitation relative to the HCHB molecule.
|-
| 2 ||-1.5 || -2.0 || -100.4 || Unreactive || [[File: bd316table2.png|200px]] || The reacting system has insufficient energy to pass over the transition state region.
|-
| 3 ||-1.5 || -2.5 || -99.0 || Reactive || [[File: bd316table3.png|200px]] || The reacting system has sufficient energy to pass over the transition state region. The resultant HBHC molecule shows a greater degree of vibrational excitation than that resulting from trajectory 1 because of the greater initial value of pCB.
|-
| 4 || -2.5 || -5.0 || -85.0 || Unreactive || [[File: bd316table4.png|200px]] || The species collide with a great deal of kinetic energy - the activation barrier is passed and re-passed, because the initially formed products HC + HB-HA have such a high degree of vibrational excitation that they recollide with sufficient kinetic energy to re-form the reactants.
|-
| 5 || -2.5 || -5.2 || -83.4 || Reactive || [[File: bd316table5.png|200px]] || The reacting species collide with even greater kinetic energy than in trajectory 4 - the activation barrier is crossed, re-crossed and crossed again, until the product species are formed with internuclear momenta such they do not re-collide.
|}

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 17:55, 7 June 2018 (BST) Good observations. For the reported total energies, are these the values taken directly from the LepsGUI? If yes, are you sure the unit is kJ mol-1?

== Using transition state theory to predict rates of reaction ==

Transition state theory provides a means of predicting the rate constant for a given reaction that passes through a transition state, based on the concepts of statistical thermodynamics.3 It is assumed that the reactants are in rapid equilibrium with the transition state complex; the products are formed when this complex decays in to the products. Upon further calculation, the rate constant of reaction is given by the Eyring equation:3

[[File: bd316eyring.png|center]]

In which kr is the reaction rate, κ is a proportionality constant between the rate of passage of the reacting system through the transition state and the vibration frequency of the reacting system along the reaction coordinate, k is the Botzmann constant, h is Planck's constant, T is the temperature and K-barcǂ is the rate of formation of the transition state.3 K-barcǂ may be determined from the partition functions of the reactants and products, which are themselves easily calculated spectroscopically. The subscript c denotes that the the motion along the reaction coordinate has been separated from the other modes of motion of the species (equivalent to the Born-Oppenheimer approximation) and treated entirely classically.1,3

In calculating the rate constant using transition state theory, three main assumptions are made:4

1. That the reacting species are in equilibrium with the transition state complex, as mentioned previously - this allows the rate constant to be determined using statistical thermodynamics expressions in which partition functions appear (is is assumed that the 'phase space' of the reactants is populated according to the Boltzmann distribution), which are derived with the assumption of equilibrium.

2. That the system can be described adequately with classical mechanics.

3. That systems which pass the transition state barrier always become products.

Each of these assumptions may cause deviations of the calculated rate constants from the experimentally determined rate constants to different extents, depending on the reality of the system considered. For our reacting system in which the molecules may repass the transition state region when they collide with sufficient kinetic energy, as in trajectory 4 in the table above, assumption three may lead to marked deviations between the experimental and calculated rate constants. Assumption two neglects the possibility of quantum tunneling over the activation barrier, underestimating the rate of reaction - as yet there is no transition state theory model that successfully takes quantum mechanical effects in to account.1,3

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 17:57, 7 June 2018 (BST) Well explained.

= Molecular reaction dynamics of an F - H - H reacting system =

In this section the dynamics of the following, slightly more complex, reaction between a free flourine atom and an H2 molecule are considered:

[[File: bd316scheme2.png|center|500px]]

=== Inspection of the potential energy surface ===
[[File: bd316figure13.png|thumb| Figure 12 a: the potential energy surface of the F-H-H reacting system showing a reactive trajectory in the exothermic direction|300px]]
[[File: bd316figure13b.png|thumb| Figure 12 b: the potential energy surface of the F-H-H reacting system from above|300px]]
[[File: bd316figure14.png|thumb| Figure 13: the contour plot of the F-H-H reacting system showing the location of the transition state|300px]]
[[File: bd316figure15.png|thumb| Figure 14: plot of internuclear momenta vs. time for the F + H2 reacting system following the trajectory illustrated in figure 13|300px]]

The reaction is exothermic in the F + H2 --> FH + H direction, because the products are lower in energy than the reactants, and endothermic in the reverse direction (figure 12a). This indicates the greater bond strength of the H-F bond relative to the H-H bond - this is confirmed by literature values.5

The activation energy of the F + H2 reaction is small - this makes the transition state difficult to locate by eye alone. Hammond's postulate states that the transition state will have the geometry closest to the species (reactants or products) that are closest in energy.6 The location of the transition state for the exothermic F + H2 reaction will therefore most closely resemble the geometry of the reactants. The location of the transition state can be found by starting from the F-H-H geometry with the H-H bond length and optimising until the potential energy surface no longer shows a trajectory with zero internuclear momentum - the location of the transition state is given in figure 13. The transition state complex obtained by this method has an F-H bond length of 1.813 Å and an H-H bond length of 0.741 Å.

=== Calculating the activation energies ===

The activation energy is defined as the difference in energy between the reactants and the transition state, i.e. the minimum energy that the reactants must have in order to react. Comparison of the energy of the transition state (-103.5 kJ mol-1) with the minimum energy of the reacting species (-133.3 kJ mol-1) gives a value of 29.8 kJ mol-1 for the activation energy of the endothermic H + HF --> F + H2 reaction. The minimum energy of H2 was found to be -104.3 kJ mol-1, hence the activation energy in the exothermic direction is 0.8 kJ mol -1.

=== Energetics of the reacting system ===

According to the first law of thermodynamics, it is not possible for energy to be created or destroyed.3 Hence, when reactants collide with insufficient energy to reach the transition state they may not react (quantum mechanical possibilities aside), and when they collide with more than sufficient energy to react, the excess energy of reaction must 'go somewhere'. This excess energy may be dissipated as heat (random motion, which results in an increase in temperature), or stored in one of the molecules translational, vibrational, rotational or electronic modes of energy (excited electronic modes are typically thermally inaccessible). The amount of energy which is dissipated as heat relative to the amount which is stored in a given species' energetic modes is determined by how many energetic modes the species has available - this is the 'heat capacity'.

This is seen in figure 12 and in many of the reactive trajectories in the previous section, where the reactive trajectory oscillates to a greater extent orthogonal to the reaction coordinate after passing the transition state region, because some of the excess energy is converted to vibrational energy of the products.3 This is also clearly illustrated in figure 14, in which it can be seen that the internuculear momentum in the product H-F molecule oscillates to a significantly greater extent than in the reactant H2 molecule because there is more energy in the H-F vibrational modes.

Calorimetry may be used to experimentally measure the flow of heat resulting from an exothermic reaction, and from the difference in this value, the activation energy and the energy of the reactants the amount of energy stored in the reacting species' energetic modes may be deduced.3

=== Polanyi's empirical rules ===

The energy that contributes to a reacting system's ability to pass an activation barrier on collision may be stored in the colliding species' vibrational, translational, rotational or electronic modes, as mentioned previously. Polanyi's rules describe the relative efficacy of energy stored in translational and vibrational modes in allowing a reacting system to pass the energy barrier on collision. These state that vibrational energy is likely to be more efficient in supporting an endothermic reaction (i.e. one with a late-stage transition state resembling the products, according to Hammond's postulate), an example of which is the H + HF reaction, and that translational energy is likely to be more efficient in supporting an exothermic reaction, an example of which is the F + H2 reaction.7

An example of a successful reactive H + HF trajectory with a high initial value of H-F vibrational energy is given in figures 15 and 16 below; figures 17 and 18 show an unreactive trajectory with a lower initial value of H-F vibrational energy and a higher value for the translational energy of the colliding hydrogen atom. This demonstrates the effect described by Polanyi's rules. However, these relative values are approximate, and it is not possible to predict with certainty the reactivity of a given trajectory based on the relative amounts of vibrational and translational energy.7

<gallery heights=200px widths=200px>
File: bd316figure16.png|Figure 15: potential energy surface of an H + HF reactive trajectory
File: bd316figure17.png|Figure 16: contour plot of an H + HF reactive trajectory
File: bd316figure18.png|Figure 17: potential energy surface of an unreactive H + HF trajectory
File: bd316figure19.png|Figure 18: contour plot of an unreactive H + HF trajectory
</gallery>

= References =

1. Steinfeld, J.I., Francisco, J.S., Hase, W.L., Chemical Kinetics and Dynamics, Prentice Hall, 1989

2. Pilling, M. J., Seakins, P. W., Reaction Kinetics, Oxford University Press, 1995

3. Atkins, P. W., Atkins' Physical Chemistry, Oxford University Press, 2014

4. Mahan, B. H., J. Chem. Educ., Activated Complex Theory of Bimolecular Reactions, University of California, 1974

5. Polanyi, J. C., Tardy, D. C., J. Chem. Phys., Energy Distribution in the Exothermic Reaction F+ H2 and the Endothermic Reaction HF + H, 1969

6. IUPAC, Compendium of Chemical Terminology, 2nd ed. (the "Gold Book"), compiled by A. D. McNaught and A. Wilkinson, Blackwell Scientific Publications, Oxford, 1997

7. Zhang, Z., Zhou, Y., Zhang, D.H, Czakó, G., Bowman, J., J. Phys. Chem. Lett., Theoretical Study of the Validity of the Polanyi Rules for the Late Barrier Cl + CHD3 Reaction, 2012

MRD:bd316

2018-06-07T16:55:47Z

Mm10114: /* Reactive and unreactive trajectories */

= Molecular reaction dynamics of an H2 + H reacting system =

In this section the reaction dynamics of the following system, whereby a free hydrogen atom collides with a H2 molecule to form a new H2 molecule and a new free H atom, are considered:

[[File: bd316scheme1.png|center|500px]]

The three atoms HC, HB and HA are constrained to a 180° angle such that a potential energy surface for the reaction may be calculated.2

=== The dynamics of the potential energy surface plot - the energy minima and transition state region ===

[[File:bd316figure1.png|thumb|Figure 1: the three-dimensional potential energy surface of the reacting system|500px]]

The trajectory of the reaction follows the path of energy minima. The transition state region is defined as the energy maximum on the path of minimum energy between the reactants and products.

Hydrogen atom A approaches hydrogen molecule BC (starting from the left hand side of figure 1) following the path of the potential energy surface along which ∂(V(rBC))/∂(rBC) = 0. Along this path the value of the gradient ∂(V(rBA))/∂(rBA) gradually increases until it reaches a maximum along the path of minimum energy - this is the transition state region. From the transition state region, the AB bond forms and the hydrogen atom C moves away from H2 molecule AB along the path along the potential energy surface on which ∂(V(rBA))/∂(rBA) = 0. The reaction coordinate continues along this path until it reaches the minimum of the potential energy surface on both axes, at which ∂(V(rBA))/∂(rBA) = ∂(V(rBC))/∂(rBC) = 0, at which point the reaction is complete.

In the transition state region, the potential energy surface has reached a saddle point, which has the minimum potential energy whereby rCB = rBA. In the transition state region, we can consider there to be a third 'axis', running diagonally from the rBC axis to the rBA axis through the saddle point (s), which defines the reaction coordinate (the view along this axis is given in figure 1). At the saddle point ∂(V(s))/∂(s)= 0; this is an energy maximum along the reaction pathway - ∂2(V(s))/∂(s)2 < 0. The 'axis' running orthogonal to the axis s, through the saddle point from the origin (o) also has ∂(V(o))/∂(o)= 0 and is an energy minimum with respect to this axis - ∂2(V(o))/∂(o)2 > 0.1 Taking the second derivative distinguishes the minima and maxima along a given axis.

[[File:bd316figure2.png|thumb|Figure 2: the contour potential energy surface of the reacting system|400px]]

=== Determining the geometry of the transition state ===

Computationally we can use the fact that, in the transition state, rBC = rBA (if this were not the case the potential energy surface in figure 1 would not be symmetrical) to determine the value of rTS. The results of this calculation are illustrated by the plot of rTS vs time in figure 3. When the transition state has zero momentum and the reaction does not proceed, its potential energy fluctuates in the transition state region and the value of rTS oscillates between approximately 0.8 to 1.0 Å.

An enlarged section of the potential energy surface plot (shown in full in figure 1) in figure 4 shows these inter-atomic vibrations. The value of rTS fluctuates because the transition state is able to interconvert between kinetic energy and potential energy, as is illustrated in figure 5. There are three vibrational degrees freedom at the saddle point - one of these corresponds to infinitesimal motion along the reaction path, and the other two are vibrational motion orthogonal to the reaction path (through the saddle point and along the energy minima).1,2

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 17:50, 7 June 2018 (BST) You correctly observe that the TS is somewhere between the values of 0.8 and 1.0 Å. However, there is only one TS at this potential energy surface. And it was the purpose of this exercise to find it. You were supposed to sample different r1=r2 until you find your best estimate. You would be able to tell this when, for example in Figure 5, you would have no oscillation for the distances and there would be no interconversion of the potential-kinetic energies. So, in the end you were supposed to report a single value for r1=r2, with your best estimate for the TS. In your current calculation, if you would increase number of steps significantly, the aggregation of numerical errors would probably result in your trajectory skewing towards reactants/products. With a good estimate for TS that would not happen.

<gallery widths=300px heights=300px>
File:bd316figure3.png|Figure 3: plot of internuclear distance vs time for the transition state
File:bd316figure4.png|Figure 4: the transition state region of the potential energy surface showing vibrational excitation of the transition state
File:bd316figure5.png|Figure 5: kinetic energy and potential energy vs time for the transition state
</gallery>

=== Calculating the reaction path ===

The minimum energy path, or reaction path, is the path that follows the energy minima from the saddle point to the products along a series of infinitesimally small steps - the vibrational degree of freedom mentioned previously (figure 7). At each of these steps, the kinetic energy is removed, such that the reacting system always has the minimum possible potential energy.1 This distinguishes the minimum energy path from the dynamic reaction path (figure 8), which gives a realistic picture of the trajectory that the reacting system will follow, vibrating orthogonally to the reaction path as the reaction coordinate progresses.1

<gallery heights=250px widths=250px>
File:bd316figure7.png|Figure 7: the minimum energy path
File:bd316figure8.png|Figure 8: the dynamic reaction path
</gallery>

Figure 9 shows the progression of the values rCB and rBA as the reaction progresses from the transition state region to the products, following the dynamic reaction trajectory. At time zero rCB = rBA = rTS. As HC leaves and the HB-HA bond forms, the value of rCB increases to infinity, and the value of rBA reaches the value of the equilibrium H-H bond length - this fluctuates between values of approximately 0.715 - 0.748 Å, with an average value of approximately 0.726 Å.

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 17:54, 7 June 2018 (BST) Very good observation for the MEP and dynamics calculations differences. However, as you have not given the value for rCB = rBA = rTS in the previous exercise, which value have you used here? Although, it does look on the plots like you're at the transition state, but without a value reported I can't be sure.

The progression of the values of interatomic momentum are shown in figure 10. As the reaction coordinate initially progresses from the transition state the values of both momenta initially increase (figure 11). The value of pCB increases sharply and plateaus at 2.5 when C and B are fully dissociated. The value of pBA first decreases as the HB-HA bond contracts on forming, then reaches the equilibrium value of approximately 1.2, oscillating between 0.95 and 1.53 as the HB-HA bond vibrates.

<gallery widths=250px heights=250px>
File:bd316figure9.png|Figure 9: plot of internuclear distance vs time for the dynamic reacting system
File:bd316figure10.png|Figure 10: plot of internuclear momentum vs time for the dynamic reacting system
File:bd316figure11.png|Figure 11: expanded region of the plot of internuclear momentum vs time for the dynamic reacting system
</gallery>

=== Reactive and unreactive trajectories ===

A trajectory may be reactive or unreactive depending on the values of pCB and pBA, which determine the kinetic energy with which the constituent species collide; they must have sufficient total energy to overcome the activation energy barrier and pass through the transition state region to the products. The reactivities of a series of trajectories of varying initial values of the momenta are compared below.

{| class="wikitable" border=1
! Trajectory !! pCB !! pBA !! Total energy (kJ mol-1) !! Reactivity!! Plot !! Comment
|-
| 1 || -1.25 || -2.5 || -99.1 || Reactive || [[File: bd316table1.png|200px]] || The reacting system has sufficient energy to pass over the transition state region. The products HC + HBHA are formed and the resultant HBHA molecule shows a greater degree of vibrational excitation relative to the HCHB molecule.
|-
| 2 ||-1.5 || -2.0 || -100.4 || Unreactive || [[File: bd316table2.png|200px]] || The reacting system has insufficient energy to pass over the transition state region.
|-
| 3 ||-1.5 || -2.5 || -99.0 || Reactive || [[File: bd316table3.png|200px]] || The reacting system has sufficient energy to pass over the transition state region. The resultant HBHC molecule shows a greater degree of vibrational excitation than that resulting from trajectory 1 because of the greater initial value of pCB.
|-
| 4 || -2.5 || -5.0 || -85.0 || Unreactive || [[File: bd316table4.png|200px]] || The species collide with a great deal of kinetic energy - the activation barrier is passed and re-passed, because the initially formed products HC + HB-HA have such a high degree of vibrational excitation that they recollide with sufficient kinetic energy to re-form the reactants.
|-
| 5 || -2.5 || -5.2 || -83.4 || Reactive || [[File: bd316table5.png|200px]] || The reacting species collide with even greater kinetic energy than in trajectory 4 - the activation barrier is crossed, re-crossed and crossed again, until the product species are formed with internuclear momenta such they do not re-collide.
|}

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 17:55, 7 June 2018 (BST) Good observations. For the reported total energies, are these the values taken directly from the LepsGUI? If yes, are you sure the unit is kJ mol-1?

== Using transition state theory to predict rates of reaction ==

Transition state theory provides a means of predicting the rate constant for a given reaction that passes through a transition state, based on the concepts of statistical thermodynamics.3 It is assumed that the reactants are in rapid equilibrium with the transition state complex; the products are formed when this complex decays in to the products. Upon further calculation, the rate constant of reaction is given by the Eyring equation:3

[[File: bd316eyring.png|center]]

In which kr is the reaction rate, κ is a proportionality constant between the rate of passage of the reacting system through the transition state and the vibration frequency of the reacting system along the reaction coordinate, k is the Botzmann constant, h is Planck's constant, T is the temperature and K-barcǂ is the rate of formation of the transition state.3 K-barcǂ may be determined from the partition functions of the reactants and products, which are themselves easily calculated spectroscopically. The subscript c denotes that the the motion along the reaction coordinate has been separated from the other modes of motion of the species (equivalent to the Born-Oppenheimer approximation) and treated entirely classically.1,3

In calculating the rate constant using transition state theory, three main assumptions are made:4

1. That the reacting species are in equilibrium with the transition state complex, as mentioned previously - this allows the rate constant to be determined using statistical thermodynamics expressions in which partition functions appear (is is assumed that the 'phase space' of the reactants is populated according to the Boltzmann distribution), which are derived with the assumption of equilibrium.

2. That the system can be described adequately with classical mechanics.

3. That systems which pass the transition state barrier always become products.

Each of these assumptions may cause deviations of the calculated rate constants from the experimentally determined rate constants to different extents, depending on the reality of the system considered. For our reacting system in which the molecules may repass the transition state region when they collide with sufficient kinetic energy, as in trajectory 4 in the table above, assumption three may lead to marked deviations between the experimental and calculated rate constants. Assumption two neglects the possibility of quantum tunneling over the activation barrier, underestimating the rate of reaction - as yet there is no transition state theory model that successfully takes quantum mechanical effects in to account.1,3

= Molecular reaction dynamics of an F - H - H reacting system =

In this section the dynamics of the following, slightly more complex, reaction between a free flourine atom and an H2 molecule are considered:

[[File: bd316scheme2.png|center|500px]]

=== Inspection of the potential energy surface ===
[[File: bd316figure13.png|thumb| Figure 12 a: the potential energy surface of the F-H-H reacting system showing a reactive trajectory in the exothermic direction|300px]]
[[File: bd316figure13b.png|thumb| Figure 12 b: the potential energy surface of the F-H-H reacting system from above|300px]]
[[File: bd316figure14.png|thumb| Figure 13: the contour plot of the F-H-H reacting system showing the location of the transition state|300px]]
[[File: bd316figure15.png|thumb| Figure 14: plot of internuclear momenta vs. time for the F + H2 reacting system following the trajectory illustrated in figure 13|300px]]

The reaction is exothermic in the F + H2 --> FH + H direction, because the products are lower in energy than the reactants, and endothermic in the reverse direction (figure 12a). This indicates the greater bond strength of the H-F bond relative to the H-H bond - this is confirmed by literature values.5

The activation energy of the F + H2 reaction is small - this makes the transition state difficult to locate by eye alone. Hammond's postulate states that the transition state will have the geometry closest to the species (reactants or products) that are closest in energy.6 The location of the transition state for the exothermic F + H2 reaction will therefore most closely resemble the geometry of the reactants. The location of the transition state can be found by starting from the F-H-H geometry with the H-H bond length and optimising until the potential energy surface no longer shows a trajectory with zero internuclear momentum - the location of the transition state is given in figure 13. The transition state complex obtained by this method has an F-H bond length of 1.813 Å and an H-H bond length of 0.741 Å.

=== Calculating the activation energies ===

The activation energy is defined as the difference in energy between the reactants and the transition state, i.e. the minimum energy that the reactants must have in order to react. Comparison of the energy of the transition state (-103.5 kJ mol-1) with the minimum energy of the reacting species (-133.3 kJ mol-1) gives a value of 29.8 kJ mol-1 for the activation energy of the endothermic H + HF --> F + H2 reaction. The minimum energy of H2 was found to be -104.3 kJ mol-1, hence the activation energy in the exothermic direction is 0.8 kJ mol -1.

=== Energetics of the reacting system ===

According to the first law of thermodynamics, it is not possible for energy to be created or destroyed.3 Hence, when reactants collide with insufficient energy to reach the transition state they may not react (quantum mechanical possibilities aside), and when they collide with more than sufficient energy to react, the excess energy of reaction must 'go somewhere'. This excess energy may be dissipated as heat (random motion, which results in an increase in temperature), or stored in one of the molecules translational, vibrational, rotational or electronic modes of energy (excited electronic modes are typically thermally inaccessible). The amount of energy which is dissipated as heat relative to the amount which is stored in a given species' energetic modes is determined by how many energetic modes the species has available - this is the 'heat capacity'.

This is seen in figure 12 and in many of the reactive trajectories in the previous section, where the reactive trajectory oscillates to a greater extent orthogonal to the reaction coordinate after passing the transition state region, because some of the excess energy is converted to vibrational energy of the products.3 This is also clearly illustrated in figure 14, in which it can be seen that the internuculear momentum in the product H-F molecule oscillates to a significantly greater extent than in the reactant H2 molecule because there is more energy in the H-F vibrational modes.

Calorimetry may be used to experimentally measure the flow of heat resulting from an exothermic reaction, and from the difference in this value, the activation energy and the energy of the reactants the amount of energy stored in the reacting species' energetic modes may be deduced.3

=== Polanyi's empirical rules ===

The energy that contributes to a reacting system's ability to pass an activation barrier on collision may be stored in the colliding species' vibrational, translational, rotational or electronic modes, as mentioned previously. Polanyi's rules describe the relative efficacy of energy stored in translational and vibrational modes in allowing a reacting system to pass the energy barrier on collision. These state that vibrational energy is likely to be more efficient in supporting an endothermic reaction (i.e. one with a late-stage transition state resembling the products, according to Hammond's postulate), an example of which is the H + HF reaction, and that translational energy is likely to be more efficient in supporting an exothermic reaction, an example of which is the F + H2 reaction.7

An example of a successful reactive H + HF trajectory with a high initial value of H-F vibrational energy is given in figures 15 and 16 below; figures 17 and 18 show an unreactive trajectory with a lower initial value of H-F vibrational energy and a higher value for the translational energy of the colliding hydrogen atom. This demonstrates the effect described by Polanyi's rules. However, these relative values are approximate, and it is not possible to predict with certainty the reactivity of a given trajectory based on the relative amounts of vibrational and translational energy.7

<gallery heights=200px widths=200px>
File: bd316figure16.png|Figure 15: potential energy surface of an H + HF reactive trajectory
File: bd316figure17.png|Figure 16: contour plot of an H + HF reactive trajectory
File: bd316figure18.png|Figure 17: potential energy surface of an unreactive H + HF trajectory
File: bd316figure19.png|Figure 18: contour plot of an unreactive H + HF trajectory
</gallery>

= References =

1. Steinfeld, J.I., Francisco, J.S., Hase, W.L., Chemical Kinetics and Dynamics, Prentice Hall, 1989

2. Pilling, M. J., Seakins, P. W., Reaction Kinetics, Oxford University Press, 1995

3. Atkins, P. W., Atkins' Physical Chemistry, Oxford University Press, 2014

4. Mahan, B. H., J. Chem. Educ., Activated Complex Theory of Bimolecular Reactions, University of California, 1974

5. Polanyi, J. C., Tardy, D. C., J. Chem. Phys., Energy Distribution in the Exothermic Reaction F+ H2 and the Endothermic Reaction HF + H, 1969

6. IUPAC, Compendium of Chemical Terminology, 2nd ed. (the "Gold Book"), compiled by A. D. McNaught and A. Wilkinson, Blackwell Scientific Publications, Oxford, 1997

7. Zhang, Z., Zhou, Y., Zhang, D.H, Czakó, G., Bowman, J., J. Phys. Chem. Lett., Theoretical Study of the Validity of the Polanyi Rules for the Late Barrier Cl + CHD3 Reaction, 2012

MRD:bd316

2018-06-07T16:54:27Z

Mm10114: /* Calculating the reaction path */

= Molecular reaction dynamics of an H2 + H reacting system =

In this section the reaction dynamics of the following system, whereby a free hydrogen atom collides with a H2 molecule to form a new H2 molecule and a new free H atom, are considered:

[[File: bd316scheme1.png|center|500px]]

The three atoms HC, HB and HA are constrained to a 180° angle such that a potential energy surface for the reaction may be calculated.2

=== The dynamics of the potential energy surface plot - the energy minima and transition state region ===

[[File:bd316figure1.png|thumb|Figure 1: the three-dimensional potential energy surface of the reacting system|500px]]

The trajectory of the reaction follows the path of energy minima. The transition state region is defined as the energy maximum on the path of minimum energy between the reactants and products.

Hydrogen atom A approaches hydrogen molecule BC (starting from the left hand side of figure 1) following the path of the potential energy surface along which ∂(V(rBC))/∂(rBC) = 0. Along this path the value of the gradient ∂(V(rBA))/∂(rBA) gradually increases until it reaches a maximum along the path of minimum energy - this is the transition state region. From the transition state region, the AB bond forms and the hydrogen atom C moves away from H2 molecule AB along the path along the potential energy surface on which ∂(V(rBA))/∂(rBA) = 0. The reaction coordinate continues along this path until it reaches the minimum of the potential energy surface on both axes, at which ∂(V(rBA))/∂(rBA) = ∂(V(rBC))/∂(rBC) = 0, at which point the reaction is complete.

In the transition state region, the potential energy surface has reached a saddle point, which has the minimum potential energy whereby rCB = rBA. In the transition state region, we can consider there to be a third 'axis', running diagonally from the rBC axis to the rBA axis through the saddle point (s), which defines the reaction coordinate (the view along this axis is given in figure 1). At the saddle point ∂(V(s))/∂(s)= 0; this is an energy maximum along the reaction pathway - ∂2(V(s))/∂(s)2 < 0. The 'axis' running orthogonal to the axis s, through the saddle point from the origin (o) also has ∂(V(o))/∂(o)= 0 and is an energy minimum with respect to this axis - ∂2(V(o))/∂(o)2 > 0.1 Taking the second derivative distinguishes the minima and maxima along a given axis.

[[File:bd316figure2.png|thumb|Figure 2: the contour potential energy surface of the reacting system|400px]]

=== Determining the geometry of the transition state ===

Computationally we can use the fact that, in the transition state, rBC = rBA (if this were not the case the potential energy surface in figure 1 would not be symmetrical) to determine the value of rTS. The results of this calculation are illustrated by the plot of rTS vs time in figure 3. When the transition state has zero momentum and the reaction does not proceed, its potential energy fluctuates in the transition state region and the value of rTS oscillates between approximately 0.8 to 1.0 Å.

An enlarged section of the potential energy surface plot (shown in full in figure 1) in figure 4 shows these inter-atomic vibrations. The value of rTS fluctuates because the transition state is able to interconvert between kinetic energy and potential energy, as is illustrated in figure 5. There are three vibrational degrees freedom at the saddle point - one of these corresponds to infinitesimal motion along the reaction path, and the other two are vibrational motion orthogonal to the reaction path (through the saddle point and along the energy minima).1,2

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 17:50, 7 June 2018 (BST) You correctly observe that the TS is somewhere between the values of 0.8 and 1.0 Å. However, there is only one TS at this potential energy surface. And it was the purpose of this exercise to find it. You were supposed to sample different r1=r2 until you find your best estimate. You would be able to tell this when, for example in Figure 5, you would have no oscillation for the distances and there would be no interconversion of the potential-kinetic energies. So, in the end you were supposed to report a single value for r1=r2, with your best estimate for the TS. In your current calculation, if you would increase number of steps significantly, the aggregation of numerical errors would probably result in your trajectory skewing towards reactants/products. With a good estimate for TS that would not happen.

<gallery widths=300px heights=300px>
File:bd316figure3.png|Figure 3: plot of internuclear distance vs time for the transition state
File:bd316figure4.png|Figure 4: the transition state region of the potential energy surface showing vibrational excitation of the transition state
File:bd316figure5.png|Figure 5: kinetic energy and potential energy vs time for the transition state
</gallery>

=== Calculating the reaction path ===

The minimum energy path, or reaction path, is the path that follows the energy minima from the saddle point to the products along a series of infinitesimally small steps - the vibrational degree of freedom mentioned previously (figure 7). At each of these steps, the kinetic energy is removed, such that the reacting system always has the minimum possible potential energy.1 This distinguishes the minimum energy path from the dynamic reaction path (figure 8), which gives a realistic picture of the trajectory that the reacting system will follow, vibrating orthogonally to the reaction path as the reaction coordinate progresses.1

<gallery heights=250px widths=250px>
File:bd316figure7.png|Figure 7: the minimum energy path
File:bd316figure8.png|Figure 8: the dynamic reaction path
</gallery>

Figure 9 shows the progression of the values rCB and rBA as the reaction progresses from the transition state region to the products, following the dynamic reaction trajectory. At time zero rCB = rBA = rTS. As HC leaves and the HB-HA bond forms, the value of rCB increases to infinity, and the value of rBA reaches the value of the equilibrium H-H bond length - this fluctuates between values of approximately 0.715 - 0.748 Å, with an average value of approximately 0.726 Å.

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 17:54, 7 June 2018 (BST) Very good observation for the MEP and dynamics calculations differences. However, as you have not given the value for rCB = rBA = rTS in the previous exercise, which value have you used here? Although, it does look on the plots like you're at the transition state, but without a value reported I can't be sure.

The progression of the values of interatomic momentum are shown in figure 10. As the reaction coordinate initially progresses from the transition state the values of both momenta initially increase (figure 11). The value of pCB increases sharply and plateaus at 2.5 when C and B are fully dissociated. The value of pBA first decreases as the HB-HA bond contracts on forming, then reaches the equilibrium value of approximately 1.2, oscillating between 0.95 and 1.53 as the HB-HA bond vibrates.

<gallery widths=250px heights=250px>
File:bd316figure9.png|Figure 9: plot of internuclear distance vs time for the dynamic reacting system
File:bd316figure10.png|Figure 10: plot of internuclear momentum vs time for the dynamic reacting system
File:bd316figure11.png|Figure 11: expanded region of the plot of internuclear momentum vs time for the dynamic reacting system
</gallery>

=== Reactive and unreactive trajectories ===

A trajectory may be reactive or unreactive depending on the values of pCB and pBA, which determine the kinetic energy with which the constituent species collide; they must have sufficient total energy to overcome the activation energy barrier and pass through the transition state region to the products. The reactivities of a series of trajectories of varying initial values of the momenta are compared below.

{| class="wikitable" border=1
! Trajectory !! pCB !! pBA !! Total energy (kJ mol-1) !! Reactivity!! Plot !! Comment
|-
| 1 || -1.25 || -2.5 || -99.1 || Reactive || [[File: bd316table1.png|200px]] || The reacting system has sufficient energy to pass over the transition state region. The products HC + HBHA are formed and the resultant HBHA molecule shows a greater degree of vibrational excitation relative to the HCHB molecule.
|-
| 2 ||-1.5 || -2.0 || -100.4 || Unreactive || [[File: bd316table2.png|200px]] || The reacting system has insufficient energy to pass over the transition state region.
|-
| 3 ||-1.5 || -2.5 || -99.0 || Reactive || [[File: bd316table3.png|200px]] || The reacting system has sufficient energy to pass over the transition state region. The resultant HBHC molecule shows a greater degree of vibrational excitation than that resulting from trajectory 1 because of the greater initial value of pCB.
|-
| 4 || -2.5 || -5.0 || -85.0 || Unreactive || [[File: bd316table4.png|200px]] || The species collide with a great deal of kinetic energy - the activation barrier is passed and re-passed, because the initially formed products HC + HB-HA have such a high degree of vibrational excitation that they recollide with sufficient kinetic energy to re-form the reactants.
|-
| 5 || -2.5 || -5.2 || -83.4 || Reactive || [[File: bd316table5.png|200px]] || The reacting species collide with even greater kinetic energy than in trajectory 4 - the activation barrier is crossed, re-crossed and crossed again, until the product species are formed with internuclear momenta such they do not re-collide.
|}

== Using transition state theory to predict rates of reaction ==

Transition state theory provides a means of predicting the rate constant for a given reaction that passes through a transition state, based on the concepts of statistical thermodynamics.3 It is assumed that the reactants are in rapid equilibrium with the transition state complex; the products are formed when this complex decays in to the products. Upon further calculation, the rate constant of reaction is given by the Eyring equation:3

[[File: bd316eyring.png|center]]

In which kr is the reaction rate, κ is a proportionality constant between the rate of passage of the reacting system through the transition state and the vibration frequency of the reacting system along the reaction coordinate, k is the Botzmann constant, h is Planck's constant, T is the temperature and K-barcǂ is the rate of formation of the transition state.3 K-barcǂ may be determined from the partition functions of the reactants and products, which are themselves easily calculated spectroscopically. The subscript c denotes that the the motion along the reaction coordinate has been separated from the other modes of motion of the species (equivalent to the Born-Oppenheimer approximation) and treated entirely classically.1,3

In calculating the rate constant using transition state theory, three main assumptions are made:4

1. That the reacting species are in equilibrium with the transition state complex, as mentioned previously - this allows the rate constant to be determined using statistical thermodynamics expressions in which partition functions appear (is is assumed that the 'phase space' of the reactants is populated according to the Boltzmann distribution), which are derived with the assumption of equilibrium.

2. That the system can be described adequately with classical mechanics.

3. That systems which pass the transition state barrier always become products.

Each of these assumptions may cause deviations of the calculated rate constants from the experimentally determined rate constants to different extents, depending on the reality of the system considered. For our reacting system in which the molecules may repass the transition state region when they collide with sufficient kinetic energy, as in trajectory 4 in the table above, assumption three may lead to marked deviations between the experimental and calculated rate constants. Assumption two neglects the possibility of quantum tunneling over the activation barrier, underestimating the rate of reaction - as yet there is no transition state theory model that successfully takes quantum mechanical effects in to account.1,3

= Molecular reaction dynamics of an F - H - H reacting system =

In this section the dynamics of the following, slightly more complex, reaction between a free flourine atom and an H2 molecule are considered:

[[File: bd316scheme2.png|center|500px]]

=== Inspection of the potential energy surface ===
[[File: bd316figure13.png|thumb| Figure 12 a: the potential energy surface of the F-H-H reacting system showing a reactive trajectory in the exothermic direction|300px]]
[[File: bd316figure13b.png|thumb| Figure 12 b: the potential energy surface of the F-H-H reacting system from above|300px]]
[[File: bd316figure14.png|thumb| Figure 13: the contour plot of the F-H-H reacting system showing the location of the transition state|300px]]
[[File: bd316figure15.png|thumb| Figure 14: plot of internuclear momenta vs. time for the F + H2 reacting system following the trajectory illustrated in figure 13|300px]]

The reaction is exothermic in the F + H2 --> FH + H direction, because the products are lower in energy than the reactants, and endothermic in the reverse direction (figure 12a). This indicates the greater bond strength of the H-F bond relative to the H-H bond - this is confirmed by literature values.5

The activation energy of the F + H2 reaction is small - this makes the transition state difficult to locate by eye alone. Hammond's postulate states that the transition state will have the geometry closest to the species (reactants or products) that are closest in energy.6 The location of the transition state for the exothermic F + H2 reaction will therefore most closely resemble the geometry of the reactants. The location of the transition state can be found by starting from the F-H-H geometry with the H-H bond length and optimising until the potential energy surface no longer shows a trajectory with zero internuclear momentum - the location of the transition state is given in figure 13. The transition state complex obtained by this method has an F-H bond length of 1.813 Å and an H-H bond length of 0.741 Å.

=== Calculating the activation energies ===

The activation energy is defined as the difference in energy between the reactants and the transition state, i.e. the minimum energy that the reactants must have in order to react. Comparison of the energy of the transition state (-103.5 kJ mol-1) with the minimum energy of the reacting species (-133.3 kJ mol-1) gives a value of 29.8 kJ mol-1 for the activation energy of the endothermic H + HF --> F + H2 reaction. The minimum energy of H2 was found to be -104.3 kJ mol-1, hence the activation energy in the exothermic direction is 0.8 kJ mol -1.

=== Energetics of the reacting system ===

According to the first law of thermodynamics, it is not possible for energy to be created or destroyed.3 Hence, when reactants collide with insufficient energy to reach the transition state they may not react (quantum mechanical possibilities aside), and when they collide with more than sufficient energy to react, the excess energy of reaction must 'go somewhere'. This excess energy may be dissipated as heat (random motion, which results in an increase in temperature), or stored in one of the molecules translational, vibrational, rotational or electronic modes of energy (excited electronic modes are typically thermally inaccessible). The amount of energy which is dissipated as heat relative to the amount which is stored in a given species' energetic modes is determined by how many energetic modes the species has available - this is the 'heat capacity'.

This is seen in figure 12 and in many of the reactive trajectories in the previous section, where the reactive trajectory oscillates to a greater extent orthogonal to the reaction coordinate after passing the transition state region, because some of the excess energy is converted to vibrational energy of the products.3 This is also clearly illustrated in figure 14, in which it can be seen that the internuculear momentum in the product H-F molecule oscillates to a significantly greater extent than in the reactant H2 molecule because there is more energy in the H-F vibrational modes.

Calorimetry may be used to experimentally measure the flow of heat resulting from an exothermic reaction, and from the difference in this value, the activation energy and the energy of the reactants the amount of energy stored in the reacting species' energetic modes may be deduced.3

=== Polanyi's empirical rules ===

The energy that contributes to a reacting system's ability to pass an activation barrier on collision may be stored in the colliding species' vibrational, translational, rotational or electronic modes, as mentioned previously. Polanyi's rules describe the relative efficacy of energy stored in translational and vibrational modes in allowing a reacting system to pass the energy barrier on collision. These state that vibrational energy is likely to be more efficient in supporting an endothermic reaction (i.e. one with a late-stage transition state resembling the products, according to Hammond's postulate), an example of which is the H + HF reaction, and that translational energy is likely to be more efficient in supporting an exothermic reaction, an example of which is the F + H2 reaction.7

An example of a successful reactive H + HF trajectory with a high initial value of H-F vibrational energy is given in figures 15 and 16 below; figures 17 and 18 show an unreactive trajectory with a lower initial value of H-F vibrational energy and a higher value for the translational energy of the colliding hydrogen atom. This demonstrates the effect described by Polanyi's rules. However, these relative values are approximate, and it is not possible to predict with certainty the reactivity of a given trajectory based on the relative amounts of vibrational and translational energy.7

<gallery heights=200px widths=200px>
File: bd316figure16.png|Figure 15: potential energy surface of an H + HF reactive trajectory
File: bd316figure17.png|Figure 16: contour plot of an H + HF reactive trajectory
File: bd316figure18.png|Figure 17: potential energy surface of an unreactive H + HF trajectory
File: bd316figure19.png|Figure 18: contour plot of an unreactive H + HF trajectory
</gallery>

= References =

1. Steinfeld, J.I., Francisco, J.S., Hase, W.L., Chemical Kinetics and Dynamics, Prentice Hall, 1989

2. Pilling, M. J., Seakins, P. W., Reaction Kinetics, Oxford University Press, 1995

3. Atkins, P. W., Atkins' Physical Chemistry, Oxford University Press, 2014

4. Mahan, B. H., J. Chem. Educ., Activated Complex Theory of Bimolecular Reactions, University of California, 1974

5. Polanyi, J. C., Tardy, D. C., J. Chem. Phys., Energy Distribution in the Exothermic Reaction F+ H2 and the Endothermic Reaction HF + H, 1969

6. IUPAC, Compendium of Chemical Terminology, 2nd ed. (the "Gold Book"), compiled by A. D. McNaught and A. Wilkinson, Blackwell Scientific Publications, Oxford, 1997

7. Zhang, Z., Zhou, Y., Zhang, D.H, Czakó, G., Bowman, J., J. Phys. Chem. Lett., Theoretical Study of the Validity of the Polanyi Rules for the Late Barrier Cl + CHD3 Reaction, 2012

MRD:bd316

2018-06-07T16:50:29Z

Mm10114: /* Determining the geometry of the transition state */

= Molecular reaction dynamics of an H2 + H reacting system =

In this section the reaction dynamics of the following system, whereby a free hydrogen atom collides with a H2 molecule to form a new H2 molecule and a new free H atom, are considered:

[[File: bd316scheme1.png|center|500px]]

The three atoms HC, HB and HA are constrained to a 180° angle such that a potential energy surface for the reaction may be calculated.2

=== The dynamics of the potential energy surface plot - the energy minima and transition state region ===

[[File:bd316figure1.png|thumb|Figure 1: the three-dimensional potential energy surface of the reacting system|500px]]

The trajectory of the reaction follows the path of energy minima. The transition state region is defined as the energy maximum on the path of minimum energy between the reactants and products.

Hydrogen atom A approaches hydrogen molecule BC (starting from the left hand side of figure 1) following the path of the potential energy surface along which ∂(V(rBC))/∂(rBC) = 0. Along this path the value of the gradient ∂(V(rBA))/∂(rBA) gradually increases until it reaches a maximum along the path of minimum energy - this is the transition state region. From the transition state region, the AB bond forms and the hydrogen atom C moves away from H2 molecule AB along the path along the potential energy surface on which ∂(V(rBA))/∂(rBA) = 0. The reaction coordinate continues along this path until it reaches the minimum of the potential energy surface on both axes, at which ∂(V(rBA))/∂(rBA) = ∂(V(rBC))/∂(rBC) = 0, at which point the reaction is complete.

In the transition state region, the potential energy surface has reached a saddle point, which has the minimum potential energy whereby rCB = rBA. In the transition state region, we can consider there to be a third 'axis', running diagonally from the rBC axis to the rBA axis through the saddle point (s), which defines the reaction coordinate (the view along this axis is given in figure 1). At the saddle point ∂(V(s))/∂(s)= 0; this is an energy maximum along the reaction pathway - ∂2(V(s))/∂(s)2 < 0. The 'axis' running orthogonal to the axis s, through the saddle point from the origin (o) also has ∂(V(o))/∂(o)= 0 and is an energy minimum with respect to this axis - ∂2(V(o))/∂(o)2 > 0.1 Taking the second derivative distinguishes the minima and maxima along a given axis.

[[File:bd316figure2.png|thumb|Figure 2: the contour potential energy surface of the reacting system|400px]]

=== Determining the geometry of the transition state ===

Computationally we can use the fact that, in the transition state, rBC = rBA (if this were not the case the potential energy surface in figure 1 would not be symmetrical) to determine the value of rTS. The results of this calculation are illustrated by the plot of rTS vs time in figure 3. When the transition state has zero momentum and the reaction does not proceed, its potential energy fluctuates in the transition state region and the value of rTS oscillates between approximately 0.8 to 1.0 Å.

An enlarged section of the potential energy surface plot (shown in full in figure 1) in figure 4 shows these inter-atomic vibrations. The value of rTS fluctuates because the transition state is able to interconvert between kinetic energy and potential energy, as is illustrated in figure 5. There are three vibrational degrees freedom at the saddle point - one of these corresponds to infinitesimal motion along the reaction path, and the other two are vibrational motion orthogonal to the reaction path (through the saddle point and along the energy minima).1,2

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 17:50, 7 June 2018 (BST) You correctly observe that the TS is somewhere between the values of 0.8 and 1.0 Å. However, there is only one TS at this potential energy surface. And it was the purpose of this exercise to find it. You were supposed to sample different r1=r2 until you find your best estimate. You would be able to tell this when, for example in Figure 5, you would have no oscillation for the distances and there would be no interconversion of the potential-kinetic energies. So, in the end you were supposed to report a single value for r1=r2, with your best estimate for the TS. In your current calculation, if you would increase number of steps significantly, the aggregation of numerical errors would probably result in your trajectory skewing towards reactants/products. With a good estimate for TS that would not happen.

<gallery widths=300px heights=300px>
File:bd316figure3.png|Figure 3: plot of internuclear distance vs time for the transition state
File:bd316figure4.png|Figure 4: the transition state region of the potential energy surface showing vibrational excitation of the transition state
File:bd316figure5.png|Figure 5: kinetic energy and potential energy vs time for the transition state
</gallery>

=== Calculating the reaction path ===

The minimum energy path, or reaction path, is the path that follows the energy minima from the saddle point to the products along a series of infinitesimally small steps - the vibrational degree of freedom mentioned previously (figure 7). At each of these steps, the kinetic energy is removed, such that the reacting system always has the minimum possible potential energy.1 This distinguishes the minimum energy path from the dynamic reaction path (figure 8), which gives a realistic picture of the trajectory that the reacting system will follow, vibrating orthogonally to the reaction path as the reaction coordinate progresses.1

<gallery heights=250px widths=250px>
File:bd316figure7.png|Figure 7: the minimum energy path
File:bd316figure8.png|Figure 8: the dynamic reaction path
</gallery>

Figure 9 shows the progression of the values rCB and rBA as the reaction progresses from the transition state region to the products, following the dynamic reaction trajectory. At time zero rCB = rBA = rTS. As HC leaves and the HB-HA bond forms, the value of rCB increases to infinity, and the value of rBA reaches the value of the equilibrium H-H bond length - this fluctuates between values of approximately 0.715 - 0.748 Å, with an average value of approximately 0.726 Å.

The progression of the values of interatomic momentum are shown in figure 10. As the reaction coordinate initially progresses from the transition state the values of both momenta initially increase (figure 11). The value of pCB increases sharply and plateaus at 2.5 when C and B are fully dissociated. The value of pBA first decreases as the HB-HA bond contracts on forming, then reaches the equilibrium value of approximately 1.2, oscillating between 0.95 and 1.53 as the HB-HA bond vibrates.

<gallery widths=250px heights=250px>
File:bd316figure9.png|Figure 9: plot of internuclear distance vs time for the dynamic reacting system
File:bd316figure10.png|Figure 10: plot of internuclear momentum vs time for the dynamic reacting system
File:bd316figure11.png|Figure 11: expanded region of the plot of internuclear momentum vs time for the dynamic reacting system
</gallery>

=== Reactive and unreactive trajectories ===

A trajectory may be reactive or unreactive depending on the values of pCB and pBA, which determine the kinetic energy with which the constituent species collide; they must have sufficient total energy to overcome the activation energy barrier and pass through the transition state region to the products. The reactivities of a series of trajectories of varying initial values of the momenta are compared below.

{| class="wikitable" border=1
! Trajectory !! pCB !! pBA !! Total energy (kJ mol-1) !! Reactivity!! Plot !! Comment
|-
| 1 || -1.25 || -2.5 || -99.1 || Reactive || [[File: bd316table1.png|200px]] || The reacting system has sufficient energy to pass over the transition state region. The products HC + HBHA are formed and the resultant HBHA molecule shows a greater degree of vibrational excitation relative to the HCHB molecule.
|-
| 2 ||-1.5 || -2.0 || -100.4 || Unreactive || [[File: bd316table2.png|200px]] || The reacting system has insufficient energy to pass over the transition state region.
|-
| 3 ||-1.5 || -2.5 || -99.0 || Reactive || [[File: bd316table3.png|200px]] || The reacting system has sufficient energy to pass over the transition state region. The resultant HBHC molecule shows a greater degree of vibrational excitation than that resulting from trajectory 1 because of the greater initial value of pCB.
|-
| 4 || -2.5 || -5.0 || -85.0 || Unreactive || [[File: bd316table4.png|200px]] || The species collide with a great deal of kinetic energy - the activation barrier is passed and re-passed, because the initially formed products HC + HB-HA have such a high degree of vibrational excitation that they recollide with sufficient kinetic energy to re-form the reactants.
|-
| 5 || -2.5 || -5.2 || -83.4 || Reactive || [[File: bd316table5.png|200px]] || The reacting species collide with even greater kinetic energy than in trajectory 4 - the activation barrier is crossed, re-crossed and crossed again, until the product species are formed with internuclear momenta such they do not re-collide.
|}

== Using transition state theory to predict rates of reaction ==

Transition state theory provides a means of predicting the rate constant for a given reaction that passes through a transition state, based on the concepts of statistical thermodynamics.3 It is assumed that the reactants are in rapid equilibrium with the transition state complex; the products are formed when this complex decays in to the products. Upon further calculation, the rate constant of reaction is given by the Eyring equation:3

[[File: bd316eyring.png|center]]

In which kr is the reaction rate, κ is a proportionality constant between the rate of passage of the reacting system through the transition state and the vibration frequency of the reacting system along the reaction coordinate, k is the Botzmann constant, h is Planck's constant, T is the temperature and K-barcǂ is the rate of formation of the transition state.3 K-barcǂ may be determined from the partition functions of the reactants and products, which are themselves easily calculated spectroscopically. The subscript c denotes that the the motion along the reaction coordinate has been separated from the other modes of motion of the species (equivalent to the Born-Oppenheimer approximation) and treated entirely classically.1,3

In calculating the rate constant using transition state theory, three main assumptions are made:4

1. That the reacting species are in equilibrium with the transition state complex, as mentioned previously - this allows the rate constant to be determined using statistical thermodynamics expressions in which partition functions appear (is is assumed that the 'phase space' of the reactants is populated according to the Boltzmann distribution), which are derived with the assumption of equilibrium.

2. That the system can be described adequately with classical mechanics.

3. That systems which pass the transition state barrier always become products.

Each of these assumptions may cause deviations of the calculated rate constants from the experimentally determined rate constants to different extents, depending on the reality of the system considered. For our reacting system in which the molecules may repass the transition state region when they collide with sufficient kinetic energy, as in trajectory 4 in the table above, assumption three may lead to marked deviations between the experimental and calculated rate constants. Assumption two neglects the possibility of quantum tunneling over the activation barrier, underestimating the rate of reaction - as yet there is no transition state theory model that successfully takes quantum mechanical effects in to account.1,3

= Molecular reaction dynamics of an F - H - H reacting system =

In this section the dynamics of the following, slightly more complex, reaction between a free flourine atom and an H2 molecule are considered:

[[File: bd316scheme2.png|center|500px]]

=== Inspection of the potential energy surface ===
[[File: bd316figure13.png|thumb| Figure 12 a: the potential energy surface of the F-H-H reacting system showing a reactive trajectory in the exothermic direction|300px]]
[[File: bd316figure13b.png|thumb| Figure 12 b: the potential energy surface of the F-H-H reacting system from above|300px]]
[[File: bd316figure14.png|thumb| Figure 13: the contour plot of the F-H-H reacting system showing the location of the transition state|300px]]
[[File: bd316figure15.png|thumb| Figure 14: plot of internuclear momenta vs. time for the F + H2 reacting system following the trajectory illustrated in figure 13|300px]]

The reaction is exothermic in the F + H2 --> FH + H direction, because the products are lower in energy than the reactants, and endothermic in the reverse direction (figure 12a). This indicates the greater bond strength of the H-F bond relative to the H-H bond - this is confirmed by literature values.5

The activation energy of the F + H2 reaction is small - this makes the transition state difficult to locate by eye alone. Hammond's postulate states that the transition state will have the geometry closest to the species (reactants or products) that are closest in energy.6 The location of the transition state for the exothermic F + H2 reaction will therefore most closely resemble the geometry of the reactants. The location of the transition state can be found by starting from the F-H-H geometry with the H-H bond length and optimising until the potential energy surface no longer shows a trajectory with zero internuclear momentum - the location of the transition state is given in figure 13. The transition state complex obtained by this method has an F-H bond length of 1.813 Å and an H-H bond length of 0.741 Å.

=== Calculating the activation energies ===

The activation energy is defined as the difference in energy between the reactants and the transition state, i.e. the minimum energy that the reactants must have in order to react. Comparison of the energy of the transition state (-103.5 kJ mol-1) with the minimum energy of the reacting species (-133.3 kJ mol-1) gives a value of 29.8 kJ mol-1 for the activation energy of the endothermic H + HF --> F + H2 reaction. The minimum energy of H2 was found to be -104.3 kJ mol-1, hence the activation energy in the exothermic direction is 0.8 kJ mol -1.

=== Energetics of the reacting system ===

According to the first law of thermodynamics, it is not possible for energy to be created or destroyed.3 Hence, when reactants collide with insufficient energy to reach the transition state they may not react (quantum mechanical possibilities aside), and when they collide with more than sufficient energy to react, the excess energy of reaction must 'go somewhere'. This excess energy may be dissipated as heat (random motion, which results in an increase in temperature), or stored in one of the molecules translational, vibrational, rotational or electronic modes of energy (excited electronic modes are typically thermally inaccessible). The amount of energy which is dissipated as heat relative to the amount which is stored in a given species' energetic modes is determined by how many energetic modes the species has available - this is the 'heat capacity'.

This is seen in figure 12 and in many of the reactive trajectories in the previous section, where the reactive trajectory oscillates to a greater extent orthogonal to the reaction coordinate after passing the transition state region, because some of the excess energy is converted to vibrational energy of the products.3 This is also clearly illustrated in figure 14, in which it can be seen that the internuculear momentum in the product H-F molecule oscillates to a significantly greater extent than in the reactant H2 molecule because there is more energy in the H-F vibrational modes.

Calorimetry may be used to experimentally measure the flow of heat resulting from an exothermic reaction, and from the difference in this value, the activation energy and the energy of the reactants the amount of energy stored in the reacting species' energetic modes may be deduced.3

=== Polanyi's empirical rules ===

The energy that contributes to a reacting system's ability to pass an activation barrier on collision may be stored in the colliding species' vibrational, translational, rotational or electronic modes, as mentioned previously. Polanyi's rules describe the relative efficacy of energy stored in translational and vibrational modes in allowing a reacting system to pass the energy barrier on collision. These state that vibrational energy is likely to be more efficient in supporting an endothermic reaction (i.e. one with a late-stage transition state resembling the products, according to Hammond's postulate), an example of which is the H + HF reaction, and that translational energy is likely to be more efficient in supporting an exothermic reaction, an example of which is the F + H2 reaction.7

An example of a successful reactive H + HF trajectory with a high initial value of H-F vibrational energy is given in figures 15 and 16 below; figures 17 and 18 show an unreactive trajectory with a lower initial value of H-F vibrational energy and a higher value for the translational energy of the colliding hydrogen atom. This demonstrates the effect described by Polanyi's rules. However, these relative values are approximate, and it is not possible to predict with certainty the reactivity of a given trajectory based on the relative amounts of vibrational and translational energy.7

<gallery heights=200px widths=200px>
File: bd316figure16.png|Figure 15: potential energy surface of an H + HF reactive trajectory
File: bd316figure17.png|Figure 16: contour plot of an H + HF reactive trajectory
File: bd316figure18.png|Figure 17: potential energy surface of an unreactive H + HF trajectory
File: bd316figure19.png|Figure 18: contour plot of an unreactive H + HF trajectory
</gallery>

= References =

1. Steinfeld, J.I., Francisco, J.S., Hase, W.L., Chemical Kinetics and Dynamics, Prentice Hall, 1989

2. Pilling, M. J., Seakins, P. W., Reaction Kinetics, Oxford University Press, 1995

3. Atkins, P. W., Atkins' Physical Chemistry, Oxford University Press, 2014

4. Mahan, B. H., J. Chem. Educ., Activated Complex Theory of Bimolecular Reactions, University of California, 1974

5. Polanyi, J. C., Tardy, D. C., J. Chem. Phys., Energy Distribution in the Exothermic Reaction F+ H2 and the Endothermic Reaction HF + H, 1969

6. IUPAC, Compendium of Chemical Terminology, 2nd ed. (the "Gold Book"), compiled by A. D. McNaught and A. Wilkinson, Blackwell Scientific Publications, Oxford, 1997

7. Zhang, Z., Zhou, Y., Zhang, D.H, Czakó, G., Bowman, J., J. Phys. Chem. Lett., Theoretical Study of the Validity of the Polanyi Rules for the Late Barrier Cl + CHD3 Reaction, 2012

Talk:MRD:AA12116

2018-06-07T16:42:24Z

Mm10114: Created page with "~~~~ Excellent report. You've fully addressed each question and exercise. One minor comment, the energy is kcal/mol and not "Kcal/mol". The 'k' for ki..."

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 17:42, 7 June 2018 (BST) Excellent report. You've fully addressed each question and exercise. One minor comment, the energy is kcal/mol and not "Kcal/mol". The 'k' for kilo should not be capitalised.

Talk:MRD:MH5015

2018-06-07T16:37:46Z

Mm10114: Created page with "~~~~ This is an excellent report, and I'm sure you're aware of that. You've done everything as stated in the wiki for the lab and more. You have a ver..."

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 17:37, 7 June 2018 (BST) This is an excellent report, and I'm sure you're aware of that. You've done everything as stated in the wiki for the lab and more. You have a very good understanding and keen eye for detail. Well done, not much to add.

MRD:MH5015

2018-06-07T16:34:04Z

Mm10114: /* H-H-H system */

This is a report from the Molecular reaction dynamics computational lab by Martin Holicky.

==Introduction==

Simulating reactions as paths on a potential energy surface (PES) can provide interesting insights into the elementary reaction rates and mechanisms. From the minima it is possible to calculate the energy difference between the reactants and the products and by finding the saddle point of the PES it is possible to both determine the activation energies and the structure of the transition state (TS). While the simulations in this case allow for any arbitrary initial positions and translational / vibrational momenta to be set, such experiments can be actually done (to some extent, of course) by manipulating the atoms with ultrafast lasers <ref>Gruebele M., Zewail A. H. Physics Today. May 1990, pp. 24-33.</ref>.

The minima, maxima and saddle point all satisfy the requirement for zero gradient of the PES:

:<math> \nabla V(x,y) = 0</math>

To distinguish between these three cases, the second derivatives have to be calculated <ref>Quirke, N. Maths 2 Lecture Notes. Imperial College (2017).</ref>, providing a Hessian matrix.
:<math>H(x,y) = \begin{bmatrix}
\dfrac{\partial^2 V}{\partial x^2} & \dfrac{\partial^2 V}{\partial x\,\partial y} \\[2.2ex]
\dfrac{\partial^2 V}{\partial y\,\partial x} & \dfrac{\partial^2 V}{\partial y^2} \\[2.2ex]
\end{bmatrix}.</math>

Then, the determinant of the Hessian has to be calculated
:<math>D(x,y)= \dfrac{\partial^2 V}{\partial x^2}\dfrac{\partial^2 V}{\partial y^2} -\left( \dfrac{\partial^2 V}{\partial x\,\partial y} \right)^2 </math>

If the determinant is negative the point is a saddle point. If the determinant is positive and <math>\dfrac{\partial^2 V}{\partial x^2}</math> is also positive, then the point is a local minimum.

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 17:31, 7 June 2018 (BST) Very detailed explanation, well done.

The collisions between three atoms were simulated using the LepsPy software, developed by T. Mackenzie and published under the GPL License. In all cases, a 180-degree collision trajectory was assumed.

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 17:31, 7 June 2018 (BST) I really like that you provide all this information.

==H-H-H system==
[[File:Mh-001-ts-hhh.png|thumb|Nuclear separation in time when the atoms are in the transition state.]]
The following reaction between three hydrogen atoms was considered:

H-H + H ⇄ H + H-H

It is clear that since all three atoms are identical the bond lengths are equal and also the potential energy surface symmetrical. As stated in the Introduction, the transition state is located at the maximum of the minimum energy path where the potential energy gradient is zero. This implies an unstable equilibrium, however, since the simulation ignores external factors (such as other atoms) it will be stable forever. By gradually adjusting the internuclear separations while keeping the momenta at zero the transition state position was found at ''rts = 0.907''. The internuclear distance did not change in time which is indeed expected from a transition state. Minor oscillations - stretches of the H-H-H bonds are visible but since these are symmetrical the do not shift the equlibrium.

A more accurate and faster way to find the TS would be by simply computing the determinants of the Hessians throughout the PES. However, after a quick inspection of the LepsPy source code, the modification of the code appeared to be more time consuming than finding it by hand, given the fact only two TS needed to be determined in this work. Nevertheless, it would be a useful feature.

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 17:34, 7 June 2018 (BST) Hard to disagree with your observations. Here, the exercises play only an introductory role into the MRD. However, I am very impressed you've taken the time to inspect the code and make these valid suggestions.

===Minimum energy paths===
[[File:Mh-003-mep_not_ts-hhh.png|thumb|Interatomic distances during the process as simulated in the MEP mode.]]By setting the speed of the atoms to zero in each step, it is possible to calculate the ''minimum energy path'' (MEP). In this simulation, the HA-HB-HC distances were set to 0.917 and 0.907 respectively, the former corresponding to a slight displacement from the transition state. This displacement resulted in the transition state collapsing into HA and HB-HC. The H-H distance ''rBC'' simply converged to the dihydrogen bond length (0.74) and ''rAB'' increased steadily (reaching a minimum at infinity).

Running the simulation in the Dynamics mode (preserving velocities between steps) produces similar results as the transition state dissociates into HA and HB-HC. However, the reaction is much faster (no surprise here, resetting the speeds in the MEP mode obviously slows the process down significantly). One interesting feature is perhaps the shape of the ''rAB'' curve. With the MEP the curve resembles a logarithmic process (an integral of the 1/r potential), in the Dynamics simulation it is approximately linear.
[[File:Mh-004-dynamicsaftermep-hhh.png|thumb|Interatomic distances during the process as simulated in the ''Dynamics'' mode.]]

Exchanging the initial distances shifts the reaction towards the other side of the equilibrium (transition state saddle point), resulting in the other set of products.
<gallery heights="220px" widths="250px">
Mh-005-dynamicsaftermep-hhh.png | The path towards H + H-H (''rAB'' longer)
Mh-005-dynamicsaftermep2-hhh.png | The path towards H-H + H (''rBC'' longer)
</gallery>

By plotting the momenta of the atoms in time, it was found that at the end of the simulation they were ''pAB = 1.230'' and ''pBC = 2.481'' (the transition state potential energy roughly divided in 2:1 ratio and converted into kinetic energy). Running the simulation backwards with these momenta (opposite signs to reverse the process) and the final positions ''rAB = 0.746'' and ''rBC = 9.006'' resulted in the molecule going into the transition state. This demonstrates the conservation of energy in the simulation.
[[File:Mh-005-reversedynamics-hhh.png|thumb|The path for the reverse process (molecule coming back towards the TS) - compare with the figures above demonstrating the dissociation of the TS.|none]]

===Reactive and unreactive pathways===

The dynamics simulations with the initial momenta as defined in the Table below with initial distances of '''r1''' = 0.74 and '''r2''' were used to investigate if simply having a high enough momentum is necessary for a succesful collision to occur.

{| class="wikitable" border=1
! p1 !! p2 !! Total energy !! Comment !! Reactive? !! Contour plot
|-
| -1.25 || -2.5 || -99.018 || A successful collision, reaction proceeds. || yes || [[File:Mh-006-react-A.png|300px]]
|-
| -1.5 || -2.0 || -100.456 || The reactants don't have a sufficient momentum to overcome the activation barrier and no reaction occurs. || no || [[File:Mh-006-react-B.png|300px]]
|-
| -1.5 || -2.5 || -98.956 ||Similar to the first case. The higher initial momentum reflects in a higher vibrational energy of the product.|| yes || [[File:Mh-006-react-C.png|300px]]
|-
| -2.5 || -5.0 || -84.956 ||The reactants cross the TS, form the products then revert back to reactants (barrier recrossing). || no || [[File:Mh-006-react-D.png|300px]]
|-
| -2.5 || -5.2 || -83.416 || The reactants cross the barrier multiple times and then continue to form the products. || yes || [[File:Mh-006-react-E.png|300px]]
|}

The Transition State Theory (TST) is used to explain the kinetics of elementary reactions and to determine the thermodynamic parameters of activation. This theory works with an assumed quasi-equilibrium between the reactants and the activated complex in the transition state corresponding to a saddle point <ref>Wikipedia: Transition State Theory. Accessed on 24/05/2018. Available online at https://en.wikipedia.org/wiki/Transition_state_theory</ref>. Then, kinetic theory can be used to determine the reaction rate. It also assumes that the nuclei behave according to classical mechanics and the transition state collapses directly into the products (no barrier recrossing). Therefore, due to the last assumption the real reaction rates would be lower than predicted by the TST. It also deviates at higher temperature due to higher vibrational states not passing the saddle point and possibly due to quantum-mechanical effects like tunneling.

== F-H-H System ==
[[File:Mh-011-FHH-ts.png|thumb|The position of the F-H-H transition state.]]Exchanging one of the hydrogen atoms for a fluorine atom, the following reaction was investigated:

F + H-H ⇄ HF + H

The transition state was found by carefully adjusting the internuclear distances while keeping the momenta to zero. The process was complicated by the fact the potential energy surface is no longer symmetric and also the forward reaction having a too small activation energy. The distances for which the TS complex remained stationary (at the TS saddle point) were ''rFHa = 1.811'' and ''rHaHb = 0.745''. This is an early transition state (with respect to the above reaction) - the H-H bond is almost intact, while the fluorine is at a large distance. Making a reference to Hammond's postulate, the structure strongly resembles the reactants.

===Energy considerations===

As shown in the table below, the energy for each stage of the reaction was found by computing the potential energy ('Initial Geometry Information') either while in the transition state (with the distances as determined above) or before/after the reaction (large separation between the species, equilibrium bond lengths set to ''rFH = 0.919'' and ''rHH = 0.740''). The equilibrium bond lengths were determined by an MEP simulation where the interatomic separation converged to the bond length value. The reaction in the forward direction required only a very small activation energy and was found to be overall exothermic.

{| class="wikitable" border=1
! Stage !! Energy
|-
| F + HH || -104.020
|-
| F-H-H (TS) || -103.743
|-
| HF + H || -134.025
|-
| Forward EA || 0.277
|-
| Reverse EA || 30.282
|-
| Forward ΔE || -30.005 (an exothermic process)
|}
<gallery widths=250px heights=250px>
[[
File:Mh-012-HH-bond.png|MEP determination of the H-H bond length.
File:Mh-012-FH-bond.png|MEP determination of the F-H bond length.
]]
</gallery>

===Reaction dynamics===

Simulating the reaction with the following parameters ''rFH = 2.4'', ''rHH = 0.740'', ''pFH = -1'' and ''pHH = -2'' provides the data and plots below. Looking at the surface plot below, it is clear that the energy is conserved since the 'vibrational peaks' reach the same levels (potential energy maxima where they have no kinetic energy) before and after the reaction. However, the vibrational energy increased significantly during the reaction, corresponding to the 'release of heat' since the reaction is exothermic. This relates a microscopic simulation to a macroscopic phenomenon, which can be easily measured using, for example, a calorimeter. Also, the excess vibrational energy can be emitted in the form of IR photons, as observed by Polanyi and others (neglected in this simulation) <ref>J. C. Polanyi, Science 236, 680 (1987).</ref>. The amplitudes of the vibrations (stretches) are especially visible on the Momenta in time plot.
<gallery widths=250px heights=250px>
[[
File:Mh-013-FHH-rxn-surface.png|Reaction potential energy surface
File:Mh-013-FHH-rxn-momenta.png|Internuclear momenta over the course of the reaction.
]]
</gallery>

====Forward Reaction Reactive and unreactive pathways====

The dynamics simulations with the initial momenta as in the Table below with initial parameters of ''rFH = 2.4'', ''rHH = 0.74'' and ''pFH = -0.5'' and a variable initial momentum were used to investigate the reactive pathways for this reaction. The sign of the HH-momentum affects if the first vibration goes 'together' or 'away' and on the contour plots are visible as 'down' and 'up' vibrations, respectively. As it can be seen from the data below, this initial setting can affect the outcome of the reaction. Notice the different vibrational amplitudes for +0.5 and -0.5 setting - this is a result of the anharmonicity of the potential.

{| class="wikitable" border=1
! pHH !! Comment !! Reactive? !! Contour plot
|-
| -3 || Reaction proceeds, the product has a significant vibrational energy. || yes || [[File:Mh-014-react-A.png|300px]]
|-
| -2.5 || Reaction proceeds, but the product has a lower vibrational energy than in the first case || yes || [[File:Mh-014-react-B.png|300px]]
|-
| -2 || The reactants don't have a sufficient momentum to overcome the activation barrier and no reaction occurs. || no || [[File:Mh-014-react-C.png|300px]]
|-
| -0.5 || The reactants don't have a sufficient momentum to overcome the activation barrier and no reaction occurs. || no || [[File:Mh-014-react-D.png|300px]]
|-
| +0.5 || The reactants don't have a sufficient momentum to overcome the activation barrier and no reaction occurs. || no || [[File:Mh-014-react-E.png|300px]]
|-
| +2 || The reaction proceeds, crossing the barrier once. || yes || [[File:Mh-014-react-F.png|300px]]
|-
| +2.5 || The reaction proceeds. || yes || [[File:Mh-014-react-G.png|300px]]
|-
| +3 || The products re-crosses the barrier and revert to reactants. || no || [[File:Mh-014-react-H.png|300px]]
|-
|}

The simulation was also performed with the following parameters: ''rFH = 2.4'', ''rHH = 0.74'', ''pFH = 0.5'' and ''pFH = -0.8''. Initially, the dihydrogen reactant has a relatively small vibrational energy but the product a much higher - this is due to the energy released in the reaction.

[[File:Mh-015-react-lowvib.png|350x350px]]

=== Reverse reaction analysis ===

Running with the following initial values: ''rFH = 0.91'', ''rHH = 2.4'', ''pFH = 0.1'' and ''pHH = -10'' produced a reactive pathway as shown in the Figure below.

[[File:Mh-016-revreact.png|frameless|350x350px]]

Starting from the transition state determined earlier (with a small deviation due to rounding errors), after 2500 steps the final parameters were found to be ''rFH = 0.89736'', ''rHH = 9.97011'', ''pFH = -5.16018'' and ''pFH = 2.06519''. When these values were fed into the software (inverting the momenta), the reactants did not even reach the transition state, even when the inputs were matched to 6 significant figures with the final position from the TS simulation.

The energy for the activation can be supplied as both vibrational and translational momenta. Polanyi's rules state<ref>Steinfeld I.J., Francisco J. S., Hase, W. L. Chemical Kinetic and Dynamics. Prentice-Hall (1998).</ref>:

''For an early barrier (forward reaction in this case) the most effective passage is with translational rather than vibrational energy. Conversely, for a late barrier (the reverse reaction), it's easier passed with vibrational rather than translational energy.''

From the results of the forward reaction, it appears that the vibrational energy had a significant effect on enhancing the forward reaction, while a high translational energy has helped with the reverse reaction - exactly the opposite as Polanyi's rules state. The results here are inconclusive and more calculations would be useful - the few calculations ran in this report might not provide a representative data set. In all cases the success of the reaction was sensitive to the initial energy values in a more complex way than simply 'high enough' - in the high energy cases barrier recrossing has often occured.

==References==

MRD:MH5015

2018-06-07T16:31:20Z

Mm10114: /* Introduction */

This is a report from the Molecular reaction dynamics computational lab by Martin Holicky.

==Introduction==

Simulating reactions as paths on a potential energy surface (PES) can provide interesting insights into the elementary reaction rates and mechanisms. From the minima it is possible to calculate the energy difference between the reactants and the products and by finding the saddle point of the PES it is possible to both determine the activation energies and the structure of the transition state (TS). While the simulations in this case allow for any arbitrary initial positions and translational / vibrational momenta to be set, such experiments can be actually done (to some extent, of course) by manipulating the atoms with ultrafast lasers <ref>Gruebele M., Zewail A. H. Physics Today. May 1990, pp. 24-33.</ref>.

The minima, maxima and saddle point all satisfy the requirement for zero gradient of the PES:

:<math> \nabla V(x,y) = 0</math>

To distinguish between these three cases, the second derivatives have to be calculated <ref>Quirke, N. Maths 2 Lecture Notes. Imperial College (2017).</ref>, providing a Hessian matrix.
:<math>H(x,y) = \begin{bmatrix}
\dfrac{\partial^2 V}{\partial x^2} & \dfrac{\partial^2 V}{\partial x\,\partial y} \\[2.2ex]
\dfrac{\partial^2 V}{\partial y\,\partial x} & \dfrac{\partial^2 V}{\partial y^2} \\[2.2ex]
\end{bmatrix}.</math>

Then, the determinant of the Hessian has to be calculated
:<math>D(x,y)= \dfrac{\partial^2 V}{\partial x^2}\dfrac{\partial^2 V}{\partial y^2} -\left( \dfrac{\partial^2 V}{\partial x\,\partial y} \right)^2 </math>

If the determinant is negative the point is a saddle point. If the determinant is positive and <math>\dfrac{\partial^2 V}{\partial x^2}</math> is also positive, then the point is a local minimum.

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 17:31, 7 June 2018 (BST) Very detailed explanation, well done.

The collisions between three atoms were simulated using the LepsPy software, developed by T. Mackenzie and published under the GPL License. In all cases, a 180-degree collision trajectory was assumed.

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 17:31, 7 June 2018 (BST) I really like that you provide all this information.

==H-H-H system==
[[File:Mh-001-ts-hhh.png|thumb|Nuclear separation in time when the atoms are in the transition state.]]
The following reaction between three hydrogen atoms was considered:

H-H + H ⇄ H + H-H

It is clear that since all three atoms are identical the bond lengths are equal and also the potential energy surface symmetrical. As stated in the Introduction, the transition state is located at the maximum of the minimum energy path where the potential energy gradient is zero. This implies an unstable equilibrium, however, since the simulation ignores external factors (such as other atoms) it will be stable forever. By gradually adjusting the internuclear separations while keeping the momenta at zero the transition state position was found at ''rts = 0.907''. The internuclear distance did not change in time which is indeed expected from a transition state. Minor oscillations - stretches of the H-H-H bonds are visible but since these are symmetrical the do not shift the equlibrium.

A more accurate and faster way to find the TS would be by simply computing the determinants of the Hessians throughout the PES. However, after a quick inspection of the LepsPy source code, the modification of the code appeared to be more time consuming than finding it by hand, given the fact only two TS needed to be determined in this work. Nevertheless, it would be a useful feature.

===Minimum energy paths===
[[File:Mh-003-mep_not_ts-hhh.png|thumb|Interatomic distances during the process as simulated in the MEP mode.]]By setting the speed of the atoms to zero in each step, it is possible to calculate the ''minimum energy path'' (MEP). In this simulation, the HA-HB-HC distances were set to 0.917 and 0.907 respectively, the former corresponding to a slight displacement from the transition state. This displacement resulted in the transition state collapsing into HA and HB-HC. The H-H distance ''rBC'' simply converged to the dihydrogen bond length (0.74) and ''rAB'' increased steadily (reaching a minimum at infinity).

Running the simulation in the Dynamics mode (preserving velocities between steps) produces similar results as the transition state dissociates into HA and HB-HC. However, the reaction is much faster (no surprise here, resetting the speeds in the MEP mode obviously slows the process down significantly). One interesting feature is perhaps the shape of the ''rAB'' curve. With the MEP the curve resembles a logarithmic process (an integral of the 1/r potential), in the Dynamics simulation it is approximately linear.
[[File:Mh-004-dynamicsaftermep-hhh.png|thumb|Interatomic distances during the process as simulated in the ''Dynamics'' mode.]]

Exchanging the initial distances shifts the reaction towards the other side of the equilibrium (transition state saddle point), resulting in the other set of products.
<gallery heights="220px" widths="250px">
Mh-005-dynamicsaftermep-hhh.png | The path towards H + H-H (''rAB'' longer)
Mh-005-dynamicsaftermep2-hhh.png | The path towards H-H + H (''rBC'' longer)
</gallery>

By plotting the momenta of the atoms in time, it was found that at the end of the simulation they were ''pAB = 1.230'' and ''pBC = 2.481'' (the transition state potential energy roughly divided in 2:1 ratio and converted into kinetic energy). Running the simulation backwards with these momenta (opposite signs to reverse the process) and the final positions ''rAB = 0.746'' and ''rBC = 9.006'' resulted in the molecule going into the transition state. This demonstrates the conservation of energy in the simulation.
[[File:Mh-005-reversedynamics-hhh.png|thumb|The path for the reverse process (molecule coming back towards the TS) - compare with the figures above demonstrating the dissociation of the TS.|none]]

===Reactive and unreactive pathways===

The dynamics simulations with the initial momenta as defined in the Table below with initial distances of '''r1''' = 0.74 and '''r2''' were used to investigate if simply having a high enough momentum is necessary for a succesful collision to occur.

{| class="wikitable" border=1
! p1 !! p2 !! Total energy !! Comment !! Reactive? !! Contour plot
|-
| -1.25 || -2.5 || -99.018 || A successful collision, reaction proceeds. || yes || [[File:Mh-006-react-A.png|300px]]
|-
| -1.5 || -2.0 || -100.456 || The reactants don't have a sufficient momentum to overcome the activation barrier and no reaction occurs. || no || [[File:Mh-006-react-B.png|300px]]
|-
| -1.5 || -2.5 || -98.956 ||Similar to the first case. The higher initial momentum reflects in a higher vibrational energy of the product.|| yes || [[File:Mh-006-react-C.png|300px]]
|-
| -2.5 || -5.0 || -84.956 ||The reactants cross the TS, form the products then revert back to reactants (barrier recrossing). || no || [[File:Mh-006-react-D.png|300px]]
|-
| -2.5 || -5.2 || -83.416 || The reactants cross the barrier multiple times and then continue to form the products. || yes || [[File:Mh-006-react-E.png|300px]]
|}

The Transition State Theory (TST) is used to explain the kinetics of elementary reactions and to determine the thermodynamic parameters of activation. This theory works with an assumed quasi-equilibrium between the reactants and the activated complex in the transition state corresponding to a saddle point <ref>Wikipedia: Transition State Theory. Accessed on 24/05/2018. Available online at https://en.wikipedia.org/wiki/Transition_state_theory</ref>. Then, kinetic theory can be used to determine the reaction rate. It also assumes that the nuclei behave according to classical mechanics and the transition state collapses directly into the products (no barrier recrossing). Therefore, due to the last assumption the real reaction rates would be lower than predicted by the TST. It also deviates at higher temperature due to higher vibrational states not passing the saddle point and possibly due to quantum-mechanical effects like tunneling.

== F-H-H System ==
[[File:Mh-011-FHH-ts.png|thumb|The position of the F-H-H transition state.]]Exchanging one of the hydrogen atoms for a fluorine atom, the following reaction was investigated:

F + H-H ⇄ HF + H

The transition state was found by carefully adjusting the internuclear distances while keeping the momenta to zero. The process was complicated by the fact the potential energy surface is no longer symmetric and also the forward reaction having a too small activation energy. The distances for which the TS complex remained stationary (at the TS saddle point) were ''rFHa = 1.811'' and ''rHaHb = 0.745''. This is an early transition state (with respect to the above reaction) - the H-H bond is almost intact, while the fluorine is at a large distance. Making a reference to Hammond's postulate, the structure strongly resembles the reactants.

===Energy considerations===

As shown in the table below, the energy for each stage of the reaction was found by computing the potential energy ('Initial Geometry Information') either while in the transition state (with the distances as determined above) or before/after the reaction (large separation between the species, equilibrium bond lengths set to ''rFH = 0.919'' and ''rHH = 0.740''). The equilibrium bond lengths were determined by an MEP simulation where the interatomic separation converged to the bond length value. The reaction in the forward direction required only a very small activation energy and was found to be overall exothermic.

{| class="wikitable" border=1
! Stage !! Energy
|-
| F + HH || -104.020
|-
| F-H-H (TS) || -103.743
|-
| HF + H || -134.025
|-
| Forward EA || 0.277
|-
| Reverse EA || 30.282
|-
| Forward ΔE || -30.005 (an exothermic process)
|}
<gallery widths=250px heights=250px>
[[
File:Mh-012-HH-bond.png|MEP determination of the H-H bond length.
File:Mh-012-FH-bond.png|MEP determination of the F-H bond length.
]]
</gallery>

===Reaction dynamics===

Simulating the reaction with the following parameters ''rFH = 2.4'', ''rHH = 0.740'', ''pFH = -1'' and ''pHH = -2'' provides the data and plots below. Looking at the surface plot below, it is clear that the energy is conserved since the 'vibrational peaks' reach the same levels (potential energy maxima where they have no kinetic energy) before and after the reaction. However, the vibrational energy increased significantly during the reaction, corresponding to the 'release of heat' since the reaction is exothermic. This relates a microscopic simulation to a macroscopic phenomenon, which can be easily measured using, for example, a calorimeter. Also, the excess vibrational energy can be emitted in the form of IR photons, as observed by Polanyi and others (neglected in this simulation) <ref>J. C. Polanyi, Science 236, 680 (1987).</ref>. The amplitudes of the vibrations (stretches) are especially visible on the Momenta in time plot.
<gallery widths=250px heights=250px>
[[
File:Mh-013-FHH-rxn-surface.png|Reaction potential energy surface
File:Mh-013-FHH-rxn-momenta.png|Internuclear momenta over the course of the reaction.
]]
</gallery>

====Forward Reaction Reactive and unreactive pathways====

The dynamics simulations with the initial momenta as in the Table below with initial parameters of ''rFH = 2.4'', ''rHH = 0.74'' and ''pFH = -0.5'' and a variable initial momentum were used to investigate the reactive pathways for this reaction. The sign of the HH-momentum affects if the first vibration goes 'together' or 'away' and on the contour plots are visible as 'down' and 'up' vibrations, respectively. As it can be seen from the data below, this initial setting can affect the outcome of the reaction. Notice the different vibrational amplitudes for +0.5 and -0.5 setting - this is a result of the anharmonicity of the potential.

{| class="wikitable" border=1
! pHH !! Comment !! Reactive? !! Contour plot
|-
| -3 || Reaction proceeds, the product has a significant vibrational energy. || yes || [[File:Mh-014-react-A.png|300px]]
|-
| -2.5 || Reaction proceeds, but the product has a lower vibrational energy than in the first case || yes || [[File:Mh-014-react-B.png|300px]]
|-
| -2 || The reactants don't have a sufficient momentum to overcome the activation barrier and no reaction occurs. || no || [[File:Mh-014-react-C.png|300px]]
|-
| -0.5 || The reactants don't have a sufficient momentum to overcome the activation barrier and no reaction occurs. || no || [[File:Mh-014-react-D.png|300px]]
|-
| +0.5 || The reactants don't have a sufficient momentum to overcome the activation barrier and no reaction occurs. || no || [[File:Mh-014-react-E.png|300px]]
|-
| +2 || The reaction proceeds, crossing the barrier once. || yes || [[File:Mh-014-react-F.png|300px]]
|-
| +2.5 || The reaction proceeds. || yes || [[File:Mh-014-react-G.png|300px]]
|-
| +3 || The products re-crosses the barrier and revert to reactants. || no || [[File:Mh-014-react-H.png|300px]]
|-
|}

The simulation was also performed with the following parameters: ''rFH = 2.4'', ''rHH = 0.74'', ''pFH = 0.5'' and ''pFH = -0.8''. Initially, the dihydrogen reactant has a relatively small vibrational energy but the product a much higher - this is due to the energy released in the reaction.

[[File:Mh-015-react-lowvib.png|350x350px]]

=== Reverse reaction analysis ===

Running with the following initial values: ''rFH = 0.91'', ''rHH = 2.4'', ''pFH = 0.1'' and ''pHH = -10'' produced a reactive pathway as shown in the Figure below.

[[File:Mh-016-revreact.png|frameless|350x350px]]

Starting from the transition state determined earlier (with a small deviation due to rounding errors), after 2500 steps the final parameters were found to be ''rFH = 0.89736'', ''rHH = 9.97011'', ''pFH = -5.16018'' and ''pFH = 2.06519''. When these values were fed into the software (inverting the momenta), the reactants did not even reach the transition state, even when the inputs were matched to 6 significant figures with the final position from the TS simulation.

The energy for the activation can be supplied as both vibrational and translational momenta. Polanyi's rules state<ref>Steinfeld I.J., Francisco J. S., Hase, W. L. Chemical Kinetic and Dynamics. Prentice-Hall (1998).</ref>:

''For an early barrier (forward reaction in this case) the most effective passage is with translational rather than vibrational energy. Conversely, for a late barrier (the reverse reaction), it's easier passed with vibrational rather than translational energy.''

From the results of the forward reaction, it appears that the vibrational energy had a significant effect on enhancing the forward reaction, while a high translational energy has helped with the reverse reaction - exactly the opposite as Polanyi's rules state. The results here are inconclusive and more calculations would be useful - the few calculations ran in this report might not provide a representative data set. In all cases the success of the reaction was sensitive to the initial energy values in a more complex way than simply 'high enough' - in the high energy cases barrier recrossing has often occured.

==References==

Talk:MRD:01193130WP

2018-06-07T16:29:36Z

Mm10114: Created page with "~~~~ Overall very good report. You show good understanding of the exercises."

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 17:29, 7 June 2018 (BST) Overall very good report. You show good understanding of the exercises.

MRD:01193130WP

2018-06-07T16:29:04Z

Mm10114:

'''''Exercise 1: H + H2 system'''''

The transition state corresponds to a saddle point on a potential energy surface. A saddle point is a stationary point (point at which gradient is equal to 0) but not an extremum point (not a minimum or a maximum). The gradient of a potential energy surface at a transition state and at a minimum is 0 but the two are not the same. We can differentiate between the minimum and transition state by taking the second derivative of the gradient: ∂2V(ri)/∂ri2 > 0 indicates a minimum and ∂2V(ri)/∂ri2 < 0 indicates a maximum (transition state).

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 17:29, 7 June 2018 (BST) Good explanation. You could have be more specific in which direction is the transition state considered a minimum, and in which direction a maximum, when considering second partial derivatives.

'''ESTIMATING THE TRANSITION STATE'''

Estimated rts = 0.908 Å. The requirement for a transition state is that it must have r1 = r2 (due to its symmetrical nature) and since the TS is a stationary point, there should be no oscillation at this point. At r1 = r2 = 0.908 Å there is only very minimal oscillation (Fig. 1).

[[File:TS est.PNG|thumb|none|350px|Figure 1: r1 = BC and r2 = AB it can be seen that at r1 = r2 = 0.908 there is no (or very minimal) oscillation.]]

'''CALCULATING THE REACTION PATH'''

The initial conditions were set such that the system was slightly displaced from the estimated transition state: r1 = 0.908 Å, r2 = 0.909 Å, p1 = p2 = 0. Calculation type: MEP (minimum energy path).

[[File:mep_contour.PNG|thumb|none|350px|Figure 2: calculating the reaction path (MEP)]]

'''TRAJECTORY'''

The same initial conditions were used with the calculation type set to 'dynamics'

[[File:dynamics_contour.PNG|thumb|none|350px|Figure 4: calculating the reaction trajectory (dynamics)]]

The MEP represents the path of lowest energy a reaction could proceed by and still go to completion, but it does not account for any motion of atoms which is an unrealistic approximation. The MEP calculation results in a straight line, with no oscillation. The dynamic calculation results in the reaction trajectory that shows oscillatory motion and is closer to what happens in reality[http://theory.cm.utexas.edu/henkelman/pubs/sheppard08_134106.pdf]

The contour plot also shows that when r1 = rts and r2 = rts + 0.01 and we start at the transition state, we fall down to the products. For the case where r1 and r2 are reversed, the reaction falls back down to the reactants.

'''REACTIVE AND UNREACTIVE TRAJECTORIES'''

{| class="wikitable" border=1
! p1 !! p2 !! total energy !! reactive/unreactive !! plot
|-
| -1.25 || -2.5 || -99.018 || reactive || [[File:react_1.PNG|200px]]
reactive trajectory - reaction path heads towards product. There is no oscillation before the transition state but oscillation after which suggests that A reacts with a stationary B-C molecule, and then the new A-B bond vibrates
|-
| -1.5 || -2.0 || -100.456 || unreactive || [[File:react_2.PNG|200px]]
this is an unreactive trajectory - reaction path heads back towards the reactants. It can be seen that A approaches an oscillating B-C molecule but energy is too low to pass through the activation barrier.
|-
| -1.5 || -2.5 || -98.956 || reactive || [[File:react_3.PNG|200px | ]]
This is a reactive trajectory - reaction path heads towards the products. There is no oscillation of B-C but the trajectory is slightly curved and suggests lengthening of the bond as A approaches. The resulting A-B molecule oscillates
|-
| -2.5 || -5.0 || -84.956 || unreactive || [[File:react_4.PNG|200px]]
reactants come together, pass through TS but the reaction paths does not go to products - falls back down to reactants since energy is too low to roll down to products from the transition state (this is an example of barrier crossing). The reaction path shows that B-C begins to oscillate as A approaches B-C
|-
| -2.5 || -5.2 || -83.416 || reactive || [[File:react_5.PNG|200px]]
reactive trajectory (another example of barrier crossing) - reaction path passes through transition state and goes to products. A approaches a non-oscillating B-C but the A-B formed in this reaction does oscillate.
|}

'''TRANSITION STATE THEORY'''[https://www.osti.gov/biblio/6251195]

The main assumptions of transition state theory (TST) are:

1. The system is in equilibrium (or close to) with the reactants and not the products

2. Reactant behaviour agrees with classical mechanics

3. The possibility of multiple reaction pathways is ignored

4. The transition state (activated complex) will be in equilibrium with the reactants

5. The reaction will always pass through the lowest energy transition state (saddle point)

6. There is negligible barrier recrossing

These are only assumptions and therefore not always true, if reactants really did behave as predicted by classical mechanics then the p1 = -2.5, p2 = -5.0 trajectory would be reactive as the assumption suggests any system that has enough energy to reach the transition state (overcome activation barrier) will go to products. That's clearly not the case in this example - the trajectory is unreactive due to barrier re-crossing which is assumed to be negligible in the TST.

Since the TST ignores barrier recrossing and other processes such as quantum mechanical tunnelling, and therefore assumes all collisions with enough energy to form the activated complex will be reactive, the rate of reaction predicted by the TST will be higher than it is in reality.

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 17:29, 7 June 2018 (BST) Very good.

'''''EXERCISE 2: F-H-H system'''''

'''PES inspection'''

The reaction F + H2 is exothermic (reactants higher in energy than products). The H-F bond formed is much stronger than the H-H bond that is broken (the H-F bond has high iconic character due to the high difference in electronegativity between H and F) - the energy released from the formation of H-F greatly exceeds energy needed to break the weaker H-H bond so energy is released into the surroundings. The reaction HF + H is endothermic (reactants lower in energy than products) due to the strength of the H-F bond - energy released in the formation of an H-H bond does not exceed the energy needed to break the strong H-F bond therefore energy is taken from the surroundings and the reaction is endothermic.

'''LOCATING THE [F-H-H]‡ (TRANSITION STATE)'''

The transition state for F + H2 and for HF + H will be the same: [F-H-H]‡. Hammond's postulate states that if a reaction is exothermic the transition state will resemble reactants and if a reaction is endothermic the transition state will resemble products (based entirely on which it is closer in energy to). Therefore for the exothermic F + H2 reaction the transition state will resemble reactants (F and H2) - using the bond length of H-H (74 pm) and optimising the rough H-F length found by looking at the value that corresponds to 74 pm H-H on a contour plot the point at which the transition state for both F + H2 and H + HF occurs was found to be at the point where H-F = 1.810 Å and H-H = 0.746 Å. The surface plot showing each transition state can be seen below (Fig. 5 and 6):

[[File:f+h2_TS.PNG|thumb|none|Figure 5: F+ H2 Transition state|350px]]

F + H2 Transition state

[[File:h+hf_TS.PNG|thumb|none|350px|Figure 6: H + HF Transition state]]

H + HF transition state

'''CALCULATING ACTIVATION ENERGIES'''

Activation energy can be found by looking at the difference between the energy at the transition state and the energy at the reactants: Ea = energy of TS - energy of reactants

F + H2: Ea = -103.741 --104 = 0.26 kcal/mol

H + HF: Ea = -103.741 --134 = 30.26 kcal/mol

Energy of reactants was determined from energy vs. time plots (MEP calculation) displacing the bond lengths slightly from the transition state lengths. The energy vs. time plots can be seen below (Fig. 7 and 8):

[[File:f+h2_PE.PNG|thumb|none|Figure 7: F+ H2 potential energy vs. time|350px]]

[[File:h+hf_PE.PNG|thumb|none|350px|Figure 8: H + HF potential energy vs. time]]

''''REACTION DYNAMICS''''

'''MECHANISM OF RELEASE OF REACTION ENERGY'''

Initial conditions were set such that a reactive F + H2 trajectory was obtained: r(F-H) = 2.3 Å, r(H-H) = 0.8, p(F-H) = -7.0, p(H-H) = -1.0

[[File:reactive_traj.PNG|thumb|none|350px|Figure: F + H2 reactive trajectory]]

An internuclear momenta vs. time graph (Fig. 9) shows that the amplitude of H-H bond vibration is much smaller than that of the H-F bond vibration. Since the F + H2 reaction is exothermic, excess energy is released into the surroundings. It can be assumed that this excess energy goes into increasing the amplitude of H-F bond (product) vibration. Since energy is conserved, a decrease in potential energy must result in an equal in magnitude increase in kinetic energy and this can be seen on an energy vs. time graph (Fig. 10) for this system.

[[File:mom_time.PNG|thumb|none|350px|Figure 9: internuclear momenta vs. time plot for F + H2 reactive trajectory]]

[[File:energy_timeeee.PNG|thumb|none|350px|Figure 10: energy vs. time plot for F + H2 reactive trajectory]]

This could be confirmed experimentally by IR- higher vibrational energy will result in more intense overtone bands.

'''POLANYI'S RULES'''[https://pubs.acs.org/doi/abs/10.1021/jz301649w]

Polanyi's rules show how different forms of energy affect reaction rates. The rules state that vibrational energy is more efficient at activating a late barrier reaction than translational energy (the reverse is true for an early barrier reaction). According to Hammond's postulate an exothermic reaction, like F + H2, has an early transition state (it is an early barrier reaction) so, in order for the reaction trajectory to be reactive, translational energy must be greater than vibrational. For example setting p1 = 2 and p2 = -5 (high translational energy and low vibrational energy) results in a reactive trajectory (Fig. 11), swapping the two p values results in an unreactive trajectory thus the F + H2 system obeys Polnayi's rules - reaction will only 'go' if translational energy is much higher than vibrational.

[[File:fh2react.PNG|thumb|none|350px|Figure 11: high translational energy, low vibrational energy - reactive trajectory for exothermic reaction]]

Since the HF + H reaction is endothermic, Hammond's postulate predicts that the transition state resembles products - late transition state (late barrier reaction). Therefore, according to Polnayi's rules, the reaction trajectory will be reactive only if vibrational energy is higher than translational energy). For example, setting p1 = -2.5 and p2 = 5 (high vibrational energy, low translational energy) results in a reactive trajectory but reversing the two values gives an unreactive trajectory. So the HF + H system also obeyes Polanyi's rules since, for an endothermic reaction, only systems with high vibrational energy should be reactive (fig. 12).

[[File:hhfreact.PNG|thumb|none|350px|Figure 12: high vibrational energy, low translational energy - reactive trajectory for endothermic reaction]]

'''REFERENCES'''

1. Daniel Sheppard, Rye Terrell and Graeme Henkelman, J. Chem. Phys, 2008, 128, 1-10 [http://theory.cm.utexas.edu/henkelman/pubs/sheppard08_134106.pdf]

2. A. C. Lasaga, Rev. Miner. States, 1981, 8 [https://www.osti.gov/biblio/6251195]

3. Z. Zhang, Y. Zhou and D. H. Zhang, J. Phys. Chem, 2012, 3, 3416-3419 [https://pubs.acs.org/doi/abs/10.1021/jz301649w]

Talk:MRD:sth16

2018-06-07T16:10:04Z

Mm10114: Created page with "~~~~ Your report is good. However, forgetting about the appropriate references is very dangerous. This is considered plagiarism. Please, be careful fo..."

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 17:10, 7 June 2018 (BST) Your report is good. However, forgetting about the appropriate references is very dangerous. This is considered plagiarism. Please, be careful for the future and reference any information that is not a result of your personal analysis/ your scientific results.

MRD:sth16

2018-06-07T16:07:13Z

Mm10114:

'''exercise 1'''

<nowiki> </nowiki>What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.


[[File:Maxisth.png]]
[[File:Minsth16.png]]

Minimum point will be a minima from all directions but for a transition structure it is a saddle point where it is a maxima along the reaction co-ordinate but a minima along the direction that is orthogonal to the reaction path.

For both minimum point and transition structure:

V stands for potential energy

r1= distance between A and B

r2 = distance between B and C

dV/dr1 = 0 dV/dr2 = 0

For minimum point : d2V/ dr1 > 0 d2V/ dr2 > 0 for all directions

The minimum will have a first derivative value equal to zero and a second derivative greater than zero along all directions and from all angles for all r (distances) .

For a saddle point (transition state) : d2V/ dr > 0 view along the way perpendicular to the reaction path for all r

d2V/ dr < 0 view along the reaction path for all

A transition structure is defined as the maxima on the minimum energy path. It is a saddle point where the maximum point of the reaction pathway meets the minimum point of the potential energy surface.
For a saddle point, transition structure in this case, it will have a first derivative equals to zero but a negative second derivative if you view the 3d plot along the reaction co-ordinates axis and a positive second derivative in orthogonal directions to the reaction path (except reaction co-ordinates) and for all r.

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 17:07, 7 June 2018 (BST) Well and detailed explanation.

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

At transition state position, it is the maximum point of the reaction pathway (therefore will not turn into product) and the minimum point of the potential energy surface (no oscillation) which means it is the stationary point and therefore will not move or oscillate.3 particles should stay at rest without any oscillation. Becasue for this system, the transition state is symmetrical means rab=rbc at transition state, so from internuclear distance vs time plot , from the crossing point of rab rbc , an estimated value can be found. Then it is impr The best estimate of transition state found is when r AB=r BC=0.908 and p AB =p BC=0. By plotting internuclear distances against time it is found that the 3 particles are almost at rest without any oscillations because the graph shows two straight lines without any fluctuations.

[[File:Qs2 best estimat sth16.png]]

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 17:07, 7 June 2018 (BST) Good.

Comment on how the mep and the trajectory you just calculated differ.
Initial conditions: rAB=0.918 rBC=0.908 pAB=pBC=0

The trajectory calculated by MEP is shorter than Dynamics uisng the same number of steps is because for MEP the velocity and momentum is set to zero for each step therefore the motion is extremely slow compared to dynamics therefore can only move a short distance. However for Dynamics it is a continuous process and the momentum is accumulated and velocity is not reset to zero for each time point therefore can move a longer distance.
The trajectory calculated by MEP is the path that has the minimum amount of potential energy which is a straight line on the floor of potential energy surface.For minimum potential energy, oscillation will not take place. For dynamics, it takes into account of atom mass and inertia of atoms and the velocity is not reset to zero for each time step therefore the atoms will oscillate in this case.

Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.
{| class="wikitable"
!
!p1
!p2
!total energy kJ/mol
!Reactive or unreactive
|-
|1
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-99.018</nowiki>
|reactive
|-
|2
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.456</nowiki>
|unreactive
|-
|3
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.956</nowiki>
|reactive
|-
|4
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.956</nowiki>
|unreactive
|-
|5
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-83.416</nowiki>
|reactive
|}
1 This is a reactive trajectory with the least extent of oscillation due to lowest momentum.

[[File:1sth16.png]]
[[File:Mom1sth.png]]

2 An unreactive trajectory with slightly bigger extent of oscillation compared to 1 due to bigger momentum. However it is unreactive because it doesn't cross the barrier

.
[[File:2sth16.png]]
[[File:Mom2sth.png]]

3 A reactive trajectory with same momentum of 2 . However in the reactant region the extent of oscillation is smaller than the product region it shows that translational energy is transferred into vibrational energy causing bigger oscillation in the product region.
[[File:3sth16.png]]
[[File:Mom3sth.png]]

4 An unreactive trajectory with barrier recrossing. After passing the transition state the trajectory turn around return and cross it again. The oscillation in the reactant region is bigger than 1,2 and 3 due to bigger momentum.

[[File:41sth16.png]]
[[File:Mom4sth.png]]

5 A reactive trajectory with barrier recrossing. After passing the transition state the trajectory turn around and return to cross it again. The oscillation in product region is much bigger than reactant region due to translational energy being converted to vibrational energy.
[[File:51sth16.png]]
[[File:Mom5sth.png]]
State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

The main assumptions:

1 The reactant is in thermal equilibrium with the transition state.

2 Born-Oppenheimer approximation is used which separates the electronic motions and nuclear motions

3 The distribution of the reactant particles (molecules,atoms etc) among the reactant state follows Boltzmann distribution.

4 Motion along the reaction coordinate can be separated from the other degrees of freedom and treated by classical mechanics.

5 Once the system attains transition state and moving in the direction of the product, it'll not turn around and move back and reform the reactants. This is the so called ' quasi-equilibrium hypothesis'

6 The transition state that is turning into product follows the Boltzsmann distribution even if the reactant and product are not in equilibrium.

The rate values predicted by transition state theory are usually overestimated compared to experimental values. This is the result of assumption 5 where it states that once crossed transition state it will not turn around and come back and cross it again. However in reality this can happen and it is not accounted in transition state theory causing overestimation. Assumption 4 also causes deviation from the experimental values because particles are quantum mechanical objects but they are treated by classical mechanical in the theory.

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 17:07, 7 June 2018 (BST) Overall good. But, where is the reference here? Without a reference for the source of the information this can be considered plagiarism.

'''exercise 2'''

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:Exosth16.png]]

F+H2 : exothermic. The energy of reactants is higher than the energy of products therefore delta H is negative.This is determined by looking at surface plot. A=F B=H C=H. H2(BC) is the reactant and from the plot the energy of it is more positive (larger) than HF (AB). Enthalpy change of reaction is calculated by energy of product minus the energy of reactants which result in a negative energy indicating the reaction is endothermic.

[[File:Endosth16.png]]

H+HF: endothermic. The energy of products is higher than energy of reactants therefore delta H is positive. This is determined by looking at surface plot. A=H B=H C=F. HF(BC) is the reactant and from the plot the energy of it is more negative (smaller) than H2 (AB). Enthalpy change of reaction is calculated by energy of product minus the energy of reactants which result in a positive energy indicating the reaction is endothermic.

The bond strength of H-F is stronger than H-H therefore it needs to take more energy in order to break the H-F bond which makes the second one endothermic.

Locate the approximate position of the transition state.

Hammond postulate is used to locate the transition state. It states that for exothermic reaction we have an early transition state which means the transition state will resemble reactants and for endothermic reaction we have a late transition state which means the transition state resembles products.

So for the F+H2 , the transition state should resemble the reactant hydrogen and has a rBC similar to the bond length of hydrogen which is around 0.74 angstrom. Then several test on rBC is made in order to locate transition state, when rBC=1.812 the transition state is located.

It was found that for F+ H2 , rAB = 1.812 rBC = 0.743. For H + HF , rAB =0.743 rBC= 1.812

[[File:Transition_h2.png]]
transition state for F + H2

[[File:Trans16.png]]
transition state for H+HF

Report the activation energy for both reactions.

Using the transition state just found , by changing the initial condition to rAB = 1.822 rBC = 0.743 for F+ H2 a MEP is calculated for 200000 steps and using the energy vs time plot the activation energy can be found. However the values are only estimates because the graph does not show the exact value and the energy of reactants may be further decreasing but it is outside the step range. From the energy vs time plot, both the energy of reactants and transition state can be found. The energy of transition state is found when time = 0s and the energy of reactants are found when the time=1000s .

Then activation energy can be found by calculating the energy difference between transition state and reactants. From the surface plot for F and H2 it may not be obvious that reactants and transition state have different energy but by zooming in on the energy vs time plot there is a slight energy difference between reactants and transition state, the activation energy is very small.

For H + HF, activation energy is found in the method but setting initial condition to rAB = 0.743 rBC = 1.812

'''For F+H2 , Ea = 0.236 kJ/mol'''
[[File:Activation_energy_f_+h2.png]]
[[File:Eafh2.png]]
MEP calculated for F+H2

'''For H+HF, Ea = 30.2 kJ/mol'''
[[File:Activation_energy_H+HF.png]]
[[File:Ea_H+HF.png]]MEP calculated for H+HF

In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?

Energy is conserved which means total energy is contant but kinetic energy and potential energy is interchanging .There are three types of kinetic energy : translational, vibrational and rotational. From the graph below which shows internuclear momentum vs time, it is obvious that the oscillation of H-F (A=F, B=H)molecule increases to a bigger extent in the product stage. The surface plot also shows the extent of oscillation for the HF molecule is bigger. This indicates that potential energy which is the energy from the chemical bond is converted into vibrational energy. For the translational energy, this can be found by looking into the internuclear distanc vs time plot. The distance B and C increases steeply wihch means C is leaving C in a high velocity which is an increase in translational energy so some potential energy is also translated into translational energy However, in this system the increase in vibrational energy is more dominant and obvious.

[[File:Momentasth.png]]
[[File:Oscialltionsth16.png]]
[[File:Translationalsth.png]]

This could be confirmed experimentally by temperature because temperature is a measure of the average kinetic energy of the particle. However this is not accurate because both vibrational and translational are kinetic energy so we won't which type of kinetic energy is increasing.

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 17:07, 7 June 2018 (BST) What is the name of experiment where we measure the change in temperature?

A better way to do this is by using IR. For an increase in vibrational energy, the intensity of the overtone will be stronger. This is due to increase in the vibrational energy will lead to the increase in the population of the first excited state which means more population can be excited from the first excited state to the second excited state. This will lead to an increase an intensity of the overtone.

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 17:07, 7 June 2018 (BST) Good.

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.

When setting the initial condition to rAB =2.643 rBC= 0.765 pAB =-1.27 pBC= 3.78 for H + HF , the trajectory is reactive.

[[File:Contsth16.png]]
[[File:Surfacesth16.png]]

Above is the contour plot for it, this is help to determine whether this endothermic reaction has late barrier or early barrier. Using the transition state value found above it was found the system has late barrier. According to Polayni's rule, for an endothermic reaction with late barrier vibrational energy is more effective in helping the system overcoming the barrier and be reactive. The contour plot and surface plot show that vibrational energy is bigger than the translation energy which corresponds to the rule stating that vibrational excitation helps to increase the efficiency of reaction

[[File:Unreact16.png]]
[[File:Unreactt16.png]]

when setting the initial condition to rAB =2.643 rBC= 0.765 pAB =-1.29 pBC= 5 the trajectory is unreactive. From the graph above, it is obvious that vibrational energy for the trajectory is bigger than the reactive one from the degree of oscillation. This indicates that it is not always the case that bigger the vibrational energy will guarantee the reactivity of the late barrier endothermic reaction. The orientation and mode of vibration is very important as well when determining whether the reaction is going to be reactive or not.

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 17:07, 7 June 2018 (BST) Very good conclusions.

Talk:MRD:vishalipala

2018-06-07T15:58:49Z

Mm10114: Created page with "~~~~ Overall your report is very good. Well done. There are few things I have comment for you to think again through. These are however very minuscule..."

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 16:58, 7 June 2018 (BST) Overall your report is very good. Well done. There are few things I have comment for you to think again through. These are however very minuscule mistakes.

MRD:vishalipala

2018-06-07T15:56:59Z

Mm10114: /* Transition State Theory */

==Exercise 1: H + H2 system==

===Relating gradient of the PES to transition states and minima===

The gradient of the potential energy surface, given by its first derivative, is zero at both the minimum and transition state. We identify each of these by considering the second derivative. Along the reaction trajectory, there is a maximum in the potential energy surface at the transition state. If we then view the surface in a direction orthogonal to the reaction pathway, there will be a minimum. This describes a saddle point and we identify this as our transition state.

This is the basis of calculating a second derivative. For a transition state, if we calculate the second derivative, it will have a negative value along the reaction coordinate, and a positive value orthogonal to the reaction coordinate. In contrast, the minimum has a positive second derivative along both the reaction coordinate and orthogonal to it.

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 16:47, 7 June 2018 (BST) Very good answer!

===Locating the transition state===

At the transition state, since the H + H2 system is symmetric, r1must equal r2. The momentum is also zero as the atoms are not moving. By varying these distances, the transition state has been identified as occurring at a H-H distance of 0.908 Angstroms. This is indicated in the plot of inter-nuclear distance vs time. The A-B and B-C distances are 0.908 Angstroms and therefore the A-C distance is double this. Over time, this distance remains constant. This can be compared to if the curve was fluctuating, indicating vibrations of the bond. At the transition state, there is no oscillation of the graph. The contour plot also depicts the position of the transition state.

[[File:VishaliTS.PNG|450px]] [[File:vishalits2.PNG|450px]]

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 16:48, 7 June 2018 (BST) Well done.

===Calculating the reaction path===

Below are plots of internuclear distance vs time for calculations run on MEP and dynamics. A reverse dynamics trajectory is also calculated.

{| class="wikitable" border="1"
|+ Comparing the results of different calculation types
! MEP !! Dynamics !! Dynamics reversed
|-
| [[File:vishalimep.PNG|450px]] || [[File:vishalidynamics.PNG|450px]]|| [[File:vishalireversefirst.PNG|450px]][[File:vishalireverseextra.PNG|450px]]
|-
| In the minimum energy path, the velocity is reset to zero after every step. The A-B and B-C distances are initially around equal. The A-B distance then decreases and reaches a constant level. This represents the new A-B bond length and since there are no oscillations, it is not vibrating. The B-C distance increases gradually. || Compared to the MEP calculation, the new A-B bond can be seen to vibrate, although not intensely. Furthermore, the departure of C occurs at a much faster rate. || We reverse the trajectory by using the previous final positions as the initial distances (r1 = 9.03, r2 = 0.74 Angstroms) and reversing the signs of the momenta. Now there is enough momentum to just go back to our initial state i.e. close to the transition state. This is shown in the top contour plot. However, there is not enough momentum to go over the transition state to product. Instead, the trajecory returns exactly where it came from. In order for the reaction to occur, a tiny more amount of momentum must be provided. This is shown on the bottom figure. Now the trajectory proceeds towards products.
|}

Swapping the initial conditions i.e. r1 = rts and r2 = rts +0.01 makes no difference to the plot. The labels for the curves have just changed.

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 16:55, 7 June 2018 (BST) Are you sure this changes nothing? You make a good observation here "The labels for the curves have just changed." Yes, the plots will be same as this is a symmetrical system, but wouldn't one scenario (r1>r2 for example) push reaction forward towards products, but the opposite scenario (r2>r1) push reaction backwards towards reactants from the transition state?

===Reactive and unreactive trajectories===

{| class="wikitable" border="1"
|+ Determining whether a trajectory is reactive
! p1 !! p2 !! Total Energy (kcal mol-1)!! Reactive? !! Internuclear distance vs time !! Contour Plot !! Comments
|-
| -1.25 || -2.5 || -99.119 || Yes || [[File:vishalireaction1.PNG|450px]]] || [[File:vishalireaction11.PNG|450px]] || B-C shows no oscillation initially and so is not vibrating (or moving). However, A-B is decreasing and therefore A is approaching BC. The transition state occurs when A-B = B-C. The B-C distance then increases as the bond has broken and the atoms separate. The A-B distance now oscillates at the previous B-C distance, indicating vibration of the newly formed bond.||
|-
| -1.5 || -2.0 || -99.119 || No || [[File:vishalireaction2.PNG|450px]]] || [[File:vishalireaction22.PNG|450px]] || The B-C distance oscillates about the bond length throughout the time period. The A-B distance initially decreases. This indicates that the hydrogen atom approaches the hydrogen molecule. However this distance then increases and the hydrogen atom moves away from the molecule again without reacting. This is seen in the contour plot as the trajectory moves back towards reactants i.e. both the entry and exit channels are near the reactant region of the plot. ||
|-
| -1.5 || -2.5 || -99.119 || Yes || [[File:vishalireaction3.PNG|450px]]] || [[File:vishalireaction33.PNG|450px]] || Again the A-B decreases as Ha approaches the HbHc molecule. After the new A-B bond is formed, hydrogen atom C moves away.||
|-
| -2.5 || -5.0 || -99.119 || No || [[File:vishalireaction4.PNG|450px]]] || [[File:vishalireaction44.PNG|450px]] || The A-B-C activated complex initially forms. However, reactants are then reformed by barrier recrossing. This BC molecule shows stronger vibrations, depicted by the larger oscillations.||
|-
| -2.5 || -5.2 || -99.119 || Yes || [[File:vishalireaction5.PNG|450px]]] || [[File:vishalireaction55.PNG|450px]] || The reaction occurs and the AB bond is formed. The exit channel occurs through the product region of the contour plot. ||
|}

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 16:56, 7 June 2018 (BST) Good. However, are you sure energies in all cases should be exactly the same? (There was an 'update' button that was supposed to be used each time you calculate new trajectory)

===Transition State Theory===

In transition state theory, the pathway from reactants to products involves a transition state which is the highest energy coordinate on the energy profile. It is based on several ideas and assumptions:

• There is a quasi-equilibrium that exists between reactants and the activated complex. This occurs even when the reactants are not in equilibrium with the products.1 For the reaction: A + BC → AB + C, we can write this as:

A + BC ⇄ ABC≠ → AB + C

• The motion of electrons is described quantum mechanically. However, we can treat the nuclear motion separately in terms of classical mechanics. In other words, the Born-Oppenheimer approximation is used.2

• Energies of the reactants are distributed according to the Boltzmann Distribution. This assumption is true at thermal equilibrium.

•If reactants are able to form an activated complex, i.e. reach the transition state, this will lead to product formation.2

All assumptions in transition state theory are not correct. For example, transition state theory would predict that if the transition state is reached, the product will be formed. The rate of formation of product depends on the rate at which the activated complex is formed. It does not take into account barrier recrossing which can lead back to reactants. This is seen in contour plot 4. In the theory however, the activated complex and products are said not to be in equilibrium with each other. Therefore the rate of reaction predicted from transition state theory will be higher than obtained experimentally.

Furthermore, quantum mechanical tunneling is not taken into account. This involves reactants tunneling through the barrier rather than crossing over it.

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 16:56, 7 June 2018 (BST) Well done.

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

===Exothermic vs Endothermic Reactions===

Below are potential energy surfaces for the reaction between H2 and F in the forward direction and the reverse formation of H2.

{| class="wikitable" border="1"
|+ Potential Energy Surfaces
! H2 + F !! H + HF
|-
| [[File:vishalih2+f.PNG]] || [[File:vishalih+hf.PNG]]||
|-
| The entrance channel is seen at higher energy on the right hand side of the graph. The AB distance initially decreases, describing the approach of F towards H2. As BC increases, this is indicative of the formation of HF and loss of H. The exit channel lies lower in energy and therefore this an exothermic reaction. This means that the H-F bond is stronger than the H-H bond. This can be attributed to the electronegativity difference between H and F, resulting in a polar covalent bond. The ionic contribution to the bond strengthens it. || The entrance channel is now on the left hand side of the graph at short BC distance. As the reaction proceeds, initially the AB distance decreases whilst the BC distance remains constant. This describes HF being approached by H. The BC distance then increases, corresponding to the departure of fluorine. The exit channel lies higher in energy. This indicates that the reaction is endothermic in this direction. Therefore the new covalent H-H bond is weaker than the polar covalent H-F bond ||
|}

===Locating the position of the transition state===

Previously, for the H2 + H system, the idea that r1 = r2 was used based on the symmetry of the transition state. This no longer applies for this system. Instead, Hammond's postulate is used to locate the position of the transition state. Transition states are energy maxima and therefore transient with no measurable lifetime. We can approximate the transition state by studying species that are close in energy to it, as these are expected to be structurally similar. Studying the forward exothermic reaction, we expect an early transition state - one that resembles reactants more closely than products. Therefore r1 (BC) was initially set to 0.74 Angstroms. AB was then varied and the contour plot was examined. Trial and error was used to find the transition state as the position at which there is no trajectory. This is shown in the figure below. The H-H and H-F bond lengths at the transition state were found to be 0.7451 and 1.81065 Angstroms respectively.

[[File:vishalitransitionstate.PNG]]

A plot of energy vs time at the transition state is also shown. The potential energy remains constant throughout.

[[File:vishalitransitionenergy.PNG]]

===Activation energies===

The activation energy was found for both the forward and reverse reaction. Since the activation energy represents the energy difference between the transition state and the reactants, it is possible to calculate it by moving slightly away from the transition state, towards the reactants. Then the difference between the starting and final energy can be found. The results are summarized in the tables below. The main finding was that the activation energies are not the same - the reverse endothermic reaction forming H2 has a higher activation energy than the forward reaction forming HF.

{| class="wikitable" border="1"
|+ Calculating activation energy for HF + H
|-
| Calculation type || MEP
|-
| Number of steps || 11,000
|-
| AB distance || 1.750
|-
|BC distance || 0.7451
|-
| TS Energy || -103.761 kcal mol-1
|-
| HF Energy ||-133.684 kcal mol-1
|-
| Activation energy for HF + H || + 30.1 kcal mol-1
|}

The following energy vs time plot represents the trajectory from the transition state to HF + H i.e. the transition state rolls back to the reactant state.

[[File:vishaliactivation2.PNG|450px]]

{| class="wikitable" border="1"
|+ Calculating activation energy for H2 + F
|-
| Calculation type || MEP
|-
| Number of steps || 11,000
|-
| AB distance || 1.950
|-
|BC distance || 0.7451
|-
| TS Energy || -103.761 kcal mol-1
|-
| H2 energy || -103.817 kcal mol-1
|-
| Activation energy for H2 + F || + 0.056 kcal mol-1
|}

Since the activation energy for the forward exothermic reaction is so small, a distance vs time plot has not been included.

===Reaction Dynamics===

====Reactive trajectory====

A reactive trajectory was identified with the following initial conditions for H2 + F: r1 = 0.75 Angstroms, r2 = 2.3 Angstroms, p1= -1.2, p2= -7.1.

This is an exothermic reaction. However the law of conservation of energy states that energy cannot be created or destroyed, only transformed from one form to another. It is possible to understand where this excess energy has ended up by studying the inter-nuclear momenta vs time plot shown below. The AB product shows very large oscillations in momentum between positive and negative values. Negative values correspond to the nuclei moving towards each other, and positive momenta corresponds to the nuclei moving away from each other. The large oscillation between the two therefore represents vibrations of the bond i.e. the excess energy is converted into the vibrational energy of the product. This can be confirmed experimentally by calorimetry which measures a change in temperature in order to indicate the enthalpy change.

[[File:vishaliexo1.PNG]]

====Application of Polanyi's Rules====

Reactions were set up on the reactant side of H2 + F and the momentum pHH was varied. Control variables are summarized in the table below.

{| class="wikitable" border="1"
| Calculation type || Dynamics
|-
| Number of steps || 500
|-
| AB Distance || 2 Angstroms
|-
| BC Distance || 0.74 Angstroms
|-
| AB Momentum || -0.5
|}

Polanyi's rules state that vibrational energy (p1) is of importance in promoting a reaction with a late transition state. On the other hand, translational energy (p2) is important in promoting a reaction with an early transition state.3 The forward reaction is exothermic with a early transition state. It is expected that if a large amount of translational energy is put into the system, the reaction will be successful.

{| class="wikitable" border="1"
|+ Reaction dynamics - contour plots
! pHH !! Contour plot !! Comments
|-
| -2.885 || [[File:vishalidynamics1.PNG|450px]]|| The reaction has occurred.
|-
| -3 || [[File:vishalidyamics2.PNG|450px]] || This change in pHH has now meant that the reaction does not occur. The transition state is reached but barrier recrossing occurs.
|-
| 3|| [[File:vishalidynamics3.PNG|450px]]|| Positive values were also tested and led to no reaction. Another example of barrier recrossing. This demonstrates the importance of energy being supplied via a particular mode of motion. Here, the vibrational energy is high, but will not be sufficient to allow the reaction to go to completion as it is in the incorrect mode.
|}

Next, the vibrational energy was reduced by changing pHH to 0.1. The translational energy was increased by changing pHF to -0.8. The reaction now occurs and obeys Polanyi's rules.

[[File:vishaliworks.PNG|450px]]

The same analysis was repeated for HF + H. Since this has a late transition state, we expect the opposite effect to above. This has been tested below.

First, the calculation was carried out with low vibrational energy, p1 = 0.1 and high translational energy, p2 = -3. The reaction was not successful. This was then repeated with p1 = 6 and p2 = -0.1 i.e. the vibrational energy was increased and the translational energy was decreased. This led to a reaction. Both contour plots are shown below.

{| class="wikitable" border="1"
|+ Reaction dynamics - contour plots
! Low 1, High p2 !! High p1, Low p2
|-
| [[File:vishalibook.PNG|450px]] ||[[File:vishalibook2.PNG|450px]]
|}

==References==

1. K. J. Laidler and M. Christine King, J.Phys. Chem., 1983, 87, 2657-2664.

2. T. Bligaard, J.K. Nørskov, in Chemical Bonding at Surfaces and Interfaces, 2008

3. Z. Zhang, Y. Zhou and D. H. Zhang, J. Phys. Chem. Lett., 2012, 3(23), 3416-3419

MRD:vishalipala

2018-06-07T15:56:33Z

Mm10114: /* Reactive and unreactive trajectories */

==Exercise 1: H + H2 system==

===Relating gradient of the PES to transition states and minima===

The gradient of the potential energy surface, given by its first derivative, is zero at both the minimum and transition state. We identify each of these by considering the second derivative. Along the reaction trajectory, there is a maximum in the potential energy surface at the transition state. If we then view the surface in a direction orthogonal to the reaction pathway, there will be a minimum. This describes a saddle point and we identify this as our transition state.

This is the basis of calculating a second derivative. For a transition state, if we calculate the second derivative, it will have a negative value along the reaction coordinate, and a positive value orthogonal to the reaction coordinate. In contrast, the minimum has a positive second derivative along both the reaction coordinate and orthogonal to it.

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 16:47, 7 June 2018 (BST) Very good answer!

===Locating the transition state===

At the transition state, since the H + H2 system is symmetric, r1must equal r2. The momentum is also zero as the atoms are not moving. By varying these distances, the transition state has been identified as occurring at a H-H distance of 0.908 Angstroms. This is indicated in the plot of inter-nuclear distance vs time. The A-B and B-C distances are 0.908 Angstroms and therefore the A-C distance is double this. Over time, this distance remains constant. This can be compared to if the curve was fluctuating, indicating vibrations of the bond. At the transition state, there is no oscillation of the graph. The contour plot also depicts the position of the transition state.

[[File:VishaliTS.PNG|450px]] [[File:vishalits2.PNG|450px]]

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 16:48, 7 June 2018 (BST) Well done.

===Calculating the reaction path===

Below are plots of internuclear distance vs time for calculations run on MEP and dynamics. A reverse dynamics trajectory is also calculated.

{| class="wikitable" border="1"
|+ Comparing the results of different calculation types
! MEP !! Dynamics !! Dynamics reversed
|-
| [[File:vishalimep.PNG|450px]] || [[File:vishalidynamics.PNG|450px]]|| [[File:vishalireversefirst.PNG|450px]][[File:vishalireverseextra.PNG|450px]]
|-
| In the minimum energy path, the velocity is reset to zero after every step. The A-B and B-C distances are initially around equal. The A-B distance then decreases and reaches a constant level. This represents the new A-B bond length and since there are no oscillations, it is not vibrating. The B-C distance increases gradually. || Compared to the MEP calculation, the new A-B bond can be seen to vibrate, although not intensely. Furthermore, the departure of C occurs at a much faster rate. || We reverse the trajectory by using the previous final positions as the initial distances (r1 = 9.03, r2 = 0.74 Angstroms) and reversing the signs of the momenta. Now there is enough momentum to just go back to our initial state i.e. close to the transition state. This is shown in the top contour plot. However, there is not enough momentum to go over the transition state to product. Instead, the trajecory returns exactly where it came from. In order for the reaction to occur, a tiny more amount of momentum must be provided. This is shown on the bottom figure. Now the trajectory proceeds towards products.
|}

Swapping the initial conditions i.e. r1 = rts and r2 = rts +0.01 makes no difference to the plot. The labels for the curves have just changed.

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 16:55, 7 June 2018 (BST) Are you sure this changes nothing? You make a good observation here "The labels for the curves have just changed." Yes, the plots will be same as this is a symmetrical system, but wouldn't one scenario (r1>r2 for example) push reaction forward towards products, but the opposite scenario (r2>r1) push reaction backwards towards reactants from the transition state?

===Reactive and unreactive trajectories===

{| class="wikitable" border="1"
|+ Determining whether a trajectory is reactive
! p1 !! p2 !! Total Energy (kcal mol-1)!! Reactive? !! Internuclear distance vs time !! Contour Plot !! Comments
|-
| -1.25 || -2.5 || -99.119 || Yes || [[File:vishalireaction1.PNG|450px]]] || [[File:vishalireaction11.PNG|450px]] || B-C shows no oscillation initially and so is not vibrating (or moving). However, A-B is decreasing and therefore A is approaching BC. The transition state occurs when A-B = B-C. The B-C distance then increases as the bond has broken and the atoms separate. The A-B distance now oscillates at the previous B-C distance, indicating vibration of the newly formed bond.||
|-
| -1.5 || -2.0 || -99.119 || No || [[File:vishalireaction2.PNG|450px]]] || [[File:vishalireaction22.PNG|450px]] || The B-C distance oscillates about the bond length throughout the time period. The A-B distance initially decreases. This indicates that the hydrogen atom approaches the hydrogen molecule. However this distance then increases and the hydrogen atom moves away from the molecule again without reacting. This is seen in the contour plot as the trajectory moves back towards reactants i.e. both the entry and exit channels are near the reactant region of the plot. ||
|-
| -1.5 || -2.5 || -99.119 || Yes || [[File:vishalireaction3.PNG|450px]]] || [[File:vishalireaction33.PNG|450px]] || Again the A-B decreases as Ha approaches the HbHc molecule. After the new A-B bond is formed, hydrogen atom C moves away.||
|-
| -2.5 || -5.0 || -99.119 || No || [[File:vishalireaction4.PNG|450px]]] || [[File:vishalireaction44.PNG|450px]] || The A-B-C activated complex initially forms. However, reactants are then reformed by barrier recrossing. This BC molecule shows stronger vibrations, depicted by the larger oscillations.||
|-
| -2.5 || -5.2 || -99.119 || Yes || [[File:vishalireaction5.PNG|450px]]] || [[File:vishalireaction55.PNG|450px]] || The reaction occurs and the AB bond is formed. The exit channel occurs through the product region of the contour plot. ||
|}

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 16:56, 7 June 2018 (BST) Good. However, are you sure energies in all cases should be exactly the same? (There was an 'update' button that was supposed to be used each time you calculate new trajectory)

===Transition State Theory===

In transition state theory, the pathway from reactants to products involves a transition state which is the highest energy coordinate on the energy profile. It is based on several ideas and assumptions:

• There is a quasi-equilibrium that exists between reactants and the activated complex. This occurs even when the reactants are not in equilibrium with the products.1 For the reaction: A + BC → AB + C, we can write this as:

A + BC ⇄ ABC≠ → AB + C

• The motion of electrons is described quantum mechanically. However, we can treat the nuclear motion separately in terms of classical mechanics. In other words, the Born-Oppenheimer approximation is used.2

• Energies of the reactants are distributed according to the Boltzmann Distribution. This assumption is true at thermal equilibrium.

•If reactants are able to form an activated complex, i.e. reach the transition state, this will lead to product formation.2

All assumptions in transition state theory are not correct. For example, transition state theory would predict that if the transition state is reached, the product will be formed. The rate of formation of product depends on the rate at which the activated complex is formed. It does not take into account barrier recrossing which can lead back to reactants. This is seen in contour plot 4. In the theory however, the activated complex and products are said not to be in equilibrium with each other. Therefore the rate of reaction predicted from transition state theory will be higher than obtained experimentally.

Furthermore, quantum mechanical tunneling is not taken into account. This involves reactants tunneling through the barrier rather than crossing over it.

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

===Exothermic vs Endothermic Reactions===

Below are potential energy surfaces for the reaction between H2 and F in the forward direction and the reverse formation of H2.

{| class="wikitable" border="1"
|+ Potential Energy Surfaces
! H2 + F !! H + HF
|-
| [[File:vishalih2+f.PNG]] || [[File:vishalih+hf.PNG]]||
|-
| The entrance channel is seen at higher energy on the right hand side of the graph. The AB distance initially decreases, describing the approach of F towards H2. As BC increases, this is indicative of the formation of HF and loss of H. The exit channel lies lower in energy and therefore this an exothermic reaction. This means that the H-F bond is stronger than the H-H bond. This can be attributed to the electronegativity difference between H and F, resulting in a polar covalent bond. The ionic contribution to the bond strengthens it. || The entrance channel is now on the left hand side of the graph at short BC distance. As the reaction proceeds, initially the AB distance decreases whilst the BC distance remains constant. This describes HF being approached by H. The BC distance then increases, corresponding to the departure of fluorine. The exit channel lies higher in energy. This indicates that the reaction is endothermic in this direction. Therefore the new covalent H-H bond is weaker than the polar covalent H-F bond ||
|}

===Locating the position of the transition state===

Previously, for the H2 + H system, the idea that r1 = r2 was used based on the symmetry of the transition state. This no longer applies for this system. Instead, Hammond's postulate is used to locate the position of the transition state. Transition states are energy maxima and therefore transient with no measurable lifetime. We can approximate the transition state by studying species that are close in energy to it, as these are expected to be structurally similar. Studying the forward exothermic reaction, we expect an early transition state - one that resembles reactants more closely than products. Therefore r1 (BC) was initially set to 0.74 Angstroms. AB was then varied and the contour plot was examined. Trial and error was used to find the transition state as the position at which there is no trajectory. This is shown in the figure below. The H-H and H-F bond lengths at the transition state were found to be 0.7451 and 1.81065 Angstroms respectively.

[[File:vishalitransitionstate.PNG]]

A plot of energy vs time at the transition state is also shown. The potential energy remains constant throughout.

[[File:vishalitransitionenergy.PNG]]

===Activation energies===

The activation energy was found for both the forward and reverse reaction. Since the activation energy represents the energy difference between the transition state and the reactants, it is possible to calculate it by moving slightly away from the transition state, towards the reactants. Then the difference between the starting and final energy can be found. The results are summarized in the tables below. The main finding was that the activation energies are not the same - the reverse endothermic reaction forming H2 has a higher activation energy than the forward reaction forming HF.

{| class="wikitable" border="1"
|+ Calculating activation energy for HF + H
|-
| Calculation type || MEP
|-
| Number of steps || 11,000
|-
| AB distance || 1.750
|-
|BC distance || 0.7451
|-
| TS Energy || -103.761 kcal mol-1
|-
| HF Energy ||-133.684 kcal mol-1
|-
| Activation energy for HF + H || + 30.1 kcal mol-1
|}

The following energy vs time plot represents the trajectory from the transition state to HF + H i.e. the transition state rolls back to the reactant state.

[[File:vishaliactivation2.PNG|450px]]

{| class="wikitable" border="1"
|+ Calculating activation energy for H2 + F
|-
| Calculation type || MEP
|-
| Number of steps || 11,000
|-
| AB distance || 1.950
|-
|BC distance || 0.7451
|-
| TS Energy || -103.761 kcal mol-1
|-
| H2 energy || -103.817 kcal mol-1
|-
| Activation energy for H2 + F || + 0.056 kcal mol-1
|}

Since the activation energy for the forward exothermic reaction is so small, a distance vs time plot has not been included.

===Reaction Dynamics===

====Reactive trajectory====

A reactive trajectory was identified with the following initial conditions for H2 + F: r1 = 0.75 Angstroms, r2 = 2.3 Angstroms, p1= -1.2, p2= -7.1.

This is an exothermic reaction. However the law of conservation of energy states that energy cannot be created or destroyed, only transformed from one form to another. It is possible to understand where this excess energy has ended up by studying the inter-nuclear momenta vs time plot shown below. The AB product shows very large oscillations in momentum between positive and negative values. Negative values correspond to the nuclei moving towards each other, and positive momenta corresponds to the nuclei moving away from each other. The large oscillation between the two therefore represents vibrations of the bond i.e. the excess energy is converted into the vibrational energy of the product. This can be confirmed experimentally by calorimetry which measures a change in temperature in order to indicate the enthalpy change.

[[File:vishaliexo1.PNG]]

====Application of Polanyi's Rules====

Reactions were set up on the reactant side of H2 + F and the momentum pHH was varied. Control variables are summarized in the table below.

{| class="wikitable" border="1"
| Calculation type || Dynamics
|-
| Number of steps || 500
|-
| AB Distance || 2 Angstroms
|-
| BC Distance || 0.74 Angstroms
|-
| AB Momentum || -0.5
|}

Polanyi's rules state that vibrational energy (p1) is of importance in promoting a reaction with a late transition state. On the other hand, translational energy (p2) is important in promoting a reaction with an early transition state.3 The forward reaction is exothermic with a early transition state. It is expected that if a large amount of translational energy is put into the system, the reaction will be successful.

{| class="wikitable" border="1"
|+ Reaction dynamics - contour plots
! pHH !! Contour plot !! Comments
|-
| -2.885 || [[File:vishalidynamics1.PNG|450px]]|| The reaction has occurred.
|-
| -3 || [[File:vishalidyamics2.PNG|450px]] || This change in pHH has now meant that the reaction does not occur. The transition state is reached but barrier recrossing occurs.
|-
| 3|| [[File:vishalidynamics3.PNG|450px]]|| Positive values were also tested and led to no reaction. Another example of barrier recrossing. This demonstrates the importance of energy being supplied via a particular mode of motion. Here, the vibrational energy is high, but will not be sufficient to allow the reaction to go to completion as it is in the incorrect mode.
|}

Next, the vibrational energy was reduced by changing pHH to 0.1. The translational energy was increased by changing pHF to -0.8. The reaction now occurs and obeys Polanyi's rules.

[[File:vishaliworks.PNG|450px]]

The same analysis was repeated for HF + H. Since this has a late transition state, we expect the opposite effect to above. This has been tested below.

First, the calculation was carried out with low vibrational energy, p1 = 0.1 and high translational energy, p2 = -3. The reaction was not successful. This was then repeated with p1 = 6 and p2 = -0.1 i.e. the vibrational energy was increased and the translational energy was decreased. This led to a reaction. Both contour plots are shown below.

{| class="wikitable" border="1"
|+ Reaction dynamics - contour plots
! Low 1, High p2 !! High p1, Low p2
|-
| [[File:vishalibook.PNG|450px]] ||[[File:vishalibook2.PNG|450px]]
|}

==References==

1. K. J. Laidler and M. Christine King, J.Phys. Chem., 1983, 87, 2657-2664.

2. T. Bligaard, J.K. Nørskov, in Chemical Bonding at Surfaces and Interfaces, 2008

3. Z. Zhang, Y. Zhou and D. H. Zhang, J. Phys. Chem. Lett., 2012, 3(23), 3416-3419

MRD:vishalipala

2018-06-07T15:55:11Z

Mm10114: /* Calculating the reaction path */

==Exercise 1: H + H2 system==

===Relating gradient of the PES to transition states and minima===

The gradient of the potential energy surface, given by its first derivative, is zero at both the minimum and transition state. We identify each of these by considering the second derivative. Along the reaction trajectory, there is a maximum in the potential energy surface at the transition state. If we then view the surface in a direction orthogonal to the reaction pathway, there will be a minimum. This describes a saddle point and we identify this as our transition state.

This is the basis of calculating a second derivative. For a transition state, if we calculate the second derivative, it will have a negative value along the reaction coordinate, and a positive value orthogonal to the reaction coordinate. In contrast, the minimum has a positive second derivative along both the reaction coordinate and orthogonal to it.

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 16:47, 7 June 2018 (BST) Very good answer!

===Locating the transition state===

At the transition state, since the H + H2 system is symmetric, r1must equal r2. The momentum is also zero as the atoms are not moving. By varying these distances, the transition state has been identified as occurring at a H-H distance of 0.908 Angstroms. This is indicated in the plot of inter-nuclear distance vs time. The A-B and B-C distances are 0.908 Angstroms and therefore the A-C distance is double this. Over time, this distance remains constant. This can be compared to if the curve was fluctuating, indicating vibrations of the bond. At the transition state, there is no oscillation of the graph. The contour plot also depicts the position of the transition state.

[[File:VishaliTS.PNG|450px]] [[File:vishalits2.PNG|450px]]

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 16:48, 7 June 2018 (BST) Well done.

===Calculating the reaction path===

Below are plots of internuclear distance vs time for calculations run on MEP and dynamics. A reverse dynamics trajectory is also calculated.

{| class="wikitable" border="1"
|+ Comparing the results of different calculation types
! MEP !! Dynamics !! Dynamics reversed
|-
| [[File:vishalimep.PNG|450px]] || [[File:vishalidynamics.PNG|450px]]|| [[File:vishalireversefirst.PNG|450px]][[File:vishalireverseextra.PNG|450px]]
|-
| In the minimum energy path, the velocity is reset to zero after every step. The A-B and B-C distances are initially around equal. The A-B distance then decreases and reaches a constant level. This represents the new A-B bond length and since there are no oscillations, it is not vibrating. The B-C distance increases gradually. || Compared to the MEP calculation, the new A-B bond can be seen to vibrate, although not intensely. Furthermore, the departure of C occurs at a much faster rate. || We reverse the trajectory by using the previous final positions as the initial distances (r1 = 9.03, r2 = 0.74 Angstroms) and reversing the signs of the momenta. Now there is enough momentum to just go back to our initial state i.e. close to the transition state. This is shown in the top contour plot. However, there is not enough momentum to go over the transition state to product. Instead, the trajecory returns exactly where it came from. In order for the reaction to occur, a tiny more amount of momentum must be provided. This is shown on the bottom figure. Now the trajectory proceeds towards products.
|}

Swapping the initial conditions i.e. r1 = rts and r2 = rts +0.01 makes no difference to the plot. The labels for the curves have just changed.

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 16:55, 7 June 2018 (BST) Are you sure this changes nothing? You make a good observation here "The labels for the curves have just changed." Yes, the plots will be same as this is a symmetrical system, but wouldn't one scenario (r1>r2 for example) push reaction forward towards products, but the opposite scenario (r2>r1) push reaction backwards towards reactants from the transition state?

===Reactive and unreactive trajectories===

{| class="wikitable" border="1"
|+ Determining whether a trajectory is reactive
! p1 !! p2 !! Total Energy (kcal mol-1)!! Reactive? !! Internuclear distance vs time !! Contour Plot !! Comments
|-
| -1.25 || -2.5 || -99.119 || Yes || [[File:vishalireaction1.PNG|450px]]] || [[File:vishalireaction11.PNG|450px]] || B-C shows no oscillation initially and so is not vibrating (or moving). However, A-B is decreasing and therefore A is approaching BC. The transition state occurs when A-B = B-C. The B-C distance then increases as the bond has broken and the atoms separate. The A-B distance now oscillates at the previous B-C distance, indicating vibration of the newly formed bond.||
|-
| -1.5 || -2.0 || -99.119 || No || [[File:vishalireaction2.PNG|450px]]] || [[File:vishalireaction22.PNG|450px]] || The B-C distance oscillates about the bond length throughout the time period. The A-B distance initially decreases. This indicates that the hydrogen atom approaches the hydrogen molecule. However this distance then increases and the hydrogen atom moves away from the molecule again without reacting. This is seen in the contour plot as the trajectory moves back towards reactants i.e. both the entry and exit channels are near the reactant region of the plot. ||
|-
| -1.5 || -2.5 || -99.119 || Yes || [[File:vishalireaction3.PNG|450px]]] || [[File:vishalireaction33.PNG|450px]] || Again the A-B decreases as Ha approaches the HbHc molecule. After the new A-B bond is formed, hydrogen atom C moves away.||
|-
| -2.5 || -5.0 || -99.119 || No || [[File:vishalireaction4.PNG|450px]]] || [[File:vishalireaction44.PNG|450px]] || The A-B-C activated complex initially forms. However, reactants are then reformed by barrier recrossing. This BC molecule shows stronger vibrations, depicted by the larger oscillations.||
|-
| -2.5 || -5.2 || -99.119 || Yes || [[File:vishalireaction5.PNG|450px]]] || [[File:vishalireaction55.PNG|450px]] || The reaction occurs and the AB bond is formed. The exit channel occurs through the product region of the contour plot. ||
|}

===Transition State Theory===

In transition state theory, the pathway from reactants to products involves a transition state which is the highest energy coordinate on the energy profile. It is based on several ideas and assumptions:

• There is a quasi-equilibrium that exists between reactants and the activated complex. This occurs even when the reactants are not in equilibrium with the products.1 For the reaction: A + BC → AB + C, we can write this as:

A + BC ⇄ ABC≠ → AB + C

• The motion of electrons is described quantum mechanically. However, we can treat the nuclear motion separately in terms of classical mechanics. In other words, the Born-Oppenheimer approximation is used.2

• Energies of the reactants are distributed according to the Boltzmann Distribution. This assumption is true at thermal equilibrium.

•If reactants are able to form an activated complex, i.e. reach the transition state, this will lead to product formation.2

All assumptions in transition state theory are not correct. For example, transition state theory would predict that if the transition state is reached, the product will be formed. The rate of formation of product depends on the rate at which the activated complex is formed. It does not take into account barrier recrossing which can lead back to reactants. This is seen in contour plot 4. In the theory however, the activated complex and products are said not to be in equilibrium with each other. Therefore the rate of reaction predicted from transition state theory will be higher than obtained experimentally.

Furthermore, quantum mechanical tunneling is not taken into account. This involves reactants tunneling through the barrier rather than crossing over it.

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

===Exothermic vs Endothermic Reactions===

Below are potential energy surfaces for the reaction between H2 and F in the forward direction and the reverse formation of H2.

{| class="wikitable" border="1"
|+ Potential Energy Surfaces
! H2 + F !! H + HF
|-
| [[File:vishalih2+f.PNG]] || [[File:vishalih+hf.PNG]]||
|-
| The entrance channel is seen at higher energy on the right hand side of the graph. The AB distance initially decreases, describing the approach of F towards H2. As BC increases, this is indicative of the formation of HF and loss of H. The exit channel lies lower in energy and therefore this an exothermic reaction. This means that the H-F bond is stronger than the H-H bond. This can be attributed to the electronegativity difference between H and F, resulting in a polar covalent bond. The ionic contribution to the bond strengthens it. || The entrance channel is now on the left hand side of the graph at short BC distance. As the reaction proceeds, initially the AB distance decreases whilst the BC distance remains constant. This describes HF being approached by H. The BC distance then increases, corresponding to the departure of fluorine. The exit channel lies higher in energy. This indicates that the reaction is endothermic in this direction. Therefore the new covalent H-H bond is weaker than the polar covalent H-F bond ||
|}

===Locating the position of the transition state===

Previously, for the H2 + H system, the idea that r1 = r2 was used based on the symmetry of the transition state. This no longer applies for this system. Instead, Hammond's postulate is used to locate the position of the transition state. Transition states are energy maxima and therefore transient with no measurable lifetime. We can approximate the transition state by studying species that are close in energy to it, as these are expected to be structurally similar. Studying the forward exothermic reaction, we expect an early transition state - one that resembles reactants more closely than products. Therefore r1 (BC) was initially set to 0.74 Angstroms. AB was then varied and the contour plot was examined. Trial and error was used to find the transition state as the position at which there is no trajectory. This is shown in the figure below. The H-H and H-F bond lengths at the transition state were found to be 0.7451 and 1.81065 Angstroms respectively.

[[File:vishalitransitionstate.PNG]]

A plot of energy vs time at the transition state is also shown. The potential energy remains constant throughout.

[[File:vishalitransitionenergy.PNG]]

===Activation energies===

The activation energy was found for both the forward and reverse reaction. Since the activation energy represents the energy difference between the transition state and the reactants, it is possible to calculate it by moving slightly away from the transition state, towards the reactants. Then the difference between the starting and final energy can be found. The results are summarized in the tables below. The main finding was that the activation energies are not the same - the reverse endothermic reaction forming H2 has a higher activation energy than the forward reaction forming HF.

{| class="wikitable" border="1"
|+ Calculating activation energy for HF + H
|-
| Calculation type || MEP
|-
| Number of steps || 11,000
|-
| AB distance || 1.750
|-
|BC distance || 0.7451
|-
| TS Energy || -103.761 kcal mol-1
|-
| HF Energy ||-133.684 kcal mol-1
|-
| Activation energy for HF + H || + 30.1 kcal mol-1
|}

The following energy vs time plot represents the trajectory from the transition state to HF + H i.e. the transition state rolls back to the reactant state.

[[File:vishaliactivation2.PNG|450px]]

{| class="wikitable" border="1"
|+ Calculating activation energy for H2 + F
|-
| Calculation type || MEP
|-
| Number of steps || 11,000
|-
| AB distance || 1.950
|-
|BC distance || 0.7451
|-
| TS Energy || -103.761 kcal mol-1
|-
| H2 energy || -103.817 kcal mol-1
|-
| Activation energy for H2 + F || + 0.056 kcal mol-1
|}

Since the activation energy for the forward exothermic reaction is so small, a distance vs time plot has not been included.

===Reaction Dynamics===

====Reactive trajectory====

A reactive trajectory was identified with the following initial conditions for H2 + F: r1 = 0.75 Angstroms, r2 = 2.3 Angstroms, p1= -1.2, p2= -7.1.

This is an exothermic reaction. However the law of conservation of energy states that energy cannot be created or destroyed, only transformed from one form to another. It is possible to understand where this excess energy has ended up by studying the inter-nuclear momenta vs time plot shown below. The AB product shows very large oscillations in momentum between positive and negative values. Negative values correspond to the nuclei moving towards each other, and positive momenta corresponds to the nuclei moving away from each other. The large oscillation between the two therefore represents vibrations of the bond i.e. the excess energy is converted into the vibrational energy of the product. This can be confirmed experimentally by calorimetry which measures a change in temperature in order to indicate the enthalpy change.

[[File:vishaliexo1.PNG]]

====Application of Polanyi's Rules====

Reactions were set up on the reactant side of H2 + F and the momentum pHH was varied. Control variables are summarized in the table below.

{| class="wikitable" border="1"
| Calculation type || Dynamics
|-
| Number of steps || 500
|-
| AB Distance || 2 Angstroms
|-
| BC Distance || 0.74 Angstroms
|-
| AB Momentum || -0.5
|}

Polanyi's rules state that vibrational energy (p1) is of importance in promoting a reaction with a late transition state. On the other hand, translational energy (p2) is important in promoting a reaction with an early transition state.3 The forward reaction is exothermic with a early transition state. It is expected that if a large amount of translational energy is put into the system, the reaction will be successful.

{| class="wikitable" border="1"
|+ Reaction dynamics - contour plots
! pHH !! Contour plot !! Comments
|-
| -2.885 || [[File:vishalidynamics1.PNG|450px]]|| The reaction has occurred.
|-
| -3 || [[File:vishalidyamics2.PNG|450px]] || This change in pHH has now meant that the reaction does not occur. The transition state is reached but barrier recrossing occurs.
|-
| 3|| [[File:vishalidynamics3.PNG|450px]]|| Positive values were also tested and led to no reaction. Another example of barrier recrossing. This demonstrates the importance of energy being supplied via a particular mode of motion. Here, the vibrational energy is high, but will not be sufficient to allow the reaction to go to completion as it is in the incorrect mode.
|}

Next, the vibrational energy was reduced by changing pHH to 0.1. The translational energy was increased by changing pHF to -0.8. The reaction now occurs and obeys Polanyi's rules.

[[File:vishaliworks.PNG|450px]]

The same analysis was repeated for HF + H. Since this has a late transition state, we expect the opposite effect to above. This has been tested below.

First, the calculation was carried out with low vibrational energy, p1 = 0.1 and high translational energy, p2 = -3. The reaction was not successful. This was then repeated with p1 = 6 and p2 = -0.1 i.e. the vibrational energy was increased and the translational energy was decreased. This led to a reaction. Both contour plots are shown below.

{| class="wikitable" border="1"
|+ Reaction dynamics - contour plots
! Low 1, High p2 !! High p1, Low p2
|-
| [[File:vishalibook.PNG|450px]] ||[[File:vishalibook2.PNG|450px]]
|}

==References==

1. K. J. Laidler and M. Christine King, J.Phys. Chem., 1983, 87, 2657-2664.

2. T. Bligaard, J.K. Nørskov, in Chemical Bonding at Surfaces and Interfaces, 2008

3. Z. Zhang, Y. Zhou and D. H. Zhang, J. Phys. Chem. Lett., 2012, 3(23), 3416-3419

MRD:vishalipala

2018-06-07T15:48:09Z

Mm10114: /* Locating the transition state */

==Exercise 1: H + H2 system==

===Relating gradient of the PES to transition states and minima===

The gradient of the potential energy surface, given by its first derivative, is zero at both the minimum and transition state. We identify each of these by considering the second derivative. Along the reaction trajectory, there is a maximum in the potential energy surface at the transition state. If we then view the surface in a direction orthogonal to the reaction pathway, there will be a minimum. This describes a saddle point and we identify this as our transition state.

This is the basis of calculating a second derivative. For a transition state, if we calculate the second derivative, it will have a negative value along the reaction coordinate, and a positive value orthogonal to the reaction coordinate. In contrast, the minimum has a positive second derivative along both the reaction coordinate and orthogonal to it.

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 16:47, 7 June 2018 (BST) Very good answer!

===Locating the transition state===

At the transition state, since the H + H2 system is symmetric, r1must equal r2. The momentum is also zero as the atoms are not moving. By varying these distances, the transition state has been identified as occurring at a H-H distance of 0.908 Angstroms. This is indicated in the plot of inter-nuclear distance vs time. The A-B and B-C distances are 0.908 Angstroms and therefore the A-C distance is double this. Over time, this distance remains constant. This can be compared to if the curve was fluctuating, indicating vibrations of the bond. At the transition state, there is no oscillation of the graph. The contour plot also depicts the position of the transition state.

[[File:VishaliTS.PNG|450px]] [[File:vishalits2.PNG|450px]]

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 16:48, 7 June 2018 (BST) Well done.

===Calculating the reaction path===

Below are plots of internuclear distance vs time for calculations run on MEP and dynamics. A reverse dynamics trajectory is also calculated.

{| class="wikitable" border="1"
|+ Comparing the results of different calculation types
! MEP !! Dynamics !! Dynamics reversed
|-
| [[File:vishalimep.PNG|450px]] || [[File:vishalidynamics.PNG|450px]]|| [[File:vishalireversefirst.PNG|450px]][[File:vishalireverseextra.PNG|450px]]
|-
| In the minimum energy path, the velocity is reset to zero after every step. The A-B and B-C distances are initially around equal. The A-B distance then decreases and reaches a constant level. This represents the new A-B bond length and since there are no oscillations, it is not vibrating. The B-C distance increases gradually. || Compared to the MEP calculation, the new A-B bond can be seen to vibrate, although not intensely. Furthermore, the departure of C occurs at a much faster rate. || We reverse the trajectory by using the previous final positions as the initial distances (r1 = 9.03, r2 = 0.74 Angstroms) and reversing the signs of the momenta. Now there is enough momentum to just go back to our initial state i.e. close to the transition state. This is shown in the top contour plot. However, there is not enough momentum to go over the transition state to product. Instead, the trajecory returns exactly where it came from. In order for the reaction to occur, a tiny more amount of momentum must be provided. This is shown on the bottom figure. Now the trajectory proceeds towards products.
|}

Swapping the initial conditions i.e. r1 = rts and r2 = rts +0.01 makes no difference to the plot. The labels for the curves have just changed.

===Reactive and unreactive trajectories===

{| class="wikitable" border="1"
|+ Determining whether a trajectory is reactive
! p1 !! p2 !! Total Energy (kcal mol-1)!! Reactive? !! Internuclear distance vs time !! Contour Plot !! Comments
|-
| -1.25 || -2.5 || -99.119 || Yes || [[File:vishalireaction1.PNG|450px]]] || [[File:vishalireaction11.PNG|450px]] || B-C shows no oscillation initially and so is not vibrating (or moving). However, A-B is decreasing and therefore A is approaching BC. The transition state occurs when A-B = B-C. The B-C distance then increases as the bond has broken and the atoms separate. The A-B distance now oscillates at the previous B-C distance, indicating vibration of the newly formed bond.||
|-
| -1.5 || -2.0 || -99.119 || No || [[File:vishalireaction2.PNG|450px]]] || [[File:vishalireaction22.PNG|450px]] || The B-C distance oscillates about the bond length throughout the time period. The A-B distance initially decreases. This indicates that the hydrogen atom approaches the hydrogen molecule. However this distance then increases and the hydrogen atom moves away from the molecule again without reacting. This is seen in the contour plot as the trajectory moves back towards reactants i.e. both the entry and exit channels are near the reactant region of the plot. ||
|-
| -1.5 || -2.5 || -99.119 || Yes || [[File:vishalireaction3.PNG|450px]]] || [[File:vishalireaction33.PNG|450px]] || Again the A-B decreases as Ha approaches the HbHc molecule. After the new A-B bond is formed, hydrogen atom C moves away.||
|-
| -2.5 || -5.0 || -99.119 || No || [[File:vishalireaction4.PNG|450px]]] || [[File:vishalireaction44.PNG|450px]] || The A-B-C activated complex initially forms. However, reactants are then reformed by barrier recrossing. This BC molecule shows stronger vibrations, depicted by the larger oscillations.||
|-
| -2.5 || -5.2 || -99.119 || Yes || [[File:vishalireaction5.PNG|450px]]] || [[File:vishalireaction55.PNG|450px]] || The reaction occurs and the AB bond is formed. The exit channel occurs through the product region of the contour plot. ||
|}

===Transition State Theory===

In transition state theory, the pathway from reactants to products involves a transition state which is the highest energy coordinate on the energy profile. It is based on several ideas and assumptions:

• There is a quasi-equilibrium that exists between reactants and the activated complex. This occurs even when the reactants are not in equilibrium with the products.1 For the reaction: A + BC → AB + C, we can write this as:

A + BC ⇄ ABC≠ → AB + C

• The motion of electrons is described quantum mechanically. However, we can treat the nuclear motion separately in terms of classical mechanics. In other words, the Born-Oppenheimer approximation is used.2

• Energies of the reactants are distributed according to the Boltzmann Distribution. This assumption is true at thermal equilibrium.

•If reactants are able to form an activated complex, i.e. reach the transition state, this will lead to product formation.2

All assumptions in transition state theory are not correct. For example, transition state theory would predict that if the transition state is reached, the product will be formed. The rate of formation of product depends on the rate at which the activated complex is formed. It does not take into account barrier recrossing which can lead back to reactants. This is seen in contour plot 4. In the theory however, the activated complex and products are said not to be in equilibrium with each other. Therefore the rate of reaction predicted from transition state theory will be higher than obtained experimentally.

Furthermore, quantum mechanical tunneling is not taken into account. This involves reactants tunneling through the barrier rather than crossing over it.

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

===Exothermic vs Endothermic Reactions===

Below are potential energy surfaces for the reaction between H2 and F in the forward direction and the reverse formation of H2.

{| class="wikitable" border="1"
|+ Potential Energy Surfaces
! H2 + F !! H + HF
|-
| [[File:vishalih2+f.PNG]] || [[File:vishalih+hf.PNG]]||
|-
| The entrance channel is seen at higher energy on the right hand side of the graph. The AB distance initially decreases, describing the approach of F towards H2. As BC increases, this is indicative of the formation of HF and loss of H. The exit channel lies lower in energy and therefore this an exothermic reaction. This means that the H-F bond is stronger than the H-H bond. This can be attributed to the electronegativity difference between H and F, resulting in a polar covalent bond. The ionic contribution to the bond strengthens it. || The entrance channel is now on the left hand side of the graph at short BC distance. As the reaction proceeds, initially the AB distance decreases whilst the BC distance remains constant. This describes HF being approached by H. The BC distance then increases, corresponding to the departure of fluorine. The exit channel lies higher in energy. This indicates that the reaction is endothermic in this direction. Therefore the new covalent H-H bond is weaker than the polar covalent H-F bond ||
|}

===Locating the position of the transition state===

Previously, for the H2 + H system, the idea that r1 = r2 was used based on the symmetry of the transition state. This no longer applies for this system. Instead, Hammond's postulate is used to locate the position of the transition state. Transition states are energy maxima and therefore transient with no measurable lifetime. We can approximate the transition state by studying species that are close in energy to it, as these are expected to be structurally similar. Studying the forward exothermic reaction, we expect an early transition state - one that resembles reactants more closely than products. Therefore r1 (BC) was initially set to 0.74 Angstroms. AB was then varied and the contour plot was examined. Trial and error was used to find the transition state as the position at which there is no trajectory. This is shown in the figure below. The H-H and H-F bond lengths at the transition state were found to be 0.7451 and 1.81065 Angstroms respectively.

[[File:vishalitransitionstate.PNG]]

A plot of energy vs time at the transition state is also shown. The potential energy remains constant throughout.

[[File:vishalitransitionenergy.PNG]]

===Activation energies===

The activation energy was found for both the forward and reverse reaction. Since the activation energy represents the energy difference between the transition state and the reactants, it is possible to calculate it by moving slightly away from the transition state, towards the reactants. Then the difference between the starting and final energy can be found. The results are summarized in the tables below. The main finding was that the activation energies are not the same - the reverse endothermic reaction forming H2 has a higher activation energy than the forward reaction forming HF.

{| class="wikitable" border="1"
|+ Calculating activation energy for HF + H
|-
| Calculation type || MEP
|-
| Number of steps || 11,000
|-
| AB distance || 1.750
|-
|BC distance || 0.7451
|-
| TS Energy || -103.761 kcal mol-1
|-
| HF Energy ||-133.684 kcal mol-1
|-
| Activation energy for HF + H || + 30.1 kcal mol-1
|}

The following energy vs time plot represents the trajectory from the transition state to HF + H i.e. the transition state rolls back to the reactant state.

[[File:vishaliactivation2.PNG|450px]]

{| class="wikitable" border="1"
|+ Calculating activation energy for H2 + F
|-
| Calculation type || MEP
|-
| Number of steps || 11,000
|-
| AB distance || 1.950
|-
|BC distance || 0.7451
|-
| TS Energy || -103.761 kcal mol-1
|-
| H2 energy || -103.817 kcal mol-1
|-
| Activation energy for H2 + F || + 0.056 kcal mol-1
|}

Since the activation energy for the forward exothermic reaction is so small, a distance vs time plot has not been included.

===Reaction Dynamics===

====Reactive trajectory====

A reactive trajectory was identified with the following initial conditions for H2 + F: r1 = 0.75 Angstroms, r2 = 2.3 Angstroms, p1= -1.2, p2= -7.1.

This is an exothermic reaction. However the law of conservation of energy states that energy cannot be created or destroyed, only transformed from one form to another. It is possible to understand where this excess energy has ended up by studying the inter-nuclear momenta vs time plot shown below. The AB product shows very large oscillations in momentum between positive and negative values. Negative values correspond to the nuclei moving towards each other, and positive momenta corresponds to the nuclei moving away from each other. The large oscillation between the two therefore represents vibrations of the bond i.e. the excess energy is converted into the vibrational energy of the product. This can be confirmed experimentally by calorimetry which measures a change in temperature in order to indicate the enthalpy change.

[[File:vishaliexo1.PNG]]

====Application of Polanyi's Rules====

Reactions were set up on the reactant side of H2 + F and the momentum pHH was varied. Control variables are summarized in the table below.

{| class="wikitable" border="1"
| Calculation type || Dynamics
|-
| Number of steps || 500
|-
| AB Distance || 2 Angstroms
|-
| BC Distance || 0.74 Angstroms
|-
| AB Momentum || -0.5
|}

Polanyi's rules state that vibrational energy (p1) is of importance in promoting a reaction with a late transition state. On the other hand, translational energy (p2) is important in promoting a reaction with an early transition state.3 The forward reaction is exothermic with a early transition state. It is expected that if a large amount of translational energy is put into the system, the reaction will be successful.

{| class="wikitable" border="1"
|+ Reaction dynamics - contour plots
! pHH !! Contour plot !! Comments
|-
| -2.885 || [[File:vishalidynamics1.PNG|450px]]|| The reaction has occurred.
|-
| -3 || [[File:vishalidyamics2.PNG|450px]] || This change in pHH has now meant that the reaction does not occur. The transition state is reached but barrier recrossing occurs.
|-
| 3|| [[File:vishalidynamics3.PNG|450px]]|| Positive values were also tested and led to no reaction. Another example of barrier recrossing. This demonstrates the importance of energy being supplied via a particular mode of motion. Here, the vibrational energy is high, but will not be sufficient to allow the reaction to go to completion as it is in the incorrect mode.
|}

Next, the vibrational energy was reduced by changing pHH to 0.1. The translational energy was increased by changing pHF to -0.8. The reaction now occurs and obeys Polanyi's rules.

[[File:vishaliworks.PNG|450px]]

The same analysis was repeated for HF + H. Since this has a late transition state, we expect the opposite effect to above. This has been tested below.

First, the calculation was carried out with low vibrational energy, p1 = 0.1 and high translational energy, p2 = -3. The reaction was not successful. This was then repeated with p1 = 6 and p2 = -0.1 i.e. the vibrational energy was increased and the translational energy was decreased. This led to a reaction. Both contour plots are shown below.

{| class="wikitable" border="1"
|+ Reaction dynamics - contour plots
! Low 1, High p2 !! High p1, Low p2
|-
| [[File:vishalibook.PNG|450px]] ||[[File:vishalibook2.PNG|450px]]
|}

==References==

1. K. J. Laidler and M. Christine King, J.Phys. Chem., 1983, 87, 2657-2664.

2. T. Bligaard, J.K. Nørskov, in Chemical Bonding at Surfaces and Interfaces, 2008

3. Z. Zhang, Y. Zhou and D. H. Zhang, J. Phys. Chem. Lett., 2012, 3(23), 3416-3419

MRD:vishalipala

2018-06-07T15:47:45Z

Mm10114: /* Relating gradient of the PES to transition states and minima */

==Exercise 1: H + H2 system==

===Relating gradient of the PES to transition states and minima===

The gradient of the potential energy surface, given by its first derivative, is zero at both the minimum and transition state. We identify each of these by considering the second derivative. Along the reaction trajectory, there is a maximum in the potential energy surface at the transition state. If we then view the surface in a direction orthogonal to the reaction pathway, there will be a minimum. This describes a saddle point and we identify this as our transition state.

This is the basis of calculating a second derivative. For a transition state, if we calculate the second derivative, it will have a negative value along the reaction coordinate, and a positive value orthogonal to the reaction coordinate. In contrast, the minimum has a positive second derivative along both the reaction coordinate and orthogonal to it.

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 16:47, 7 June 2018 (BST) Very good answer!

===Locating the transition state===

At the transition state, since the H + H2 system is symmetric, r1must equal r2. The momentum is also zero as the atoms are not moving. By varying these distances, the transition state has been identified as occurring at a H-H distance of 0.908 Angstroms. This is indicated in the plot of inter-nuclear distance vs time. The A-B and B-C distances are 0.908 Angstroms and therefore the A-C distance is double this. Over time, this distance remains constant. This can be compared to if the curve was fluctuating, indicating vibrations of the bond. At the transition state, there is no oscillation of the graph. The contour plot also depicts the position of the transition state.

[[File:VishaliTS.PNG|450px]] [[File:vishalits2.PNG|450px]]

===Calculating the reaction path===

Below are plots of internuclear distance vs time for calculations run on MEP and dynamics. A reverse dynamics trajectory is also calculated.

{| class="wikitable" border="1"
|+ Comparing the results of different calculation types
! MEP !! Dynamics !! Dynamics reversed
|-
| [[File:vishalimep.PNG|450px]] || [[File:vishalidynamics.PNG|450px]]|| [[File:vishalireversefirst.PNG|450px]][[File:vishalireverseextra.PNG|450px]]
|-
| In the minimum energy path, the velocity is reset to zero after every step. The A-B and B-C distances are initially around equal. The A-B distance then decreases and reaches a constant level. This represents the new A-B bond length and since there are no oscillations, it is not vibrating. The B-C distance increases gradually. || Compared to the MEP calculation, the new A-B bond can be seen to vibrate, although not intensely. Furthermore, the departure of C occurs at a much faster rate. || We reverse the trajectory by using the previous final positions as the initial distances (r1 = 9.03, r2 = 0.74 Angstroms) and reversing the signs of the momenta. Now there is enough momentum to just go back to our initial state i.e. close to the transition state. This is shown in the top contour plot. However, there is not enough momentum to go over the transition state to product. Instead, the trajecory returns exactly where it came from. In order for the reaction to occur, a tiny more amount of momentum must be provided. This is shown on the bottom figure. Now the trajectory proceeds towards products.
|}

Swapping the initial conditions i.e. r1 = rts and r2 = rts +0.01 makes no difference to the plot. The labels for the curves have just changed.

===Reactive and unreactive trajectories===

{| class="wikitable" border="1"
|+ Determining whether a trajectory is reactive
! p1 !! p2 !! Total Energy (kcal mol-1)!! Reactive? !! Internuclear distance vs time !! Contour Plot !! Comments
|-
| -1.25 || -2.5 || -99.119 || Yes || [[File:vishalireaction1.PNG|450px]]] || [[File:vishalireaction11.PNG|450px]] || B-C shows no oscillation initially and so is not vibrating (or moving). However, A-B is decreasing and therefore A is approaching BC. The transition state occurs when A-B = B-C. The B-C distance then increases as the bond has broken and the atoms separate. The A-B distance now oscillates at the previous B-C distance, indicating vibration of the newly formed bond.||
|-
| -1.5 || -2.0 || -99.119 || No || [[File:vishalireaction2.PNG|450px]]] || [[File:vishalireaction22.PNG|450px]] || The B-C distance oscillates about the bond length throughout the time period. The A-B distance initially decreases. This indicates that the hydrogen atom approaches the hydrogen molecule. However this distance then increases and the hydrogen atom moves away from the molecule again without reacting. This is seen in the contour plot as the trajectory moves back towards reactants i.e. both the entry and exit channels are near the reactant region of the plot. ||
|-
| -1.5 || -2.5 || -99.119 || Yes || [[File:vishalireaction3.PNG|450px]]] || [[File:vishalireaction33.PNG|450px]] || Again the A-B decreases as Ha approaches the HbHc molecule. After the new A-B bond is formed, hydrogen atom C moves away.||
|-
| -2.5 || -5.0 || -99.119 || No || [[File:vishalireaction4.PNG|450px]]] || [[File:vishalireaction44.PNG|450px]] || The A-B-C activated complex initially forms. However, reactants are then reformed by barrier recrossing. This BC molecule shows stronger vibrations, depicted by the larger oscillations.||
|-
| -2.5 || -5.2 || -99.119 || Yes || [[File:vishalireaction5.PNG|450px]]] || [[File:vishalireaction55.PNG|450px]] || The reaction occurs and the AB bond is formed. The exit channel occurs through the product region of the contour plot. ||
|}

===Transition State Theory===

In transition state theory, the pathway from reactants to products involves a transition state which is the highest energy coordinate on the energy profile. It is based on several ideas and assumptions:

• There is a quasi-equilibrium that exists between reactants and the activated complex. This occurs even when the reactants are not in equilibrium with the products.1 For the reaction: A + BC → AB + C, we can write this as:

A + BC ⇄ ABC≠ → AB + C

• The motion of electrons is described quantum mechanically. However, we can treat the nuclear motion separately in terms of classical mechanics. In other words, the Born-Oppenheimer approximation is used.2

• Energies of the reactants are distributed according to the Boltzmann Distribution. This assumption is true at thermal equilibrium.

•If reactants are able to form an activated complex, i.e. reach the transition state, this will lead to product formation.2

All assumptions in transition state theory are not correct. For example, transition state theory would predict that if the transition state is reached, the product will be formed. The rate of formation of product depends on the rate at which the activated complex is formed. It does not take into account barrier recrossing which can lead back to reactants. This is seen in contour plot 4. In the theory however, the activated complex and products are said not to be in equilibrium with each other. Therefore the rate of reaction predicted from transition state theory will be higher than obtained experimentally.

Furthermore, quantum mechanical tunneling is not taken into account. This involves reactants tunneling through the barrier rather than crossing over it.

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

===Exothermic vs Endothermic Reactions===

Below are potential energy surfaces for the reaction between H2 and F in the forward direction and the reverse formation of H2.

{| class="wikitable" border="1"
|+ Potential Energy Surfaces
! H2 + F !! H + HF
|-
| [[File:vishalih2+f.PNG]] || [[File:vishalih+hf.PNG]]||
|-
| The entrance channel is seen at higher energy on the right hand side of the graph. The AB distance initially decreases, describing the approach of F towards H2. As BC increases, this is indicative of the formation of HF and loss of H. The exit channel lies lower in energy and therefore this an exothermic reaction. This means that the H-F bond is stronger than the H-H bond. This can be attributed to the electronegativity difference between H and F, resulting in a polar covalent bond. The ionic contribution to the bond strengthens it. || The entrance channel is now on the left hand side of the graph at short BC distance. As the reaction proceeds, initially the AB distance decreases whilst the BC distance remains constant. This describes HF being approached by H. The BC distance then increases, corresponding to the departure of fluorine. The exit channel lies higher in energy. This indicates that the reaction is endothermic in this direction. Therefore the new covalent H-H bond is weaker than the polar covalent H-F bond ||
|}

===Locating the position of the transition state===

Previously, for the H2 + H system, the idea that r1 = r2 was used based on the symmetry of the transition state. This no longer applies for this system. Instead, Hammond's postulate is used to locate the position of the transition state. Transition states are energy maxima and therefore transient with no measurable lifetime. We can approximate the transition state by studying species that are close in energy to it, as these are expected to be structurally similar. Studying the forward exothermic reaction, we expect an early transition state - one that resembles reactants more closely than products. Therefore r1 (BC) was initially set to 0.74 Angstroms. AB was then varied and the contour plot was examined. Trial and error was used to find the transition state as the position at which there is no trajectory. This is shown in the figure below. The H-H and H-F bond lengths at the transition state were found to be 0.7451 and 1.81065 Angstroms respectively.

[[File:vishalitransitionstate.PNG]]

A plot of energy vs time at the transition state is also shown. The potential energy remains constant throughout.

[[File:vishalitransitionenergy.PNG]]

===Activation energies===

The activation energy was found for both the forward and reverse reaction. Since the activation energy represents the energy difference between the transition state and the reactants, it is possible to calculate it by moving slightly away from the transition state, towards the reactants. Then the difference between the starting and final energy can be found. The results are summarized in the tables below. The main finding was that the activation energies are not the same - the reverse endothermic reaction forming H2 has a higher activation energy than the forward reaction forming HF.

{| class="wikitable" border="1"
|+ Calculating activation energy for HF + H
|-
| Calculation type || MEP
|-
| Number of steps || 11,000
|-
| AB distance || 1.750
|-
|BC distance || 0.7451
|-
| TS Energy || -103.761 kcal mol-1
|-
| HF Energy ||-133.684 kcal mol-1
|-
| Activation energy for HF + H || + 30.1 kcal mol-1
|}

The following energy vs time plot represents the trajectory from the transition state to HF + H i.e. the transition state rolls back to the reactant state.

[[File:vishaliactivation2.PNG|450px]]

{| class="wikitable" border="1"
|+ Calculating activation energy for H2 + F
|-
| Calculation type || MEP
|-
| Number of steps || 11,000
|-
| AB distance || 1.950
|-
|BC distance || 0.7451
|-
| TS Energy || -103.761 kcal mol-1
|-
| H2 energy || -103.817 kcal mol-1
|-
| Activation energy for H2 + F || + 0.056 kcal mol-1
|}

Since the activation energy for the forward exothermic reaction is so small, a distance vs time plot has not been included.

===Reaction Dynamics===

====Reactive trajectory====

A reactive trajectory was identified with the following initial conditions for H2 + F: r1 = 0.75 Angstroms, r2 = 2.3 Angstroms, p1= -1.2, p2= -7.1.

This is an exothermic reaction. However the law of conservation of energy states that energy cannot be created or destroyed, only transformed from one form to another. It is possible to understand where this excess energy has ended up by studying the inter-nuclear momenta vs time plot shown below. The AB product shows very large oscillations in momentum between positive and negative values. Negative values correspond to the nuclei moving towards each other, and positive momenta corresponds to the nuclei moving away from each other. The large oscillation between the two therefore represents vibrations of the bond i.e. the excess energy is converted into the vibrational energy of the product. This can be confirmed experimentally by calorimetry which measures a change in temperature in order to indicate the enthalpy change.

[[File:vishaliexo1.PNG]]

====Application of Polanyi's Rules====

Reactions were set up on the reactant side of H2 + F and the momentum pHH was varied. Control variables are summarized in the table below.

{| class="wikitable" border="1"
| Calculation type || Dynamics
|-
| Number of steps || 500
|-
| AB Distance || 2 Angstroms
|-
| BC Distance || 0.74 Angstroms
|-
| AB Momentum || -0.5
|}

Polanyi's rules state that vibrational energy (p1) is of importance in promoting a reaction with a late transition state. On the other hand, translational energy (p2) is important in promoting a reaction with an early transition state.3 The forward reaction is exothermic with a early transition state. It is expected that if a large amount of translational energy is put into the system, the reaction will be successful.

{| class="wikitable" border="1"
|+ Reaction dynamics - contour plots
! pHH !! Contour plot !! Comments
|-
| -2.885 || [[File:vishalidynamics1.PNG|450px]]|| The reaction has occurred.
|-
| -3 || [[File:vishalidyamics2.PNG|450px]] || This change in pHH has now meant that the reaction does not occur. The transition state is reached but barrier recrossing occurs.
|-
| 3|| [[File:vishalidynamics3.PNG|450px]]|| Positive values were also tested and led to no reaction. Another example of barrier recrossing. This demonstrates the importance of energy being supplied via a particular mode of motion. Here, the vibrational energy is high, but will not be sufficient to allow the reaction to go to completion as it is in the incorrect mode.
|}

Next, the vibrational energy was reduced by changing pHH to 0.1. The translational energy was increased by changing pHF to -0.8. The reaction now occurs and obeys Polanyi's rules.

[[File:vishaliworks.PNG|450px]]

The same analysis was repeated for HF + H. Since this has a late transition state, we expect the opposite effect to above. This has been tested below.

First, the calculation was carried out with low vibrational energy, p1 = 0.1 and high translational energy, p2 = -3. The reaction was not successful. This was then repeated with p1 = 6 and p2 = -0.1 i.e. the vibrational energy was increased and the translational energy was decreased. This led to a reaction. Both contour plots are shown below.

{| class="wikitable" border="1"
|+ Reaction dynamics - contour plots
! Low 1, High p2 !! High p1, Low p2
|-
| [[File:vishalibook.PNG|450px]] ||[[File:vishalibook2.PNG|450px]]
|}

==References==

1. K. J. Laidler and M. Christine King, J.Phys. Chem., 1983, 87, 2657-2664.

2. T. Bligaard, J.K. Nørskov, in Chemical Bonding at Surfaces and Interfaces, 2008

3. Z. Zhang, Y. Zhou and D. H. Zhang, J. Phys. Chem. Lett., 2012, 3(23), 3416-3419

Talk:MRD:zz3116

2018-06-07T15:44:29Z

Mm10114: Created page with "~~~~ Your report is sufficient and you have addressed all the questions. However, your report is very scarce in information at best. You don't write h..."

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 16:44, 7 June 2018 (BST) Your report is sufficient and you have addressed all the questions. However, your report is very scarce in information at best. You don't write how did you obtain some of the values, just provide them. This is not enough, each question should have a description of how did you obtained the values, with some observations followed by conclusions, showing your understanding of each of the exercise. You forget references in question 5, which is a very bad practice. See the in-line comments that should help you make better reports in the future.

MRD:zz3116

2018-06-07T15:41:45Z

Mm10114: /* 10 */

===1===
'''What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.'''

The gradients in different directions at minimum and at transition state are all 0. Saddle point fits the situation: the second differentiation is smaller than 0 in one direction, and larger than 0 in the other.

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 16:09, 7 June 2018 (BST) These are valid observations. However, in which direction is the second differentiation smaller than 0 and larger than 0 at the saddle point? You need to explain (surely not any random directions).

[[File:Saddle_point.PNG]]
The point T represents the transition point which means the minimum of potential energy and the maximum of reaction energy.

===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.'''

rts=0.9075 , as can be seen ,flat lines means no vibrations.

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 16:11, 7 June 2018 (BST) Your rts estimate is correct. However, you need to explain what these flat lines mean. You need to be more descriptive in your answers, showing the understanding, describing which line gives a value for what etc.

[[File:Zzh_q2.PNG]]

===3===
'''Comment on how the mep and the trajectory you just calculated differ.'''

MEP graph:

[[File:ZZH_3_mep.PNG]]

DYNAMIC graph:

[[File:ZZH_3_dynamic.PNG]]

Potential energy graph:

[[File:ZZH_3_potential.PNG]]

the reaction goes towards the product. MEP means the lowest point in the oscillating energy in reaction path. We did not see any oscillations in the graph. The trajectory shows the actual energy changes in the reaction including any oscillations. P1=P2=0 means no initial kinetic energy. HB-HC distance is longer than that of HA-HB gives initial potential energy for the molecule. therefore, HB-HC bond oscillates along the potential energy surface.

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 16:17, 7 June 2018 (BST) Why is the reaction going towards the products? ("product"? are you sure is there only one?) You need to explain why is it that we see no oscillation in MEP. Overall, your answer is sufficient, but it is very scarce. Some more in-depth explanation would be necessary here (cause -> effect), not just stating what is in the graphs, also you need to use the same references/naming for the bond distances as in the plots (AB/BC, not HA/HB/HC, or you need to explain how these relate to these on the plots).

===4===
'''Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.'''

{| class="wikitable" border="1"
! p1 !! p2 !! Total Energy !! Reactive or Unreactive !! Trajectory !! Description
|-
| -1.25 || -2.5 || -99.018 ||Reactive ||[[File:ZZH_4.1.PNG]] || Reactive trajectory starts from the reactants and passes through the transition state and towards the products.
|-
| -1.5 || -2.0 || -100.456 || Unreactive || [[File:ZZH_4.2.PNG]] || Unreactive trajectory starts from the reactants and returns at transition state.
|-
| -1.5 || -2.5 || -98.956 || Reactive || [[File:Zzh_4.3.PNG ]] || Reactive trajectory starts from the reactants and passes through the transition state and towards the products.
|-
| -2.5 || -5.0 ||-84.956 || Unreactive|| [[File:ZZH_4.4.PNG]] ||Unreactive trajectory starts from the reactants and crosses the transition state, bonds formed first and then go back to the reactants.
|-
| -2.5 || -5.2 ||-83.416|| Reactive || [[File:Zzh_4.5.PNG]] ||Unreactive trajectory starts from the reactants and crosses the transition state, bonds formed first and then go back to the reactants, finally moves towards products.
|}

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 16:19, 7 June 2018 (BST) Your answers are sufficient. The purpose of this exercise is to compare different initial conditions for p1 and p2, thus also making comments why in some cases these are reactive or not.

===5===
'''State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''

{{fontcolor1|gray|(You are quoting this from somewhere. You must cite your sources, otherwise you are incurring into plagiarism. [[User:Jbettenc|João]] ([[User talk:Jbettenc|talk]]) 12:40, 2 June 2018 (BST))}}

Transition state theory assumes:
1.Quantum-tunneling effects are assumed negligible

2.The born-oppenheimer approximation is invoked

3.the energy of atoms in the reactant is considered as Boltzman distributed

4.Once the system attains the transition state, the reaction would not go back to the reactant side again.

Comparison of experiment value and the TST:

1.in the actual experiment, the reaction can go back to the reactant partially.but the TST theory says the system would not reenter the reactant region after reaching the transition state. thus the experimental rate of reaction would be slower than the theoretical value.

2.the quantum tunneling takes place in the actual experiment. the energy of an atom decreases over the barrier rather than bounce. actual energy of each atom may not reach so high as the theory.

===Exercise 2===
'''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?'''

potential surface of F+HH reaction:

[[File:ZZH_6_FHH.PNG ]]

potential surface of H+HF reaction:

[[File:Zzh_6_HHF.PNG]]

F+H2 reaction is exothermic, and H+HF reaction is endothermic. HF bond energy is stronger than HH bond as the dissociation energy of HF is larger than that of HH. HF is more stable.

===7===
'''Locate the approximate position of the transition state.'''

the position of transition state is r(FH)=1.8114, r(HH)=0.7445. there are no vibrations at the transition state.

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 16:23, 7 June 2018 (BST) How did you locate this transition state? Was it a lucky guess?

[[File:ZZH_7.PNG]]

===8===
'''Report the activation energy for both reactions.'''
the reaction energy of F +H2 at transition state:

[[File:ZZH_8.1.PNG ]]

the reaction energy of F+H2 :

[[File:ZZH_8.2.PNG]]

Ea of F+H2 =-103.752-(-104.011)=+0.259 kcal/mol.

the reaction of H+HF at transition state:

[[File:Zzh_8.3.PNG ‎]]

Ea of H+HF=-103.752-(-133.756)=30.004 Kcal/mol.

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 16:28, 7 June 2018 (BST) Again, your values are correct. But, you need to be much more verbal about how you've obtained them. The initial conditions do not tell me much, without the appropriate justification and additional plots for distances or potential energy surfaces.

===9===
''' In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?'''

Initial conditions:

[[File:ZZH_9_initial_conditions.PNG ]]

Animation:

[[File:Zzh_9_ani1.PNG]]

[[File:Zzh_ani_2.PNG]]

[[File:Zzh_9_ani3.PNG ]]

Internuclear momenta vs time:

[[File:ZZH_9_momenta_time.PNG ]]

after the formation of HF bond, energy is released since it is an exothermic reaction, and this energy converts to the kinetic energy of H atom which moves away. H atom then collides with other things. Therefore, kinetic energy is converted to thermal energy. this can be confirmed by measuring the temperature.

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 16:38, 7 June 2018 (BST) Very good observations. However, "H atom then collides with other things"? What other things? You need to be specific.

'''Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 0.74, with a momentum pFH = -0.5, and explore several values of pHH in the range -3 to 3 (explore values also close to these limits). What do you observe?'''

figures with changing P2:

[[File:Zzh_10.1.PNG]]

[[File:Zzh_10.2.PNG]]

[[File:Zzh_10.3.PNG]]

[[File:Zzh_10.4.PNG]]

[[File:Zzh_10.5.PNG]]

[[File:Zzh_10.6.PNG]]

[[File:Zzh_10.7.PNG]]

The momenta in the reaction provide main energy to activation. when the momentum of BC is set -3 or 3, it is reactive, otherwise it's not.

'''For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now?'''

[[File:Zzh_change_10.PNG]]

the trajectory recrosses in the reaction but forms products finally. so it is reactive.

===10===
'''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.'''
F+H2 is exothermic with an early barrier, the translation has effect but vibration does not. For the early transition state, the vibration can provide the energy to get to the transition state, but it leads that reaction can recross the reactant region rapidly.
H+HF is endothermic with a late barrier, the vibration has effect but transition does not.

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 16:41, 7 June 2018 (BST) This is good. Have you come to these conclusions by yourself, or are you citing here Polanyi's rules? At the end, you made a mistake by saying transition instead of 'translation'. Be more careful.

MRD:zz3116

2018-06-07T15:38:58Z

Mm10114: /* 9 */

MRD:EM23416

2018-05-31T19:19:28Z

Mm10114: /* Reaction Dynamics */

=Exercise 1: H + H2 system=
==Distinguishing between Transition States and Minima==

The transition state is a saddle point on the potential energy surface, at the maximum point along the minimum energy path between the reactants and products.

At both the transition state and the minimum points corresponding to the reactants and products the potential energy gradients with respect to both reaction coordinates will be zero, so ∂V(r1)/∂r1=∂V(r2)/∂r2=0.
The transition state can be distinguished from the minimum point by taking the second derivatives of the potential energy gradient.
This can be achieved by examining the Hessian matrix for the transition state, which gives a matrix

At the transition state, (∂V(r1)/∂r1)(∂V(r2)/∂r2)=0, and
whereas at the minimum points (∂2V(r1)/∂r12)>0.
(∂2V(r2)/∂r22)>0.

==Locating the transition state==

'''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.'''

The transition state position is estimated as AB=BC=0.9077. As seen on the below diagram, at this position there is no oscillation in the AB or BC bond distances as only a single straight line can be seen on the graph at this value. This shows that the atoms are not oscillating at this position and therefore that there is little to no kinetic energy in the molecule, and thus the transition state has been located.
[[File: Emck_internuclear_distance_ts.PNG|300px]]

==MEPs and reaction paths==

The minimum energy pathway (mep) is the pathway taken by the molecules when the momentum is consistently at a minimum value close to zero and molecules move infinitely slowly, thus kinetic energy is kept at almost zero.
[[File: Emck_hhh_mep_con.png|300px]]
This trajectory shows the actual pathway taken by the molecules as they move with kinetic energy and inertial motion. The momentum does not remain near zero in this pathway, as shown by the oscillation in the reaction path as the reaction moves away from the transition state, which is not present in the mep trajectory shown abovee.
[[File: Emck_hhh_trajectory_con.png|300px]]

The reverse of this trajectory with using its final geometries is the e
[[File: Emck_hhh_revtrajectory_con.png|300px]]

The reverse of this trajectory, using its final geometries of r1= 5.2810, r2= 0.7455, p1=-2.4810, p2= -1.5492, is its exact opposite pathway.

==Reactive and Unreactive trajectories==

{| class="wikitable" border=1
! Reaction !! p1 !! p2 !! Total energy (kJmol-1) !! Trajectory !! Reactivity
|-
| a || -1.25 || -2.5 || -99.119 ||[[File: Emck_hhh_n2.5_n1.25_con.png|200px]]|| This trajectory is reactive, and passes from reactants to products cleanly. The lack of oscillation visible prior to the transition state indicates that the H2 molecule has little vibrational momentum. The oscillation after the transition state shows that the H-H bond formed vibrates.
|-
| b || -1.5 || -2.0 || -100.456 ||[[File: Emck_hhh_n2_n1.5_con.png|200px]]|| This trajectory is unreactive, as the reaction does not progress into the product channel. The colliding molecules do not have sufficient energy to overcome the activation barrier.
|-
| c || -1.5 || -2.5 || -98.956 ||[[File: Emck_hhh_n2.5_n1.5_con.png|200px]]|| This trajectory is reactive, and the increase in oscillation visible before the transition state relative to reaction trajectory '''a''' corresponds to the increase in vibrational momentum of the starting H2 molecule.
|-
| d || -2.5 || -5.0 || -84.956 ||[[File: Emck_hhh_n5_n2.5_con.png|200px]]|| This trajectory is unreactive. The H2 molecule has alot of vibrational energy prior to reaction and thus oscillates strongly prior to the transition state, however the energy of the system is not enough to move fully from the transition state to the products, and thus it recrosses the barrier and the reaction doesn't progress.
|-
| e || -2.5 || -5.2 || -83.416 ||[[File: Emck_hhh_n5.2_n2.5_con.png|200px]]|| This trajectory is reactive, as the higher translational energy means that after recrossing the energy barrier, as seen in reaction '''d''', the molecule collides with the wall of the energy surface and is able to recross the barrier again to form the products.
|}

Transition state theory (TST) makes 3 main assumptions:
1) The activated complex is in equilibrium with the reactants, but not the products, thus recrossing of the barrier is negligible.
2) The reactant nuclei behave as predicted by classical mechanics.
3) The reaction path will pass through the lowest energy saddle point on the potential energy surface, so multiple reaction pathways are ignored.

The key assumption of transition state theory (TST) is that the activated complex is in equilibrium with the reactants, but not the products. This means that atoms colliding with sufficient energy to overcome the activation barrier and form the transition
structure will always react to form products.
TST ignores all quantum mechanical behaviour of the reactant nuclei, which becomes especially prevalent in systems of light atoms such as hydrogen. This means that the values predicted by TST don't consider the tunnelling of reactant nuclei sufficient energy to overcome the activation barrier. It also means that the values won't account for particles in energy levels above the bottom of the potential well, resulting in additional zero-point energy. Both of these quantum effects effectively lower the activation barrier for the reaction and will cause experimental values for the reaction rate in this system to be higher than those predicted by TST.

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 20:08, 31 May 2018 (BST) What is your source for the TST assumptions? You're missing a reference.

=Exercise 2: F-H-H system=

==Reaction energetics for F + H2 and H + HF==

From the potential energy surface, it can be seen that the F + H2 reaction is exothermic. This is due to the stability of the H-F bond forming during the reaction, which has a large ionic character due to the difference in the electronegativities of H and F and this strengthens it. This means that the energy released during formation of the H-F bond is greater than the energy consumed in the breaking of the H-H bond during the reaction, causing energy to be released overall. Conversely, the H + HF reaction is endothermic, as the energy released during formation of the new H-H bond can't compensate for that consumed in breaking the strong H-F bond. This confirms that the bond strength of H-F is larger than that of H-H, as seen in literature as 569.7 and 435.8, respectively.'''(REF)'''

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 20:17, 31 May 2018 (BST) You are right. However, you report some values, what are these? Without unit these values are meaningless.

Hammond's postulate states that the transition state will be similar in structure to the reactants or products depending on which is closest to it in energy. Therefore for the early transition state of the exothermic reaction it will resemble the reactants and for the late transition state of the endothermic reaction it will resemble the products. Thus it was predicted that the transition state would be similar in structure to F + H2.
The transition state was found to be at rFH= 1.8107 Å and rHH=0.7449 Å.

[[File:Emck_f_h2_mep_energy_vs_time.png]]
[[File:Emck_f_h2_mep_contour.png]]

[[File:Emck_h_hf_mep_energy_vs_time.png]]
[[File:Emck_h_hf_mep_contour.png]]

The activation energies were found by performing an mep to either side of the transition state, and analysing the energy vs time graphs to extract the energy differences.
The activation energies were as follows:
F + H2 = 0.176 kJmol-1
H + HF = 30.201 kJmol-1

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 20:17, 31 May 2018 (BST) Are you sure about the unit here? Or was it kcal mol-1 instead?

==Reaction Dynamics==

A reactive trajectory was identified for the F + H2 reaction with the following conditions, and is shown below.

rFH=2.0
rHH=0.74
pFH=-1.4
pHH=-0.75

[[File: emck_f2_n1.4_n0.75_con.png|250px]]
[[File: emck_f2_n1.4_n0.75_mom.png|250px]]

As this reaction is exothermic, it can be expected that, due to the law of conservation of energy, the excess energy released during the reaction is converted into other forms such as heat flow to the surroundings and vibrational energy in the products. This can be seen in the internuclear momentum vs time graph for the trajectory, as the amplitude of oscillations in the H-F product molecule is much larger than in the H-H reactant, indicating that energy has been converted. The appearance of this oscillation in a graph of the momentum of the molecules also shows that the vibrational momentum and energy are closely related.
This conversion of energy during the reaction could be observed experimentally via calorimetry to study the heat flow of the reaction, and thus the heat released. The difference in vibrational energy of the reactant and product molecules could be observed by IR spectroscopy, which would display overtone bands for molecules in higher vibrational states.

Polonyi's empirical rules state that vibrational energy is more efficient than translational energy in overcoming a late transition state energy barrier of an endothermic reaction, whereas translational energy is more effecient than vibrational energy for an endothermic reaction.

The graph below shows that significant vibrational momentum is needed to overcome the late transition barrier of the H + HF reaction, whereas translational energy is negligible.

[[File:emck_hf_n0.1_n10_con.png|300px]]

The graph below shows that translational energy is better at promoting the exothermic F + H2 reaction, as when the translational momentum increases the reaction becomes much more feasible.

NOTE TO MARKER: Very sorry, I just forgot to add the image file links. Its been a long day.
[[File:Emck_f2_n0.5_1_con.png|300px]]
[[File:emck_f2_n0.8_0.1_con.png|300px]]

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 20:19, 31 May 2018 (BST) Overall good observations and stating the Polanyi's rules. However, where is the reference to the source you've taken the Polanyi's rules from? And the 'NOTE TO MARKER' completely unnecessary.

MRD:EM23416

2018-05-31T19:17:31Z

Mm10114: /* Exercise 2: F-H-H system */

MRD:EM23416

2018-05-31T19:08:35Z

Mm10114: /* Exercise 1: H + H2 system */

=Exercise 1: H + H2 system=
==Distinguishing between Transition States and Minima==

The transition state is a saddle point on the potential energy surface, at the maximum point along the minimum energy path between the reactants and products.

At both the transition state and the minimum points corresponding to the reactants and products the potential energy gradients with respect to both reaction coordinates will be zero, so ∂V(r1)/∂r1=∂V(r2)/∂r2=0.
The transition state can be distinguished from the minimum point by taking the second derivatives of the potential energy gradient.
This can be achieved by examining the Hessian matrix for the transition state, which gives a matrix

At the transition state, (∂V(r1)/∂r1)(∂V(r2)/∂r2)=0, and
whereas at the minimum points (∂2V(r1)/∂r12)>0.
(∂2V(r2)/∂r22)>0.

==Locating the transition state==

'''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.'''

The transition state position is estimated as AB=BC=0.9077. As seen on the below diagram, at this position there is no oscillation in the AB or BC bond distances as only a single straight line can be seen on the graph at this value. This shows that the atoms are not oscillating at this position and therefore that there is little to no kinetic energy in the molecule, and thus the transition state has been located.
[[File: Emck_internuclear_distance_ts.PNG|300px]]

==MEPs and reaction paths==

The minimum energy pathway (mep) is the pathway taken by the molecules when the momentum is consistently at a minimum value close to zero and molecules move infinitely slowly, thus kinetic energy is kept at almost zero.
[[File: Emck_hhh_mep_con.png|300px]]
This trajectory shows the actual pathway taken by the molecules as they move with kinetic energy and inertial motion. The momentum does not remain near zero in this pathway, as shown by the oscillation in the reaction path as the reaction moves away from the transition state, which is not present in the mep trajectory shown abovee.
[[File: Emck_hhh_trajectory_con.png|300px]]

The reverse of this trajectory with using its final geometries is the e
[[File: Emck_hhh_revtrajectory_con.png|300px]]

The reverse of this trajectory, using its final geometries of r1= 5.2810, r2= 0.7455, p1=-2.4810, p2= -1.5492, is its exact opposite pathway.

==Reactive and Unreactive trajectories==

{| class="wikitable" border=1
! Reaction !! p1 !! p2 !! Total energy (kJmol-1) !! Trajectory !! Reactivity
|-
| a || -1.25 || -2.5 || -99.119 ||[[File: Emck_hhh_n2.5_n1.25_con.png|200px]]|| This trajectory is reactive, and passes from reactants to products cleanly. The lack of oscillation visible prior to the transition state indicates that the H2 molecule has little vibrational momentum. The oscillation after the transition state shows that the H-H bond formed vibrates.
|-
| b || -1.5 || -2.0 || -100.456 ||[[File: Emck_hhh_n2_n1.5_con.png|200px]]|| This trajectory is unreactive, as the reaction does not progress into the product channel. The colliding molecules do not have sufficient energy to overcome the activation barrier.
|-
| c || -1.5 || -2.5 || -98.956 ||[[File: Emck_hhh_n2.5_n1.5_con.png|200px]]|| This trajectory is reactive, and the increase in oscillation visible before the transition state relative to reaction trajectory '''a''' corresponds to the increase in vibrational momentum of the starting H2 molecule.
|-
| d || -2.5 || -5.0 || -84.956 ||[[File: Emck_hhh_n5_n2.5_con.png|200px]]|| This trajectory is unreactive. The H2 molecule has alot of vibrational energy prior to reaction and thus oscillates strongly prior to the transition state, however the energy of the system is not enough to move fully from the transition state to the products, and thus it recrosses the barrier and the reaction doesn't progress.
|-
| e || -2.5 || -5.2 || -83.416 ||[[File: Emck_hhh_n5.2_n2.5_con.png|200px]]|| This trajectory is reactive, as the higher translational energy means that after recrossing the energy barrier, as seen in reaction '''d''', the molecule collides with the wall of the energy surface and is able to recross the barrier again to form the products.
|}

Transition state theory (TST) makes 3 main assumptions:
1) The activated complex is in equilibrium with the reactants, but not the products, thus recrossing of the barrier is negligible.
2) The reactant nuclei behave as predicted by classical mechanics.
3) The reaction path will pass through the lowest energy saddle point on the potential energy surface, so multiple reaction pathways are ignored.

The key assumption of transition state theory (TST) is that the activated complex is in equilibrium with the reactants, but not the products. This means that atoms colliding with sufficient energy to overcome the activation barrier and form the transition
structure will always react to form products.
TST ignores all quantum mechanical behaviour of the reactant nuclei, which becomes especially prevalent in systems of light atoms such as hydrogen. This means that the values predicted by TST don't consider the tunnelling of reactant nuclei sufficient energy to overcome the activation barrier. It also means that the values won't account for particles in energy levels above the bottom of the potential well, resulting in additional zero-point energy. Both of these quantum effects effectively lower the activation barrier for the reaction and will cause experimental values for the reaction rate in this system to be higher than those predicted by TST.

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 20:08, 31 May 2018 (BST) What is your source for the TST assumptions? You're missing a reference.

=Exercise 2: F-H-H system=

==Reaction energetics for F + H2 and H + HF==

From the potential energy surface, it can be seen that the F + H2 reaction is exothermic. This is due to the stability of the H-F bond forming during the reaction, which has a large ionic character due to the difference in the electronegativities of H and F and this strengthens it. This means that the energy released during formation of the H-F bond is greater than the energy consumed in the breaking of the H-H bond during the reaction, causing energy to be released overall. Conversely, the H + HF reaction is endothermic, as the energy released during formation of the new H-H bond can't compensate for that consumed in breaking the strong H-F bond. This confirms that the bond strength of H-F is larger than that of H-H, as seen in literature as 569.7 and 435.8, respectively.'''(REF)'''

Hammond's postulate states that the transition state will be similar in structure to the reactants or products depending on which is closest to it in energy. Therefore for the early transition state of the exothermic reaction it will resemble the reactants and for the late transition state of the endothermic reaction it will resemble the products. Thus it was predicted that the transition state would be similar in structure to F + H2.
The transition state was found to be at rFH= 1.8107 Å and rHH=0.7449 Å.

[[File:Emck_f_h2_mep_energy_vs_time.png]]
[[File:Emck_f_h2_mep_contour.png]]

[[File:Emck_h_hf_mep_energy_vs_time.png]]
[[File:Emck_h_hf_mep_contour.png]]

The activation energies were found by performing an mep to either side of the transition state, and analysing the energy vs time graphs to extract the energy differences.
The activation energies were as follows:
F + H2 = 0.176 kJmol-1
H + HF = 30.201 kJmol-1

==Reaction Dynamics==

A reactive trajectory was identified for the F + H2 reaction with the following conditions, and is shown below.

rFH=2.0
rHH=0.74
pFH=-1.4
pHH=-0.75

[[File: emck_f2_n1.4_n0.75_con.png|250px]]
[[File: emck_f2_n1.4_n0.75_mom.png|250px]]

As this reaction is exothermic, it can be expected that, due to the law of conservation of energy, the excess energy released during the reaction is converted into other forms such as heat flow to the surroundings and vibrational energy in the products. This can be seen in the internuclear momentum vs time graph for the trajectory, as the amplitude of oscillations in the H-F product molecule is much larger than in the H-H reactant, indicating that energy has been converted. The appearance of this oscillation in a graph of the momentum of the molecules also shows that the vibrational momentum and energy are closely related.
This conversion of energy during the reaction could be observed experimentally via calorimetry to study the heat flow of the reaction, and thus the heat released. The difference in vibrational energy of the reactant and product molecules could be observed by IR spectroscopy, which would display overtone bands for molecules in higher vibrational states.

Polonyi's empirical rules state that vibrational energy is more efficient than translational energy in overcoming a late transition state energy barrier of an endothermic reaction, whereas translational energy is more effecient than vibrational energy for an endothermic reaction.

The graph below shows that significant vibrational momentum is needed to overcome the late transition barrier of the H + HF reaction, whereas translational energy is negligible.

[[File:emck_hf_n0.1_n10_con.png|300px]]

The graph below shows that translational energy is better at promoting the exothermic F + H2 reaction, as when the translational momentum increases the reaction becomes much more feasible.

NOTE TO MARKER: Very sorry, I just forgot to add the image file links. Its been a long day.
[[File:Emck_f2_n0.5_1_con.png|300px]]
[[File:emck_f2_n0.8_0.1_con.png|300px]]

Talk:MRD:lh1316

2018-05-31T19:04:13Z

Mm10114: Created page with "~~~~ Your report is good. You've answered most of the questions. Some answers could have some more in-depth conclusions. Small mistakes like mixing t..."

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 20:04, 31 May 2018 (BST) Your report is good. You've answered most of the questions. Some answers could have some more in-depth conclusions. Small mistakes like mixing the energy unit (could've been confirmed with a demonstrator).

MRD:lh1316

2018-05-31T19:02:07Z

Mm10114: /* Reaction dynamics */

== EXERCISE 1: H + H2 system ==
=== Dynamics from the transition state region ===
1.What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minimal and transition structures can be distinguished using the curvature of the potential energy surface.

The gradient of the potential energy surface(PES) at a minimum and at the transition structure(saddle point) are both zero. However, Second Partial Derivative test can be used to distinguish if a critical point of a function is a local minimum, maximum or saddle point. The Hessian matrix H of f is the 2 × 2 matrix of partial derivatives of f, if:
1. det(H(x,y))>0 and fxx(x,y)>0 then (x,y) is a local minimum of f
2. det(H(x,y))>0 and fxx(x,y)<0 then (x,y) is a local maximum of f
3. det(H(x,y))<0 then (x,y) is a saddle point of f
4. det(H(x,y))=0 then this test is inconclusive <ref name="Second Partial Derivative test" />

=== Trajectories from r1 = r2: locating the transition state ===
2.Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Inter nuclear Distances vs Time” plot for a relevant trajectory.

[[File:Lh1316_h3_Surface_Plot_0.9076.png|thumb|400px|none|Fig.2.1 Surface plot of estimate of the transition state position ]]
[[File:Lh1316_h3_Inter_nuclear_Distances.png|thumb|400px|none|Fig.2.2 “Inter nuclear Distances vs Time” plot of estimate of the transition state position ]]

The best estimate of the transition state position(rts) was found when the value of the distances between AB and BC are both equal to 0.9076 Å. The initial setting of both distance are both 0.9076 Å and both momentum are both zero. From Fig.2.1 and Fig.2.2, it is obvious that the trajectory is just a black point locating on the ridge and never move. If that is not the correct position of the transition sate, then the trajectory will simply fall off and transfer all the potential energy to the kinetic energy.

=== Trajectories from r1 = rts+δ, r2 = rts ===
3.Comment on how the mep and the trajectory you just calculated differ.
[[File:Lh1316_h3_Surface_Plot_mep.png|thumb|400px|none|Fig.3.1 Surface Plot with calculation type:MEP]]
[[File:Lh1316_h3_Surface_Plot_0.9176+0.9076.png|thumb|400px|none|Fig.3.2 Surface Plot with calculation type:Dynamics]]

For MEP calculation, a very special trajectory is produced as the figure3.1 shows. Since the momentum is always reset to zero in each step, this trajectory moves in a extremely slow rate towards the direction with lower potential energy. Therefore, in this case, the trajectory moves along the minimum points of the PES tunnel and looks like a straight line. On the contrary, for dynamic calculation, the trajectory has oscillation and thus looks like wave.

4.What would change if we used the initial conditions r1 = rts and r2 = rts+0.01 instead?
[[File:Lh1316_h3_Inter_nuclear_Distances_r10.9176.png|thumb|400px|none|Fig.4.1“Inter nuclear Distances vs Time” plot when r2 = rts and r1 = rts+0.01]]
[[File:Lh1316_h3_Inter_nuclear_Distances_r20.9176.png|thumb|400px|none|Fig.4.2 “Inter nuclear Distances vs Time” plot when r1 = rts and r2 = rts+0.01 ]]
[[File:Lh1316_h3_Inter_nuclear_momenta_r10.9176.png|thumb|400px|none|Fig.4.3 “Inter nuclear Momenta vs Time” plot when r2 = rts and r1 = rts+0.01 ]]
[[File:Lh1316_h3_Inter_nuclear_momenta_r20.9176.png|thumb|400px|none|Fig.4.4 “Inter nuclear Momenta vs Time” plot when r2 = rts and r1 = rts+0.01 ]]

{| class="wikitable" border="1"
|+ Table.4.1Final values obtained form 4 pictures
! !! r1 = rts+0.01 !! r2 = rts+0.01
|-
| r1/Å || 9.005 || 0.740
|-
| r2/Å || 0.740 || 9.005
|-
| p1/kg⋅ms-1 || 2.481 || 0.913
|-
| p2/kg⋅ms-1 || 0.913 || 2.481
|}
The first and third pictures are “Inter nuclear Distances vs Time” and “Inter nuclear Momenta vs Time” with conditions of r1 = rts+0.01, r2 = rts. On the contrast The second and fourth pictures are those with initial conditions of r1 = rts and r2 = rts+0.01 instead. The final values of distance and momentum are summarized in the table. It's obvious that after conditions switch, the final values of distance and momentum switch as well.

5.Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.What do you observe? 
[[File:Lh1316_h3_surface_plot_r20.9176.png|thumb|400px|none|Fig.5.1 Surface plot when r2 = rts and r1 = rts+0.01]]
[[File:Lh1316_h3_surface_plot_r2_0.9176_reverse.png|thumb|400px|none|Fig.5.2 Surface plot with reverse condition]]
When the initial condition is r1 = rts and r2 = rts+0.01 with both momentum of 0, the trajectory starts from the transition state to the end of BC potential energy curve. In this case, the potential energy converts to the kinetic energy which enables atom A and B approach each other and atom C moves away from atom A and B.
When the calculation resets as the question describes, the new trajectory is simply the reverse of the original one. It starts at the ends of last calculation and ends at the starts of the last calculation due to energy conservation.

===Reactive and nonreactive trajectories===
6.Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.

{| class="wikitable" border="1"
|+ Table 6.1 Reactive and nonreactive trajectories 
! p1/kg⋅ms-1 !! p2/kg⋅ms-1 !! kinetic energy/kJmol-1 !! total energy/kJmol-1 !! active trajectory?
|-
| -1.25 || -2.5 || +4.687 || -99.119 || yes
|-
| -1.5 || -2.0 || +3.250 || -100.456 || no
|-
| -1.5 || -2.5 || +4.750 || -98.956 || yes
|-
| -2.5 || -5.0 || +18.750 || -84.956 || no
|-
| -2.5 || -5.2 || +20.290 || -83.416 || yes
|}
[[File:Lh1316_h3_contour_1.png|thumb|400px|none|Fig.6.1 Contour plot of first initial condition]]
These 5 combinations have the same initial positions r1 = 0.74 Å and r2 = 2.0 Å leading to the same potential energy since they start at the same position of potential energy surface.
For the first combination, the reactants have enough kinetic energy to overcome the activation barrier. Hence after atom C approaches molecule A-B, atom C form bond with atom B. Then atom A and molecule B-C move away from each other. The products have higher vibration energy than that of the reactants as the graph above shows.

[[File:Lh1316_h3_contour_2.png|thumb|400px|none|Fig.6.2 Contour plot of second initial condition]]
For the second combination atom C and molecule A-B moves towards each other initially. When the distance between atom C and atom B nearly approaches 0.9076 Å, atom C and molecule A-B moves away from each other along the original path because in this case atom C and molecule A-B do not have enough kinetic energy to overcome the activation barrier.

[[File:Lh1316_h3_contour_3.png|thumb|400px|none|Fig.6.3 Contour plot of third initial condition]]
The third combination is similar to the first one, but the reactants pass the activation barrier with kinetic energy that is higher than the energy required. Moreover, the atoms move quicker.

[[File:Lh1316_h3_contour_4.png|thumb|400px|none|Fig.6.4 Contour plot of fourth initial condition]]
[[file:Lh1316_h3_surface_4.png|thumb|400px|none|Fig.6.5 Surface plot of fourth initial condition]]
In this case, the reactants have higher oscillating energy and pass transition region twice. The trajectory reverts back immediately after crossing the transition state. In the Animation, despite that atom B and C form bond initially, it dissociates later when the distance between atom A and B become close.

[[File:Lh1316_h3_contour_5.png|thumb|400px|none|Fig.6.6 Contour plot of fourth initial condition]]
[[file:Lh1316_h3_surface_5.png|thumb|400px|none|Fig.6.7 Surface plot of fourth initial condition]]
In this case, the reactants also have high oscillating energy but pass transition energy three times. The trajectory reverts back twice and finally goes towards the product side. In the Animation, Atom B and C form bond initially, it dissociates later when the distance between atom A and B become close. The bond between atom A and B also dissociates immediately when the distance between atom A and B become close. Finally molecule B-C is reformed and atom A, molecule B-C move away from each other.

7.State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

There are 3 main assumptions of Transition State Theory:
1.Quantum-tunneling effects are assumed negligible and the Born-Oppenheimer approximation is applied.
2. the atoms in the reactant state have energies that obeys Boltzmann distribution.
3. Once the system attains the transition state, with a velocity towards the product configuration, it will not reenter the initial state region again.<ref name="Transition State Theory" />

In reality, the third assumption breaks down which can be approved by the fourth and fifth examples in Question 6. Therefore, the experimental trajectory should be longer than what it's expected leading to a slower reaction rate.

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 19:56, 31 May 2018 (BST) I don't quite follow what you mean by that last sentence.

== EXERCISE 2: F - H - H system ==

=== PES inspection ===
8.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? Locate the approximate position of the transition state. 

[[File:Lh1316_ex2_surface_setting_1.PNG|400px|thumb|none|Fig.8.1 Inital setting of F+ H2 reaction]]
[[File:Lh1316_ex2_surface1.png|400px|thumb|none|Fig.8.2 surface plot of F+ H2 reaction]]

In this case, atom A is F and atom B and C are H. The initial conditions are setup as figure 8.1 shows,therefore the reaction is from F + H2 model to H + HF model. In the Figure 8.2, the left PES belongs to F + H2 model while the right PES refers to H + HF model. The reaction starts form the left hand side and oscillates strongly towards the product side. Since the reactants have higher potential energy, this reaction is exothermic. As the literature shows, the bond energies of H2 and HF are 435 kJmol-1 and 569 kJmol-1, which obeys the calculated trajectory. <ref name="Enthapies of HH and HF" />

[[File:Lh1316_ex2_surface_setting_2.PNG|400px|thumb|none|Fig.8.3 Setting of finding transition state]]
[[File:Lh1316_ex2_distanceplot_2.png|400px|thumb|none|Fig.8.4 Inter nuclear distance VS time]]
[[File:Lh1316_ex2_surface2.png|400px|thumb|none|Fig.8.5 Surface plot]]

The transition state is found when the initial setting is as the figure 8.3 shows. Figure 8.4 and figure 8.5 proves the success of finding the transition state of this reaction. Therefore, the transition state is where AB distance equals 1.810 Å and BC distance equals 0.746 Å.

For the H + HF reaction, it is simply the reverse of the F + H2 reaction.

9.Report the activation energy for both reactions. 

[[File:Lh1316_ex2_setting_f+h2.PNG|400px|thumb|none|Fig.9.1 Setting for finding activation energy of F+H2 model(MEP)]]
[[File:Lh1316_ex2_surface_f+h2.png|400px|thumb|none|Fig.9.2 Surface plot of finding activation energy of F+H2 model(MEP)]]
[[File:Lh1316_ex2_f+h2+energy.png|400px|thumb|none|Fig.9.3 Energy VS Time for F+H2 model(MEP)]]

There are two ways to get the activation energy. MEP calculation is performed in this case. The structure with this initial condition (as figure9.1 shows) resembles to the structure of the transition state. The position of the starting point is located at the left hand side of transition state, which is towards the reactant tunnel,like figure 9.2 shows. Since there is no momentum at the start, the trajectory can't cross the transition state and just simply fall down from the initial position(near the transition state) towards the reactant position(F+H2 model). Therefore the activation energy of F+H2 reaction is the difference between the initial energy and final energy in the fig.9.3, which is summarized in table9.1.

[[File:Lh1316_ex2_setting_hf+h.PNG|400px|thumb|none|Fig.9.4 Setting for finding activation energy of HF+H model(MEP)]]
[[File:Lh1316_ex2_surface_hf+h.png|400px|thumb|none|Fig.9.5 Surface plot of finding activation energy of HF+H model(MEP)]]
[[File:Lh1316_ex2_hf_h_energy.png|400px|thumb|none|Fig.9.6 Energy VS Time for HF+H model(MEP)]]

By making HF bond shorter, the position of the starting point is now switched to the right hand side of transition state, which is towards the product tunnel,like figure 9.5 shows. Fig.9.6 illustrates how energy drops when the trajectory moves to the product side(HF+H model). Since the HF+H reaction is simply the reverse of H2+F, its activation energy equals the final energy minus the initial energy of Fig.9.6.

{| class="wikitable" border="1"
|+ Table9.1 Summary of energies of F + H2 and H + HF reactions
! !! F + H2 reaction(MEP) !! H + HF reaction(MEP) !! F + H2 reaction(dynamics) !! H + HF reaction(dynamics)
|-
| potential energy/kJmol-1 || -103.995 || -133.992 || -104.005 || -133.995
|-
| Transition energy/kJmol-1 || -103.751 || -103.752 || -103.751 || -103.751
|-
| Activation energy/kJmol-1 || 0.244 || 30.240 || 0.254 || 30.244
|}
Another method is via the Dynamics calculation. Since We already found the position of the transition state, We can obtain the exact value of the energy at the transition state which is -103.751 kJmol-1. Then we set the initial condition which is far away from the transition state and calculate its potential energy. The activation energy is simply the difference between these two energies. The value is summarized in the fourth and fifth columns of the table9.1.
Theoretically the activation energies calculated via MEP method should be a bit smaller than that calculated via Dynamics method, because the transition energy used is the energy of structure neighboring the transition state.

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 19:59, 31 May 2018 (BST) Good job. However, are you sure the energy unit, in the lespgui software, is kJ mol-1? It should be kcal mol-1.

===Reaction dynamics===
10.In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?

[[File:Lh1316_ex2_2_setting.PNG|400px|thumb|none|Fig.10.1 Initial setting of F+ H2 reaction]]
[[File:Lh1316_ex2_2_inter_momentum.png|400px|thumb|none|Fig.10.2 Inter nuclear momenta VS time plot]]
[[File:Lh1316_ex2_2_energy.png|400px|thumb|none|Fig.10.3 Energy VS time plot]]
A set of initial conditions leading to a reactive trajectory for the F + H2 can be found as Fig10.1 shows and the value of p2 can be changed from 7 without upper limit. In this case, molecule BC(H2) is approaching atom A(F) until atom B(H) collide with atom A(F). During the process, the potential energy is converted to kinetic energy and molecule AB(HF) is excited to its excited vibrational state as Fig.10.2 shows. As Fig10.3 shows, After the absorption of the reaction energy, the energy of the system is interchanging between the potential energy and the kinetic energy.

[[User:Mm10114|Mm10114]] ([[User talk:Mm10114|talk]]) 20:02, 31 May 2018 (BST) You didn't provide an answer to the second question.

11.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.

In a typical chemical reaction with an energetic barrier, there is a saddle point that the reactants must cross in order to reach
the product side. Which form of energy (translational or vibrational) initially deposited in reactants is more efficacious in surmounting the barrier is one of the central topics in the field of reaction dynamics. Polanyi's empirical rules state that vibrational energy is more efficient in promoting a late-barrier reaction( a transition state resembling the products) whereas transnational energy is proficient in promoting an early barrier reaction( a transition state resembling the reactants). <ref name="Principle def" />

[[File:Lh1316_setting_0.1.PNG|400px|thumb|none|Fig.11.1 Initial setting of HF+H reaction:low vibration]]
[[File:Lh1316_contour_0.1.png|400px|thumb|none|Fig.11.2 Counter plot of HF+H reaction:low vibration]]
[[File:Lh1316_a_0.1.png|400px|thumb|none|Fig.11.6 Skew plot of HF+H reaction:low vibration]]
[[File:Lh1316_setting_2.PNG|400px|thumb|none|Fig.11.4 Initial setting of HF+H reaction:high vibration]]
[[File:Lh1316_contour_2.png|400px|thumb|none|Fig.11.5 Counter plot of HF+H reaction:high vibration]]
[[File:Lh1316_a_2.png|400px|thumb|none|Fig.11.6 Skew plot of HF+H reaction:high vibration]]

In the calculation, since the HF molecule(AB) is bounded, all the AB momentum is contributed to vibrational energy. The value of BC momentum refers to the translational energy that H atom(C) can collide with molecule HF(AB). The value of vibration energy can be proved by the extent of oscillation on the contour graph(the number of contour lines that the trajectory has passed)

== Reference ==
<references>

<ref name="Second Partial Derivative test">https://en.wikipedia.org/wiki/Second_partial_derivative_test</ref>
<ref name="Transition State Theory">T. Bligaard, J.K. Nørskov, in Chemical Bonding at Surfaces and Interfaces, 2008</ref>
<ref name="Enthapies of HH and HF">http://www4.ncsu.edu/~shultz/Common_Bond_enthalpies.pdf.</ref>
<ref name="Principle def">1.Z. Zhang, Y. Zhou, D. Zhang, G. Czakó and J. Bowman, The Journal of Physical Chemistry Letters, 2012, 3, 3416-3419.</ref>

</references>