ChemWiki - User contributions [en]

MRD:thl3318

2020-05-15T14:20:30Z

Thl3318: /* Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? */

== EXERCISE 1: H + H2 system ==
===On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?===

On a potential energy surface diagram, the transition state can be mathematically defined as a saddle point (which is also an unstable equilibrium). It can be identified by determining the point where the partial derivatives of the function potential with respect to the internuclear distances are equal to 0.

<math> \frac{\partial V}{\partial r_1} =0</math> and <math> \frac{\partial V}{\partial r_2} =0</math>

However, simply looking at the results for the first derivative does not give any information with regards to whether the point is a local minimum, a local maximum, or a saddle point. Analyzing the 2nd derivative then allows for the determination of whether the point is a saddle point. This can be done using the following equation:

<math> D=\frac{\partial^2 V}{\partial r_1^2} \frac{\partial^2 V}{\partial r_2^2} - (\frac{\partial^2 V}{\partial r_1\partial r_2})^2 </math>

If D<0, the point is therefore a saddle point and also the transition state.

===Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.===
My best estimate for the transition state position is when the A-B and B-C distance is equal to 90.8pm. This point in the "Internuclear Distance vs Time" plot can be rationed as the transition state position because when the bonds have no momentum the bond distances remain constant.

[[File:Surface_Plot_HH_TS_thl3318.png|thumb|300px|none|Internuclear Distance vs Time plot at Transition State]]

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

The mep and dynamics trajectory differ in many different ways. The first difference is the time it takes for the reaction to take place (in this case, for the distance to follow to the valley floor of the potential graph). In the mep trajectory, it takes around 30 steps (15 fs) for the reaction to proceed whereas in the dynamics trajectory this number is around 140 fs. The second difference is that the internuclear distance between the H2 molecule and the lone H atom increases logarithmically in the mep calculation but increases linearly in the dynamics calculation.

===Complete the table below by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?===
{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 ||-414.280 KJ.mol-1||Yes|| Plot shows trajectory going from reactant, going over the minimum energy path via the transition state and finally to form the product.
||[[File:Surface Plot 1 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -4.1 ||-420.077 KJ.mol-1||No||Plot shows the trajectory going up the potential slope towards the transition state. However it does not have enough momentum to go over the curve and thus the trajectory returns to its original position. The reaction does not proceed and the reactants remain.||[[File:Surface Plot 2 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -5.1 ||-413.997 KJ.mol-1||Yes||Plot shows that increasing the momentum of the H-H bond in the reactants results in more vibrational energy hence a more distinct wave pattern appears in the trajectory. Similar to the first plot, the trajectory shows the reactant forming the product by going over the minimum energy path via the transition state.||[[File:Surface Plot 3 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.1 ||-357.277 KJ.mol-1||Yes||Plot shows the trajectory having an "excess" amount of momentum causing the H atom in the middle to bounce back and forth between molecules. The trajectory eventually goes to the product but does not go through the transition state. ||[[File:Surface Plot 4 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.6 ||-349.477 KJ.mol-1||Yes||Plot is similar to that above in that the trajectory has an "excess" amount of momentum. The trajectory eventually goes to the product but does not go through the transition state. ||[[File:Surface Plot 5 thl3318.png|thumb|300px|none|]]
|}

From the table above, I can conclude that whether or not a trajectory results in a reaction depends on the momentum p2. Changing the momentum p1 does not affect the result of the trajectory but if momentum p2 is not great enough the result will be unreactive.

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

Transition State Theory predictions will overestimate rate values when compared with experimental values. There are many assumptions that are made in the Transition State Theory but the most important of them all is that multiple crossing do not occur on the potential surface. This assumption therefore assumes that the entire trajectory across the surface contributes to the reaction rate, failing to take into account the possibility of recrossings which actually lowers the reaction rate from reactants to products. As a result, Transition State Theory will overestimate reaction rate values.<ref>K. J. Laidler, Chemical Kinetics 3rd ed., Harper-Collins, 1987</ref>

== EXERCISE 2: F + H2 system ==
===By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?===

The F + H2 reaction is exothermic whereas the H + HF reaction is endothermic. This means that the H-F bond strength is greater than the H-H bond strength and is lower in energy.

===Locate the approximate position of the transition state.===
H-H distance = 77pm

H-F distance = 179.5pm

[[File:Surface_Plot_TS_3_thl3318.png|thumb|300px|none|Surface Plot showing Transition State with 500 steps at step size 0.21 fs ]]

===Report the activation energy for both reactions.===
transition state energy = -433.020

H-H energy = -435.100

H-F energy = -560.700

Activation energy for F + H2 --> HF + H = 2.080

Activation energy for H + HF --> H2 + F = 127.68

[[File:TS_enegy_vs_time_1_thl3318.png|thumb|300px|none|Energy vs Time plot showing TS energy and HF energy (step=51fs) ]]
[[File:TS_enegy_vs_time_2_thl3318.png|thumb|300px|none|Energy vs Time plot showing TS energy and H2 energy (step=51fs) ]]

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

Although energy is conserved in the reaction, there is a change from kinetic energy to potential energy during the collision. Upon the H2 molecule colliding with the F atom, an elastic collision occurs and the kinetic energy of the H2 molecule (in the form of translational motion) is converted into potential energy of the HF molecule (in the form of vibrational motion). This can be confirmed experimentally by measuring whether there is an increase in vibrational motion by using IR spectroscopy.

[[File:Momenta_plot_thl3318.png|thumb|300px|none|Plot showing momentum of bonds before and after collision where A and B are Hydrogen atoms and C is a Fluorine atom ]]

===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.===

For reactions with an early transition state (reactant-like), the efficiency of the reaction is most enhanced by the relative translation energy of the two atoms/molecules colliding. However for reactions with a late transition state (product-like), the efficiency of the reaction is most enhanced by vibrational excitation has a higher efficiency in enhancing the reactivity.<ref>H. Guo and K. Liu, Chem. Sci., 2016, 7, 3992-4003</ref>

==Reference==
<references />

MRD:thl3318

2020-05-15T14:18:56Z

Thl3318: /* Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? */

== EXERCISE 1: H + H2 system ==
===On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?===

On a potential energy surface diagram, the transition state can be mathematically defined as a saddle point (which is also an unstable equilibrium). It can be identified by determining the point where the partial derivatives of the function potential with respect to the internuclear distances are equal to 0.

<math> \frac{\partial V}{\partial r_1} =0</math> and <math> \frac{\partial V}{\partial r_2} =0</math>

However, simply looking at the results for the first derivative does not give any information with regards to whether the point is a local minimum, a local maximum, or a saddle point. Analyzing the 2nd derivative then allows for the determination of whether the point is a saddle point. This can be done using the following equation:

<math> D=\frac{\partial^2 V}{\partial r_1^2} \frac{\partial^2 V}{\partial r_2^2} - (\frac{\partial^2 V}{\partial r_1\partial r_2})^2 </math>

If D<0, the point is therefore a saddle point and also the transition state.

===Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.===
My best estimate for the transition state position is when the A-B and B-C distance is equal to 90.8pm. This point in the "Internuclear Distance vs Time" plot can be rationed as the transition state position because when the bonds have no momentum the bond distances remain constant.

[[File:Surface_Plot_HH_TS_thl3318.png|thumb|300px|none|Internuclear Distance vs Time plot at Transition State]]

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

The mep and dynamics trajectory differ in many different ways. The first difference is the time it takes for the reaction to take place (in this case, for the distance to follow to the valley floor of the potential graph). In the mep trajectory, it takes around 30 steps (15 fs) for the reaction to proceed whereas in the dynamics trajectory this number is around 140 fs. The second difference is that the internuclear distance between the H2 molecule and the lone H atom increases logarithmically in the mep calculation but increases linearly in the dynamics calculation.

===Complete the table below by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?===
{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 ||-414.280 KJ.mol-1||Yes|| Plot shows trajectory going from reactant, going over the minimum energy path via the transition state and finally to form the product.
||[[File:Surface Plot 1 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -4.1 ||-420.077 KJ.mol-1||No||Plot shows the trajectory going up the potential slope towards the transition state. However it does not have enough momentum to go over the curve and thus the trajectory returns to its original position. The reaction does not proceed and the reactants remain.||[[File:Surface Plot 2 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -5.1 ||-413.997 KJ.mol-1||Yes||Plot shows that increasing the momentum of the H-H bond in the reactants results in more vibrational energy hence a more distinct wave pattern appears in the trajectory. Similar to the first plot, the trajectory shows the reactant forming the product by going over the minimum energy path via the transition state.||[[File:Surface Plot 3 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.1 ||-357.277 KJ.mol-1||Yes||Plot shows the trajectory having an "excess" amount of momentum causing the H atom in the middle to bounce back and forth between molecules. The trajectory eventually goes to the product but does not go through the transition state. ||[[File:Surface Plot 4 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.6 ||-349.477 KJ.mol-1||Yes||Plot is similar to that above in that the trajectory has an "excess" amount of momentum. The trajectory eventually goes to the product but does not go through the transition state. ||[[File:Surface Plot 5 thl3318.png|thumb|300px|none|]]
|}

From the table above, I can conclude that whether or not a trajectory results in a reaction depends on the momentum p2. Changing the momentum p1 does not affect the result of the trajectory but if momentum p2 is not great enough the result will be unreactive.

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

Transition State Theory predictions will overestimate rate values when compared with experimental values. There are many assumptions that are made in the Transition State Theory but the most important of them all is that multiple crossing do not occur on the potential surface. This assumption therefore assumes that the entire trajectory across the surface contributes to the reaction rate, failing to take into account the possibility of recrossings which actually lowers the reaction rate from reactants to products. As a result, Transition State Theory will overestimate reaction rate values.

== EXERCISE 2: F + H2 system ==
===By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?===

The F + H2 reaction is exothermic whereas the H + HF reaction is endothermic. This means that the H-F bond strength is greater than the H-H bond strength and is lower in energy.

===Locate the approximate position of the transition state.===
H-H distance = 77pm

H-F distance = 179.5pm

[[File:Surface_Plot_TS_3_thl3318.png|thumb|300px|none|Surface Plot showing Transition State with 500 steps at step size 0.21 fs ]]

===Report the activation energy for both reactions.===
transition state energy = -433.020

H-H energy = -435.100

H-F energy = -560.700

Activation energy for F + H2 --> HF + H = 2.080

Activation energy for H + HF --> H2 + F = 127.68

[[File:TS_enegy_vs_time_1_thl3318.png|thumb|300px|none|Energy vs Time plot showing TS energy and HF energy (step=51fs) ]]
[[File:TS_enegy_vs_time_2_thl3318.png|thumb|300px|none|Energy vs Time plot showing TS energy and H2 energy (step=51fs) ]]

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

Although energy is conserved in the reaction, there is a change from kinetic energy to potential energy during the collision. Upon the H2 molecule colliding with the F atom, an elastic collision occurs and the kinetic energy of the H2 molecule (in the form of translational motion) is converted into potential energy of the HF molecule (in the form of vibrational motion). This can be confirmed experimentally by measuring whether there is an increase in vibrational motion by using IR spectroscopy.

[[File:Momenta_plot_thl3318.png|thumb|300px|none|Plot showing momentum of bonds before and after collision where A and B are Hydrogen atoms and C is a Fluorine atom ]]

===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.===

For reactions with an early transition state (reactant-like), the efficiency of the reaction is most enhanced by the relative translation energy of the two atoms/molecules colliding. However for reactions with a late transition state (product-like), the efficiency of the reaction is most enhanced by vibrational excitation has a higher efficiency in enhancing the reactivity.<ref>H. Guo and K. Liu, Chem. Sci., 2016, 7, 3992-4003</ref>

==Reference==
<references />

MRD:thl3318

2020-05-15T13:20:10Z

Thl3318: /* Reference */

== EXERCISE 1: H + H2 system ==
===On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?===

On a potential energy surface diagram, the transition state can be mathematically defined as a saddle point (which is also an unstable equilibrium). It can be identified by determining the point where the partial derivatives of the function potential with respect to the internuclear distances are equal to 0.

<math> \frac{\partial V}{\partial r_1} =0</math> and <math> \frac{\partial V}{\partial r_2} =0</math>

However, simply looking at the results for the first derivative does not give any information with regards to whether the point is a local minimum, a local maximum, or a saddle point. Analyzing the 2nd derivative then allows for the determination of whether the point is a saddle point. This can be done using the following equation:

<math> D=\frac{\partial^2 V}{\partial r_1^2} \frac{\partial^2 V}{\partial r_2^2} - (\frac{\partial^2 V}{\partial r_1\partial r_2})^2 </math>

If D<0, the point is therefore a saddle point and also the transition state.

===Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.===
My best estimate for the transition state position is when the A-B and B-C distance is equal to 90.8pm. This point in the "Internuclear Distance vs Time" plot can be rationed as the transition state position because when the bonds have no momentum the bond distances remain constant.

[[File:Surface_Plot_HH_TS_thl3318.png|thumb|300px|none|Internuclear Distance vs Time plot at Transition State]]

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

The mep and dynamics trajectory differ in many different ways. The first difference is the time it takes for the reaction to take place (in this case, for the distance to follow to the valley floor of the potential graph). In the mep trajectory, it takes around 30 steps (15 fs) for the reaction to proceed whereas in the dynamics trajectory this number is around 140 fs. The second difference is that the internuclear distance between the H2 molecule and the lone H atom increases logarithmically in the mep calculation but increases linearly in the dynamics calculation.

===Complete the table below by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?===
{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 ||-414.280 KJ.mol-1||Yes|| Plot shows trajectory going from reactant, going over the minimum energy path via the transition state and finally to form the product.
||[[File:Surface Plot 1 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -4.1 ||-420.077 KJ.mol-1||No||Plot shows the trajectory going up the potential slope towards the transition state. However it does not have enough momentum to go over the curve and thus the trajectory returns to its original position. The reaction does not proceed and the reactants remain.||[[File:Surface Plot 2 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -5.1 ||-413.997 KJ.mol-1||Yes||Plot shows that increasing the momentum of the H-H bond in the reactants results in more vibrational energy hence a more distinct wave pattern appears in the trajectory. Similar to the first plot, the trajectory shows the reactant forming the product by going over the minimum energy path via the transition state.||[[File:Surface Plot 3 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.1 ||-357.277 KJ.mol-1||Yes||Plot shows the trajectory having an "excess" amount of momentum causing the H atom in the middle to bounce back and forth between molecules. The trajectory eventually goes to the product but does not go through the transition state. ||[[File:Surface Plot 4 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.6 ||-349.477 KJ.mol-1||Yes||Plot is similar to that above in that the trajectory has an "excess" amount of momentum. The trajectory eventually goes to the product but does not go through the transition state. ||[[File:Surface Plot 5 thl3318.png|thumb|300px|none|]]
|}

From the table above, I can conclude that whether or not a trajectory results in a reaction depends on the momentum p2. Changing the momentum p1 does not affect the result of the trajectory but if momentum p2 is not great enough the result will be unreactive.

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

Transition State Theory predictions will overestimate rate values when compared with experimental values. The first assumption of TST is that atoms behave according to classical mechanics (hence no quantum tunneling). The theory also assumes that the reaction system will past through the lowest saddle point (the transition state) which may not be true when higher energy levels begin to be populated at higher temperatures. Most importantly the reaction assumes that as long as the collision kinetic energy is greater than the activation energy the reaction will proceed, which is not true for all cases.

== EXERCISE 2: F + H2 system ==
===By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?===

The F + H2 reaction is exothermic whereas the H + HF reaction is endothermic. This means that the H-F bond strength is greater than the H-H bond strength and is lower in energy.

===Locate the approximate position of the transition state.===
H-H distance = 77pm

H-F distance = 179.5pm

[[File:Surface_Plot_TS_3_thl3318.png|thumb|300px|none|Surface Plot showing Transition State with 500 steps at step size 0.21 fs ]]

===Report the activation energy for both reactions.===
transition state energy = -433.020

H-H energy = -435.100

H-F energy = -560.700

Activation energy for F + H2 --> HF + H = 2.080

Activation energy for H + HF --> H2 + F = 127.68

[[File:TS_enegy_vs_time_1_thl3318.png|thumb|300px|none|Energy vs Time plot showing TS energy and HF energy (step=51fs) ]]
[[File:TS_enegy_vs_time_2_thl3318.png|thumb|300px|none|Energy vs Time plot showing TS energy and H2 energy (step=51fs) ]]

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

Although energy is conserved in the reaction, there is a change from kinetic energy to potential energy during the collision. Upon the H2 molecule colliding with the F atom, an elastic collision occurs and the kinetic energy of the H2 molecule (in the form of translational motion) is converted into potential energy of the HF molecule (in the form of vibrational motion). This can be confirmed experimentally by measuring whether there is an increase in vibrational motion by using IR spectroscopy.

[[File:Momenta_plot_thl3318.png|thumb|300px|none|Plot showing momentum of bonds before and after collision where A and B are Hydrogen atoms and C is a Fluorine atom ]]

===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.===

For reactions with an early transition state (reactant-like), the efficiency of the reaction is most enhanced by the relative translation energy of the two atoms/molecules colliding. However for reactions with a late transition state (product-like), the efficiency of the reaction is most enhanced by vibrational excitation has a higher efficiency in enhancing the reactivity.<ref>H. Guo and K. Liu, Chem. Sci., 2016, 7, 3992-4003</ref>

==Reference==
<references />

MRD:thl3318

2020-05-15T13:17:14Z

Thl3318: /* Reference */

== EXERCISE 1: H + H2 system ==
===On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?===

On a potential energy surface diagram, the transition state can be mathematically defined as a saddle point (which is also an unstable equilibrium). It can be identified by determining the point where the partial derivatives of the function potential with respect to the internuclear distances are equal to 0.

<math> \frac{\partial V}{\partial r_1} =0</math> and <math> \frac{\partial V}{\partial r_2} =0</math>

However, simply looking at the results for the first derivative does not give any information with regards to whether the point is a local minimum, a local maximum, or a saddle point. Analyzing the 2nd derivative then allows for the determination of whether the point is a saddle point. This can be done using the following equation:

<math> D=\frac{\partial^2 V}{\partial r_1^2} \frac{\partial^2 V}{\partial r_2^2} - (\frac{\partial^2 V}{\partial r_1\partial r_2})^2 </math>

If D<0, the point is therefore a saddle point and also the transition state.

===Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.===
My best estimate for the transition state position is when the A-B and B-C distance is equal to 90.8pm. This point in the "Internuclear Distance vs Time" plot can be rationed as the transition state position because when the bonds have no momentum the bond distances remain constant.

[[File:Surface_Plot_HH_TS_thl3318.png|thumb|300px|none|Internuclear Distance vs Time plot at Transition State]]

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

The mep and dynamics trajectory differ in many different ways. The first difference is the time it takes for the reaction to take place (in this case, for the distance to follow to the valley floor of the potential graph). In the mep trajectory, it takes around 30 steps (15 fs) for the reaction to proceed whereas in the dynamics trajectory this number is around 140 fs. The second difference is that the internuclear distance between the H2 molecule and the lone H atom increases logarithmically in the mep calculation but increases linearly in the dynamics calculation.

===Complete the table below by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?===
{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 ||-414.280 KJ.mol-1||Yes|| Plot shows trajectory going from reactant, going over the minimum energy path via the transition state and finally to form the product.
||[[File:Surface Plot 1 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -4.1 ||-420.077 KJ.mol-1||No||Plot shows the trajectory going up the potential slope towards the transition state. However it does not have enough momentum to go over the curve and thus the trajectory returns to its original position. The reaction does not proceed and the reactants remain.||[[File:Surface Plot 2 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -5.1 ||-413.997 KJ.mol-1||Yes||Plot shows that increasing the momentum of the H-H bond in the reactants results in more vibrational energy hence a more distinct wave pattern appears in the trajectory. Similar to the first plot, the trajectory shows the reactant forming the product by going over the minimum energy path via the transition state.||[[File:Surface Plot 3 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.1 ||-357.277 KJ.mol-1||Yes||Plot shows the trajectory having an "excess" amount of momentum causing the H atom in the middle to bounce back and forth between molecules. The trajectory eventually goes to the product but does not go through the transition state. ||[[File:Surface Plot 4 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.6 ||-349.477 KJ.mol-1||Yes||Plot is similar to that above in that the trajectory has an "excess" amount of momentum. The trajectory eventually goes to the product but does not go through the transition state. ||[[File:Surface Plot 5 thl3318.png|thumb|300px|none|]]
|}

From the table above, I can conclude that whether or not a trajectory results in a reaction depends on the momentum p2. Changing the momentum p1 does not affect the result of the trajectory but if momentum p2 is not great enough the result will be unreactive.

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

Transition State Theory predictions will overestimate rate values when compared with experimental values. The first assumption of TST is that atoms behave according to classical mechanics (hence no quantum tunneling). The theory also assumes that the reaction system will past through the lowest saddle point (the transition state) which may not be true when higher energy levels begin to be populated at higher temperatures. Most importantly the reaction assumes that as long as the collision kinetic energy is greater than the activation energy the reaction will proceed, which is not true for all cases.

== EXERCISE 2: F + H2 system ==
===By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?===

The F + H2 reaction is exothermic whereas the H + HF reaction is endothermic. This means that the H-F bond strength is greater than the H-H bond strength and is lower in energy.

===Locate the approximate position of the transition state.===
H-H distance = 77pm

H-F distance = 179.5pm

[[File:Surface_Plot_TS_3_thl3318.png|thumb|300px|none|Surface Plot showing Transition State with 500 steps at step size 0.21 fs ]]

===Report the activation energy for both reactions.===
transition state energy = -433.020

H-H energy = -435.100

H-F energy = -560.700

Activation energy for F + H2 --> HF + H = 2.080

Activation energy for H + HF --> H2 + F = 127.68

[[File:TS_enegy_vs_time_1_thl3318.png|thumb|300px|none|Energy vs Time plot showing TS energy and HF energy (step=51fs) ]]
[[File:TS_enegy_vs_time_2_thl3318.png|thumb|300px|none|Energy vs Time plot showing TS energy and H2 energy (step=51fs) ]]

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

Although energy is conserved in the reaction, there is a change from kinetic energy to potential energy during the collision. Upon the H2 molecule colliding with the F atom, an elastic collision occurs and the kinetic energy of the H2 molecule (in the form of translational motion) is converted into potential energy of the HF molecule (in the form of vibrational motion). This can be confirmed experimentally by measuring whether there is an increase in vibrational motion by using IR spectroscopy.

[[File:Momenta_plot_thl3318.png|thumb|300px|none|Plot showing momentum of bonds before and after collision where A and B are Hydrogen atoms and C is a Fluorine atom ]]

===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.===

For reactions with an early transition state (reactant-like), the efficiency of the reaction is most enhanced by the relative translation energy of the two atoms/molecules colliding. However for reactions with a late transition state (product-like), the efficiency of the reaction is most enhanced by vibrational excitation has a higher efficiency in enhancing the reactivity.<ref>H. Guo and K. Liu, Chem. Sci., 2016, 7, 3992-4003</ref>

===Reference===
<references />

MRD:thl3318

2020-05-15T13:17:02Z

Thl3318: /* References */

== EXERCISE 1: H + H2 system ==
===On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?===

On a potential energy surface diagram, the transition state can be mathematically defined as a saddle point (which is also an unstable equilibrium). It can be identified by determining the point where the partial derivatives of the function potential with respect to the internuclear distances are equal to 0.

<math> \frac{\partial V}{\partial r_1} =0</math> and <math> \frac{\partial V}{\partial r_2} =0</math>

However, simply looking at the results for the first derivative does not give any information with regards to whether the point is a local minimum, a local maximum, or a saddle point. Analyzing the 2nd derivative then allows for the determination of whether the point is a saddle point. This can be done using the following equation:

<math> D=\frac{\partial^2 V}{\partial r_1^2} \frac{\partial^2 V}{\partial r_2^2} - (\frac{\partial^2 V}{\partial r_1\partial r_2})^2 </math>

If D<0, the point is therefore a saddle point and also the transition state.

===Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.===
My best estimate for the transition state position is when the A-B and B-C distance is equal to 90.8pm. This point in the "Internuclear Distance vs Time" plot can be rationed as the transition state position because when the bonds have no momentum the bond distances remain constant.

[[File:Surface_Plot_HH_TS_thl3318.png|thumb|300px|none|Internuclear Distance vs Time plot at Transition State]]

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

The mep and dynamics trajectory differ in many different ways. The first difference is the time it takes for the reaction to take place (in this case, for the distance to follow to the valley floor of the potential graph). In the mep trajectory, it takes around 30 steps (15 fs) for the reaction to proceed whereas in the dynamics trajectory this number is around 140 fs. The second difference is that the internuclear distance between the H2 molecule and the lone H atom increases logarithmically in the mep calculation but increases linearly in the dynamics calculation.

===Complete the table below by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?===
{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 ||-414.280 KJ.mol-1||Yes|| Plot shows trajectory going from reactant, going over the minimum energy path via the transition state and finally to form the product.
||[[File:Surface Plot 1 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -4.1 ||-420.077 KJ.mol-1||No||Plot shows the trajectory going up the potential slope towards the transition state. However it does not have enough momentum to go over the curve and thus the trajectory returns to its original position. The reaction does not proceed and the reactants remain.||[[File:Surface Plot 2 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -5.1 ||-413.997 KJ.mol-1||Yes||Plot shows that increasing the momentum of the H-H bond in the reactants results in more vibrational energy hence a more distinct wave pattern appears in the trajectory. Similar to the first plot, the trajectory shows the reactant forming the product by going over the minimum energy path via the transition state.||[[File:Surface Plot 3 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.1 ||-357.277 KJ.mol-1||Yes||Plot shows the trajectory having an "excess" amount of momentum causing the H atom in the middle to bounce back and forth between molecules. The trajectory eventually goes to the product but does not go through the transition state. ||[[File:Surface Plot 4 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.6 ||-349.477 KJ.mol-1||Yes||Plot is similar to that above in that the trajectory has an "excess" amount of momentum. The trajectory eventually goes to the product but does not go through the transition state. ||[[File:Surface Plot 5 thl3318.png|thumb|300px|none|]]
|}

From the table above, I can conclude that whether or not a trajectory results in a reaction depends on the momentum p2. Changing the momentum p1 does not affect the result of the trajectory but if momentum p2 is not great enough the result will be unreactive.

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

Transition State Theory predictions will overestimate rate values when compared with experimental values. The first assumption of TST is that atoms behave according to classical mechanics (hence no quantum tunneling). The theory also assumes that the reaction system will past through the lowest saddle point (the transition state) which may not be true when higher energy levels begin to be populated at higher temperatures. Most importantly the reaction assumes that as long as the collision kinetic energy is greater than the activation energy the reaction will proceed, which is not true for all cases.

== EXERCISE 2: F + H2 system ==
===By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?===

The F + H2 reaction is exothermic whereas the H + HF reaction is endothermic. This means that the H-F bond strength is greater than the H-H bond strength and is lower in energy.

===Locate the approximate position of the transition state.===
H-H distance = 77pm

H-F distance = 179.5pm

[[File:Surface_Plot_TS_3_thl3318.png|thumb|300px|none|Surface Plot showing Transition State with 500 steps at step size 0.21 fs ]]

===Report the activation energy for both reactions.===
transition state energy = -433.020

H-H energy = -435.100

H-F energy = -560.700

Activation energy for F + H2 --> HF + H = 2.080

Activation energy for H + HF --> H2 + F = 127.68

[[File:TS_enegy_vs_time_1_thl3318.png|thumb|300px|none|Energy vs Time plot showing TS energy and HF energy (step=51fs) ]]
[[File:TS_enegy_vs_time_2_thl3318.png|thumb|300px|none|Energy vs Time plot showing TS energy and H2 energy (step=51fs) ]]

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

Although energy is conserved in the reaction, there is a change from kinetic energy to potential energy during the collision. Upon the H2 molecule colliding with the F atom, an elastic collision occurs and the kinetic energy of the H2 molecule (in the form of translational motion) is converted into potential energy of the HF molecule (in the form of vibrational motion). This can be confirmed experimentally by measuring whether there is an increase in vibrational motion by using IR spectroscopy.

[[File:Momenta_plot_thl3318.png|thumb|300px|none|Plot showing momentum of bonds before and after collision where A and B are Hydrogen atoms and C is a Fluorine atom ]]

===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.===

For reactions with an early transition state (reactant-like), the efficiency of the reaction is most enhanced by the relative translation energy of the two atoms/molecules colliding. However for reactions with a late transition state (product-like), the efficiency of the reaction is most enhanced by vibrational excitation has a higher efficiency in enhancing the reactivity.<ref>H. Guo and K. Liu, Chem. Sci., 2016, 7, 3992-4003</ref>

===Reference===
#
<references />

MRD:thl3318

2020-05-15T13:15:00Z

Thl3318: /* 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. */

== EXERCISE 1: H + H2 system ==
===On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?===

On a potential energy surface diagram, the transition state can be mathematically defined as a saddle point (which is also an unstable equilibrium). It can be identified by determining the point where the partial derivatives of the function potential with respect to the internuclear distances are equal to 0.

<math> \frac{\partial V}{\partial r_1} =0</math> and <math> \frac{\partial V}{\partial r_2} =0</math>

However, simply looking at the results for the first derivative does not give any information with regards to whether the point is a local minimum, a local maximum, or a saddle point. Analyzing the 2nd derivative then allows for the determination of whether the point is a saddle point. This can be done using the following equation:

<math> D=\frac{\partial^2 V}{\partial r_1^2} \frac{\partial^2 V}{\partial r_2^2} - (\frac{\partial^2 V}{\partial r_1\partial r_2})^2 </math>

If D<0, the point is therefore a saddle point and also the transition state.

===Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.===
My best estimate for the transition state position is when the A-B and B-C distance is equal to 90.8pm. This point in the "Internuclear Distance vs Time" plot can be rationed as the transition state position because when the bonds have no momentum the bond distances remain constant.

[[File:Surface_Plot_HH_TS_thl3318.png|thumb|300px|none|Internuclear Distance vs Time plot at Transition State]]

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

The mep and dynamics trajectory differ in many different ways. The first difference is the time it takes for the reaction to take place (in this case, for the distance to follow to the valley floor of the potential graph). In the mep trajectory, it takes around 30 steps (15 fs) for the reaction to proceed whereas in the dynamics trajectory this number is around 140 fs. The second difference is that the internuclear distance between the H2 molecule and the lone H atom increases logarithmically in the mep calculation but increases linearly in the dynamics calculation.

===Complete the table below by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?===
{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 ||-414.280 KJ.mol-1||Yes|| Plot shows trajectory going from reactant, going over the minimum energy path via the transition state and finally to form the product.
||[[File:Surface Plot 1 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -4.1 ||-420.077 KJ.mol-1||No||Plot shows the trajectory going up the potential slope towards the transition state. However it does not have enough momentum to go over the curve and thus the trajectory returns to its original position. The reaction does not proceed and the reactants remain.||[[File:Surface Plot 2 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -5.1 ||-413.997 KJ.mol-1||Yes||Plot shows that increasing the momentum of the H-H bond in the reactants results in more vibrational energy hence a more distinct wave pattern appears in the trajectory. Similar to the first plot, the trajectory shows the reactant forming the product by going over the minimum energy path via the transition state.||[[File:Surface Plot 3 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.1 ||-357.277 KJ.mol-1||Yes||Plot shows the trajectory having an "excess" amount of momentum causing the H atom in the middle to bounce back and forth between molecules. The trajectory eventually goes to the product but does not go through the transition state. ||[[File:Surface Plot 4 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.6 ||-349.477 KJ.mol-1||Yes||Plot is similar to that above in that the trajectory has an "excess" amount of momentum. The trajectory eventually goes to the product but does not go through the transition state. ||[[File:Surface Plot 5 thl3318.png|thumb|300px|none|]]
|}

From the table above, I can conclude that whether or not a trajectory results in a reaction depends on the momentum p2. Changing the momentum p1 does not affect the result of the trajectory but if momentum p2 is not great enough the result will be unreactive.

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

Transition State Theory predictions will overestimate rate values when compared with experimental values. The first assumption of TST is that atoms behave according to classical mechanics (hence no quantum tunneling). The theory also assumes that the reaction system will past through the lowest saddle point (the transition state) which may not be true when higher energy levels begin to be populated at higher temperatures. Most importantly the reaction assumes that as long as the collision kinetic energy is greater than the activation energy the reaction will proceed, which is not true for all cases.

== EXERCISE 2: F + H2 system ==
===By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?===

The F + H2 reaction is exothermic whereas the H + HF reaction is endothermic. This means that the H-F bond strength is greater than the H-H bond strength and is lower in energy.

===Locate the approximate position of the transition state.===
H-H distance = 77pm

H-F distance = 179.5pm

[[File:Surface_Plot_TS_3_thl3318.png|thumb|300px|none|Surface Plot showing Transition State with 500 steps at step size 0.21 fs ]]

===Report the activation energy for both reactions.===
transition state energy = -433.020

H-H energy = -435.100

H-F energy = -560.700

Activation energy for F + H2 --> HF + H = 2.080

Activation energy for H + HF --> H2 + F = 127.68

[[File:TS_enegy_vs_time_1_thl3318.png|thumb|300px|none|Energy vs Time plot showing TS energy and HF energy (step=51fs) ]]
[[File:TS_enegy_vs_time_2_thl3318.png|thumb|300px|none|Energy vs Time plot showing TS energy and H2 energy (step=51fs) ]]

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

Although energy is conserved in the reaction, there is a change from kinetic energy to potential energy during the collision. Upon the H2 molecule colliding with the F atom, an elastic collision occurs and the kinetic energy of the H2 molecule (in the form of translational motion) is converted into potential energy of the HF molecule (in the form of vibrational motion). This can be confirmed experimentally by measuring whether there is an increase in vibrational motion by using IR spectroscopy.

[[File:Momenta_plot_thl3318.png|thumb|300px|none|Plot showing momentum of bonds before and after collision where A and B are Hydrogen atoms and C is a Fluorine atom ]]

===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.===

For reactions with an early transition state (reactant-like), the efficiency of the reaction is most enhanced by the relative translation energy of the two atoms/molecules colliding. However for reactions with a late transition state (product-like), the efficiency of the reaction is most enhanced by vibrational excitation has a higher efficiency in enhancing the reactivity.

https://pubs.rsc.org/en/content/articlehtml/2016/sc/c6sc01066k?fbclid=IwAR2ewmOVzAbLMfnguNOtvv55flPVkD64qLPVAZMOgJvMcmIFE9WrEYM19Lo
DOI: 10.1039/C6SC01066K (Perspective) Chem. Sci., 2016, 7, 3992-4003

===References===
H. Guo and K. Liu, Chem. Sci., 2016, 7, 3992-4003

MRD:thl3318

2020-05-14T17:46:22Z

Thl3318: /* Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? */

== EXERCISE 1: H + H2 system ==
===On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?===

On a potential energy surface diagram, the transition state can be mathematically defined as a saddle point (which is also an unstable equilibrium). It can be identified by determining the point where the partial derivatives of the function potential with respect to the internuclear distances are equal to 0.

<math> \frac{\partial V}{\partial r_1} =0</math> and <math> \frac{\partial V}{\partial r_2} =0</math>

However, simply looking at the results for the first derivative does not give any information with regards to whether the point is a local minimum, a local maximum, or a saddle point. Analyzing the 2nd derivative then allows for the determination of whether the point is a saddle point. This can be done using the following equation:

<math> D=\frac{\partial^2 V}{\partial r_1^2} \frac{\partial^2 V}{\partial r_2^2} - (\frac{\partial^2 V}{\partial r_1\partial r_2})^2 </math>

If D<0, the point is therefore a saddle point and also the transition state.

===Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.===
My best estimate for the transition state position is when the A-B and B-C distance is equal to 90.8pm. This point in the "Internuclear Distance vs Time" plot can be rationed as the transition state position because when the bonds have no momentum the bond distances remain constant.

[[File:Surface_Plot_HH_TS_thl3318.png|thumb|300px|none|Internuclear Distance vs Time plot at Transition State]]

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

The mep and dynamics trajectory differ in many different ways. The first difference is the time it takes for the reaction to take place (in this case, for the distance to follow to the valley floor of the potential graph). In the mep trajectory, it takes around 30 steps (15 fs) for the reaction to proceed whereas in the dynamics trajectory this number is around 140 fs. The second difference is that the internuclear distance between the H2 molecule and the lone H atom increases logarithmically in the mep calculation but increases linearly in the dynamics calculation.

===Complete the table below by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?===
{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 ||-414.280 KJ.mol-1||Yes|| Plot shows trajectory going from reactant, going over the minimum energy path via the transition state and finally to form the product.
||[[File:Surface Plot 1 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -4.1 ||-420.077 KJ.mol-1||No||Plot shows the trajectory going up the potential slope towards the transition state. However it does not have enough momentum to go over the curve and thus the trajectory returns to its original position. The reaction does not proceed and the reactants remain.||[[File:Surface Plot 2 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -5.1 ||-413.997 KJ.mol-1||Yes||Plot shows that increasing the momentum of the H-H bond in the reactants results in more vibrational energy hence a more distinct wave pattern appears in the trajectory. Similar to the first plot, the trajectory shows the reactant forming the product by going over the minimum energy path via the transition state.||[[File:Surface Plot 3 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.1 ||-357.277 KJ.mol-1||Yes||Plot shows the trajectory having an "excess" amount of momentum causing the H atom in the middle to bounce back and forth between molecules. The trajectory eventually goes to the product but does not go through the transition state. ||[[File:Surface Plot 4 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.6 ||-349.477 KJ.mol-1||Yes||Plot is similar to that above in that the trajectory has an "excess" amount of momentum. The trajectory eventually goes to the product but does not go through the transition state. ||[[File:Surface Plot 5 thl3318.png|thumb|300px|none|]]
|}

From the table above, I can conclude that whether or not a trajectory results in a reaction depends on the momentum p2. Changing the momentum p1 does not affect the result of the trajectory but if momentum p2 is not great enough the result will be unreactive.

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

Transition State Theory predictions will overestimate rate values when compared with experimental values. The first assumption of TST is that atoms behave according to classical mechanics (hence no quantum tunneling). The theory also assumes that the reaction system will past through the lowest saddle point (the transition state) which may not be true when higher energy levels begin to be populated at higher temperatures. Most importantly the reaction assumes that as long as the collision kinetic energy is greater than the activation energy the reaction will proceed, which is not true for all cases.

== EXERCISE 2: F + H2 system ==
===By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?===

The F + H2 reaction is exothermic whereas the H + HF reaction is endothermic. This means that the H-F bond strength is greater than the H-H bond strength and is lower in energy.

===Locate the approximate position of the transition state.===
H-H distance = 77pm

H-F distance = 179.5pm

[[File:Surface_Plot_TS_3_thl3318.png|thumb|300px|none|Surface Plot showing Transition State with 500 steps at step size 0.21 fs ]]

===Report the activation energy for both reactions.===
transition state energy = -433.020

H-H energy = -435.100

H-F energy = -560.700

Activation energy for F + H2 --> HF + H = 2.080

Activation energy for H + HF --> H2 + F = 127.68

[[File:TS_enegy_vs_time_1_thl3318.png|thumb|300px|none|Energy vs Time plot showing TS energy and HF energy (step=51fs) ]]
[[File:TS_enegy_vs_time_2_thl3318.png|thumb|300px|none|Energy vs Time plot showing TS energy and H2 energy (step=51fs) ]]

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

Although energy is conserved in the reaction, there is a change from kinetic energy to potential energy during the collision. Upon the H2 molecule colliding with the F atom, an elastic collision occurs and the kinetic energy of the H2 molecule (in the form of translational motion) is converted into potential energy of the HF molecule (in the form of vibrational motion). This can be confirmed experimentally by measuring whether there is an increase in vibrational motion by using IR spectroscopy.

[[File:Momenta_plot_thl3318.png|thumb|300px|none|Plot showing momentum of bonds before and after collision where A and B are Hydrogen atoms and C is a Fluorine atom ]]

===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.===

For reactions with an early transition state (reactant-like), the efficiency of the reaction is most enhanced by the relative translation energy of the two atoms/molecules colliding. However for reactions with a late transition state (product-like), the efficiency of the reaction is most enhanced by vibrational excitation has a higher efficiency in enhancing the reactivity.

https://pubs.rsc.org/en/content/articlehtml/2016/sc/c6sc01066k?fbclid=IwAR2ewmOVzAbLMfnguNOtvv55flPVkD64qLPVAZMOgJvMcmIFE9WrEYM19Lo
DOI: 10.1039/C6SC01066K (Perspective) Chem. Sci., 2016, 7, 3992-4003

MRD:thl3318

2020-05-14T17:13:22Z

Thl3318: /* 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. */

== EXERCISE 1: H + H2 system ==
===On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?===

On a potential energy surface diagram, the transition state can be mathematically defined as a saddle point (which is also an unstable equilibrium). It can be identified by determining the point where the partial derivatives of the function potential with respect to the internuclear distances are equal to 0.

<math> \frac{\partial V}{\partial r_1} =0</math> and <math> \frac{\partial V}{\partial r_2} =0</math>

However, simply looking at the results for the first derivative does not give any information with regards to whether the point is a local minimum, a local maximum, or a saddle point. Analyzing the 2nd derivative then allows for the determination of whether the point is a saddle point. This can be done using the following equation:

<math> D=\frac{\partial^2 V}{\partial r_1^2} \frac{\partial^2 V}{\partial r_2^2} - (\frac{\partial^2 V}{\partial r_1\partial r_2})^2 </math>

If D<0, the point is therefore a saddle point and also the transition state.

===Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.===
My best estimate for the transition state position is when the A-B and B-C distance is equal to 90.8pm. This point in the "Internuclear Distance vs Time" plot can be rationed as the transition state position because when the bonds have no momentum the bond distances remain constant.

[[File:Surface_Plot_HH_TS_thl3318.png|thumb|300px|none|Internuclear Distance vs Time plot at Transition State]]

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

The mep and dynamics trajectory differ in many different ways. The first difference is the time it takes for the reaction to take place (in this case, for the distance to follow to the valley floor of the potential graph). In the mep trajectory, it takes around 30 steps (15 fs) for the reaction to proceed whereas in the dynamics trajectory this number is around 140 fs. The second difference is that the internuclear distance between the H2 molecule and the lone H atom increases logarithmically in the mep calculation but increases linearly in the dynamics calculation.

===Complete the table below by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?===
{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 ||-414.280 KJ.mol-1||Yes|| Plot shows trajectory going from reactant, going over the minimum energy path via the transition state and finally to form the product.
||[[File:Surface Plot 1 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -4.1 ||-420.077 KJ.mol-1||No||Plot shows the trajectory going up the potential slope towards the transition state. However it does not have enough momentum to go over the curve and thus the trajectory returns to its original position. The reaction does not proceed and the reactants remain.||[[File:Surface Plot 2 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -5.1 ||-413.997 KJ.mol-1||Yes||Plot shows that increasing the momentum of the H-H bond in the reactants results in more vibrational energy hence a more distinct wave pattern appears in the trajectory. Similar to the first plot, the trajectory shows the reactant forming the product by going over the minimum energy path via the transition state.||[[File:Surface Plot 3 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.1 ||-357.277 KJ.mol-1||Yes||Plot shows the trajectory having an "excess" amount of momentum causing the H atom in the middle to bounce back and forth between molecules. The trajectory eventually goes to the product but does not go through the transition state. ||[[File:Surface Plot 4 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.6 ||-349.477 KJ.mol-1||Yes||Plot is similar to that above in that the trajectory has an "excess" amount of momentum. The trajectory eventually goes to the product but does not go through the transition state. ||[[File:Surface Plot 5 thl3318.png|thumb|300px|none|]]
|}

From the table above, I can conclude that whether or not a trajectory results in a reaction depends on the momentum p2. Changing the momentum p1 does not affect the result of the trajectory but if momentum p2 is not great enough the result will be unreactive.

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

Experimental value will be greater than the reaction rate the transition state theory predicts because it does not take into account quantum tunneling which is more significant at low temperatures. Transition State Theory states that as long as the energy is greater than the activation energy the reaction will proceed to the products but this does not take into account the possibility of recrossing back to the reactants and thus the reaction not proceeding.

== EXERCISE 2: F + H2 system ==
===By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?===

The F + H2 reaction is exothermic whereas the H + HF reaction is endothermic. This means that the H-F bond strength is greater than the H-H bond strength and is lower in energy.

===Locate the approximate position of the transition state.===
H-H distance = 77pm

H-F distance = 179.5pm

[[File:Surface_Plot_TS_3_thl3318.png|thumb|300px|none|Surface Plot showing Transition State with 500 steps at step size 0.21 fs ]]

===Report the activation energy for both reactions.===
transition state energy = -433.020

H-H energy = -435.100

H-F energy = -560.700

Activation energy for F + H2 --> HF + H = 2.080

Activation energy for H + HF --> H2 + F = 127.68

[[File:TS_enegy_vs_time_1_thl3318.png|thumb|300px|none|Energy vs Time plot showing TS energy and HF energy (step=51fs) ]]
[[File:TS_enegy_vs_time_2_thl3318.png|thumb|300px|none|Energy vs Time plot showing TS energy and H2 energy (step=51fs) ]]

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

Although energy is conserved in the reaction, there is a change from kinetic energy to potential energy during the collision. Upon the H2 molecule colliding with the F atom, an elastic collision occurs and the kinetic energy of the H2 molecule (in the form of translational motion) is converted into potential energy of the HF molecule (in the form of vibrational motion). This can be confirmed experimentally by measuring whether there is an increase in vibrational motion by using IR spectroscopy.

[[File:Momenta_plot_thl3318.png|thumb|300px|none|Plot showing momentum of bonds before and after collision where A and B are Hydrogen atoms and C is a Fluorine atom ]]

===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.===

For reactions with an early transition state (reactant-like), the efficiency of the reaction is most enhanced by the relative translation energy of the two atoms/molecules colliding. However for reactions with a late transition state (product-like), the efficiency of the reaction is most enhanced by vibrational excitation has a higher efficiency in enhancing the reactivity.

https://pubs.rsc.org/en/content/articlehtml/2016/sc/c6sc01066k?fbclid=IwAR2ewmOVzAbLMfnguNOtvv55flPVkD64qLPVAZMOgJvMcmIFE9WrEYM19Lo
DOI: 10.1039/C6SC01066K (Perspective) Chem. Sci., 2016, 7, 3992-4003

MRD:thl3318

2020-05-14T17:12:38Z

Thl3318: /* Complete the table below by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? */

== EXERCISE 1: H + H2 system ==
===On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?===

On a potential energy surface diagram, the transition state can be mathematically defined as a saddle point (which is also an unstable equilibrium). It can be identified by determining the point where the partial derivatives of the function potential with respect to the internuclear distances are equal to 0.

<math> \frac{\partial V}{\partial r_1} =0</math> and <math> \frac{\partial V}{\partial r_2} =0</math>

However, simply looking at the results for the first derivative does not give any information with regards to whether the point is a local minimum, a local maximum, or a saddle point. Analyzing the 2nd derivative then allows for the determination of whether the point is a saddle point. This can be done using the following equation:

<math> D=\frac{\partial^2 V}{\partial r_1^2} \frac{\partial^2 V}{\partial r_2^2} - (\frac{\partial^2 V}{\partial r_1\partial r_2})^2 </math>

If D<0, the point is therefore a saddle point and also the transition state.

===Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.===
My best estimate for the transition state position is when the A-B and B-C distance is equal to 90.8pm. This point in the "Internuclear Distance vs Time" plot can be rationed as the transition state position because when the bonds have no momentum the bond distances remain constant.

Surface_Plot_HH_TS_thl3318.png

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

The mep and dynamics trajectory differ in many different ways. The first difference is the time it takes for the reaction to take place (in this case, for the distance to follow to the valley floor of the potential graph). In the mep trajectory, it takes around 30 steps (15 fs) for the reaction to proceed whereas in the dynamics trajectory this number is around 140 fs. The second difference is that the internuclear distance between the H2 molecule and the lone H atom increases logarithmically in the mep calculation but increases linearly in the dynamics calculation.

===Complete the table below by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?===
{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 ||-414.280 KJ.mol-1||Yes|| Plot shows trajectory going from reactant, going over the minimum energy path via the transition state and finally to form the product.
||[[File:Surface Plot 1 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -4.1 ||-420.077 KJ.mol-1||No||Plot shows the trajectory going up the potential slope towards the transition state. However it does not have enough momentum to go over the curve and thus the trajectory returns to its original position. The reaction does not proceed and the reactants remain.||[[File:Surface Plot 2 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -5.1 ||-413.997 KJ.mol-1||Yes||Plot shows that increasing the momentum of the H-H bond in the reactants results in more vibrational energy hence a more distinct wave pattern appears in the trajectory. Similar to the first plot, the trajectory shows the reactant forming the product by going over the minimum energy path via the transition state.||[[File:Surface Plot 3 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.1 ||-357.277 KJ.mol-1||Yes||Plot shows the trajectory having an "excess" amount of momentum causing the H atom in the middle to bounce back and forth between molecules. The trajectory eventually goes to the product but does not go through the transition state. ||[[File:Surface Plot 4 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.6 ||-349.477 KJ.mol-1||Yes||Plot is similar to that above in that the trajectory has an "excess" amount of momentum. The trajectory eventually goes to the product but does not go through the transition state. ||[[File:Surface Plot 5 thl3318.png|thumb|300px|none|]]
|}

From the table above, I can conclude that whether or not a trajectory results in a reaction depends on the momentum p2. Changing the momentum p1 does not affect the result of the trajectory but if momentum p2 is not great enough the result will be unreactive.

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

Experimental value will be greater than the reaction rate the transition state theory predicts because it does not take into account quantum tunneling which is more significant at low temperatures. Transition State Theory states that as long as the energy is greater than the activation energy the reaction will proceed to the products but this does not take into account the possibility of recrossing back to the reactants and thus the reaction not proceeding.

== EXERCISE 2: F + H2 system ==
===By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?===

The F + H2 reaction is exothermic whereas the H + HF reaction is endothermic. This means that the H-F bond strength is greater than the H-H bond strength and is lower in energy.

===Locate the approximate position of the transition state.===
H-H distance = 77pm

H-F distance = 179.5pm

[[File:Surface_Plot_TS_3_thl3318.png|thumb|300px|none|Surface Plot showing Transition State with 500 steps at step size 0.21 fs ]]

===Report the activation energy for both reactions.===
transition state energy = -433.020

H-H energy = -435.100

H-F energy = -560.700

Activation energy for F + H2 --> HF + H = 2.080

Activation energy for H + HF --> H2 + F = 127.68

[[File:TS_enegy_vs_time_1_thl3318.png|thumb|300px|none|Energy vs Time plot showing TS energy and HF energy (step=51fs) ]]
[[File:TS_enegy_vs_time_2_thl3318.png|thumb|300px|none|Energy vs Time plot showing TS energy and H2 energy (step=51fs) ]]

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

Although energy is conserved in the reaction, there is a change from kinetic energy to potential energy during the collision. Upon the H2 molecule colliding with the F atom, an elastic collision occurs and the kinetic energy of the H2 molecule (in the form of translational motion) is converted into potential energy of the HF molecule (in the form of vibrational motion). This can be confirmed experimentally by measuring whether there is an increase in vibrational motion by using IR spectroscopy.

[[File:Momenta_plot_thl3318.png|thumb|300px|none|Plot showing momentum of bonds before and after collision where A and B are Hydrogen atoms and C is a Fluorine atom ]]

===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.===

For reactions with an early transition state (reactant-like), the efficiency of the reaction is most enhanced by the relative translation energy of the two atoms/molecules colliding. However for reactions with a late transition state (product-like), the efficiency of the reaction is most enhanced by vibrational excitation has a higher efficiency in enhancing the reactivity.

https://pubs.rsc.org/en/content/articlehtml/2016/sc/c6sc01066k?fbclid=IwAR2ewmOVzAbLMfnguNOtvv55flPVkD64qLPVAZMOgJvMcmIFE9WrEYM19Lo
DOI: 10.1039/C6SC01066K (Perspective) Chem. Sci., 2016, 7, 3992-4003

MRD:thl3318

2020-05-14T17:12:19Z

Thl3318: /* 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. */

== EXERCISE 1: H + H2 system ==
===On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?===

On a potential energy surface diagram, the transition state can be mathematically defined as a saddle point (which is also an unstable equilibrium). It can be identified by determining the point where the partial derivatives of the function potential with respect to the internuclear distances are equal to 0.

<math> \frac{\partial V}{\partial r_1} =0</math> and <math> \frac{\partial V}{\partial r_2} =0</math>

However, simply looking at the results for the first derivative does not give any information with regards to whether the point is a local minimum, a local maximum, or a saddle point. Analyzing the 2nd derivative then allows for the determination of whether the point is a saddle point. This can be done using the following equation:

<math> D=\frac{\partial^2 V}{\partial r_1^2} \frac{\partial^2 V}{\partial r_2^2} - (\frac{\partial^2 V}{\partial r_1\partial r_2})^2 </math>

If D<0, the point is therefore a saddle point and also the transition state.

===Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.===
My best estimate for the transition state position is when the A-B and B-C distance is equal to 90.8pm. This point in the "Internuclear Distance vs Time" plot can be rationed as the transition state position because when the bonds have no momentum the bond distances remain constant.

Surface_Plot_HH_TS_thl3318.png

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

The mep and dynamics trajectory differ in many different ways. The first difference is the time it takes for the reaction to take place (in this case, for the distance to follow to the valley floor of the potential graph). In the mep trajectory, it takes around 30 steps (15 fs) for the reaction to proceed whereas in the dynamics trajectory this number is around 140 fs. The second difference is that the internuclear distance between the H2 molecule and the lone H atom increases logarithmically in the mep calculation but increases linearly in the dynamics calculation.

===Complete the table below by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?===
{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 ||-414.280 KJ.mol-1||Yes|| Plot shows trajectory going from reactant, going over the minimum energy path via the transition state and finally to form the product.
||[[File:Surface Plot 1 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -4.1 ||-420.077 KJ.mol-1||No||Plot shows the trajectory going up the potential slope towards the transition state. However it does not have enough momentum to go over the curve and thus the trajectory returns to its original position. The reaction does not proceed and the reactants remain.||[[File:Surface Plot 2 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -5.1 ||-413.997 KJ.mol-1||Yes||Plot shows that increasing the momentum of the H-H bond in the reactants results in more vibrational energy hence a more distinct wave pattern appears in the trajectory. Similar to the first plot, the trajectory shows the reactant forming the product by going over the minimum energy path via the transition state.||[[File:Surface Plot 3 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.1 ||-357.277 KJ.mol-1||Yes||Plot shows the trajectory having an "excess" amount of momentum causing the H atom in the middle to bounce back and forth between molecules. The trajectory eventually goes to the product but does not go through the transition state. ||[[File:Surface Plot 4 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.6 ||-349.477 KJ.mol-1||Yes||Plot is similar to that above in that the trajectory has an "excess" amount of momentum. The trajectory eventually goes to the product but does not go through the transition state. ||[[File:Surface Plot 5 thl3318.png|thumb|300px|none|]]
|}

From the table above, I can conclude that whether or not a trajectory results in a reaction depends on the momentum p2. Changing the momentum p1 does not affect the result of the trajectory but if momentum p2 is not great enough the result will be unreactive.

[[File:Y2C11.png|thumb|600px|none|Trajectory that re-crosses the TS region]]

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

Experimental value will be greater than the reaction rate the transition state theory predicts because it does not take into account quantum tunneling which is more significant at low temperatures. Transition State Theory states that as long as the energy is greater than the activation energy the reaction will proceed to the products but this does not take into account the possibility of recrossing back to the reactants and thus the reaction not proceeding.

== EXERCISE 2: F + H2 system ==
===By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?===

The F + H2 reaction is exothermic whereas the H + HF reaction is endothermic. This means that the H-F bond strength is greater than the H-H bond strength and is lower in energy.

===Locate the approximate position of the transition state.===
H-H distance = 77pm

H-F distance = 179.5pm

[[File:Surface_Plot_TS_3_thl3318.png|thumb|300px|none|Surface Plot showing Transition State with 500 steps at step size 0.21 fs ]]

===Report the activation energy for both reactions.===
transition state energy = -433.020

H-H energy = -435.100

H-F energy = -560.700

Activation energy for F + H2 --> HF + H = 2.080

Activation energy for H + HF --> H2 + F = 127.68

[[File:TS_enegy_vs_time_1_thl3318.png|thumb|300px|none|Energy vs Time plot showing TS energy and HF energy (step=51fs) ]]
[[File:TS_enegy_vs_time_2_thl3318.png|thumb|300px|none|Energy vs Time plot showing TS energy and H2 energy (step=51fs) ]]

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

Although energy is conserved in the reaction, there is a change from kinetic energy to potential energy during the collision. Upon the H2 molecule colliding with the F atom, an elastic collision occurs and the kinetic energy of the H2 molecule (in the form of translational motion) is converted into potential energy of the HF molecule (in the form of vibrational motion). This can be confirmed experimentally by measuring whether there is an increase in vibrational motion by using IR spectroscopy.

[[File:Momenta_plot_thl3318.png|thumb|300px|none|Plot showing momentum of bonds before and after collision where A and B are Hydrogen atoms and C is a Fluorine atom ]]

===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.===

For reactions with an early transition state (reactant-like), the efficiency of the reaction is most enhanced by the relative translation energy of the two atoms/molecules colliding. However for reactions with a late transition state (product-like), the efficiency of the reaction is most enhanced by vibrational excitation has a higher efficiency in enhancing the reactivity.

https://pubs.rsc.org/en/content/articlehtml/2016/sc/c6sc01066k?fbclid=IwAR2ewmOVzAbLMfnguNOtvv55flPVkD64qLPVAZMOgJvMcmIFE9WrEYM19Lo
DOI: 10.1039/C6SC01066K (Perspective) Chem. Sci., 2016, 7, 3992-4003

File:Surface Plot HH TS thl3318.png

2020-05-14T17:12:06Z

Thl3318:

MRD:thl3318

2020-05-14T17:11:22Z

Thl3318: /* 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. */

== EXERCISE 1: H + H2 system ==
===On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?===

On a potential energy surface diagram, the transition state can be mathematically defined as a saddle point (which is also an unstable equilibrium). It can be identified by determining the point where the partial derivatives of the function potential with respect to the internuclear distances are equal to 0.

<math> \frac{\partial V}{\partial r_1} =0</math> and <math> \frac{\partial V}{\partial r_2} =0</math>

However, simply looking at the results for the first derivative does not give any information with regards to whether the point is a local minimum, a local maximum, or a saddle point. Analyzing the 2nd derivative then allows for the determination of whether the point is a saddle point. This can be done using the following equation:

<math> D=\frac{\partial^2 V}{\partial r_1^2} \frac{\partial^2 V}{\partial r_2^2} - (\frac{\partial^2 V}{\partial r_1\partial r_2})^2 </math>

If D<0, the point is therefore a saddle point and also the transition state.

===Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.===
My best estimate for the transition state position is when the A-B and B-C distance is equal to 90.8pm. This point in the "Internuclear Distance vs Time" plot can be rationed as the transition state position because when the bonds have no momentum the bond distances remain constant.

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

The mep and dynamics trajectory differ in many different ways. The first difference is the time it takes for the reaction to take place (in this case, for the distance to follow to the valley floor of the potential graph). In the mep trajectory, it takes around 30 steps (15 fs) for the reaction to proceed whereas in the dynamics trajectory this number is around 140 fs. The second difference is that the internuclear distance between the H2 molecule and the lone H atom increases logarithmically in the mep calculation but increases linearly in the dynamics calculation.

===Complete the table below by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?===
{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 ||-414.280 KJ.mol-1||Yes|| Plot shows trajectory going from reactant, going over the minimum energy path via the transition state and finally to form the product.
||[[File:Surface Plot 1 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -4.1 ||-420.077 KJ.mol-1||No||Plot shows the trajectory going up the potential slope towards the transition state. However it does not have enough momentum to go over the curve and thus the trajectory returns to its original position. The reaction does not proceed and the reactants remain.||[[File:Surface Plot 2 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -5.1 ||-413.997 KJ.mol-1||Yes||Plot shows that increasing the momentum of the H-H bond in the reactants results in more vibrational energy hence a more distinct wave pattern appears in the trajectory. Similar to the first plot, the trajectory shows the reactant forming the product by going over the minimum energy path via the transition state.||[[File:Surface Plot 3 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.1 ||-357.277 KJ.mol-1||Yes||Plot shows the trajectory having an "excess" amount of momentum causing the H atom in the middle to bounce back and forth between molecules. The trajectory eventually goes to the product but does not go through the transition state. ||[[File:Surface Plot 4 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.6 ||-349.477 KJ.mol-1||Yes||Plot is similar to that above in that the trajectory has an "excess" amount of momentum. The trajectory eventually goes to the product but does not go through the transition state. ||[[File:Surface Plot 5 thl3318.png|thumb|300px|none|]]
|}

From the table above, I can conclude that whether or not a trajectory results in a reaction depends on the momentum p2. Changing the momentum p1 does not affect the result of the trajectory but if momentum p2 is not great enough the result will be unreactive.

[[File:Y2C11.png|thumb|600px|none|Trajectory that re-crosses the TS region]]

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

Experimental value will be greater than the reaction rate the transition state theory predicts because it does not take into account quantum tunneling which is more significant at low temperatures. Transition State Theory states that as long as the energy is greater than the activation energy the reaction will proceed to the products but this does not take into account the possibility of recrossing back to the reactants and thus the reaction not proceeding.

== EXERCISE 2: F + H2 system ==
===By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?===

The F + H2 reaction is exothermic whereas the H + HF reaction is endothermic. This means that the H-F bond strength is greater than the H-H bond strength and is lower in energy.

===Locate the approximate position of the transition state.===
H-H distance = 77pm

H-F distance = 179.5pm

[[File:Surface_Plot_TS_3_thl3318.png|thumb|300px|none|Surface Plot showing Transition State with 500 steps at step size 0.21 fs ]]

===Report the activation energy for both reactions.===
transition state energy = -433.020

H-H energy = -435.100

H-F energy = -560.700

Activation energy for F + H2 --> HF + H = 2.080

Activation energy for H + HF --> H2 + F = 127.68

[[File:TS_enegy_vs_time_1_thl3318.png|thumb|300px|none|Energy vs Time plot showing TS energy and HF energy (step=51fs) ]]
[[File:TS_enegy_vs_time_2_thl3318.png|thumb|300px|none|Energy vs Time plot showing TS energy and H2 energy (step=51fs) ]]

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

Although energy is conserved in the reaction, there is a change from kinetic energy to potential energy during the collision. Upon the H2 molecule colliding with the F atom, an elastic collision occurs and the kinetic energy of the H2 molecule (in the form of translational motion) is converted into potential energy of the HF molecule (in the form of vibrational motion). This can be confirmed experimentally by measuring whether there is an increase in vibrational motion by using IR spectroscopy.

[[File:Momenta_plot_thl3318.png|thumb|300px|none|Plot showing momentum of bonds before and after collision where A and B are Hydrogen atoms and C is a Fluorine atom ]]

===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.===

For reactions with an early transition state (reactant-like), the efficiency of the reaction is most enhanced by the relative translation energy of the two atoms/molecules colliding. However for reactions with a late transition state (product-like), the efficiency of the reaction is most enhanced by vibrational excitation has a higher efficiency in enhancing the reactivity.

https://pubs.rsc.org/en/content/articlehtml/2016/sc/c6sc01066k?fbclid=IwAR2ewmOVzAbLMfnguNOtvv55flPVkD64qLPVAZMOgJvMcmIFE9WrEYM19Lo
DOI: 10.1039/C6SC01066K (Perspective) Chem. Sci., 2016, 7, 3992-4003

MRD:thl3318

2020-05-14T17:06:17Z

== EXERCISE 1: H + H2 system ==
===On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?===

On a potential energy surface diagram, the transition state can be mathematically defined as a saddle point (which is also an unstable equilibrium). It can be identified by determining the point where the partial derivatives of the function potential with respect to the internuclear distances are equal to 0.

<math> \frac{\partial V}{\partial r_1} =0</math> and <math> \frac{\partial V}{\partial r_2} =0</math>

However, simply looking at the results for the first derivative does not give any information with regards to whether the point is a local minimum, a local maximum, or a saddle point. Analyzing the 2nd derivative then allows for the determination of whether the point is a saddle point. This can be done using the following equation:

<math> D=\frac{\partial^2 V}{\partial r_1^2} \frac{\partial^2 V}{\partial r_2^2} - (\frac{\partial^2 V}{\partial r_1\partial r_2})^2 </math>

If D<0, the point is therefore a saddle point and also the transition state.

===Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.===
My best estimate for the transition state position is at time=20fs when the A-B distance is equal to the B-C distance. The A-B and B-C distance at this point is determined to be 90.8pm. This point in the "Internuclear Distance vs Time" plot can be rationed as the transition state position

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

The mep and dynamics trajectory differ in many different ways. The first difference is the time it takes for the reaction to take place (in this case, for the distance to follow to the valley floor of the potential graph). In the mep trajectory, it takes around 30 steps (15 fs) for the reaction to proceed whereas in the dynamics trajectory this number is around 140 fs. The second difference is that the internuclear distance between the H2 molecule and the lone H atom increases logarithmically in the mep calculation but increases linearly in the dynamics calculation.

===Complete the table below by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?===
{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 ||-414.280 KJ.mol-1||Yes|| Plot shows trajectory going from reactant, going over the minimum energy path via the transition state and finally to form the product.
||[[File:Surface Plot 1 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -4.1 ||-420.077 KJ.mol-1||No||Plot shows the trajectory going up the potential slope towards the transition state. However it does not have enough momentum to go over the curve and thus the trajectory returns to its original position. The reaction does not proceed and the reactants remain.||[[File:Surface Plot 2 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -5.1 ||-413.997 KJ.mol-1||Yes||Plot shows that increasing the momentum of the H-H bond in the reactants results in more vibrational energy hence a more distinct wave pattern appears in the trajectory. Similar to the first plot, the trajectory shows the reactant forming the product by going over the minimum energy path via the transition state.||[[File:Surface Plot 3 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.1 ||-357.277 KJ.mol-1||Yes||Plot shows the trajectory having an "excess" amount of momentum causing the H atom in the middle to bounce back and forth between molecules. The trajectory eventually goes to the product but does not go through the transition state. ||[[File:Surface Plot 4 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.6 ||-349.477 KJ.mol-1||Yes||Plot is similar to that above in that the trajectory has an "excess" amount of momentum. The trajectory eventually goes to the product but does not go through the transition state. ||[[File:Surface Plot 5 thl3318.png|thumb|300px|none|]]
|}

From the table above, I can conclude that whether or not a trajectory results in a reaction depends on the momentum p2. Changing the momentum p1 does not affect the result of the trajectory but if momentum p2 is not great enough the result will be unreactive.

[[File:Y2C11.png|thumb|600px|none|Trajectory that re-crosses the TS region]]

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

Experimental value will be greater than the reaction rate the transition state theory predicts because it does not take into account quantum tunneling which is more significant at low temperatures. Transition State Theory states that as long as the energy is greater than the activation energy the reaction will proceed to the products but this does not take into account the possibility of recrossing back to the reactants and thus the reaction not proceeding.

== EXERCISE 2: F + H2 system ==
===By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?===

The F + H2 reaction is exothermic whereas the H + HF reaction is endothermic. This means that the H-F bond strength is greater than the H-H bond strength and is lower in energy.

===Locate the approximate position of the transition state.===
H-H distance = 77pm

H-F distance = 179.5pm

[[File:Surface_Plot_TS_3_thl3318.png|thumb|300px|none|Surface Plot showing Transition State with 500 steps at step size 0.21 fs ]]

===Report the activation energy for both reactions.===
transition state energy = -433.020

H-H energy = -435.100

H-F energy = -560.700

Activation energy for F + H2 --> HF + H = 2.080

Activation energy for H + HF --> H2 + F = 127.68

[[File:TS_enegy_vs_time_1_thl3318.png|thumb|300px|none|Energy vs Time plot showing TS energy and HF energy (step=51fs) ]]
[[File:TS_enegy_vs_time_2_thl3318.png|thumb|300px|none|Energy vs Time plot showing TS energy and H2 energy (step=51fs) ]]

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

Although energy is conserved in the reaction, there is a change from kinetic energy to potential energy during the collision. Upon the H2 molecule colliding with the F atom, an elastic collision occurs and the kinetic energy of the H2 molecule (in the form of translational motion) is converted into potential energy of the HF molecule (in the form of vibrational motion). This can be confirmed experimentally by measuring whether there is an increase in vibrational motion by using IR spectroscopy.

[[File:Momenta_plot_thl3318.png|thumb|300px|none|Plot showing momentum of bonds before and after collision where A and B are Hydrogen atoms and C is a Fluorine atom ]]

===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.===

For reactions with an early transition state (reactant-like), the efficiency of the reaction is most enhanced by the relative translation energy of the two atoms/molecules colliding. However for reactions with a late transition state (product-like), the efficiency of the reaction is most enhanced by vibrational excitation has a higher efficiency in enhancing the reactivity.

https://pubs.rsc.org/en/content/articlehtml/2016/sc/c6sc01066k?fbclid=IwAR2ewmOVzAbLMfnguNOtvv55flPVkD64qLPVAZMOgJvMcmIFE9WrEYM19Lo
DOI: 10.1039/C6SC01066K (Perspective) Chem. Sci., 2016, 7, 3992-4003

MRD:thl3318

2020-05-14T16:54:22Z

Thl3318: /* Report the activation energy for both reactions. */

== EXERCISE 1: H + H2 system ==
===On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?===

On a potential energy surface diagram, the transition state can be mathematically defined as a saddle point (which is also an unstable equilibrium). It can be identified by determining the point where the partial derivatives of the function potential with respect to the internuclear distances are equal to 0.

<math> \frac{\partial V}{\partial r_1} =0</math> and <math> \frac{\partial V}{\partial r_2} =0</math>

However, simply looking at the results for the first derivative does not give any information with regards to whether the point is a local minimum, a local maximum, or a saddle point. Analyzing the 2nd derivative then allows for the determination of whether the point is a saddle point. This can be done using the following equation:

<math> D=\frac{\partial^2 V}{\partial r_1^2} \frac{\partial^2 V}{\partial r_2^2} - (\frac{\partial^2 V}{\partial r_1\partial r_2})^2 </math>

If D<0, the point is therefore a saddle point and also the transition state.

===Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.===
My best estimate for the transition state position is at time=20fs when the A-B distance is equal to the B-C distance. The A-B and B-C distance at this point is determined to be 90.8pm. This point in the "Internuclear Distance vs Time" plot can be rationed as the transition state position

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

The mep and dynamics trajectory differ in many different ways. The first difference is the time it takes for the reaction to take place (in this case, for the distance to follow to the valley floor of the potential graph). In the mep trajectory, it takes around 30 steps (15 fs) for the reaction to proceed whereas in the dynamics trajectory this number is around 140 fs. The second difference is that the internuclear distance between the H2 molecule and the lone H atom increases logarithmically in the mep calculation but increases linearly in the dynamics calculation.

===Complete the table below by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?===
{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 ||-414.280 KJ.mol-1||Yes|| Plot shows trajectory going from reactant, going over the minimum energy path via the transition state and finally to form the product.
||[[File:Surface Plot 1 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -4.1 ||-420.077 KJ.mol-1||No||Plot shows the trajectory going up the potential slope towards the transition state. However it does not have enough momentum to go over the curve and thus the trajectory returns to its original position. The reaction does not proceed and the reactants remain.||[[File:Surface Plot 2 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -5.1 ||-413.997 KJ.mol-1||Yes||Plot shows that increasing the momentum of the H-H bond in the reactants results in more vibrational energy hence a more distinct wave pattern appears in the trajectory. Similar to the first plot, the trajectory shows the reactant forming the product by going over the minimum energy path via the transition state.||[[File:Surface Plot 3 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.1 ||-357.277 KJ.mol-1||Yes|| ||[[File:Surface Plot 4 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.6 ||-349.477 KJ.mol-1||Yes|| ||[[File:Surface Plot 5 thl3318.png|thumb|300px|none|]]
|}

From the table above, I can conclude that whether or not a trajectory results in a reaction depends on the momentum p2. Changing the momentum p1 does not affect the result of the trajectory but if momentum p2 is not great enough the result will be unreactive.

[[File:Y2C11.png|thumb|600px|none|Trajectory that re-crosses the TS region]]

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

Experimental value will be greater than the reaction rate the transition state theory predicts because it does not take into account quantum tunneling which is more significant at low temperatures. Transition State Theory states that as long as the energy is greater than the activation energy the reaction will proceed to the products but this does not take into account the possibility of recrossing back to the reactants and thus the reaction not proceeding.

== EXERCISE 2: F + H2 system ==
===By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?===

The F + H2 reaction is exothermic whereas the H + HF reaction is endothermic. This means that the H-F bond strength is greater than the H-H bond strength and is lower in energy.

===Locate the approximate position of the transition state.===
H-H distance = 77pm

H-F distance = 179.5pm

[[File:Surface_Plot_TS_3_thl3318.png|thumb|300px|none|Surface Plot showing Transition State with 500 steps at step size 0.21 fs ]]

===Report the activation energy for both reactions.===
transition state energy = -433.020

H-H energy = -435.100

H-F energy = -560.700

Activation energy for F + H2 --> HF + H = 2.080

Activation energy for H + HF --> H2 + F = 127.68

[[File:TS_enegy_vs_time_1_thl3318.png|thumb|300px|none|Energy vs Time plot showing TS energy and HF energy (step=51fs) ]]
[[File:TS_enegy_vs_time_2_thl3318.png|thumb|300px|none|Energy vs Time plot showing TS energy and H2 energy (step=51fs) ]]

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

Although energy is conserved in the reaction, there is a change from kinetic energy to potential energy during the collision. Upon the H2 molecule colliding with the F atom, an elastic collision occurs and the kinetic energy of the H2 molecule (in the form of translational motion) is converted into potential energy of the HF molecule (in the form of vibrational motion). This can be confirmed experimentally by measuring whether there is an increase in vibrational motion by using IR spectroscopy.

[[File:Momenta_plot_thl3318.png|thumb|300px|none|Plot showing momentum of bonds before and after collision where A and B are Hydrogen atoms and C is a Fluorine atom ]]

===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.===

For reactions with an early transition state (reactant-like), the efficiency of the reaction is most enhanced by the relative translation energy of the two atoms/molecules colliding. However for reactions with a late transition state (product-like), the efficiency of the reaction is most enhanced by vibrational excitation has a higher efficiency in enhancing the reactivity.

https://pubs.rsc.org/en/content/articlehtml/2016/sc/c6sc01066k?fbclid=IwAR2ewmOVzAbLMfnguNOtvv55flPVkD64qLPVAZMOgJvMcmIFE9WrEYM19Lo
DOI: 10.1039/C6SC01066K (Perspective) Chem. Sci., 2016, 7, 3992-4003

MRD:thl3318

2020-05-14T16:52:59Z

Thl3318: /* Report the activation energy for both reactions. */

== EXERCISE 1: H + H2 system ==
===On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?===

On a potential energy surface diagram, the transition state can be mathematically defined as a saddle point (which is also an unstable equilibrium). It can be identified by determining the point where the partial derivatives of the function potential with respect to the internuclear distances are equal to 0.

<math> \frac{\partial V}{\partial r_1} =0</math> and <math> \frac{\partial V}{\partial r_2} =0</math>

However, simply looking at the results for the first derivative does not give any information with regards to whether the point is a local minimum, a local maximum, or a saddle point. Analyzing the 2nd derivative then allows for the determination of whether the point is a saddle point. This can be done using the following equation:

<math> D=\frac{\partial^2 V}{\partial r_1^2} \frac{\partial^2 V}{\partial r_2^2} - (\frac{\partial^2 V}{\partial r_1\partial r_2})^2 </math>

If D<0, the point is therefore a saddle point and also the transition state.

===Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.===
My best estimate for the transition state position is at time=20fs when the A-B distance is equal to the B-C distance. The A-B and B-C distance at this point is determined to be 90.8pm. This point in the "Internuclear Distance vs Time" plot can be rationed as the transition state position

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

The mep and dynamics trajectory differ in many different ways. The first difference is the time it takes for the reaction to take place (in this case, for the distance to follow to the valley floor of the potential graph). In the mep trajectory, it takes around 30 steps (15 fs) for the reaction to proceed whereas in the dynamics trajectory this number is around 140 fs. The second difference is that the internuclear distance between the H2 molecule and the lone H atom increases logarithmically in the mep calculation but increases linearly in the dynamics calculation.

===Complete the table below by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?===
{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 ||-414.280 KJ.mol-1||Yes|| Plot shows trajectory going from reactant, going over the minimum energy path via the transition state and finally to form the product.
||[[File:Surface Plot 1 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -4.1 ||-420.077 KJ.mol-1||No||Plot shows the trajectory going up the potential slope towards the transition state. However it does not have enough momentum to go over the curve and thus the trajectory returns to its original position. The reaction does not proceed and the reactants remain.||[[File:Surface Plot 2 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -5.1 ||-413.997 KJ.mol-1||Yes||Plot shows that increasing the momentum of the H-H bond in the reactants results in more vibrational energy hence a more distinct wave pattern appears in the trajectory. Similar to the first plot, the trajectory shows the reactant forming the product by going over the minimum energy path via the transition state.||[[File:Surface Plot 3 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.1 ||-357.277 KJ.mol-1||Yes|| ||[[File:Surface Plot 4 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.6 ||-349.477 KJ.mol-1||Yes|| ||[[File:Surface Plot 5 thl3318.png|thumb|300px|none|]]
|}

From the table above, I can conclude that whether or not a trajectory results in a reaction depends on the momentum p2. Changing the momentum p1 does not affect the result of the trajectory but if momentum p2 is not great enough the result will be unreactive.

[[File:Y2C11.png|thumb|600px|none|Trajectory that re-crosses the TS region]]

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

Experimental value will be greater than the reaction rate the transition state theory predicts because it does not take into account quantum tunneling which is more significant at low temperatures. Transition State Theory states that as long as the energy is greater than the activation energy the reaction will proceed to the products but this does not take into account the possibility of recrossing back to the reactants and thus the reaction not proceeding.

== EXERCISE 2: F + H2 system ==
===By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?===

The F + H2 reaction is exothermic whereas the H + HF reaction is endothermic. This means that the H-F bond strength is greater than the H-H bond strength and is lower in energy.

===Locate the approximate position of the transition state.===
H-H distance = 77pm

H-F distance = 179.5pm

[[File:Surface_Plot_TS_3_thl3318.png|thumb|300px|none|Surface Plot showing Transition State with 500 steps at step size 0.21 fs ]]

===Report the activation energy for both reactions.===
transition state energy = -433.020

H-H energy = -435.100

H-F energy = -560.700

Activation energy for F + H2 --> HF + H = 2.080

Activation energy for H + HF --> H2 + F = 127.68

[[File:TS_enegy_vs_time_1_thl3318.png|thumb|300px|none|Plot showing TS energy and HF energy ]]
[[File:TS_enegy_vs_time_2_thl3318.png|thumb|300px|none|Plot showing TS energy and H2 energy ]]

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

Although energy is conserved in the reaction, there is a change from kinetic energy to potential energy during the collision. Upon the H2 molecule colliding with the F atom, an elastic collision occurs and the kinetic energy of the H2 molecule (in the form of translational motion) is converted into potential energy of the HF molecule (in the form of vibrational motion). This can be confirmed experimentally by measuring whether there is an increase in vibrational motion by using IR spectroscopy.

[[File:Momenta_plot_thl3318.png|thumb|300px|none|Plot showing momentum of bonds before and after collision where A and B are Hydrogen atoms and C is a Fluorine atom ]]

===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.===

For reactions with an early transition state (reactant-like), the efficiency of the reaction is most enhanced by the relative translation energy of the two atoms/molecules colliding. However for reactions with a late transition state (product-like), the efficiency of the reaction is most enhanced by vibrational excitation has a higher efficiency in enhancing the reactivity.

https://pubs.rsc.org/en/content/articlehtml/2016/sc/c6sc01066k?fbclid=IwAR2ewmOVzAbLMfnguNOtvv55flPVkD64qLPVAZMOgJvMcmIFE9WrEYM19Lo
DOI: 10.1039/C6SC01066K (Perspective) Chem. Sci., 2016, 7, 3992-4003

MRD:thl3318

2020-05-14T16:51:27Z

Thl3318: /* Report the activation energy for both reactions. */

== EXERCISE 1: H + H2 system ==
===On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?===

On a potential energy surface diagram, the transition state can be mathematically defined as a saddle point (which is also an unstable equilibrium). It can be identified by determining the point where the partial derivatives of the function potential with respect to the internuclear distances are equal to 0.

<math> \frac{\partial V}{\partial r_1} =0</math> and <math> \frac{\partial V}{\partial r_2} =0</math>

However, simply looking at the results for the first derivative does not give any information with regards to whether the point is a local minimum, a local maximum, or a saddle point. Analyzing the 2nd derivative then allows for the determination of whether the point is a saddle point. This can be done using the following equation:

<math> D=\frac{\partial^2 V}{\partial r_1^2} \frac{\partial^2 V}{\partial r_2^2} - (\frac{\partial^2 V}{\partial r_1\partial r_2})^2 </math>

If D<0, the point is therefore a saddle point and also the transition state.

===Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.===
My best estimate for the transition state position is at time=20fs when the A-B distance is equal to the B-C distance. The A-B and B-C distance at this point is determined to be 90.8pm. This point in the "Internuclear Distance vs Time" plot can be rationed as the transition state position

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

The mep and dynamics trajectory differ in many different ways. The first difference is the time it takes for the reaction to take place (in this case, for the distance to follow to the valley floor of the potential graph). In the mep trajectory, it takes around 30 steps (15 fs) for the reaction to proceed whereas in the dynamics trajectory this number is around 140 fs. The second difference is that the internuclear distance between the H2 molecule and the lone H atom increases logarithmically in the mep calculation but increases linearly in the dynamics calculation.

===Complete the table below by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?===
{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 ||-414.280 KJ.mol-1||Yes|| Plot shows trajectory going from reactant, going over the minimum energy path via the transition state and finally to form the product.
||[[File:Surface Plot 1 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -4.1 ||-420.077 KJ.mol-1||No||Plot shows the trajectory going up the potential slope towards the transition state. However it does not have enough momentum to go over the curve and thus the trajectory returns to its original position. The reaction does not proceed and the reactants remain.||[[File:Surface Plot 2 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -5.1 ||-413.997 KJ.mol-1||Yes||Plot shows that increasing the momentum of the H-H bond in the reactants results in more vibrational energy hence a more distinct wave pattern appears in the trajectory. Similar to the first plot, the trajectory shows the reactant forming the product by going over the minimum energy path via the transition state.||[[File:Surface Plot 3 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.1 ||-357.277 KJ.mol-1||Yes|| ||[[File:Surface Plot 4 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.6 ||-349.477 KJ.mol-1||Yes|| ||[[File:Surface Plot 5 thl3318.png|thumb|300px|none|]]
|}

From the table above, I can conclude that whether or not a trajectory results in a reaction depends on the momentum p2. Changing the momentum p1 does not affect the result of the trajectory but if momentum p2 is not great enough the result will be unreactive.

[[File:Y2C11.png|thumb|600px|none|Trajectory that re-crosses the TS region]]

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

Experimental value will be greater than the reaction rate the transition state theory predicts because it does not take into account quantum tunneling which is more significant at low temperatures. Transition State Theory states that as long as the energy is greater than the activation energy the reaction will proceed to the products but this does not take into account the possibility of recrossing back to the reactants and thus the reaction not proceeding.

== EXERCISE 2: F + H2 system ==
===By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?===

The F + H2 reaction is exothermic whereas the H + HF reaction is endothermic. This means that the H-F bond strength is greater than the H-H bond strength and is lower in energy.

===Locate the approximate position of the transition state.===
H-H distance = 77pm

H-F distance = 179.5pm

[[File:Surface_Plot_TS_3_thl3318.png|thumb|300px|none|Surface Plot showing Transition State with 500 steps at step size 0.21 fs ]]

===Report the activation energy for both reactions.===
transition state energy = -433.020

H-H energy = -435.100

H-F energy = -560.700

Activation energy for F + H2 --> HF + H = 2.080

Activation energy for H + HF --> H2 + F = 127.68

TS_enegy_vs_time_1_thl3318.png

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

Although energy is conserved in the reaction, there is a change from kinetic energy to potential energy during the collision. Upon the H2 molecule colliding with the F atom, an elastic collision occurs and the kinetic energy of the H2 molecule (in the form of translational motion) is converted into potential energy of the HF molecule (in the form of vibrational motion). This can be confirmed experimentally by measuring whether there is an increase in vibrational motion by using IR spectroscopy.

[[File:Momenta_plot_thl3318.png|thumb|300px|none|Plot showing momentum of bonds before and after collision where A and B are Hydrogen atoms and C is a Fluorine atom ]]

===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.===

For reactions with an early transition state (reactant-like), the efficiency of the reaction is most enhanced by the relative translation energy of the two atoms/molecules colliding. However for reactions with a late transition state (product-like), the efficiency of the reaction is most enhanced by vibrational excitation has a higher efficiency in enhancing the reactivity.

https://pubs.rsc.org/en/content/articlehtml/2016/sc/c6sc01066k?fbclid=IwAR2ewmOVzAbLMfnguNOtvv55flPVkD64qLPVAZMOgJvMcmIFE9WrEYM19Lo
DOI: 10.1039/C6SC01066K (Perspective) Chem. Sci., 2016, 7, 3992-4003

File:TS enegy vs time 2 thl3318.png

2020-05-14T16:51:00Z

Thl3318:

File:TS enegy vs time 1 thl3318.png

2020-05-14T16:50:41Z

Thl3318:

MRD:thl3318

2020-05-14T16:43:46Z

== EXERCISE 1: H + H2 system ==
===On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?===

On a potential energy surface diagram, the transition state can be mathematically defined as a saddle point (which is also an unstable equilibrium). It can be identified by determining the point where the partial derivatives of the function potential with respect to the internuclear distances are equal to 0.

<math> \frac{\partial V}{\partial r_1} =0</math> and <math> \frac{\partial V}{\partial r_2} =0</math>

However, simply looking at the results for the first derivative does not give any information with regards to whether the point is a local minimum, a local maximum, or a saddle point. Analyzing the 2nd derivative then allows for the determination of whether the point is a saddle point. This can be done using the following equation:

<math> D=\frac{\partial^2 V}{\partial r_1^2} \frac{\partial^2 V}{\partial r_2^2} - (\frac{\partial^2 V}{\partial r_1\partial r_2})^2 </math>

If D<0, the point is therefore a saddle point and also the transition state.

===Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.===
My best estimate for the transition state position is at time=20fs when the A-B distance is equal to the B-C distance. The A-B and B-C distance at this point is determined to be 90.8pm. This point in the "Internuclear Distance vs Time" plot can be rationed as the transition state position

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

The mep and dynamics trajectory differ in many different ways. The first difference is the time it takes for the reaction to take place (in this case, for the distance to follow to the valley floor of the potential graph). In the mep trajectory, it takes around 30 steps (15 fs) for the reaction to proceed whereas in the dynamics trajectory this number is around 140 fs. The second difference is that the internuclear distance between the H2 molecule and the lone H atom increases logarithmically in the mep calculation but increases linearly in the dynamics calculation.

===Complete the table below by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?===
{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 ||-414.280 KJ.mol-1||Yes|| Plot shows trajectory going from reactant, going over the minimum energy path via the transition state and finally to form the product.
||[[File:Surface Plot 1 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -4.1 ||-420.077 KJ.mol-1||No||Plot shows the trajectory going up the potential slope towards the transition state. However it does not have enough momentum to go over the curve and thus the trajectory returns to its original position. The reaction does not proceed and the reactants remain.||[[File:Surface Plot 2 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -5.1 ||-413.997 KJ.mol-1||Yes||Plot shows that increasing the momentum of the H-H bond in the reactants results in more vibrational energy hence a more distinct wave pattern appears in the trajectory. Similar to the first plot, the trajectory shows the reactant forming the product by going over the minimum energy path via the transition state.||[[File:Surface Plot 3 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.1 ||-357.277 KJ.mol-1||Yes|| ||[[File:Surface Plot 4 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.6 ||-349.477 KJ.mol-1||Yes|| ||[[File:Surface Plot 5 thl3318.png|thumb|300px|none|]]
|}

From the table above, I can conclude that whether or not a trajectory results in a reaction depends on the momentum p2. Changing the momentum p1 does not affect the result of the trajectory but if momentum p2 is not great enough the result will be unreactive.

[[File:Y2C11.png|thumb|600px|none|Trajectory that re-crosses the TS region]]

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

Experimental value will be greater than the reaction rate the transition state theory predicts because it does not take into account quantum tunneling which is more significant at low temperatures. Transition State Theory states that as long as the energy is greater than the activation energy the reaction will proceed to the products but this does not take into account the possibility of recrossing back to the reactants and thus the reaction not proceeding.

== EXERCISE 2: F + H2 system ==
===By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?===

The F + H2 reaction is exothermic whereas the H + HF reaction is endothermic. This means that the H-F bond strength is greater than the H-H bond strength and is lower in energy.

===Locate the approximate position of the transition state.===
H-H distance = 77pm

H-F distance = 179.5pm

[[File:Surface_Plot_TS_3_thl3318.png|thumb|300px|none|Surface Plot showing Transition State with 500 steps at step size 0.21 fs ]]

===Report the activation energy for both reactions.===
transition state energy = -433.020

H-H energy = -435.100

H-F energy = -560.700

Activation energy for F + H2 --> HF + H = 2.080

Activation energy for H + HF --> H2 + F = 127.68

You will be able to report a reasonable estimate by performing a mep (with a sufficient number of steps) from a structure neighbouring the transition state, and choosing to plot the appropriate quantity as a function of "time".

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

Although energy is conserved in the reaction, there is a change from kinetic energy to potential energy during the collision. Upon the H2 molecule colliding with the F atom, an elastic collision occurs and the kinetic energy of the H2 molecule (in the form of translational motion) is converted into potential energy of the HF molecule (in the form of vibrational motion). This can be confirmed experimentally by measuring whether there is an increase in vibrational motion by using IR spectroscopy.

[[File:Momenta_plot_thl3318.png|thumb|300px|none|Plot showing momentum of bonds before and after collision where A and B are Hydrogen atoms and C is a Fluorine atom ]]

===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.===

For reactions with an early transition state (reactant-like), the efficiency of the reaction is most enhanced by the relative translation energy of the two atoms/molecules colliding. However for reactions with a late transition state (product-like), the efficiency of the reaction is most enhanced by vibrational excitation has a higher efficiency in enhancing the reactivity.

https://pubs.rsc.org/en/content/articlehtml/2016/sc/c6sc01066k?fbclid=IwAR2ewmOVzAbLMfnguNOtvv55flPVkD64qLPVAZMOgJvMcmIFE9WrEYM19Lo
DOI: 10.1039/C6SC01066K (Perspective) Chem. Sci., 2016, 7, 3992-4003

MRD:thl3318

2020-05-14T16:28:13Z

Thl3318: /* Locate the approximate position of the transition state. */

== EXERCISE 1: H + H2 system ==
===On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?===

On a potential energy surface diagram, the transition state can be mathematically defined as a saddle point (which is also an unstable equilibrium). It can be identified by determining the point where the partial derivatives of the function potential with respect to the internuclear distances are equal to 0.

<math> \frac{\partial V}{\partial r_1} =0</math> and <math> \frac{\partial V}{\partial r_2} =0</math>

However, simply looking at the results for the first derivative does not give any information with regards to whether the point is a local minimum, a local maximum, or a saddle point. Analyzing the 2nd derivative then allows for the determination of whether the point is a saddle point. This can be done using the following equation:

<math> D=\frac{\partial^2 V}{\partial r_1^2} \frac{\partial^2 V}{\partial r_2^2} - (\frac{\partial^2 V}{\partial r_1\partial r_2})^2 </math>

If D<0, the point is therefore a saddle point and also the transition state.

===Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.===
My best estimate for the transition state position is at time=20fs when the A-B distance is equal to the B-C distance. The A-B and B-C distance at this point is determined to be 90.8pm. This point in the "Internuclear Distance vs Time" plot can be rationed as the transition state position

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

The mep and dynamics trajectory differ in many different ways. The first difference is the time it takes for the reaction to take place (in this case, for the distance to follow to the valley floor of the potential graph). In the mep trajectory, it takes around 30 steps (15 fs) for the reaction to proceed whereas in the dynamics trajectory this number is around 140 fs. The second difference is that the internuclear distance between the H2 molecule and the lone H atom increases logarithmically in the mep calculation but increases linearly in the dynamics calculation.

===Complete the table below by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?===
{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 ||-414.280 KJ.mol-1||Yes|| Plot shows trajectory going from reactant, going over the minimum energy path via the transition state and finally to form the product.
||[[File:Surface Plot 1 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -4.1 ||-420.077 KJ.mol-1||No||Plot shows the trajectory going up the potential slope towards the transition state. However it does not have enough momentum to go over the curve and thus the trajectory returns to its original position. The reaction does not proceed and the reactants remain.||[[File:Surface Plot 2 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -5.1 ||-413.997 KJ.mol-1||Yes||Plot shows that increasing the momentum of the H-H bond in the reactants results in more vibrational energy hence a more distinct wave pattern appears in the trajectory. Similar to the first plot, the trajectory shows the reactant forming the product by going over the minimum energy path via the transition state.||[[File:Surface Plot 3 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.1 ||-357.277 KJ.mol-1||Yes|| ||[[File:Surface Plot 4 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.6 ||-349.477 KJ.mol-1||Yes|| ||[[File:Surface Plot 5 thl3318.png|thumb|300px|none|]]
|}

From the table above, I can conclude that whether or not a trajectory results in a reaction depends on the momentum p2. Changing the momentum p1 does not affect the result of the trajectory but if momentum p2 is not great enough the result will be unreactive.

[[File:Y2C11.png|thumb|600px|none|Trajectory that re-crosses the TS region]]

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

Experimental value will be greater than the reaction rate the transition state theory predicts because it does not take into account quantum tunneling which is more significant at low temperatures. Transition State Theory states that as long as the energy is greater than the activation energy the reaction will proceed to the products but this does not take into account the possibility of recrossing back to the reactants and thus the reaction not proceeding.

== EXERCISE 2: F + H2 system ==
===By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?===

The F + H2 reaction is exothermic whereas the H + HF reaction is endothermic. This means that the H-F bond strength is greater than the H-H bond strength and is lower in energy.

===Locate the approximate position of the transition state.===
H-H distance = 77pm

H-F distance = 179.5pm

[[File:Surface_Plot_TS_3_thl3318.png|thumb|300px|none|Surface Plot showing Transition State with 500 steps at step size 0.21 fs ]]

===Report the activation energy for both reactions.===
transition state energy = -433.020

H-H energy = -435.100

H-F energy = -560.700

Activation energy for F + H2 --> HF + H = 2.080

Activation energy for H + HF --> H2 + F = 127.68

You will be able to report a reasonable estimate by performing a mep (with a sufficient number of steps) from a structure neighbouring the transition state, and choosing to plot the appropriate quantity as a function of "time".

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

Although energy is conserved in the reaction, there is a change from kinetic energy to potential energy during the collision. Upon the H2 molecule colliding with the F atom, an elastic collision occurs and the kinetic energy of the H2 molecule (in the form of translational motion) is converted into potential energy of the HF molecule (in the form of vibrational motion). This can be confirmed experimentally by measuring whether there is an increase in vibrational motion by using IR spectroscopy.

[[File:Momenta_plot_thl3318.png|thumb|300px|none|Plot showing momentum of bonds before and after collision where A and B are Hydrogen atoms and C is a Fluorine atom ]]

===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:thl3318

2020-05-14T16:26:53Z

Thl3318: /* Locate the approximate position of the transition state. */

== EXERCISE 1: H + H2 system ==
===On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?===

On a potential energy surface diagram, the transition state can be mathematically defined as a saddle point (which is also an unstable equilibrium). It can be identified by determining the point where the partial derivatives of the function potential with respect to the internuclear distances are equal to 0.

<math> \frac{\partial V}{\partial r_1} =0</math> and <math> \frac{\partial V}{\partial r_2} =0</math>

However, simply looking at the results for the first derivative does not give any information with regards to whether the point is a local minimum, a local maximum, or a saddle point. Analyzing the 2nd derivative then allows for the determination of whether the point is a saddle point. This can be done using the following equation:

<math> D=\frac{\partial^2 V}{\partial r_1^2} \frac{\partial^2 V}{\partial r_2^2} - (\frac{\partial^2 V}{\partial r_1\partial r_2})^2 </math>

If D<0, the point is therefore a saddle point and also the transition state.

===Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.===
My best estimate for the transition state position is at time=20fs when the A-B distance is equal to the B-C distance. The A-B and B-C distance at this point is determined to be 90.8pm. This point in the "Internuclear Distance vs Time" plot can be rationed as the transition state position

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

The mep and dynamics trajectory differ in many different ways. The first difference is the time it takes for the reaction to take place (in this case, for the distance to follow to the valley floor of the potential graph). In the mep trajectory, it takes around 30 steps (15 fs) for the reaction to proceed whereas in the dynamics trajectory this number is around 140 fs. The second difference is that the internuclear distance between the H2 molecule and the lone H atom increases logarithmically in the mep calculation but increases linearly in the dynamics calculation.

===Complete the table below by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?===
{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 ||-414.280 KJ.mol-1||Yes|| Plot shows trajectory going from reactant, going over the minimum energy path via the transition state and finally to form the product.
||[[File:Surface Plot 1 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -4.1 ||-420.077 KJ.mol-1||No||Plot shows the trajectory going up the potential slope towards the transition state. However it does not have enough momentum to go over the curve and thus the trajectory returns to its original position. The reaction does not proceed and the reactants remain.||[[File:Surface Plot 2 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -5.1 ||-413.997 KJ.mol-1||Yes||Plot shows that increasing the momentum of the H-H bond in the reactants results in more vibrational energy hence a more distinct wave pattern appears in the trajectory. Similar to the first plot, the trajectory shows the reactant forming the product by going over the minimum energy path via the transition state.||[[File:Surface Plot 3 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.1 ||-357.277 KJ.mol-1||Yes|| ||[[File:Surface Plot 4 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.6 ||-349.477 KJ.mol-1||Yes|| ||[[File:Surface Plot 5 thl3318.png|thumb|300px|none|]]
|}

From the table above, I can conclude that whether or not a trajectory results in a reaction depends on the momentum p2. Changing the momentum p1 does not affect the result of the trajectory but if momentum p2 is not great enough the result will be unreactive.

[[File:Y2C11.png|thumb|600px|none|Trajectory that re-crosses the TS region]]

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

Experimental value will be greater than the reaction rate the transition state theory predicts because it does not take into account quantum tunneling which is more significant at low temperatures. Transition State Theory states that as long as the energy is greater than the activation energy the reaction will proceed to the products but this does not take into account the possibility of recrossing back to the reactants and thus the reaction not proceeding.

== EXERCISE 2: F + H2 system ==
===By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?===

The F + H2 reaction is exothermic whereas the H + HF reaction is endothermic. This means that the H-F bond strength is greater than the H-H bond strength and is lower in energy.

===Locate the approximate position of the transition state.===
H-H distance = 77pm

H-F distance = 179.5pm

step size = 0.51

Surface_Plot_TS_3_thl3318.png

===Report the activation energy for both reactions.===
transition state energy = -433.020

H-H energy = -435.100

H-F energy = -560.700

Activation energy for F + H2 --> HF + H = 2.080

Activation energy for H + HF --> H2 + F = 127.68

You will be able to report a reasonable estimate by performing a mep (with a sufficient number of steps) from a structure neighbouring the transition state, and choosing to plot the appropriate quantity as a function of "time".

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

Although energy is conserved in the reaction, there is a change from kinetic energy to potential energy during the collision. Upon the H2 molecule colliding with the F atom, an elastic collision occurs and the kinetic energy of the H2 molecule (in the form of translational motion) is converted into potential energy of the HF molecule (in the form of vibrational motion). This can be confirmed experimentally by measuring whether there is an increase in vibrational motion by using IR spectroscopy.

[[File:Momenta_plot_thl3318.png|thumb|300px|none|Plot showing momentum of bonds before and after collision where A and B are Hydrogen atoms and C is a Fluorine atom ]]

===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.===

File:Surface Plot TS 3 thl3318.png

2020-05-14T16:26:31Z

Thl3318:

MRD:thl3318

2020-05-14T16:24:10Z

Thl3318: /* Locate the approximate position of the transition state. */

== EXERCISE 1: H + H2 system ==
===On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?===

On a potential energy surface diagram, the transition state can be mathematically defined as a saddle point (which is also an unstable equilibrium). It can be identified by determining the point where the partial derivatives of the function potential with respect to the internuclear distances are equal to 0.

<math> \frac{\partial V}{\partial r_1} =0</math> and <math> \frac{\partial V}{\partial r_2} =0</math>

However, simply looking at the results for the first derivative does not give any information with regards to whether the point is a local minimum, a local maximum, or a saddle point. Analyzing the 2nd derivative then allows for the determination of whether the point is a saddle point. This can be done using the following equation:

<math> D=\frac{\partial^2 V}{\partial r_1^2} \frac{\partial^2 V}{\partial r_2^2} - (\frac{\partial^2 V}{\partial r_1\partial r_2})^2 </math>

If D<0, the point is therefore a saddle point and also the transition state.

===Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.===
My best estimate for the transition state position is at time=20fs when the A-B distance is equal to the B-C distance. The A-B and B-C distance at this point is determined to be 90.8pm. This point in the "Internuclear Distance vs Time" plot can be rationed as the transition state position

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

The mep and dynamics trajectory differ in many different ways. The first difference is the time it takes for the reaction to take place (in this case, for the distance to follow to the valley floor of the potential graph). In the mep trajectory, it takes around 30 steps (15 fs) for the reaction to proceed whereas in the dynamics trajectory this number is around 140 fs. The second difference is that the internuclear distance between the H2 molecule and the lone H atom increases logarithmically in the mep calculation but increases linearly in the dynamics calculation.

===Complete the table below by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?===
{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 ||-414.280 KJ.mol-1||Yes|| Plot shows trajectory going from reactant, going over the minimum energy path via the transition state and finally to form the product.
||[[File:Surface Plot 1 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -4.1 ||-420.077 KJ.mol-1||No||Plot shows the trajectory going up the potential slope towards the transition state. However it does not have enough momentum to go over the curve and thus the trajectory returns to its original position. The reaction does not proceed and the reactants remain.||[[File:Surface Plot 2 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -5.1 ||-413.997 KJ.mol-1||Yes||Plot shows that increasing the momentum of the H-H bond in the reactants results in more vibrational energy hence a more distinct wave pattern appears in the trajectory. Similar to the first plot, the trajectory shows the reactant forming the product by going over the minimum energy path via the transition state.||[[File:Surface Plot 3 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.1 ||-357.277 KJ.mol-1||Yes|| ||[[File:Surface Plot 4 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.6 ||-349.477 KJ.mol-1||Yes|| ||[[File:Surface Plot 5 thl3318.png|thumb|300px|none|]]
|}

From the table above, I can conclude that whether or not a trajectory results in a reaction depends on the momentum p2. Changing the momentum p1 does not affect the result of the trajectory but if momentum p2 is not great enough the result will be unreactive.

[[File:Y2C11.png|thumb|600px|none|Trajectory that re-crosses the TS region]]

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

Experimental value will be greater than the reaction rate the transition state theory predicts because it does not take into account quantum tunneling which is more significant at low temperatures. Transition State Theory states that as long as the energy is greater than the activation energy the reaction will proceed to the products but this does not take into account the possibility of recrossing back to the reactants and thus the reaction not proceeding.

== EXERCISE 2: F + H2 system ==
===By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?===

The F + H2 reaction is exothermic whereas the H + HF reaction is endothermic. This means that the H-F bond strength is greater than the H-H bond strength and is lower in energy.

===Locate the approximate position of the transition state.===
H-H distance = 77pm

H-F distance = 179.5pm

step size = 0.51

===Report the activation energy for both reactions.===
transition state energy = -433.020

H-H energy = -435.100

H-F energy = -560.700

Activation energy for F + H2 --> HF + H = 2.080

Activation energy for H + HF --> H2 + F = 127.68

You will be able to report a reasonable estimate by performing a mep (with a sufficient number of steps) from a structure neighbouring the transition state, and choosing to plot the appropriate quantity as a function of "time".

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

Although energy is conserved in the reaction, there is a change from kinetic energy to potential energy during the collision. Upon the H2 molecule colliding with the F atom, an elastic collision occurs and the kinetic energy of the H2 molecule (in the form of translational motion) is converted into potential energy of the HF molecule (in the form of vibrational motion). This can be confirmed experimentally by measuring whether there is an increase in vibrational motion by using IR spectroscopy.

[[File:Momenta_plot_thl3318.png|thumb|300px|none|Plot showing momentum of bonds before and after collision where A and B are Hydrogen atoms and C is a Fluorine atom ]]

===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:thl3318

2020-05-14T16:23:43Z

Thl3318: /* Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? */

== EXERCISE 1: H + H2 system ==
===On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?===

On a potential energy surface diagram, the transition state can be mathematically defined as a saddle point (which is also an unstable equilibrium). It can be identified by determining the point where the partial derivatives of the function potential with respect to the internuclear distances are equal to 0.

<math> \frac{\partial V}{\partial r_1} =0</math> and <math> \frac{\partial V}{\partial r_2} =0</math>

However, simply looking at the results for the first derivative does not give any information with regards to whether the point is a local minimum, a local maximum, or a saddle point. Analyzing the 2nd derivative then allows for the determination of whether the point is a saddle point. This can be done using the following equation:

<math> D=\frac{\partial^2 V}{\partial r_1^2} \frac{\partial^2 V}{\partial r_2^2} - (\frac{\partial^2 V}{\partial r_1\partial r_2})^2 </math>

If D<0, the point is therefore a saddle point and also the transition state.

===Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.===
My best estimate for the transition state position is at time=20fs when the A-B distance is equal to the B-C distance. The A-B and B-C distance at this point is determined to be 90.8pm. This point in the "Internuclear Distance vs Time" plot can be rationed as the transition state position

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

The mep and dynamics trajectory differ in many different ways. The first difference is the time it takes for the reaction to take place (in this case, for the distance to follow to the valley floor of the potential graph). In the mep trajectory, it takes around 30 steps (15 fs) for the reaction to proceed whereas in the dynamics trajectory this number is around 140 fs. The second difference is that the internuclear distance between the H2 molecule and the lone H atom increases logarithmically in the mep calculation but increases linearly in the dynamics calculation.

===Complete the table below by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?===
{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 ||-414.280 KJ.mol-1||Yes|| Plot shows trajectory going from reactant, going over the minimum energy path via the transition state and finally to form the product.
||[[File:Surface Plot 1 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -4.1 ||-420.077 KJ.mol-1||No||Plot shows the trajectory going up the potential slope towards the transition state. However it does not have enough momentum to go over the curve and thus the trajectory returns to its original position. The reaction does not proceed and the reactants remain.||[[File:Surface Plot 2 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -5.1 ||-413.997 KJ.mol-1||Yes||Plot shows that increasing the momentum of the H-H bond in the reactants results in more vibrational energy hence a more distinct wave pattern appears in the trajectory. Similar to the first plot, the trajectory shows the reactant forming the product by going over the minimum energy path via the transition state.||[[File:Surface Plot 3 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.1 ||-357.277 KJ.mol-1||Yes|| ||[[File:Surface Plot 4 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.6 ||-349.477 KJ.mol-1||Yes|| ||[[File:Surface Plot 5 thl3318.png|thumb|300px|none|]]
|}

From the table above, I can conclude that whether or not a trajectory results in a reaction depends on the momentum p2. Changing the momentum p1 does not affect the result of the trajectory but if momentum p2 is not great enough the result will be unreactive.

[[File:Y2C11.png|thumb|600px|none|Trajectory that re-crosses the TS region]]

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

Experimental value will be greater than the reaction rate the transition state theory predicts because it does not take into account quantum tunneling which is more significant at low temperatures. Transition State Theory states that as long as the energy is greater than the activation energy the reaction will proceed to the products but this does not take into account the possibility of recrossing back to the reactants and thus the reaction not proceeding.

== EXERCISE 2: F + H2 system ==
===By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?===

The F + H2 reaction is exothermic whereas the H + HF reaction is endothermic. This means that the H-F bond strength is greater than the H-H bond strength and is lower in energy.

===Locate the approximate position of the transition state.===
H-H distance = 77pm
H-F distance = 179.5pm 1step size = 0.51

===Report the activation energy for both reactions.===
transition state energy = -433.020

H-H energy = -435.100

H-F energy = -560.700

Activation energy for F + H2 --> HF + H = 2.080

Activation energy for H + HF --> H2 + F = 127.68

You will be able to report a reasonable estimate by performing a mep (with a sufficient number of steps) from a structure neighbouring the transition state, and choosing to plot the appropriate quantity as a function of "time".

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

Although energy is conserved in the reaction, there is a change from kinetic energy to potential energy during the collision. Upon the H2 molecule colliding with the F atom, an elastic collision occurs and the kinetic energy of the H2 molecule (in the form of translational motion) is converted into potential energy of the HF molecule (in the form of vibrational motion). This can be confirmed experimentally by measuring whether there is an increase in vibrational motion by using IR spectroscopy.

[[File:Momenta_plot_thl3318.png|thumb|300px|none|Plot showing momentum of bonds before and after collision where A and B are Hydrogen atoms and C is a Fluorine atom ]]

===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:thl3318

2020-05-14T16:18:25Z

Thl3318: /* In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally. */

== EXERCISE 1: H + H2 system ==
===On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?===

On a potential energy surface diagram, the transition state can be mathematically defined as a saddle point (which is also an unstable equilibrium). It can be identified by determining the point where the partial derivatives of the function potential with respect to the internuclear distances are equal to 0.

<math> \frac{\partial V}{\partial r_1} =0</math> and <math> \frac{\partial V}{\partial r_2} =0</math>

However, simply looking at the results for the first derivative does not give any information with regards to whether the point is a local minimum, a local maximum, or a saddle point. Analyzing the 2nd derivative then allows for the determination of whether the point is a saddle point. This can be done using the following equation:

<math> D=\frac{\partial^2 V}{\partial r_1^2} \frac{\partial^2 V}{\partial r_2^2} - (\frac{\partial^2 V}{\partial r_1\partial r_2})^2 </math>

If D<0, the point is therefore a saddle point and also the transition state.

===Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.===
My best estimate for the transition state position is at time=20fs when the A-B distance is equal to the B-C distance. The A-B and B-C distance at this point is determined to be 90.8pm. This point in the "Internuclear Distance vs Time" plot can be rationed as the transition state position

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

The mep and dynamics trajectory differ in many different ways. The first difference is the time it takes for the reaction to take place (in this case, for the distance to follow to the valley floor of the potential graph). In the mep trajectory, it takes around 30 steps (15 fs) for the reaction to proceed whereas in the dynamics trajectory this number is around 140 fs. The second difference is that the internuclear distance between the H2 molecule and the lone H atom increases logarithmically in the mep calculation but increases linearly in the dynamics calculation.

===Complete the table below by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?===
{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 ||-414.280 KJ.mol-1||Yes|| Plot shows trajectory going from reactant, going over the minimum energy path via the transition state and finally to form the product.
||[[File:Surface Plot 1 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -4.1 ||-420.077 KJ.mol-1||No||Plot shows the trajectory going up the potential slope towards the transition state. However it does not have enough momentum to go over the curve and thus the trajectory returns to its original position. The reaction does not proceed and the reactants remain.||[[File:Surface Plot 2 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -5.1 ||-413.997 KJ.mol-1||Yes||Plot shows that increasing the momentum of the H-H bond in the reactants results in more vibrational energy hence a more distinct wave pattern appears in the trajectory. Similar to the first plot, the trajectory shows the reactant forming the product by going over the minimum energy path via the transition state.||[[File:Surface Plot 3 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.1 ||-357.277 KJ.mol-1||Yes|| ||[[File:Surface Plot 4 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.6 ||-349.477 KJ.mol-1||Yes|| ||[[File:Surface Plot 5 thl3318.png|thumb|300px|none|]]
|}

From the table above, I can conclude that whether or not a trajectory results in a reaction depends on the momentum p2. Changing the momentum p1 does not affect the result of the trajectory but if momentum p2 is not great enough the result will be unreactive.

[[File:Y2C11.png|thumb|600px|none|Trajectory that re-crosses the TS region]]

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

Experimental value will be greater than the reaction rate the transition state theory predicts because it does not take into account quantum tunneling which is more significant at low temperatures. Transition State Theory states that as long as the energy is greater than the activation energy the reaction will proceed to the products but this does not take into account the possibility of the "rebounding" and thus the reaction not proceeding.

== EXERCISE 2: F + H2 system ==
===By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?===

The F + H2 reaction is exothermic whereas the H + HF reaction is endothermic. This means that the H-F bond strength is greater than the H-H bond strength and is lower in energy.

===Locate the approximate position of the transition state.===
H-H distance = 77pm
H-F distance = 179.5pm 1step size = 0.51

===Report the activation energy for both reactions.===
transition state energy = -433.020

H-H energy = -435.100

H-F energy = -560.700

Activation energy for F + H2 --> HF + H = 2.080

Activation energy for H + HF --> H2 + F = 127.68

You will be able to report a reasonable estimate by performing a mep (with a sufficient number of steps) from a structure neighbouring the transition state, and choosing to plot the appropriate quantity as a function of "time".

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

Although energy is conserved in the reaction, there is a change from kinetic energy to potential energy during the collision. Upon the H2 molecule colliding with the F atom, an elastic collision occurs and the kinetic energy of the H2 molecule (in the form of translational motion) is converted into potential energy of the HF molecule (in the form of vibrational motion). This can be confirmed experimentally by measuring whether there is an increase in vibrational motion by using IR spectroscopy.

[[File:Momenta_plot_thl3318.png|thumb|300px|none|Plot showing momentum of bonds before and after collision where A and B are Hydrogen atoms and C is a Fluorine atom ]]

===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:thl3318

2020-05-14T16:18:11Z

Thl3318: /* In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally. */

== EXERCISE 1: H + H2 system ==
===On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?===

On a potential energy surface diagram, the transition state can be mathematically defined as a saddle point (which is also an unstable equilibrium). It can be identified by determining the point where the partial derivatives of the function potential with respect to the internuclear distances are equal to 0.

<math> \frac{\partial V}{\partial r_1} =0</math> and <math> \frac{\partial V}{\partial r_2} =0</math>

However, simply looking at the results for the first derivative does not give any information with regards to whether the point is a local minimum, a local maximum, or a saddle point. Analyzing the 2nd derivative then allows for the determination of whether the point is a saddle point. This can be done using the following equation:

<math> D=\frac{\partial^2 V}{\partial r_1^2} \frac{\partial^2 V}{\partial r_2^2} - (\frac{\partial^2 V}{\partial r_1\partial r_2})^2 </math>

If D<0, the point is therefore a saddle point and also the transition state.

===Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.===
My best estimate for the transition state position is at time=20fs when the A-B distance is equal to the B-C distance. The A-B and B-C distance at this point is determined to be 90.8pm. This point in the "Internuclear Distance vs Time" plot can be rationed as the transition state position

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

The mep and dynamics trajectory differ in many different ways. The first difference is the time it takes for the reaction to take place (in this case, for the distance to follow to the valley floor of the potential graph). In the mep trajectory, it takes around 30 steps (15 fs) for the reaction to proceed whereas in the dynamics trajectory this number is around 140 fs. The second difference is that the internuclear distance between the H2 molecule and the lone H atom increases logarithmically in the mep calculation but increases linearly in the dynamics calculation.

===Complete the table below by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?===
{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 ||-414.280 KJ.mol-1||Yes|| Plot shows trajectory going from reactant, going over the minimum energy path via the transition state and finally to form the product.
||[[File:Surface Plot 1 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -4.1 ||-420.077 KJ.mol-1||No||Plot shows the trajectory going up the potential slope towards the transition state. However it does not have enough momentum to go over the curve and thus the trajectory returns to its original position. The reaction does not proceed and the reactants remain.||[[File:Surface Plot 2 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -5.1 ||-413.997 KJ.mol-1||Yes||Plot shows that increasing the momentum of the H-H bond in the reactants results in more vibrational energy hence a more distinct wave pattern appears in the trajectory. Similar to the first plot, the trajectory shows the reactant forming the product by going over the minimum energy path via the transition state.||[[File:Surface Plot 3 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.1 ||-357.277 KJ.mol-1||Yes|| ||[[File:Surface Plot 4 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.6 ||-349.477 KJ.mol-1||Yes|| ||[[File:Surface Plot 5 thl3318.png|thumb|300px|none|]]
|}

From the table above, I can conclude that whether or not a trajectory results in a reaction depends on the momentum p2. Changing the momentum p1 does not affect the result of the trajectory but if momentum p2 is not great enough the result will be unreactive.

[[File:Y2C11.png|thumb|600px|none|Trajectory that re-crosses the TS region]]

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

Experimental value will be greater than the reaction rate the transition state theory predicts because it does not take into account quantum tunneling which is more significant at low temperatures. Transition State Theory states that as long as the energy is greater than the activation energy the reaction will proceed to the products but this does not take into account the possibility of the "rebounding" and thus the reaction not proceeding.

== EXERCISE 2: F + H2 system ==
===By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?===

The F + H2 reaction is exothermic whereas the H + HF reaction is endothermic. This means that the H-F bond strength is greater than the H-H bond strength and is lower in energy.

===Locate the approximate position of the transition state.===
H-H distance = 77pm
H-F distance = 179.5pm 1step size = 0.51

===Report the activation energy for both reactions.===
transition state energy = -433.020

H-H energy = -435.100

H-F energy = -560.700

Activation energy for F + H2 --> HF + H = 2.080

Activation energy for H + HF --> H2 + F = 127.68

You will be able to report a reasonable estimate by performing a mep (with a sufficient number of steps) from a structure neighbouring the transition state, and choosing to plot the appropriate quantity as a function of "time".

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

Although energy is conserved in the reaction, there is a change from kinetic energy to potential energy during the collision. Upon the H2 molecule colliding with the F atom, an elastic collision occurs and the kinetic energy of the H2 molecule (in the form of translational motion) is converted into potential energy of the HF molecule (in the form of vibrational motion). This can be confirmed experimentally by measuring whether there is an increase in vibrational motion by using IR spectroscopy.

[[File:Momenta_plot_thl3318.png|thumb|300px|none|Plot showing momentum of bonds before and after collision where A and B are Hydrogen atoms and C is a Fluorine atom ]]

===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:thl3318

2020-05-14T16:17:53Z

Thl3318: /* In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally. */

== EXERCISE 1: H + H2 system ==
===On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?===

On a potential energy surface diagram, the transition state can be mathematically defined as a saddle point (which is also an unstable equilibrium). It can be identified by determining the point where the partial derivatives of the function potential with respect to the internuclear distances are equal to 0.

<math> \frac{\partial V}{\partial r_1} =0</math> and <math> \frac{\partial V}{\partial r_2} =0</math>

However, simply looking at the results for the first derivative does not give any information with regards to whether the point is a local minimum, a local maximum, or a saddle point. Analyzing the 2nd derivative then allows for the determination of whether the point is a saddle point. This can be done using the following equation:

<math> D=\frac{\partial^2 V}{\partial r_1^2} \frac{\partial^2 V}{\partial r_2^2} - (\frac{\partial^2 V}{\partial r_1\partial r_2})^2 </math>

If D<0, the point is therefore a saddle point and also the transition state.

===Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.===
My best estimate for the transition state position is at time=20fs when the A-B distance is equal to the B-C distance. The A-B and B-C distance at this point is determined to be 90.8pm. This point in the "Internuclear Distance vs Time" plot can be rationed as the transition state position

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

The mep and dynamics trajectory differ in many different ways. The first difference is the time it takes for the reaction to take place (in this case, for the distance to follow to the valley floor of the potential graph). In the mep trajectory, it takes around 30 steps (15 fs) for the reaction to proceed whereas in the dynamics trajectory this number is around 140 fs. The second difference is that the internuclear distance between the H2 molecule and the lone H atom increases logarithmically in the mep calculation but increases linearly in the dynamics calculation.

===Complete the table below by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?===
{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 ||-414.280 KJ.mol-1||Yes|| Plot shows trajectory going from reactant, going over the minimum energy path via the transition state and finally to form the product.
||[[File:Surface Plot 1 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -4.1 ||-420.077 KJ.mol-1||No||Plot shows the trajectory going up the potential slope towards the transition state. However it does not have enough momentum to go over the curve and thus the trajectory returns to its original position. The reaction does not proceed and the reactants remain.||[[File:Surface Plot 2 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -5.1 ||-413.997 KJ.mol-1||Yes||Plot shows that increasing the momentum of the H-H bond in the reactants results in more vibrational energy hence a more distinct wave pattern appears in the trajectory. Similar to the first plot, the trajectory shows the reactant forming the product by going over the minimum energy path via the transition state.||[[File:Surface Plot 3 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.1 ||-357.277 KJ.mol-1||Yes|| ||[[File:Surface Plot 4 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.6 ||-349.477 KJ.mol-1||Yes|| ||[[File:Surface Plot 5 thl3318.png|thumb|300px|none|]]
|}

From the table above, I can conclude that whether or not a trajectory results in a reaction depends on the momentum p2. Changing the momentum p1 does not affect the result of the trajectory but if momentum p2 is not great enough the result will be unreactive.

[[File:Y2C11.png|thumb|600px|none|Trajectory that re-crosses the TS region]]

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

Experimental value will be greater than the reaction rate the transition state theory predicts because it does not take into account quantum tunneling which is more significant at low temperatures. Transition State Theory states that as long as the energy is greater than the activation energy the reaction will proceed to the products but this does not take into account the possibility of the "rebounding" and thus the reaction not proceeding.

== EXERCISE 2: F + H2 system ==
===By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?===

The F + H2 reaction is exothermic whereas the H + HF reaction is endothermic. This means that the H-F bond strength is greater than the H-H bond strength and is lower in energy.

===Locate the approximate position of the transition state.===
H-H distance = 77pm
H-F distance = 179.5pm 1step size = 0.51

===Report the activation energy for both reactions.===
transition state energy = -433.020

H-H energy = -435.100

H-F energy = -560.700

Activation energy for F + H2 --> HF + H = 2.080

Activation energy for H + HF --> H2 + F = 127.68

You will be able to report a reasonable estimate by performing a mep (with a sufficient number of steps) from a structure neighbouring the transition state, and choosing to plot the appropriate quantity as a function of "time".

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

Although energy is conserved in the reaction, there is a change from kinetic energy to potential energy during the collision. Upon the H2 molecule colliding with the F atom, an elastic collision occurs and the kinetic energy of the H2 molecule (in the form of translational motion) is converted into potential energy of the HF molecule (in the form of vibrational motion). This can be confirmed experimentally by measuring whether there is an increase in vibrational motion by using IR spectroscopy.

[[File:Momenta_plot_thl3318.png|thumb|600px|none|Plot showing momentum of bonds before and after collision where A and B are Hydrogen atoms and C is a Fluorine atom ]]

===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:thl3318

2020-05-14T16:16:07Z

Thl3318: /* In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally. */

== EXERCISE 1: H + H2 system ==
===On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?===

On a potential energy surface diagram, the transition state can be mathematically defined as a saddle point (which is also an unstable equilibrium). It can be identified by determining the point where the partial derivatives of the function potential with respect to the internuclear distances are equal to 0.

<math> \frac{\partial V}{\partial r_1} =0</math> and <math> \frac{\partial V}{\partial r_2} =0</math>

However, simply looking at the results for the first derivative does not give any information with regards to whether the point is a local minimum, a local maximum, or a saddle point. Analyzing the 2nd derivative then allows for the determination of whether the point is a saddle point. This can be done using the following equation:

<math> D=\frac{\partial^2 V}{\partial r_1^2} \frac{\partial^2 V}{\partial r_2^2} - (\frac{\partial^2 V}{\partial r_1\partial r_2})^2 </math>

If D<0, the point is therefore a saddle point and also the transition state.

===Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.===
My best estimate for the transition state position is at time=20fs when the A-B distance is equal to the B-C distance. The A-B and B-C distance at this point is determined to be 90.8pm. This point in the "Internuclear Distance vs Time" plot can be rationed as the transition state position

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

The mep and dynamics trajectory differ in many different ways. The first difference is the time it takes for the reaction to take place (in this case, for the distance to follow to the valley floor of the potential graph). In the mep trajectory, it takes around 30 steps (15 fs) for the reaction to proceed whereas in the dynamics trajectory this number is around 140 fs. The second difference is that the internuclear distance between the H2 molecule and the lone H atom increases logarithmically in the mep calculation but increases linearly in the dynamics calculation.

===Complete the table below by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?===
{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 ||-414.280 KJ.mol-1||Yes|| Plot shows trajectory going from reactant, going over the minimum energy path via the transition state and finally to form the product.
||[[File:Surface Plot 1 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -4.1 ||-420.077 KJ.mol-1||No||Plot shows the trajectory going up the potential slope towards the transition state. However it does not have enough momentum to go over the curve and thus the trajectory returns to its original position. The reaction does not proceed and the reactants remain.||[[File:Surface Plot 2 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -5.1 ||-413.997 KJ.mol-1||Yes||Plot shows that increasing the momentum of the H-H bond in the reactants results in more vibrational energy hence a more distinct wave pattern appears in the trajectory. Similar to the first plot, the trajectory shows the reactant forming the product by going over the minimum energy path via the transition state.||[[File:Surface Plot 3 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.1 ||-357.277 KJ.mol-1||Yes|| ||[[File:Surface Plot 4 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.6 ||-349.477 KJ.mol-1||Yes|| ||[[File:Surface Plot 5 thl3318.png|thumb|300px|none|]]
|}

From the table above, I can conclude that whether or not a trajectory results in a reaction depends on the momentum p2. Changing the momentum p1 does not affect the result of the trajectory but if momentum p2 is not great enough the result will be unreactive.

[[File:Y2C11.png|thumb|600px|none|Trajectory that re-crosses the TS region]]

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

Experimental value will be greater than the reaction rate the transition state theory predicts because it does not take into account quantum tunneling which is more significant at low temperatures. Transition State Theory states that as long as the energy is greater than the activation energy the reaction will proceed to the products but this does not take into account the possibility of the "rebounding" and thus the reaction not proceeding.

== EXERCISE 2: F + H2 system ==
===By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?===

The F + H2 reaction is exothermic whereas the H + HF reaction is endothermic. This means that the H-F bond strength is greater than the H-H bond strength and is lower in energy.

===Locate the approximate position of the transition state.===
H-H distance = 77pm
H-F distance = 179.5pm 1step size = 0.51

===Report the activation energy for both reactions.===
transition state energy = -433.020

H-H energy = -435.100

H-F energy = -560.700

Activation energy for F + H2 --> HF + H = 2.080

Activation energy for H + HF --> H2 + F = 127.68

You will be able to report a reasonable estimate by performing a mep (with a sufficient number of steps) from a structure neighbouring the transition state, and choosing to plot the appropriate quantity as a function of "time".

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

Although energy is conserved in the reaction, there is a change from kinetic energy to potential energy during the collision. Upon the H2 molecule colliding with the F atom, an elastic collision occurs and the kinetic energy of the H2 molecule (in the form of translational motion) is converted into potential energy of the HF molecule (in the form of vibrational motion). This can be confirmed experimentally by measuring whether there is an increase in vibrational motion by using IR spectroscopy.

Momenta_plot_thl3318.png

===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.===

File:Momenta plot thl3318.png

2020-05-14T16:15:44Z

Thl3318:

MRD:thl3318

2020-05-14T16:12:27Z

Thl3318: /* In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally. */

== EXERCISE 1: H + H2 system ==
===On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?===

On a potential energy surface diagram, the transition state can be mathematically defined as a saddle point (which is also an unstable equilibrium). It can be identified by determining the point where the partial derivatives of the function potential with respect to the internuclear distances are equal to 0.

<math> \frac{\partial V}{\partial r_1} =0</math> and <math> \frac{\partial V}{\partial r_2} =0</math>

However, simply looking at the results for the first derivative does not give any information with regards to whether the point is a local minimum, a local maximum, or a saddle point. Analyzing the 2nd derivative then allows for the determination of whether the point is a saddle point. This can be done using the following equation:

<math> D=\frac{\partial^2 V}{\partial r_1^2} \frac{\partial^2 V}{\partial r_2^2} - (\frac{\partial^2 V}{\partial r_1\partial r_2})^2 </math>

If D<0, the point is therefore a saddle point and also the transition state.

===Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.===
My best estimate for the transition state position is at time=20fs when the A-B distance is equal to the B-C distance. The A-B and B-C distance at this point is determined to be 90.8pm. This point in the "Internuclear Distance vs Time" plot can be rationed as the transition state position

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

The mep and dynamics trajectory differ in many different ways. The first difference is the time it takes for the reaction to take place (in this case, for the distance to follow to the valley floor of the potential graph). In the mep trajectory, it takes around 30 steps (15 fs) for the reaction to proceed whereas in the dynamics trajectory this number is around 140 fs. The second difference is that the internuclear distance between the H2 molecule and the lone H atom increases logarithmically in the mep calculation but increases linearly in the dynamics calculation.

===Complete the table below by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?===
{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 ||-414.280 KJ.mol-1||Yes|| Plot shows trajectory going from reactant, going over the minimum energy path via the transition state and finally to form the product.
||[[File:Surface Plot 1 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -4.1 ||-420.077 KJ.mol-1||No||Plot shows the trajectory going up the potential slope towards the transition state. However it does not have enough momentum to go over the curve and thus the trajectory returns to its original position. The reaction does not proceed and the reactants remain.||[[File:Surface Plot 2 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -5.1 ||-413.997 KJ.mol-1||Yes||Plot shows that increasing the momentum of the H-H bond in the reactants results in more vibrational energy hence a more distinct wave pattern appears in the trajectory. Similar to the first plot, the trajectory shows the reactant forming the product by going over the minimum energy path via the transition state.||[[File:Surface Plot 3 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.1 ||-357.277 KJ.mol-1||Yes|| ||[[File:Surface Plot 4 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.6 ||-349.477 KJ.mol-1||Yes|| ||[[File:Surface Plot 5 thl3318.png|thumb|300px|none|]]
|}

From the table above, I can conclude that whether or not a trajectory results in a reaction depends on the momentum p2. Changing the momentum p1 does not affect the result of the trajectory but if momentum p2 is not great enough the result will be unreactive.

[[File:Y2C11.png|thumb|600px|none|Trajectory that re-crosses the TS region]]

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

Experimental value will be greater than the reaction rate the transition state theory predicts because it does not take into account quantum tunneling which is more significant at low temperatures. Transition State Theory states that as long as the energy is greater than the activation energy the reaction will proceed to the products but this does not take into account the possibility of the "rebounding" and thus the reaction not proceeding.

== EXERCISE 2: F + H2 system ==
===By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?===

The F + H2 reaction is exothermic whereas the H + HF reaction is endothermic. This means that the H-F bond strength is greater than the H-H bond strength and is lower in energy.

===Locate the approximate position of the transition state.===
H-H distance = 77pm
H-F distance = 179.5pm 1step size = 0.51

===Report the activation energy for both reactions.===
transition state energy = -433.020

H-H energy = -435.100

H-F energy = -560.700

Activation energy for F + H2 --> HF + H = 2.080

Activation energy for H + HF --> H2 + F = 127.68

You will be able to report a reasonable estimate by performing a mep (with a sufficient number of steps) from a structure neighbouring the transition state, and choosing to plot the appropriate quantity as a function of "time".

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

Although energy is conserved in the reaction, there is a change from kinetic energy to potential energy during the collision. Upon the H2 molecule colliding with the F atom, an elastic collision occurs and the kinetic energy of the H2 molecule (in the form of translational motion) is converted into potential energy of the HF molecule (in the form of vibrational motion). This can be confirmed experimentally by measuring whether there is an increase in vibrational motion by using IR spectroscopy.

===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:thl3318

2020-05-14T15:23:31Z

Thl3318: /* In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally. */

== EXERCISE 1: H + H2 system ==
===On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?===

On a potential energy surface diagram, the transition state can be mathematically defined as a saddle point (which is also an unstable equilibrium). It can be identified by determining the point where the partial derivatives of the function potential with respect to the internuclear distances are equal to 0.

<math> \frac{\partial V}{\partial r_1} =0</math> and <math> \frac{\partial V}{\partial r_2} =0</math>

However, simply looking at the results for the first derivative does not give any information with regards to whether the point is a local minimum, a local maximum, or a saddle point. Analyzing the 2nd derivative then allows for the determination of whether the point is a saddle point. This can be done using the following equation:

<math> D=\frac{\partial^2 V}{\partial r_1^2} \frac{\partial^2 V}{\partial r_2^2} - (\frac{\partial^2 V}{\partial r_1\partial r_2})^2 </math>

If D<0, the point is therefore a saddle point and also the transition state.

===Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.===
My best estimate for the transition state position is at time=20fs when the A-B distance is equal to the B-C distance. The A-B and B-C distance at this point is determined to be 90.8pm. This point in the "Internuclear Distance vs Time" plot can be rationed as the transition state position

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

The mep and dynamics trajectory differ in many different ways. The first difference is the time it takes for the reaction to take place (in this case, for the distance to follow to the valley floor of the potential graph). In the mep trajectory, it takes around 30 steps (15 fs) for the reaction to proceed whereas in the dynamics trajectory this number is around 140 fs. The second difference is that the internuclear distance between the H2 molecule and the lone H atom increases logarithmically in the mep calculation but increases linearly in the dynamics calculation.

===Complete the table below by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?===
{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 ||-414.280 KJ.mol-1||Yes|| Plot shows trajectory going from reactant, going over the minimum energy path via the transition state and finally to form the product.
||[[File:Surface Plot 1 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -4.1 ||-420.077 KJ.mol-1||No||Plot shows the trajectory going up the potential slope towards the transition state. However it does not have enough momentum to go over the curve and thus the trajectory returns to its original position. The reaction does not proceed and the reactants remain.||[[File:Surface Plot 2 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -5.1 ||-413.997 KJ.mol-1||Yes||Plot shows that increasing the momentum of the H-H bond in the reactants results in more vibrational energy hence a more distinct wave pattern appears in the trajectory. Similar to the first plot, the trajectory shows the reactant forming the product by going over the minimum energy path via the transition state.||[[File:Surface Plot 3 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.1 ||-357.277 KJ.mol-1||Yes|| ||[[File:Surface Plot 4 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.6 ||-349.477 KJ.mol-1||Yes|| ||[[File:Surface Plot 5 thl3318.png|thumb|300px|none|]]
|}

From the table above, I can conclude that whether or not a trajectory results in a reaction depends on the momentum p2. Changing the momentum p1 does not affect the result of the trajectory but if momentum p2 is not great enough the result will be unreactive.

[[File:Y2C11.png|thumb|600px|none|Trajectory that re-crosses the TS region]]

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

Experimental value will be greater than the reaction rate the transition state theory predicts because it does not take into account quantum tunneling which is more significant at low temperatures. Transition State Theory states that as long as the energy is greater than the activation energy the reaction will proceed to the products but this does not take into account the possibility of the "rebounding" and thus the reaction not proceeding.

== EXERCISE 2: F + H2 system ==
===By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?===

The F + H2 reaction is exothermic whereas the H + HF reaction is endothermic. This means that the H-F bond strength is greater than the H-H bond strength and is lower in energy.

===Locate the approximate position of the transition state.===
H-H distance = 77pm
H-F distance = 179.5pm 1step size = 0.51

===Report the activation energy for both reactions.===
transition state energy = -433.020

H-H energy = -435.100

H-F energy = -560.700

Activation energy for F + H2 --> HF + H = 2.080

Activation energy for H + HF --> H2 + F = 127.68

You will be able to report a reasonable estimate by performing a mep (with a sufficient number of steps) from a structure neighbouring the transition state, and choosing to plot the appropriate quantity as a function of "time".

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

Reaction energy is released in the form of translational motion also known as heat. This can be confirmed experimentally by measuring the temperatures before and after the reaction. In the case of the reaction of F + H2, this can be confirmed by an increase in temperature due to the reaction being exothermic. In the case of the reaction H + HF, this can be confirmed by a decrease in the temperature due to the reaction being endothermic.

===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:thl3318

2020-05-14T15:16:02Z

Thl3318: /* In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally. */

== EXERCISE 1: H + H2 system ==
===On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?===

On a potential energy surface diagram, the transition state can be mathematically defined as a saddle point (which is also an unstable equilibrium). It can be identified by determining the point where the partial derivatives of the function potential with respect to the internuclear distances are equal to 0.

<math> \frac{\partial V}{\partial r_1} =0</math> and <math> \frac{\partial V}{\partial r_2} =0</math>

However, simply looking at the results for the first derivative does not give any information with regards to whether the point is a local minimum, a local maximum, or a saddle point. Analyzing the 2nd derivative then allows for the determination of whether the point is a saddle point. This can be done using the following equation:

<math> D=\frac{\partial^2 V}{\partial r_1^2} \frac{\partial^2 V}{\partial r_2^2} - (\frac{\partial^2 V}{\partial r_1\partial r_2})^2 </math>

If D<0, the point is therefore a saddle point and also the transition state.

===Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.===
My best estimate for the transition state position is at time=20fs when the A-B distance is equal to the B-C distance. The A-B and B-C distance at this point is determined to be 90.8pm. This point in the "Internuclear Distance vs Time" plot can be rationed as the transition state position

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

The mep and dynamics trajectory differ in many different ways. The first difference is the time it takes for the reaction to take place (in this case, for the distance to follow to the valley floor of the potential graph). In the mep trajectory, it takes around 30 steps (15 fs) for the reaction to proceed whereas in the dynamics trajectory this number is around 140 fs. The second difference is that the internuclear distance between the H2 molecule and the lone H atom increases logarithmically in the mep calculation but increases linearly in the dynamics calculation.

===Complete the table below by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?===
{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 ||-414.280 KJ.mol-1||Yes|| Plot shows trajectory going from reactant, going over the minimum energy path via the transition state and finally to form the product.
||[[File:Surface Plot 1 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -4.1 ||-420.077 KJ.mol-1||No||Plot shows the trajectory going up the potential slope towards the transition state. However it does not have enough momentum to go over the curve and thus the trajectory returns to its original position. The reaction does not proceed and the reactants remain.||[[File:Surface Plot 2 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -5.1 ||-413.997 KJ.mol-1||Yes||Plot shows that increasing the momentum of the H-H bond in the reactants results in more vibrational energy hence a more distinct wave pattern appears in the trajectory. Similar to the first plot, the trajectory shows the reactant forming the product by going over the minimum energy path via the transition state.||[[File:Surface Plot 3 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.1 ||-357.277 KJ.mol-1||Yes|| ||[[File:Surface Plot 4 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.6 ||-349.477 KJ.mol-1||Yes|| ||[[File:Surface Plot 5 thl3318.png|thumb|300px|none|]]
|}

From the table above, I can conclude that whether or not a trajectory results in a reaction depends on the momentum p2. Changing the momentum p1 does not affect the result of the trajectory but if momentum p2 is not great enough the result will be unreactive.

[[File:Y2C11.png|thumb|600px|none|Trajectory that re-crosses the TS region]]

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

Experimental value will be greater than the reaction rate the transition state theory predicts because it does not take into account quantum tunneling which is more significant at low temperatures. Transition State Theory states that as long as the energy is greater than the activation energy the reaction will proceed to the products but this does not take into account the possibility of the "rebounding" and thus the reaction not proceeding.

== EXERCISE 2: F + H2 system ==
===By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?===

The F + H2 reaction is exothermic whereas the H + HF reaction is endothermic. This means that the H-F bond strength is greater than the H-H bond strength and is lower in energy.

===Locate the approximate position of the transition state.===
H-H distance = 77pm
H-F distance = 179.5pm 1step size = 0.51

===Report the activation energy for both reactions.===
transition state energy = -433.020

H-H energy = -435.100

H-F energy = -560.700

Activation energy for F + H2 --> HF + H = 2.080

Activation energy for H + HF --> H2 + F = 127.68

You will be able to report a reasonable estimate by performing a mep (with a sufficient number of steps) from a structure neighbouring the transition state, and choosing to plot the appropriate quantity as a function of "time".

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

Reaction energy is released in the form of translational motion also known as heat. This can be confirmed experimentally by measuring the temperatures before and after the reaction.

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

MRD:thl3318

2020-05-14T15:08:39Z

Thl3318: /* Locate the approximate position of the transition state. */

== EXERCISE 1: H + H2 system ==
===On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?===

On a potential energy surface diagram, the transition state can be mathematically defined as a saddle point (which is also an unstable equilibrium). It can be identified by determining the point where the partial derivatives of the function potential with respect to the internuclear distances are equal to 0.

<math> \frac{\partial V}{\partial r_1} =0</math> and <math> \frac{\partial V}{\partial r_2} =0</math>

However, simply looking at the results for the first derivative does not give any information with regards to whether the point is a local minimum, a local maximum, or a saddle point. Analyzing the 2nd derivative then allows for the determination of whether the point is a saddle point. This can be done using the following equation:

<math> D=\frac{\partial^2 V}{\partial r_1^2} \frac{\partial^2 V}{\partial r_2^2} - (\frac{\partial^2 V}{\partial r_1\partial r_2})^2 </math>

If D<0, the point is therefore a saddle point and also the transition state.

===Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.===
My best estimate for the transition state position is at time=20fs when the A-B distance is equal to the B-C distance. The A-B and B-C distance at this point is determined to be 90.8pm. This point in the "Internuclear Distance vs Time" plot can be rationed as the transition state position

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

The mep and dynamics trajectory differ in many different ways. The first difference is the time it takes for the reaction to take place (in this case, for the distance to follow to the valley floor of the potential graph). In the mep trajectory, it takes around 30 steps (15 fs) for the reaction to proceed whereas in the dynamics trajectory this number is around 140 fs. The second difference is that the internuclear distance between the H2 molecule and the lone H atom increases logarithmically in the mep calculation but increases linearly in the dynamics calculation.

===Complete the table below by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?===
{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 ||-414.280 KJ.mol-1||Yes|| Plot shows trajectory going from reactant, going over the minimum energy path via the transition state and finally to form the product.
||[[File:Surface Plot 1 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -4.1 ||-420.077 KJ.mol-1||No||Plot shows the trajectory going up the potential slope towards the transition state. However it does not have enough momentum to go over the curve and thus the trajectory returns to its original position. The reaction does not proceed and the reactants remain.||[[File:Surface Plot 2 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -5.1 ||-413.997 KJ.mol-1||Yes||Plot shows that increasing the momentum of the H-H bond in the reactants results in more vibrational energy hence a more distinct wave pattern appears in the trajectory. Similar to the first plot, the trajectory shows the reactant forming the product by going over the minimum energy path via the transition state.||[[File:Surface Plot 3 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.1 ||-357.277 KJ.mol-1||Yes|| ||[[File:Surface Plot 4 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.6 ||-349.477 KJ.mol-1||Yes|| ||[[File:Surface Plot 5 thl3318.png|thumb|300px|none|]]
|}

From the table above, I can conclude that whether or not a trajectory results in a reaction depends on the momentum p2. Changing the momentum p1 does not affect the result of the trajectory but if momentum p2 is not great enough the result will be unreactive.

[[File:Y2C11.png|thumb|600px|none|Trajectory that re-crosses the TS region]]

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

Experimental value will be greater than the reaction rate the transition state theory predicts because it does not take into account quantum tunneling which is more significant at low temperatures. Transition State Theory states that as long as the energy is greater than the activation energy the reaction will proceed to the products but this does not take into account the possibility of the "rebounding" and thus the reaction not proceeding.

== EXERCISE 2: F + H2 system ==
===By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?===

The F + H2 reaction is exothermic whereas the H + HF reaction is endothermic. This means that the H-F bond strength is greater than the H-H bond strength and is lower in energy.

===Locate the approximate position of the transition state.===
H-H distance = 77pm
H-F distance = 179.5pm 1step size = 0.51

===Report the activation energy for both reactions.===
transition state energy = -433.020

H-H energy = -435.100

H-F energy = -560.700

Activation energy for F + H2 --> HF + H = 2.080

Activation energy for H + HF --> H2 + F = 127.68

You will be able to report a reasonable estimate by performing a mep (with a sufficient number of steps) from a structure neighbouring the transition state, and choosing to plot the appropriate quantity as a function of "time".

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

===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:thl3318

2020-05-14T15:07:20Z

Thl3318: /* EXERCISE 2: F + H2 system */

== EXERCISE 1: H + H2 system ==
===On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?===

On a potential energy surface diagram, the transition state can be mathematically defined as a saddle point (which is also an unstable equilibrium). It can be identified by determining the point where the partial derivatives of the function potential with respect to the internuclear distances are equal to 0.

<math> \frac{\partial V}{\partial r_1} =0</math> and <math> \frac{\partial V}{\partial r_2} =0</math>

However, simply looking at the results for the first derivative does not give any information with regards to whether the point is a local minimum, a local maximum, or a saddle point. Analyzing the 2nd derivative then allows for the determination of whether the point is a saddle point. This can be done using the following equation:

<math> D=\frac{\partial^2 V}{\partial r_1^2} \frac{\partial^2 V}{\partial r_2^2} - (\frac{\partial^2 V}{\partial r_1\partial r_2})^2 </math>

If D<0, the point is therefore a saddle point and also the transition state.

===Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.===
My best estimate for the transition state position is at time=20fs when the A-B distance is equal to the B-C distance. The A-B and B-C distance at this point is determined to be 90.8pm. This point in the "Internuclear Distance vs Time" plot can be rationed as the transition state position

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

The mep and dynamics trajectory differ in many different ways. The first difference is the time it takes for the reaction to take place (in this case, for the distance to follow to the valley floor of the potential graph). In the mep trajectory, it takes around 30 steps (15 fs) for the reaction to proceed whereas in the dynamics trajectory this number is around 140 fs. The second difference is that the internuclear distance between the H2 molecule and the lone H atom increases logarithmically in the mep calculation but increases linearly in the dynamics calculation.

===Complete the table below by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?===
{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 ||-414.280 KJ.mol-1||Yes|| Plot shows trajectory going from reactant, going over the minimum energy path via the transition state and finally to form the product.
||[[File:Surface Plot 1 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -4.1 ||-420.077 KJ.mol-1||No||Plot shows the trajectory going up the potential slope towards the transition state. However it does not have enough momentum to go over the curve and thus the trajectory returns to its original position. The reaction does not proceed and the reactants remain.||[[File:Surface Plot 2 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -5.1 ||-413.997 KJ.mol-1||Yes||Plot shows that increasing the momentum of the H-H bond in the reactants results in more vibrational energy hence a more distinct wave pattern appears in the trajectory. Similar to the first plot, the trajectory shows the reactant forming the product by going over the minimum energy path via the transition state.||[[File:Surface Plot 3 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.1 ||-357.277 KJ.mol-1||Yes|| ||[[File:Surface Plot 4 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.6 ||-349.477 KJ.mol-1||Yes|| ||[[File:Surface Plot 5 thl3318.png|thumb|300px|none|]]
|}

From the table above, I can conclude that whether or not a trajectory results in a reaction depends on the momentum p2. Changing the momentum p1 does not affect the result of the trajectory but if momentum p2 is not great enough the result will be unreactive.

[[File:Y2C11.png|thumb|600px|none|Trajectory that re-crosses the TS region]]

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

Experimental value will be greater than the reaction rate the transition state theory predicts because it does not take into account quantum tunneling which is more significant at low temperatures. Transition State Theory states that as long as the energy is greater than the activation energy the reaction will proceed to the products but this does not take into account the possibility of the "rebounding" and thus the reaction not proceeding.

== EXERCISE 2: F + H2 system ==
===By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?===

The F + H2 reaction is exothermic whereas the H + HF reaction is endothermic. This means that the H-F bond strength is greater than the H-H bond strength and is lower in energy.

===Locate the approximate position of the transition state.===
H-H distance = 77pm
H-F distance = 179.5pm 1step size = 0.51

===Report the activation energy for both reactions.===
transition state energy = -433.020

H-H energy = -435.100

H-F energy = -560.700

Activation energy for F + H2 --> HF + H = 2.080

Activation energy for H + HF --> H2 + F = 127.68

You will be able to report a reasonable estimate by performing a mep (with a sufficient number of steps) from a structure neighbouring the transition state, and choosing to plot the appropriate quantity as a function of "time".

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

===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:thl3318

2020-05-14T15:03:30Z

Thl3318: /* EXERCISE 2: F + H2 system */

== EXERCISE 1: H + H2 system ==
===On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?===

On a potential energy surface diagram, the transition state can be mathematically defined as a saddle point (which is also an unstable equilibrium). It can be identified by determining the point where the partial derivatives of the function potential with respect to the internuclear distances are equal to 0.

<math> \frac{\partial V}{\partial r_1} =0</math> and <math> \frac{\partial V}{\partial r_2} =0</math>

However, simply looking at the results for the first derivative does not give any information with regards to whether the point is a local minimum, a local maximum, or a saddle point. Analyzing the 2nd derivative then allows for the determination of whether the point is a saddle point. This can be done using the following equation:

<math> D=\frac{\partial^2 V}{\partial r_1^2} \frac{\partial^2 V}{\partial r_2^2} - (\frac{\partial^2 V}{\partial r_1\partial r_2})^2 </math>

If D<0, the point is therefore a saddle point and also the transition state.

===Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.===
My best estimate for the transition state position is at time=20fs when the A-B distance is equal to the B-C distance. The A-B and B-C distance at this point is determined to be 90.8pm. This point in the "Internuclear Distance vs Time" plot can be rationed as the transition state position

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

The mep and dynamics trajectory differ in many different ways. The first difference is the time it takes for the reaction to take place (in this case, for the distance to follow to the valley floor of the potential graph). In the mep trajectory, it takes around 30 steps (15 fs) for the reaction to proceed whereas in the dynamics trajectory this number is around 140 fs. The second difference is that the internuclear distance between the H2 molecule and the lone H atom increases logarithmically in the mep calculation but increases linearly in the dynamics calculation.

===Complete the table below by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?===
{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 ||-414.280 KJ.mol-1||Yes|| Plot shows trajectory going from reactant, going over the minimum energy path via the transition state and finally to form the product.
||[[File:Surface Plot 1 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -4.1 ||-420.077 KJ.mol-1||No||Plot shows the trajectory going up the potential slope towards the transition state. However it does not have enough momentum to go over the curve and thus the trajectory returns to its original position. The reaction does not proceed and the reactants remain.||[[File:Surface Plot 2 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -5.1 ||-413.997 KJ.mol-1||Yes||Plot shows that increasing the momentum of the H-H bond in the reactants results in more vibrational energy hence a more distinct wave pattern appears in the trajectory. Similar to the first plot, the trajectory shows the reactant forming the product by going over the minimum energy path via the transition state.||[[File:Surface Plot 3 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.1 ||-357.277 KJ.mol-1||Yes|| ||[[File:Surface Plot 4 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.6 ||-349.477 KJ.mol-1||Yes|| ||[[File:Surface Plot 5 thl3318.png|thumb|300px|none|]]
|}

From the table above, I can conclude that whether or not a trajectory results in a reaction depends on the momentum p2. Changing the momentum p1 does not affect the result of the trajectory but if momentum p2 is not great enough the result will be unreactive.

[[File:Y2C11.png|thumb|600px|none|Trajectory that re-crosses the TS region]]

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

Experimental value will be greater than the reaction rate the transition state theory predicts because it does not take into account quantum tunneling which is more significant at low temperatures. Transition State Theory states that as long as the energy is greater than the activation energy the reaction will proceed to the products but this does not take into account the possibility of the "rebounding" and thus the reaction not proceeding.

== EXERCISE 2: F + H2 system ==
{{fontcolor1|blue|By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?}}

The F + H2 reaction is exothermic whereas the H + HF reaction is endothermic. This means that the H-F bond strength is greater than the H-H bond strength and is lower in energy.

{{fontcolor1|blue|Locate the approximate position of the transition state.}}
H-H 77pm
H-F 179.5pm 1step size = 0.51

transition state energy = -433.020
H-H energy = -435.100
H-F energy = -560.700

Activation energy for F + H2 --> HF + H = 2.080
Activation energy for H + HF --> H2 + F = 127.68

{{fontcolor1|blue|Report the activation energy for both reactions.}}

You will be able to report a reasonable estimate by performing a mep (with a sufficient number of steps) from a structure neighbouring the transition state, and choosing to plot the appropriate quantity as a function of "time".

MRD:thl3318

2020-05-14T15:01:17Z

Thl3318: /* EXERCISE 2: F + H2 system */

== EXERCISE 1: H + H2 system ==
===On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?===

On a potential energy surface diagram, the transition state can be mathematically defined as a saddle point (which is also an unstable equilibrium). It can be identified by determining the point where the partial derivatives of the function potential with respect to the internuclear distances are equal to 0.

<math> \frac{\partial V}{\partial r_1} =0</math> and <math> \frac{\partial V}{\partial r_2} =0</math>

However, simply looking at the results for the first derivative does not give any information with regards to whether the point is a local minimum, a local maximum, or a saddle point. Analyzing the 2nd derivative then allows for the determination of whether the point is a saddle point. This can be done using the following equation:

<math> D=\frac{\partial^2 V}{\partial r_1^2} \frac{\partial^2 V}{\partial r_2^2} - (\frac{\partial^2 V}{\partial r_1\partial r_2})^2 </math>

If D<0, the point is therefore a saddle point and also the transition state.

===Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.===
My best estimate for the transition state position is at time=20fs when the A-B distance is equal to the B-C distance. The A-B and B-C distance at this point is determined to be 90.8pm. This point in the "Internuclear Distance vs Time" plot can be rationed as the transition state position

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

The mep and dynamics trajectory differ in many different ways. The first difference is the time it takes for the reaction to take place (in this case, for the distance to follow to the valley floor of the potential graph). In the mep trajectory, it takes around 30 steps (15 fs) for the reaction to proceed whereas in the dynamics trajectory this number is around 140 fs. The second difference is that the internuclear distance between the H2 molecule and the lone H atom increases logarithmically in the mep calculation but increases linearly in the dynamics calculation.

===Complete the table below by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?===
{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 ||-414.280 KJ.mol-1||Yes|| Plot shows trajectory going from reactant, going over the minimum energy path via the transition state and finally to form the product.
||[[File:Surface Plot 1 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -4.1 ||-420.077 KJ.mol-1||No||Plot shows the trajectory going up the potential slope towards the transition state. However it does not have enough momentum to go over the curve and thus the trajectory returns to its original position. The reaction does not proceed and the reactants remain.||[[File:Surface Plot 2 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -5.1 ||-413.997 KJ.mol-1||Yes||Plot shows that increasing the momentum of the H-H bond in the reactants results in more vibrational energy hence a more distinct wave pattern appears in the trajectory. Similar to the first plot, the trajectory shows the reactant forming the product by going over the minimum energy path via the transition state.||[[File:Surface Plot 3 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.1 ||-357.277 KJ.mol-1||Yes|| ||[[File:Surface Plot 4 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.6 ||-349.477 KJ.mol-1||Yes|| ||[[File:Surface Plot 5 thl3318.png|thumb|300px|none|]]
|}

From the table above, I can conclude that whether or not a trajectory results in a reaction depends on the momentum p2. Changing the momentum p1 does not affect the result of the trajectory but if momentum p2 is not great enough the result will be unreactive.

[[File:Y2C11.png|thumb|600px|none|Trajectory that re-crosses the TS region]]

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

Experimental value will be greater than the reaction rate the transition state theory predicts because it does not take into account quantum tunneling which is more significant at low temperatures. Transition State Theory states that as long as the energy is greater than the activation energy the reaction will proceed to the products but this does not take into account the possibility of the "rebounding" and thus the reaction not proceeding.

== EXERCISE 2: F + H2 system ==
{{fontcolor1|blue|By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?}}

The F + H2 reaction is exothermic whereas the H + HF reaction is endothermic. This means that the H-F bond strength is greater than the H-H bond strength and is lower in energy.

{{fontcolor1|blue|Locate the approximate position of the transition state.}}
H-H 77pm
H-F 179.5pm 1step size = 0.51

transition state energy = -433.020
H-H energy = -435.100
H-F energy = -560.700

Activation energy for F + H2 --> HF + H = 2.080
Activation energy for H + HF --> H2 + F = 127.68

xx

{{fontcolor1|blue|Report the activation energy for both reactions.}}

You will be able to report a reasonable estimate by performing a mep (with a sufficient number of steps) from a structure neighbouring the transition state, and choosing to plot the appropriate quantity as a function of "time".

MRD:thl3318

2020-05-14T14:54:24Z

Thl3318: /* EXERCISE 2: F + H2 system */

== EXERCISE 1: H + H2 system ==
===On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?===

On a potential energy surface diagram, the transition state can be mathematically defined as a saddle point (which is also an unstable equilibrium). It can be identified by determining the point where the partial derivatives of the function potential with respect to the internuclear distances are equal to 0.

<math> \frac{\partial V}{\partial r_1} =0</math> and <math> \frac{\partial V}{\partial r_2} =0</math>

However, simply looking at the results for the first derivative does not give any information with regards to whether the point is a local minimum, a local maximum, or a saddle point. Analyzing the 2nd derivative then allows for the determination of whether the point is a saddle point. This can be done using the following equation:

<math> D=\frac{\partial^2 V}{\partial r_1^2} \frac{\partial^2 V}{\partial r_2^2} - (\frac{\partial^2 V}{\partial r_1\partial r_2})^2 </math>

If D<0, the point is therefore a saddle point and also the transition state.

===Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.===
My best estimate for the transition state position is at time=20fs when the A-B distance is equal to the B-C distance. The A-B and B-C distance at this point is determined to be 90.8pm. This point in the "Internuclear Distance vs Time" plot can be rationed as the transition state position

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

The mep and dynamics trajectory differ in many different ways. The first difference is the time it takes for the reaction to take place (in this case, for the distance to follow to the valley floor of the potential graph). In the mep trajectory, it takes around 30 steps (15 fs) for the reaction to proceed whereas in the dynamics trajectory this number is around 140 fs. The second difference is that the internuclear distance between the H2 molecule and the lone H atom increases logarithmically in the mep calculation but increases linearly in the dynamics calculation.

===Complete the table below by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?===
{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 ||-414.280 KJ.mol-1||Yes|| Plot shows trajectory going from reactant, going over the minimum energy path via the transition state and finally to form the product.
||[[File:Surface Plot 1 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -4.1 ||-420.077 KJ.mol-1||No||Plot shows the trajectory going up the potential slope towards the transition state. However it does not have enough momentum to go over the curve and thus the trajectory returns to its original position. The reaction does not proceed and the reactants remain.||[[File:Surface Plot 2 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -5.1 ||-413.997 KJ.mol-1||Yes||Plot shows that increasing the momentum of the H-H bond in the reactants results in more vibrational energy hence a more distinct wave pattern appears in the trajectory. Similar to the first plot, the trajectory shows the reactant forming the product by going over the minimum energy path via the transition state.||[[File:Surface Plot 3 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.1 ||-357.277 KJ.mol-1||Yes|| ||[[File:Surface Plot 4 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.6 ||-349.477 KJ.mol-1||Yes|| ||[[File:Surface Plot 5 thl3318.png|thumb|300px|none|]]
|}

From the table above, I can conclude that whether or not a trajectory results in a reaction depends on the momentum p2. Changing the momentum p1 does not affect the result of the trajectory but if momentum p2 is not great enough the result will be unreactive.

[[File:Y2C11.png|thumb|600px|none|Trajectory that re-crosses the TS region]]

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

Experimental value will be greater than the reaction rate the transition state theory predicts because it does not take into account quantum tunneling which is more significant at low temperatures. Transition State Theory states that as long as the energy is greater than the activation energy the reaction will proceed to the products but this does not take into account the possibility of the "rebounding" and thus the reaction not proceeding.

== EXERCISE 2: F + H2 system ==
{{fontcolor1|blue|By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?}}

The F + H2 reaction is exothermic whereas the H + HF reaction is endothermic. This means that the H-F bond strength is greater than the H-H bond strength and is lower in energy.

{{fontcolor1|blue|Locate the approximate position of the transition state.}}
H-H 77pm
H-F 179.5pm 1step size = 0.51

transition state energy = -433.020
H-H energy = -435.100
H-F energy = -560.700

Activation energy for F + H2 --> HF + H = 2.080
Activation energy for H + HF --> H2 + F = 127.68

{{fontcolor1|blue|Report the activation energy for both reactions.}}

You will be able to report a reasonable estimate by performing a mep (with a sufficient number of steps) from a structure neighbouring the transition state, and choosing to plot the appropriate quantity as a function of "time".

MRD:thl3318

2020-05-14T14:41:53Z

Thl3318: /* EXERCISE 2: F + H2 system */

== EXERCISE 1: H + H2 system ==
===On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?===

On a potential energy surface diagram, the transition state can be mathematically defined as a saddle point (which is also an unstable equilibrium). It can be identified by determining the point where the partial derivatives of the function potential with respect to the internuclear distances are equal to 0.

<math> \frac{\partial V}{\partial r_1} =0</math> and <math> \frac{\partial V}{\partial r_2} =0</math>

However, simply looking at the results for the first derivative does not give any information with regards to whether the point is a local minimum, a local maximum, or a saddle point. Analyzing the 2nd derivative then allows for the determination of whether the point is a saddle point. This can be done using the following equation:

<math> D=\frac{\partial^2 V}{\partial r_1^2} \frac{\partial^2 V}{\partial r_2^2} - (\frac{\partial^2 V}{\partial r_1\partial r_2})^2 </math>

If D<0, the point is therefore a saddle point and also the transition state.

===Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.===
My best estimate for the transition state position is at time=20fs when the A-B distance is equal to the B-C distance. The A-B and B-C distance at this point is determined to be 90.8pm. This point in the "Internuclear Distance vs Time" plot can be rationed as the transition state position

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

The mep and dynamics trajectory differ in many different ways. The first difference is the time it takes for the reaction to take place (in this case, for the distance to follow to the valley floor of the potential graph). In the mep trajectory, it takes around 30 steps (15 fs) for the reaction to proceed whereas in the dynamics trajectory this number is around 140 fs. The second difference is that the internuclear distance between the H2 molecule and the lone H atom increases logarithmically in the mep calculation but increases linearly in the dynamics calculation.

===Complete the table below by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?===
{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 ||-414.280 KJ.mol-1||Yes|| Plot shows trajectory going from reactant, going over the minimum energy path via the transition state and finally to form the product.
||[[File:Surface Plot 1 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -4.1 ||-420.077 KJ.mol-1||No||Plot shows the trajectory going up the potential slope towards the transition state. However it does not have enough momentum to go over the curve and thus the trajectory returns to its original position. The reaction does not proceed and the reactants remain.||[[File:Surface Plot 2 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -5.1 ||-413.997 KJ.mol-1||Yes||Plot shows that increasing the momentum of the H-H bond in the reactants results in more vibrational energy hence a more distinct wave pattern appears in the trajectory. Similar to the first plot, the trajectory shows the reactant forming the product by going over the minimum energy path via the transition state.||[[File:Surface Plot 3 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.1 ||-357.277 KJ.mol-1||Yes|| ||[[File:Surface Plot 4 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.6 ||-349.477 KJ.mol-1||Yes|| ||[[File:Surface Plot 5 thl3318.png|thumb|300px|none|]]
|}

From the table above, I can conclude that whether or not a trajectory results in a reaction depends on the momentum p2. Changing the momentum p1 does not affect the result of the trajectory but if momentum p2 is not great enough the result will be unreactive.

[[File:Y2C11.png|thumb|600px|none|Trajectory that re-crosses the TS region]]

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

Experimental value will be greater than the reaction rate the transition state theory predicts because it does not take into account quantum tunneling which is more significant at low temperatures. Transition State Theory states that as long as the energy is greater than the activation energy the reaction will proceed to the products but this does not take into account the possibility of the "rebounding" and thus the reaction not proceeding.

== EXERCISE 2: F + H2 system ==
{{fontcolor1|blue|By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?}}

The F + H2 reaction is exothermic whereas the H + HF reaction is endothermic. This means that the H-F bond strength is greater than the H-H bond strength and is lower in energy.

{{fontcolor1|blue|Locate the approximate position of the transition state.}}
H-H 77pm
H-F 179.5pm 1step size = 0.51

transition state energy = -433.020
H-H energy = -433.639
H-F energy = -560.700

Activation energy for F + H2 --> HF + H = 0.619
Activation energy for H + HF --> H2 + F = 127.68

{{fontcolor1|blue|Report the activation energy for both reactions.}}

You will be able to report a reasonable estimate by performing a mep (with a sufficient number of steps) from a structure neighbouring the transition state, and choosing to plot the appropriate quantity as a function of "time".

MRD:thl3318

2020-05-14T14:38:09Z

Thl3318: /* EXERCISE 2: F + H2 system */

== EXERCISE 1: H + H2 system ==
===On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?===

On a potential energy surface diagram, the transition state can be mathematically defined as a saddle point (which is also an unstable equilibrium). It can be identified by determining the point where the partial derivatives of the function potential with respect to the internuclear distances are equal to 0.

<math> \frac{\partial V}{\partial r_1} =0</math> and <math> \frac{\partial V}{\partial r_2} =0</math>

However, simply looking at the results for the first derivative does not give any information with regards to whether the point is a local minimum, a local maximum, or a saddle point. Analyzing the 2nd derivative then allows for the determination of whether the point is a saddle point. This can be done using the following equation:

<math> D=\frac{\partial^2 V}{\partial r_1^2} \frac{\partial^2 V}{\partial r_2^2} - (\frac{\partial^2 V}{\partial r_1\partial r_2})^2 </math>

If D<0, the point is therefore a saddle point and also the transition state.

===Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.===
My best estimate for the transition state position is at time=20fs when the A-B distance is equal to the B-C distance. The A-B and B-C distance at this point is determined to be 90.8pm. This point in the "Internuclear Distance vs Time" plot can be rationed as the transition state position

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

The mep and dynamics trajectory differ in many different ways. The first difference is the time it takes for the reaction to take place (in this case, for the distance to follow to the valley floor of the potential graph). In the mep trajectory, it takes around 30 steps (15 fs) for the reaction to proceed whereas in the dynamics trajectory this number is around 140 fs. The second difference is that the internuclear distance between the H2 molecule and the lone H atom increases logarithmically in the mep calculation but increases linearly in the dynamics calculation.

===Complete the table below by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?===
{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 ||-414.280 KJ.mol-1||Yes|| Plot shows trajectory going from reactant, going over the minimum energy path via the transition state and finally to form the product.
||[[File:Surface Plot 1 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -4.1 ||-420.077 KJ.mol-1||No||Plot shows the trajectory going up the potential slope towards the transition state. However it does not have enough momentum to go over the curve and thus the trajectory returns to its original position. The reaction does not proceed and the reactants remain.||[[File:Surface Plot 2 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -5.1 ||-413.997 KJ.mol-1||Yes||Plot shows that increasing the momentum of the H-H bond in the reactants results in more vibrational energy hence a more distinct wave pattern appears in the trajectory. Similar to the first plot, the trajectory shows the reactant forming the product by going over the minimum energy path via the transition state.||[[File:Surface Plot 3 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.1 ||-357.277 KJ.mol-1||Yes|| ||[[File:Surface Plot 4 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.6 ||-349.477 KJ.mol-1||Yes|| ||[[File:Surface Plot 5 thl3318.png|thumb|300px|none|]]
|}

From the table above, I can conclude that whether or not a trajectory results in a reaction depends on the momentum p2. Changing the momentum p1 does not affect the result of the trajectory but if momentum p2 is not great enough the result will be unreactive.

[[File:Y2C11.png|thumb|600px|none|Trajectory that re-crosses the TS region]]

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

Experimental value will be greater than the reaction rate the transition state theory predicts because it does not take into account quantum tunneling which is more significant at low temperatures. Transition State Theory states that as long as the energy is greater than the activation energy the reaction will proceed to the products but this does not take into account the possibility of the "rebounding" and thus the reaction not proceeding.

== EXERCISE 2: F + H2 system ==
{{fontcolor1|blue|By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?}}

The F + H2 reaction is exothermic whereas the H + HF reaction is endothermic. This means that the H-F bond strength is greater than the H-H bond strength and is lower in energy.

{{fontcolor1|blue|Locate the approximate position of the transition state.}}
H-H 77pm
H-F 179.5pm 1step size = 0.51

transition state energy = -433.020
H-H energy = -433.639
H-F energy = -560.700

Activation energy for F + H2 = 0.619
Activation energy for H + HF = 127.68

{{fontcolor1|blue|Report the activation energy for both reactions.}}

You will be able to report a reasonable estimate by performing a mep (with a sufficient number of steps) from a structure neighbouring the transition state, and choosing to plot the appropriate quantity as a function of "time".

MRD:thl3318

2020-05-14T14:17:03Z

Thl3318: /* EXERCISE 2: F + H2 system */

== EXERCISE 1: H + H2 system ==
===On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?===

On a potential energy surface diagram, the transition state can be mathematically defined as a saddle point (which is also an unstable equilibrium). It can be identified by determining the point where the partial derivatives of the function potential with respect to the internuclear distances are equal to 0.

<math> \frac{\partial V}{\partial r_1} =0</math> and <math> \frac{\partial V}{\partial r_2} =0</math>

However, simply looking at the results for the first derivative does not give any information with regards to whether the point is a local minimum, a local maximum, or a saddle point. Analyzing the 2nd derivative then allows for the determination of whether the point is a saddle point. This can be done using the following equation:

<math> D=\frac{\partial^2 V}{\partial r_1^2} \frac{\partial^2 V}{\partial r_2^2} - (\frac{\partial^2 V}{\partial r_1\partial r_2})^2 </math>

If D<0, the point is therefore a saddle point and also the transition state.

===Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.===
My best estimate for the transition state position is at time=20fs when the A-B distance is equal to the B-C distance. The A-B and B-C distance at this point is determined to be 90.8pm. This point in the "Internuclear Distance vs Time" plot can be rationed as the transition state position

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

The mep and dynamics trajectory differ in many different ways. The first difference is the time it takes for the reaction to take place (in this case, for the distance to follow to the valley floor of the potential graph). In the mep trajectory, it takes around 30 steps (15 fs) for the reaction to proceed whereas in the dynamics trajectory this number is around 140 fs. The second difference is that the internuclear distance between the H2 molecule and the lone H atom increases logarithmically in the mep calculation but increases linearly in the dynamics calculation.

===Complete the table below by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?===
{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 ||-414.280 KJ.mol-1||Yes|| Plot shows trajectory going from reactant, going over the minimum energy path via the transition state and finally to form the product.
||[[File:Surface Plot 1 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -4.1 ||-420.077 KJ.mol-1||No||Plot shows the trajectory going up the potential slope towards the transition state. However it does not have enough momentum to go over the curve and thus the trajectory returns to its original position. The reaction does not proceed and the reactants remain.||[[File:Surface Plot 2 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -5.1 ||-413.997 KJ.mol-1||Yes||Plot shows that increasing the momentum of the H-H bond in the reactants results in more vibrational energy hence a more distinct wave pattern appears in the trajectory. Similar to the first plot, the trajectory shows the reactant forming the product by going over the minimum energy path via the transition state.||[[File:Surface Plot 3 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.1 ||-357.277 KJ.mol-1||Yes|| ||[[File:Surface Plot 4 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.6 ||-349.477 KJ.mol-1||Yes|| ||[[File:Surface Plot 5 thl3318.png|thumb|300px|none|]]
|}

From the table above, I can conclude that whether or not a trajectory results in a reaction depends on the momentum p2. Changing the momentum p1 does not affect the result of the trajectory but if momentum p2 is not great enough the result will be unreactive.

[[File:Y2C11.png|thumb|600px|none|Trajectory that re-crosses the TS region]]

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

Experimental value will be greater than the reaction rate the transition state theory predicts because it does not take into account quantum tunneling which is more significant at low temperatures. Transition State Theory states that as long as the energy is greater than the activation energy the reaction will proceed to the products but this does not take into account the possibility of the "rebounding" and thus the reaction not proceeding.

== EXERCISE 2: F + H2 system ==
{{fontcolor1|blue|By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?}}

The F + H2 reaction is exothermic whereas the H + HF reaction is endothermic. This means that the H-F bond strength is greater than the H-H bond strength and is lower in energy.

{{fontcolor1|blue|Locate the approximate position of the transition state.}}
H-H 77pm
H-F 179.5pm

Because the activation energy for one of the reactions is so small, it is not easy to locate the transition state immediately. Use the Hammond postulate to guide your search.

{{fontcolor1|blue|Report the activation energy for both reactions.}}

You will be able to report a reasonable estimate by performing a mep (with a sufficient number of steps) from a structure neighbouring the transition state, and choosing to plot the appropriate quantity as a function of "time".

MRD:thl3318

2020-05-14T06:20:45Z

Thl3318: /* Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? */

== EXERCISE 1: H + H2 system ==
===On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?===

On a potential energy surface diagram, the transition state can be mathematically defined as a saddle point (which is also an unstable equilibrium). It can be identified by determining the point where the partial derivatives of the function potential with respect to the internuclear distances are equal to 0.

<math> \frac{\partial V}{\partial r_1} =0</math> and <math> \frac{\partial V}{\partial r_2} =0</math>

However, simply looking at the results for the first derivative does not give any information with regards to whether the point is a local minimum, a local maximum, or a saddle point. Analyzing the 2nd derivative then allows for the determination of whether the point is a saddle point. This can be done using the following equation:

<math> D=\frac{\partial^2 V}{\partial r_1^2} \frac{\partial^2 V}{\partial r_2^2} - (\frac{\partial^2 V}{\partial r_1\partial r_2})^2 </math>

If D<0, the point is therefore a saddle point and also the transition state.

===Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.===
My best estimate for the transition state position is at time=20fs when the A-B distance is equal to the B-C distance. The A-B and B-C distance at this point is determined to be 90.8pm. This point in the "Internuclear Distance vs Time" plot can be rationed as the transition state position

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

The mep and dynamics trajectory differ in many different ways. The first difference is the time it takes for the reaction to take place (in this case, for the distance to follow to the valley floor of the potential graph). In the mep trajectory, it takes around 30 steps (15 fs) for the reaction to proceed whereas in the dynamics trajectory this number is around 140 fs. The second difference is that the internuclear distance between the H2 molecule and the lone H atom increases logarithmically in the mep calculation but increases linearly in the dynamics calculation.

===Complete the table below by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?===
{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 ||-414.280 KJ.mol-1||Yes|| Plot shows trajectory going from reactant, going over the minimum energy path via the transition state and finally to form the product.
||[[File:Surface Plot 1 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -4.1 ||-420.077 KJ.mol-1||No||Plot shows the trajectory going up the potential slope towards the transition state. However it does not have enough momentum to go over the curve and thus the trajectory returns to its original position. The reaction does not proceed and the reactants remain.||[[File:Surface Plot 2 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -5.1 ||-413.997 KJ.mol-1||Yes||Plot shows that increasing the momentum of the H-H bond in the reactants results in more vibrational energy hence a more distinct wave pattern appears in the trajectory. Similar to the first plot, the trajectory shows the reactant forming the product by going over the minimum energy path via the transition state.||[[File:Surface Plot 3 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.1 ||-357.277 KJ.mol-1||Yes|| ||[[File:Surface Plot 4 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.6 ||-349.477 KJ.mol-1||Yes|| ||[[File:Surface Plot 5 thl3318.png|thumb|300px|none|]]
|}

From the table above, I can conclude that whether or not a trajectory results in a reaction depends on the momentum p2. Changing the momentum p1 does not affect the result of the trajectory but if momentum p2 is not great enough the result will be unreactive.

[[File:Y2C11.png|thumb|600px|none|Trajectory that re-crosses the TS region]]

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

Experimental value will be greater than the reaction rate the transition state theory predicts because it does not take into account quantum tunneling which is more significant at low temperatures. Transition State Theory states that as long as the energy is greater than the activation energy the reaction will proceed to the products but this does not take into account the possibility of the "rebounding" and thus the reaction not proceeding.

== EXERCISE 2: F + H2 system ==
{{fontcolor1|blue|By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?}}

The F + H2 reaction is exothermic whereas the H + HF reaction is endothermic. This means that the H-F bond strength is greater than the H-H bond strength and is lower in energy.

{{fontcolor1|blue|Locate the approximate position of the transition state.}}
H-H 77pm
H-F 105pm

Because the activation energy for one of the reactions is so small, it is not easy to locate the transition state immediately. Use the Hammond postulate to guide your search.

{{fontcolor1|blue|Report the activation energy for both reactions.}}

You will be able to report a reasonable estimate by performing a mep (with a sufficient number of steps) from a structure neighbouring the transition state, and choosing to plot the appropriate quantity as a function of "time".

MRD:thl3318

2020-05-13T10:33:00Z

MRD:thl3318

2020-05-13T10:15:19Z

Thl3318: /* 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? */

MRD:thl3318

2020-05-13T10:14:08Z

MRD:thl3318

2020-05-13T10:07:40Z

Thl3318: /* EXERCISE 1: H + H2 system */

== EXERCISE 1: H + H2 system ==
====On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?====

On a potential energy surface diagram, the transition state can be mathematically defined as a saddle point (which is also an unstable equilibrium). It can be identified by determining the point where the partial derivatives of the function potential with respect to the internuclear distances are equal to 0.

<math> \frac{\partial V}{\partial r_1} =0</math> and <math> \frac{\partial V}{\partial r_2} =0</math>

However, simply looking at the results for the first derivative does not give any information with regards to whether the point is a local minimum, a local maximum, or a saddle point. Analyzing the 2nd derivative then allows for the determination of whether the point is a saddle point. This can be done using the following equation:

<math> D=\frac{\partial^2 V}{\partial r_1^2} \frac{\partial^2 V}{\partial r_2^2} - (\frac{\partial^2 V}{\partial r_1\partial r_2})^2 </math>

If D<0, the point is therefore a saddle point and also the transition state.

{{fontcolor|blue|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.}}
My best estimate for the transition state position is at time=20fs when the A-B distance is equal to the B-C distance. The A-B and B-C distance at this point is determined to be 90.8pm. This point in the "Internuclear Distance vs Time" plot can be rationed as the transition state position

{{fontcolor1|blue|Comment on how the ''mep'' and the trajectory you just calculated differ.}}

The mep and dynamics trajectory differ in many different ways. The first difference is the time it takes for the reaction to take place (in this case, for the distance to follow to the valley floor of the potential graph). In the mep trajectory, it takes around 30 steps (15 fs) for the reaction to proceed whereas in the dynamics trajectory this number is around 140 fs. The second difference is that the internuclear distance between the H2 molecule and the lone H atom increases logarithmically in the mep calculation but increases linearly in the dynamics calculation.

{{fontcolor1|blue|Complete the table below by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?}}
{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 ||-414.280 KJ.mol-1||Yes|| ||[[File:Surface Plot 1 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -4.1 ||-420.077 KJ.mol-1||No|| ||[[File:Surface Plot 2 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -5.1 ||-413.997 KJ.mol-1||Yes|| ||[[File:Surface Plot 3 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.1 ||-357.277 KJ.mol-1||Yes|| ||[[File:Surface Plot 4 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.6 ||-349.477 KJ.mol-1||Yes|| ||[[File:Surface Plot 5 thl3318.png|thumb|300px|none|]]
|}

From the table above, I can conclude that whether or not a trajectory results in a reaction depends on the momentum p2. Changing the momentum p1 does not affect the result of the trajectory but if momentum p2 is not great enough the result will be unreactive.

[[File:Y2C11.png|thumb|600px|none|Trajectory that re-crosses the TS region]]

{{fontcolor|blue|Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}

== EXERCISE 2: F + H2 system ==
{{fontcolor1|blue|By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?}}

The F + H2 reaction is exothermic whereas the H + HF reaction is endothermic. This means that the H-F bond strength is greater than the H-H bond strength and is lower in energy.

{{fontcolor1|blue|Locate the approximate position of the transition state.}}
H-H 77pm
H-F 105pm

Because the activation energy for one of the reactions is so small, it is not easy to locate the transition state immediately. Use the Hammond postulate to guide your search.

{{fontcolor1|blue|Report the activation energy for both reactions.}}

You will be able to report a reasonable estimate by performing a mep (with a sufficient number of steps) from a structure neighbouring the transition state, and choosing to plot the appropriate quantity as a function of "time".

MRD:thl3318

2020-05-13T10:06:43Z

Thl3318: /* EXERCISE 1: H + H2 system */

== EXERCISE 1: H + H2 system ==
===={{fontcolor|blue|On a potential energy surface diagram, how is the transition state mathematically defined?
How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?}}====

On a potential energy surface diagram, the transition state can be mathematically defined as a saddle point (which is also an unstable equilibrium). It can be identified by determining the point where the partial derivatives of the function potential with respect to the internuclear distances are equal to 0.

<math> \frac{\partial V}{\partial r_1} =0</math> and <math> \frac{\partial V}{\partial r_2} =0</math>

However, simply looking at the results for the first derivative does not give any information with regards to whether the point is a local minimum, a local maximum, or a saddle point. Analyzing the 2nd derivative then allows for the determination of whether the point is a saddle point. This can be done using the following equation:

<math> D=\frac{\partial^2 V}{\partial r_1^2} \frac{\partial^2 V}{\partial r_2^2} - (\frac{\partial^2 V}{\partial r_1\partial r_2})^2 </math>

If D<0, the point is therefore a saddle point and also the transition state.

{{fontcolor|blue|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.}}
My best estimate for the transition state position is at time=20fs when the A-B distance is equal to the B-C distance. The A-B and B-C distance at this point is determined to be 90.8pm. This point in the "Internuclear Distance vs Time" plot can be rationed as the transition state position

{{fontcolor1|blue|Comment on how the ''mep'' and the trajectory you just calculated differ.}}

The mep and dynamics trajectory differ in many different ways. The first difference is the time it takes for the reaction to take place (in this case, for the distance to follow to the valley floor of the potential graph). In the mep trajectory, it takes around 30 steps (15 fs) for the reaction to proceed whereas in the dynamics trajectory this number is around 140 fs. The second difference is that the internuclear distance between the H2 molecule and the lone H atom increases logarithmically in the mep calculation but increases linearly in the dynamics calculation.

{{fontcolor1|blue|Complete the table below by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?}}
{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 ||-414.280 KJ.mol-1||Yes|| ||[[File:Surface Plot 1 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -4.1 ||-420.077 KJ.mol-1||No|| ||[[File:Surface Plot 2 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -5.1 ||-413.997 KJ.mol-1||Yes|| ||[[File:Surface Plot 3 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.1 ||-357.277 KJ.mol-1||Yes|| ||[[File:Surface Plot 4 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.6 ||-349.477 KJ.mol-1||Yes|| ||[[File:Surface Plot 5 thl3318.png|thumb|300px|none|]]
|}

From the table above, I can conclude that whether or not a trajectory results in a reaction depends on the momentum p2. Changing the momentum p1 does not affect the result of the trajectory but if momentum p2 is not great enough the result will be unreactive.

[[File:Y2C11.png|thumb|600px|none|Trajectory that re-crosses the TS region]]

{{fontcolor|blue|Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}

== EXERCISE 2: F + H2 system ==
{{fontcolor1|blue|By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?}}

The F + H2 reaction is exothermic whereas the H + HF reaction is endothermic. This means that the H-F bond strength is greater than the H-H bond strength and is lower in energy.

{{fontcolor1|blue|Locate the approximate position of the transition state.}}
H-H 77pm
H-F 105pm

Because the activation energy for one of the reactions is so small, it is not easy to locate the transition state immediately. Use the Hammond postulate to guide your search.

{{fontcolor1|blue|Report the activation energy for both reactions.}}

You will be able to report a reasonable estimate by performing a mep (with a sufficient number of steps) from a structure neighbouring the transition state, and choosing to plot the appropriate quantity as a function of "time".

MRD:thl3318

2020-05-13T09:55:02Z

Thl3318: /* EXERCISE 1: H + H2 system */

== EXERCISE 1: H + H2 system ==
{{fontcolor|blue|On a potential energy surface diagram, how is the transition state mathematically defined?
How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?}}

On a potential energy surface diagram, the transition state can be mathematically defined as a saddle point (which is also an unstable equilibrium). It can be identified by determining the point where the partial derivatives of the function potential with respect to the internuclear distances are equal to 0.

<math> \frac{\partial V}{\partial r_1} =0</math> and <math> \frac{\partial V}{\partial r_2} =0</math>

However, simply looking at the results for the first derivative does not give any information with regards to whether the point is a local minimum, a local maximum, or a saddle point. Analyzing the 2nd derivative then allows for the determination of whether the point is a saddle point. This can be done using the following equation:

<math> D=\frac{\partial^2 V}{\partial r_1^2} \frac{\partial^2 V}{\partial r_2^2} - (\frac{\partial^2 V}{\partial r_1\partial r_2})^2 </math>

If D<0, the point is therefore a saddle point and also the transition state.

{{fontcolor|blue|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.}}
My best estimate for the transition state position is at time=20fs when the A-B distance is equal to the B-C distance. The A-B and B-C distance at this point is determined to be 90.8pm. This point in the "Internuclear Distance vs Time" plot can be rationed as the transition state position

{{fontcolor1|blue|Comment on how the ''mep'' and the trajectory you just calculated differ.}}

The mep and dynamics trajectory differ in many different ways. The first difference is the time it takes for the reaction to take place (in this case, for the distance to follow to the valley floor of the potential graph). In the mep trajectory, it takes around 30 steps (15 fs) for the reaction to proceed whereas in the dynamics trajectory this number is around 140 fs. The second difference is that the internuclear distance between the H2 molecule and the lone H atom increases logarithmically in the mep calculation but increases linearly in the dynamics calculation.

{{fontcolor1|blue|Complete the table below by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?}}
{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 ||-414.280 KJ.mol-1||Yes|| ||[[File:Surface Plot 1 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -4.1 ||-420.077 KJ.mol-1||No|| ||[[File:Surface Plot 2 thl3318.png|thumb|300px|none|]]
|-
| -3.1 || -5.1 ||-413.997 KJ.mol-1||Yes|| ||[[File:Surface Plot 3 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.1 ||-357.277 KJ.mol-1||Yes|| ||[[File:Surface Plot 4 thl3318.png|thumb|300px|none|]]
|-
| -5.1 || -10.6 ||-349.477 KJ.mol-1||Yes|| ||[[File:Surface Plot 5 thl3318.png|thumb|300px|none|]]
|}

From the table above, I can conclude that whether or not a trajectory results in a reaction depends on the momentum p2. Changing the momentum p1 does not affect the result of the trajectory but if momentum p2 is not great enough the result will be unreactive.

[[File:Y2C11.png|thumb|600px|none|Trajectory that re-crosses the TS region]]

{{fontcolor|blue|Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}

== EXERCISE 2: F + H2 system ==
{{fontcolor1|blue|By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?}}

The F + H2 reaction is exothermic whereas the H + HF reaction is endothermic. This means that the H-F bond strength is greater than the H-H bond strength and is lower in energy.

{{fontcolor1|blue|Locate the approximate position of the transition state.}}
H-H 77pm
H-F 105pm

Because the activation energy for one of the reactions is so small, it is not easy to locate the transition state immediately. Use the Hammond postulate to guide your search.

{{fontcolor1|blue|Report the activation energy for both reactions.}}

You will be able to report a reasonable estimate by performing a mep (with a sufficient number of steps) from a structure neighbouring the transition state, and choosing to plot the appropriate quantity as a function of "time".

MRD:thl3318

2020-05-13T09:53:52Z

Thl3318: /* EXERCISE 1: H + H2 system */

== EXERCISE 1: H + H2 system ==
{{fontcolor|blue|On a potential energy surface diagram, how is the transition state mathematically defined?
How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?}}

On a potential energy surface diagram, the transition state can be mathematically defined as a saddle point (which is also an unstable equilibrium). It can be identified by determining the point where the partial derivatives of the function potential with respect to the internuclear distances are equal to 0.

<math> \frac{\partial V}{\partial r_1} =0</math> and <math> \frac{\partial V}{\partial r_2} =0</math>

However, simply looking at the results for the first derivative does not give any information with regards to whether the point is a local minimum, a local maximum, or a saddle point. Analyzing the 2nd derivative then allows for the determination of whether the point is a saddle point. This can be done using the following equation:

<math> D=\frac{\partial^2 V}{\partial r_1^2} \frac{\partial^2 V}{\partial r_2^2} - (\frac{\partial^2 V}{\partial r_1\partial r_2})^2 </math>

If D<0, the point is therefore a saddle point and also the transition state.

{{fontcolor|blue|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.}}
My best estimate for the transition state position is at time=20fs when the A-B distance is equal to the B-C distance. The A-B and B-C distance at this point is determined to be 90.8pm. This point in the "Internuclear Distance vs Time" plot can be rationed as the transition state position

{{fontcolor1|blue|Comment on how the ''mep'' and the trajectory you just calculated differ.}}

The mep and dynamics trajectory differ in many different ways. The first difference is the time it takes for the reaction to take place (in this case, for the distance to follow to the valley floor of the potential graph). In the mep trajectory, it takes around 30 steps (15 fs) for the reaction to proceed whereas in the dynamics trajectory this number is around 140 fs. The second difference is that the internuclear distance between the H2 molecule and the lone H atom increases logarithmically in the mep calculation but increases linearly in the dynamics calculation.

{{fontcolor1|blue|Complete the table below by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?}}
{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 ||-414.280 KJ.mol-1||Yes|| ||
|-
| -3.1 || -4.1 ||-420.077 KJ.mol-1||No|| ||
|-
| -3.1 || -5.1 ||-413.997 KJ.mol-1||Yes|| ||
|-
| -5.1 || -10.1 ||-357.277 KJ.mol-1||Yes|| ||
|-
| -5.1 || -10.6 ||-349.477 KJ.mol-1||Yes||
[[File:Surface Plot 5 thl3318.png|thumb|600px|none|Trajectory that re-crosses the TS region]] ||
|}

From the table above, I can conclude that whether or not a trajectory results in a reaction depends on the momentum p2. Changing the momentum p1 does not affect the result of the trajectory but if momentum p2 is not great enough the result will be unreactive.

[[File:Y2C11.png|thumb|600px|none|Trajectory that re-crosses the TS region]]

{{fontcolor|blue|Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}

== EXERCISE 2: F + H2 system ==
{{fontcolor1|blue|By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?}}

The F + H2 reaction is exothermic whereas the H + HF reaction is endothermic. This means that the H-F bond strength is greater than the H-H bond strength and is lower in energy.

{{fontcolor1|blue|Locate the approximate position of the transition state.}}
H-H 77pm
H-F 105pm

Because the activation energy for one of the reactions is so small, it is not easy to locate the transition state immediately. Use the Hammond postulate to guide your search.

{{fontcolor1|blue|Report the activation energy for both reactions.}}

You will be able to report a reasonable estimate by performing a mep (with a sufficient number of steps) from a structure neighbouring the transition state, and choosing to plot the appropriate quantity as a function of "time".

File:Surface Plot 5 thl3318.png

2020-05-13T09:53:04Z

Thl3318: Molecular Reaction Dynamics Question 4 Plot 5 (thl3318)

Molecular Reaction Dynamics Question 4 Plot 5 (thl3318)

File:Surface Plot 4 thl3318.png

2020-05-13T09:52:38Z

Thl3318: Molecular Reaction Dynamics Question 4 Plot 4 (thl3318)

Molecular Reaction Dynamics Question 4 Plot 4 (thl3318)