MRD:Kemicompwiki

2018-05-25T13:26:32Z

== Exercise 1 ==

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

The potential energy surface plot shown below is composed of both the AB distance and the BC distance, and the relative energy of each state. At the start of the reaction, on the left of the graph the AB distance is short, and the black dotted line shows small fluctuations in energy are due to the vibrations.

The transition state is shown by where the potential energy reaches a maxima and the kinetic energy reaches a minima. This is known as the saddle point, at approximately 0.75,0.75. To find the transition state, you need to differentiate both the potential energy and the kinetic energy and find out where both values are stationary. You then need to partially differentiate both to see where they cross, and this is known to be the saddle point.

[[File:Q1 pic jko cl.png]]

=== ''' A-C=B-C Trajectories ''' ===

Below is the closest estimate I could find for locating the transition state. The AB distance is equal to the BC at 0.9076Ǎ. In the internuclear distance vs Time graph below, both lines are close to flat and any oscillations have been minimised. By setting the momentum to zero, the parameters were optimised as in the transition state there is no oscillation so this is the best way to model it theoretically, both the kinetic and potential energy are at a stationary point and the atoms do not oscillate.

[[File:Q2 pic jko cl.png]]

===''' Calculating the reaction path '''===

{|class="wikitable" border=1
|r2=rts and r1=rts+δ, dynamics vs mep
|-
|[[File:Contour plot dynamics q2 jko.PNG|350px]]|| [[File:Ind vs time dynamics q2 jko.PNG|350px]]|| [[File:Inm vs time dynamics q2 jko.PNG|350px]]
|-
|[[File:Contour plot mep q2 jko.PNG|350px]]
|[[File:Ind vs time mep q2 jko.PNG|350px]]
|[[File:Inm vs time mep q2 jko.PNG|350px]]
|}

There are some clear differences between the MEP and dynamics plots. On the contour plots with the dynamics graph the energy shows oscillation in the energy well while in MEP the line is a smooth curve. Similar trends are observed when looking at the internuclear distance and the internulcear momentum plots. In the MEP by resetting the velocity to zero, you lose all information about the vibrational energy. In the dynamic plots no information is lost and as a result the oscillations are only seen in the dynamic plots. The MEP just shows the lowest energy path taken in the reaction.

If you were to change the r1=rts and the r2=rts + δ then the graphs would simply be a reflection of what is shown above. All plots would go in the opposite direction to the one that they are going in currently.

=== Reactive and Unreactive Trajectories ===
{| class="wikitable" border="1"
|+ caption
|-
| P1 || P2 || reactive or unreactive? || Contour plot ||Total Energy(kcalories) || Comments
|-
| -1.25 || -2.5 || Reactive|| [[File:Rvur plot 1 contour jko.PNG|400px]]|| -99.018|| With this specified momentum the reaction proceeds smoothly.
|-
| -1.5 || -2.0 || Unreactive || [[File:Rvur plot 2 contour jko.PNG|400px]]|| -100.455|| With these specified parameters the reaction does not occur. The reaction approaches the transition state then drops down as when the particles collide they do not have enough energy to react.
|-
| -1.5 || -2.5 || Reactive || [[File:Rvur plot 3 contour jko.PNG|400px]]|| -98.995|| The reaction occurs less smoothly that in the first reaction as the particles oscillate however they are still able to pass through the transition state and form the products.
|-
| -2.5 || -5.0 || Uneactive || [[File:Rvur plot 4 contour jko.PNG|400px]]|| -84.954|| The particles collide and are close to the activation energy however do not have enough to surpass is, meaning they go back to the reactants.
|-
| -2.5 || -5.2 || Reactive || [[File:Rvur plot 5 contour jko.PNG|400px]]|| -83.414|| With the momenta stated the particles are now able to surpass the activation energy and form the products, as shown by the plot. The molecules collide once without enough energy to surpass the transition state however the cross again, in a process called recrossing and form the products.
|}

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

The transition state theory states that in order to reach the the products the reactants must first overcome an energy barrier that aligns with the saddle point. The products must align in a specific configuration in order to be converted into the products. From the data above it is clear that this is not the case, as the reactants were able to overcome the energy barrier using more than one graph, as shown by the first third and fifth contour plots that were all able to react with the correct momentum, even if they had differetn geometries when passing through the transition state.

Transition state theory is based solely upon a classical approach, therefore cannot account for effects like tunneling and recrossing. It can be used to obtain a good estimate for whether or not a reaction has enough energy to react however once the activation energy is reached it becomes less accurate as you drift further from the classical picture.

== Exercise 2 ==

When colliding a flourine atome with a H2 molecule the reaction is exothermic. As shown in the surface plot below the transition state is passed through easily and the product channel has a lower energy than the reactant channel, which is evidence that it is an exothermic reaction. If the reaction were done in reverse it should be endothermic as the reactants would be at a lower energy than the products, making it an energy consuming process. From this it can also be deduced that the H-F bond is much stronger than the H-H bond as more energy is required to break it.

[[File:Hh f surface plot jko exercise 2.PNG|500px]]

When HF and another hydrogen atom collide with each other the only possible reaction is a proton transfer, technically no chemical reaction takes place as the reactants are equivalent to the products. If the momentum is high enough this can take place, as shown in the surface plot below. both the reactants and products are the same energy, as they are effectively the same.
[[File:Hfh surface plot jko exercise 2.PNG|500px]]

=== Locating the transition state ===

In the transition state it was determined that the H-H distance was 0.747Å the H-F distance was 1.810Å. The H-H distance is very close to that of the H-H bond, and the H-F distance is similar to that of the original one. This shows that the transition state energy lies closer to the reactants than the products.

[[File:Ts exercise 2 jko.PNG]]

===[[Reporting the activation energy for the reaction]]===

To calculate the activation energy an MEP plot was taken around the transition state calculated previously, and the change in energy has been assumed to be the activation energy.

[[File:Activation energy 1 jko exercise 2.PNG]]

The activation energy for the reaction of H2 reacting with a fluorine atom was measured to be approximately 0.2 kcal/mol.

[[File:Activation energy 2 jko exercise 2.PNG]]

The activation energy for the reaction of HF with H was calculated to be approximately 30kcal/mol.

The activation energy for the reaction between the fluorine and hydrogen is much lower than that between hydrogen fluoride and the hydrogen atom, which seems logical as the H-F bond is stronger than that of the H-H bond so more energy is required.

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

The set of initial conditions that resulted in a reactive trajectory were: A=H, B=H, C=F, AB distance = 0.74, BC distance = 2.3, AB momentum = -1.0, BC momentum = -1.5, shown in the contour plot and internuclear momenta vs time plot below.

[[File:Reaction dynamics contour jko.PNG]]

[[File:Reaction dynamics momenta plot jko.PNG]]

In the Internuclear momenta vs time graph, after 1 nanosecond there is a significant increase in momenta which is representative of the release of kinetic energy. This is an example of the first law on thermodynamics, where energy cannot be gained or lost only transferred. This energy has been released by breaking the H-H bond. The high oscillations could be a result of the H-F bond formation being very high energy. To investigate this experimentally you could measure the temperature change observed as the reaction proceeds.

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

MRD:Kemicompwiki

2018-05-25T08:43:03Z

Jko116: /* In Light of the Fact that energy is conserved, discuss the mechanism of the release of the reaction energy. How could this be confirmed experimentally? */

== Exercise 1 ==

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

The potential energy surface plot shown below is composed of both the AB distance and the BC distance, and the relative energy of each state. At the start of the reaction, on the left of the graph the AB distance is short, and the black dotted line shows small fluctuations in energy are due to the vibrations.

The transition state is shown by where the potential energy reaches a maxima and the kinetic energy reaches a minima. This is known as the saddle point, at approximately 0.75,0.75. To find the transition state, you need to differentiate both the potential energy and the kinetic energy and find out where both values are stationary. You then need to partially differentiate both to see where they cross, and this is known to be the saddle point.

[[File:Q1 pic jko cl.png]]

=== ''' A-C=B-C Trajectories ''' ===

Below is the closest estimate I could find for locating the transition state. The AB distance is equal to the BC at 0.9076Ǎ. In the internuclear distance vs Time graph below, both lines are close to flat and any oscillations have been minimised. By setting the momentum to zero, the parameters were optimised as in the transition state there is no oscillation so this is the best way to model it theoretically, both the kinetic and potential energy are at a stationary point and the atoms do not oscillate.

[[File:Q2 pic jko cl.png]]

===''' Calculating the reaction path '''===

{|class="wikitable" border=1
|r2=rts and r1=rts+δ, dynamics vs mep
|-
|[[File:Contour plot dynamics q2 jko.PNG|350px]]|| [[File:Ind vs time dynamics q2 jko.PNG|350px]]|| [[File:Inm vs time dynamics q2 jko.PNG|350px]]
|-
|[[File:Contour plot mep q2 jko.PNG|350px]]
|[[File:Ind vs time mep q2 jko.PNG|350px]]
|[[File:Inm vs time mep q2 jko.PNG|350px]]
|}

There are some clear differences between the MEP and dynamics plots. On the contour plots with the dynamics graph the energy shows oscillation in the energy well while in MEP the line is a smooth curve. Similar trends are observed when looking at the internuclear distance and the internulcear momentum plots. In the MEP by resetting the velocity to zero, you lose all information about the vibrational energy. In the dynamic plots no information is lost and as a result the oscillations are only seen in the dynamic plots. The MEP just shows the lowest energy path taken in the reaction.

If you were to change the r1=rts and the r2=rts + δ then the graphs would simply be a reflection of what is shown above. All plots would go in the opposite direction to the one that they are going in currently.

=== Reactive and Unreactive Trajectories ===
{| class="wikitable" border="1"
|+ caption
|-
| P1 || P2 || reactive or unreactive? || Contour plot ||Total Energy(kcalories) || Comments
|-
| -1.25 || -2.5 || Reactive|| [[File:Rvur plot 1 contour jko.PNG|400px]]|| -99.018|| With this specified momentum the reaction proceeds smoothly.
|-
| -1.5 || -2.0 || Unreactive || [[File:Rvur plot 2 contour jko.PNG|400px]]|| -100.455|| With these specified parameters the reaction does not occur. The reaction approaches the transition state then drops down as when the particles collide they do not have enough energy to react.
|-
| -1.5 || -2.5 || Reactive || [[File:Rvur plot 3 contour jko.PNG|400px]]|| -98.995|| The reaction occurs less smoothly that in the first reaction as the particles oscillate however they are still able to pass through the transition state and form the products.
|-
| -2.5 || -5.0 || Uneactive || [[File:Rvur plot 4 contour jko.PNG|400px]]|| -84.954|| The particles collide and are close to the activation energy however do not have enough to surpass is, meaning they go back to the reactants.
|-
| -2.5 || -5.2 || Reactive || [[File:Rvur plot 5 contour jko.PNG|400px]]|| -83.414|| With the momenta stated the particles are now able to surpass the activation energy and form the products, as shown by the plot. The molecules collide once without enough energy to surpass the transition state however the cross again, in a process called recrossing and form the products.
|}

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

The transition state theory states that in order to reach the the products the reactants must first overcome an energy barrier that aligns with the saddle point. The products must align in a specific configuration in order to be converted into the products. From the data above it is clear that this is not the case, as the reactants were able to overcome the energy barrier using more than one graph, as shown by the first third and fifth contour plots that were all able to react with the correct momentum, even if they had differetn geometries when passing through the transition state.

Transition state theory is based solely upon a classical approach, therefore cannot account for effects like tunneling and recrossing. It can be used to obtain a good estimate for whether or not a reaction has enough energy to react however once the activation energy is reached it becomes less accurate as you drift further from the classical picture.

== Exercise 2 ==

When colliding a flourine atome with a H2 molecule the reaction is exothermic. As shown in the surface plot below the transition state is passed through easily and the product channel has a lower energy than the reactant channel, which is evidence that it is an exothermic reaction. If the reaction were done in reverse it should be endothermic as the reactants would be at a lower energy than the products, making it an energy consuming process. From this it can also be deduced that the H-F bond is much stronger than the H-H bond as more energy is required to break it.

[[File:Hh f surface plot jko exercise 2.PNG|500px]]

When HF and another hydrogen atom collide with each other the only possible reaction is a proton transfer, technically no chemical reaction takes place as the reactants are equivalent to the products. If the momentum is high enough this can take place, as shown in the surface plot below. both the reactants and products are the same energy, as they are effectively the same.
[[File:Hfh surface plot jko exercise 2.PNG|500px]]

=== Locating the transition state ===

In the transition state it was determined that the H-H distance was 0.747Å the H-F distance was 1.810Å. The H-H distance is very close to that of the H-H bond, and the H-F distance is similar to that of the original one. This shows that the transition state energy lies closer to the reactants than the products.

[[File:Ts exercise 2 jko.PNG]]

===[[Reporting the activation energy for the reaction]]===

To calculate the activation energy an MEP plot was taken around the transition state calculated previously, and the change in energy has been assumed to be the activation energy.

[[File:Activation energy 1 jko exercise 2.PNG]]

The activation energy for the reaction of H2 reacting with a fluorine atom was measured to be approximately 0.2 kcal/mol.

[[File:Activation energy 2 jko exercise 2.PNG]]

The activation energy for the reaction of HF with H was calculated to be approximately 30kcal/mol.

The activation energy for the reaction between the fluorine and hydrogen is much lower than that between hydrogen fluoride and the hydrogen atom, which seems logical as the H-F bond is stronger than that of the H-H bond so more energy is required.

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

The set of initial conditions that resulted in a reactive trajectory were: A=H, B=H, C=F, AB distance = 0.74, BC distance = 2.3, AB momentum = -1.0, BC momentum = -1.5, shown in the contour plot and internuclear momenta vs time plot below.

[[File:Reaction dynamics contour jko.PNG]]

[[File:Reaction dynamics momenta plot jko.PNG]]

In the Internuclear momenta vs time graph, after 1 nanosecond there is a significant increase in momenta which is representative of the release of kinetic energy. This is an example of the first law on thermodynamics, where energy cannot be gained or lost only transferred. This energy has been released by breaking the H-H bond. The high oscillations could be a result of the H-F bond formation being very high energy. To investigate this experimentally you could measure the temperature change observed as the reaction proceeds.

MRD:Kemicompwiki

2018-05-24T22:11:06Z

Jko116: /* In Light of the Fact that energy is conserved, discuss the mechanism of the release of the reaction energy. How could this be confirmed experimentally? */

== Exercise 1 ==

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

The potential energy surface plot shown below is composed of both the AB distance and the BC distance, and the relative energy of each state. At the start of the reaction, on the left of the graph the AB distance is short, and the black dotted line shows small fluctuations in energy are due to the vibrations.

The transition state is shown by where the potential energy reaches a maxima and the kinetic energy reaches a minima. This is known as the saddle point, at approximately 0.75,0.75. To find the transition state, you need to differentiate both the potential energy and the kinetic energy and find out where both values are stationary. You then need to partially differentiate both to see where they cross, and this is known to be the saddle point.

[[File:Q1 pic jko cl.png]]

=== ''' A-C=B-C Trajectories ''' ===

Below is the closest estimate I could find for locating the transition state. The AB distance is equal to the BC at 0.9076Ǎ. In the internuclear distance vs Time graph below, both lines are close to flat and any oscillations have been minimised. By setting the momentum to zero, the parameters were optimised as in the transition state there is no oscillation so this is the best way to model it theoretically, both the kinetic and potential energy are at a stationary point and the atoms do not oscillate.

[[File:Q2 pic jko cl.png]]

===''' Calculating the reaction path '''===

{|class="wikitable" border=1
|r2=rts and r1=rts+δ, dynamics vs mep
|-
|[[File:Contour plot dynamics q2 jko.PNG|350px]]|| [[File:Ind vs time dynamics q2 jko.PNG|350px]]|| [[File:Inm vs time dynamics q2 jko.PNG|350px]]
|-
|[[File:Contour plot mep q2 jko.PNG|350px]]
|[[File:Ind vs time mep q2 jko.PNG|350px]]
|[[File:Inm vs time mep q2 jko.PNG|350px]]
|}

There are some clear differences between the MEP and dynamics plots. On the contour plots with the dynamics graph the energy shows oscillation in the energy well while in MEP the line is a smooth curve. Similar trends are observed when looking at the internuclear distance and the internulcear momentum plots. In the MEP by resetting the velocity to zero, you lose all information about the vibrational energy. In the dynamic plots no information is lost and as a result the oscillations are only seen in the dynamic plots. The MEP just shows the lowest energy path taken in the reaction.

If you were to change the r1=rts and the r2=rts + δ then the graphs would simply be a reflection of what is shown above. All plots would go in the opposite direction to the one that they are going in currently.

=== Reactive and Unreactive Trajectories ===
{| class="wikitable" border="1"
|+ caption
|-
| P1 || P2 || reactive or unreactive? || Contour plot ||Total Energy(kcalories) || Comments
|-
| -1.25 || -2.5 || Reactive|| [[File:Rvur plot 1 contour jko.PNG|400px]]|| -99.018|| With this specified momentum the reaction proceeds smoothly.
|-
| -1.5 || -2.0 || Unreactive || [[File:Rvur plot 2 contour jko.PNG|400px]]|| -100.455|| With these specified parameters the reaction does not occur. The reaction approaches the transition state then drops down as when the particles collide they do not have enough energy to react.
|-
| -1.5 || -2.5 || Reactive || [[File:Rvur plot 3 contour jko.PNG|400px]]|| -98.995|| The reaction occurs less smoothly that in the first reaction as the particles oscillate however they are still able to pass through the transition state and form the products.
|-
| -2.5 || -5.0 || Uneactive || [[File:Rvur plot 4 contour jko.PNG|400px]]|| -84.954|| The particles collide and are close to the activation energy however do not have enough to surpass is, meaning they go back to the reactants.
|-
| -2.5 || -5.2 || Reactive || [[File:Rvur plot 5 contour jko.PNG|400px]]|| -83.414|| With the momenta stated the particles are now able to surpass the activation energy and form the products, as shown by the plot. The molecules collide once without enough energy to surpass the transition state however the cross again, in a process called recrossing and form the products.
|}

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

The transition state theory states that in order to reach the the products the reactants must first overcome an energy barrier that aligns with the saddle point. The products must align in a specific configuration in order to be converted into the products. From the data above it is clear that this is not the case, as the reactants were able to overcome the energy barrier using more than one graph, as shown by the first third and fifth contour plots that were all able to react with the correct momentum, even if they had differetn geometries when passing through the transition state.

Transition state theory is based solely upon a classical approach, therefore cannot account for effects like tunneling and recrossing. It can be used to obtain a good estimate for whether or not a reaction has enough energy to react however once the activation energy is reached it becomes less accurate as you drift further from the classical picture.

== Exercise 2 ==

When colliding a flourine atome with a H2 molecule the reaction is exothermic. As shown in the surface plot below the transition state is passed through easily and the product channel has a lower energy than the reactant channel, which is evidence that it is an exothermic reaction. If the reaction were done in reverse it should be endothermic as the reactants would be at a lower energy than the products, making it an energy consuming process. From this it can also be deduced that the H-F bond is much stronger than the H-H bond as more energy is required to break it.

[[File:Hh f surface plot jko exercise 2.PNG|500px]]

When HF and another hydrogen atom collide with each other the only possible reaction is a proton transfer, technically no chemical reaction takes place as the reactants are equivalent to the products. If the momentum is high enough this can take place, as shown in the surface plot below. both the reactants and products are the same energy, as they are effectively the same.
[[File:Hfh surface plot jko exercise 2.PNG|500px]]

=== Locating the transition state ===

In the transition state it was determined that the H-H distance was 0.747Å the H-F distance was 1.810Å. The H-H distance is very close to that of the H-H bond, and the H-F distance is similar to that of the original one. This shows that the transition state energy lies closer to the reactants than the products.

[[File:Ts exercise 2 jko.PNG]]

===[[Reporting the activation energy for the reaction]]===

To calculate the activation energy an MEP plot was taken around the transition state calculated previously, and the change in energy has been assumed to be the activation energy.

[[File:Activation energy 1 jko exercise 2.PNG]]

The activation energy for the reaction of H2 reacting with a fluorine atom was measured to be approximately 0.2 kcal/mol.

[[File:Activation energy 2 jko exercise 2.PNG]]

The activation energy for the reaction of HF with H was calculated to be approximately 30kcal/mol.

The activation energy for the reaction between the fluorine and hydrogen is much lower than that between hydrogen fluoride and the hydrogen atom, which seems logical as the H-F bond is stronger than that of the H-H bond so more energy is required.

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

The set of initial conditions that resulted in a reactive trajectory were: A=H, B=H, C=F, AB distance = 0.74, BC distance = 2.3, AB momentum = -1.0, BC momentum = -1.5, shown in the contour plot and internuclear momenta vs time plot below.

[[File:Reaction dynamics contour jko.PNG]]

[[File:Reaction dynamics momenta plot jko.PNG]]

In the Internuclear momenta vs time graph, after 1 nanosecond there is a significant increase in momenta which is representative of the release of kinetic energy. This is an example of the first law on thermodynamics, where energy cannot be gained or lost only transferred. This energy has been released by breaking the H-H bond. The high oscillations could be a result of the H-F bond formation being very high energy.

MRD:Kemicompwiki

2018-05-24T21:58:38Z

Jko116: /* In Light of the Fact that energy is conserved, discuss the mechanism of the release of the reaction energy. How could this be confirmed experimentally? */

== Exercise 1 ==

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

The potential energy surface plot shown below is composed of both the AB distance and the BC distance, and the relative energy of each state. At the start of the reaction, on the left of the graph the AB distance is short, and the black dotted line shows small fluctuations in energy are due to the vibrations.

The transition state is shown by where the potential energy reaches a maxima and the kinetic energy reaches a minima. This is known as the saddle point, at approximately 0.75,0.75. To find the transition state, you need to differentiate both the potential energy and the kinetic energy and find out where both values are stationary. You then need to partially differentiate both to see where they cross, and this is known to be the saddle point.

[[File:Q1 pic jko cl.png]]

=== ''' A-C=B-C Trajectories ''' ===

Below is the closest estimate I could find for locating the transition state. The AB distance is equal to the BC at 0.9076Ǎ. In the internuclear distance vs Time graph below, both lines are close to flat and any oscillations have been minimised. By setting the momentum to zero, the parameters were optimised as in the transition state there is no oscillation so this is the best way to model it theoretically, both the kinetic and potential energy are at a stationary point and the atoms do not oscillate.

[[File:Q2 pic jko cl.png]]

===''' Calculating the reaction path '''===

{|class="wikitable" border=1
|r2=rts and r1=rts+δ, dynamics vs mep
|-
|[[File:Contour plot dynamics q2 jko.PNG|350px]]|| [[File:Ind vs time dynamics q2 jko.PNG|350px]]|| [[File:Inm vs time dynamics q2 jko.PNG|350px]]
|-
|[[File:Contour plot mep q2 jko.PNG|350px]]
|[[File:Ind vs time mep q2 jko.PNG|350px]]
|[[File:Inm vs time mep q2 jko.PNG|350px]]
|}

There are some clear differences between the MEP and dynamics plots. On the contour plots with the dynamics graph the energy shows oscillation in the energy well while in MEP the line is a smooth curve. Similar trends are observed when looking at the internuclear distance and the internulcear momentum plots. In the MEP by resetting the velocity to zero, you lose all information about the vibrational energy. In the dynamic plots no information is lost and as a result the oscillations are only seen in the dynamic plots. The MEP just shows the lowest energy path taken in the reaction.

If you were to change the r1=rts and the r2=rts + δ then the graphs would simply be a reflection of what is shown above. All plots would go in the opposite direction to the one that they are going in currently.

=== Reactive and Unreactive Trajectories ===
{| class="wikitable" border="1"
|+ caption
|-
| P1 || P2 || reactive or unreactive? || Contour plot ||Total Energy(kcalories) || Comments
|-
| -1.25 || -2.5 || Reactive|| [[File:Rvur plot 1 contour jko.PNG|400px]]|| -99.018|| With this specified momentum the reaction proceeds smoothly.
|-
| -1.5 || -2.0 || Unreactive || [[File:Rvur plot 2 contour jko.PNG|400px]]|| -100.455|| With these specified parameters the reaction does not occur. The reaction approaches the transition state then drops down as when the particles collide they do not have enough energy to react.
|-
| -1.5 || -2.5 || Reactive || [[File:Rvur plot 3 contour jko.PNG|400px]]|| -98.995|| The reaction occurs less smoothly that in the first reaction as the particles oscillate however they are still able to pass through the transition state and form the products.
|-
| -2.5 || -5.0 || Uneactive || [[File:Rvur plot 4 contour jko.PNG|400px]]|| -84.954|| The particles collide and are close to the activation energy however do not have enough to surpass is, meaning they go back to the reactants.
|-
| -2.5 || -5.2 || Reactive || [[File:Rvur plot 5 contour jko.PNG|400px]]|| -83.414|| With the momenta stated the particles are now able to surpass the activation energy and form the products, as shown by the plot. The molecules collide once without enough energy to surpass the transition state however the cross again, in a process called recrossing and form the products.
|}

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

The transition state theory states that in order to reach the the products the reactants must first overcome an energy barrier that aligns with the saddle point. The products must align in a specific configuration in order to be converted into the products. From the data above it is clear that this is not the case, as the reactants were able to overcome the energy barrier using more than one graph, as shown by the first third and fifth contour plots that were all able to react with the correct momentum, even if they had differetn geometries when passing through the transition state.

Transition state theory is based solely upon a classical approach, therefore cannot account for effects like tunneling and recrossing. It can be used to obtain a good estimate for whether or not a reaction has enough energy to react however once the activation energy is reached it becomes less accurate as you drift further from the classical picture.

== Exercise 2 ==

When colliding a flourine atome with a H2 molecule the reaction is exothermic. As shown in the surface plot below the transition state is passed through easily and the product channel has a lower energy than the reactant channel, which is evidence that it is an exothermic reaction. If the reaction were done in reverse it should be endothermic as the reactants would be at a lower energy than the products, making it an energy consuming process. From this it can also be deduced that the H-F bond is much stronger than the H-H bond as more energy is required to break it.

[[File:Hh f surface plot jko exercise 2.PNG|500px]]

When HF and another hydrogen atom collide with each other the only possible reaction is a proton transfer, technically no chemical reaction takes place as the reactants are equivalent to the products. If the momentum is high enough this can take place, as shown in the surface plot below. both the reactants and products are the same energy, as they are effectively the same.
[[File:Hfh surface plot jko exercise 2.PNG|500px]]

=== Locating the transition state ===

In the transition state it was determined that the H-H distance was 0.747Å the H-F distance was 1.810Å. The H-H distance is very close to that of the H-H bond, and the H-F distance is similar to that of the original one. This shows that the transition state energy lies closer to the reactants than the products.

[[File:Ts exercise 2 jko.PNG]]

===[[Reporting the activation energy for the reaction]]===

To calculate the activation energy an MEP plot was taken around the transition state calculated previously, and the change in energy has been assumed to be the activation energy.

[[File:Activation energy 1 jko exercise 2.PNG]]

The activation energy for the reaction of H2 reacting with a fluorine atom was measured to be approximately 0.2 kcal/mol.

[[File:Activation energy 2 jko exercise 2.PNG]]

The activation energy for the reaction of HF with H was calculated to be approximately 30kcal/mol.

The activation energy for the reaction between the fluorine and hydrogen is much lower than that between hydrogen fluoride and the hydrogen atom, which seems logical as the H-F bond is stronger than that of the H-H bond so more energy is required.

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

The set of initial conditions that resulted in a reactive trajectory were: A=H, B=H, C=F, AB distance = 0.74, BC distance = 2.3, AB momentum = -1.0, BC momentum = -1.5, shown in the contour plot and internuclear momenta vs time plot below.

[[File:Reaction dynamics contour jko.PNG]]

[[File:Reaction dynamics momenta plot jko.PNG]]

In the Internuclear momenta vs time graph, after 1 nanosecond there is a significant increase in momenta which is representative of the release of kinetic energy.

MRD:Kemicompwiki

2018-05-24T21:58:28Z

Jko116: /* In Light of the Fact that energy is conserved, discuss the mechanism of the release of the reaction energy. How could this be confirmed experimentally? */

== Exercise 1 ==

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

The potential energy surface plot shown below is composed of both the AB distance and the BC distance, and the relative energy of each state. At the start of the reaction, on the left of the graph the AB distance is short, and the black dotted line shows small fluctuations in energy are due to the vibrations.

The transition state is shown by where the potential energy reaches a maxima and the kinetic energy reaches a minima. This is known as the saddle point, at approximately 0.75,0.75. To find the transition state, you need to differentiate both the potential energy and the kinetic energy and find out where both values are stationary. You then need to partially differentiate both to see where they cross, and this is known to be the saddle point.

[[File:Q1 pic jko cl.png]]

=== ''' A-C=B-C Trajectories ''' ===

Below is the closest estimate I could find for locating the transition state. The AB distance is equal to the BC at 0.9076Ǎ. In the internuclear distance vs Time graph below, both lines are close to flat and any oscillations have been minimised. By setting the momentum to zero, the parameters were optimised as in the transition state there is no oscillation so this is the best way to model it theoretically, both the kinetic and potential energy are at a stationary point and the atoms do not oscillate.

[[File:Q2 pic jko cl.png]]

===''' Calculating the reaction path '''===

{|class="wikitable" border=1
|r2=rts and r1=rts+δ, dynamics vs mep
|-
|[[File:Contour plot dynamics q2 jko.PNG|350px]]|| [[File:Ind vs time dynamics q2 jko.PNG|350px]]|| [[File:Inm vs time dynamics q2 jko.PNG|350px]]
|-
|[[File:Contour plot mep q2 jko.PNG|350px]]
|[[File:Ind vs time mep q2 jko.PNG|350px]]
|[[File:Inm vs time mep q2 jko.PNG|350px]]
|}

There are some clear differences between the MEP and dynamics plots. On the contour plots with the dynamics graph the energy shows oscillation in the energy well while in MEP the line is a smooth curve. Similar trends are observed when looking at the internuclear distance and the internulcear momentum plots. In the MEP by resetting the velocity to zero, you lose all information about the vibrational energy. In the dynamic plots no information is lost and as a result the oscillations are only seen in the dynamic plots. The MEP just shows the lowest energy path taken in the reaction.

If you were to change the r1=rts and the r2=rts + δ then the graphs would simply be a reflection of what is shown above. All plots would go in the opposite direction to the one that they are going in currently.

=== Reactive and Unreactive Trajectories ===
{| class="wikitable" border="1"
|+ caption
|-
| P1 || P2 || reactive or unreactive? || Contour plot ||Total Energy(kcalories) || Comments
|-
| -1.25 || -2.5 || Reactive|| [[File:Rvur plot 1 contour jko.PNG|400px]]|| -99.018|| With this specified momentum the reaction proceeds smoothly.
|-
| -1.5 || -2.0 || Unreactive || [[File:Rvur plot 2 contour jko.PNG|400px]]|| -100.455|| With these specified parameters the reaction does not occur. The reaction approaches the transition state then drops down as when the particles collide they do not have enough energy to react.
|-
| -1.5 || -2.5 || Reactive || [[File:Rvur plot 3 contour jko.PNG|400px]]|| -98.995|| The reaction occurs less smoothly that in the first reaction as the particles oscillate however they are still able to pass through the transition state and form the products.
|-
| -2.5 || -5.0 || Uneactive || [[File:Rvur plot 4 contour jko.PNG|400px]]|| -84.954|| The particles collide and are close to the activation energy however do not have enough to surpass is, meaning they go back to the reactants.
|-
| -2.5 || -5.2 || Reactive || [[File:Rvur plot 5 contour jko.PNG|400px]]|| -83.414|| With the momenta stated the particles are now able to surpass the activation energy and form the products, as shown by the plot. The molecules collide once without enough energy to surpass the transition state however the cross again, in a process called recrossing and form the products.
|}

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

The transition state theory states that in order to reach the the products the reactants must first overcome an energy barrier that aligns with the saddle point. The products must align in a specific configuration in order to be converted into the products. From the data above it is clear that this is not the case, as the reactants were able to overcome the energy barrier using more than one graph, as shown by the first third and fifth contour plots that were all able to react with the correct momentum, even if they had differetn geometries when passing through the transition state.

Transition state theory is based solely upon a classical approach, therefore cannot account for effects like tunneling and recrossing. It can be used to obtain a good estimate for whether or not a reaction has enough energy to react however once the activation energy is reached it becomes less accurate as you drift further from the classical picture.

== Exercise 2 ==

When colliding a flourine atome with a H2 molecule the reaction is exothermic. As shown in the surface plot below the transition state is passed through easily and the product channel has a lower energy than the reactant channel, which is evidence that it is an exothermic reaction. If the reaction were done in reverse it should be endothermic as the reactants would be at a lower energy than the products, making it an energy consuming process. From this it can also be deduced that the H-F bond is much stronger than the H-H bond as more energy is required to break it.

[[File:Hh f surface plot jko exercise 2.PNG|500px]]

When HF and another hydrogen atom collide with each other the only possible reaction is a proton transfer, technically no chemical reaction takes place as the reactants are equivalent to the products. If the momentum is high enough this can take place, as shown in the surface plot below. both the reactants and products are the same energy, as they are effectively the same.
[[File:Hfh surface plot jko exercise 2.PNG|500px]]

=== Locating the transition state ===

In the transition state it was determined that the H-H distance was 0.747Å the H-F distance was 1.810Å. The H-H distance is very close to that of the H-H bond, and the H-F distance is similar to that of the original one. This shows that the transition state energy lies closer to the reactants than the products.

[[File:Ts exercise 2 jko.PNG]]

===[[Reporting the activation energy for the reaction]]===

To calculate the activation energy an MEP plot was taken around the transition state calculated previously, and the change in energy has been assumed to be the activation energy.

[[File:Activation energy 1 jko exercise 2.PNG]]

The activation energy for the reaction of H2 reacting with a fluorine atom was measured to be approximately 0.2 kcal/mol.

[[File:Activation energy 2 jko exercise 2.PNG]]

The activation energy for the reaction of HF with H was calculated to be approximately 30kcal/mol.

The activation energy for the reaction between the fluorine and hydrogen is much lower than that between hydrogen fluoride and the hydrogen atom, which seems logical as the H-F bond is stronger than that of the H-H bond so more energy is required.

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

The set of initial conditions that resulted in a reactive trajectory were: A=H, B=H, C=F, AB distance = 0.74, BC distance = 2.3, AB momentum = -1.0, BC momentum = -1.5, shown in the contour plot and internuclear momenta vs time plot below.

[[File:Reaction dynamics contour jko.PNG]]

[[File:Reaction dynamics momenta plot jko2.PNG]]

In the Internuclear momenta vs time graph, after 1 nanosecond there is a significant increase in momenta which is representative of the release of kinetic energy.

File:Reaction dynamics momenta plot jko.PNG

2018-05-24T21:57:05Z

Jko116: Jko116 uploaded a new version of File:Reaction dynamics momenta plot jko.PNG

File:Reaction dynamics momenta plot jko.PNG

2018-05-24T21:56:31Z

Jko116: Jko116 uploaded a new version of File:Reaction dynamics momenta plot jko.PNG

MRD:Kemicompwiki

2018-05-24T21:55:42Z

Jko116: /* In Light of the Fact that energy is conserved, discuss the mechanism of the release of the reaction energy. How could this be confirmed experimentally? */

== Exercise 1 ==

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

The potential energy surface plot shown below is composed of both the AB distance and the BC distance, and the relative energy of each state. At the start of the reaction, on the left of the graph the AB distance is short, and the black dotted line shows small fluctuations in energy are due to the vibrations.

The transition state is shown by where the potential energy reaches a maxima and the kinetic energy reaches a minima. This is known as the saddle point, at approximately 0.75,0.75. To find the transition state, you need to differentiate both the potential energy and the kinetic energy and find out where both values are stationary. You then need to partially differentiate both to see where they cross, and this is known to be the saddle point.

[[File:Q1 pic jko cl.png]]

=== ''' A-C=B-C Trajectories ''' ===

Below is the closest estimate I could find for locating the transition state. The AB distance is equal to the BC at 0.9076Ǎ. In the internuclear distance vs Time graph below, both lines are close to flat and any oscillations have been minimised. By setting the momentum to zero, the parameters were optimised as in the transition state there is no oscillation so this is the best way to model it theoretically, both the kinetic and potential energy are at a stationary point and the atoms do not oscillate.

[[File:Q2 pic jko cl.png]]

===''' Calculating the reaction path '''===

{|class="wikitable" border=1
|r2=rts and r1=rts+δ, dynamics vs mep
|-
|[[File:Contour plot dynamics q2 jko.PNG|350px]]|| [[File:Ind vs time dynamics q2 jko.PNG|350px]]|| [[File:Inm vs time dynamics q2 jko.PNG|350px]]
|-
|[[File:Contour plot mep q2 jko.PNG|350px]]
|[[File:Ind vs time mep q2 jko.PNG|350px]]
|[[File:Inm vs time mep q2 jko.PNG|350px]]
|}

There are some clear differences between the MEP and dynamics plots. On the contour plots with the dynamics graph the energy shows oscillation in the energy well while in MEP the line is a smooth curve. Similar trends are observed when looking at the internuclear distance and the internulcear momentum plots. In the MEP by resetting the velocity to zero, you lose all information about the vibrational energy. In the dynamic plots no information is lost and as a result the oscillations are only seen in the dynamic plots. The MEP just shows the lowest energy path taken in the reaction.

If you were to change the r1=rts and the r2=rts + δ then the graphs would simply be a reflection of what is shown above. All plots would go in the opposite direction to the one that they are going in currently.

=== Reactive and Unreactive Trajectories ===
{| class="wikitable" border="1"
|+ caption
|-
| P1 || P2 || reactive or unreactive? || Contour plot ||Total Energy(kcalories) || Comments
|-
| -1.25 || -2.5 || Reactive|| [[File:Rvur plot 1 contour jko.PNG|400px]]|| -99.018|| With this specified momentum the reaction proceeds smoothly.
|-
| -1.5 || -2.0 || Unreactive || [[File:Rvur plot 2 contour jko.PNG|400px]]|| -100.455|| With these specified parameters the reaction does not occur. The reaction approaches the transition state then drops down as when the particles collide they do not have enough energy to react.
|-
| -1.5 || -2.5 || Reactive || [[File:Rvur plot 3 contour jko.PNG|400px]]|| -98.995|| The reaction occurs less smoothly that in the first reaction as the particles oscillate however they are still able to pass through the transition state and form the products.
|-
| -2.5 || -5.0 || Uneactive || [[File:Rvur plot 4 contour jko.PNG|400px]]|| -84.954|| The particles collide and are close to the activation energy however do not have enough to surpass is, meaning they go back to the reactants.
|-
| -2.5 || -5.2 || Reactive || [[File:Rvur plot 5 contour jko.PNG|400px]]|| -83.414|| With the momenta stated the particles are now able to surpass the activation energy and form the products, as shown by the plot. The molecules collide once without enough energy to surpass the transition state however the cross again, in a process called recrossing and form the products.
|}

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

The transition state theory states that in order to reach the the products the reactants must first overcome an energy barrier that aligns with the saddle point. The products must align in a specific configuration in order to be converted into the products. From the data above it is clear that this is not the case, as the reactants were able to overcome the energy barrier using more than one graph, as shown by the first third and fifth contour plots that were all able to react with the correct momentum, even if they had differetn geometries when passing through the transition state.

Transition state theory is based solely upon a classical approach, therefore cannot account for effects like tunneling and recrossing. It can be used to obtain a good estimate for whether or not a reaction has enough energy to react however once the activation energy is reached it becomes less accurate as you drift further from the classical picture.

== Exercise 2 ==

When colliding a flourine atome with a H2 molecule the reaction is exothermic. As shown in the surface plot below the transition state is passed through easily and the product channel has a lower energy than the reactant channel, which is evidence that it is an exothermic reaction. If the reaction were done in reverse it should be endothermic as the reactants would be at a lower energy than the products, making it an energy consuming process. From this it can also be deduced that the H-F bond is much stronger than the H-H bond as more energy is required to break it.

[[File:Hh f surface plot jko exercise 2.PNG|500px]]

When HF and another hydrogen atom collide with each other the only possible reaction is a proton transfer, technically no chemical reaction takes place as the reactants are equivalent to the products. If the momentum is high enough this can take place, as shown in the surface plot below. both the reactants and products are the same energy, as they are effectively the same.
[[File:Hfh surface plot jko exercise 2.PNG|500px]]

=== Locating the transition state ===

In the transition state it was determined that the H-H distance was 0.747Å the H-F distance was 1.810Å. The H-H distance is very close to that of the H-H bond, and the H-F distance is similar to that of the original one. This shows that the transition state energy lies closer to the reactants than the products.

[[File:Ts exercise 2 jko.PNG]]

===[[Reporting the activation energy for the reaction]]===

To calculate the activation energy an MEP plot was taken around the transition state calculated previously, and the change in energy has been assumed to be the activation energy.

[[File:Activation energy 1 jko exercise 2.PNG]]

The activation energy for the reaction of H2 reacting with a fluorine atom was measured to be approximately 0.2 kcal/mol.

[[File:Activation energy 2 jko exercise 2.PNG]]

The activation energy for the reaction of HF with H was calculated to be approximately 30kcal/mol.

The activation energy for the reaction between the fluorine and hydrogen is much lower than that between hydrogen fluoride and the hydrogen atom, which seems logical as the H-F bond is stronger than that of the H-H bond so more energy is required.

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

The set of initial conditions that resulted in a reactive trajectory were: A=H, B=H, C=F, AB distance = 0.74, BC distance = 2.3, AB momentum = -1.0, BC momentum = -1.5, shown in the contour plot and internuclear momenta vs time plot below.

[[File:Reaction dynamics contour jko.PNG]]

[[File:Reaction dynamics momenta plot jko.PNG]]

In the Internuclear momenta vs time graph, after 1 nanosecond there is a significant increase in momenta which is representative of the release of kinetic energy.

MRD:Kemicompwiki

2018-05-24T21:35:47Z

Jko116: /* In Light of the Fact that energy is conserved, discuss the mechanism of the release of the reaction energy. How could this be confirmed experimentally? */

== Exercise 1 ==

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

The potential energy surface plot shown below is composed of both the AB distance and the BC distance, and the relative energy of each state. At the start of the reaction, on the left of the graph the AB distance is short, and the black dotted line shows small fluctuations in energy are due to the vibrations.

The transition state is shown by where the potential energy reaches a maxima and the kinetic energy reaches a minima. This is known as the saddle point, at approximately 0.75,0.75. To find the transition state, you need to differentiate both the potential energy and the kinetic energy and find out where both values are stationary. You then need to partially differentiate both to see where they cross, and this is known to be the saddle point.

[[File:Q1 pic jko cl.png]]

=== ''' A-C=B-C Trajectories ''' ===

Below is the closest estimate I could find for locating the transition state. The AB distance is equal to the BC at 0.9076Ǎ. In the internuclear distance vs Time graph below, both lines are close to flat and any oscillations have been minimised. By setting the momentum to zero, the parameters were optimised as in the transition state there is no oscillation so this is the best way to model it theoretically, both the kinetic and potential energy are at a stationary point and the atoms do not oscillate.

[[File:Q2 pic jko cl.png]]

===''' Calculating the reaction path '''===

{|class="wikitable" border=1
|r2=rts and r1=rts+δ, dynamics vs mep
|-
|[[File:Contour plot dynamics q2 jko.PNG|350px]]|| [[File:Ind vs time dynamics q2 jko.PNG|350px]]|| [[File:Inm vs time dynamics q2 jko.PNG|350px]]
|-
|[[File:Contour plot mep q2 jko.PNG|350px]]
|[[File:Ind vs time mep q2 jko.PNG|350px]]
|[[File:Inm vs time mep q2 jko.PNG|350px]]
|}

There are some clear differences between the MEP and dynamics plots. On the contour plots with the dynamics graph the energy shows oscillation in the energy well while in MEP the line is a smooth curve. Similar trends are observed when looking at the internuclear distance and the internulcear momentum plots. In the MEP by resetting the velocity to zero, you lose all information about the vibrational energy. In the dynamic plots no information is lost and as a result the oscillations are only seen in the dynamic plots. The MEP just shows the lowest energy path taken in the reaction.

If you were to change the r1=rts and the r2=rts + δ then the graphs would simply be a reflection of what is shown above. All plots would go in the opposite direction to the one that they are going in currently.

=== Reactive and Unreactive Trajectories ===
{| class="wikitable" border="1"
|+ caption
|-
| P1 || P2 || reactive or unreactive? || Contour plot ||Total Energy(kcalories) || Comments
|-
| -1.25 || -2.5 || Reactive|| [[File:Rvur plot 1 contour jko.PNG|400px]]|| -99.018|| With this specified momentum the reaction proceeds smoothly.
|-
| -1.5 || -2.0 || Unreactive || [[File:Rvur plot 2 contour jko.PNG|400px]]|| -100.455|| With these specified parameters the reaction does not occur. The reaction approaches the transition state then drops down as when the particles collide they do not have enough energy to react.
|-
| -1.5 || -2.5 || Reactive || [[File:Rvur plot 3 contour jko.PNG|400px]]|| -98.995|| The reaction occurs less smoothly that in the first reaction as the particles oscillate however they are still able to pass through the transition state and form the products.
|-
| -2.5 || -5.0 || Uneactive || [[File:Rvur plot 4 contour jko.PNG|400px]]|| -84.954|| The particles collide and are close to the activation energy however do not have enough to surpass is, meaning they go back to the reactants.
|-
| -2.5 || -5.2 || Reactive || [[File:Rvur plot 5 contour jko.PNG|400px]]|| -83.414|| With the momenta stated the particles are now able to surpass the activation energy and form the products, as shown by the plot. The molecules collide once without enough energy to surpass the transition state however the cross again, in a process called recrossing and form the products.
|}

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

The transition state theory states that in order to reach the the products the reactants must first overcome an energy barrier that aligns with the saddle point. The products must align in a specific configuration in order to be converted into the products. From the data above it is clear that this is not the case, as the reactants were able to overcome the energy barrier using more than one graph, as shown by the first third and fifth contour plots that were all able to react with the correct momentum, even if they had differetn geometries when passing through the transition state.

Transition state theory is based solely upon a classical approach, therefore cannot account for effects like tunneling and recrossing. It can be used to obtain a good estimate for whether or not a reaction has enough energy to react however once the activation energy is reached it becomes less accurate as you drift further from the classical picture.

== Exercise 2 ==

When colliding a flourine atome with a H2 molecule the reaction is exothermic. As shown in the surface plot below the transition state is passed through easily and the product channel has a lower energy than the reactant channel, which is evidence that it is an exothermic reaction. If the reaction were done in reverse it should be endothermic as the reactants would be at a lower energy than the products, making it an energy consuming process. From this it can also be deduced that the H-F bond is much stronger than the H-H bond as more energy is required to break it.

[[File:Hh f surface plot jko exercise 2.PNG|500px]]

When HF and another hydrogen atom collide with each other the only possible reaction is a proton transfer, technically no chemical reaction takes place as the reactants are equivalent to the products. If the momentum is high enough this can take place, as shown in the surface plot below. both the reactants and products are the same energy, as they are effectively the same.
[[File:Hfh surface plot jko exercise 2.PNG|500px]]

=== Locating the transition state ===

In the transition state it was determined that the H-H distance was 0.747Å the H-F distance was 1.810Å. The H-H distance is very close to that of the H-H bond, and the H-F distance is similar to that of the original one. This shows that the transition state energy lies closer to the reactants than the products.

[[File:Ts exercise 2 jko.PNG]]

===[[Reporting the activation energy for the reaction]]===

To calculate the activation energy an MEP plot was taken around the transition state calculated previously, and the change in energy has been assumed to be the activation energy.

[[File:Activation energy 1 jko exercise 2.PNG]]

The activation energy for the reaction of H2 reacting with a fluorine atom was measured to be approximately 0.2 kcal/mol.

[[File:Activation energy 2 jko exercise 2.PNG]]

The activation energy for the reaction of HF with H was calculated to be approximately 30kcal/mol.

The activation energy for the reaction between the fluorine and hydrogen is much lower than that between hydrogen fluoride and the hydrogen atom, which seems logical as the H-F bond is stronger than that of the H-H bond so more energy is required.

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

The set of initial conditions that resulted in a reactive trajectory were: A=H, B=H, C=F, AB distance = 0.74, BC distance = 2.3, AB momentum = -1.0, BC momentum = -1.5, shown in the contour plot and internuclear momenta vs time plot below.

[[File:Reaction dynamics contour jko.PNG]]

[[File:Reaction dynamics momenta plot jko.PNG]]

MRD:Kemicompwiki

2018-05-24T21:34:52Z

Jko116: /* In Light of the Fact that energy is conserved, discuss the mechanism of the release of the reaction energy. How could this be confirmed experimentally? */

== Exercise 1 ==

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

The potential energy surface plot shown below is composed of both the AB distance and the BC distance, and the relative energy of each state. At the start of the reaction, on the left of the graph the AB distance is short, and the black dotted line shows small fluctuations in energy are due to the vibrations.

The transition state is shown by where the potential energy reaches a maxima and the kinetic energy reaches a minima. This is known as the saddle point, at approximately 0.75,0.75. To find the transition state, you need to differentiate both the potential energy and the kinetic energy and find out where both values are stationary. You then need to partially differentiate both to see where they cross, and this is known to be the saddle point.

[[File:Q1 pic jko cl.png]]

=== ''' A-C=B-C Trajectories ''' ===

Below is the closest estimate I could find for locating the transition state. The AB distance is equal to the BC at 0.9076Ǎ. In the internuclear distance vs Time graph below, both lines are close to flat and any oscillations have been minimised. By setting the momentum to zero, the parameters were optimised as in the transition state there is no oscillation so this is the best way to model it theoretically, both the kinetic and potential energy are at a stationary point and the atoms do not oscillate.

[[File:Q2 pic jko cl.png]]

===''' Calculating the reaction path '''===

{|class="wikitable" border=1
|r2=rts and r1=rts+δ, dynamics vs mep
|-
|[[File:Contour plot dynamics q2 jko.PNG|350px]]|| [[File:Ind vs time dynamics q2 jko.PNG|350px]]|| [[File:Inm vs time dynamics q2 jko.PNG|350px]]
|-
|[[File:Contour plot mep q2 jko.PNG|350px]]
|[[File:Ind vs time mep q2 jko.PNG|350px]]
|[[File:Inm vs time mep q2 jko.PNG|350px]]
|}

There are some clear differences between the MEP and dynamics plots. On the contour plots with the dynamics graph the energy shows oscillation in the energy well while in MEP the line is a smooth curve. Similar trends are observed when looking at the internuclear distance and the internulcear momentum plots. In the MEP by resetting the velocity to zero, you lose all information about the vibrational energy. In the dynamic plots no information is lost and as a result the oscillations are only seen in the dynamic plots. The MEP just shows the lowest energy path taken in the reaction.

If you were to change the r1=rts and the r2=rts + δ then the graphs would simply be a reflection of what is shown above. All plots would go in the opposite direction to the one that they are going in currently.

=== Reactive and Unreactive Trajectories ===
{| class="wikitable" border="1"
|+ caption
|-
| P1 || P2 || reactive or unreactive? || Contour plot ||Total Energy(kcalories) || Comments
|-
| -1.25 || -2.5 || Reactive|| [[File:Rvur plot 1 contour jko.PNG|400px]]|| -99.018|| With this specified momentum the reaction proceeds smoothly.
|-
| -1.5 || -2.0 || Unreactive || [[File:Rvur plot 2 contour jko.PNG|400px]]|| -100.455|| With these specified parameters the reaction does not occur. The reaction approaches the transition state then drops down as when the particles collide they do not have enough energy to react.
|-
| -1.5 || -2.5 || Reactive || [[File:Rvur plot 3 contour jko.PNG|400px]]|| -98.995|| The reaction occurs less smoothly that in the first reaction as the particles oscillate however they are still able to pass through the transition state and form the products.
|-
| -2.5 || -5.0 || Uneactive || [[File:Rvur plot 4 contour jko.PNG|400px]]|| -84.954|| The particles collide and are close to the activation energy however do not have enough to surpass is, meaning they go back to the reactants.
|-
| -2.5 || -5.2 || Reactive || [[File:Rvur plot 5 contour jko.PNG|400px]]|| -83.414|| With the momenta stated the particles are now able to surpass the activation energy and form the products, as shown by the plot. The molecules collide once without enough energy to surpass the transition state however the cross again, in a process called recrossing and form the products.
|}

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

The transition state theory states that in order to reach the the products the reactants must first overcome an energy barrier that aligns with the saddle point. The products must align in a specific configuration in order to be converted into the products. From the data above it is clear that this is not the case, as the reactants were able to overcome the energy barrier using more than one graph, as shown by the first third and fifth contour plots that were all able to react with the correct momentum, even if they had differetn geometries when passing through the transition state.

Transition state theory is based solely upon a classical approach, therefore cannot account for effects like tunneling and recrossing. It can be used to obtain a good estimate for whether or not a reaction has enough energy to react however once the activation energy is reached it becomes less accurate as you drift further from the classical picture.

== Exercise 2 ==

When colliding a flourine atome with a H2 molecule the reaction is exothermic. As shown in the surface plot below the transition state is passed through easily and the product channel has a lower energy than the reactant channel, which is evidence that it is an exothermic reaction. If the reaction were done in reverse it should be endothermic as the reactants would be at a lower energy than the products, making it an energy consuming process. From this it can also be deduced that the H-F bond is much stronger than the H-H bond as more energy is required to break it.

[[File:Hh f surface plot jko exercise 2.PNG|500px]]

When HF and another hydrogen atom collide with each other the only possible reaction is a proton transfer, technically no chemical reaction takes place as the reactants are equivalent to the products. If the momentum is high enough this can take place, as shown in the surface plot below. both the reactants and products are the same energy, as they are effectively the same.
[[File:Hfh surface plot jko exercise 2.PNG|500px]]

=== Locating the transition state ===

In the transition state it was determined that the H-H distance was 0.747Å the H-F distance was 1.810Å. The H-H distance is very close to that of the H-H bond, and the H-F distance is similar to that of the original one. This shows that the transition state energy lies closer to the reactants than the products.

[[File:Ts exercise 2 jko.PNG]]

===[[Reporting the activation energy for the reaction]]===

To calculate the activation energy an MEP plot was taken around the transition state calculated previously, and the change in energy has been assumed to be the activation energy.

[[File:Activation energy 1 jko exercise 2.PNG]]

The activation energy for the reaction of H2 reacting with a fluorine atom was measured to be approximately 0.2 kcal/mol.

[[File:Activation energy 2 jko exercise 2.PNG]]

The activation energy for the reaction of HF with H was calculated to be approximately 30kcal/mol.

The activation energy for the reaction between the fluorine and hydrogen is much lower than that between hydrogen fluoride and the hydrogen atom, which seems logical as the H-F bond is stronger than that of the H-H bond so more energy is required.

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

The set of initial conditions that resulted in a reactive trajectory were: A=H, B=H, C=F, AB distance = 0.74, BC distance = 2.3, AB momentum = -1.0, BC momentum = -1.5, shown in the contour plot and internuclear momenta vs time plot below.

[[File:Reaction dynamics contour jko.PNG]]

File:Reaction dynamics momenta plot jko.PNG

File:Reaction dynamics momenta plot jko.PNG

2018-05-24T21:34:10Z

Jko116:

MRD:Kemicompwiki

2018-05-24T21:31:24Z

Jko116: /* In Light of the Fact that energy is conserved, discuss the mechanism of the release of the reaction energy. How could this be confirmed experimentally? */

File:Reaction dynamics contour jko.PNG

2018-05-24T21:30:30Z

Jko116: