ChemWiki - User contributions [en]

MRD:ld2416

2019-05-24T16:35:51Z

Ld2416: /* Activation E HF + H */

==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?
[[File:Contour plot opf H H2.PNG]]

The transition state is when the H2 is equally distant to H1 and H3, for the contour plot this is approximately 0.93 A.

[[File:SUrface plot.PNG]]

The transition state can be seen at the saddle point, where the transition state is a minimum for the frank condon plot for the reaction. The transition state is a maximum for the reaction trajectory.
Mathematically, the TS has a second derivative greater than 0 in the Q1 direction (minimum) and TS has a second derivative less than 0 in the Q2 direction (maximum).

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.

From the contour plot we see that the transition state occurs when the r1=r2. Therefore to estimate the transition state position distances AB must be equal to BC.

[[File:Ab DIS.PNG]]

From this we can estimate the transition position from the contour plot:
[[File:Contour TS.PNG]]

This gives a value of 0.9075 A

Looking at this on a distance/time graph:
[[File:Bad ds TS.PNG]]

The momentum is set to 0 because the transition state is stationary and the distance between the 2 are equal.

Dynamic Surface:
[[File:Dynamic surface.PNG]]

MEP surface diagram:

[[File:MEP surface.PNG]]

Differences can be seen in the length and oscillations in the reaction trajectory. For the minimum energy pathway (MEP) there is no oscillations as there it neglects the molecules vibration modes and it moves infinitely slowly. The line is shorter as the reaction coordinate isn't as long due to the reaction.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 ||-99.0|| Yes||H2 initially not vibrating, upon reaction the new H2 has a high vibrational energy||[[File:P=-1.25,-2.5.PNG]]|
|-
| -1.5 || -2.0 ||-100.5||No|| H approaching is repelled, causing the HH bond to vibrate ||[[File:P=-1.5,-2.PNG]]|
|-
| -1.5 || -2.5 ||-99.0||Yes||Slow approach of H and H2 after reaction stronger vibrations||[[File:P=-1.5,-2.5.PNG]]|
|-
| -2.5 || -5.0 ||-85.0||No||H approaches H2 causes a the H-H bond to weaken almost up to a transition state, not enough energy to form a new HH bond| |[[File:P=-2.5,-5.PNG]]|
|-
| -2.5 || -5.2 ||-83.4||Yes||H on approach caused increase in vibration for H2 bond until it broke and new HH bond formed with the approaching H||[[File:P=-2.5,-5.2.PNG]]|
|}

For the penultimate conditions, the reaction reaches the transition state yet its still falls back to the reactants despite the high momentum from the approaching H. The difference for the reaction to occur is the momentum to the approaching H to the molecule. It can be seen that the both the species in the reactants need the required momentum in order to reach transition state and to form the products.

== F-H-H system ==

H2 + F potential energy surface:
[[File:H2 + F potential surface.PNG]]

Hf + H potential surface:
[[File:HF+H surface plot.PNG]]

From the potential surface we see that the products; H2 + F are higher in energy than the product F-H this suggests that the F-H bond is stronger than the H-H bond. This reaction is exothermic.
The reverse reaction must then be endothermic for H + HF

==Transition state position H2 + F ==
[[File:Energy and configuration for h2 + F.PNG]]

To find the transition state, initial conditions were set and momentum is set to 0. From this it is clear that it isnt optimised as there are forces present between HH and F.

Setting HH distance to 0.744 and HF distance to 1.811, the forces are approximately 0 and so this is where the transition state exists.

== Activation Energy H2 + F ==
[[File:Energy v time Activation energy for Fh2.PNG]]

Looking at the energy v time graph the difference between the total energy and the potential energy results in the activation

Activation energy = 104.019 - 103.749 = 0.270

==Transition state position HF + H ==

[[File:Rstolihgrtiuhg.PNG]]

Transition state occurs at 0.74 A this resembles the H-H bond length in molecular H2

==Activation E HF + H ==
The activation energy for this reaction can be calculated by the difference between the TS and the energy of the reactants HF + H
[[File:Energy HF + H.PNG]]

133-103 = 30

===F + H2 reaction trajectory: ===

[[File:Config for animation for h2 + f reaction path.PNG]]

These are the initial conditions selected that gave a reaction trajectory

[[File:Conservation of energy H2 +F.PNG]]

The momentum v time plot shows B-C (H-H) vibrating as F has little momentum. Once reaction happens, BC momentum stops oscillating and AB (HF) starts oscillating showing the conservation of energy as the vibration transfers from HH to HF.
The mechanism of release:

rHH=0.74, rHF=2 and pHF = -0.5 for the following table. The pHH ranges from -3 and 3 to see if a reaction occurs with these initial conditions.
{| class="wikitable" border=1
! pHH !! Reaction?!! Contour Plot!
|-
| 3.0 || No|| [[File:H2+F, p=3.PNG]]
|-
| 2.0 || No|| [[File:H2+F,p=2.PNG]]
|-
| 1.0 || No|| [[File:H2+F,p=1.PNG]]
|-
| 0 ||No || [[File:H2+F,p=0.PNG]]
|-
| -1.0 || No||[[File:H2+F,p=-1.PNG]]
|-
| -2.0 ||Yes || [[File:H2+F,p=-2.PNG]]
|-
| -3.0 || No|| [[File:H2+F, p=-3.PNG]]
|}

For F having momentum -0.5, reaction only occurs when the HH bond has -2 momentum. For reaction to complete is not enough to have a large energy in the system. Each both H2 and F require specific energy in order to react with each other therefore changing either energy values will require the other species to have a complimentary energy.

[[File:H2+F,p=-0.8,0.1.PNG]]
Reaction occurs even when H2 has a small momentum. F having an increased momentum allowed this reaction to occur therefore changes to F compared to changes in momentum for H2 have a greater weighting on the progress of the reaction.

===H + HF reaction trajectory: ===

Setting the conditions:
r1 = 0.94, r2 = 2
{| class="wikitable" border=1
! pHH !! pHF !! Etot !! Description of the dynamics !! Illustration of the trajectory
|-
| -10 || -0.5 || -85.6 || No reaction. The system does reach the transition state before falling back to the products||[[File:-10,-0.5.PNG]]
|-
| -8 || -1 || -106.8 ||No reaction. H approach increases vibration frequency of HF ||[[File:-8,-1.PNG]]|
|-
| -1 || -12 ||-68.7 || Goes to transition state but does not react ||[[File:HF+H,P=-1,-12.PNG]]|
|-
|-1|| -13 ||-56.5 ||High HF vibrations cause reaction to occur. || [[File:HF+H reaction.PNG]]||
|}

[[File:HF+H reaction.PNG]]

Polanyi's empirical rules state there is a linear relationship between the activation energy and the position of the transition state. For H + HF there is a late transition state and therefore a higher Ea is needed compared to H2 + F. For late transition states the reactants require high vibration energy, which can be changed by altering the momentum. For other high energy reactions for HF + H the reaction was unsuccessful due to the composition of momentum being vibrational and translation energy. For the unsuccessful reactions the composition of vibrational and translation energy did not have enought vibrational energy for the reaction to go past the TS. On the other hand early stage TS need more translation energy.

File:Energy HF + H.PNG

2019-05-24T16:34:31Z

Ld2416:

MRD:ld2416

2019-05-24T16:27:45Z

Ld2416: /* Transition state position HF + H */

==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?
[[File:Contour plot opf H H2.PNG]]

The transition state is when the H2 is equally distant to H1 and H3, for the contour plot this is approximately 0.93 A.

[[File:SUrface plot.PNG]]

The transition state can be seen at the saddle point, where the transition state is a minimum for the frank condon plot for the reaction. The transition state is a maximum for the reaction trajectory.
Mathematically, the TS has a second derivative greater than 0 in the Q1 direction (minimum) and TS has a second derivative less than 0 in the Q2 direction (maximum).

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.

From the contour plot we see that the transition state occurs when the r1=r2. Therefore to estimate the transition state position distances AB must be equal to BC.

[[File:Ab DIS.PNG]]

From this we can estimate the transition position from the contour plot:
[[File:Contour TS.PNG]]

This gives a value of 0.9075 A

Looking at this on a distance/time graph:
[[File:Bad ds TS.PNG]]

The momentum is set to 0 because the transition state is stationary and the distance between the 2 are equal.

Dynamic Surface:
[[File:Dynamic surface.PNG]]

MEP surface diagram:

[[File:MEP surface.PNG]]

Differences can be seen in the length and oscillations in the reaction trajectory. For the minimum energy pathway (MEP) there is no oscillations as there it neglects the molecules vibration modes and it moves infinitely slowly. The line is shorter as the reaction coordinate isn't as long due to the reaction.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 ||-99.0|| Yes||H2 initially not vibrating, upon reaction the new H2 has a high vibrational energy||[[File:P=-1.25,-2.5.PNG]]|
|-
| -1.5 || -2.0 ||-100.5||No|| H approaching is repelled, causing the HH bond to vibrate ||[[File:P=-1.5,-2.PNG]]|
|-
| -1.5 || -2.5 ||-99.0||Yes||Slow approach of H and H2 after reaction stronger vibrations||[[File:P=-1.5,-2.5.PNG]]|
|-
| -2.5 || -5.0 ||-85.0||No||H approaches H2 causes a the H-H bond to weaken almost up to a transition state, not enough energy to form a new HH bond| |[[File:P=-2.5,-5.PNG]]|
|-
| -2.5 || -5.2 ||-83.4||Yes||H on approach caused increase in vibration for H2 bond until it broke and new HH bond formed with the approaching H||[[File:P=-2.5,-5.2.PNG]]|
|}

For the penultimate conditions, the reaction reaches the transition state yet its still falls back to the reactants despite the high momentum from the approaching H. The difference for the reaction to occur is the momentum to the approaching H to the molecule. It can be seen that the both the species in the reactants need the required momentum in order to reach transition state and to form the products.

== F-H-H system ==

H2 + F potential energy surface:
[[File:H2 + F potential surface.PNG]]

Hf + H potential surface:
[[File:HF+H surface plot.PNG]]

From the potential surface we see that the products; H2 + F are higher in energy than the product F-H this suggests that the F-H bond is stronger than the H-H bond. This reaction is exothermic.
The reverse reaction must then be endothermic for H + HF

==Transition state position H2 + F ==
[[File:Energy and configuration for h2 + F.PNG]]

To find the transition state, initial conditions were set and momentum is set to 0. From this it is clear that it isnt optimised as there are forces present between HH and F.

Setting HH distance to 0.744 and HF distance to 1.811, the forces are approximately 0 and so this is where the transition state exists.

== Activation Energy H2 + F ==
[[File:Energy v time Activation energy for Fh2.PNG]]

Looking at the energy v time graph the difference between the total energy and the potential energy results in the activation

Activation energy = 104.019 - 103.749 = 0.270

==Transition state position HF + H ==

[[File:Rstolihgrtiuhg.PNG]]

Transition state occurs at 0.74 A this resembles the H-H bond length in molecular H2

==Activation E HF + H ==

30

===F + H2 reaction trajectory: ===

[[File:Config for animation for h2 + f reaction path.PNG]]

These are the initial conditions selected that gave a reaction trajectory

[[File:Conservation of energy H2 +F.PNG]]

The momentum v time plot shows B-C (H-H) vibrating as F has little momentum. Once reaction happens, BC momentum stops oscillating and AB (HF) starts oscillating showing the conservation of energy as the vibration transfers from HH to HF.
The mechanism of release:

rHH=0.74, rHF=2 and pHF = -0.5 for the following table. The pHH ranges from -3 and 3 to see if a reaction occurs with these initial conditions.
{| class="wikitable" border=1
! pHH !! Reaction?!! Contour Plot!
|-
| 3.0 || No|| [[File:H2+F, p=3.PNG]]
|-
| 2.0 || No|| [[File:H2+F,p=2.PNG]]
|-
| 1.0 || No|| [[File:H2+F,p=1.PNG]]
|-
| 0 ||No || [[File:H2+F,p=0.PNG]]
|-
| -1.0 || No||[[File:H2+F,p=-1.PNG]]
|-
| -2.0 ||Yes || [[File:H2+F,p=-2.PNG]]
|-
| -3.0 || No|| [[File:H2+F, p=-3.PNG]]
|}

For F having momentum -0.5, reaction only occurs when the HH bond has -2 momentum. For reaction to complete is not enough to have a large energy in the system. Each both H2 and F require specific energy in order to react with each other therefore changing either energy values will require the other species to have a complimentary energy.

[[File:H2+F,p=-0.8,0.1.PNG]]
Reaction occurs even when H2 has a small momentum. F having an increased momentum allowed this reaction to occur therefore changes to F compared to changes in momentum for H2 have a greater weighting on the progress of the reaction.

===H + HF reaction trajectory: ===

Setting the conditions:
r1 = 0.94, r2 = 2
{| class="wikitable" border=1
! pHH !! pHF !! Etot !! Description of the dynamics !! Illustration of the trajectory
|-
| -10 || -0.5 || -85.6 || No reaction. The system does reach the transition state before falling back to the products||[[File:-10,-0.5.PNG]]
|-
| -8 || -1 || -106.8 ||No reaction. H approach increases vibration frequency of HF ||[[File:-8,-1.PNG]]|
|-
| -1 || -12 ||-68.7 || Goes to transition state but does not react ||[[File:HF+H,P=-1,-12.PNG]]|
|-
|-1|| -13 ||-56.5 ||High HF vibrations cause reaction to occur. || [[File:HF+H reaction.PNG]]||
|}

[[File:HF+H reaction.PNG]]

Polanyi's empirical rules state there is a linear relationship between the activation energy and the position of the transition state. For H + HF there is a late transition state and therefore a higher Ea is needed compared to H2 + F. For late transition states the reactants require high vibration energy, which can be changed by altering the momentum. For other high energy reactions for HF + H the reaction was unsuccessful due to the composition of momentum being vibrational and translation energy. For the unsuccessful reactions the composition of vibrational and translation energy did not have enought vibrational energy for the reaction to go past the TS. On the other hand early stage TS need more translation energy.

File:Rstolihgrtiuhg.PNG

2019-05-24T16:27:14Z

Ld2416:

MRD:ld2416

2019-05-24T16:22:01Z

Ld2416: /* Transition state position HF + H */

==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?
[[File:Contour plot opf H H2.PNG]]

The transition state is when the H2 is equally distant to H1 and H3, for the contour plot this is approximately 0.93 A.

[[File:SUrface plot.PNG]]

The transition state can be seen at the saddle point, where the transition state is a minimum for the frank condon plot for the reaction. The transition state is a maximum for the reaction trajectory.
Mathematically, the TS has a second derivative greater than 0 in the Q1 direction (minimum) and TS has a second derivative less than 0 in the Q2 direction (maximum).

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.

From the contour plot we see that the transition state occurs when the r1=r2. Therefore to estimate the transition state position distances AB must be equal to BC.

[[File:Ab DIS.PNG]]

From this we can estimate the transition position from the contour plot:
[[File:Contour TS.PNG]]

This gives a value of 0.9075 A

Looking at this on a distance/time graph:
[[File:Bad ds TS.PNG]]

The momentum is set to 0 because the transition state is stationary and the distance between the 2 are equal.

Dynamic Surface:
[[File:Dynamic surface.PNG]]

MEP surface diagram:

[[File:MEP surface.PNG]]

Differences can be seen in the length and oscillations in the reaction trajectory. For the minimum energy pathway (MEP) there is no oscillations as there it neglects the molecules vibration modes and it moves infinitely slowly. The line is shorter as the reaction coordinate isn't as long due to the reaction.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 ||-99.0|| Yes||H2 initially not vibrating, upon reaction the new H2 has a high vibrational energy||[[File:P=-1.25,-2.5.PNG]]|
|-
| -1.5 || -2.0 ||-100.5||No|| H approaching is repelled, causing the HH bond to vibrate ||[[File:P=-1.5,-2.PNG]]|
|-
| -1.5 || -2.5 ||-99.0||Yes||Slow approach of H and H2 after reaction stronger vibrations||[[File:P=-1.5,-2.5.PNG]]|
|-
| -2.5 || -5.0 ||-85.0||No||H approaches H2 causes a the H-H bond to weaken almost up to a transition state, not enough energy to form a new HH bond| |[[File:P=-2.5,-5.PNG]]|
|-
| -2.5 || -5.2 ||-83.4||Yes||H on approach caused increase in vibration for H2 bond until it broke and new HH bond formed with the approaching H||[[File:P=-2.5,-5.2.PNG]]|
|}

For the penultimate conditions, the reaction reaches the transition state yet its still falls back to the reactants despite the high momentum from the approaching H. The difference for the reaction to occur is the momentum to the approaching H to the molecule. It can be seen that the both the species in the reactants need the required momentum in order to reach transition state and to form the products.

== F-H-H system ==

H2 + F potential energy surface:
[[File:H2 + F potential surface.PNG]]

Hf + H potential surface:
[[File:HF+H surface plot.PNG]]

From the potential surface we see that the products; H2 + F are higher in energy than the product F-H this suggests that the F-H bond is stronger than the H-H bond. This reaction is exothermic.
The reverse reaction must then be endothermic for H + HF

==Transition state position H2 + F ==
[[File:Energy and configuration for h2 + F.PNG]]

To find the transition state, initial conditions were set and momentum is set to 0. From this it is clear that it isnt optimised as there are forces present between HH and F.

Setting HH distance to 0.744 and HF distance to 1.811, the forces are approximately 0 and so this is where the transition state exists.

== Activation Energy H2 + F ==
[[File:Energy v time Activation energy for Fh2.PNG]]

Looking at the energy v time graph the difference between the total energy and the potential energy results in the activation

Activation energy = 104.019 - 103.749 = 0.270

==Transition state position HF + H ==

[[File:Ts reverse 0.74.PNG]]

Transition state occurs at 0.74 A this resembles the H-H bond length in molecular H2

==Activation E HF + H ==

30

===F + H2 reaction trajectory: ===

[[File:Config for animation for h2 + f reaction path.PNG]]

These are the initial conditions selected that gave a reaction trajectory

[[File:Conservation of energy H2 +F.PNG]]

The momentum v time plot shows B-C (H-H) vibrating as F has little momentum. Once reaction happens, BC momentum stops oscillating and AB (HF) starts oscillating showing the conservation of energy as the vibration transfers from HH to HF.
The mechanism of release:

rHH=0.74, rHF=2 and pHF = -0.5 for the following table. The pHH ranges from -3 and 3 to see if a reaction occurs with these initial conditions.
{| class="wikitable" border=1
! pHH !! Reaction?!! Contour Plot!
|-
| 3.0 || No|| [[File:H2+F, p=3.PNG]]
|-
| 2.0 || No|| [[File:H2+F,p=2.PNG]]
|-
| 1.0 || No|| [[File:H2+F,p=1.PNG]]
|-
| 0 ||No || [[File:H2+F,p=0.PNG]]
|-
| -1.0 || No||[[File:H2+F,p=-1.PNG]]
|-
| -2.0 ||Yes || [[File:H2+F,p=-2.PNG]]
|-
| -3.0 || No|| [[File:H2+F, p=-3.PNG]]
|}

For F having momentum -0.5, reaction only occurs when the HH bond has -2 momentum. For reaction to complete is not enough to have a large energy in the system. Each both H2 and F require specific energy in order to react with each other therefore changing either energy values will require the other species to have a complimentary energy.

[[File:H2+F,p=-0.8,0.1.PNG]]
Reaction occurs even when H2 has a small momentum. F having an increased momentum allowed this reaction to occur therefore changes to F compared to changes in momentum for H2 have a greater weighting on the progress of the reaction.

===H + HF reaction trajectory: ===

Setting the conditions:
r1 = 0.94, r2 = 2
{| class="wikitable" border=1
! pHH !! pHF !! Etot !! Description of the dynamics !! Illustration of the trajectory
|-
| -10 || -0.5 || -85.6 || No reaction. The system does reach the transition state before falling back to the products||[[File:-10,-0.5.PNG]]
|-
| -8 || -1 || -106.8 ||No reaction. H approach increases vibration frequency of HF ||[[File:-8,-1.PNG]]|
|-
| -1 || -12 ||-68.7 || Goes to transition state but does not react ||[[File:HF+H,P=-1,-12.PNG]]|
|-
|-1|| -13 ||-56.5 ||High HF vibrations cause reaction to occur. || [[File:HF+H reaction.PNG]]||
|}

[[File:HF+H reaction.PNG]]

Polanyi's empirical rules state there is a linear relationship between the activation energy and the position of the transition state. For H + HF there is a late transition state and therefore a higher Ea is needed compared to H2 + F. For late transition states the reactants require high vibration energy, which can be changed by altering the momentum. For other high energy reactions for HF + H the reaction was unsuccessful due to the composition of momentum being vibrational and translation energy. For the unsuccessful reactions the composition of vibrational and translation energy did not have enought vibrational energy for the reaction to go past the TS. On the other hand early stage TS need more translation energy.

File:Ts reverse 0.74.PNG

2019-05-24T16:21:17Z

Ld2416:

MRD:ld2416

2019-05-24T16:18:05Z

Ld2416: /* H + HF reaction trajectory: */

==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?
[[File:Contour plot opf H H2.PNG]]

The transition state is when the H2 is equally distant to H1 and H3, for the contour plot this is approximately 0.93 A.

[[File:SUrface plot.PNG]]

The transition state can be seen at the saddle point, where the transition state is a minimum for the frank condon plot for the reaction. The transition state is a maximum for the reaction trajectory.
Mathematically, the TS has a second derivative greater than 0 in the Q1 direction (minimum) and TS has a second derivative less than 0 in the Q2 direction (maximum).

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.

From the contour plot we see that the transition state occurs when the r1=r2. Therefore to estimate the transition state position distances AB must be equal to BC.

[[File:Ab DIS.PNG]]

From this we can estimate the transition position from the contour plot:
[[File:Contour TS.PNG]]

This gives a value of 0.9075 A

Looking at this on a distance/time graph:
[[File:Bad ds TS.PNG]]

The momentum is set to 0 because the transition state is stationary and the distance between the 2 are equal.

Dynamic Surface:
[[File:Dynamic surface.PNG]]

MEP surface diagram:

[[File:MEP surface.PNG]]

Differences can be seen in the length and oscillations in the reaction trajectory. For the minimum energy pathway (MEP) there is no oscillations as there it neglects the molecules vibration modes and it moves infinitely slowly. The line is shorter as the reaction coordinate isn't as long due to the reaction.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 ||-99.0|| Yes||H2 initially not vibrating, upon reaction the new H2 has a high vibrational energy||[[File:P=-1.25,-2.5.PNG]]|
|-
| -1.5 || -2.0 ||-100.5||No|| H approaching is repelled, causing the HH bond to vibrate ||[[File:P=-1.5,-2.PNG]]|
|-
| -1.5 || -2.5 ||-99.0||Yes||Slow approach of H and H2 after reaction stronger vibrations||[[File:P=-1.5,-2.5.PNG]]|
|-
| -2.5 || -5.0 ||-85.0||No||H approaches H2 causes a the H-H bond to weaken almost up to a transition state, not enough energy to form a new HH bond| |[[File:P=-2.5,-5.PNG]]|
|-
| -2.5 || -5.2 ||-83.4||Yes||H on approach caused increase in vibration for H2 bond until it broke and new HH bond formed with the approaching H||[[File:P=-2.5,-5.2.PNG]]|
|}

For the penultimate conditions, the reaction reaches the transition state yet its still falls back to the reactants despite the high momentum from the approaching H. The difference for the reaction to occur is the momentum to the approaching H to the molecule. It can be seen that the both the species in the reactants need the required momentum in order to reach transition state and to form the products.

== F-H-H system ==

H2 + F potential energy surface:
[[File:H2 + F potential surface.PNG]]

Hf + H potential surface:
[[File:HF+H surface plot.PNG]]

From the potential surface we see that the products; H2 + F are higher in energy than the product F-H this suggests that the F-H bond is stronger than the H-H bond. This reaction is exothermic.
The reverse reaction must then be endothermic for H + HF

==Transition state position H2 + F ==
[[File:Energy and configuration for h2 + F.PNG]]

To find the transition state, initial conditions were set and momentum is set to 0. From this it is clear that it isnt optimised as there are forces present between HH and F.

Setting HH distance to 0.744 and HF distance to 1.811, the forces are approximately 0 and so this is where the transition state exists.

== Activation Energy H2 + F ==
[[File:Energy v time Activation energy for Fh2.PNG]]

Looking at the energy v time graph the difference between the total energy and the potential energy results in the activation

Activation energy = 104.019 - 103.749 = 0.270

==Transition state position HF + H ==
0.74 this resembles the H-H bond length in molecular H2
==Activation E HF + H ==

30

===F + H2 reaction trajectory: ===

[[File:Config for animation for h2 + f reaction path.PNG]]

These are the initial conditions selected that gave a reaction trajectory

[[File:Conservation of energy H2 +F.PNG]]

The momentum v time plot shows B-C (H-H) vibrating as F has little momentum. Once reaction happens, BC momentum stops oscillating and AB (HF) starts oscillating showing the conservation of energy as the vibration transfers from HH to HF.
The mechanism of release:

rHH=0.74, rHF=2 and pHF = -0.5 for the following table. The pHH ranges from -3 and 3 to see if a reaction occurs with these initial conditions.
{| class="wikitable" border=1
! pHH !! Reaction?!! Contour Plot!
|-
| 3.0 || No|| [[File:H2+F, p=3.PNG]]
|-
| 2.0 || No|| [[File:H2+F,p=2.PNG]]
|-
| 1.0 || No|| [[File:H2+F,p=1.PNG]]
|-
| 0 ||No || [[File:H2+F,p=0.PNG]]
|-
| -1.0 || No||[[File:H2+F,p=-1.PNG]]
|-
| -2.0 ||Yes || [[File:H2+F,p=-2.PNG]]
|-
| -3.0 || No|| [[File:H2+F, p=-3.PNG]]
|}

For F having momentum -0.5, reaction only occurs when the HH bond has -2 momentum. For reaction to complete is not enough to have a large energy in the system. Each both H2 and F require specific energy in order to react with each other therefore changing either energy values will require the other species to have a complimentary energy.

[[File:H2+F,p=-0.8,0.1.PNG]]
Reaction occurs even when H2 has a small momentum. F having an increased momentum allowed this reaction to occur therefore changes to F compared to changes in momentum for H2 have a greater weighting on the progress of the reaction.

===H + HF reaction trajectory: ===

Setting the conditions:
r1 = 0.94, r2 = 2
{| class="wikitable" border=1
! pHH !! pHF !! Etot !! Description of the dynamics !! Illustration of the trajectory
|-
| -10 || -0.5 || -85.6 || No reaction. The system does reach the transition state before falling back to the products||[[File:-10,-0.5.PNG]]
|-
| -8 || -1 || -106.8 ||No reaction. H approach increases vibration frequency of HF ||[[File:-8,-1.PNG]]|
|-
| -1 || -12 ||-68.7 || Goes to transition state but does not react ||[[File:HF+H,P=-1,-12.PNG]]|
|-
|-1|| -13 ||-56.5 ||High HF vibrations cause reaction to occur. || [[File:HF+H reaction.PNG]]||
|}

[[File:HF+H reaction.PNG]]

Polanyi's empirical rules state there is a linear relationship between the activation energy and the position of the transition state. For H + HF there is a late transition state and therefore a higher Ea is needed compared to H2 + F. For late transition states the reactants require high vibration energy, which can be changed by altering the momentum. For other high energy reactions for HF + H the reaction was unsuccessful due to the composition of momentum being vibrational and translation energy. For the unsuccessful reactions the composition of vibrational and translation energy did not have enought vibrational energy for the reaction to go past the TS. On the other hand early stage TS need more translation energy.

MRD:ld2416

2019-05-24T16:11:57Z

Ld2416: /* H + HF reaction trajectory: */

==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?
[[File:Contour plot opf H H2.PNG]]

The transition state is when the H2 is equally distant to H1 and H3, for the contour plot this is approximately 0.93 A.

[[File:SUrface plot.PNG]]

The transition state can be seen at the saddle point, where the transition state is a minimum for the frank condon plot for the reaction. The transition state is a maximum for the reaction trajectory.
Mathematically, the TS has a second derivative greater than 0 in the Q1 direction (minimum) and TS has a second derivative less than 0 in the Q2 direction (maximum).

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.

From the contour plot we see that the transition state occurs when the r1=r2. Therefore to estimate the transition state position distances AB must be equal to BC.

[[File:Ab DIS.PNG]]

From this we can estimate the transition position from the contour plot:
[[File:Contour TS.PNG]]

This gives a value of 0.9075 A

Looking at this on a distance/time graph:
[[File:Bad ds TS.PNG]]

The momentum is set to 0 because the transition state is stationary and the distance between the 2 are equal.

Dynamic Surface:
[[File:Dynamic surface.PNG]]

MEP surface diagram:

[[File:MEP surface.PNG]]

Differences can be seen in the length and oscillations in the reaction trajectory. For the minimum energy pathway (MEP) there is no oscillations as there it neglects the molecules vibration modes and it moves infinitely slowly. The line is shorter as the reaction coordinate isn't as long due to the reaction.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 ||-99.0|| Yes||H2 initially not vibrating, upon reaction the new H2 has a high vibrational energy||[[File:P=-1.25,-2.5.PNG]]|
|-
| -1.5 || -2.0 ||-100.5||No|| H approaching is repelled, causing the HH bond to vibrate ||[[File:P=-1.5,-2.PNG]]|
|-
| -1.5 || -2.5 ||-99.0||Yes||Slow approach of H and H2 after reaction stronger vibrations||[[File:P=-1.5,-2.5.PNG]]|
|-
| -2.5 || -5.0 ||-85.0||No||H approaches H2 causes a the H-H bond to weaken almost up to a transition state, not enough energy to form a new HH bond| |[[File:P=-2.5,-5.PNG]]|
|-
| -2.5 || -5.2 ||-83.4||Yes||H on approach caused increase in vibration for H2 bond until it broke and new HH bond formed with the approaching H||[[File:P=-2.5,-5.2.PNG]]|
|}

For the penultimate conditions, the reaction reaches the transition state yet its still falls back to the reactants despite the high momentum from the approaching H. The difference for the reaction to occur is the momentum to the approaching H to the molecule. It can be seen that the both the species in the reactants need the required momentum in order to reach transition state and to form the products.

== F-H-H system ==

H2 + F potential energy surface:
[[File:H2 + F potential surface.PNG]]

Hf + H potential surface:
[[File:HF+H surface plot.PNG]]

From the potential surface we see that the products; H2 + F are higher in energy than the product F-H this suggests that the F-H bond is stronger than the H-H bond. This reaction is exothermic.
The reverse reaction must then be endothermic for H + HF

==Transition state position H2 + F ==
[[File:Energy and configuration for h2 + F.PNG]]

To find the transition state, initial conditions were set and momentum is set to 0. From this it is clear that it isnt optimised as there are forces present between HH and F.

Setting HH distance to 0.744 and HF distance to 1.811, the forces are approximately 0 and so this is where the transition state exists.

== Activation Energy H2 + F ==
[[File:Energy v time Activation energy for Fh2.PNG]]

Looking at the energy v time graph the difference between the total energy and the potential energy results in the activation

Activation energy = 104.019 - 103.749 = 0.270

==Transition state position HF + H ==
0.74 this resembles the H-H bond length in molecular H2
==Activation E HF + H ==

30

===F + H2 reaction trajectory: ===

[[File:Config for animation for h2 + f reaction path.PNG]]

These are the initial conditions selected that gave a reaction trajectory

[[File:Conservation of energy H2 +F.PNG]]

The momentum v time plot shows B-C (H-H) vibrating as F has little momentum. Once reaction happens, BC momentum stops oscillating and AB (HF) starts oscillating showing the conservation of energy as the vibration transfers from HH to HF.
The mechanism of release:

rHH=0.74, rHF=2 and pHF = -0.5 for the following table. The pHH ranges from -3 and 3 to see if a reaction occurs with these initial conditions.
{| class="wikitable" border=1
! pHH !! Reaction?!! Contour Plot!
|-
| 3.0 || No|| [[File:H2+F, p=3.PNG]]
|-
| 2.0 || No|| [[File:H2+F,p=2.PNG]]
|-
| 1.0 || No|| [[File:H2+F,p=1.PNG]]
|-
| 0 ||No || [[File:H2+F,p=0.PNG]]
|-
| -1.0 || No||[[File:H2+F,p=-1.PNG]]
|-
| -2.0 ||Yes || [[File:H2+F,p=-2.PNG]]
|-
| -3.0 || No|| [[File:H2+F, p=-3.PNG]]
|}

For F having momentum -0.5, reaction only occurs when the HH bond has -2 momentum. For reaction to complete is not enough to have a large energy in the system. Each both H2 and F require specific energy in order to react with each other therefore changing either energy values will require the other species to have a complimentary energy.

[[File:H2+F,p=-0.8,0.1.PNG]]
Reaction occurs even when H2 has a small momentum. F having an increased momentum allowed this reaction to occur therefore changes to F compared to changes in momentum for H2 have a greater weighting on the progress of the reaction.

===H + HF reaction trajectory: ===

Setting the conditions:
r1 = 0.94, r2 = 2
{| class="wikitable" border=1
! pHH !! pHF !! Etot !! Description of the dynamics !! Illustration of the trajectory
|-
| -10 || -0.5 || -85.6 || No reaction. The system does reach the transition state before falling back to the products||[[File:-10,-0.5.PNG]]
|-
| -8 || -1 || -106.8 ||No reaction. H approach increases vibration frequency of HF ||[[File:-8,-1.PNG]]|
|-
| -1 || -12 ||-68.7 || Goes to transition state but does not react ||[[File:HF+H,P=-1,-12.PNG]]|
|-
|-1|| -13 ||-56.5 ||High HF vibrations cause reaction to occur. || [[File:HF+H reaction.PNG]]||
|}

High HF vibrations cause reaction to occur. || [[File:HF+H reaction.PNG]]

Polanyi's empirical rules state there is a linear relationship between the activation energy and the position of the transition state. For H + HF there is a late transition state and therefore a higher Ea is needed compared to H2 + F. For late transition states the reactants require high vibration energy, which can be changed by altering the momentum.

MRD:ld2416

2019-05-24T16:05:51Z

Ld2416: /* H + HF reaction trajectory: */

==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?
[[File:Contour plot opf H H2.PNG]]

The transition state is when the H2 is equally distant to H1 and H3, for the contour plot this is approximately 0.93 A.

[[File:SUrface plot.PNG]]

The transition state can be seen at the saddle point, where the transition state is a minimum for the frank condon plot for the reaction. The transition state is a maximum for the reaction trajectory.
Mathematically, the TS has a second derivative greater than 0 in the Q1 direction (minimum) and TS has a second derivative less than 0 in the Q2 direction (maximum).

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.

From the contour plot we see that the transition state occurs when the r1=r2. Therefore to estimate the transition state position distances AB must be equal to BC.

[[File:Ab DIS.PNG]]

From this we can estimate the transition position from the contour plot:
[[File:Contour TS.PNG]]

This gives a value of 0.9075 A

Looking at this on a distance/time graph:
[[File:Bad ds TS.PNG]]

The momentum is set to 0 because the transition state is stationary and the distance between the 2 are equal.

Dynamic Surface:
[[File:Dynamic surface.PNG]]

MEP surface diagram:

[[File:MEP surface.PNG]]

Differences can be seen in the length and oscillations in the reaction trajectory. For the minimum energy pathway (MEP) there is no oscillations as there it neglects the molecules vibration modes and it moves infinitely slowly. The line is shorter as the reaction coordinate isn't as long due to the reaction.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 ||-99.0|| Yes||H2 initially not vibrating, upon reaction the new H2 has a high vibrational energy||[[File:P=-1.25,-2.5.PNG]]|
|-
| -1.5 || -2.0 ||-100.5||No|| H approaching is repelled, causing the HH bond to vibrate ||[[File:P=-1.5,-2.PNG]]|
|-
| -1.5 || -2.5 ||-99.0||Yes||Slow approach of H and H2 after reaction stronger vibrations||[[File:P=-1.5,-2.5.PNG]]|
|-
| -2.5 || -5.0 ||-85.0||No||H approaches H2 causes a the H-H bond to weaken almost up to a transition state, not enough energy to form a new HH bond| |[[File:P=-2.5,-5.PNG]]|
|-
| -2.5 || -5.2 ||-83.4||Yes||H on approach caused increase in vibration for H2 bond until it broke and new HH bond formed with the approaching H||[[File:P=-2.5,-5.2.PNG]]|
|}

For the penultimate conditions, the reaction reaches the transition state yet its still falls back to the reactants despite the high momentum from the approaching H. The difference for the reaction to occur is the momentum to the approaching H to the molecule. It can be seen that the both the species in the reactants need the required momentum in order to reach transition state and to form the products.

== F-H-H system ==

H2 + F potential energy surface:
[[File:H2 + F potential surface.PNG]]

Hf + H potential surface:
[[File:HF+H surface plot.PNG]]

From the potential surface we see that the products; H2 + F are higher in energy than the product F-H this suggests that the F-H bond is stronger than the H-H bond. This reaction is exothermic.
The reverse reaction must then be endothermic for H + HF

==Transition state position H2 + F ==
[[File:Energy and configuration for h2 + F.PNG]]

To find the transition state, initial conditions were set and momentum is set to 0. From this it is clear that it isnt optimised as there are forces present between HH and F.

Setting HH distance to 0.744 and HF distance to 1.811, the forces are approximately 0 and so this is where the transition state exists.

== Activation Energy H2 + F ==
[[File:Energy v time Activation energy for Fh2.PNG]]

Looking at the energy v time graph the difference between the total energy and the potential energy results in the activation

Activation energy = 104.019 - 103.749 = 0.270

==Transition state position HF + H ==
0.74 this resembles the H-H bond length in molecular H2
==Activation E HF + H ==

30

===F + H2 reaction trajectory: ===

[[File:Config for animation for h2 + f reaction path.PNG]]

These are the initial conditions selected that gave a reaction trajectory

[[File:Conservation of energy H2 +F.PNG]]

The momentum v time plot shows B-C (H-H) vibrating as F has little momentum. Once reaction happens, BC momentum stops oscillating and AB (HF) starts oscillating showing the conservation of energy as the vibration transfers from HH to HF.
The mechanism of release:

rHH=0.74, rHF=2 and pHF = -0.5 for the following table. The pHH ranges from -3 and 3 to see if a reaction occurs with these initial conditions.
{| class="wikitable" border=1
! pHH !! Reaction?!! Contour Plot!
|-
| 3.0 || No|| [[File:H2+F, p=3.PNG]]
|-
| 2.0 || No|| [[File:H2+F,p=2.PNG]]
|-
| 1.0 || No|| [[File:H2+F,p=1.PNG]]
|-
| 0 ||No || [[File:H2+F,p=0.PNG]]
|-
| -1.0 || No||[[File:H2+F,p=-1.PNG]]
|-
| -2.0 ||Yes || [[File:H2+F,p=-2.PNG]]
|-
| -3.0 || No|| [[File:H2+F, p=-3.PNG]]
|}

For F having momentum -0.5, reaction only occurs when the HH bond has -2 momentum. For reaction to complete is not enough to have a large energy in the system. Each both H2 and F require specific energy in order to react with each other therefore changing either energy values will require the other species to have a complimentary energy.

[[File:H2+F,p=-0.8,0.1.PNG]]
Reaction occurs even when H2 has a small momentum. F having an increased momentum allowed this reaction to occur therefore changes to F compared to changes in momentum for H2 have a greater weighting on the progress of the reaction.

===H + HF reaction trajectory: ===

Setting the conditions:
r1 = 0.94, r2 = 2
{| class="wikitable" border=1
! pHH !! pHF !! Etot !! Description of the dynamics !! Illustration of the trajectory
|-
| -10 || -0.5 || -85.6 || No reaction. The system does reach the transition state before falling back to the products||[[File:-10,-0.5.PNG]]
|-
| -8 || -1 || -106.8 ||No reaction. H approach increases vibration frequency of HF ||[[File:-8,-1.PNG]]|
|-
| -1 || -12 ||-68.7 || Goes to transition state but does not react ||[[File:HF+H,P=-1,-12.PNG]]|
|-
|-1|| -13 ||-56.5 ||High HF vibrations cause reaction to occur. || [[File:HF+H reaction.PNG]]||
|}

Polanyi's empirical rules

MRD:ld2416

2019-05-24T16:05:16Z

Ld2416: /* H + HF reaction trajectory: */

==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?
[[File:Contour plot opf H H2.PNG]]

The transition state is when the H2 is equally distant to H1 and H3, for the contour plot this is approximately 0.93 A.

[[File:SUrface plot.PNG]]

The transition state can be seen at the saddle point, where the transition state is a minimum for the frank condon plot for the reaction. The transition state is a maximum for the reaction trajectory.
Mathematically, the TS has a second derivative greater than 0 in the Q1 direction (minimum) and TS has a second derivative less than 0 in the Q2 direction (maximum).

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.

From the contour plot we see that the transition state occurs when the r1=r2. Therefore to estimate the transition state position distances AB must be equal to BC.

[[File:Ab DIS.PNG]]

From this we can estimate the transition position from the contour plot:
[[File:Contour TS.PNG]]

This gives a value of 0.9075 A

Looking at this on a distance/time graph:
[[File:Bad ds TS.PNG]]

The momentum is set to 0 because the transition state is stationary and the distance between the 2 are equal.

Dynamic Surface:
[[File:Dynamic surface.PNG]]

MEP surface diagram:

[[File:MEP surface.PNG]]

Differences can be seen in the length and oscillations in the reaction trajectory. For the minimum energy pathway (MEP) there is no oscillations as there it neglects the molecules vibration modes and it moves infinitely slowly. The line is shorter as the reaction coordinate isn't as long due to the reaction.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 ||-99.0|| Yes||H2 initially not vibrating, upon reaction the new H2 has a high vibrational energy||[[File:P=-1.25,-2.5.PNG]]|
|-
| -1.5 || -2.0 ||-100.5||No|| H approaching is repelled, causing the HH bond to vibrate ||[[File:P=-1.5,-2.PNG]]|
|-
| -1.5 || -2.5 ||-99.0||Yes||Slow approach of H and H2 after reaction stronger vibrations||[[File:P=-1.5,-2.5.PNG]]|
|-
| -2.5 || -5.0 ||-85.0||No||H approaches H2 causes a the H-H bond to weaken almost up to a transition state, not enough energy to form a new HH bond| |[[File:P=-2.5,-5.PNG]]|
|-
| -2.5 || -5.2 ||-83.4||Yes||H on approach caused increase in vibration for H2 bond until it broke and new HH bond formed with the approaching H||[[File:P=-2.5,-5.2.PNG]]|
|}

For the penultimate conditions, the reaction reaches the transition state yet its still falls back to the reactants despite the high momentum from the approaching H. The difference for the reaction to occur is the momentum to the approaching H to the molecule. It can be seen that the both the species in the reactants need the required momentum in order to reach transition state and to form the products.

== F-H-H system ==

H2 + F potential energy surface:
[[File:H2 + F potential surface.PNG]]

Hf + H potential surface:
[[File:HF+H surface plot.PNG]]

From the potential surface we see that the products; H2 + F are higher in energy than the product F-H this suggests that the F-H bond is stronger than the H-H bond. This reaction is exothermic.
The reverse reaction must then be endothermic for H + HF

==Transition state position H2 + F ==
[[File:Energy and configuration for h2 + F.PNG]]

To find the transition state, initial conditions were set and momentum is set to 0. From this it is clear that it isnt optimised as there are forces present between HH and F.

Setting HH distance to 0.744 and HF distance to 1.811, the forces are approximately 0 and so this is where the transition state exists.

== Activation Energy H2 + F ==
[[File:Energy v time Activation energy for Fh2.PNG]]

Looking at the energy v time graph the difference between the total energy and the potential energy results in the activation

Activation energy = 104.019 - 103.749 = 0.270

==Transition state position HF + H ==
0.74 this resembles the H-H bond length in molecular H2
==Activation E HF + H ==

30

===F + H2 reaction trajectory: ===

[[File:Config for animation for h2 + f reaction path.PNG]]

These are the initial conditions selected that gave a reaction trajectory

[[File:Conservation of energy H2 +F.PNG]]

The momentum v time plot shows B-C (H-H) vibrating as F has little momentum. Once reaction happens, BC momentum stops oscillating and AB (HF) starts oscillating showing the conservation of energy as the vibration transfers from HH to HF.
The mechanism of release:

rHH=0.74, rHF=2 and pHF = -0.5 for the following table. The pHH ranges from -3 and 3 to see if a reaction occurs with these initial conditions.
{| class="wikitable" border=1
! pHH !! Reaction?!! Contour Plot!
|-
| 3.0 || No|| [[File:H2+F, p=3.PNG]]
|-
| 2.0 || No|| [[File:H2+F,p=2.PNG]]
|-
| 1.0 || No|| [[File:H2+F,p=1.PNG]]
|-
| 0 ||No || [[File:H2+F,p=0.PNG]]
|-
| -1.0 || No||[[File:H2+F,p=-1.PNG]]
|-
| -2.0 ||Yes || [[File:H2+F,p=-2.PNG]]
|-
| -3.0 || No|| [[File:H2+F, p=-3.PNG]]
|}

For F having momentum -0.5, reaction only occurs when the HH bond has -2 momentum. For reaction to complete is not enough to have a large energy in the system. Each both H2 and F require specific energy in order to react with each other therefore changing either energy values will require the other species to have a complimentary energy.

[[File:H2+F,p=-0.8,0.1.PNG]]
Reaction occurs even when H2 has a small momentum. F having an increased momentum allowed this reaction to occur therefore changes to F compared to changes in momentum for H2 have a greater weighting on the progress of the reaction.

===H + HF reaction trajectory: ===

Setting the conditions:
r1 = 0.94, r2 = 2
{| class="wikitable" border=1
! pHH !! pHF !! Etot !! Description of the dynamics !! Illustration of the trajectory
|-
| -10 || -0.5 || -85.6 || No reaction. The system does reach the transition state before falling back to the products||[[File:-10,-0.5.PNG]]
|-
| -8 || -1 || -106.8 ||No reaction. H approach increases vibration frequency of HF ||[[File:-8,-1.PNG]]|
|-
| -1 || -12 ||-68.7 || Goes to transition state but does not react ||[[File:HF+H,P=-1,-12.PNG]]|
|-
|-1|| -13 ||-56.5 ||High HF vibrations cause reaction to occur. || [[File:HF+H reaction.PNG]]||
|-

|}

Polanyi's empirical rules

MRD:ld2416

2019-05-24T16:05:02Z

Ld2416: /* H + HF reaction trajectory: */

==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?
[[File:Contour plot opf H H2.PNG]]

The transition state is when the H2 is equally distant to H1 and H3, for the contour plot this is approximately 0.93 A.

[[File:SUrface plot.PNG]]

The transition state can be seen at the saddle point, where the transition state is a minimum for the frank condon plot for the reaction. The transition state is a maximum for the reaction trajectory.
Mathematically, the TS has a second derivative greater than 0 in the Q1 direction (minimum) and TS has a second derivative less than 0 in the Q2 direction (maximum).

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.

From the contour plot we see that the transition state occurs when the r1=r2. Therefore to estimate the transition state position distances AB must be equal to BC.

[[File:Ab DIS.PNG]]

From this we can estimate the transition position from the contour plot:
[[File:Contour TS.PNG]]

This gives a value of 0.9075 A

Looking at this on a distance/time graph:
[[File:Bad ds TS.PNG]]

The momentum is set to 0 because the transition state is stationary and the distance between the 2 are equal.

Dynamic Surface:
[[File:Dynamic surface.PNG]]

MEP surface diagram:

[[File:MEP surface.PNG]]

Differences can be seen in the length and oscillations in the reaction trajectory. For the minimum energy pathway (MEP) there is no oscillations as there it neglects the molecules vibration modes and it moves infinitely slowly. The line is shorter as the reaction coordinate isn't as long due to the reaction.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 ||-99.0|| Yes||H2 initially not vibrating, upon reaction the new H2 has a high vibrational energy||[[File:P=-1.25,-2.5.PNG]]|
|-
| -1.5 || -2.0 ||-100.5||No|| H approaching is repelled, causing the HH bond to vibrate ||[[File:P=-1.5,-2.PNG]]|
|-
| -1.5 || -2.5 ||-99.0||Yes||Slow approach of H and H2 after reaction stronger vibrations||[[File:P=-1.5,-2.5.PNG]]|
|-
| -2.5 || -5.0 ||-85.0||No||H approaches H2 causes a the H-H bond to weaken almost up to a transition state, not enough energy to form a new HH bond| |[[File:P=-2.5,-5.PNG]]|
|-
| -2.5 || -5.2 ||-83.4||Yes||H on approach caused increase in vibration for H2 bond until it broke and new HH bond formed with the approaching H||[[File:P=-2.5,-5.2.PNG]]|
|}

For the penultimate conditions, the reaction reaches the transition state yet its still falls back to the reactants despite the high momentum from the approaching H. The difference for the reaction to occur is the momentum to the approaching H to the molecule. It can be seen that the both the species in the reactants need the required momentum in order to reach transition state and to form the products.

== F-H-H system ==

H2 + F potential energy surface:
[[File:H2 + F potential surface.PNG]]

Hf + H potential surface:
[[File:HF+H surface plot.PNG]]

From the potential surface we see that the products; H2 + F are higher in energy than the product F-H this suggests that the F-H bond is stronger than the H-H bond. This reaction is exothermic.
The reverse reaction must then be endothermic for H + HF

==Transition state position H2 + F ==
[[File:Energy and configuration for h2 + F.PNG]]

To find the transition state, initial conditions were set and momentum is set to 0. From this it is clear that it isnt optimised as there are forces present between HH and F.

Setting HH distance to 0.744 and HF distance to 1.811, the forces are approximately 0 and so this is where the transition state exists.

== Activation Energy H2 + F ==
[[File:Energy v time Activation energy for Fh2.PNG]]

Looking at the energy v time graph the difference between the total energy and the potential energy results in the activation

Activation energy = 104.019 - 103.749 = 0.270

==Transition state position HF + H ==
0.74 this resembles the H-H bond length in molecular H2
==Activation E HF + H ==

30

===F + H2 reaction trajectory: ===

[[File:Config for animation for h2 + f reaction path.PNG]]

These are the initial conditions selected that gave a reaction trajectory

[[File:Conservation of energy H2 +F.PNG]]

The momentum v time plot shows B-C (H-H) vibrating as F has little momentum. Once reaction happens, BC momentum stops oscillating and AB (HF) starts oscillating showing the conservation of energy as the vibration transfers from HH to HF.
The mechanism of release:

rHH=0.74, rHF=2 and pHF = -0.5 for the following table. The pHH ranges from -3 and 3 to see if a reaction occurs with these initial conditions.
{| class="wikitable" border=1
! pHH !! Reaction?!! Contour Plot!
|-
| 3.0 || No|| [[File:H2+F, p=3.PNG]]
|-
| 2.0 || No|| [[File:H2+F,p=2.PNG]]
|-
| 1.0 || No|| [[File:H2+F,p=1.PNG]]
|-
| 0 ||No || [[File:H2+F,p=0.PNG]]
|-
| -1.0 || No||[[File:H2+F,p=-1.PNG]]
|-
| -2.0 ||Yes || [[File:H2+F,p=-2.PNG]]
|-
| -3.0 || No|| [[File:H2+F, p=-3.PNG]]
|}

For F having momentum -0.5, reaction only occurs when the HH bond has -2 momentum. For reaction to complete is not enough to have a large energy in the system. Each both H2 and F require specific energy in order to react with each other therefore changing either energy values will require the other species to have a complimentary energy.

[[File:H2+F,p=-0.8,0.1.PNG]]
Reaction occurs even when H2 has a small momentum. F having an increased momentum allowed this reaction to occur therefore changes to F compared to changes in momentum for H2 have a greater weighting on the progress of the reaction.

===H + HF reaction trajectory: ===

Setting the conditions:
r1 = 0.94, r2 = 2
{| class="wikitable" border=1
! pHH !! pHF !! Etot !! Description of the dynamics !! Illustration of the trajectory
|-
| -10 || -0.5 || -85.6 || No reaction. The system does reach the transition state before falling back to the products||[[File:-10,-0.5.PNG]]|
|-
| -8 || -1 || -106.8 ||No reaction. H approach increases vibration frequency of HF ||[[File:-8,-1.PNG]]|
|-
| -1 || -12 ||-68.7 || Goes to transition state but does not react ||[[File:HF+H,P=-1,-12.PNG]]||
|-
|-1|| -13 ||-56.5 ||High HF vibrations cause reaction to occur. || [[File:HF+H reaction.PNG]]||
|-

|}

Polanyi's empirical rules

MRD:ld2416

2019-05-24T16:02:56Z

Ld2416: /* H + HF reaction trajectory: */

==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?
[[File:Contour plot opf H H2.PNG]]

The transition state is when the H2 is equally distant to H1 and H3, for the contour plot this is approximately 0.93 A.

[[File:SUrface plot.PNG]]

The transition state can be seen at the saddle point, where the transition state is a minimum for the frank condon plot for the reaction. The transition state is a maximum for the reaction trajectory.
Mathematically, the TS has a second derivative greater than 0 in the Q1 direction (minimum) and TS has a second derivative less than 0 in the Q2 direction (maximum).

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.

From the contour plot we see that the transition state occurs when the r1=r2. Therefore to estimate the transition state position distances AB must be equal to BC.

[[File:Ab DIS.PNG]]

From this we can estimate the transition position from the contour plot:
[[File:Contour TS.PNG]]

This gives a value of 0.9075 A

Looking at this on a distance/time graph:
[[File:Bad ds TS.PNG]]

The momentum is set to 0 because the transition state is stationary and the distance between the 2 are equal.

Dynamic Surface:
[[File:Dynamic surface.PNG]]

MEP surface diagram:

[[File:MEP surface.PNG]]

Differences can be seen in the length and oscillations in the reaction trajectory. For the minimum energy pathway (MEP) there is no oscillations as there it neglects the molecules vibration modes and it moves infinitely slowly. The line is shorter as the reaction coordinate isn't as long due to the reaction.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 ||-99.0|| Yes||H2 initially not vibrating, upon reaction the new H2 has a high vibrational energy||[[File:P=-1.25,-2.5.PNG]]|
|-
| -1.5 || -2.0 ||-100.5||No|| H approaching is repelled, causing the HH bond to vibrate ||[[File:P=-1.5,-2.PNG]]|
|-
| -1.5 || -2.5 ||-99.0||Yes||Slow approach of H and H2 after reaction stronger vibrations||[[File:P=-1.5,-2.5.PNG]]|
|-
| -2.5 || -5.0 ||-85.0||No||H approaches H2 causes a the H-H bond to weaken almost up to a transition state, not enough energy to form a new HH bond| |[[File:P=-2.5,-5.PNG]]|
|-
| -2.5 || -5.2 ||-83.4||Yes||H on approach caused increase in vibration for H2 bond until it broke and new HH bond formed with the approaching H||[[File:P=-2.5,-5.2.PNG]]|
|}

For the penultimate conditions, the reaction reaches the transition state yet its still falls back to the reactants despite the high momentum from the approaching H. The difference for the reaction to occur is the momentum to the approaching H to the molecule. It can be seen that the both the species in the reactants need the required momentum in order to reach transition state and to form the products.

== F-H-H system ==

H2 + F potential energy surface:
[[File:H2 + F potential surface.PNG]]

Hf + H potential surface:
[[File:HF+H surface plot.PNG]]

From the potential surface we see that the products; H2 + F are higher in energy than the product F-H this suggests that the F-H bond is stronger than the H-H bond. This reaction is exothermic.
The reverse reaction must then be endothermic for H + HF

==Transition state position H2 + F ==
[[File:Energy and configuration for h2 + F.PNG]]

To find the transition state, initial conditions were set and momentum is set to 0. From this it is clear that it isnt optimised as there are forces present between HH and F.

Setting HH distance to 0.744 and HF distance to 1.811, the forces are approximately 0 and so this is where the transition state exists.

== Activation Energy H2 + F ==
[[File:Energy v time Activation energy for Fh2.PNG]]

Looking at the energy v time graph the difference between the total energy and the potential energy results in the activation

Activation energy = 104.019 - 103.749 = 0.270

==Transition state position HF + H ==
0.74 this resembles the H-H bond length in molecular H2
==Activation E HF + H ==

30

===F + H2 reaction trajectory: ===

[[File:Config for animation for h2 + f reaction path.PNG]]

These are the initial conditions selected that gave a reaction trajectory

[[File:Conservation of energy H2 +F.PNG]]

The momentum v time plot shows B-C (H-H) vibrating as F has little momentum. Once reaction happens, BC momentum stops oscillating and AB (HF) starts oscillating showing the conservation of energy as the vibration transfers from HH to HF.
The mechanism of release:

rHH=0.74, rHF=2 and pHF = -0.5 for the following table. The pHH ranges from -3 and 3 to see if a reaction occurs with these initial conditions.
{| class="wikitable" border=1
! pHH !! Reaction?!! Contour Plot!
|-
| 3.0 || No|| [[File:H2+F, p=3.PNG]]
|-
| 2.0 || No|| [[File:H2+F,p=2.PNG]]
|-
| 1.0 || No|| [[File:H2+F,p=1.PNG]]
|-
| 0 ||No || [[File:H2+F,p=0.PNG]]
|-
| -1.0 || No||[[File:H2+F,p=-1.PNG]]
|-
| -2.0 ||Yes || [[File:H2+F,p=-2.PNG]]
|-
| -3.0 || No|| [[File:H2+F, p=-3.PNG]]
|}

For F having momentum -0.5, reaction only occurs when the HH bond has -2 momentum. For reaction to complete is not enough to have a large energy in the system. Each both H2 and F require specific energy in order to react with each other therefore changing either energy values will require the other species to have a complimentary energy.

[[File:H2+F,p=-0.8,0.1.PNG]]
Reaction occurs even when H2 has a small momentum. F having an increased momentum allowed this reaction to occur therefore changes to F compared to changes in momentum for H2 have a greater weighting on the progress of the reaction.

===H + HF reaction trajectory: ===

Setting the conditions:
r1 = 0.94, r2 = 2
{| class="wikitable" border=1
! pHH !! pHF !! Etot !! Description of the dynamics !! Illustration of the trajectory
|-
| -10 || -0.5 || -85.6 || No reaction. The system does reach the transition state before falling back to the products||[[File:-10,-0.5.PNG]]|
|-
| -8 || -1 || -106.8 ||No reaction. H approach increases vibration frequency of HF ||[[File:-8,-1.PNG]]|
|-
| -1 || -12 ||-68.7 || Goes to transition state but does not react ||[[File:HF+H,P=-1,-12.PNG]]||
|-
|-1|| -13 ||-56.5 ||High HF vibrations cause reaction to occur. || [[File:HF+H reaction.PNG]]||
|-
|}

Polanyi's empirical rules

MRD:ld2416

2019-05-24T16:01:20Z

Ld2416: /* H + HF reaction trajectory: */

==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?
[[File:Contour plot opf H H2.PNG]]

The transition state is when the H2 is equally distant to H1 and H3, for the contour plot this is approximately 0.93 A.

[[File:SUrface plot.PNG]]

The transition state can be seen at the saddle point, where the transition state is a minimum for the frank condon plot for the reaction. The transition state is a maximum for the reaction trajectory.
Mathematically, the TS has a second derivative greater than 0 in the Q1 direction (minimum) and TS has a second derivative less than 0 in the Q2 direction (maximum).

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.

From the contour plot we see that the transition state occurs when the r1=r2. Therefore to estimate the transition state position distances AB must be equal to BC.

[[File:Ab DIS.PNG]]

From this we can estimate the transition position from the contour plot:
[[File:Contour TS.PNG]]

This gives a value of 0.9075 A

Looking at this on a distance/time graph:
[[File:Bad ds TS.PNG]]

The momentum is set to 0 because the transition state is stationary and the distance between the 2 are equal.

Dynamic Surface:
[[File:Dynamic surface.PNG]]

MEP surface diagram:

[[File:MEP surface.PNG]]

Differences can be seen in the length and oscillations in the reaction trajectory. For the minimum energy pathway (MEP) there is no oscillations as there it neglects the molecules vibration modes and it moves infinitely slowly. The line is shorter as the reaction coordinate isn't as long due to the reaction.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 ||-99.0|| Yes||H2 initially not vibrating, upon reaction the new H2 has a high vibrational energy||[[File:P=-1.25,-2.5.PNG]]|
|-
| -1.5 || -2.0 ||-100.5||No|| H approaching is repelled, causing the HH bond to vibrate ||[[File:P=-1.5,-2.PNG]]|
|-
| -1.5 || -2.5 ||-99.0||Yes||Slow approach of H and H2 after reaction stronger vibrations||[[File:P=-1.5,-2.5.PNG]]|
|-
| -2.5 || -5.0 ||-85.0||No||H approaches H2 causes a the H-H bond to weaken almost up to a transition state, not enough energy to form a new HH bond| |[[File:P=-2.5,-5.PNG]]|
|-
| -2.5 || -5.2 ||-83.4||Yes||H on approach caused increase in vibration for H2 bond until it broke and new HH bond formed with the approaching H||[[File:P=-2.5,-5.2.PNG]]|
|}

For the penultimate conditions, the reaction reaches the transition state yet its still falls back to the reactants despite the high momentum from the approaching H. The difference for the reaction to occur is the momentum to the approaching H to the molecule. It can be seen that the both the species in the reactants need the required momentum in order to reach transition state and to form the products.

== F-H-H system ==

H2 + F potential energy surface:
[[File:H2 + F potential surface.PNG]]

Hf + H potential surface:
[[File:HF+H surface plot.PNG]]

From the potential surface we see that the products; H2 + F are higher in energy than the product F-H this suggests that the F-H bond is stronger than the H-H bond. This reaction is exothermic.
The reverse reaction must then be endothermic for H + HF

==Transition state position H2 + F ==
[[File:Energy and configuration for h2 + F.PNG]]

To find the transition state, initial conditions were set and momentum is set to 0. From this it is clear that it isnt optimised as there are forces present between HH and F.

Setting HH distance to 0.744 and HF distance to 1.811, the forces are approximately 0 and so this is where the transition state exists.

== Activation Energy H2 + F ==
[[File:Energy v time Activation energy for Fh2.PNG]]

Looking at the energy v time graph the difference between the total energy and the potential energy results in the activation

Activation energy = 104.019 - 103.749 = 0.270

==Transition state position HF + H ==
0.74 this resembles the H-H bond length in molecular H2
==Activation E HF + H ==

30

===F + H2 reaction trajectory: ===

[[File:Config for animation for h2 + f reaction path.PNG]]

These are the initial conditions selected that gave a reaction trajectory

[[File:Conservation of energy H2 +F.PNG]]

The momentum v time plot shows B-C (H-H) vibrating as F has little momentum. Once reaction happens, BC momentum stops oscillating and AB (HF) starts oscillating showing the conservation of energy as the vibration transfers from HH to HF.
The mechanism of release:

rHH=0.74, rHF=2 and pHF = -0.5 for the following table. The pHH ranges from -3 and 3 to see if a reaction occurs with these initial conditions.
{| class="wikitable" border=1
! pHH !! Reaction?!! Contour Plot!
|-
| 3.0 || No|| [[File:H2+F, p=3.PNG]]
|-
| 2.0 || No|| [[File:H2+F,p=2.PNG]]
|-
| 1.0 || No|| [[File:H2+F,p=1.PNG]]
|-
| 0 ||No || [[File:H2+F,p=0.PNG]]
|-
| -1.0 || No||[[File:H2+F,p=-1.PNG]]
|-
| -2.0 ||Yes || [[File:H2+F,p=-2.PNG]]
|-
| -3.0 || No|| [[File:H2+F, p=-3.PNG]]
|}

For F having momentum -0.5, reaction only occurs when the HH bond has -2 momentum. For reaction to complete is not enough to have a large energy in the system. Each both H2 and F require specific energy in order to react with each other therefore changing either energy values will require the other species to have a complimentary energy.

[[File:H2+F,p=-0.8,0.1.PNG]]
Reaction occurs even when H2 has a small momentum. F having an increased momentum allowed this reaction to occur therefore changes to F compared to changes in momentum for H2 have a greater weighting on the progress of the reaction.

===H + HF reaction trajectory: ===

Setting the conditions:
r1 = 0.94, r2 = 2
{| class="wikitable" border=1
! pHH !! pHF !! Etot !! Description of the dynamics !! Illustration of the trajectory
|-
| -10 || -0.5 || -85.6 || No reaction. The system does reach the transition state before falling back to the products||[[File:-10,-0.5.PNG]]|
|-
| -8 || -1 || -106.8 ||No reaction. H approach increases vibration frequency of HF ||[[File:-8,-1.PNG]]|
|-
| -1 || -12 ||-68.7 || Goes to transition state but does not react ||[[File:HF+H,P=-1,-12.PNG]]||
|-
|-1|| -13 ||-56.5 ||High HF vibrations cause reaction to occur. || [[File:HF+H reaction.PNG]]||
|-
|}

MRD:ld2416

2019-05-24T16:01:05Z

Ld2416: /* H + HF reaction trajectory: */

==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?
[[File:Contour plot opf H H2.PNG]]

The transition state is when the H2 is equally distant to H1 and H3, for the contour plot this is approximately 0.93 A.

[[File:SUrface plot.PNG]]

The transition state can be seen at the saddle point, where the transition state is a minimum for the frank condon plot for the reaction. The transition state is a maximum for the reaction trajectory.
Mathematically, the TS has a second derivative greater than 0 in the Q1 direction (minimum) and TS has a second derivative less than 0 in the Q2 direction (maximum).

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.

From the contour plot we see that the transition state occurs when the r1=r2. Therefore to estimate the transition state position distances AB must be equal to BC.

[[File:Ab DIS.PNG]]

From this we can estimate the transition position from the contour plot:
[[File:Contour TS.PNG]]

This gives a value of 0.9075 A

Looking at this on a distance/time graph:
[[File:Bad ds TS.PNG]]

The momentum is set to 0 because the transition state is stationary and the distance between the 2 are equal.

Dynamic Surface:
[[File:Dynamic surface.PNG]]

MEP surface diagram:

[[File:MEP surface.PNG]]

Differences can be seen in the length and oscillations in the reaction trajectory. For the minimum energy pathway (MEP) there is no oscillations as there it neglects the molecules vibration modes and it moves infinitely slowly. The line is shorter as the reaction coordinate isn't as long due to the reaction.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 ||-99.0|| Yes||H2 initially not vibrating, upon reaction the new H2 has a high vibrational energy||[[File:P=-1.25,-2.5.PNG]]|
|-
| -1.5 || -2.0 ||-100.5||No|| H approaching is repelled, causing the HH bond to vibrate ||[[File:P=-1.5,-2.PNG]]|
|-
| -1.5 || -2.5 ||-99.0||Yes||Slow approach of H and H2 after reaction stronger vibrations||[[File:P=-1.5,-2.5.PNG]]|
|-
| -2.5 || -5.0 ||-85.0||No||H approaches H2 causes a the H-H bond to weaken almost up to a transition state, not enough energy to form a new HH bond| |[[File:P=-2.5,-5.PNG]]|
|-
| -2.5 || -5.2 ||-83.4||Yes||H on approach caused increase in vibration for H2 bond until it broke and new HH bond formed with the approaching H||[[File:P=-2.5,-5.2.PNG]]|
|}

For the penultimate conditions, the reaction reaches the transition state yet its still falls back to the reactants despite the high momentum from the approaching H. The difference for the reaction to occur is the momentum to the approaching H to the molecule. It can be seen that the both the species in the reactants need the required momentum in order to reach transition state and to form the products.

== F-H-H system ==

H2 + F potential energy surface:
[[File:H2 + F potential surface.PNG]]

Hf + H potential surface:
[[File:HF+H surface plot.PNG]]

From the potential surface we see that the products; H2 + F are higher in energy than the product F-H this suggests that the F-H bond is stronger than the H-H bond. This reaction is exothermic.
The reverse reaction must then be endothermic for H + HF

==Transition state position H2 + F ==
[[File:Energy and configuration for h2 + F.PNG]]

To find the transition state, initial conditions were set and momentum is set to 0. From this it is clear that it isnt optimised as there are forces present between HH and F.

Setting HH distance to 0.744 and HF distance to 1.811, the forces are approximately 0 and so this is where the transition state exists.

== Activation Energy H2 + F ==
[[File:Energy v time Activation energy for Fh2.PNG]]

Looking at the energy v time graph the difference between the total energy and the potential energy results in the activation

Activation energy = 104.019 - 103.749 = 0.270

==Transition state position HF + H ==
0.74 this resembles the H-H bond length in molecular H2
==Activation E HF + H ==

30

===F + H2 reaction trajectory: ===

[[File:Config for animation for h2 + f reaction path.PNG]]

These are the initial conditions selected that gave a reaction trajectory

[[File:Conservation of energy H2 +F.PNG]]

The momentum v time plot shows B-C (H-H) vibrating as F has little momentum. Once reaction happens, BC momentum stops oscillating and AB (HF) starts oscillating showing the conservation of energy as the vibration transfers from HH to HF.
The mechanism of release:

rHH=0.74, rHF=2 and pHF = -0.5 for the following table. The pHH ranges from -3 and 3 to see if a reaction occurs with these initial conditions.
{| class="wikitable" border=1
! pHH !! Reaction?!! Contour Plot!
|-
| 3.0 || No|| [[File:H2+F, p=3.PNG]]
|-
| 2.0 || No|| [[File:H2+F,p=2.PNG]]
|-
| 1.0 || No|| [[File:H2+F,p=1.PNG]]
|-
| 0 ||No || [[File:H2+F,p=0.PNG]]
|-
| -1.0 || No||[[File:H2+F,p=-1.PNG]]
|-
| -2.0 ||Yes || [[File:H2+F,p=-2.PNG]]
|-
| -3.0 || No|| [[File:H2+F, p=-3.PNG]]
|}

For F having momentum -0.5, reaction only occurs when the HH bond has -2 momentum. For reaction to complete is not enough to have a large energy in the system. Each both H2 and F require specific energy in order to react with each other therefore changing either energy values will require the other species to have a complimentary energy.

[[File:H2+F,p=-0.8,0.1.PNG]]
Reaction occurs even when H2 has a small momentum. F having an increased momentum allowed this reaction to occur therefore changes to F compared to changes in momentum for H2 have a greater weighting on the progress of the reaction.

===H + HF reaction trajectory: ===

Setting the conditions:
r1 = 0.94, r2 = 2
{| class="wikitable" border=1
! pHH !! pHF !! Etot !! Description of the dynamics !! Illustration of the trajectory
|-
| -10 || -0.5 || -85.6 || No reaction. The system does reach the transition state before falling back to the products||[[File:-10,-0.5.PNG]]|
|-
| -8 || -1 || -106.8 ||No reaction. H approach increases vibration frequency of HF ||[[File:-8,-1.PNG]]|
|-
| -1 || -12 ||-68.7 || Goes to transition state but does not react ||[[File:HF+H,P=-1,-12.PNG]]||
|-
|-1|| -13 ||-56.5 ||High HF vibrations cause reaction to occur. || [[File:HF+H reaction.PNG]]||
|}

MRD:ld2416

2019-05-24T16:00:28Z

Ld2416: /* H + HF reaction trajectory: */

==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?
[[File:Contour plot opf H H2.PNG]]

The transition state is when the H2 is equally distant to H1 and H3, for the contour plot this is approximately 0.93 A.

[[File:SUrface plot.PNG]]

The transition state can be seen at the saddle point, where the transition state is a minimum for the frank condon plot for the reaction. The transition state is a maximum for the reaction trajectory.
Mathematically, the TS has a second derivative greater than 0 in the Q1 direction (minimum) and TS has a second derivative less than 0 in the Q2 direction (maximum).

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.

From the contour plot we see that the transition state occurs when the r1=r2. Therefore to estimate the transition state position distances AB must be equal to BC.

[[File:Ab DIS.PNG]]

From this we can estimate the transition position from the contour plot:
[[File:Contour TS.PNG]]

This gives a value of 0.9075 A

Looking at this on a distance/time graph:
[[File:Bad ds TS.PNG]]

The momentum is set to 0 because the transition state is stationary and the distance between the 2 are equal.

Dynamic Surface:
[[File:Dynamic surface.PNG]]

MEP surface diagram:

[[File:MEP surface.PNG]]

Differences can be seen in the length and oscillations in the reaction trajectory. For the minimum energy pathway (MEP) there is no oscillations as there it neglects the molecules vibration modes and it moves infinitely slowly. The line is shorter as the reaction coordinate isn't as long due to the reaction.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 ||-99.0|| Yes||H2 initially not vibrating, upon reaction the new H2 has a high vibrational energy||[[File:P=-1.25,-2.5.PNG]]|
|-
| -1.5 || -2.0 ||-100.5||No|| H approaching is repelled, causing the HH bond to vibrate ||[[File:P=-1.5,-2.PNG]]|
|-
| -1.5 || -2.5 ||-99.0||Yes||Slow approach of H and H2 after reaction stronger vibrations||[[File:P=-1.5,-2.5.PNG]]|
|-
| -2.5 || -5.0 ||-85.0||No||H approaches H2 causes a the H-H bond to weaken almost up to a transition state, not enough energy to form a new HH bond| |[[File:P=-2.5,-5.PNG]]|
|-
| -2.5 || -5.2 ||-83.4||Yes||H on approach caused increase in vibration for H2 bond until it broke and new HH bond formed with the approaching H||[[File:P=-2.5,-5.2.PNG]]|
|}

For the penultimate conditions, the reaction reaches the transition state yet its still falls back to the reactants despite the high momentum from the approaching H. The difference for the reaction to occur is the momentum to the approaching H to the molecule. It can be seen that the both the species in the reactants need the required momentum in order to reach transition state and to form the products.

== F-H-H system ==

H2 + F potential energy surface:
[[File:H2 + F potential surface.PNG]]

Hf + H potential surface:
[[File:HF+H surface plot.PNG]]

From the potential surface we see that the products; H2 + F are higher in energy than the product F-H this suggests that the F-H bond is stronger than the H-H bond. This reaction is exothermic.
The reverse reaction must then be endothermic for H + HF

==Transition state position H2 + F ==
[[File:Energy and configuration for h2 + F.PNG]]

To find the transition state, initial conditions were set and momentum is set to 0. From this it is clear that it isnt optimised as there are forces present between HH and F.

Setting HH distance to 0.744 and HF distance to 1.811, the forces are approximately 0 and so this is where the transition state exists.

== Activation Energy H2 + F ==
[[File:Energy v time Activation energy for Fh2.PNG]]

Looking at the energy v time graph the difference between the total energy and the potential energy results in the activation

Activation energy = 104.019 - 103.749 = 0.270

==Transition state position HF + H ==
0.74 this resembles the H-H bond length in molecular H2
==Activation E HF + H ==

30

===F + H2 reaction trajectory: ===

[[File:Config for animation for h2 + f reaction path.PNG]]

These are the initial conditions selected that gave a reaction trajectory

[[File:Conservation of energy H2 +F.PNG]]

The momentum v time plot shows B-C (H-H) vibrating as F has little momentum. Once reaction happens, BC momentum stops oscillating and AB (HF) starts oscillating showing the conservation of energy as the vibration transfers from HH to HF.
The mechanism of release:

rHH=0.74, rHF=2 and pHF = -0.5 for the following table. The pHH ranges from -3 and 3 to see if a reaction occurs with these initial conditions.
{| class="wikitable" border=1
! pHH !! Reaction?!! Contour Plot!
|-
| 3.0 || No|| [[File:H2+F, p=3.PNG]]
|-
| 2.0 || No|| [[File:H2+F,p=2.PNG]]
|-
| 1.0 || No|| [[File:H2+F,p=1.PNG]]
|-
| 0 ||No || [[File:H2+F,p=0.PNG]]
|-
| -1.0 || No||[[File:H2+F,p=-1.PNG]]
|-
| -2.0 ||Yes || [[File:H2+F,p=-2.PNG]]
|-
| -3.0 || No|| [[File:H2+F, p=-3.PNG]]
|}

For F having momentum -0.5, reaction only occurs when the HH bond has -2 momentum. For reaction to complete is not enough to have a large energy in the system. Each both H2 and F require specific energy in order to react with each other therefore changing either energy values will require the other species to have a complimentary energy.

[[File:H2+F,p=-0.8,0.1.PNG]]
Reaction occurs even when H2 has a small momentum. F having an increased momentum allowed this reaction to occur therefore changes to F compared to changes in momentum for H2 have a greater weighting on the progress of the reaction.

===H + HF reaction trajectory: ===

Setting the conditions:
r1 = 0.94, r2 = 2
{| class="wikitable" border=1
! pHH !! pHF !! Etot !! Description of the dynamics !! Illustration of the trajectory
|-
| -10 || -0.5 || -85.6 || No reaction. The system does reach the transition state before falling back to the products||[[File:-10,-0.5.PNG]]|
|-
| -8 || -1 || -106.8 ||No reaction. H approach increases vibration frequency of HF ||[[File:-8,-1.PNG]]|
|-
| -1 || -12 ||-68.7 || Goes to transition state but does not react ||[[File:HF+H,P=-1,-12.PNG]]||
|-
|-1|| -13 ||-56.5 ||High HF vibrations cause reaction to occur. || [[File:HF+H reaction.PNG]] |
|}

File:HF+H reaction.PNG

2019-05-24T15:59:31Z

Ld2416:

File:HF+H,P=-1,-12.PNG

2019-05-24T15:48:15Z

Ld2416:

MRD:ld2416

2019-05-24T15:30:05Z

Ld2416: /* F + H2 reaction trajectory: */

==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?
[[File:Contour plot opf H H2.PNG]]

The transition state is when the H2 is equally distant to H1 and H3, for the contour plot this is approximately 0.93 A.

[[File:SUrface plot.PNG]]

The transition state can be seen at the saddle point, where the transition state is a minimum for the frank condon plot for the reaction. The transition state is a maximum for the reaction trajectory.
Mathematically, the TS has a second derivative greater than 0 in the Q1 direction (minimum) and TS has a second derivative less than 0 in the Q2 direction (maximum).

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.

From the contour plot we see that the transition state occurs when the r1=r2. Therefore to estimate the transition state position distances AB must be equal to BC.

[[File:Ab DIS.PNG]]

From this we can estimate the transition position from the contour plot:
[[File:Contour TS.PNG]]

This gives a value of 0.9075 A

Looking at this on a distance/time graph:
[[File:Bad ds TS.PNG]]

The momentum is set to 0 because the transition state is stationary and the distance between the 2 are equal.

Dynamic Surface:
[[File:Dynamic surface.PNG]]

MEP surface diagram:

[[File:MEP surface.PNG]]

Differences can be seen in the length and oscillations in the reaction trajectory. For the minimum energy pathway (MEP) there is no oscillations as there it neglects the molecules vibration modes and it moves infinitely slowly. The line is shorter as the reaction coordinate isn't as long due to the reaction.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 ||-99.0|| Yes||H2 initially not vibrating, upon reaction the new H2 has a high vibrational energy||[[File:P=-1.25,-2.5.PNG]]|
|-
| -1.5 || -2.0 ||-100.5||No|| H approaching is repelled, causing the HH bond to vibrate ||[[File:P=-1.5,-2.PNG]]|
|-
| -1.5 || -2.5 ||-99.0||Yes||Slow approach of H and H2 after reaction stronger vibrations||[[File:P=-1.5,-2.5.PNG]]|
|-
| -2.5 || -5.0 ||-85.0||No||H approaches H2 causes a the H-H bond to weaken almost up to a transition state, not enough energy to form a new HH bond| |[[File:P=-2.5,-5.PNG]]|
|-
| -2.5 || -5.2 ||-83.4||Yes||H on approach caused increase in vibration for H2 bond until it broke and new HH bond formed with the approaching H||[[File:P=-2.5,-5.2.PNG]]|
|}

For the penultimate conditions, the reaction reaches the transition state yet its still falls back to the reactants despite the high momentum from the approaching H. The difference for the reaction to occur is the momentum to the approaching H to the molecule. It can be seen that the both the species in the reactants need the required momentum in order to reach transition state and to form the products.

== F-H-H system ==

H2 + F potential energy surface:
[[File:H2 + F potential surface.PNG]]

Hf + H potential surface:
[[File:HF+H surface plot.PNG]]

From the potential surface we see that the products; H2 + F are higher in energy than the product F-H this suggests that the F-H bond is stronger than the H-H bond. This reaction is exothermic.
The reverse reaction must then be endothermic for H + HF

==Transition state position H2 + F ==
[[File:Energy and configuration for h2 + F.PNG]]

To find the transition state, initial conditions were set and momentum is set to 0. From this it is clear that it isnt optimised as there are forces present between HH and F.

Setting HH distance to 0.744 and HF distance to 1.811, the forces are approximately 0 and so this is where the transition state exists.

== Activation Energy H2 + F ==
[[File:Energy v time Activation energy for Fh2.PNG]]

Looking at the energy v time graph the difference between the total energy and the potential energy results in the activation

Activation energy = 104.019 - 103.749 = 0.270

==Transition state position HF + H ==
0.74 this resembles the H-H bond length in molecular H2
==Activation E HF + H ==

30

===F + H2 reaction trajectory: ===

[[File:Config for animation for h2 + f reaction path.PNG]]

These are the initial conditions selected that gave a reaction trajectory

[[File:Conservation of energy H2 +F.PNG]]

The momentum v time plot shows B-C (H-H) vibrating as F has little momentum. Once reaction happens, BC momentum stops oscillating and AB (HF) starts oscillating showing the conservation of energy as the vibration transfers from HH to HF.
The mechanism of release:

rHH=0.74, rHF=2 and pHF = -0.5 for the following table. The pHH ranges from -3 and 3 to see if a reaction occurs with these initial conditions.
{| class="wikitable" border=1
! pHH !! Reaction?!! Contour Plot!
|-
| 3.0 || No|| [[File:H2+F, p=3.PNG]]
|-
| 2.0 || No|| [[File:H2+F,p=2.PNG]]
|-
| 1.0 || No|| [[File:H2+F,p=1.PNG]]
|-
| 0 ||No || [[File:H2+F,p=0.PNG]]
|-
| -1.0 || No||[[File:H2+F,p=-1.PNG]]
|-
| -2.0 ||Yes || [[File:H2+F,p=-2.PNG]]
|-
| -3.0 || No|| [[File:H2+F, p=-3.PNG]]
|}

For F having momentum -0.5, reaction only occurs when the HH bond has -2 momentum. For reaction to complete is not enough to have a large energy in the system. Each both H2 and F require specific energy in order to react with each other therefore changing either energy values will require the other species to have a complimentary energy.

[[File:H2+F,p=-0.8,0.1.PNG]]
Reaction occurs even when H2 has a small momentum. F having an increased momentum allowed this reaction to occur therefore changes to F compared to changes in momentum for H2 have a greater weighting on the progress of the reaction.

===H + HF reaction trajectory: ===

Setting the conditions:
r1 = 0.94, r2 = 2
{| class="wikitable" border=1
! pHH !! pHF !! Etot !! Description of the dynamics !! Illustration of the trajectory
|-
| -10 || -0.5 || -85.6 || No reaction. The system does reach the transition state before falling back to the products||[[File:-10,-0.5.PNG]]|
|-
| -8 || -1 || -106.8 ||No reaction. H approach increases vibration frequency of HF ||[[File:-8,-1.PNG]]|
|-
| -1.5 || -2.5 || || ||
|-
| -2.5 || -5.0 || || ||
|-
| -2.5 || -5.2 || || ||
|}

File:H2+F,p=-0.8,0.1.PNG

2019-05-24T15:20:50Z

Ld2416:

MRD:ld2416

2019-05-24T15:19:55Z

Ld2416: /* F + H2 reaction trajectory: */

==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?
[[File:Contour plot opf H H2.PNG]]

The transition state is when the H2 is equally distant to H1 and H3, for the contour plot this is approximately 0.93 A.

[[File:SUrface plot.PNG]]

The transition state can be seen at the saddle point, where the transition state is a minimum for the frank condon plot for the reaction. The transition state is a maximum for the reaction trajectory.
Mathematically, the TS has a second derivative greater than 0 in the Q1 direction (minimum) and TS has a second derivative less than 0 in the Q2 direction (maximum).

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.

From the contour plot we see that the transition state occurs when the r1=r2. Therefore to estimate the transition state position distances AB must be equal to BC.

[[File:Ab DIS.PNG]]

From this we can estimate the transition position from the contour plot:
[[File:Contour TS.PNG]]

This gives a value of 0.9075 A

Looking at this on a distance/time graph:
[[File:Bad ds TS.PNG]]

The momentum is set to 0 because the transition state is stationary and the distance between the 2 are equal.

Dynamic Surface:
[[File:Dynamic surface.PNG]]

MEP surface diagram:

[[File:MEP surface.PNG]]

Differences can be seen in the length and oscillations in the reaction trajectory. For the minimum energy pathway (MEP) there is no oscillations as there it neglects the molecules vibration modes and it moves infinitely slowly. The line is shorter as the reaction coordinate isn't as long due to the reaction.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 ||-99.0|| Yes||H2 initially not vibrating, upon reaction the new H2 has a high vibrational energy||[[File:P=-1.25,-2.5.PNG]]|
|-
| -1.5 || -2.0 ||-100.5||No|| H approaching is repelled, causing the HH bond to vibrate ||[[File:P=-1.5,-2.PNG]]|
|-
| -1.5 || -2.5 ||-99.0||Yes||Slow approach of H and H2 after reaction stronger vibrations||[[File:P=-1.5,-2.5.PNG]]|
|-
| -2.5 || -5.0 ||-85.0||No||H approaches H2 causes a the H-H bond to weaken almost up to a transition state, not enough energy to form a new HH bond| |[[File:P=-2.5,-5.PNG]]|
|-
| -2.5 || -5.2 ||-83.4||Yes||H on approach caused increase in vibration for H2 bond until it broke and new HH bond formed with the approaching H||[[File:P=-2.5,-5.2.PNG]]|
|}

For the penultimate conditions, the reaction reaches the transition state yet its still falls back to the reactants despite the high momentum from the approaching H. The difference for the reaction to occur is the momentum to the approaching H to the molecule. It can be seen that the both the species in the reactants need the required momentum in order to reach transition state and to form the products.

== F-H-H system ==

H2 + F potential energy surface:
[[File:H2 + F potential surface.PNG]]

Hf + H potential surface:
[[File:HF+H surface plot.PNG]]

From the potential surface we see that the products; H2 + F are higher in energy than the product F-H this suggests that the F-H bond is stronger than the H-H bond. This reaction is exothermic.
The reverse reaction must then be endothermic for H + HF

==Transition state position H2 + F ==
[[File:Energy and configuration for h2 + F.PNG]]

To find the transition state, initial conditions were set and momentum is set to 0. From this it is clear that it isnt optimised as there are forces present between HH and F.

Setting HH distance to 0.744 and HF distance to 1.811, the forces are approximately 0 and so this is where the transition state exists.

== Activation Energy H2 + F ==
[[File:Energy v time Activation energy for Fh2.PNG]]

Looking at the energy v time graph the difference between the total energy and the potential energy results in the activation

Activation energy = 104.019 - 103.749 = 0.270

==Transition state position HF + H ==
0.74 this resembles the H-H bond length in molecular H2
==Activation E HF + H ==

30

===F + H2 reaction trajectory: ===

[[File:Config for animation for h2 + f reaction path.PNG]]

These are the initial conditions selected that gave a reaction trajectory

[[File:Conservation of energy H2 +F.PNG]]

The momentum v time plot shows B-C (H-H) vibrating as F has little momentum. Once reaction happens, BC momentum stops oscillating and AB (HF) starts oscillating showing the conservation of energy as the vibration transfers from HH to HF.
The mechanism of release:

rHH=0.74, rHF=2 and pHF = -0.5 for the following table. The pHH ranges from -3 and 3 to see if a reaction occurs with these initial conditions.
{| class="wikitable" border=1
! pHH !! Reaction?!! Contour Plot!
|-
| 3.0 || No|| [[File:H2+F, p=3.PNG]]
|-
| 2.0 || No|| [[File:H2+F,p=2.PNG]]
|-
| 1.0 || No|| [[File:H2+F,p=1.PNG]]
|-
| 0 ||No || [[File:H2+F,p=0.PNG]]
|-
| -1.0 || ||[[File:H2+F,p=-1.PNG]]
|-
| -2.0 ||Yes || [[File:H2+F,p=-2.PNG]]
|-
| -3.0 || || [[File:H2+F, p=-3.PNG]]
|}

As PHH approaches 3, the reaction has a larger vibration frequency when the product have formed hence the

[[File:0.1,-.08 contour.PNG]]
Vibrations in the products are significantly reduced and shows that the products are favourable, even past the transition state position calculated the

===H + HF reaction trajectory: ===

Setting the conditions:
r1 = 0.94, r2 = 2
{| class="wikitable" border=1
! pHH !! pHF !! Etot !! Description of the dynamics !! Illustration of the trajectory
|-
| -10 || -0.5 || -85.6 || No reaction. The system does reach the transition state before falling back to the products||[[File:-10,-0.5.PNG]]|
|-
| -8 || -1 || -106.8 ||No reaction. H approach increases vibration frequency of HF ||[[File:-8,-1.PNG]]|
|-
| -1.5 || -2.5 || || ||
|-
| -2.5 || -5.0 || || ||
|-
| -2.5 || -5.2 || || ||
|}

File:H2+F,p=-1.PNG

2019-05-24T15:19:10Z

Ld2416:

File:H2+F,p=0.PNG

2019-05-24T15:18:01Z

Ld2416:

File:H2+F, p=-3.PNG

2019-05-24T15:14:33Z

Ld2416:

MRD:ld2416

2019-05-24T15:12:13Z

Ld2416: /* F + H2 reaction trajectory: */

==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?
[[File:Contour plot opf H H2.PNG]]

The transition state is when the H2 is equally distant to H1 and H3, for the contour plot this is approximately 0.93 A.

[[File:SUrface plot.PNG]]

The transition state can be seen at the saddle point, where the transition state is a minimum for the frank condon plot for the reaction. The transition state is a maximum for the reaction trajectory.
Mathematically, the TS has a second derivative greater than 0 in the Q1 direction (minimum) and TS has a second derivative less than 0 in the Q2 direction (maximum).

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.

From the contour plot we see that the transition state occurs when the r1=r2. Therefore to estimate the transition state position distances AB must be equal to BC.

[[File:Ab DIS.PNG]]

From this we can estimate the transition position from the contour plot:
[[File:Contour TS.PNG]]

This gives a value of 0.9075 A

Looking at this on a distance/time graph:
[[File:Bad ds TS.PNG]]

The momentum is set to 0 because the transition state is stationary and the distance between the 2 are equal.

Dynamic Surface:
[[File:Dynamic surface.PNG]]

MEP surface diagram:

[[File:MEP surface.PNG]]

Differences can be seen in the length and oscillations in the reaction trajectory. For the minimum energy pathway (MEP) there is no oscillations as there it neglects the molecules vibration modes and it moves infinitely slowly. The line is shorter as the reaction coordinate isn't as long due to the reaction.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 ||-99.0|| Yes||H2 initially not vibrating, upon reaction the new H2 has a high vibrational energy||[[File:P=-1.25,-2.5.PNG]]|
|-
| -1.5 || -2.0 ||-100.5||No|| H approaching is repelled, causing the HH bond to vibrate ||[[File:P=-1.5,-2.PNG]]|
|-
| -1.5 || -2.5 ||-99.0||Yes||Slow approach of H and H2 after reaction stronger vibrations||[[File:P=-1.5,-2.5.PNG]]|
|-
| -2.5 || -5.0 ||-85.0||No||H approaches H2 causes a the H-H bond to weaken almost up to a transition state, not enough energy to form a new HH bond| |[[File:P=-2.5,-5.PNG]]|
|-
| -2.5 || -5.2 ||-83.4||Yes||H on approach caused increase in vibration for H2 bond until it broke and new HH bond formed with the approaching H||[[File:P=-2.5,-5.2.PNG]]|
|}

For the penultimate conditions, the reaction reaches the transition state yet its still falls back to the reactants despite the high momentum from the approaching H. The difference for the reaction to occur is the momentum to the approaching H to the molecule. It can be seen that the both the species in the reactants need the required momentum in order to reach transition state and to form the products.

== F-H-H system ==

H2 + F potential energy surface:
[[File:H2 + F potential surface.PNG]]

Hf + H potential surface:
[[File:HF+H surface plot.PNG]]

From the potential surface we see that the products; H2 + F are higher in energy than the product F-H this suggests that the F-H bond is stronger than the H-H bond. This reaction is exothermic.
The reverse reaction must then be endothermic for H + HF

==Transition state position H2 + F ==
[[File:Energy and configuration for h2 + F.PNG]]

To find the transition state, initial conditions were set and momentum is set to 0. From this it is clear that it isnt optimised as there are forces present between HH and F.

Setting HH distance to 0.744 and HF distance to 1.811, the forces are approximately 0 and so this is where the transition state exists.

== Activation Energy H2 + F ==
[[File:Energy v time Activation energy for Fh2.PNG]]

Looking at the energy v time graph the difference between the total energy and the potential energy results in the activation

Activation energy = 104.019 - 103.749 = 0.270

==Transition state position HF + H ==
0.74 this resembles the H-H bond length in molecular H2
==Activation E HF + H ==

30

===F + H2 reaction trajectory: ===

[[File:Config for animation for h2 + f reaction path.PNG]]

These are the initial conditions selected that gave a reaction trajectory

[[File:Conservation of energy H2 +F.PNG]]

The momentum v time plot shows B-C (H-H) vibrating as F has little momentum. Once reaction happens, BC momentum stops oscillating and AB (HF) starts oscillating showing the conservation of energy as the vibration transfers from HH to HF.
The mechanism of release:

rHH=0.74, rHF=2 and pHF = -0.5 for the following table. The pHH ranges from -3 and 3 to see if a reaction occurs with these initial conditions.
{| class="wikitable" border=1
! pHH !! Reaction?!! Contour Plot!
|-
| 3.0 || No|| [[File:H2+F, p=3.PNG]]
|-
| 2.0 || No|| [[File:H2+F,p=2.PNG]]
|-
| 1.0 || No|| [[File:H2+F,p=1.PNG]]
|-
| 0 || || [[]]
|-
| -1.0 || ||[[]]
|-
| -2.0 ||Yes || [[File:H2+F,p=-2.PNG]]
|-
| -3.0 || || [[]]
|}

As PHH approaches 3, the reaction has a larger vibration frequency when the product have formed hence the

[[File:0.1,-.08 contour.PNG]]
Vibrations in the products are significantly reduced and shows that the products are favourable, even past the transition state position calculated the

===H + HF reaction trajectory: ===

Setting the conditions:
r1 = 0.94, r2 = 2
{| class="wikitable" border=1
! pHH !! pHF !! Etot !! Description of the dynamics !! Illustration of the trajectory
|-
| -10 || -0.5 || -85.6 || No reaction. The system does reach the transition state before falling back to the products||[[File:-10,-0.5.PNG]]|
|-
| -8 || -1 || -106.8 ||No reaction. H approach increases vibration frequency of HF ||[[File:-8,-1.PNG]]|
|-
| -1.5 || -2.5 || || ||
|-
| -2.5 || -5.0 || || ||
|-
| -2.5 || -5.2 || || ||
|}

File:H2+F,p=-2.PNG

2019-05-24T15:03:19Z

Ld2416:

File:H2+F,p=1.PNG

2019-05-24T15:00:50Z

Ld2416:

File:H2+F,p=2.PNG

2019-05-24T14:59:34Z

Ld2416:

File:H2+F, p=3.PNG

2019-05-24T14:58:04Z

Ld2416:

MRD:ld2416

2019-05-24T14:30:36Z

Ld2416: /* H + HF reaction trajectory: */

==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?
[[File:Contour plot opf H H2.PNG]]

The transition state is when the H2 is equally distant to H1 and H3, for the contour plot this is approximately 0.93 A.

[[File:SUrface plot.PNG]]

The transition state can be seen at the saddle point, where the transition state is a minimum for the frank condon plot for the reaction. The transition state is a maximum for the reaction trajectory.
Mathematically, the TS has a second derivative greater than 0 in the Q1 direction (minimum) and TS has a second derivative less than 0 in the Q2 direction (maximum).

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.

From the contour plot we see that the transition state occurs when the r1=r2. Therefore to estimate the transition state position distances AB must be equal to BC.

[[File:Ab DIS.PNG]]

From this we can estimate the transition position from the contour plot:
[[File:Contour TS.PNG]]

This gives a value of 0.9075 A

Looking at this on a distance/time graph:
[[File:Bad ds TS.PNG]]

The momentum is set to 0 because the transition state is stationary and the distance between the 2 are equal.

Dynamic Surface:
[[File:Dynamic surface.PNG]]

MEP surface diagram:

[[File:MEP surface.PNG]]

Differences can be seen in the length and oscillations in the reaction trajectory. For the minimum energy pathway (MEP) there is no oscillations as there it neglects the molecules vibration modes and it moves infinitely slowly. The line is shorter as the reaction coordinate isn't as long due to the reaction.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 ||-99.0|| Yes||H2 initially not vibrating, upon reaction the new H2 has a high vibrational energy||[[File:P=-1.25,-2.5.PNG]]|
|-
| -1.5 || -2.0 ||-100.5||No|| H approaching is repelled, causing the HH bond to vibrate ||[[File:P=-1.5,-2.PNG]]|
|-
| -1.5 || -2.5 ||-99.0||Yes||Slow approach of H and H2 after reaction stronger vibrations||[[File:P=-1.5,-2.5.PNG]]|
|-
| -2.5 || -5.0 ||-85.0||No||H approaches H2 causes a the H-H bond to weaken almost up to a transition state, not enough energy to form a new HH bond| |[[File:P=-2.5,-5.PNG]]|
|-
| -2.5 || -5.2 ||-83.4||Yes||H on approach caused increase in vibration for H2 bond until it broke and new HH bond formed with the approaching H||[[File:P=-2.5,-5.2.PNG]]|
|}

For the penultimate conditions, the reaction reaches the transition state yet its still falls back to the reactants despite the high momentum from the approaching H. The difference for the reaction to occur is the momentum to the approaching H to the molecule. It can be seen that the both the species in the reactants need the required momentum in order to reach transition state and to form the products.

== F-H-H system ==

H2 + F potential energy surface:
[[File:H2 + F potential surface.PNG]]

Hf + H potential surface:
[[File:HF+H surface plot.PNG]]

From the potential surface we see that the products; H2 + F are higher in energy than the product F-H this suggests that the F-H bond is stronger than the H-H bond. This reaction is exothermic.
The reverse reaction must then be endothermic for H + HF

==Transition state position H2 + F ==
[[File:Energy and configuration for h2 + F.PNG]]

To find the transition state, initial conditions were set and momentum is set to 0. From this it is clear that it isnt optimised as there are forces present between HH and F.

Setting HH distance to 0.744 and HF distance to 1.811, the forces are approximately 0 and so this is where the transition state exists.

== Activation Energy H2 + F ==
[[File:Energy v time Activation energy for Fh2.PNG]]

Looking at the energy v time graph the difference between the total energy and the potential energy results in the activation

Activation energy = 104.019 - 103.749 = 0.270

==Transition state position HF + H ==
0.74 this resembles the H-H bond length in molecular H2
==Activation E HF + H ==

30

===F + H2 reaction trajectory: ===

[[File:Config for animation for h2 + f reaction path.PNG]]

These are the initial conditions selected that gave a reaction trajectory

[[File:Conservation of energy H2 +F.PNG]]

The momentum v time plot shows B-C (H-H) vibrating as F has little momentum. Once reaction happens, BC momentum stops oscillating and AB (HF) starts oscillating showing the conservation of energy as the vibration transfers from HH to HF.
The mechanism of release:

{| class="wikitable" border=1
! pHH !! Contour Plot!
|-
| 3.0 || [[File:Phh=3 contor.PNG]]
|-
| 2.0 || [[File:PPH2.PNG]]
|-
| 1.0 || [[File:PPH1.PNG]]
|-
| 0 || [[File:PPH0.PNG]]
|-
| -1.0 ||[[File:PPH-1.PNG]]
|-
| -2.0 || [[File:Pph-2.PNG]]
|-
| -3.0 || [[File:Pph-3.PNG]]
|}

As PHH approaches 3, the reaction has a larger vibration frequency when the product have formed hence the

[[File:0.1,-.08 contour.PNG]]
Vibrations in the products are significantly reduced and shows that the products are favourable, even past the transition state position calculated the

===H + HF reaction trajectory: ===

Setting the conditions:
r1 = 0.94, r2 = 2
{| class="wikitable" border=1
! pHH !! pHF !! Etot !! Description of the dynamics !! Illustration of the trajectory
|-
| -10 || -0.5 || -85.6 || No reaction. The system does reach the transition state before falling back to the products||[[File:-10,-0.5.PNG]]|
|-
| -8 || -1 || -106.8 ||No reaction. H approach increases vibration frequency of HF ||[[File:-8,-1.PNG]]|
|-
| -1.5 || -2.5 || || ||
|-
| -2.5 || -5.0 || || ||
|-
| -2.5 || -5.2 || || ||
|}

MRD:ld2416

2019-05-24T14:29:56Z

Ld2416: /* H + HF reaction trajectory: */

==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?
[[File:Contour plot opf H H2.PNG]]

The transition state is when the H2 is equally distant to H1 and H3, for the contour plot this is approximately 0.93 A.

[[File:SUrface plot.PNG]]

The transition state can be seen at the saddle point, where the transition state is a minimum for the frank condon plot for the reaction. The transition state is a maximum for the reaction trajectory.
Mathematically, the TS has a second derivative greater than 0 in the Q1 direction (minimum) and TS has a second derivative less than 0 in the Q2 direction (maximum).

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.

From the contour plot we see that the transition state occurs when the r1=r2. Therefore to estimate the transition state position distances AB must be equal to BC.

[[File:Ab DIS.PNG]]

From this we can estimate the transition position from the contour plot:
[[File:Contour TS.PNG]]

This gives a value of 0.9075 A

Looking at this on a distance/time graph:
[[File:Bad ds TS.PNG]]

The momentum is set to 0 because the transition state is stationary and the distance between the 2 are equal.

Dynamic Surface:
[[File:Dynamic surface.PNG]]

MEP surface diagram:

[[File:MEP surface.PNG]]

Differences can be seen in the length and oscillations in the reaction trajectory. For the minimum energy pathway (MEP) there is no oscillations as there it neglects the molecules vibration modes and it moves infinitely slowly. The line is shorter as the reaction coordinate isn't as long due to the reaction.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 ||-99.0|| Yes||H2 initially not vibrating, upon reaction the new H2 has a high vibrational energy||[[File:P=-1.25,-2.5.PNG]]|
|-
| -1.5 || -2.0 ||-100.5||No|| H approaching is repelled, causing the HH bond to vibrate ||[[File:P=-1.5,-2.PNG]]|
|-
| -1.5 || -2.5 ||-99.0||Yes||Slow approach of H and H2 after reaction stronger vibrations||[[File:P=-1.5,-2.5.PNG]]|
|-
| -2.5 || -5.0 ||-85.0||No||H approaches H2 causes a the H-H bond to weaken almost up to a transition state, not enough energy to form a new HH bond| |[[File:P=-2.5,-5.PNG]]|
|-
| -2.5 || -5.2 ||-83.4||Yes||H on approach caused increase in vibration for H2 bond until it broke and new HH bond formed with the approaching H||[[File:P=-2.5,-5.2.PNG]]|
|}

For the penultimate conditions, the reaction reaches the transition state yet its still falls back to the reactants despite the high momentum from the approaching H. The difference for the reaction to occur is the momentum to the approaching H to the molecule. It can be seen that the both the species in the reactants need the required momentum in order to reach transition state and to form the products.

== F-H-H system ==

H2 + F potential energy surface:
[[File:H2 + F potential surface.PNG]]

Hf + H potential surface:
[[File:HF+H surface plot.PNG]]

From the potential surface we see that the products; H2 + F are higher in energy than the product F-H this suggests that the F-H bond is stronger than the H-H bond. This reaction is exothermic.
The reverse reaction must then be endothermic for H + HF

==Transition state position H2 + F ==
[[File:Energy and configuration for h2 + F.PNG]]

To find the transition state, initial conditions were set and momentum is set to 0. From this it is clear that it isnt optimised as there are forces present between HH and F.

Setting HH distance to 0.744 and HF distance to 1.811, the forces are approximately 0 and so this is where the transition state exists.

== Activation Energy H2 + F ==
[[File:Energy v time Activation energy for Fh2.PNG]]

Looking at the energy v time graph the difference between the total energy and the potential energy results in the activation

Activation energy = 104.019 - 103.749 = 0.270

==Transition state position HF + H ==
0.74 this resembles the H-H bond length in molecular H2
==Activation E HF + H ==

30

===F + H2 reaction trajectory: ===

[[File:Config for animation for h2 + f reaction path.PNG]]

These are the initial conditions selected that gave a reaction trajectory

[[File:Conservation of energy H2 +F.PNG]]

The momentum v time plot shows B-C (H-H) vibrating as F has little momentum. Once reaction happens, BC momentum stops oscillating and AB (HF) starts oscillating showing the conservation of energy as the vibration transfers from HH to HF.
The mechanism of release:

{| class="wikitable" border=1
! pHH !! Contour Plot!
|-
| 3.0 || [[File:Phh=3 contor.PNG]]
|-
| 2.0 || [[File:PPH2.PNG]]
|-
| 1.0 || [[File:PPH1.PNG]]
|-
| 0 || [[File:PPH0.PNG]]
|-
| -1.0 ||[[File:PPH-1.PNG]]
|-
| -2.0 || [[File:Pph-2.PNG]]
|-
| -3.0 || [[File:Pph-3.PNG]]
|}

As PHH approaches 3, the reaction has a larger vibration frequency when the product have formed hence the

[[File:0.1,-.08 contour.PNG]]
Vibrations in the products are significantly reduced and shows that the products are favourable, even past the transition state position calculated the

===H + HF reaction trajectory: ===

Setting the conditions:
r1 = 0.94, r2 = 2
{| class="wikitable" border=1
! pHH !! pHF !! Etot !! Description of the dynamics !! Illustration of the trajectory
|-
| -10 || -0.5 || -85.6 || No reaction. The system does reach the transition state before falling back to the products|[[File:-10,-0.5.PNG]]|
|-
| -8 || -1 || -106.8 ||No reaction. H approach increases vibration frequency of HF ||[[File:-8,-1.PNG]]
|-
| -1.5 || -2.5 || || ||
|-
| -2.5 || -5.0 || || ||
|-
| -2.5 || -5.2 || || ||
|}

File:-8,-1.PNG

2019-05-24T14:29:37Z

Ld2416:

File:-10,-0.5.PNG

2019-05-24T14:23:51Z

Ld2416:

MRD:ld2416

2019-05-24T14:15:25Z

Ld2416: /* H + H2 System */

==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?
[[File:Contour plot opf H H2.PNG]]

The transition state is when the H2 is equally distant to H1 and H3, for the contour plot this is approximately 0.93 A.

[[File:SUrface plot.PNG]]

The transition state can be seen at the saddle point, where the transition state is a minimum for the frank condon plot for the reaction. The transition state is a maximum for the reaction trajectory.
Mathematically, the TS has a second derivative greater than 0 in the Q1 direction (minimum) and TS has a second derivative less than 0 in the Q2 direction (maximum).

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.

From the contour plot we see that the transition state occurs when the r1=r2. Therefore to estimate the transition state position distances AB must be equal to BC.

[[File:Ab DIS.PNG]]

From this we can estimate the transition position from the contour plot:
[[File:Contour TS.PNG]]

This gives a value of 0.9075 A

Looking at this on a distance/time graph:
[[File:Bad ds TS.PNG]]

The momentum is set to 0 because the transition state is stationary and the distance between the 2 are equal.

Dynamic Surface:
[[File:Dynamic surface.PNG]]

MEP surface diagram:

[[File:MEP surface.PNG]]

Differences can be seen in the length and oscillations in the reaction trajectory. For the minimum energy pathway (MEP) there is no oscillations as there it neglects the molecules vibration modes and it moves infinitely slowly. The line is shorter as the reaction coordinate isn't as long due to the reaction.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 ||-99.0|| Yes||H2 initially not vibrating, upon reaction the new H2 has a high vibrational energy||[[File:P=-1.25,-2.5.PNG]]|
|-
| -1.5 || -2.0 ||-100.5||No|| H approaching is repelled, causing the HH bond to vibrate ||[[File:P=-1.5,-2.PNG]]|
|-
| -1.5 || -2.5 ||-99.0||Yes||Slow approach of H and H2 after reaction stronger vibrations||[[File:P=-1.5,-2.5.PNG]]|
|-
| -2.5 || -5.0 ||-85.0||No||H approaches H2 causes a the H-H bond to weaken almost up to a transition state, not enough energy to form a new HH bond| |[[File:P=-2.5,-5.PNG]]|
|-
| -2.5 || -5.2 ||-83.4||Yes||H on approach caused increase in vibration for H2 bond until it broke and new HH bond formed with the approaching H||[[File:P=-2.5,-5.2.PNG]]|
|}

For the penultimate conditions, the reaction reaches the transition state yet its still falls back to the reactants despite the high momentum from the approaching H. The difference for the reaction to occur is the momentum to the approaching H to the molecule. It can be seen that the both the species in the reactants need the required momentum in order to reach transition state and to form the products.

== F-H-H system ==

H2 + F potential energy surface:
[[File:H2 + F potential surface.PNG]]

Hf + H potential surface:
[[File:HF+H surface plot.PNG]]

From the potential surface we see that the products; H2 + F are higher in energy than the product F-H this suggests that the F-H bond is stronger than the H-H bond. This reaction is exothermic.
The reverse reaction must then be endothermic for H + HF

==Transition state position H2 + F ==
[[File:Energy and configuration for h2 + F.PNG]]

To find the transition state, initial conditions were set and momentum is set to 0. From this it is clear that it isnt optimised as there are forces present between HH and F.

Setting HH distance to 0.744 and HF distance to 1.811, the forces are approximately 0 and so this is where the transition state exists.

== Activation Energy H2 + F ==
[[File:Energy v time Activation energy for Fh2.PNG]]

Looking at the energy v time graph the difference between the total energy and the potential energy results in the activation

Activation energy = 104.019 - 103.749 = 0.270

==Transition state position HF + H ==
0.74 this resembles the H-H bond length in molecular H2
==Activation E HF + H ==

30

===F + H2 reaction trajectory: ===

[[File:Config for animation for h2 + f reaction path.PNG]]

These are the initial conditions selected that gave a reaction trajectory

[[File:Conservation of energy H2 +F.PNG]]

The momentum v time plot shows B-C (H-H) vibrating as F has little momentum. Once reaction happens, BC momentum stops oscillating and AB (HF) starts oscillating showing the conservation of energy as the vibration transfers from HH to HF.
The mechanism of release:

{| class="wikitable" border=1
! pHH !! Contour Plot!
|-
| 3.0 || [[File:Phh=3 contor.PNG]]
|-
| 2.0 || [[File:PPH2.PNG]]
|-
| 1.0 || [[File:PPH1.PNG]]
|-
| 0 || [[File:PPH0.PNG]]
|-
| -1.0 ||[[File:PPH-1.PNG]]
|-
| -2.0 || [[File:Pph-2.PNG]]
|-
| -3.0 || [[File:Pph-3.PNG]]
|}

As PHH approaches 3, the reaction has a larger vibration frequency when the product have formed hence the

[[File:0.1,-.08 contour.PNG]]
Vibrations in the products are significantly reduced and shows that the products are favourable, even past the transition state position calculated the

===H + HF reaction trajectory: ===

{| class="wikitable" border=1
! pHH !! pHF !! Etot !! Description of the dynamics !! Illustration of the trajectory
|-
| -10 || -0.5 || || ||
|-
| -9 || -2.0 || || ||
|-
| -1.5 || -2.5 || || ||
|-
| -2.5 || -5.0 || || ||
|-
| -2.5 || -5.2 || || ||
|}

MRD:ld2416

2019-05-24T14:06:35Z

Ld2416: /* H + H2 System */

==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?
[[File:Contour plot opf H H2.PNG]]

The transition state is when the H2 is equally distant to H1 and H3, for the contour plot this is approximately 0.93 A.

[[File:SUrface plot.PNG]]

The transition state can be seen at the saddle point, where the transition state is a minimum for the frank condon plot for the reaction. The transition state is a maximum for the reaction trajectory.
Mathematically, the TS has a second derivative greater than 0 in the Q1 direction (minimum) and TS has a second derivative less than 0 in the Q2 direction (maximum).

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.

From the contour plot we see that the transition state occurs when the r1=r2. Therefore to estimate the transition state position distances AB must be equal to BC.

[[File:Ab DIS.PNG]]

From this we can estimate the transition position from the contour plot:
[[File:Contour TS.PNG]]

This gives a value of 0.9075 A

Looking at this on a distance/time graph:
[[File:Bad ds TS.PNG]]

The momentum is set to 0 because the transition state is stationary and the distance between the 2 are equal.

Dynamic Surface:
[[File:Dynamic surface.PNG]]

MEP surface diagram:

[[File:MEP surface.PNG]]

Differences can be seen in the length and oscillations in the reaction trajectory. For the minimum energy pathway (MEP) there is no oscillations as there it neglects the molecules vibration modes and it moves infinitely slowly. The line is shorter as the reaction coordinate isn't as long due to the reaction.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 ||-99.0|| Yes||H2 initially not vibrating, upon reaction the new H2 has a high vibrational energy||[[File:P=-1.25,-2.5.PNG]]|
|-
| -1.5 || -2.0 ||-100.5||No|| H approaching is repelled, causing the HH bond to vibrate ||[[File:P=-1.5,-2.PNG]]|
|-
| -1.5 || -2.5 ||-99.0||Yes||Slow approach of H and H2 after reaction stronger vibrations||[[File:P=-1.5,-2.5.PNG]]|
|-
| -2.5 || -5.0 ||-85.0||No||H approaches H2 causes a the H-H bond to weaken almost up to a transition state, not enough energy to form a new HH bond| |[[File:P=-2.5,-5.PNG]]|
|-
| -2.5 || -5.2 ||-83.4||Yes||H on approach caused increase in vibration for H2 bond until it broke and new HH bond formed with the approaching H||[[File:P=-2.5,-5.2.PNG]]|
|}

== F-H-H system ==

H2 + F potential energy surface:
[[File:H2 + F potential surface.PNG]]

Hf + H potential surface:
[[File:HF+H surface plot.PNG]]

From the potential surface we see that the products; H2 + F are higher in energy than the product F-H this suggests that the F-H bond is stronger than the H-H bond. This reaction is exothermic.
The reverse reaction must then be endothermic for H + HF

==Transition state position H2 + F ==
[[File:Energy and configuration for h2 + F.PNG]]

To find the transition state, initial conditions were set and momentum is set to 0. From this it is clear that it isnt optimised as there are forces present between HH and F.

Setting HH distance to 0.744 and HF distance to 1.811, the forces are approximately 0 and so this is where the transition state exists.

== Activation Energy H2 + F ==
[[File:Energy v time Activation energy for Fh2.PNG]]

Looking at the energy v time graph the difference between the total energy and the potential energy results in the activation

Activation energy = 104.019 - 103.749 = 0.270

==Transition state position HF + H ==
0.74 this resembles the H-H bond length in molecular H2
==Activation E HF + H ==

30

===F + H2 reaction trajectory: ===

[[File:Config for animation for h2 + f reaction path.PNG]]

These are the initial conditions selected that gave a reaction trajectory

[[File:Conservation of energy H2 +F.PNG]]

The momentum v time plot shows B-C (H-H) vibrating as F has little momentum. Once reaction happens, BC momentum stops oscillating and AB (HF) starts oscillating showing the conservation of energy as the vibration transfers from HH to HF.
The mechanism of release:

{| class="wikitable" border=1
! pHH !! Contour Plot!
|-
| 3.0 || [[File:Phh=3 contor.PNG]]
|-
| 2.0 || [[File:PPH2.PNG]]
|-
| 1.0 || [[File:PPH1.PNG]]
|-
| 0 || [[File:PPH0.PNG]]
|-
| -1.0 ||[[File:PPH-1.PNG]]
|-
| -2.0 || [[File:Pph-2.PNG]]
|-
| -3.0 || [[File:Pph-3.PNG]]
|}

As PHH approaches 3, the reaction has a larger vibration frequency when the product have formed hence the

[[File:0.1,-.08 contour.PNG]]
Vibrations in the products are significantly reduced and shows that the products are favourable, even past the transition state position calculated the

===H + HF reaction trajectory: ===

{| class="wikitable" border=1
! pHH !! pHF !! Etot !! Description of the dynamics !! Illustration of the trajectory
|-
| -10 || -0.5 || || ||
|-
| -9 || -2.0 || || ||
|-
| -1.5 || -2.5 || || ||
|-
| -2.5 || -5.0 || || ||
|-
| -2.5 || -5.2 || || ||
|}

MRD:ld2416

2019-05-24T14:05:27Z

Ld2416: /* H + H2 System */

==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?
[[File:Contour plot opf H H2.PNG]]

The transition state is when the H2 is equally distant to H1 and H3, for the contour plot this is approximately 0.93 A.

[[File:SUrface plot.PNG]]

The transition state can be seen at the saddle point, where the transition state is a minimum for the frank condon plot for the reaction. The transition state is a maximum for the reaction trajectory.
Mathematically, the TS has a second derivative greater than 0 in the Q1 direction (minimum) and TS has a second derivative less than 0 in the Q2 direction (maximum).

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.

From the contour plot we see that the transition state occurs when the r1=r2. Therefore to estimate the transition state position distances AB must be equal to BC.

[[File:Ab DIS.PNG]]

From this we can estimate the transition position from the contour plot:
[[File:Contour TS.PNG]]

This gives a value of 0.9075 A

Looking at this on a distance/time graph:
[[File:Bad ds TS.PNG]]

The momentum is set to 0 because the transition state is stationary and the distance between the 2 are equal.

Dynamic Surface:
[[File:Dynamic surface.PNG]]

MEP surface diagram:

[[File:MEP surface.PNG]]

Differences can be seen in the length and oscillations in the reaction trajectory. For the minimum energy pathway (MEP) there is no oscillations as there it neglects the molecules vibration modes and it moves infinitely slowly. The line is shorter as the reaction coordinate isn't as long due to the reaction.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 ||-99.0|| Yes||H2 initially not vibrating, upon reaction the new H2 has a high vibrational energy||[[File:P=-1.25,-2.5.PNG]]|
|-
| -1.5 || -2.0 ||-100.5||No|| H approaching is repelled, causing the HH bond to vibrate ||[[File:P=-1.5,-2.PNG]]|
|-
| -1.5 || -2.5 ||-99.0||Yes||Slow approach of H and H2 after reaction stronger vibrations||[[File:P=-1.5,-2.5.PNG]]|
|-
| -2.5 || -5.0 ||-85.0||No||H approaches H2 causes a the H-H bond to weaken almost up to a transition state, not enough energy to form a new HH bond| |[[File:P=-2.5,-5.PNG]]|
|-
| -2.5 || -5.2 ||-83.4||Yes||H on approach caused increase in vibration for H2 bond until it broke and new HH bond formed with the approaching H| |[[File:P=-2.5,-5.2.PNG]]|
|}

== F-H-H system ==

H2 + F potential energy surface:
[[File:H2 + F potential surface.PNG]]

Hf + H potential surface:
[[File:HF+H surface plot.PNG]]

From the potential surface we see that the products; H2 + F are higher in energy than the product F-H this suggests that the F-H bond is stronger than the H-H bond. This reaction is exothermic.
The reverse reaction must then be endothermic for H + HF

==Transition state position H2 + F ==
[[File:Energy and configuration for h2 + F.PNG]]

To find the transition state, initial conditions were set and momentum is set to 0. From this it is clear that it isnt optimised as there are forces present between HH and F.

Setting HH distance to 0.744 and HF distance to 1.811, the forces are approximately 0 and so this is where the transition state exists.

== Activation Energy H2 + F ==
[[File:Energy v time Activation energy for Fh2.PNG]]

Looking at the energy v time graph the difference between the total energy and the potential energy results in the activation

Activation energy = 104.019 - 103.749 = 0.270

==Transition state position HF + H ==
0.74 this resembles the H-H bond length in molecular H2
==Activation E HF + H ==

30

===F + H2 reaction trajectory: ===

[[File:Config for animation for h2 + f reaction path.PNG]]

These are the initial conditions selected that gave a reaction trajectory

[[File:Conservation of energy H2 +F.PNG]]

The momentum v time plot shows B-C (H-H) vibrating as F has little momentum. Once reaction happens, BC momentum stops oscillating and AB (HF) starts oscillating showing the conservation of energy as the vibration transfers from HH to HF.
The mechanism of release:

{| class="wikitable" border=1
! pHH !! Contour Plot!
|-
| 3.0 || [[File:Phh=3 contor.PNG]]
|-
| 2.0 || [[File:PPH2.PNG]]
|-
| 1.0 || [[File:PPH1.PNG]]
|-
| 0 || [[File:PPH0.PNG]]
|-
| -1.0 ||[[File:PPH-1.PNG]]
|-
| -2.0 || [[File:Pph-2.PNG]]
|-
| -3.0 || [[File:Pph-3.PNG]]
|}

As PHH approaches 3, the reaction has a larger vibration frequency when the product have formed hence the

[[File:0.1,-.08 contour.PNG]]
Vibrations in the products are significantly reduced and shows that the products are favourable, even past the transition state position calculated the

===H + HF reaction trajectory: ===

{| class="wikitable" border=1
! pHH !! pHF !! Etot !! Description of the dynamics !! Illustration of the trajectory
|-
| -10 || -0.5 || || ||
|-
| -9 || -2.0 || || ||
|-
| -1.5 || -2.5 || || ||
|-
| -2.5 || -5.0 || || ||
|-
| -2.5 || -5.2 || || ||
|}

MRD:ld2416

2019-05-24T14:04:12Z

Ld2416: /* H + H2 System */

==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?
[[File:Contour plot opf H H2.PNG]]

The transition state is when the H2 is equally distant to H1 and H3, for the contour plot this is approximately 0.93 A.

[[File:SUrface plot.PNG]]

The transition state can be seen at the saddle point, where the transition state is a minimum for the frank condon plot for the reaction. The transition state is a maximum for the reaction trajectory.
Mathematically, the TS has a second derivative greater than 0 in the Q1 direction (minimum) and TS has a second derivative less than 0 in the Q2 direction (maximum).

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.

From the contour plot we see that the transition state occurs when the r1=r2. Therefore to estimate the transition state position distances AB must be equal to BC.

[[File:Ab DIS.PNG]]

From this we can estimate the transition position from the contour plot:
[[File:Contour TS.PNG]]

This gives a value of 0.9075 A

Looking at this on a distance/time graph:
[[File:Bad ds TS.PNG]]

The momentum is set to 0 because the transition state is stationary and the distance between the 2 are equal.

Dynamic Surface:
[[File:Dynamic surface.PNG]]

MEP surface diagram:

[[File:MEP surface.PNG]]

Differences can be seen in the length and oscillations in the reaction trajectory. For the minimum energy pathway (MEP) there is no oscillations as there it neglects the molecules vibration modes and it moves infinitely slowly. The line is shorter as the reaction coordinate isn't as long due to the reaction.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 ||-99.0|| Yes||H2 initially not vibrating, upon reaction the new H2 has a high vibrational energy||[[File:P=-1.25,-2.5.PNG]]|
|-
| -1.5 || -2.0 ||-100.5||No|| H approaching is repelled, causing the HH bond to vibrate ||[[File:P=-1.5,-2.PNG]]|
|-
| -1.5 || -2.5 ||-99.0||Yes||Slow approach of H and H2 after reaction stronger vibrations||[[File:P=-1.5,-2.5.PNG]]|
|-
| -2.5 || -5.0 ||-85.0||No| |H approaches H2 causes a the H-H bond to weaken almost up to a transition state, not enough energy to form a new HH bond| |[[File:P=-2.5,-5.PNG]]|
|-
| -2.5 || -5.2 ||-83.4||Yes| |H on approach caused increase in vibration for H2 bond until it broke and new HH bond formed with the approaching H| |[[File:P=-2.5,-5.2.PNG]]|
|}

== F-H-H system ==

H2 + F potential energy surface:
[[File:H2 + F potential surface.PNG]]

Hf + H potential surface:
[[File:HF+H surface plot.PNG]]

From the potential surface we see that the products; H2 + F are higher in energy than the product F-H this suggests that the F-H bond is stronger than the H-H bond. This reaction is exothermic.
The reverse reaction must then be endothermic for H + HF

==Transition state position H2 + F ==
[[File:Energy and configuration for h2 + F.PNG]]

To find the transition state, initial conditions were set and momentum is set to 0. From this it is clear that it isnt optimised as there are forces present between HH and F.

Setting HH distance to 0.744 and HF distance to 1.811, the forces are approximately 0 and so this is where the transition state exists.

== Activation Energy H2 + F ==
[[File:Energy v time Activation energy for Fh2.PNG]]

Looking at the energy v time graph the difference between the total energy and the potential energy results in the activation

Activation energy = 104.019 - 103.749 = 0.270

==Transition state position HF + H ==
0.74 this resembles the H-H bond length in molecular H2
==Activation E HF + H ==

30

===F + H2 reaction trajectory: ===

[[File:Config for animation for h2 + f reaction path.PNG]]

These are the initial conditions selected that gave a reaction trajectory

[[File:Conservation of energy H2 +F.PNG]]

The momentum v time plot shows B-C (H-H) vibrating as F has little momentum. Once reaction happens, BC momentum stops oscillating and AB (HF) starts oscillating showing the conservation of energy as the vibration transfers from HH to HF.
The mechanism of release:

{| class="wikitable" border=1
! pHH !! Contour Plot!
|-
| 3.0 || [[File:Phh=3 contor.PNG]]
|-
| 2.0 || [[File:PPH2.PNG]]
|-
| 1.0 || [[File:PPH1.PNG]]
|-
| 0 || [[File:PPH0.PNG]]
|-
| -1.0 ||[[File:PPH-1.PNG]]
|-
| -2.0 || [[File:Pph-2.PNG]]
|-
| -3.0 || [[File:Pph-3.PNG]]
|}

As PHH approaches 3, the reaction has a larger vibration frequency when the product have formed hence the

[[File:0.1,-.08 contour.PNG]]
Vibrations in the products are significantly reduced and shows that the products are favourable, even past the transition state position calculated the

===H + HF reaction trajectory: ===

{| class="wikitable" border=1
! pHH !! pHF !! Etot !! Description of the dynamics !! Illustration of the trajectory
|-
| -10 || -0.5 || || ||
|-
| -9 || -2.0 || || ||
|-
| -1.5 || -2.5 || || ||
|-
| -2.5 || -5.0 || || ||
|-
| -2.5 || -5.2 || || ||
|}

MRD:ld2416

2019-05-24T13:58:25Z

Ld2416: /* H + H2 collision */

File:P=-2.5,-5.PNG

2019-05-24T13:58:08Z

Ld2416:

File:P=-2.5,-5.2.PNG

2019-05-24T13:57:52Z

Ld2416:

File:P=-1.25,-2.5.PNG

2019-05-24T13:56:52Z

Ld2416:

File:P=-1.5,-2.PNG

2019-05-24T13:56:25Z

Ld2416:

File:P=-1.5,-2.5.PNG

2019-05-24T13:55:54Z

Ld2416:

File:P1=-1.5,p2=-2.5.PNG

2019-05-24T13:37:01Z

Ld2416:

File:P1=-1.5,p2=-2.0.PNG

2019-05-24T13:32:45Z

Ld2416:

File:P1=1.25,p2=-2.5.PNG

2019-05-24T13:31:12Z

Ld2416:

MRD:ld2416

2019-05-24T13:11:52Z

Ld2416: /* F + H2 reaction trajectory: */

==H + H2 collision==

On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?
[[File:Contour plot opf H H2.PNG]]

The transition state is when the H2 is equally distant to H1 and H3, for the contour plot this is approximately 0.93 A.

[[File:SUrface plot.PNG]]

The transition state can be seen at the saddle point, where the transition state is a minimum for the frank condon plot for the reaction. The transition state is a maximum for the reaction trajectory.
Mathematically, the TS has a second derivative greater than 0 in the Q1 direction (minimum) and TS has a second derivative less than 0 in the Q2 direction (maximum).

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.

From the contour plot we see that the transition state occurs when the r1=r2. Therefore to estimate the transition state position distances AB must be equal to BC.

[[File:Ab DIS.PNG]]

From this we can estimate the transition position from the contour plot:
[[File:Contour TS.PNG]]

This gives a value of 0.9075 A

Looking at this on a distance/time graph:
[[File:Bad ds TS.PNG]]

The momentum is set to 0 because the transition state is stationary and the distance between the 2 are equal.

Dynamic Surface:
[[File:Dynamic surface.PNG]]

MEP surface diagram:

[[File:MEP surface.PNG]]

Differences can be seen in the length and oscillations in the reaction trajectory. For the minimum energy pathway (MEP) there is no oscillations as there it neglects the molecules vibration modes and it moves infinitely slowly. The line is shorter as the reaction coordinate isn't as long due to the reaction.

== F-H-H system ==

H2 + F potential energy surface:
[[File:H2 + F potential surface.PNG]]

Hf + H potential surface:
[[File:HF+H surface plot.PNG]]

From the potential surface we see that the products; H2 + F are higher in energy than the product F-H this suggests that the F-H bond is stronger than the H-H bond. This reaction is exothermic.
The reverse reaction must then be endothermic for H + HF

==Transition state position H2 + F ==
[[File:Energy and configuration for h2 + F.PNG]]

To find the transition state, initial conditions were set and momentum is set to 0. From this it is clear that it isnt optimised as there are forces present between HH and F.

Setting HH distance to 0.744 and HF distance to 1.811, the forces are approximately 0 and so this is where the transition state exists.

== Activation Energy H2 + F ==
[[File:Energy v time Activation energy for Fh2.PNG]]

Looking at the energy v time graph the difference between the total energy and the potential energy results in the activation

Activation energy = 104.019 - 103.749 = 0.270

==Transition state position HF + H ==
0.74 this resembles the H-H bond length in molecular H2
==Activation E HF + H ==

30

===F + H2 reaction trajectory: ===

[[File:Config for animation for h2 + f reaction path.PNG]]

These are the initial conditions selected that gave a reaction trajectory

[[File:Conservation of energy H2 +F.PNG]]

The momentum v time plot shows B-C (H-H) vibrating as F has little momentum. Once reaction happens, BC momentum stops oscillating and AB (HF) starts oscillating showing the conservation of energy as the vibration transfers from HH to HF.
The mechanism of release:

{| class="wikitable" border=1
! pHH !! Contour Plot!
|-
| 3.0 || [[File:Phh=3 contor.PNG]]
|-
| 2.0 || [[File:PPH2.PNG]]
|-
| 1.0 || [[File:PPH1.PNG]]
|-
| 0 || [[File:PPH0.PNG]]
|-
| -1.0 ||[[File:PPH-1.PNG]]
|-
| -2.0 || [[File:Pph-2.PNG]]
|-
| -3.0 || [[File:Pph-3.PNG]]
|}

As PHH approaches 3, the reaction has a larger vibration frequency when the product have formed hence the

[[File:0.1,-.08 contour.PNG]]
Vibrations in the products are significantly reduced and shows that the products are favourable, even past the transition state position calculated the

===H + HF reaction trajectory: ===

{| class="wikitable" border=1
! pHH !! pHF !! Etot !! Description of the dynamics !! Illustration of the trajectory
|-
| -10 || -0.5 || || ||
|-
| -9 || -2.0 || || ||
|-
| -1.5 || -2.5 || || ||
|-
| -2.5 || -5.0 || || ||
|-
| -2.5 || -5.2 || || ||
|}

MRD:ld2416

2019-05-23T17:34:06Z

Ld2416: /* F-H-H system */

MRD:ld2416

2019-05-23T17:19:57Z

Ld2416: /* F-H-H system */

==H + H2 collision==

On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?
[[File:Contour plot opf H H2.PNG]]

The transition state is when the H2 is equally distant to H1 and H3, for the contour plot this is approximately 0.93 A.

[[File:SUrface plot.PNG]]

The transition state can be seen at the saddle point, where the transition state is a minimum for the frank condon plot for the reaction. The transition state is a maximum for the reaction trajectory.
Mathematically, the TS has a second derivative greater than 0 in the Q1 direction (minimum) and TS has a second derivative less than 0 in the Q2 direction (maximum).

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.

From the contour plot we see that the transition state occurs when the r1=r2. Therefore to estimate the transition state position distances AB must be equal to BC.

[[File:Ab DIS.PNG]]

From this we can estimate the transition position from the contour plot:
[[File:Contour TS.PNG]]

This gives a value of 0.9075 A

Looking at this on a distance/time graph:
[[File:Bad ds TS.PNG]]

The momentum is set to 0 because the transition state is stationary and the distance between the 2 are equal.

Dynamic Surface:
[[File:Dynamic surface.PNG]]

MEP surface diagram:

[[File:MEP surface.PNG]]

Differences can be seen in the length and oscillations in the reaction trajectory. For the minimum energy pathway (MEP) there is no oscillations as there it neglects the molecules vibration modes and it moves infinitely slowly. The line is shorter as the reaction coordinate isn't as long due to the reaction.

== F-H-H system ==

H2 + F potential energy surface:
[[File:H2 + F potential surface.PNG]]

Hf + H potential surface:
[[File:HF+H surface plot.PNG]]

From the potential surface we see that the products; H2 + F are higher in energy than the product F-H this suggests that the F-H bond is stronger than the H-H bond. This reaction is exothermic.
The reverse reaction must then be endothermic for H + HF

[[File:Energy and configuration for h2 + F.PNG]]

To find the transition state, initial conditions were set and momentum is set to 0. From this it is clear that it isnt optimised as there are forces present between HH and F.

Setting HH distance to 0.744 and HF distance to 1.811, the forces are approximately 0 and so this is where the transition state exists.

[[File:Energy v time Activation energy for Fh2.PNG]]

Looking at the energy v time graph the difference between the total energy and the potential energy results in the activation

Activation energy = 104.019 - 103.749 = 0.270

[[File:Config for animation for h2 + f reaction path.PNG]]

These are the initial conditions selected that gave a reaction trajectory

[[File:Conservation of energy H2 +F.PNG]]

The momentum v time plot shows B-C (H-H) vibrating as F has little momentum. Once reaction happens, BC momentum stops oscillating and AB (HF) starts oscillating showing the conservation of energy as the vibration transfers from HH to HF.
The mechanism of release:

===F + H2 reaction trajectory: ===

{| class="wikitable" border=1
! pHH !! Contour Plot!
|-
| 3.0 || [[File:Phh=3 contor.PNG]]
|-
| 2.0 || [[File:PPH2.PNG]]
|-
| 1.0 || [[File:PPH1.PNG]]
|-
| 0 || [[File:PPH0.PNG]]
|-
| -1.0 ||[[File:PPH-1.PNG]]
|-
| -2.0 || [[File:Pph-2.PNG]]
|-
| -3.0 || [[File:Pph-3.PNG]]
|}

As PHH approaches 3, the reaction has a larger vibration frequency when the product have formed hence the

[[File:0.1,-.08 contour.PNG]]
Vibrations in the products are significantly reduced and shows that the products are favourable, even past the transition state position calculated the

File:HF+H surface plot.PNG

2019-05-23T17:19:11Z

Ld2416:

MRD:ld2416

2019-05-23T17:12:15Z

Ld2416: /* F + H2 reaction trajectory: */

==H + H2 collision==

On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?
[[File:Contour plot opf H H2.PNG]]

The transition state is when the H2 is equally distant to H1 and H3, for the contour plot this is approximately 0.93 A.

[[File:SUrface plot.PNG]]

The transition state can be seen at the saddle point, where the transition state is a minimum for the frank condon plot for the reaction. The transition state is a maximum for the reaction trajectory.
Mathematically, the TS has a second derivative greater than 0 in the Q1 direction (minimum) and TS has a second derivative less than 0 in the Q2 direction (maximum).

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.

From the contour plot we see that the transition state occurs when the r1=r2. Therefore to estimate the transition state position distances AB must be equal to BC.

[[File:Ab DIS.PNG]]

From this we can estimate the transition position from the contour plot:
[[File:Contour TS.PNG]]

This gives a value of 0.9075 A

Looking at this on a distance/time graph:
[[File:Bad ds TS.PNG]]

The momentum is set to 0 because the transition state is stationary and the distance between the 2 are equal.

Dynamic Surface:
[[File:Dynamic surface.PNG]]

MEP surface diagram:

[[File:MEP surface.PNG]]

Differences can be seen in the length and oscillations in the reaction trajectory. For the minimum energy pathway (MEP) there is no oscillations as there it neglects the molecules vibration modes and it moves infinitely slowly. The line is shorter as the reaction coordinate isn't as long due to the reaction.

== F-H-H system ==

H2 + F potential energy surface:
[[File:H2 + F potential surface.PNG]]

From the potential surface we see that the products; H2 + F are higher in energy than the product F-H this suggests that the F-H bond is stronger than the H-H bond. This reaction is exothermic.
The reverse reaction must then be endothermic for H + HF
[[File:Energy and configuration for h2 + F.PNG]]

To find the transition state, initial conditions were set and momentum is set to 0. From this it is clear that it isnt optimised as there are forces present between HH and F.

Setting HH distance to 0.744 and HF distance to 1.811, the forces are approximately 0 and so this is where the transition state exists.

[[File:Energy v time Activation energy for Fh2.PNG]]

Looking at the energy v time graph the difference between the total energy and the potential energy results in the activation

Activation energy = 104.019 - 103.749 = 0.270

[[File:Config for animation for h2 + f reaction path.PNG]]

These are the initial conditions selected that gave a reaction trajectory

[[File:Conservation of energy H2 +F.PNG]]

The momentum v time plot shows B-C (H-H) vibrating as F has little momentum. Once reaction happens, BC momentum stops oscillating and AB (HF) starts oscillating showing the conservation of energy as the vibration transfers from HH to HF.
The mechanism of release:

===F + H2 reaction trajectory: ===

{| class="wikitable" border=1
! pHH !! Contour Plot!
|-
| 3.0 || [[File:Phh=3 contor.PNG]]
|-
| 2.0 || [[File:PPH2.PNG]]
|-
| 1.0 || [[File:PPH1.PNG]]
|-
| 0 || [[File:PPH0.PNG]]
|-
| -1.0 ||[[File:PPH-1.PNG]]
|-
| -2.0 || [[File:Pph-2.PNG]]
|-
| -3.0 || [[File:Pph-3.PNG]]
|}

As PHH approaches 3, the reaction has a larger vibration frequency when the product have formed hence the

[[File:0.1,-.08 contour.PNG]]
Vibrations in the products are significantly reduced and shows that the products are favourable, even past the transition state position calculated the