2020-05-22T20:43:04Z

Sl13318: /* Reaction dynamics */

== Exercise 1: H + H2 system ==
[[File:01514934_distance_vs_time.png|250px|thumb|right|Figure 1. The "internuclear Distances vs Time" plot of H + H2 system at the transition state position.]]
* ''On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?''
On a potential energy surface, the transition state is defined as the saddle point or the local maximum, where the derivative of potential energy '''V''' with respect to the coordinate '''ri''' is zero (i.e. ∂V(ri)/∂ri=0). To distinguish the local maximum from the local minimums, a second derivative can be calculated. If the second derivative is larger than 0, it means that the stationary point is a local minimum. If it is smaller than 0, it indicates a local maximumn thus 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 estimated transition state position ('''rts''') is at 90.77 pm. At the transition state of this system, the distances of AB and BC would be the same and the forces acting on the atoms (the first derivatives) would be zero, meaning that the atoms are stationary. In the "internuclear distance vs time" plot (Figure 1) , the distances of AB, BC and AC remain unchanged and the distances of AB ('''r2''') and BC ('''r1''') have the same values. Therefore, 90.77 pm is the transition state position.
[[File:01514934_mep.png|250px|thumb|right|Figure 2. The contour plot of MEP calculation.]] [[File:01514934_dynamics.png|250px|thumb|right|Figure 3. The contour plot of DYNAMICS calculation.]]

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

The mep does not include kinetic energy so it will not show the atom motions. In contrast, the dynamics trajectory includes kinetic energy so the distance of AB oscillates slightly, which is shown in Figure 3, whereas mep shows a linear trajectory in Figure 2.

· Look at the “Internuclear Distances vs Time” and “Momenta vs Time”. What would change if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead?

If the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm are used, the distance AB ('''r2''') is the one that increases instead of distance BC ('''r1'''), and '''p2 '''is the one that oscillates instead of '''p1'''.

The last geometry is: '''r2'''=74.05 , '''p2 '''=3.20, '''r1'''=352.59, '''p1'''=5.07.

· 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.

The motion of atoms are reversed. Before they were moving oppositely. Now they approach each other.
* ''Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?''
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.28</nowiki>
|Yes
|The trajectory starts with atom A and molecule BC, passes through the transition state and ends with products atom C and molecule AB.
|[[File:01514934_table_1.png|200px|center]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|The trajectory starts with atom A and molecule BC, but it cannot reach the transition structure, so it bounces back to regenerate atom A and molecule BC.
|[[File:01514934_table_2.png|200px|center]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|The trajectory starts with atom A and molecule BC, passes through the transition state and ends with products atom C and molecule AB.
|[[File:01514934_table_3.png|200px|center]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|The trajectory starts with atom A and molecule BC, crosses the transition state region but goes back the reactants atom A and molecule BC.
|[[File:01514934_table_4.png|200px|center]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|The trajectory starts with atom A and molecule BC, passes through the transition state and ends with products atom C and molecule AB. This trajectory oscillates the most strongly compared to those above as it has more kinetic energy.
|[[File:01514934_table_5.png|200px|center]]
|}
The total energy includes the kinetic energy and potential energy of the system. For the trajectory to be reactive, the total energy must be distributed correctly so that the kinetic energy is higher than the activation energy. Even though the requirement for kinetic energy is fulfilled, it is still possible that the trajectory is unreactive like the second last case, which suggests that the transition state region can be recrossed. Therefore, energy is not the only factor that decides whether the trajectory is reactive or not.
* ''Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?''
The predictions of Transition State Theory is overestimated compared to the experimental values. In TST, once the reactants have enough kinetic energy along the reaction coordinates to cross the transition state, it cannot go back to the reactants, which contributes largely to the overestimation of the reaction rate values. If recrossing is available, the reaction rates would decrease. TST ignores the possibility of quantum tunneling, which would increases the reaction rates but the effect is negligible.1

== Exercise 2: F-H-F system ==
[[File:01514934_endo_exo.png|200px|right|thumb|Figure 4. The potential energy surface of F-H-H system.]]
=== PES inspection ===
* ''By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?''
The PSE of F-H-F system (Figure 4) shows that the F + H2 reaction is exothermic because the reactants have higher energy than the products. This means that the formation of H-F bond of the product releases more energy than the energy required to break the H-H bond of the reactant. The H + HF reaction is endothermic because the reactants are lower in energy than the products. This means that the breaking of H-F bond requires more energy than the energy released by formation of the H-H bond of the product. Therefore, it can be concluded that the H-F bond is stronger than the H-H bond.
* ''Locate the approximate position of the transition state.''
For exothermic reactions, the transition state resembles the reactants more so it has an early transition state. For endothermic reactions, the transition state resembles the products more so it has a late transition state. Along with the fact that the Hessian matrix has one positive eigenvalue and an negative one and the forces are zero at transition state, it can be estimated that the position of the transition state is '''r(FAHB)ts = '''180. pm,''' r(HBHC)ts = '''74.5 pm.
* ''Report the activation energy for both reactions.''
The energy of the transition state is -433.981 kJ mol-1. The energy of the reactant was found by displacing the transition state slightly towards the reactants. The energy of H + HF is -560.141 kJ mol-1. The energy of F + H2 is -435.100 kJ mol-1. Therefore, the activation energy of the F + H2 reaction is 1.119 kJ mol-1. The activation energy of H + HF reaction is 126.16 kJ mol-1.

=== Reaction dynamics ===
* ''In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.''
For the exothermic F + H2 reaction, once the product is formed, the release of reaction energy causes the vibrational excitation of the product. The IR chemiluminescence can be used to confirm the mechanism because as the vibration excited product relax back to the ground state, it emits photons, which can be detected by this technique.2 
* ''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.''
Endotheric reactions have a late transition state. For late transition barrier, the vibrational energy is better at promoting the reaction than the translational energy because if it's translational energy, the system would hit the barrier and bounces back to the reactants. In contrast, if the transition state is early for exothermic reaction, the translational energy would be more effective.3 The Polanyi's empirical rules
{| class="wikitable"
!r(FH)/pm
!r(HH)/pm
!pFH/ g.mol-1.pm.fs-1
!pHH/ g.mol-1.pm.fs-1
!Kinetic energy/kJ mol-1
!Reactive?
!Contour plot
|-
|230
|74
|<nowiki>-1</nowiki>
|3
|12.526
|No
|[[File:01514934_exo_unreactive.png|200px|center]]
|-
|230
|74
|<nowiki>-1.6</nowiki>
|0.2
|1.707
|Yes
|[[File:01514934_exo_reactive.png|200px|center]]
|-
|
|
|
|
|
|
|
|}

== References ==
1 J. I. Steinfeld, in ''Chemical kinetics and dynamics'', 2nd., 1989, pp. 287–321.

2 J. C. Poianyi, ''Science'', 1987, 236, 680–690.

3 K. J. Laidler, in ''Chemical Kinetics'', 3rd., 1987, pp. 460–471.

MRD:01514934

2020-05-22T20:41:13Z

Sl13318: /* Reaction dynamics */

== Exercise 1: H + H2 system ==
[[File:01514934_distance_vs_time.png|250px|thumb|right|Figure 1. The "internuclear Distances vs Time" plot of H + H2 system at the transition state position.]]
* ''On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?''
On a potential energy surface, the transition state is defined as the saddle point or the local maximum, where the derivative of potential energy '''V''' with respect to the coordinate '''ri''' is zero (i.e. ∂V(ri)/∂ri=0). To distinguish the local maximum from the local minimums, a second derivative can be calculated. If the second derivative is larger than 0, it means that the stationary point is a local minimum. If it is smaller than 0, it indicates a local maximumn thus 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 estimated transition state position ('''rts''') is at 90.77 pm. At the transition state of this system, the distances of AB and BC would be the same and the forces acting on the atoms (the first derivatives) would be zero, meaning that the atoms are stationary. In the "internuclear distance vs time" plot (Figure 1) , the distances of AB, BC and AC remain unchanged and the distances of AB ('''r2''') and BC ('''r1''') have the same values. Therefore, 90.77 pm is the transition state position.
[[File:01514934_mep.png|250px|thumb|right|Figure 2. The contour plot of MEP calculation.]] [[File:01514934_dynamics.png|250px|thumb|right|Figure 3. The contour plot of DYNAMICS calculation.]]

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

The mep does not include kinetic energy so it will not show the atom motions. In contrast, the dynamics trajectory includes kinetic energy so the distance of AB oscillates slightly, which is shown in Figure 3, whereas mep shows a linear trajectory in Figure 2.

· Look at the “Internuclear Distances vs Time” and “Momenta vs Time”. What would change if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead?

If the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm are used, the distance AB ('''r2''') is the one that increases instead of distance BC ('''r1'''), and '''p2 '''is the one that oscillates instead of '''p1'''.

The last geometry is: '''r2'''=74.05 , '''p2 '''=3.20, '''r1'''=352.59, '''p1'''=5.07.

· 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.

The motion of atoms are reversed. Before they were moving oppositely. Now they approach each other.
* ''Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?''
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.28</nowiki>
|Yes
|The trajectory starts with atom A and molecule BC, passes through the transition state and ends with products atom C and molecule AB.
|[[File:01514934_table_1.png|200px|center]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|The trajectory starts with atom A and molecule BC, but it cannot reach the transition structure, so it bounces back to regenerate atom A and molecule BC.
|[[File:01514934_table_2.png|200px|center]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|The trajectory starts with atom A and molecule BC, passes through the transition state and ends with products atom C and molecule AB.
|[[File:01514934_table_3.png|200px|center]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|The trajectory starts with atom A and molecule BC, crosses the transition state region but goes back the reactants atom A and molecule BC.
|[[File:01514934_table_4.png|200px|center]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|The trajectory starts with atom A and molecule BC, passes through the transition state and ends with products atom C and molecule AB. This trajectory oscillates the most strongly compared to those above as it has more kinetic energy.
|[[File:01514934_table_5.png|200px|center]]
|}
The total energy includes the kinetic energy and potential energy of the system. For the trajectory to be reactive, the total energy must be distributed correctly so that the kinetic energy is higher than the activation energy. Even though the requirement for kinetic energy is fulfilled, it is still possible that the trajectory is unreactive like the second last case, which suggests that the transition state region can be recrossed. Therefore, energy is not the only factor that decides whether the trajectory is reactive or not.
* ''Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?''
The predictions of Transition State Theory is overestimated compared to the experimental values. In TST, once the reactants have enough kinetic energy along the reaction coordinates to cross the transition state, it cannot go back to the reactants, which contributes largely to the overestimation of the reaction rate values. If recrossing is available, the reaction rates would decrease. TST ignores the possibility of quantum tunneling, which would increases the reaction rates but the effect is negligible.1

== Exercise 2: F-H-F system ==
[[File:01514934_endo_exo.png|200px|right|thumb|Figure 4. The potential energy surface of F-H-H system.]]
=== PES inspection ===
* ''By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?''
The PSE of F-H-F system (Figure 4) shows that the F + H2 reaction is exothermic because the reactants have higher energy than the products. This means that the formation of H-F bond of the product releases more energy than the energy required to break the H-H bond of the reactant. The H + HF reaction is endothermic because the reactants are lower in energy than the products. This means that the breaking of H-F bond requires more energy than the energy released by formation of the H-H bond of the product. Therefore, it can be concluded that the H-F bond is stronger than the H-H bond.
* ''Locate the approximate position of the transition state.''
For exothermic reactions, the transition state resembles the reactants more so it has an early transition state. For endothermic reactions, the transition state resembles the products more so it has a late transition state. Along with the fact that the Hessian matrix has one positive eigenvalue and an negative one and the forces are zero at transition state, it can be estimated that the position of the transition state is '''r(FAHB)ts = '''180. pm,''' r(HBHC)ts = '''74.5 pm.
* ''Report the activation energy for both reactions.''
The energy of the transition state is -433.981 kJ mol-1. The energy of the reactant was found by displacing the transition state slightly towards the reactants. The energy of H + HF is -560.141 kJ mol-1. The energy of F + H2 is -435.100 kJ mol-1. Therefore, the activation energy of the F + H2 reaction is 1.119 kJ mol-1. The activation energy of H + HF reaction is 126.16 kJ mol-1.

=== Reaction dynamics ===
* ''In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.''
For the exothermic F + H2 reaction, once the product is formed, the release of reaction energy causes the vibrational excitation of the product. The IR chemiluminescence can be used to confirm the mechanism because as the vibration excited product relax back to the ground state, it emits photons, which can be detected by this technique.2 
* ''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.''
Endotheric reactions have a late transition state. For late transition barrier, the vibrational energy is better at promoting the reaction than the translational energy because if it's translational energy, the system would hit the barrier and bounces back to the reactants. In contrast, if the transition state is early for exothermic reaction, the translational energy would be more effective.3 The Polanyi's empirical rules
{| class="wikitable"
!r(FH)/pm
!r(HH)/pm
!pFH/ g.mol-1.pm.fs-1
!pHH/ g.mol-1.pm.fs-1
!Kinetic energy/kJ mol-1
!Reactive?
!Contour plot
|-
|230
|74
|<nowiki>-1</nowiki>
|3
|12.526
|No
|[[File:01514934_exo_unreactive.png|200px|center]]
|-
|
|
|
|
|
|
|[[File:01514934_exo_reactive.png|200px|center]]
|-
|
|
|
|
|
|
|
|}

== References ==
1 J. I. Steinfeld, in ''Chemical kinetics and dynamics'', 2nd., 1989, pp. 287–321.

2 J. C. Poianyi, ''Science'', 1987, 236, 680–690.

3 K. J. Laidler, in ''Chemical Kinetics'', 3rd., 1987, pp. 460–471.

File:01514934 exo reactive.png

2020-05-22T20:40:52Z

Sl13318:

MRD:01514934

2020-05-22T20:40:22Z

Sl13318: /* Exercise 2: F-H-F system */

== Exercise 1: H + H2 system ==
[[File:01514934_distance_vs_time.png|250px|thumb|right|Figure 1. The "internuclear Distances vs Time" plot of H + H2 system at the transition state position.]]
* ''On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?''
On a potential energy surface, the transition state is defined as the saddle point or the local maximum, where the derivative of potential energy '''V''' with respect to the coordinate '''ri''' is zero (i.e. ∂V(ri)/∂ri=0). To distinguish the local maximum from the local minimums, a second derivative can be calculated. If the second derivative is larger than 0, it means that the stationary point is a local minimum. If it is smaller than 0, it indicates a local maximumn thus 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 estimated transition state position ('''rts''') is at 90.77 pm. At the transition state of this system, the distances of AB and BC would be the same and the forces acting on the atoms (the first derivatives) would be zero, meaning that the atoms are stationary. In the "internuclear distance vs time" plot (Figure 1) , the distances of AB, BC and AC remain unchanged and the distances of AB ('''r2''') and BC ('''r1''') have the same values. Therefore, 90.77 pm is the transition state position.
[[File:01514934_mep.png|250px|thumb|right|Figure 2. The contour plot of MEP calculation.]] [[File:01514934_dynamics.png|250px|thumb|right|Figure 3. The contour plot of DYNAMICS calculation.]]

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

The mep does not include kinetic energy so it will not show the atom motions. In contrast, the dynamics trajectory includes kinetic energy so the distance of AB oscillates slightly, which is shown in Figure 3, whereas mep shows a linear trajectory in Figure 2.

· Look at the “Internuclear Distances vs Time” and “Momenta vs Time”. What would change if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead?

If the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm are used, the distance AB ('''r2''') is the one that increases instead of distance BC ('''r1'''), and '''p2 '''is the one that oscillates instead of '''p1'''.

The last geometry is: '''r2'''=74.05 , '''p2 '''=3.20, '''r1'''=352.59, '''p1'''=5.07.

· 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.

The motion of atoms are reversed. Before they were moving oppositely. Now they approach each other.
* ''Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?''
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.28</nowiki>
|Yes
|The trajectory starts with atom A and molecule BC, passes through the transition state and ends with products atom C and molecule AB.
|[[File:01514934_table_1.png|200px|center]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|The trajectory starts with atom A and molecule BC, but it cannot reach the transition structure, so it bounces back to regenerate atom A and molecule BC.
|[[File:01514934_table_2.png|200px|center]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|The trajectory starts with atom A and molecule BC, passes through the transition state and ends with products atom C and molecule AB.
|[[File:01514934_table_3.png|200px|center]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|The trajectory starts with atom A and molecule BC, crosses the transition state region but goes back the reactants atom A and molecule BC.
|[[File:01514934_table_4.png|200px|center]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|The trajectory starts with atom A and molecule BC, passes through the transition state and ends with products atom C and molecule AB. This trajectory oscillates the most strongly compared to those above as it has more kinetic energy.
|[[File:01514934_table_5.png|200px|center]]
|}
The total energy includes the kinetic energy and potential energy of the system. For the trajectory to be reactive, the total energy must be distributed correctly so that the kinetic energy is higher than the activation energy. Even though the requirement for kinetic energy is fulfilled, it is still possible that the trajectory is unreactive like the second last case, which suggests that the transition state region can be recrossed. Therefore, energy is not the only factor that decides whether the trajectory is reactive or not.
* ''Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?''
The predictions of Transition State Theory is overestimated compared to the experimental values. In TST, once the reactants have enough kinetic energy along the reaction coordinates to cross the transition state, it cannot go back to the reactants, which contributes largely to the overestimation of the reaction rate values. If recrossing is available, the reaction rates would decrease. TST ignores the possibility of quantum tunneling, which would increases the reaction rates but the effect is negligible.1

== Exercise 2: F-H-F system ==
[[File:01514934_endo_exo.png|200px|right|thumb|Figure 4. The potential energy surface of F-H-H system.]]
=== PES inspection ===
* ''By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?''
The PSE of F-H-F system (Figure 4) shows that the F + H2 reaction is exothermic because the reactants have higher energy than the products. This means that the formation of H-F bond of the product releases more energy than the energy required to break the H-H bond of the reactant. The H + HF reaction is endothermic because the reactants are lower in energy than the products. This means that the breaking of H-F bond requires more energy than the energy released by formation of the H-H bond of the product. Therefore, it can be concluded that the H-F bond is stronger than the H-H bond.
* ''Locate the approximate position of the transition state.''
For exothermic reactions, the transition state resembles the reactants more so it has an early transition state. For endothermic reactions, the transition state resembles the products more so it has a late transition state. Along with the fact that the Hessian matrix has one positive eigenvalue and an negative one and the forces are zero at transition state, it can be estimated that the position of the transition state is '''r(FAHB)ts = '''180. pm,''' r(HBHC)ts = '''74.5 pm.
* ''Report the activation energy for both reactions.''
The energy of the transition state is -433.981 kJ mol-1. The energy of the reactant was found by displacing the transition state slightly towards the reactants. The energy of H + HF is -560.141 kJ mol-1. The energy of F + H2 is -435.100 kJ mol-1. Therefore, the activation energy of the F + H2 reaction is 1.119 kJ mol-1. The activation energy of H + HF reaction is 126.16 kJ mol-1.

=== Reaction dynamics ===
* ''In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.''
For the exothermic F + H2 reaction, once the product is formed, the release of reaction energy causes the vibrational excitation of the product. The IR chemiluminescence can be used to confirm the mechanism because as the vibration excited product relax back to the ground state, it emits photons, which can be detected by this technique.2 
* ''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.''
Endotheric reactions have a late transition state. For late transition barrier, the vibrational energy is better at promoting the reaction than the translational energy because if it's translational energy, the system would hit the barrier and bounces back to the reactants. In contrast, if the transition state is early for exothermic reaction, the translational energy would be more effective.3 The Polanyi's empirical rules
{| class="wikitable"
!r(FH)/pm
!r(HH)/pm
!pFH/ g.mol-1.pm.fs-1
!pHH/ g.mol-1.pm.fs-1
!Kinetic energy/kJ mol-1
!Reactive?
!Contour plot
|-
|230
|74
|<nowiki>-1</nowiki>
|3
|12.526
|No
|[[File:01514934_exo_unreactive.png|200px|center]]
|-
|
|
|
|
|
|
|
|-
|
|
|
|
|
|
|
|}

== References ==
1 J. I. Steinfeld, in ''Chemical kinetics and dynamics'', 2nd., 1989, pp. 287–321.

2 J. C. Poianyi, ''Science'', 1987, 236, 680–690.

3 K. J. Laidler, in ''Chemical Kinetics'', 3rd., 1987, pp. 460–471.

MRD:01514934

2020-05-22T20:39:24Z

Sl13318: /* Reaction dynamics */

== Exercise 1: H + H2 system ==
[[File:01514934_distance_vs_time.png|250px|thumb|right|Figure 1. The "internuclear Distances vs Time" plot of H + H2 system at the transition state position.]]
* ''On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?''
On a potential energy surface, the transition state is defined as the saddle point or the local maximum, where the derivative of potential energy '''V''' with respect to the coordinate '''ri''' is zero (i.e. ∂V(ri)/∂ri=0). To distinguish the local maximum from the local minimums, a second derivative can be calculated. If the second derivative is larger than 0, it means that the stationary point is a local minimum. If it is smaller than 0, it indicates a local maximumn thus 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 estimated transition state position ('''rts''') is at 90.77 pm. At the transition state of this system, the distances of AB and BC would be the same and the forces acting on the atoms (the first derivatives) would be zero, meaning that the atoms are stationary. In the "internuclear distance vs time" plot (Figure 1) , the distances of AB, BC and AC remain unchanged and the distances of AB ('''r2''') and BC ('''r1''') have the same values. Therefore, 90.77 pm is the transition state position.
[[File:01514934_mep.png|250px|thumb|right|Figure 2. The contour plot of MEP calculation.]] [[File:01514934_dynamics.png|250px|thumb|right|Figure 3. The contour plot of DYNAMICS calculation.]]

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

The mep does not include kinetic energy so it will not show the atom motions. In contrast, the dynamics trajectory includes kinetic energy so the distance of AB oscillates slightly, which is shown in Figure 3, whereas mep shows a linear trajectory in Figure 2.

· Look at the “Internuclear Distances vs Time” and “Momenta vs Time”. What would change if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead?

If the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm are used, the distance AB ('''r2''') is the one that increases instead of distance BC ('''r1'''), and '''p2 '''is the one that oscillates instead of '''p1'''.

The last geometry is: '''r2'''=74.05 , '''p2 '''=3.20, '''r1'''=352.59, '''p1'''=5.07.

· 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.

The motion of atoms are reversed. Before they were moving oppositely. Now they approach each other.
* ''Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?''
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.28</nowiki>
|Yes
|The trajectory starts with atom A and molecule BC, passes through the transition state and ends with products atom C and molecule AB.
|[[File:01514934_table_1.png|200px|center]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|The trajectory starts with atom A and molecule BC, but it cannot reach the transition structure, so it bounces back to regenerate atom A and molecule BC.
|[[File:01514934_table_2.png|200px|center]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|The trajectory starts with atom A and molecule BC, passes through the transition state and ends with products atom C and molecule AB.
|[[File:01514934_table_3.png|200px|center]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|The trajectory starts with atom A and molecule BC, crosses the transition state region but goes back the reactants atom A and molecule BC.
|[[File:01514934_table_4.png|200px|center]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|The trajectory starts with atom A and molecule BC, passes through the transition state and ends with products atom C and molecule AB. This trajectory oscillates the most strongly compared to those above as it has more kinetic energy.
|[[File:01514934_table_5.png|200px|center]]
|}
The total energy includes the kinetic energy and potential energy of the system. For the trajectory to be reactive, the total energy must be distributed correctly so that the kinetic energy is higher than the activation energy. Even though the requirement for kinetic energy is fulfilled, it is still possible that the trajectory is unreactive like the second last case, which suggests that the transition state region can be recrossed. Therefore, energy is not the only factor that decides whether the trajectory is reactive or not.
* ''Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?''
The predictions of Transition State Theory is overestimated compared to the experimental values. In TST, once the reactants have enough kinetic energy along the reaction coordinates to cross the transition state, it cannot go back to the reactants, which contributes largely to the overestimation of the reaction rate values. If recrossing is available, the reaction rates would decrease. TST ignores the possibility of quantum tunneling, which would increases the reaction rates but the effect is negligible.1

== Exercise 2: F-H-F system ==
[[File:01514934_endo_exo.png|200px|right|thumb|Figure 4. The potential energy surface of F-H-H system.]]
=== PES inspection ===
* ''By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?''
The PSE of F-H-F system (Figure 4) shows that the F + H2 reaction is exothermic because the reactants have higher energy than the products. This means that the formation of H-F bond of the product releases more energy than the energy required to break the H-H bond of the reactant. The H + HF reaction is endothermic because the reactants are lower in energy than the products. This means that the breaking of H-F bond requires more energy than the energy released by formation of the H-H bond of the product. Therefore, it can be concluded that the H-F bond is stronger than the H-H bond.
* ''Locate the approximate position of the transition state.''
For exothermic reactions, the transition state resembles the reactants more so it has an early transition state. For endothermic reactions, the transition state resembles the products more so it has a late transition state. Along with the fact that the Hessian matrix has one positive eigenvalue and an negative one and the forces are zero at transition state, it can be estimated that the position of the transition state is '''r(FAHB)ts = '''180. pm,''' r(HBHC)ts = '''74.5 pm.
* ''Report the activation energy for both reactions.''
The energy of the transition state is -433.981 kJ mol-1. The energy of the reactant was found by displacing the transition state slightly towards the reactants. The energy of H + HF is -560.141 kJ mol-1. The energy of F + H2 is -435.100 kJ mol-1. Therefore, the activation energy of the F + H2 reaction is 1.119 kJ mol-1. The activation energy of H + HF reaction is 126.16 kJ mol-1.

=== Reaction dynamics ===
* ''In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.''
For the exothermic F + H2 reaction, once the product is formed, the release of reaction energy causes the vibrational excitation of the product. The IR chemiluminescence can be used to confirm the mechanism because as the vibration excited product relax back to the ground state, it emits photons, which can be detected by this technique.2 
* ''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.''
Endotheric reactions have a late transition state. For late transition barrier, the vibrational energy is better at promoting the reaction than the translational energy because if it's translational energy, the system would hit the barrier and bounces back to the reactants. In contrast, if the transition state is early for exothermic reaction, the translational energy would be more effective.3 The Polanyi's empirical rules
{| class="wikitable"
!r(FH)/pm
!r(HH)/pm
!pFH/ g.mol-1.pm.fs-1
!pHH/ g.mol-1.pm.fs-1
!Kinetic energy/kJ mol-1
!Reactive?
!Contour plot
|-
|230
|74
|<nowiki>-1</nowiki>
|3
|12.526
|No
|
|-
|
|
|
|
|
|
|
|-
|
|
|
|
|
|
|
|}

== References ==
1 J. I. Steinfeld, in ''Chemical kinetics and dynamics'', 2nd., 1989, pp. 287–321.

2 J. C. Poianyi, ''Science'', 1987, 236, 680–690.

3 K. J. Laidler, in ''Chemical Kinetics'', 3rd., 1987, pp. 460–471.

MRD:01514934

2020-05-21T13:58:23Z

Sl13318: /* Exercise 1: H + H2 system */

== Exercise 1: H + H2 system ==
* On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?
On a potential energy surface, the transition state is defined as the saddle point or the local maximum, where the derivative of potential energy '''V''' with respect to the coordinate '''ri''' is zero (i.e. ∂V(ri)/∂ri=0). To distinguish the local maximum from the local minimums, a second derivative can be calculated. If the second derivative is larger than 0, it means that the stationary point is a local minimum. If it is smaller than 0, it indicates a local maximumn thus 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 estimated transition state position (rts) is at 90.772 pm. At the transition state of this system, the distances of AB and BC would be the same and the forces acting on the atoms (the first derivatives) would be zero, meaning that the atoms are stationary. In the "internuclear distance vs time" plot (Figure 1) below, the distances of AB, BC and AC remain unchanged and the distances of AB and BC have the same values. Therefore, 90.772 pm is the transition state position.
[[File:01514934_distance_vs_time.png|thumb|center|Figure 1. The "internuclear Distances vs Time" plot of H + H2 system at the transition state position.]]
* Comment on how the mep and the trajectory you just calculated differ.
The mep does not have kinetic energy so it doesn not include the atomic motions whereas the trajectory from dynamics does.
[[File:01514934_mep.png|thumb|Figure 2. The contour plot of MEP calculation.]] [[File:01514934_dynamics.png|thumb|Figure 3. The contour plot of DYNAMICS calculation.]]
· Look at the “Internuclear Distances vs Time” and “Momenta vs Time”. What would change if we used the initial conditions r1 = rts and r2 = rts+1 pm instead?

The distances of AB and BC are no longer constant. The distance of BC increases and the distance of AB oscillates slightly. The momentum of BC increases and reaches a plateau. The momentum of AB increases and then oscillates.

Last geometry: AB=74.04585280873204 pm, momentum of AB=3.202634283122685

BC=352.5914093624435 pm, momentum of BC=5.068046969200336

·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.

The motion of atoms are reversed. Before they were moving oppositely. Now they are approaching each other.

· Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?

p1/ g.mol-1.pm.fs-1
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.28</nowiki>
|Yes
|
|
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|
|
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|
|
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|
|
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|
|
|}
The total energy includes the kinetic energy and potential energy of the system. For the trajectory to be reactive, the total energy must be distributed correctly so that the kinetic energy is higher than the activation energy.

· Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

The predictions of Transition State Theory will be higher thant the experimental values. TST is a

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

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