File:H-H-H transition state point.png 01503585

2020-05-22T21:13:53Z

Yc14518:

MRD:01503585 yulong

2020-05-22T21:13:43Z

Yc14518: /* Question 3: Comment on how the mep and the trajectory you just calculated differ. */

== EXERCISE 1: H + H2 system ==

=== Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===
The transition state is a saddle point on a potential energy diagram, it is a critical point but is not the local extremum of the function. It can be identified by finding a point which its first and second derivative are both zero. The difference between the local minimum and transition state is that the local minimum has a non zero second derivative (positive derivative).

=== Question 2: Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ===
The estimate of mine is that the distance between AB and BC are both 90.78 pm. Based on the definition of transition state, at the transition state, the atoms should have a fixed position. Based on the “Internuclear Distances vs Time”, the distances between atoms do not change with time, which means it is a good estimate of transition state.

=== Question 3: Comment on how the ''mep'' and the trajectory you just calculated differ. ===
The mep just include the minimum energy of the reaction passed, and it doesn't calculate the vibration of atoms while them moving, which its trajectory changes smoothly without fluctuation.
[[File:Distance vs time r1=r2=rts 01503585.png|thumb|Internuclear Distances vs Time when r1=r2=rts|300x300px|none]][[File:D_V_time_r1=rts_r2=rts+1.png_01503585|thumb|300x300px|
D V time r1=rts r2=rts+1
|none]]By comparing the "internuclear distances vs time" plots of r1=r2=rts and r1=rts, r2=rts+1, it is easy to obtain that when r1=r2=rts, the distances between atoms do not change. But as we change the starting point to r2=rts+1, the distance of between atoms A and C, and the distance between atoms A and B start to increase with similar rate. While the distance between B and C starts to decrease. This difference between two graphs shows that once the reaction pass the transition state, new molecule BC starts to form and at the same time atom A goes away from the molecule.[[File:M V time r1=r2=rts.png 01503585|thumb|300x300px|M V time r1=r2=rts|none]][[File:M V time r1=rts r2=rts+1.png 01503585|thumb|300x300px|M V time r1=rts, r2=rts+1|none]]By comparing the "momenta vs time" plots of r1=r2=rts and r1=rts, r2=rts+1, we see that the momentum of B-C is fluctuating between 0.0024 g.mol-1.pm.fs-1 and -0.0024 g.mol-1.pm.fs-1. However, the real momenta of B-C and AB should both be zero at the transition state. The none zero values are caused by approximated input. For r1=rts, r2=rts+1, the momenta of A-B and B-C are both increasing. The momentum of A-B increases faster as A is going away from molecule BC, and reach a plateau. While the momentum B-C starts to fluctuate after increasing since they reach a vibrating status after forming a molecule.

=== Question 4: 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>-414.280</nowiki>
|Reactive
|At the start the distance between A and B is large and the distance between B and C is small since BC is a molecule. The momenta given are enough to let the reaction happen. When A collide molecule BC, new molecule AB is formed and molecule BC is broke. The kinetic energy for reaction is 19.508 kJ.mol-1.
|[[File:Reactive group 1.png 01503585|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|Unreactive
|The distances between atoms at the starting point are similar with group one, but the molecule atoms have a larger momentum which causes the molecule has a greater vibration mode. The result of this is that the system has a lower kinetic energy to complete the reaction. The kinetic energy for reaction is 13.710 kJ.mol-1.
|[[File:Reactive group 2.png 01503585|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Reactive
|The momentum of molecule is same as group 2, but the atom A has a greater momentum which causes the reaction has enough kinetic energy to complete. The kinetic energy for reaction is 19.801 kJ.mol-1.
|[[File:Reactive group 3.png 01503585|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|Unreactive
|Under this condition, the momenta of both molecule and atom are so high which causes the bond between B-C breaks and A-B bond forms. However the reaction does not stops here, the bond between A-B breaks again and B-C bond forms again, which cause this reaction not completed. The kinetic energy for reaction is 76.510 kJ.mol-1.
|[[File:Reactive group 4.png 01503585|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Reactive
|For this one, the starting momenta are similar to group 4. However, the small excess of momentum of atom A causes the system has enough kinetic energy to break the B-C bond again and finally form the A-B bond and leads the reaction to completion. The kinetic energy for reaction is 84.310 kJ.mol-1.
|[[File:Reactive group 5.png 01503585|thumb]]
|}
The hypothesis is "All trajectories starting with the same positions but with higher values of momenta (higher kinetic energy) would be reactive, as they have enough kinetic energy to overcome the activation barrier". Based on the result collected, this hypothesis is only partially correct. For some combination of starting momenta, the kinetic energy is too high to let the product stay stable. The extra kinetic energy will cause the reaction fall back to reactants.

=== Question 5: Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===
Based on the result obtained, the transition state theory predictions for rate values are overestimated.

According to the TST, systems with great enough energy will let the reaction happen. However, when the system with very high energy, the high energy vibration mode of it will cause the reaction goes to an unexpected result. Like the energy combination group 4 in question 4, the bonds between atoms form and break more than we need. Under this situation, some of the reaction predicted happen will actually not happen, which cause the predicted reaction rates higher than the real ones.

In addition, TST only involves classical mechanics but not quantum mechanics, which it does not concern the effect of "quantum tunnelling effect". This effect causes some reaction happens even it does not reach the activation energy. In this way, if the quantum tunnelling effect is concerned, the actual reaction rates will slightly higher than the ones predicted. However this effect is minor which the overall TST predictions for rate values are still overestimated.

== EXERCISE 2: F - H - H system ==

=== Question 6: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===
[[File:F+H2 reaction.png 01503585|centre|thumb|F-H-H system]]For this part, the atom A now is Fluorine, and atoms B and C are hydrogen.

The bond energy of H-F bond is 565 kJ/mol, and the bond energy of H-H bond is 432 kJ/mol. The HF molecule has a higher bond energy, which means more energy is needed to break the bond of HF than the bond of H-H. In this way, the reaction of F + H2 is an exothermic reaction and the reaction of H + HF is an endothermic reaction. The reason why HF has a stonger bond energy is that HF molecule is polar and it has partial ionic character.

=== Question 7: Locate the approximate position of the transition state. ===
{| class="wikitable"
|[[File:F-H-H transition state point.png 01503585|thumb|422x422px|F-H-H transition state point]]
|[[File:F-H-H D V T.png|thumb|421x421px|Internuclear distance vs time plot]]
|}
The approximate position of the transition state of this reaction is the point at rBC=74.49 pm and rAB=181.1 pm. Start simulation from this point, the distances between atoms are remains unchanged. In this way, this is a good approximation of transition state point.

=== Question 8: Report the activation energy for both reactions.kJ/mol ===
Set the calculation type as MEP, by changing the atom distances, we can obtain the activation energy for two reactions. At the transition point, the total energy of the system is -433.981 kJ/mol.

For F + H2, set the distance between atom F and H2 rAB=1000pm, and keep rBC=74.49 pm:

The enegy obtained when time is long enough is -435.059 kJ/mol, and the acrivation energy for this reaction thus is -433.981 kJ/mol-(-435.059 kJ/mol)=1.078 kJ/mol.

For H + HF, set the distance between atom H and HF rBC=1000pm, and keep rAB=181.1 pm:

The enegy obtained when time is long enough is -560.7 kJ/mol, and the acrivation energy for this reaction thus is -433.981 kJ/mol-(-560.7 kJ/mol)=126.719 kJ/mol

=== Question 9: 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 reaction F + H2 ====
The distances set are rHH = 74 pm, rFH=200 pm.
{| class="wikitable"
!pFH/ g.mol-1.pm.fs-1
!pHH/ g.mol-1.pm.fs-1
!Kinetic Energy
!Reactivity
!Distance vs time plot
|-
|<nowiki>-1.0</nowiki>
|6.1
|43.836
|Unreactive
|[[File:First combination.png 01503585|thumb]]
|-
|<nowiki>-1.0</nowiki>
|3
|12.526
|Reactive
|[[File:Second combination.png 01503585|thumb]]
|-
|<nowiki>-1.0</nowiki>
|0
|0.526
|Reactive
|[[File:Third combination.png 01503585|thumb]]
|-
|<nowiki>-1.0</nowiki>
|<nowiki>-3</nowiki>
|6.526
|Unreactive
|[[File:Fourth combination.png 01503585|thumb]]
|-
|<nowiki>-1.0</nowiki>
|<nowiki>-6.1</nowiki>
|31.636
|Unreactive
|[[File:Fifth combination.png 01503585|thumb]]
|-
|<nowiki>-1.6</nowiki>
|0.2
|1.707
|Unreactive
|[[File:Sixth combination.png 01503585|thumb]]
|}
For diagram 1, the kinetic energy is quite high at the begining. After the H2 molecule and Fluorine atom react, the bond between H and F forms and breaks serveral times. The kinetic energy and potential energy converse to each other several times. In the end, since the kinetic energy is too high, the high vibration mode causes the bond of HF breaks and then the reaction is unreactive.

For diagram 2, the kinetic energy is enough to let the reaction happen and not break the HF bond forms in the end. Some kinetic energy stores in HF bond and some potential energy stored in HH bond turns into kinetic energy.

Diagram 3 is similar to diagram 2, and diagrams 4 & 5 is similar to diagram 1.

For the combination of pFH= -1.6 g.mol-1.pm.fs-1 and pHH= 0.2 g.mol-1.pm.fs-1, the reaction is unreactive since the momentum of the system is too low to overcome the barrier of the reaction. Thus the reaction does not happen. 

==== For reaction H + HF ====
[[File:Question nine last.png|left|thumb]]

For the reaction of H + HF, one possible initial condition found is rAB = 92 pm, rBC = 200 pm, and the momenta are pFH= 0 g.mol-1.pm.fs-1 and pHH= 0.2 g.mol-1.pm.fs-1. The energy of the system is -270.348 kJ/mol. This reaction can happen is because that the H atom has enough energy to break the HF bond and form a new HH bond. Since the bond energy of HF is very high, the H atom needs more momentum to start the reaction.

To confirm the observations experimentally, controlling temperature of the reaction is a good way. By increasing the temperature, the momenta of molecules and atoms increases. So the reaction starts to change from unreactive to reactive. However, if the temperature is too high, which causes extra momenta in the reaction, the high energy vibration mode will cause the product fall back to reactant.

=== Question 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. ===
Based on Polanyi's empirical rules, the vibrational energy more likely to promote more late-barrier reaction and the translational energy more likely to promote early-barrier reaction.

For the reaction of F + H2, it is an exothermic reaction with an early transition state, which is an early-barrier reaction. In this way, it is more likely to be promoted by translational energy. Thus when pHF (leads to translational energy)is relatively larger, the reaction is more likely to happen.

For the reaction of H + HF, it is an endothermic reaction with an late transition state, which is an late-barrier reaction. In this way, it is more likely to be promoted by vibrational energy. Thus when pHF (leads to vibrational energy) is relatively larger, the reaction is more likely to happen.

MRD:01503585 yulong

2020-05-22T20:39:03Z

Yc14518: /* For reaction H + HF */

== EXERCISE 1: H + H2 system ==

=== Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===
The transition state is a saddle point on a potential energy diagram, it is a critical point but is not the local extremum of the function. It can be identified by finding a point which its first and second derivative are both zero. The difference between the local minimum and transition state is that the local minimum has a non zero second derivative (positive derivative).

=== Question 2: Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ===
The estimate of mine is that the distance between AB and BC are both 90.78 pm. Based on the definition of transition state, at the transition state, the atoms should have a fixed position. Based on the “Internuclear Distances vs Time”, the distances between atoms do not change with time, which means it is a good estimate of transition state.

=== Question 3: Comment on how the ''mep'' and the trajectory you just calculated differ. ===
The mep just include the minimum energy of the reaction passed, and it doesn't calculate the vibration of atoms while them moving, which its trajectory changes smoothly without fluctuation.
[[File:Distance vs time r1=r2=rts 01503585.png|thumb|Internuclear Distances vs Time when r1=r2=rts|206x206px|none]][[File:D_V_time_r1=rts_r2=rts+1.png_01503585|thumb|205x205px|
D V time r1=rts r2=rts+1
|none]]By comparing the "internuclear distances vs time" plots of r1=r2=rts and r1=rts, r2=rts+1, it is easy to obtain that when r1=r2=rts, the distances between atoms do not change. But as we change the starting point to r2=rts+1, the distance of between atoms A and C, and the distance between atoms A and B start to increase with similar rate. While the distance between B and C starts to decrease. This difference between two graphs shows that once the reaction pass the transition state, new molecule BC starts to form and at the same time atom A goes away from the molecule.[[File:M V time r1=r2=rts.png 01503585|thumb|206x206px|M V time r1=r2=rts|none]][[File:M V time r1=rts r2=rts+1.png 01503585|thumb|205x205px|M V time r1=rts, r2=rts+1|none]]By comparing the "momenta vs time" plots of r1=r2=rts and r1=rts, r2=rts+1, we see that the momentum of B-C is fluctuating between 0.0024 g.mol-1.pm.fs-1 and -0.0024 g.mol-1.pm.fs-1. However, the real momenta of B-C and AB should both be zero at the transition state. The none zero values are caused by approximated input. For r1=rts, r2=rts+1, the momenta of A-B and B-C are both increasing. The momentum of A-B increases faster as A is going away from molecule BC, and reach a plateau. While the momentum B-C starts to fluctuate after increasing since they reach a vibrating status after forming a molecule.

=== Question 4: 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>-414.280</nowiki>
|Reactive
|At the start the distance between A and B is large and the distance between B and C is small since BC is a molecule. The momenta given are enough to let the reaction happen. When A collide molecule BC, new molecule AB is formed and molecule BC is broke. The kinetic energy for reaction is 19.508 kJ.mol-1.
|[[File:Reactive group 1.png 01503585|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|Unreactive
|The distances between atoms at the starting point are similar with group one, but the molecule atoms have a larger momentum which causes the molecule has a greater vibration mode. The result of this is that the system has a lower kinetic energy to complete the reaction. The kinetic energy for reaction is 13.710 kJ.mol-1.
|[[File:Reactive group 2.png 01503585|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Reactive
|The momentum of molecule is same as group 2, but the atom A has a greater momentum which causes the reaction has enough kinetic energy to complete. The kinetic energy for reaction is 19.801 kJ.mol-1.
|[[File:Reactive group 3.png 01503585|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|Unreactive
|Under this condition, the momenta of both molecule and atom are so high which causes the bond between B-C breaks and A-B bond forms. However the reaction does not stops here, the bond between A-B breaks again and B-C bond forms again, which cause this reaction not completed. The kinetic energy for reaction is 76.510 kJ.mol-1.
|[[File:Reactive group 4.png 01503585|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Reactive
|For this one, the starting momenta are similar to group 4. However, the small excess of momentum of atom A causes the system has enough kinetic energy to break the B-C bond again and finally form the A-B bond and leads the reaction to completion. The kinetic energy for reaction is 84.310 kJ.mol-1.
|[[File:Reactive group 5.png 01503585|thumb]]
|}
The hypothesis is "All trajectories starting with the same positions but with higher values of momenta (higher kinetic energy) would be reactive, as they have enough kinetic energy to overcome the activation barrier". Based on the result collected, this hypothesis is only partially correct. For some combination of starting momenta, the kinetic energy is too high to let the product stay stable. The extra kinetic energy will cause the reaction fall back to reactants.

=== Question 5: Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===
Based on the result obtained, the transition state theory predictions for rate values are overestimated.

According to the TST, systems with great enough energy will let the reaction happen. However, when the system with very high energy, the high energy vibration mode of it will cause the reaction goes to an unexpected result. Like the energy combination group 4 in question 4, the bonds between atoms form and break more than we need. Under this situation, some of the reaction predicted happen will actually not happen, which cause the predicted reaction rates higher than the real ones.

In addition, TST only involves classical mechanics but not quantum mechanics, which it does not concern the effect of "quantum tunnelling effect". This effect causes some reaction happens even it does not reach the activation energy. In this way, if the quantum tunnelling effect is concerned, the actual reaction rates will slightly higher than the ones predicted. However this effect is minor which the overall TST predictions for rate values are still overestimated.

== EXERCISE 2: F - H - H system ==

=== Question 6: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===
[[File:F+H2 reaction.png 01503585|centre|thumb|F-H-H system]]For this part, the atom A now is Fluorine, and atoms B and C are hydrogen.

The bond energy of H-F bond is 565 kJ/mol, and the bond energy of H-H bond is 432 kJ/mol. The HF molecule has a higher bond energy, which means more energy is needed to break the bond of HF than the bond of H-H. In this way, the reaction of F + H2 is an exothermic reaction and the reaction of H + HF is an endothermic reaction. The reason why HF has a stonger bond energy is that HF molecule is polar and it has partial ionic character.

=== Question 7: Locate the approximate position of the transition state. ===
{| class="wikitable"
|[[File:F-H-H transition state point.png 01503585|thumb|422x422px|F-H-H transition state point]]
|[[File:F-H-H D V T.png|thumb|421x421px|Internuclear distance vs time plot]]
|}
The approximate position of the transition state of this reaction is the point at rBC=74.49 pm and rAB=181.1 pm. Start simulation from this point, the distances between atoms are remains unchanged. In this way, this is a good approximation of transition state point.

=== Question 8: Report the activation energy for both reactions.kJ/mol ===
Set the calculation type as MEP, by changing the atom distances, we can obtain the activation energy for two reactions. At the transition point, the total energy of the system is -433.981 kJ/mol.

For F + H2, set the distance between atom F and H2 rAB=1000pm, and keep rBC=74.49 pm:

The enegy obtained when time is long enough is -435.059 kJ/mol, and the acrivation energy for this reaction thus is -433.981 kJ/mol-(-435.059 kJ/mol)=1.078 kJ/mol.

For H + HF, set the distance between atom H and HF rBC=1000pm, and keep rAB=181.1 pm:

The enegy obtained when time is long enough is -560.7 kJ/mol, and the acrivation energy for this reaction thus is -433.981 kJ/mol-(-560.7 kJ/mol)=126.719 kJ/mol

=== Question 9: 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 reaction F + H2 ====
The distances set are rHH = 74 pm, rFH=200 pm.
{| class="wikitable"
!pFH/ g.mol-1.pm.fs-1
!pHH/ g.mol-1.pm.fs-1
!Kinetic Energy
!Reactivity
!Distance vs time plot
|-
|<nowiki>-1.0</nowiki>
|6.1
|43.836
|Unreactive
|[[File:First combination.png 01503585|thumb]]
|-
|<nowiki>-1.0</nowiki>
|3
|12.526
|Reactive
|[[File:Second combination.png 01503585|thumb]]
|-
|<nowiki>-1.0</nowiki>
|0
|0.526
|Reactive
|[[File:Third combination.png 01503585|thumb]]
|-
|<nowiki>-1.0</nowiki>
|<nowiki>-3</nowiki>
|6.526
|Unreactive
|[[File:Fourth combination.png 01503585|thumb]]
|-
|<nowiki>-1.0</nowiki>
|<nowiki>-6.1</nowiki>
|31.636
|Unreactive
|[[File:Fifth combination.png 01503585|thumb]]
|-
|<nowiki>-1.6</nowiki>
|0.2
|1.707
|Unreactive
|[[File:Sixth combination.png 01503585|thumb]]
|}
For diagram 1, the kinetic energy is quite high at the begining. After the H2 molecule and Fluorine atom react, the bond between H and F forms and breaks serveral times. The kinetic energy and potential energy converse to each other several times. In the end, since the kinetic energy is too high, the high vibration mode causes the bond of HF breaks and then the reaction is unreactive.

For diagram 2, the kinetic energy is enough to let the reaction happen and not break the HF bond forms in the end. Some kinetic energy stores in HF bond and some potential energy stored in HH bond turns into kinetic energy.

Diagram 3 is similar to diagram 2, and diagrams 4 & 5 is similar to diagram 1.

For the combination of pFH= -1.6 g.mol-1.pm.fs-1 and pHH= 0.2 g.mol-1.pm.fs-1, the reaction is unreactive since the momentum of the system is too low to overcome the barrier of the reaction. Thus the reaction does not happen. 

==== For reaction H + HF ====
[[File:Question nine last.png|left|thumb]]

For the reaction of H + HF, one possible initial condition found is rAB = 92 pm, rBC = 200 pm, and the momenta are pFH= 0 g.mol-1.pm.fs-1 and pHH= 0.2 g.mol-1.pm.fs-1. The energy of the system is -270.348 kJ/mol. This reaction can happen is because that the H atom has enough energy to break the HF bond and form a new HH bond. Since the bond energy of HF is very high, the H atom needs more momentum to start the reaction.

To confirm the observations experimentally, controlling temperature of the reaction is a good way. By increasing the temperature, the momenta of molecules and atoms increases. So the reaction starts to change from unreactive to reactive. However, if the temperature is too high, which causes extra momenta in the reaction, the high energy vibration mode will cause the product fall back to reactant.

=== Question 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. ===
Based on Polanyi's empirical rules, the vibrational energy more likely to promote more late-barrier reaction and the translational energy more likely to promote early-barrier reaction.

For the reaction of F + H2, it is an exothermic reaction with an early transition state, which is an early-barrier reaction. In this way, it is more likely to be promoted by translational energy. Thus when pHF (leads to translational energy)is relatively larger, the reaction is more likely to happen.

For the reaction of H + HF, it is an endothermic reaction with an late transition state, which is an late-barrier reaction. In this way, it is more likely to be promoted by vibrational energy. Thus when pHF (leads to vibrational energy) is relatively larger, the reaction is more likely to happen.

MRD:01503585 yulong

2020-05-22T19:51:59Z

Yc14518: /* For reaction F + H2 */

== EXERCISE 1: H + H2 system ==

=== Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===
The transition state is a saddle point on a potential energy diagram, it is a critical point but is not the local extremum of the function. It can be identified by finding a point which its first and second derivative are both zero. The difference between the local minimum and transition state is that the local minimum has a non zero second derivative (positive derivative).

=== Question 2: Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ===
The estimate of mine is that the distance between AB and BC are both 90.78 pm. Based on the definition of transition state, at the transition state, the atoms should have a fixed position. Based on the “Internuclear Distances vs Time”, the distances between atoms do not change with time, which means it is a good estimate of transition state.

=== Question 3: Comment on how the ''mep'' and the trajectory you just calculated differ. ===
The mep just include the minimum energy of the reaction passed, and it doesn't calculate the vibration of atoms while them moving, which its trajectory changes smoothly without fluctuation.
[[File:Distance vs time r1=r2=rts 01503585.png|thumb|Internuclear Distances vs Time when r1=r2=rts|206x206px|none]][[File:D_V_time_r1=rts_r2=rts+1.png_01503585|thumb|205x205px|
D V time r1=rts r2=rts+1
|none]]By comparing the "internuclear distances vs time" plots of r1=r2=rts and r1=rts, r2=rts+1, it is easy to obtain that when r1=r2=rts, the distances between atoms do not change. But as we change the starting point to r2=rts+1, the distance of between atoms A and C, and the distance between atoms A and B start to increase with similar rate. While the distance between B and C starts to decrease. This difference between two graphs shows that once the reaction pass the transition state, new molecule BC starts to form and at the same time atom A goes away from the molecule.[[File:M V time r1=r2=rts.png 01503585|thumb|206x206px|M V time r1=r2=rts|none]][[File:M V time r1=rts r2=rts+1.png 01503585|thumb|205x205px|M V time r1=rts, r2=rts+1|none]]By comparing the "momenta vs time" plots of r1=r2=rts and r1=rts, r2=rts+1, we see that the momentum of B-C is fluctuating between 0.0024 g.mol-1.pm.fs-1 and -0.0024 g.mol-1.pm.fs-1. However, the real momenta of B-C and AB should both be zero at the transition state. The none zero values are caused by approximated input. For r1=rts, r2=rts+1, the momenta of A-B and B-C are both increasing. The momentum of A-B increases faster as A is going away from molecule BC, and reach a plateau. While the momentum B-C starts to fluctuate after increasing since they reach a vibrating status after forming a molecule.

=== Question 4: 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>-414.280</nowiki>
|Reactive
|At the start the distance between A and B is large and the distance between B and C is small since BC is a molecule. The momenta given are enough to let the reaction happen. When A collide molecule BC, new molecule AB is formed and molecule BC is broke. The kinetic energy for reaction is 19.508 kJ.mol-1.
|[[File:Reactive group 1.png 01503585|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|Unreactive
|The distances between atoms at the starting point are similar with group one, but the molecule atoms have a larger momentum which causes the molecule has a greater vibration mode. The result of this is that the system has a lower kinetic energy to complete the reaction. The kinetic energy for reaction is 13.710 kJ.mol-1.
|[[File:Reactive group 2.png 01503585|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Reactive
|The momentum of molecule is same as group 2, but the atom A has a greater momentum which causes the reaction has enough kinetic energy to complete. The kinetic energy for reaction is 19.801 kJ.mol-1.
|[[File:Reactive group 3.png 01503585|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|Unreactive
|Under this condition, the momenta of both molecule and atom are so high which causes the bond between B-C breaks and A-B bond forms. However the reaction does not stops here, the bond between A-B breaks again and B-C bond forms again, which cause this reaction not completed. The kinetic energy for reaction is 76.510 kJ.mol-1.
|[[File:Reactive group 4.png 01503585|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Reactive
|For this one, the starting momenta are similar to group 4. However, the small excess of momentum of atom A causes the system has enough kinetic energy to break the B-C bond again and finally form the A-B bond and leads the reaction to completion. The kinetic energy for reaction is 84.310 kJ.mol-1.
|[[File:Reactive group 5.png 01503585|thumb]]
|}
The hypothesis is "All trajectories starting with the same positions but with higher values of momenta (higher kinetic energy) would be reactive, as they have enough kinetic energy to overcome the activation barrier". Based on the result collected, this hypothesis is only partially correct. For some combination of starting momenta, the kinetic energy is too high to let the product stay stable. The extra kinetic energy will cause the reaction fall back to reactants.

=== Question 5: Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===
Based on the result obtained, the transition state theory predictions for rate values are overestimated.

According to the TST, systems with great enough energy will let the reaction happen. However, when the system with very high energy, the high energy vibration mode of it will cause the reaction goes to an unexpected result. Like the energy combination group 4 in question 4, the bonds between atoms form and break more than we need. Under this situation, some of the reaction predicted happen will actually not happen, which cause the predicted reaction rates higher than the real ones.

In addition, TST only involves classical mechanics but not quantum mechanics, which it does not concern the effect of "quantum tunnelling effect". This effect causes some reaction happens even it does not reach the activation energy. In this way, if the quantum tunnelling effect is concerned, the actual reaction rates will slightly higher than the ones predicted. However this effect is minor which the overall TST predictions for rate values are still overestimated.

== EXERCISE 2: F - H - H system ==

=== Question 6: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===
[[File:F+H2 reaction.png 01503585|centre|thumb|F-H-H system]]For this part, the atom A now is Fluorine, and atoms B and C are hydrogen.

The bond energy of H-F bond is 565 kJ/mol, and the bond energy of H-H bond is 432 kJ/mol. The HF molecule has a higher bond energy, which means more energy is needed to break the bond of HF than the bond of H-H. In this way, the reaction of F + H2 is an exothermic reaction and the reaction of H + HF is an endothermic reaction. The reason why HF has a stonger bond energy is that HF molecule is polar and it has partial ionic character.

=== Question 7: Locate the approximate position of the transition state. ===
{| class="wikitable"
|[[File:F-H-H transition state point.png 01503585|thumb|422x422px|F-H-H transition state point]]
|[[File:F-H-H D V T.png|thumb|421x421px|Internuclear distance vs time plot]]
|}
The approximate position of the transition state of this reaction is the point at rBC=74.49 pm and rAB=181.1 pm. Start simulation from this point, the distances between atoms are remains unchanged. In this way, this is a good approximation of transition state point.

=== Question 8: Report the activation energy for both reactions.kJ/mol ===
Set the calculation type as MEP, by changing the atom distances, we can obtain the activation energy for two reactions. At the transition point, the total energy of the system is -433.981 kJ/mol.

For F + H2, set the distance between atom F and H2 rAB=1000pm, and keep rBC=74.49 pm:

The enegy obtained when time is long enough is -435.059 kJ/mol, and the acrivation energy for this reaction thus is -433.981 kJ/mol-(-435.059 kJ/mol)=1.078 kJ/mol.

For H + HF, set the distance between atom H and HF rBC=1000pm, and keep rAB=181.1 pm:

The enegy obtained when time is long enough is -560.7 kJ/mol, and the acrivation energy for this reaction thus is -433.981 kJ/mol-(-560.7 kJ/mol)=126.719 kJ/mol

=== Question 9: 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 reaction F + H2 ====
The distances set are rHH = 74 pm, rFH=200 pm.
{| class="wikitable"
!pFH/ g.mol-1.pm.fs-1
!pHH/ g.mol-1.pm.fs-1
!Kinetic Energy
!Reactivity
!Distance vs time plot
|-
|<nowiki>-1.0</nowiki>
|6.1
|43.836
|Unreactive
|[[File:First combination.png 01503585|thumb]]
|-
|<nowiki>-1.0</nowiki>
|3
|12.526
|Reactive
|[[File:Second combination.png 01503585|thumb]]
|-
|<nowiki>-1.0</nowiki>
|0
|0.526
|Reactive
|[[File:Third combination.png 01503585|thumb]]
|-
|<nowiki>-1.0</nowiki>
|<nowiki>-3</nowiki>
|6.526
|Unreactive
|[[File:Fourth combination.png 01503585|thumb]]
|-
|<nowiki>-1.0</nowiki>
|<nowiki>-6.1</nowiki>
|31.636
|Unreactive
|[[File:Fifth combination.png 01503585|thumb]]
|-
|<nowiki>-1.6</nowiki>
|0.2
|1.707
|Unreactive
|[[File:Sixth combination.png 01503585|thumb]]
|}
For diagram 1, the kinetic energy is quite high at the begining. After the H2 molecule and Fluorine atom react, the bond between H and F forms and breaks serveral times. The kinetic energy and potential energy converse to each other several times. In the end, since the kinetic energy is too high, the high vibration mode causes the bond of HF breaks and then the reaction is unreactive.

For diagram 2, the kinetic energy is enough to let the reaction happen and not break the HF bond forms in the end. Some kinetic energy stores in HF bond and some potential energy stored in HH bond turns into kinetic energy.

Diagram 3 is similar to diagram 2, and diagrams 4 & 5 is similar to diagram 1.

For the combination of pFH= -1.6 g.mol-1.pm.fs-1 and pHH= 0.2 g.mol-1.pm.fs-1, the reaction is unreactive since the momentum of the system is too low to overcome the barrier of the reaction. Thus the reaction does not happen. 

==== For reaction H + HF ====
[[File:Question nine last.png|left|thumb]]

For the reaction of H + HF, one possible initial condition found is rAB = 92 pm, rBC = 200 pm, and the momenta are pFH= 0 g.mol-1.pm.fs-1 and pHH= 0.2 g.mol-1.pm.fs-1. The energy of the system is -270.348 kJ/mol. This reaction can happen is because that the H atom has enough energy to break the HF bond and form a new HH bond. Since the bond energy of HF is very high, the H atom needs more momentum to start the reaction.

To confirm the observations experimentally, controlling temperature of the reaction is a good way. By increasing the temperature, the momenta of molecules and atoms increases. So the reaction starts to change from unreactive to reactive. However, if the temperature is too high, which causes extra momenta in the reaction, the high energy vibration mode will cause the product fall back to reactant.

File:Question nine last.png

2020-05-22T19:38:54Z

Yc14518:

MRD:01503585 yulong

2020-05-22T19:17:09Z

Yc14518: /* For reaction F + H2 */

== EXERCISE 1: H + H2 system ==

=== Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===
The transition state is a saddle point on a potential energy diagram, it is a critical point but is not the local extremum of the function. It can be identified by finding a point which its first and second derivative are both zero. The difference between the local minimum and transition state is that the local minimum has a non zero second derivative (positive derivative).

=== Question 2: Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ===
The estimate of mine is that the distance between AB and BC are both 90.78 pm. Based on the definition of transition state, at the transition state, the atoms should have a fixed position. Based on the “Internuclear Distances vs Time”, the distances between atoms do not change with time, which means it is a good estimate of transition state.

=== Question 3: Comment on how the ''mep'' and the trajectory you just calculated differ. ===
The mep just include the minimum energy of the reaction passed, and it doesn't calculate the vibration of atoms while them moving, which its trajectory changes smoothly without fluctuation.
[[File:Distance vs time r1=r2=rts 01503585.png|thumb|Internuclear Distances vs Time when r1=r2=rts|206x206px|none]][[File:D_V_time_r1=rts_r2=rts+1.png_01503585|thumb|205x205px|
D V time r1=rts r2=rts+1
|none]]By comparing the "internuclear distances vs time" plots of r1=r2=rts and r1=rts, r2=rts+1, it is easy to obtain that when r1=r2=rts, the distances between atoms do not change. But as we change the starting point to r2=rts+1, the distance of between atoms A and C, and the distance between atoms A and B start to increase with similar rate. While the distance between B and C starts to decrease. This difference between two graphs shows that once the reaction pass the transition state, new molecule BC starts to form and at the same time atom A goes away from the molecule.[[File:M V time r1=r2=rts.png 01503585|thumb|206x206px|M V time r1=r2=rts|none]][[File:M V time r1=rts r2=rts+1.png 01503585|thumb|205x205px|M V time r1=rts, r2=rts+1|none]]By comparing the "momenta vs time" plots of r1=r2=rts and r1=rts, r2=rts+1, we see that the momentum of B-C is fluctuating between 0.0024 g.mol-1.pm.fs-1 and -0.0024 g.mol-1.pm.fs-1. However, the real momenta of B-C and AB should both be zero at the transition state. The none zero values are caused by approximated input. For r1=rts, r2=rts+1, the momenta of A-B and B-C are both increasing. The momentum of A-B increases faster as A is going away from molecule BC, and reach a plateau. While the momentum B-C starts to fluctuate after increasing since they reach a vibrating status after forming a molecule.

=== Question 4: 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>-414.280</nowiki>
|Reactive
|At the start the distance between A and B is large and the distance between B and C is small since BC is a molecule. The momenta given are enough to let the reaction happen. When A collide molecule BC, new molecule AB is formed and molecule BC is broke. The kinetic energy for reaction is 19.508 kJ.mol-1.
|[[File:Reactive group 1.png 01503585|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|Unreactive
|The distances between atoms at the starting point are similar with group one, but the molecule atoms have a larger momentum which causes the molecule has a greater vibration mode. The result of this is that the system has a lower kinetic energy to complete the reaction. The kinetic energy for reaction is 13.710 kJ.mol-1.
|[[File:Reactive group 2.png 01503585|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Reactive
|The momentum of molecule is same as group 2, but the atom A has a greater momentum which causes the reaction has enough kinetic energy to complete. The kinetic energy for reaction is 19.801 kJ.mol-1.
|[[File:Reactive group 3.png 01503585|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|Unreactive
|Under this condition, the momenta of both molecule and atom are so high which causes the bond between B-C breaks and A-B bond forms. However the reaction does not stops here, the bond between A-B breaks again and B-C bond forms again, which cause this reaction not completed. The kinetic energy for reaction is 76.510 kJ.mol-1.
|[[File:Reactive group 4.png 01503585|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Reactive
|For this one, the starting momenta are similar to group 4. However, the small excess of momentum of atom A causes the system has enough kinetic energy to break the B-C bond again and finally form the A-B bond and leads the reaction to completion. The kinetic energy for reaction is 84.310 kJ.mol-1.
|[[File:Reactive group 5.png 01503585|thumb]]
|}
The hypothesis is "All trajectories starting with the same positions but with higher values of momenta (higher kinetic energy) would be reactive, as they have enough kinetic energy to overcome the activation barrier". Based on the result collected, this hypothesis is only partially correct. For some combination of starting momenta, the kinetic energy is too high to let the product stay stable. The extra kinetic energy will cause the reaction fall back to reactants.

=== Question 5: Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===
Based on the result obtained, the transition state theory predictions for rate values are overestimated.

According to the TST, systems with great enough energy will let the reaction happen. However, when the system with very high energy, the high energy vibration mode of it will cause the reaction goes to an unexpected result. Like the energy combination group 4 in question 4, the bonds between atoms form and break more than we need. Under this situation, some of the reaction predicted happen will actually not happen, which cause the predicted reaction rates higher than the real ones.

In addition, TST only involves classical mechanics but not quantum mechanics, which it does not concern the effect of "quantum tunnelling effect". This effect causes some reaction happens even it does not reach the activation energy. In this way, if the quantum tunnelling effect is concerned, the actual reaction rates will slightly higher than the ones predicted. However this effect is minor which the overall TST predictions for rate values are still overestimated.

== EXERCISE 2: F - H - H system ==

=== Question 6: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===
[[File:F+H2 reaction.png 01503585|centre|thumb|F-H-H system]]For this part, the atom A now is Fluorine, and atoms B and C are hydrogen.

The bond energy of H-F bond is 565 kJ/mol, and the bond energy of H-H bond is 432 kJ/mol. The HF molecule has a higher bond energy, which means more energy is needed to break the bond of HF than the bond of H-H. In this way, the reaction of F + H2 is an exothermic reaction and the reaction of H + HF is an endothermic reaction. The reason why HF has a stonger bond energy is that HF molecule is polar and it has partial ionic character.

=== Question 7: Locate the approximate position of the transition state. ===
{| class="wikitable"
|[[File:F-H-H transition state point.png 01503585|thumb|422x422px|F-H-H transition state point]]
|[[File:F-H-H D V T.png|thumb|421x421px|Internuclear distance vs time plot]]
|}
The approximate position of the transition state of this reaction is the point at rBC=74.49 pm and rAB=181.1 pm. Start simulation from this point, the distances between atoms are remains unchanged. In this way, this is a good approximation of transition state point.

=== Question 8: Report the activation energy for both reactions.kJ/mol ===
Set the calculation type as MEP, by changing the atom distances, we can obtain the activation energy for two reactions. At the transition point, the total energy of the system is -433.981 kJ/mol.

For F + H2, set the distance between atom F and H2 rAB=1000pm, and keep rBC=74.49 pm:

The enegy obtained when time is long enough is -435.059 kJ/mol, and the acrivation energy for this reaction thus is -433.981 kJ/mol-(-435.059 kJ/mol)=1.078 kJ/mol.

For H + HF, set the distance between atom H and HF rBC=1000pm, and keep rAB=181.1 pm:

The enegy obtained when time is long enough is -560.7 kJ/mol, and the acrivation energy for this reaction thus is -433.981 kJ/mol-(-560.7 kJ/mol)=126.719 kJ/mol

=== Question 9: 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 reaction F + H2 ====
The distances set are rHH = 74 pm, rFH=200 pm.
{| class="wikitable"
!pFH/ g.mol-1.pm.fs-1
!pHH/ g.mol-1.pm.fs-1
!Kinetic Energy
!Reactivity
!Distance vs time plot
|-
|<nowiki>-1.0</nowiki>
|6.1
|43.836
|Unreactive
|[[File:First combination.png 01503585|thumb]]
|-
|<nowiki>-1.0</nowiki>
|3
|12.526
|Reactive
|[[File:Second combination.png 01503585|thumb]]
|-
|<nowiki>-1.0</nowiki>
|0
|0.526
|Reactive
|[[File:Third combination.png 01503585|thumb]]
|-
|<nowiki>-1.0</nowiki>
|<nowiki>-3</nowiki>
|6.526
|Unreactive
|[[File:Fourth combination.png 01503585|thumb]]
|-
|<nowiki>-1.0</nowiki>
|<nowiki>-6.1</nowiki>
|31.636
|Unreactive
|[[File:Fifth combination.png 01503585|thumb]]
|-
|<nowiki>-1.6</nowiki>
|0.2
|1.707
|Unreactive
|[[File:Sixth combination.png 01503585|thumb]]
|}
For diagram 1, the kinetic energy is quite high at the begining. After the H2 molecule and Fluorine atom react, the bond between H and F forms and breaks serveral times. The kinetic energy and potential energy converse to each other several times. In the end, since the kinetic energy is too high, the high vibration mode causes the bond of HF breaks and then the reaction is unreactive.

For diagram 2, the kinetic energy is enough to let the reaction happen and not break the HF bond forms in the end. Some kinetic energy stores in HF bond and some potential energy stored in HH bond turns into kinetic energy.

Diagram 3 is similar to diagram 2, and diagrams 4 & 5 is similar to diagram 1.

For the combination of pFH= -1.6 g.mol-1.pm.fs-1 and pHH= 0.2 g.mol-1.pm.fs-1, the reaction is unreactive since the momentum of the system is too low to overcome the barrier of the reaction. Thus the reaction does not happen.

MRD:01503585 yulong

2020-05-22T19:00:01Z

Yc14518: /* For reaction F + H2 */

== EXERCISE 1: H + H2 system ==

=== Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===
The transition state is a saddle point on a potential energy diagram, it is a critical point but is not the local extremum of the function. It can be identified by finding a point which its first and second derivative are both zero. The difference between the local minimum and transition state is that the local minimum has a non zero second derivative (positive derivative).

=== Question 2: Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ===
The estimate of mine is that the distance between AB and BC are both 90.78 pm. Based on the definition of transition state, at the transition state, the atoms should have a fixed position. Based on the “Internuclear Distances vs Time”, the distances between atoms do not change with time, which means it is a good estimate of transition state.

=== Question 3: Comment on how the ''mep'' and the trajectory you just calculated differ. ===
The mep just include the minimum energy of the reaction passed, and it doesn't calculate the vibration of atoms while them moving, which its trajectory changes smoothly without fluctuation.
[[File:Distance vs time r1=r2=rts 01503585.png|thumb|Internuclear Distances vs Time when r1=r2=rts|206x206px|none]][[File:D_V_time_r1=rts_r2=rts+1.png_01503585|thumb|205x205px|
D V time r1=rts r2=rts+1
|none]]By comparing the "internuclear distances vs time" plots of r1=r2=rts and r1=rts, r2=rts+1, it is easy to obtain that when r1=r2=rts, the distances between atoms do not change. But as we change the starting point to r2=rts+1, the distance of between atoms A and C, and the distance between atoms A and B start to increase with similar rate. While the distance between B and C starts to decrease. This difference between two graphs shows that once the reaction pass the transition state, new molecule BC starts to form and at the same time atom A goes away from the molecule.[[File:M V time r1=r2=rts.png 01503585|thumb|206x206px|M V time r1=r2=rts|none]][[File:M V time r1=rts r2=rts+1.png 01503585|thumb|205x205px|M V time r1=rts, r2=rts+1|none]]By comparing the "momenta vs time" plots of r1=r2=rts and r1=rts, r2=rts+1, we see that the momentum of B-C is fluctuating between 0.0024 g.mol-1.pm.fs-1 and -0.0024 g.mol-1.pm.fs-1. However, the real momenta of B-C and AB should both be zero at the transition state. The none zero values are caused by approximated input. For r1=rts, r2=rts+1, the momenta of A-B and B-C are both increasing. The momentum of A-B increases faster as A is going away from molecule BC, and reach a plateau. While the momentum B-C starts to fluctuate after increasing since they reach a vibrating status after forming a molecule.

=== Question 4: 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>-414.280</nowiki>
|Reactive
|At the start the distance between A and B is large and the distance between B and C is small since BC is a molecule. The momenta given are enough to let the reaction happen. When A collide molecule BC, new molecule AB is formed and molecule BC is broke. The kinetic energy for reaction is 19.508 kJ.mol-1.
|[[File:Reactive group 1.png 01503585|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|Unreactive
|The distances between atoms at the starting point are similar with group one, but the molecule atoms have a larger momentum which causes the molecule has a greater vibration mode. The result of this is that the system has a lower kinetic energy to complete the reaction. The kinetic energy for reaction is 13.710 kJ.mol-1.
|[[File:Reactive group 2.png 01503585|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Reactive
|The momentum of molecule is same as group 2, but the atom A has a greater momentum which causes the reaction has enough kinetic energy to complete. The kinetic energy for reaction is 19.801 kJ.mol-1.
|[[File:Reactive group 3.png 01503585|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|Unreactive
|Under this condition, the momenta of both molecule and atom are so high which causes the bond between B-C breaks and A-B bond forms. However the reaction does not stops here, the bond between A-B breaks again and B-C bond forms again, which cause this reaction not completed. The kinetic energy for reaction is 76.510 kJ.mol-1.
|[[File:Reactive group 4.png 01503585|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Reactive
|For this one, the starting momenta are similar to group 4. However, the small excess of momentum of atom A causes the system has enough kinetic energy to break the B-C bond again and finally form the A-B bond and leads the reaction to completion. The kinetic energy for reaction is 84.310 kJ.mol-1.
|[[File:Reactive group 5.png 01503585|thumb]]
|}
The hypothesis is "All trajectories starting with the same positions but with higher values of momenta (higher kinetic energy) would be reactive, as they have enough kinetic energy to overcome the activation barrier". Based on the result collected, this hypothesis is only partially correct. For some combination of starting momenta, the kinetic energy is too high to let the product stay stable. The extra kinetic energy will cause the reaction fall back to reactants.

=== Question 5: Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===
Based on the result obtained, the transition state theory predictions for rate values are overestimated.

According to the TST, systems with great enough energy will let the reaction happen. However, when the system with very high energy, the high energy vibration mode of it will cause the reaction goes to an unexpected result. Like the energy combination group 4 in question 4, the bonds between atoms form and break more than we need. Under this situation, some of the reaction predicted happen will actually not happen, which cause the predicted reaction rates higher than the real ones.

In addition, TST only involves classical mechanics but not quantum mechanics, which it does not concern the effect of "quantum tunnelling effect". This effect causes some reaction happens even it does not reach the activation energy. In this way, if the quantum tunnelling effect is concerned, the actual reaction rates will slightly higher than the ones predicted. However this effect is minor which the overall TST predictions for rate values are still overestimated.

== EXERCISE 2: F - H - H system ==

=== Question 6: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===
[[File:F+H2 reaction.png 01503585|centre|thumb|F-H-H system]]For this part, the atom A now is Fluorine, and atoms B and C are hydrogen.

The bond energy of H-F bond is 565 kJ/mol, and the bond energy of H-H bond is 432 kJ/mol. The HF molecule has a higher bond energy, which means more energy is needed to break the bond of HF than the bond of H-H. In this way, the reaction of F + H2 is an exothermic reaction and the reaction of H + HF is an endothermic reaction. The reason why HF has a stonger bond energy is that HF molecule is polar and it has partial ionic character.

=== Question 7: Locate the approximate position of the transition state. ===
{| class="wikitable"
|[[File:F-H-H transition state point.png 01503585|thumb|422x422px|F-H-H transition state point]]
|[[File:F-H-H D V T.png|thumb|421x421px|Internuclear distance vs time plot]]
|}
The approximate position of the transition state of this reaction is the point at rBC=74.49 pm and rAB=181.1 pm. Start simulation from this point, the distances between atoms are remains unchanged. In this way, this is a good approximation of transition state point.

=== Question 8: Report the activation energy for both reactions.kJ/mol ===
Set the calculation type as MEP, by changing the atom distances, we can obtain the activation energy for two reactions. At the transition point, the total energy of the system is -433.981 kJ/mol.

For F + H2, set the distance between atom F and H2 rAB=1000pm, and keep rBC=74.49 pm:

The enegy obtained when time is long enough is -435.059 kJ/mol, and the acrivation energy for this reaction thus is -433.981 kJ/mol-(-435.059 kJ/mol)=1.078 kJ/mol.

For H + HF, set the distance between atom H and HF rBC=1000pm, and keep rAB=181.1 pm:

The enegy obtained when time is long enough is -560.7 kJ/mol, and the acrivation energy for this reaction thus is -433.981 kJ/mol-(-560.7 kJ/mol)=126.719 kJ/mol

=== Question 9: 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 reaction F + H2 ====
The distances set are rHH = 74 pm, rFH=200 pm.
{| class="wikitable"
!pFH/ g.mol-1.pm.fs-1
!pHH/ g.mol-1.pm.fs-1
!Kinetic Energy
!Reactivity
!Distance vs time plot
|-
|<nowiki>-1.0</nowiki>
|6.1
|43.836
|Unreactive
|[[File:First combination.png 01503585|thumb]]
|-
|<nowiki>-1.0</nowiki>
|3
|12.526
|Reactive
|[[File:Second combination.png 01503585|thumb]]
|-
|<nowiki>-1.0</nowiki>
|0
|0.526
|Reactive
|[[File:Third combination.png 01503585|thumb]]
|-
|<nowiki>-1.0</nowiki>
|<nowiki>-3</nowiki>
|6.526
|Unreactive
|[[File:Fourth combination.png 01503585|thumb]]
|-
|<nowiki>-1.0</nowiki>
|<nowiki>-6.1</nowiki>
|31.636
|Unreactive
|[[File:Fifth combination.png 01503585|thumb]]
|-
|<nowiki>-1.6</nowiki>
|0.2
|1.707
|Unreactive
|[[File:Sixth combination.png 01503585|thumb]]
|}
For diagram 1, the kinetic energy is quite high at the begining. After the H2 molecule and Fluorine atom react, the bond between H and F forms and breaks serveral times. The kinetic energy and potential energy converse to each other several times. In the end, since the kinetic energy is too high, the high vibration mode causes the bond of HF breaks and then the reaction is unreactive.

For diagram 2, the kinetic energy is enough to let the reaction happen and not break the HF bond forms in the end. Some kinetic energy stores in HF bond and some potential energy stored in HH bond turns into kinetic energy.

File:Sixth combination.png 01503585

2020-05-22T18:50:20Z

Yc14518:

MRD:01503585 yulong

2020-05-22T18:38:57Z

Yc14518: /* For reaction F + H2 */

== EXERCISE 1: H + H2 system ==

=== Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===
The transition state is a saddle point on a potential energy diagram, it is a critical point but is not the local extremum of the function. It can be identified by finding a point which its first and second derivative are both zero. The difference between the local minimum and transition state is that the local minimum has a non zero second derivative (positive derivative).

=== Question 2: Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ===
The estimate of mine is that the distance between AB and BC are both 90.78 pm. Based on the definition of transition state, at the transition state, the atoms should have a fixed position. Based on the “Internuclear Distances vs Time”, the distances between atoms do not change with time, which means it is a good estimate of transition state.

=== Question 3: Comment on how the ''mep'' and the trajectory you just calculated differ. ===
The mep just include the minimum energy of the reaction passed, and it doesn't calculate the vibration of atoms while them moving, which its trajectory changes smoothly without fluctuation.
[[File:Distance vs time r1=r2=rts 01503585.png|thumb|Internuclear Distances vs Time when r1=r2=rts|206x206px|none]][[File:D_V_time_r1=rts_r2=rts+1.png_01503585|thumb|205x205px|
D V time r1=rts r2=rts+1
|none]]By comparing the "internuclear distances vs time" plots of r1=r2=rts and r1=rts, r2=rts+1, it is easy to obtain that when r1=r2=rts, the distances between atoms do not change. But as we change the starting point to r2=rts+1, the distance of between atoms A and C, and the distance between atoms A and B start to increase with similar rate. While the distance between B and C starts to decrease. This difference between two graphs shows that once the reaction pass the transition state, new molecule BC starts to form and at the same time atom A goes away from the molecule.[[File:M V time r1=r2=rts.png 01503585|thumb|206x206px|M V time r1=r2=rts|none]][[File:M V time r1=rts r2=rts+1.png 01503585|thumb|205x205px|M V time r1=rts, r2=rts+1|none]]By comparing the "momenta vs time" plots of r1=r2=rts and r1=rts, r2=rts+1, we see that the momentum of B-C is fluctuating between 0.0024 g.mol-1.pm.fs-1 and -0.0024 g.mol-1.pm.fs-1. However, the real momenta of B-C and AB should both be zero at the transition state. The none zero values are caused by approximated input. For r1=rts, r2=rts+1, the momenta of A-B and B-C are both increasing. The momentum of A-B increases faster as A is going away from molecule BC, and reach a plateau. While the momentum B-C starts to fluctuate after increasing since they reach a vibrating status after forming a molecule.

=== Question 4: 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>-414.280</nowiki>
|Reactive
|At the start the distance between A and B is large and the distance between B and C is small since BC is a molecule. The momenta given are enough to let the reaction happen. When A collide molecule BC, new molecule AB is formed and molecule BC is broke. The kinetic energy for reaction is 19.508 kJ.mol-1.
|[[File:Reactive group 1.png 01503585|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|Unreactive
|The distances between atoms at the starting point are similar with group one, but the molecule atoms have a larger momentum which causes the molecule has a greater vibration mode. The result of this is that the system has a lower kinetic energy to complete the reaction. The kinetic energy for reaction is 13.710 kJ.mol-1.
|[[File:Reactive group 2.png 01503585|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Reactive
|The momentum of molecule is same as group 2, but the atom A has a greater momentum which causes the reaction has enough kinetic energy to complete. The kinetic energy for reaction is 19.801 kJ.mol-1.
|[[File:Reactive group 3.png 01503585|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|Unreactive
|Under this condition, the momenta of both molecule and atom are so high which causes the bond between B-C breaks and A-B bond forms. However the reaction does not stops here, the bond between A-B breaks again and B-C bond forms again, which cause this reaction not completed. The kinetic energy for reaction is 76.510 kJ.mol-1.
|[[File:Reactive group 4.png 01503585|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Reactive
|For this one, the starting momenta are similar to group 4. However, the small excess of momentum of atom A causes the system has enough kinetic energy to break the B-C bond again and finally form the A-B bond and leads the reaction to completion. The kinetic energy for reaction is 84.310 kJ.mol-1.
|[[File:Reactive group 5.png 01503585|thumb]]
|}
The hypothesis is "All trajectories starting with the same positions but with higher values of momenta (higher kinetic energy) would be reactive, as they have enough kinetic energy to overcome the activation barrier". Based on the result collected, this hypothesis is only partially correct. For some combination of starting momenta, the kinetic energy is too high to let the product stay stable. The extra kinetic energy will cause the reaction fall back to reactants.

=== Question 5: Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===
Based on the result obtained, the transition state theory predictions for rate values are overestimated.

According to the TST, systems with great enough energy will let the reaction happen. However, when the system with very high energy, the high energy vibration mode of it will cause the reaction goes to an unexpected result. Like the energy combination group 4 in question 4, the bonds between atoms form and break more than we need. Under this situation, some of the reaction predicted happen will actually not happen, which cause the predicted reaction rates higher than the real ones.

In addition, TST only involves classical mechanics but not quantum mechanics, which it does not concern the effect of "quantum tunnelling effect". This effect causes some reaction happens even it does not reach the activation energy. In this way, if the quantum tunnelling effect is concerned, the actual reaction rates will slightly higher than the ones predicted. However this effect is minor which the overall TST predictions for rate values are still overestimated.

== EXERCISE 2: F - H - H system ==

=== Question 6: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===
[[File:F+H2 reaction.png 01503585|centre|thumb|F-H-H system]]For this part, the atom A now is Fluorine, and atoms B and C are hydrogen.

The bond energy of H-F bond is 565 kJ/mol, and the bond energy of H-H bond is 432 kJ/mol. The HF molecule has a higher bond energy, which means more energy is needed to break the bond of HF than the bond of H-H. In this way, the reaction of F + H2 is an exothermic reaction and the reaction of H + HF is an endothermic reaction. The reason why HF has a stonger bond energy is that HF molecule is polar and it has partial ionic character.

=== Question 7: Locate the approximate position of the transition state. ===
{| class="wikitable"
|[[File:F-H-H transition state point.png 01503585|thumb|422x422px|F-H-H transition state point]]
|[[File:F-H-H D V T.png|thumb|421x421px|Internuclear distance vs time plot]]
|}
The approximate position of the transition state of this reaction is the point at rBC=74.49 pm and rAB=181.1 pm. Start simulation from this point, the distances between atoms are remains unchanged. In this way, this is a good approximation of transition state point.

=== Question 8: Report the activation energy for both reactions.kJ/mol ===
Set the calculation type as MEP, by changing the atom distances, we can obtain the activation energy for two reactions. At the transition point, the total energy of the system is -433.981 kJ/mol.

For F + H2, set the distance between atom F and H2 rAB=1000pm, and keep rBC=74.49 pm:

The enegy obtained when time is long enough is -435.059 kJ/mol, and the acrivation energy for this reaction thus is -433.981 kJ/mol-(-435.059 kJ/mol)=1.078 kJ/mol.

For H + HF, set the distance between atom H and HF rBC=1000pm, and keep rAB=181.1 pm:

The enegy obtained when time is long enough is -560.7 kJ/mol, and the acrivation energy for this reaction thus is -433.981 kJ/mol-(-560.7 kJ/mol)=126.719 kJ/mol

=== Question 9: 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 reaction F + H2 ====
The distances set are rHH = 74 pm, rFH=200 pm.
{| class="wikitable"
!pFH/ g.mol-1.pm.fs-1
!pHH/ g.mol-1.pm.fs-1
!Kinetic Energy
!Reactivity
!Distance vs time plot
|-
|<nowiki>-1.0</nowiki>
|6.1
|43.836
|Unreactive
|[[File:First combination.png 01503585|thumb]]
|-
|<nowiki>-1.0</nowiki>
|3
|12.526
|Reactive
|[[File:Second combination.png 01503585|thumb]]
|-
|<nowiki>-1.0</nowiki>
|0
|0.526
|Reactive
|[[File:Third combination.png 01503585|thumb]]
|-
|<nowiki>-1.0</nowiki>
|<nowiki>-3</nowiki>
|6.526
|Unreactive
|[[File:Fourth combination.png 01503585|thumb]]
|-
|<nowiki>-1.0</nowiki>
|<nowiki>-6.1</nowiki>
|31.636
|Unreactive
|[[File:Fifth combination.png 01503585|thumb]]
|}

File:Fifth combination.png 01503585

2020-05-22T18:38:31Z

Yc14518:

File:Fourth combination.png 01503585

2020-05-22T18:36:57Z

Yc14518:

MRD:01503585 yulong

File:F-H-H D V T.png

2020-05-22T15:41:17Z

Yc14518:

File:F-H-H transition state point.png 01503585

2020-05-22T15:39:43Z

Yc14518:

2020-05-22T11:31:44Z

File:Reactive group 4.png 01503585

2020-05-22T11:29:38Z

2020-05-22T10:57:21Z

File:Reactive group 1.png 01503585

2020-05-22T10:52:41Z

Yc14518: