ChemWiki - User contributions [en]

MRD:aja4117

2019-05-17T16:13:47Z

Aja4117: /* EXERCISE 2: F - H - H system */

= [[Molecular reaction dynamics|'''Molecular reaction dynamics''']] =

== 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?''''' ===
The transition state is a saddle point which is defined as the point which is both a minimum along one vector and a maximum along an orthogonal vector. Along the reaction path, the transition state is a maximum and along the second vector, which is perpendicular to our initial vector it is a minimum. This can be visualised in the diagram below.

For a co linear system, as given for the H + H2 system there are two orthogonal degrees of freedom with the coordinates given as follows:

'''Z = r2 + r1'''
 

<math>\frac {\partial V}{\partial Z} = 0 </math>, <math>\frac {\partial^2 V}{\partial Z^2}>0 </math>

 

'''S = r2 - r1'''

 

<math>\frac {\partial V}{\partial S} = 0 </math>, <math>\frac {\partial^2 V}{\partial S^2}<0 </math>

 

Note that these are the orthogonal vectors.

The saddle point has the configuration in which r1 = r2 =r0 and this will be the point at which these two lines intersect on the PES. This is the transition state. S is the tangent to the reaction pathway at the maximum whilst Z bisects the reaction pathway at the same point, but will be seen as a minimum.

[[File:Aja4117Question 1.png]]

=== ''Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'' ===
The best estimate was AB = 0.9077425 and BC = 0.9077425. There are only two lines when looking at internuclear distances vs time as seen below. This indicates that two lines are at the same distance and one is covering the other. 

To locate the transition state, is to begin the reaction trajectories close to the transition state. Due to the transition state being a global maxima, the trajectories will either form the reactants or products in order to find the lowest energy configuration. Whether the product or reactants are formed is directly influenced by the trajectories. If the trajectories are closer to the reactants then the reactants will form and if they are closer to the products then the products will form.

[[File:Aja4117INT for HH.png]]

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

===== Trajectories from r1 = rts+δ, r2 = rts =====
Along the mep, because the atoms are moving infinitely slowly and the momenta and the velocities always reset to zero, there is no oscillation and hence no vibration. This is seen along the mep where there is no wavy lines. In contrast, when looking at the dynamics, there is oscillation as post collision there is transfer of kinetic energy that causes the molecule to vibrate. 

When considering the inter nuclear distance vs time, A-C and A-B distances increase rapidly over the first 250 steps after which the rate of increase of the distance begins to decrease. As the number of steps increases the rate of increase of distance begins to plateau as the gradient flattens. This can be rationalised when looking at the animation where post collision, the molecules initially rapidly move away from one another but the speed at which they move away from each other falls off with time. When considering the dynamic situation, A-C and A-B distances increase at a constant gradient over the course of the plot. The reasoning is that in the dynamic situation, the velocities and momenta are not reset to zero at each step.

[[File:Aja4117(dynamics type)Mep vs dynamics distance v time 1.png]]
[[File:Aja4117Distance vs time 1.png]]

For the same reason as stated as above, the momenta and velocities are reset to zero at each point this means that the momentum vs time plot is a straight line for the mep . However, when running the dynamic situation we see an initial sharp increase in the momentum followed by a constant oscillation of momentum that is due to the momentum being retained within the vibrational modes. For A-B, there is a sharp increase in the momentum which plateaus at the same instance as B-C begins to undergo constant oscillation. This is because A-B refers to the momentum of the system A-B. Once both reach constant momentum the combined system now has a set defined level of momentum with no way to lose it.

[[File:Aja4117(dynamics type)Mep vs dynamics momenta vs time1.png]]
[[File:Aja4117dynamics mom vs time1.png]]

===== Trajectories from r1 = rts, r2 = rts +δ =====
The result of switching the distances is that

<nowiki> </nowiki>Take note of the final values of the positions '''r1'''(t) '''r2'''(t) and '''p1'''(t) '''p2'''(t) for your trajectory for large enough t.

== Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed. ==

=== Reactive and unreactive trajectories ===
{| class="wikitable" border="1"
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics
!Plots of trajectory
|-
| -1.25 || -2.5 ||-99.018||Yes||Reaction proceeds with minimal vibration||[[File:-1.25, -2.5.png]]
|
|-
| -1.5 || -2.0 ||-100.456||No||P2 does not have enough momentum to reach P1||[[File:-1.5, -2.0.png]]
|
|-
| -1.5 || -2.5 ||-98.956||Yes||Transfer of vibrational energy||[[File:-1.5, -2.5.png]]
|
|-
| -2.5 || -5.0 || -84.956
||No||Product forms but contains too much vibrational energy so dissociates||[[File:-2.5, -5.0.png ]]
|
|-
| -2.5 || -5.2 || -83.416
||Yes||Reaction does occur but recrosses the transition state barrier||[[File:-2.5,5.2.png]]
|
|}
The fact that the last experiment recrosses the transition state barrier is evidence of the breakdown of Transition State Theory, which is presented below. this states that once products are formed, the system cannot travel back through the transition state to the reactants.

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

There are two basic assumptions for Transition State Theory, the two most basic are the separation of the electronic and nuclear motions which is the same as the Born-Oppenheimer approximation from quantum mechanics. The second basic assumption is that the reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution.

There are however, additional assumptions and they are as follows:
# Molecular systems that have crossed the transition state in the direction of the products cannot turn around and reform reactants.
# In the transition state, motion along the reaction coordinate may be separated from the other motions and be treated classically as a translation.
# Even in the absence of an equilibrium between the reactant and product molecules, the transition states that are become products are distributed among their states according to the Maxwell-Boltzmann laws.

''Given that Transition State theory neglects quantum mechanical contributions such as influence on and between between electronic and nuclear motions and the idea that molecules passing through the transition state through to the products cannot pass back through to the reactants, this would suggest that there would be deviation between experimental and calculated values using Transition State theory.''

== EXERCISE 2: F - H - H 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?'' ====

Looking at the PES of F + H2 this indicated the reaction is exothermic because there is a decrease in the energy going from the reactants to products. This can be seen along the reaction path for this reaction where there is a dip in energy on the path to the formation of the products. This makes intuitive sense as the driving force of the reaction will be the formation of a strong F-H (565 Kj/mol) bond, in place of a weaker H-H (464 Kj/mol) bond. Additional evidence that the reaction is exothermic is that when a mep calculation was run with RAB = 1.812 and RBC = 0.74 with zero momentum for each case, the reaction proceeded and the products formed with no additional input of energy.

For the H + HF reaction, the reaction is surmised to be endothermic due to as there is no decrease in energy on forming the products. This is in line with intuition considering the strength of the H-F bond. The very high electronegativity of fluorine would make it so that significant energy input would be required to cause this reaction to proceed. This can be seen in the pes below for the H + Hf system where there is a difference in height corresponding to the energies of the reactants and products.

==== ''Locate the approximate position of the transition state.'' ====
{| class="wikitable"
!Reaction
!rAB
!rBC
!Activation Energy/ Kj/mol
|-
|F + H2
|1.8108757
|0.7448759
|0.2673
|-
|H + HF
|0.7448756
|1.8108759
|30.251
|}
The positions of the transition states were located through experimentation of the initial AB and BC distances, until the forces acting along AB and BC were zero. The reasoning behind this comes from the notion that

<math>F = - \frac {\partial V}{\partial r} </math>

where F is the force

and seeing as we are at a saddle point, <math> \frac {\partial V}{\partial r} = 0 </math> and hence

<math> F = 0 </math>

 
[[File:Aja4117Surface Plot f + h2.png]]
[[File:Aja4117Surface Plot h + HF.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.'' ====

The set of reaction conditions that were used were:

Atom A: F

Atom B: H

Atom C: H

Distance AB: 2.25

Distance BC: 0.74

Momentum AB = -0.8

Momentum BC = 0.1

These conditions yielded a set of successful reactions, and post reaction and upon formation of F-H, there is transfer of energy in the form of kinetic energy to both the Hydrogen that is travelling away from the reaction complex and also kinetic energy in the bond vibration of F-H. This can be visualised in the Momenta vs Time graph below. Monitoring the emission spectrum of the product and observing electromagnetic radiation in the IR region would allow identification of the product.

[[File:Aja4117Surface Plot FH2 momenta vs time.png]]

For rBC = 0.74, with a momentum pAB = -0.5, and several values of pBC in the range -3 to 3 were explored. For pBC = -3, the reaction did proceed and there was significant transfer of kinetic energy to the product F-H. What is interesting to note is that the F-H bond forms twice. On the first formation, the bond breaks due to what is assumed to be, too high a vibrational energy. This leads to the reformation of the H-H bond, upon which there is still significant vibrational energy and combined with the electronegativity of the F atom causes the H-H bond to break and the F-H bond to form for the second time. The F-H bond is now stable, although still containing significant vibrational energy. For pBC = 3, the same phenomena occurs as for pBC = -3. 

==== ''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.'' ====
A transition state for an attractive potential energy surface occurs early in the reaction coordinate and involves high translational kinetic energy (evidenced by the random motion for the reactants) and results in a vibrationally excited product. This translational energy runs up the side of the valley and falls from side to side, showing that there is significant vibrational motion. We can therefore conclude that reactions with attractive potential energy surfaces proceed more efficiently if the energy is in relative translational motion.

In contrast, reactions with repulsive potential surfaces experience a late transition state and are expected to proceed with greater efficiency if the excess energy is present as vibrations. The energy is in the vibration of the reactant molecule and the motion causes the trajectory to move from side to side up the valley as it approaches the transition state. This energy is enough to tip the reactant molecules through the transition state and into the products, it is therefore apparent that the vibrational energy on the repulsive potential energy surface is key to the reactivity and formation of the products.

MRD:aja4117

2019-05-16T15:20:33Z

Aja4117: /* EXERCISE 2: F - H - H system */

= [[Molecular reaction dynamics|'''Molecular reaction dynamics''']] =

== 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?''''' ===
The transition state is a saddle point which is defined as the point which is both a minimum along one vector and a maximum along an orthogonal vector. Along the reaction path, the transition state is a maximum and along the second vector, which is perpendicular to our initial vector it is a minimum. This can be visualised in the diagram below.

For a co linear system, as given for the H + H2 system there are two orthogonal degrees of freedom with the coordinates given as follows:

'''Z = r2 + r1'''
 

<math>\frac {\partial V}{\partial Z} = 0 </math>, <math>\frac {\partial^2 V}{\partial Z^2}>0 </math>

 

'''S = r2 - r1'''

 

<math>\frac {\partial V}{\partial S} = 0 </math>, <math>\frac {\partial^2 V}{\partial S^2}<0 </math>

 

Note that these are the orthogonal vectors.

The saddle point has the configuration in which r1 = r2 =r0 and this will be the point at which these two lines intersect on the PES. This is the transition state. S is the tangent to the reaction pathway at the maximum whilst Z bisects the reaction pathway at the same point, but will be seen as a minimum.

[[File:Aja4117Question 1.png]]

=== ''Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'' ===
The best estimate was AB = 0.9077425 and BC = 0.9077425. There are only two lines when looking at internuclear distances vs time as seen below. This indicates that two lines are at the same distance and one is covering the other. 

To locate the transition state, is to begin the reaction trajectories close to the transition state. Due to the transition state being a global maxima, the trajectories will either form the reactants or products in order to find the lowest energy configuration. Whether the product or reactants are formed is directly influenced by the trajectories. If the trajectories are closer to the reactants then the reactants will form and if they are closer to the products then the products will form.

[[File:Aja4117INT for HH.png]]

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

===== Trajectories from r1 = rts+δ, r2 = rts =====
Along the mep, because the atoms are moving infinitely slowly and the momenta and the velocities always reset to zero, there is no oscillation and hence no vibration. This is seen along the mep where there is no wavy lines. In contrast, when looking at the dynamics, there is oscillation as post collision there is transfer of kinetic energy that causes the molecule to vibrate. 

When considering the inter nuclear distance vs time, A-C and A-B distances increase rapidly over the first 250 steps after which the rate of increase of the distance begins to decrease. As the number of steps increases the rate of increase of distance begins to plateau as the gradient flattens. This can be rationalised when looking at the animation where post collision, the molecules initially rapidly move away from one another but the speed at which they move away from each other falls off with time. When considering the dynamic situation, A-C and A-B distances increase at a constant gradient over the course of the plot. The reasoning is that in the dynamic situation, the velocities and momenta are not reset to zero at each step.

[[File:Aja4117(dynamics type)Mep vs dynamics distance v time 1.png]]
[[File:Aja4117Distance vs time 1.png]]

For the same reason as stated as above, the momenta and velocities are reset to zero at each point this means that the momentum vs time plot is a straight line for the mep . However, when running the dynamic situation we see an initial sharp increase in the momentum followed by a constant oscillation of momentum that is due to the momentum being retained within the vibrational modes. For A-B, there is a sharp increase in the momentum which plateaus at the same instance as B-C begins to undergo constant oscillation. This is because A-B refers to the momentum of the system A-B. Once both reach constant momentum the combined system now has a set defined level of momentum with no way to lose it.

[[File:Aja4117(dynamics type)Mep vs dynamics momenta vs time1.png]]
[[File:Aja4117dynamics mom vs time1.png]]

===== Trajectories from r1 = rts, r2 = rts +δ =====
The result of switching the distances is that

<nowiki> </nowiki>Take note of the final values of the positions '''r1'''(t) '''r2'''(t) and '''p1'''(t) '''p2'''(t) for your trajectory for large enough t.

== Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed. ==

=== Reactive and unreactive trajectories ===
{| class="wikitable" border="1"
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics
!Plots of trajectory
|-
| -1.25 || -2.5 ||-99.018||Yes||Reaction proceeds with minimal vibration||[[File:-1.25, -2.5.png]]
|
|-
| -1.5 || -2.0 ||-100.456||No||P2 does not have enough momentum to reach P1||[[File:-1.5, -2.0.png]]
|
|-
| -1.5 || -2.5 ||-98.956||Yes||Transfer of vibrational energy||[[File:-1.5, -2.5.png]]
|
|-
| -2.5 || -5.0 || -84.956
||No||Product forms but contains too much vibrational energy so dissociates||[[File:-2.5, -5.0.png ]]
|
|-
| -2.5 || -5.2 || -83.416
||Yes||Reaction does occur but recrosses the transition state barrier||[[File:-2.5,5.2.png]]
|
|}
The fact that the last experiment recrosses the transition state barrier is evidence of the breakdown of Transition State Theory, which is presented below. this states that once products are formed, the system cannot travel back through the transition state to the reactants.

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

There are two basic assumptions for Transition State Theory, the two most basic are the separation of the electronic and nuclear motions which is the same as the Born-Oppenheimer approximation from quantum mechanics. The second basic assumption is that the reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution.

There are however, additional assumptions and they are as follows:
# Molecular systems that have crossed the transition state in the direction of the products cannot turn around and reform reactants.
# In the transition state, motion along the reaction coordinate may be separated from the other motions and be treated classically as a translation.
# Even in the absence of an equilibrium between the reactant and product molecules, the transition states that are become products are distributed among their states according to the Maxwell-Boltzmann laws.

''Given that Transition State theory neglects quantum mechanical contributions such as influence on and between between electronic and nuclear motions and the idea that molecules passing through the transition state through to the products cannot pass back through to the reactants, this would suggest that there would be deviation between experimental and calculated values using Transition State theory.''

== EXERCISE 2: F - H - H 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?'' ====

Looking at the PES of F + H2 this indicated the reaction is exothermic because there is a decrease in the energy going from the reactants to products. This can be seen along the reaction path for this reaction where there is a dip in energy on the path to the formation of the products. This makes intuitive sense as the driving force of the reaction will be the formation of a strong F-H bond, in place of a weaker H-H bond. Additional evidence that the reaction is exothermic is that when a mep calculation was run with RAB = 1.812 and RBC = 0.74 with zero momentum for each case, the reaction proceeded and the products formed with no additional input of energy.

For the H + HF reaction, the reaction is surmised to be endothermic due to as there is no decrease in energy on forming the products. This is in line with intuition considering the strength of the H-F bond. The very high electronegativity of fluorine would make it so that significant energy input would be required to cause this reaction to proceed. This can be seen in the pes below for the H + Hf system where there is a difference in height corresponding to the energies of the reactants and products.

==== ''Locate the approximate position of the transition state.'' ====
{| class="wikitable"
!Reaction
!rAB
!rBC
!Activation Energy/ Kj/mol
|-
|F + H2
|1.8108757
|0.7448759
|0.2673
|-
|H + HF
|0.7448756
|1.8108759
|30.251
|}
The positions of the transition states were located through experimentation of the initial AB and BC distances, until the forces acting along AB and BC were zero. The reasoning behind this comes from the notion that

<math>F = - \frac {\partial V}{\partial r} </math>

where F is the force

and seeing as we are at a saddle point, <math> \frac {\partial V}{\partial r} = 0 </math> and hence

<math> F = 0 </math>

 
[[File:Aja4117Surface Plot f + h2.png]]
[[File:Aja4117Surface Plot h + HF.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.'' ====

The set of reaction conditions that were used were:

Atom A: F

Atom B: H

Atom C: H

Distance AB: 2.25

Distance BC: 0.74

Momentum AB = -0.8

Momentum BC = 0.1

These conditions yielded a set of successful reactions, and post reaction and upon formation of F-H, there is transfer of energy in the form of kinetic energy to both the Hydrogen that is travelling away from the reaction complex and also kinetic energy in the bond vibration of F-H. This can be visualised in the Momenta vs Time graph below. Monitoring the emission spectrum of the product and observing electromagnetic radiation in the IR region would allow identification of the product.

[[File:Aja4117Surface Plot FH2 momenta vs time.png]]

For rBC = 0.74, with a momentum pAB = -0.5, and several values of pBC in the range -3 to 3 were explored. For pBC = -3, the reaction did proceed and there was significant transfer of kinetic energy to the product F-H. What is interesting to note is that the F-H bond forms twice. On the first formation, the bond breaks due to what is assumed to be, too high a vibrational energy. This leads to the reformation of the H-H bond, upon which there is still significant vibrational energy and combined with the electronegativity of the F atom causes the H-H bond to break and the F-H bond to form for the second time. The F-H bond is now stable, although still containing significant vibrational energy. For pBC = 3, the same phenomena occurs as for pBC = -3. 

==== ''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.'' ====
A transition state for an attractive potential energy surface occurs early in the reaction coordinate and involves high translational kinetic energy (evidenced by the random motion for the reactants) and results in a vibrationally excited product. This translational energy runs up the side of the valley and falls from side to side, showing that there is significant vibrational motion. We can therefore conclude that reactions with attractive potential energy surfaces proceed more efficiently if the energy is in relative translational motion.

In contrast, reactions with repulsive potential surfaces experience a late transition state and are expected to proceed with greater efficiency if the excess energy is present as vibrations. The energy is in the vibration of the reactant molecule and the motion causes the trajectory to move from side to side up the valley as it approaches the transition state. This energy is enough to tip the reactant molecules through the transition state and into the products, it is therefore apparent that the vibrational energy on the repulsive potential energy surface is key to the reactivity and formation of the products.

MRD:aja4117

2019-05-16T15:15:03Z

Aja4117: /* What do you observe? */

= [[Molecular reaction dynamics|'''Molecular reaction dynamics''']] =

== 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?''''' ===
The transition state is a saddle point which is defined as the point which is both a minimum along one vector and a maximum along an orthogonal vector. Along the reaction path, the transition state is a maximum and along the second vector, which is perpendicular to our initial vector it is a minimum. This can be visualised in the diagram below.

For a co linear system, as given for the H + H2 system there are two orthogonal degrees of freedom with the coordinates given as follows:

'''Z = r2 + r1'''
 

<math>\frac {\partial V}{\partial Z} = 0 </math>, <math>\frac {\partial^2 V}{\partial Z^2}>0 </math>

 

'''S = r2 - r1'''

 

<math>\frac {\partial V}{\partial S} = 0 </math>, <math>\frac {\partial^2 V}{\partial S^2}<0 </math>

 

Note that these are the orthogonal vectors.

The saddle point has the configuration in which r1 = r2 =r0 and this will be the point at which these two lines intersect on the PES. This is the transition state. S is the tangent to the reaction pathway at the maximum whilst Z bisects the reaction pathway at the same point, but will be seen as a minimum.

[[File:Aja4117Question 1.png]]

=== ''Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'' ===
The best estimate was AB = 0.9077425 and BC = 0.9077425. There are only two lines when looking at internuclear distances vs time as seen below. This indicates that two lines are at the same distance and one is covering the other. 

To locate the transition state, is to begin the reaction trajectories close to the transition state. Due to the transition state being a global maxima, the trajectories will either form the reactants or products in order to find the lowest energy configuration. Whether the product or reactants are formed is directly influenced by the trajectories. If the trajectories are closer to the reactants then the reactants will form and if they are closer to the products then the products will form.

[[File:Aja4117INT for HH.png]]

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

===== Trajectories from r1 = rts+δ, r2 = rts =====
Along the mep, because the atoms are moving infinitely slowly and the momenta and the velocities always reset to zero, there is no oscillation and hence no vibration. This is seen along the mep where there is no wavy lines. In contrast, when looking at the dynamics, there is oscillation as post collision there is transfer of kinetic energy that causes the molecule to vibrate. 

When considering the inter nuclear distance vs time, A-C and A-B distances increase rapidly over the first 250 steps after which the rate of increase of the distance begins to decrease. As the number of steps increases the rate of increase of distance begins to plateau as the gradient flattens. This can be rationalised when looking at the animation where post collision, the molecules initially rapidly move away from one another but the speed at which they move away from each other falls off with time. When considering the dynamic situation, A-C and A-B distances increase at a constant gradient over the course of the plot. The reasoning is that in the dynamic situation, the velocities and momenta are not reset to zero at each step.

[[File:Aja4117(dynamics type)Mep vs dynamics distance v time 1.png]]
[[File:Aja4117Distance vs time 1.png]]

For the same reason as stated as above, the momenta and velocities are reset to zero at each point this means that the momentum vs time plot is a straight line for the mep . However, when running the dynamic situation we see an initial sharp increase in the momentum followed by a constant oscillation of momentum that is due to the momentum being retained within the vibrational modes. For A-B, there is a sharp increase in the momentum which plateaus at the same instance as B-C begins to undergo constant oscillation. This is because A-B refers to the momentum of the system A-B. Once both reach constant momentum the combined system now has a set defined level of momentum with no way to lose it.

[[File:Aja4117(dynamics type)Mep vs dynamics momenta vs time1.png]]
[[File:Aja4117dynamics mom vs time1.png]]

===== Trajectories from r1 = rts, r2 = rts +δ =====
The result of switching the distances is that

<nowiki> </nowiki>Take note of the final values of the positions '''r1'''(t) '''r2'''(t) and '''p1'''(t) '''p2'''(t) for your trajectory for large enough t.

== Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed. ==

=== Reactive and unreactive trajectories ===
{| class="wikitable" border="1"
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics
!Plots of trajectory
|-
| -1.25 || -2.5 ||-99.018||Yes||Reaction proceeds with minimal vibration||[[File:-1.25, -2.5.png]]
|
|-
| -1.5 || -2.0 ||-100.456||No||P2 does not have enough momentum to reach P1||[[File:-1.5, -2.0.png]]
|
|-
| -1.5 || -2.5 ||-98.956||Yes||Transfer of vibrational energy||[[File:-1.5, -2.5.png]]
|
|-
| -2.5 || -5.0 || -84.956
||No||Product forms but contains too much vibrational energy so dissociates||[[File:-2.5, -5.0.png ]]
|
|-
| -2.5 || -5.2 || -83.416
||Yes||Reaction does occur but recrosses the transition state barrier||[[File:-2.5,5.2.png]]
|
|}
The fact that the last experiment recrosses the transition state barrier is evidence of the breakdown of Transition State Theory, which is presented below. this states that once products are formed, the system cannot travel back through the transition state to the reactants.

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

There are two basic assumptions for Transition State Theory, the two most basic are the separation of the electronic and nuclear motions which is the same as the Born-Oppenheimer approximation from quantum mechanics. The second basic assumption is that the reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution.

There are however, additional assumptions and they are as follows:
# Molecular systems that have crossed the transition state in the direction of the products cannot turn around and reform reactants.
# In the transition state, motion along the reaction coordinate may be separated from the other motions and be treated classically as a translation.
# Even in the absence of an equilibrium between the reactant and product molecules, the transition states that are become products are distributed among their states according to the Maxwell-Boltzmann laws.

''Given that Transition State theory neglects quantum mechanical contributions such as influence on and between between electronic and nuclear motions and the idea that molecules passing through the transition state through to the products cannot pass back through to the reactants, this would suggest that there would be deviation between experimental and calculated values using Transition State theory.''

== EXERCISE 2: F - H - H 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?'' ====
Looking at the PES of F + H2 this indicated the reaction is exothermic because there is a decrease in the energy going from the reactants to products. This can be seen along the reaction path for this reaction where there is a dip in energy on the path to the formation of the products. This makes intuitive sense as the driving force of the reaction will be the formation of a strong F-H bond, in place of a weaker H-H bond. Additional evidence that the reaction is exothermic is that when a mep calculation was run with RAB = 1.812 and RBC = 0.74 with zero momentum for each case, the reaction proceeded and the products formed with no additional input of energy.

For the H + HF reaction, '''the reaction is surmised to be endothermic due to as there is no decrease in energy on forming the products. This is in line with intuition considering the strength of the H-F bond. The very high electronegativity of fluorine would make it so that significant energy would be required to cause this reaction to proceed. '''

==== ''Locate the approximate position of the transition state.'' ====
{| class="wikitable"
!Reaction
!rAB
!rBC
!Activation Energy/ Kj/mol
|-
|F + H2
|1.8108757
|0.7448759
|0.2673
|-
|H + HF
|0.7448756
|1.8108759
|30.251
|}
The positions of the transition states were located through experimentation of the initial AB and BC distances, until the forces acting along AB and BC were zero. The reasoning behind this comes from the notion that

<math>F = - \frac {\partial V}{\partial r} </math>

where F is the force

and seeing as we are at a saddle point, <math> \frac {\partial V}{\partial r} = 0 </math> and hence

<math> F = 0 </math>

 
[[File:Aja4117Surface Plot f + h2.png]]
[[File:Aja4117Surface Plot h + HF.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.'' ====

The set of reaction conditions that were used were:

Atom A: F

Atom B: H

Atom C: H

Distance AB: 2.25

Distance BC: 0.74

Momentum AB = -0.8

Momentum BC = 0.1

These conditions yielded a set of successful reactions, and post reaction and upon formation of F-H, there is transfer of energy in the form of kinetic energy to both the Hydrogen that is travelling away from the reaction complex and also kinetic energy in the bond vibration of F-H. This can be visualised in the Momenta vs Time graph below. Monitoring the emission spectrum of the product and observing electromagnetic radiation in the IR region would allow identification of the product.

[[File:Aja4117Surface Plot FH2 momenta vs time.png]]

For rBC = 0.74, with a momentum pAB = -0.5, and several values of pBC in the range -3 to 3 were explored. For pBC = -3, the reaction did proceed and there was significant transfer of kinetic energy to the product F-H. What is interesting to note is that the F-H bond forms twice. On the first formation, the bond breaks due to what is assumed to be, too high a vibrational energy. This leads to the reformation of the H-H bond, upon which there is still significant vibrational energy and combined with the electronegativity of the F atom causes the H-H bond to break and the F-H bond to form for the second time. The F-H bond is now stable, although still containing significant vibrational energy. For pBC = 3, the same phenomena occurs as for pBC = -3. 

==== ''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.'' ====
A transition state for an attractive potential energy surface occurs early in the reaction coordinate and involves high translational kinetic energy (evidenced by the random motion for the reactants) and results in a vibrationally excited product. This translational energy runs up the side of the valley and falls from side to side, showing that there is significant vibrational motion. We can therefore conclude that reactions with attractive potential energy surfaces proceed more efficiently if the energy is in relative translational motion.

In contrast, reactions with repulsive potential surfaces experience a late transition state and are expected to proceed with greater efficiency if the excess energy is present as vibrations. The energy is in the vibration of the reactant molecule and the motion causes the trajectory to move from side to side up the valley as it approaches the transition state. This energy is enough to tip the reactant molecules through the transition state and into the products, it is therefore apparent that the vibrational energy on the repulsive potential energy surface is key to the reactivity and formation of the products.

MRD:aja4117

2019-05-16T15:14:45Z

Aja4117: /* What do you observe? STILL TO DO */

= [[Molecular reaction dynamics|'''Molecular reaction dynamics''']] =

== 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?''''' ===
The transition state is a saddle point which is defined as the point which is both a minimum along one vector and a maximum along an orthogonal vector. Along the reaction path, the transition state is a maximum and along the second vector, which is perpendicular to our initial vector it is a minimum. This can be visualised in the diagram below.

For a co linear system, as given for the H + H2 system there are two orthogonal degrees of freedom with the coordinates given as follows:

'''Z = r2 + r1'''
 

<math>\frac {\partial V}{\partial Z} = 0 </math>, <math>\frac {\partial^2 V}{\partial Z^2}>0 </math>

 

'''S = r2 - r1'''

 

<math>\frac {\partial V}{\partial S} = 0 </math>, <math>\frac {\partial^2 V}{\partial S^2}<0 </math>

 

Note that these are the orthogonal vectors.

The saddle point has the configuration in which r1 = r2 =r0 and this will be the point at which these two lines intersect on the PES. This is the transition state. S is the tangent to the reaction pathway at the maximum whilst Z bisects the reaction pathway at the same point, but will be seen as a minimum.

[[File:Aja4117Question 1.png]]

=== ''Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'' ===
The best estimate was AB = 0.9077425 and BC = 0.9077425. There are only two lines when looking at internuclear distances vs time as seen below. This indicates that two lines are at the same distance and one is covering the other. 

To locate the transition state, is to begin the reaction trajectories close to the transition state. Due to the transition state being a global maxima, the trajectories will either form the reactants or products in order to find the lowest energy configuration. Whether the product or reactants are formed is directly influenced by the trajectories. If the trajectories are closer to the reactants then the reactants will form and if they are closer to the products then the products will form.

[[File:Aja4117INT for HH.png]]

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

===== Trajectories from r1 = rts+δ, r2 = rts =====
Along the mep, because the atoms are moving infinitely slowly and the momenta and the velocities always reset to zero, there is no oscillation and hence no vibration. This is seen along the mep where there is no wavy lines. In contrast, when looking at the dynamics, there is oscillation as post collision there is transfer of kinetic energy that causes the molecule to vibrate. 

When considering the inter nuclear distance vs time, A-C and A-B distances increase rapidly over the first 250 steps after which the rate of increase of the distance begins to decrease. As the number of steps increases the rate of increase of distance begins to plateau as the gradient flattens. This can be rationalised when looking at the animation where post collision, the molecules initially rapidly move away from one another but the speed at which they move away from each other falls off with time. When considering the dynamic situation, A-C and A-B distances increase at a constant gradient over the course of the plot. The reasoning is that in the dynamic situation, the velocities and momenta are not reset to zero at each step.

[[File:Aja4117(dynamics type)Mep vs dynamics distance v time 1.png]]
[[File:Aja4117Distance vs time 1.png]]

For the same reason as stated as above, the momenta and velocities are reset to zero at each point this means that the momentum vs time plot is a straight line for the mep . However, when running the dynamic situation we see an initial sharp increase in the momentum followed by a constant oscillation of momentum that is due to the momentum being retained within the vibrational modes. For A-B, there is a sharp increase in the momentum which plateaus at the same instance as B-C begins to undergo constant oscillation. This is because A-B refers to the momentum of the system A-B. Once both reach constant momentum the combined system now has a set defined level of momentum with no way to lose it.

[[File:Aja4117(dynamics type)Mep vs dynamics momenta vs time1.png]]
[[File:Aja4117dynamics mom vs time1.png]]

===== Trajectories from r1 = rts, r2 = rts +δ =====
The result of switching the distances is that

<nowiki> </nowiki>Take note of the final values of the positions '''r1'''(t) '''r2'''(t) and '''p1'''(t) '''p2'''(t) for your trajectory for large enough t.

== Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed. ==

== What do you observe? ==

=== Reactive and unreactive trajectories ===
{| class="wikitable" border="1"
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics
!Plots of trajectory
|-
| -1.25 || -2.5 ||-99.018||Yes||Reaction proceeds with minimal vibration||[[File:-1.25, -2.5.png]]
|
|-
| -1.5 || -2.0 ||-100.456||No||P2 does not have enough momentum to reach P1||[[File:-1.5, -2.0.png]]
|
|-
| -1.5 || -2.5 ||-98.956||Yes||Transfer of vibrational energy||[[File:-1.5, -2.5.png]]
|
|-
| -2.5 || -5.0 || -84.956
||No||Product forms but contains too much vibrational energy so dissociates||[[File:-2.5, -5.0.png ]]
|
|-
| -2.5 || -5.2 || -83.416
||Yes||Reaction does occur but recrosses the transition state barrier||[[File:-2.5,5.2.png]]
|
|}
The fact that the last experiment recrosses the transition state barrier is evidence of the breakdown of Transition State Theory, which is presented below. this states that once products are formed, the system cannot travel back through the transition state to the reactants.

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

There are two basic assumptions for Transition State Theory, the two most basic are the separation of the electronic and nuclear motions which is the same as the Born-Oppenheimer approximation from quantum mechanics. The second basic assumption is that the reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution.

There are however, additional assumptions and they are as follows:
# Molecular systems that have crossed the transition state in the direction of the products cannot turn around and reform reactants.
# In the transition state, motion along the reaction coordinate may be separated from the other motions and be treated classically as a translation.
# Even in the absence of an equilibrium between the reactant and product molecules, the transition states that are become products are distributed among their states according to the Maxwell-Boltzmann laws.

''Given that Transition State theory neglects quantum mechanical contributions such as influence on and between between electronic and nuclear motions and the idea that molecules passing through the transition state through to the products cannot pass back through to the reactants, this would suggest that there would be deviation between experimental and calculated values using Transition State theory.''

== EXERCISE 2: F - H - H 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?'' ====
Looking at the PES of F + H2 this indicated the reaction is exothermic because there is a decrease in the energy going from the reactants to products. This can be seen along the reaction path for this reaction where there is a dip in energy on the path to the formation of the products. This makes intuitive sense as the driving force of the reaction will be the formation of a strong F-H bond, in place of a weaker H-H bond. Additional evidence that the reaction is exothermic is that when a mep calculation was run with RAB = 1.812 and RBC = 0.74 with zero momentum for each case, the reaction proceeded and the products formed with no additional input of energy.

For the H + HF reaction, '''the reaction is surmised to be endothermic due to as there is no decrease in energy on forming the products. This is in line with intuition considering the strength of the H-F bond. The very high electronegativity of fluorine would make it so that significant energy would be required to cause this reaction to proceed. '''

==== ''Locate the approximate position of the transition state.'' ====
{| class="wikitable"
!Reaction
!rAB
!rBC
!Activation Energy/ Kj/mol
|-
|F + H2
|1.8108757
|0.7448759
|0.2673
|-
|H + HF
|0.7448756
|1.8108759
|30.251
|}
The positions of the transition states were located through experimentation of the initial AB and BC distances, until the forces acting along AB and BC were zero. The reasoning behind this comes from the notion that

<math>F = - \frac {\partial V}{\partial r} </math>

where F is the force

and seeing as we are at a saddle point, <math> \frac {\partial V}{\partial r} = 0 </math> and hence

<math> F = 0 </math>

 
[[File:Aja4117Surface Plot f + h2.png]]
[[File:Aja4117Surface Plot h + HF.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.'' ====

The set of reaction conditions that were used were:

Atom A: F

Atom B: H

Atom C: H

Distance AB: 2.25

Distance BC: 0.74

Momentum AB = -0.8

Momentum BC = 0.1

These conditions yielded a set of successful reactions, and post reaction and upon formation of F-H, there is transfer of energy in the form of kinetic energy to both the Hydrogen that is travelling away from the reaction complex and also kinetic energy in the bond vibration of F-H. This can be visualised in the Momenta vs Time graph below. Monitoring the emission spectrum of the product and observing electromagnetic radiation in the IR region would allow identification of the product.

[[File:Aja4117Surface Plot FH2 momenta vs time.png]]

For rBC = 0.74, with a momentum pAB = -0.5, and several values of pBC in the range -3 to 3 were explored. For pBC = -3, the reaction did proceed and there was significant transfer of kinetic energy to the product F-H. What is interesting to note is that the F-H bond forms twice. On the first formation, the bond breaks due to what is assumed to be, too high a vibrational energy. This leads to the reformation of the H-H bond, upon which there is still significant vibrational energy and combined with the electronegativity of the F atom causes the H-H bond to break and the F-H bond to form for the second time. The F-H bond is now stable, although still containing significant vibrational energy. For pBC = 3, the same phenomena occurs as for pBC = -3. 

==== ''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.'' ====
A transition state for an attractive potential energy surface occurs early in the reaction coordinate and involves high translational kinetic energy (evidenced by the random motion for the reactants) and results in a vibrationally excited product. This translational energy runs up the side of the valley and falls from side to side, showing that there is significant vibrational motion. We can therefore conclude that reactions with attractive potential energy surfaces proceed more efficiently if the energy is in relative translational motion.

In contrast, reactions with repulsive potential surfaces experience a late transition state and are expected to proceed with greater efficiency if the excess energy is present as vibrations. The energy is in the vibration of the reactant molecule and the motion causes the trajectory to move from side to side up the valley as it approaches the transition state. This energy is enough to tip the reactant molecules through the transition state and into the products, it is therefore apparent that the vibrational energy on the repulsive potential energy surface is key to the reactivity and formation of the products.

MRD:aja4117

2019-05-16T15:11:22Z

Aja4117: /* Locate the approximate position of the transition state. */

= [[Molecular reaction dynamics|'''Molecular reaction dynamics''']] =

== 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?''''' ===
The transition state is a saddle point which is defined as the point which is both a minimum along one vector and a maximum along an orthogonal vector. Along the reaction path, the transition state is a maximum and along the second vector, which is perpendicular to our initial vector it is a minimum. This can be visualised in the diagram below.

For a co linear system, as given for the H + H2 system there are two orthogonal degrees of freedom with the coordinates given as follows:

'''Z = r2 + r1'''
 

<math>\frac {\partial V}{\partial Z} = 0 </math>, <math>\frac {\partial^2 V}{\partial Z^2}>0 </math>

 

'''S = r2 - r1'''

 

<math>\frac {\partial V}{\partial S} = 0 </math>, <math>\frac {\partial^2 V}{\partial S^2}<0 </math>

 

Note that these are the orthogonal vectors.

The saddle point has the configuration in which r1 = r2 =r0 and this will be the point at which these two lines intersect on the PES. This is the transition state. S is the tangent to the reaction pathway at the maximum whilst Z bisects the reaction pathway at the same point, but will be seen as a minimum.

[[File:Aja4117Question 1.png]]

=== ''Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'' ===
The best estimate was AB = 0.9077425 and BC = 0.9077425. There are only two lines when looking at internuclear distances vs time as seen below. This indicates that two lines are at the same distance and one is covering the other. 

To locate the transition state, is to begin the reaction trajectories close to the transition state. Due to the transition state being a global maxima, the trajectories will either form the reactants or products in order to find the lowest energy configuration. Whether the product or reactants are formed is directly influenced by the trajectories. If the trajectories are closer to the reactants then the reactants will form and if they are closer to the products then the products will form.

[[File:Aja4117INT for HH.png]]

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

===== Trajectories from r1 = rts+δ, r2 = rts =====
Along the mep, because the atoms are moving infinitely slowly and the momenta and the velocities always reset to zero, there is no oscillation and hence no vibration. This is seen along the mep where there is no wavy lines. In contrast, when looking at the dynamics, there is oscillation as post collision there is transfer of kinetic energy that causes the molecule to vibrate. 

When considering the inter nuclear distance vs time, A-C and A-B distances increase rapidly over the first 250 steps after which the rate of increase of the distance begins to decrease. As the number of steps increases the rate of increase of distance begins to plateau as the gradient flattens. This can be rationalised when looking at the animation where post collision, the molecules initially rapidly move away from one another but the speed at which they move away from each other falls off with time. When considering the dynamic situation, A-C and A-B distances increase at a constant gradient over the course of the plot. The reasoning is that in the dynamic situation, the velocities and momenta are not reset to zero at each step.

[[File:Aja4117(dynamics type)Mep vs dynamics distance v time 1.png]]
[[File:Aja4117Distance vs time 1.png]]

For the same reason as stated as above, the momenta and velocities are reset to zero at each point this means that the momentum vs time plot is a straight line for the mep . However, when running the dynamic situation we see an initial sharp increase in the momentum followed by a constant oscillation of momentum that is due to the momentum being retained within the vibrational modes. For A-B, there is a sharp increase in the momentum which plateaus at the same instance as B-C begins to undergo constant oscillation. This is because A-B refers to the momentum of the system A-B. Once both reach constant momentum the combined system now has a set defined level of momentum with no way to lose it.

[[File:Aja4117(dynamics type)Mep vs dynamics momenta vs time1.png]]
[[File:Aja4117dynamics mom vs time1.png]]

===== Trajectories from r1 = rts, r2 = rts +δ =====
The result of switching the distances is that

<nowiki> </nowiki>Take note of the final values of the positions '''r1'''(t) '''r2'''(t) and '''p1'''(t) '''p2'''(t) for your trajectory for large enough t.

== Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed. ==

== What do you observe? STILL TO DO ==

=== Reactive and unreactive trajectories ===
{| class="wikitable" border="1"
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics
!Plots of trajectory
|-
| -1.25 || -2.5 ||-99.018||Yes||Reaction proceeds with minimal vibration||[[File:-1.25, -2.5.png]]
|
|-
| -1.5 || -2.0 ||-100.456||No||P2 does not have enough momentum to reach P1||[[File:-1.5, -2.0.png]]
|
|-
| -1.5 || -2.5 ||-98.956||Yes||Transfer of vibrational energy||[[File:-1.5, -2.5.png]]
|
|-
| -2.5 || -5.0 || -84.956
||No||Product forms but contains too much vibrational energy so dissociates||[[File:-2.5, -5.0.png ]]
|
|-
| -2.5 || -5.2 || -83.416
||Yes||Reaction does occur but recrosses the transition state barrier||[[File:-2.5,5.2.png]]
|
|}
The fact that the last experiment recrosses the transition state barrier is evidence of the breakdown of Transition State Theory, which is presented below. this states that once products are formed, the system cannot travel back through the transition state to the reactants.

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

There are two basic assumptions for Transition State Theory, the two most basic are the separation of the electronic and nuclear motions which is the same as the Born-Oppenheimer approximation from quantum mechanics. The second basic assumption is that the reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution.

There are however, additional assumptions and they are as follows:
# Molecular systems that have crossed the transition state in the direction of the products cannot turn around and reform reactants.
# In the transition state, motion along the reaction coordinate may be separated from the other motions and be treated classically as a translation.
# Even in the absence of an equilibrium between the reactant and product molecules, the transition states that are become products are distributed among their states according to the Maxwell-Boltzmann laws.

''Given that Transition State theory neglects quantum mechanical contributions such as influence on and between between electronic and nuclear motions and the idea that molecules passing through the transition state through to the products cannot pass back through to the reactants, this would suggest that there would be deviation between experimental and calculated values using Transition State theory.''

== EXERCISE 2: F - H - H 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?'' ====
Looking at the PES of F + H2 this indicated the reaction is exothermic because there is a decrease in the energy going from the reactants to products. This can be seen along the reaction path for this reaction where there is a dip in energy on the path to the formation of the products. This makes intuitive sense as the driving force of the reaction will be the formation of a strong F-H bond, in place of a weaker H-H bond. Additional evidence that the reaction is exothermic is that when a mep calculation was run with RAB = 1.812 and RBC = 0.74 with zero momentum for each case, the reaction proceeded and the products formed with no additional input of energy.

For the H + HF reaction, '''the reaction is surmised to be endothermic due to as there is no decrease in energy on forming the products. This is in line with intuition considering the strength of the H-F bond. The very high electronegativity of fluorine would make it so that significant energy would be required to cause this reaction to proceed. '''

==== ''Locate the approximate position of the transition state.'' ====
{| class="wikitable"
!Reaction
!rAB
!rBC
!Activation Energy/ Kj/mol
|-
|F + H2
|1.8108757
|0.7448759
|0.2673
|-
|H + HF
|0.7448756
|1.8108759
|30.251
|}
The positions of the transition states were located through experimentation of the initial AB and BC distances, until the forces acting along AB and BC were zero. The reasoning behind this comes from the notion that

<math>F = - \frac {\partial V}{\partial r} </math>

where F is the force

and seeing as we are at a saddle point, <math> \frac {\partial V}{\partial r} = 0 </math> and hence

<math> F = 0 </math>

 
[[File:Aja4117Surface Plot f + h2.png]]
[[File:Aja4117Surface Plot h + HF.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.'' ====

The set of reaction conditions that were used were:

Atom A: F

Atom B: H

Atom C: H

Distance AB: 2.25

Distance BC: 0.74

Momentum AB = -0.8

Momentum BC = 0.1

These conditions yielded a set of successful reactions, and post reaction and upon formation of F-H, there is transfer of energy in the form of kinetic energy to both the Hydrogen that is travelling away from the reaction complex and also kinetic energy in the bond vibration of F-H. This can be visualised in the Momenta vs Time graph below. Monitoring the emission spectrum of the product and observing electromagnetic radiation in the IR region would allow identification of the product.

[[File:Aja4117Surface Plot FH2 momenta vs time.png]]

For rBC = 0.74, with a momentum pAB = -0.5, and several values of pBC in the range -3 to 3 were explored. For pBC = -3, the reaction did proceed and there was significant transfer of kinetic energy to the product F-H. What is interesting to note is that the F-H bond forms twice. On the first formation, the bond breaks due to what is assumed to be, too high a vibrational energy. This leads to the reformation of the H-H bond, upon which there is still significant vibrational energy and combined with the electronegativity of the F atom causes the H-H bond to break and the F-H bond to form for the second time. The F-H bond is now stable, although still containing significant vibrational energy. For pBC = 3, the same phenomena occurs as for pBC = -3. 

==== ''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.'' ====
A transition state for an attractive potential energy surface occurs early in the reaction coordinate and involves high translational kinetic energy (evidenced by the random motion for the reactants) and results in a vibrationally excited product. This translational energy runs up the side of the valley and falls from side to side, showing that there is significant vibrational motion. We can therefore conclude that reactions with attractive potential energy surfaces proceed more efficiently if the energy is in relative translational motion.

In contrast, reactions with repulsive potential surfaces experience a late transition state and are expected to proceed with greater efficiency if the excess energy is present as vibrations. The energy is in the vibration of the reactant molecule and the motion causes the trajectory to move from side to side up the valley as it approaches the transition state. This energy is enough to tip the reactant molecules through the transition state and into the products, it is therefore apparent that the vibrational energy on the repulsive potential energy surface is key to the reactivity and formation of the products.

MRD:aja4117

2019-05-16T15:11:06Z

Aja4117: /* Locate the approximate position of the transition state. */

= [[Molecular reaction dynamics|'''Molecular reaction dynamics''']] =

== 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?''''' ===
The transition state is a saddle point which is defined as the point which is both a minimum along one vector and a maximum along an orthogonal vector. Along the reaction path, the transition state is a maximum and along the second vector, which is perpendicular to our initial vector it is a minimum. This can be visualised in the diagram below.

For a co linear system, as given for the H + H2 system there are two orthogonal degrees of freedom with the coordinates given as follows:

'''Z = r2 + r1'''
 

<math>\frac {\partial V}{\partial Z} = 0 </math>, <math>\frac {\partial^2 V}{\partial Z^2}>0 </math>

 

'''S = r2 - r1'''

 

<math>\frac {\partial V}{\partial S} = 0 </math>, <math>\frac {\partial^2 V}{\partial S^2}<0 </math>

 

Note that these are the orthogonal vectors.

The saddle point has the configuration in which r1 = r2 =r0 and this will be the point at which these two lines intersect on the PES. This is the transition state. S is the tangent to the reaction pathway at the maximum whilst Z bisects the reaction pathway at the same point, but will be seen as a minimum.

[[File:Aja4117Question 1.png]]

=== ''Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'' ===
The best estimate was AB = 0.9077425 and BC = 0.9077425. There are only two lines when looking at internuclear distances vs time as seen below. This indicates that two lines are at the same distance and one is covering the other. 

To locate the transition state, is to begin the reaction trajectories close to the transition state. Due to the transition state being a global maxima, the trajectories will either form the reactants or products in order to find the lowest energy configuration. Whether the product or reactants are formed is directly influenced by the trajectories. If the trajectories are closer to the reactants then the reactants will form and if they are closer to the products then the products will form.

[[File:Aja4117INT for HH.png]]

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

===== Trajectories from r1 = rts+δ, r2 = rts =====
Along the mep, because the atoms are moving infinitely slowly and the momenta and the velocities always reset to zero, there is no oscillation and hence no vibration. This is seen along the mep where there is no wavy lines. In contrast, when looking at the dynamics, there is oscillation as post collision there is transfer of kinetic energy that causes the molecule to vibrate. 

When considering the inter nuclear distance vs time, A-C and A-B distances increase rapidly over the first 250 steps after which the rate of increase of the distance begins to decrease. As the number of steps increases the rate of increase of distance begins to plateau as the gradient flattens. This can be rationalised when looking at the animation where post collision, the molecules initially rapidly move away from one another but the speed at which they move away from each other falls off with time. When considering the dynamic situation, A-C and A-B distances increase at a constant gradient over the course of the plot. The reasoning is that in the dynamic situation, the velocities and momenta are not reset to zero at each step.

[[File:Aja4117(dynamics type)Mep vs dynamics distance v time 1.png]]
[[File:Aja4117Distance vs time 1.png]]

For the same reason as stated as above, the momenta and velocities are reset to zero at each point this means that the momentum vs time plot is a straight line for the mep . However, when running the dynamic situation we see an initial sharp increase in the momentum followed by a constant oscillation of momentum that is due to the momentum being retained within the vibrational modes. For A-B, there is a sharp increase in the momentum which plateaus at the same instance as B-C begins to undergo constant oscillation. This is because A-B refers to the momentum of the system A-B. Once both reach constant momentum the combined system now has a set defined level of momentum with no way to lose it.

[[File:Aja4117(dynamics type)Mep vs dynamics momenta vs time1.png]]
[[File:Aja4117dynamics mom vs time1.png]]

===== Trajectories from r1 = rts, r2 = rts +δ =====
The result of switching the distances is that

<nowiki> </nowiki>Take note of the final values of the positions '''r1'''(t) '''r2'''(t) and '''p1'''(t) '''p2'''(t) for your trajectory for large enough t.

== Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed. ==

== What do you observe? STILL TO DO ==

=== Reactive and unreactive trajectories ===
{| class="wikitable" border="1"
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics
!Plots of trajectory
|-
| -1.25 || -2.5 ||-99.018||Yes||Reaction proceeds with minimal vibration||[[File:-1.25, -2.5.png]]
|
|-
| -1.5 || -2.0 ||-100.456||No||P2 does not have enough momentum to reach P1||[[File:-1.5, -2.0.png]]
|
|-
| -1.5 || -2.5 ||-98.956||Yes||Transfer of vibrational energy||[[File:-1.5, -2.5.png]]
|
|-
| -2.5 || -5.0 || -84.956
||No||Product forms but contains too much vibrational energy so dissociates||[[File:-2.5, -5.0.png ]]
|
|-
| -2.5 || -5.2 || -83.416
||Yes||Reaction does occur but recrosses the transition state barrier||[[File:-2.5,5.2.png]]
|
|}
The fact that the last experiment recrosses the transition state barrier is evidence of the breakdown of Transition State Theory, which is presented below. this states that once products are formed, the system cannot travel back through the transition state to the reactants.

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

There are two basic assumptions for Transition State Theory, the two most basic are the separation of the electronic and nuclear motions which is the same as the Born-Oppenheimer approximation from quantum mechanics. The second basic assumption is that the reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution.

There are however, additional assumptions and they are as follows:
# Molecular systems that have crossed the transition state in the direction of the products cannot turn around and reform reactants.
# In the transition state, motion along the reaction coordinate may be separated from the other motions and be treated classically as a translation.
# Even in the absence of an equilibrium between the reactant and product molecules, the transition states that are become products are distributed among their states according to the Maxwell-Boltzmann laws.

''Given that Transition State theory neglects quantum mechanical contributions such as influence on and between between electronic and nuclear motions and the idea that molecules passing through the transition state through to the products cannot pass back through to the reactants, this would suggest that there would be deviation between experimental and calculated values using Transition State theory.''

== EXERCISE 2: F - H - H 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?'' ====
Looking at the PES of F + H2 this indicated the reaction is exothermic because there is a decrease in the energy going from the reactants to products. This can be seen along the reaction path for this reaction where there is a dip in energy on the path to the formation of the products. This makes intuitive sense as the driving force of the reaction will be the formation of a strong F-H bond, in place of a weaker H-H bond. Additional evidence that the reaction is exothermic is that when a mep calculation was run with RAB = 1.812 and RBC = 0.74 with zero momentum for each case, the reaction proceeded and the products formed with no additional input of energy.

For the H + HF reaction, '''the reaction is surmised to be endothermic due to as there is no decrease in energy on forming the products. This is in line with intuition considering the strength of the H-F bond. The very high electronegativity of fluorine would make it so that significant energy would be required to cause this reaction to proceed. '''

==== ''Locate the approximate position of the transition state.'' ====
{| class="wikitable"
!Reaction
!rAB
!rBC
!Activation Energy/ Kj/mol
|-
|F + H2
|1.8108757
|0.7448759
|0.2673
|-
|H + HF
|0.7448756
|1.8108759
|30.251
|}
The positions of the transition states were located through experimentation of the initial AB and BC distances, until the forces acting along AB and BC were zero. The reasoning behind this comes from the notion that

<math>F = - \frac {\partial V}{\partial r} </math>

where F is the force

and seeing as we are at a saddle point, <math> \frac {\partial V}{\partial r} = 0 </math> and hence

<math> F = 0 </math>

 

[[File:Aja4117Surface Plot f + h2.png]]
[[File:Aja4117Surface Plot h + HF.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.'' ====

The set of reaction conditions that were used were:

Atom A: F

Atom B: H

Atom C: H

Distance AB: 2.25

Distance BC: 0.74

Momentum AB = -0.8

Momentum BC = 0.1

These conditions yielded a set of successful reactions, and post reaction and upon formation of F-H, there is transfer of energy in the form of kinetic energy to both the Hydrogen that is travelling away from the reaction complex and also kinetic energy in the bond vibration of F-H. This can be visualised in the Momenta vs Time graph below. Monitoring the emission spectrum of the product and observing electromagnetic radiation in the IR region would allow identification of the product.

[[File:Aja4117Surface Plot FH2 momenta vs time.png]]

For rBC = 0.74, with a momentum pAB = -0.5, and several values of pBC in the range -3 to 3 were explored. For pBC = -3, the reaction did proceed and there was significant transfer of kinetic energy to the product F-H. What is interesting to note is that the F-H bond forms twice. On the first formation, the bond breaks due to what is assumed to be, too high a vibrational energy. This leads to the reformation of the H-H bond, upon which there is still significant vibrational energy and combined with the electronegativity of the F atom causes the H-H bond to break and the F-H bond to form for the second time. The F-H bond is now stable, although still containing significant vibrational energy. For pBC = 3, the same phenomena occurs as for pBC = -3. 

==== ''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.'' ====
A transition state for an attractive potential energy surface occurs early in the reaction coordinate and involves high translational kinetic energy (evidenced by the random motion for the reactants) and results in a vibrationally excited product. This translational energy runs up the side of the valley and falls from side to side, showing that there is significant vibrational motion. We can therefore conclude that reactions with attractive potential energy surfaces proceed more efficiently if the energy is in relative translational motion.

In contrast, reactions with repulsive potential surfaces experience a late transition state and are expected to proceed with greater efficiency if the excess energy is present as vibrations. The energy is in the vibration of the reactant molecule and the motion causes the trajectory to move from side to side up the valley as it approaches the transition state. This energy is enough to tip the reactant molecules through the transition state and into the products, it is therefore apparent that the vibrational energy on the repulsive potential energy surface is key to the reactivity and formation of the products.

MRD:aja4117

2019-05-16T15:10:47Z

Aja4117: /* Locate the approximate position of the transition state. */

= [[Molecular reaction dynamics|'''Molecular reaction dynamics''']] =

== 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?''''' ===
The transition state is a saddle point which is defined as the point which is both a minimum along one vector and a maximum along an orthogonal vector. Along the reaction path, the transition state is a maximum and along the second vector, which is perpendicular to our initial vector it is a minimum. This can be visualised in the diagram below.

For a co linear system, as given for the H + H2 system there are two orthogonal degrees of freedom with the coordinates given as follows:

'''Z = r2 + r1'''
 

<math>\frac {\partial V}{\partial Z} = 0 </math>, <math>\frac {\partial^2 V}{\partial Z^2}>0 </math>

 

'''S = r2 - r1'''

 

<math>\frac {\partial V}{\partial S} = 0 </math>, <math>\frac {\partial^2 V}{\partial S^2}<0 </math>

 

Note that these are the orthogonal vectors.

The saddle point has the configuration in which r1 = r2 =r0 and this will be the point at which these two lines intersect on the PES. This is the transition state. S is the tangent to the reaction pathway at the maximum whilst Z bisects the reaction pathway at the same point, but will be seen as a minimum.

[[File:Aja4117Question 1.png]]

=== ''Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'' ===
The best estimate was AB = 0.9077425 and BC = 0.9077425. There are only two lines when looking at internuclear distances vs time as seen below. This indicates that two lines are at the same distance and one is covering the other. 

To locate the transition state, is to begin the reaction trajectories close to the transition state. Due to the transition state being a global maxima, the trajectories will either form the reactants or products in order to find the lowest energy configuration. Whether the product or reactants are formed is directly influenced by the trajectories. If the trajectories are closer to the reactants then the reactants will form and if they are closer to the products then the products will form.

[[File:Aja4117INT for HH.png]]

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

===== Trajectories from r1 = rts+δ, r2 = rts =====
Along the mep, because the atoms are moving infinitely slowly and the momenta and the velocities always reset to zero, there is no oscillation and hence no vibration. This is seen along the mep where there is no wavy lines. In contrast, when looking at the dynamics, there is oscillation as post collision there is transfer of kinetic energy that causes the molecule to vibrate. 

When considering the inter nuclear distance vs time, A-C and A-B distances increase rapidly over the first 250 steps after which the rate of increase of the distance begins to decrease. As the number of steps increases the rate of increase of distance begins to plateau as the gradient flattens. This can be rationalised when looking at the animation where post collision, the molecules initially rapidly move away from one another but the speed at which they move away from each other falls off with time. When considering the dynamic situation, A-C and A-B distances increase at a constant gradient over the course of the plot. The reasoning is that in the dynamic situation, the velocities and momenta are not reset to zero at each step.

[[File:Aja4117(dynamics type)Mep vs dynamics distance v time 1.png]]
[[File:Aja4117Distance vs time 1.png]]

For the same reason as stated as above, the momenta and velocities are reset to zero at each point this means that the momentum vs time plot is a straight line for the mep . However, when running the dynamic situation we see an initial sharp increase in the momentum followed by a constant oscillation of momentum that is due to the momentum being retained within the vibrational modes. For A-B, there is a sharp increase in the momentum which plateaus at the same instance as B-C begins to undergo constant oscillation. This is because A-B refers to the momentum of the system A-B. Once both reach constant momentum the combined system now has a set defined level of momentum with no way to lose it.

[[File:Aja4117(dynamics type)Mep vs dynamics momenta vs time1.png]]
[[File:Aja4117dynamics mom vs time1.png]]

===== Trajectories from r1 = rts, r2 = rts +δ =====
The result of switching the distances is that

<nowiki> </nowiki>Take note of the final values of the positions '''r1'''(t) '''r2'''(t) and '''p1'''(t) '''p2'''(t) for your trajectory for large enough t.

== Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed. ==

== What do you observe? STILL TO DO ==

=== Reactive and unreactive trajectories ===
{| class="wikitable" border="1"
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics
!Plots of trajectory
|-
| -1.25 || -2.5 ||-99.018||Yes||Reaction proceeds with minimal vibration||[[File:-1.25, -2.5.png]]
|
|-
| -1.5 || -2.0 ||-100.456||No||P2 does not have enough momentum to reach P1||[[File:-1.5, -2.0.png]]
|
|-
| -1.5 || -2.5 ||-98.956||Yes||Transfer of vibrational energy||[[File:-1.5, -2.5.png]]
|
|-
| -2.5 || -5.0 || -84.956
||No||Product forms but contains too much vibrational energy so dissociates||[[File:-2.5, -5.0.png ]]
|
|-
| -2.5 || -5.2 || -83.416
||Yes||Reaction does occur but recrosses the transition state barrier||[[File:-2.5,5.2.png]]
|
|}
The fact that the last experiment recrosses the transition state barrier is evidence of the breakdown of Transition State Theory, which is presented below. this states that once products are formed, the system cannot travel back through the transition state to the reactants.

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

There are two basic assumptions for Transition State Theory, the two most basic are the separation of the electronic and nuclear motions which is the same as the Born-Oppenheimer approximation from quantum mechanics. The second basic assumption is that the reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution.

There are however, additional assumptions and they are as follows:
# Molecular systems that have crossed the transition state in the direction of the products cannot turn around and reform reactants.
# In the transition state, motion along the reaction coordinate may be separated from the other motions and be treated classically as a translation.
# Even in the absence of an equilibrium between the reactant and product molecules, the transition states that are become products are distributed among their states according to the Maxwell-Boltzmann laws.

''Given that Transition State theory neglects quantum mechanical contributions such as influence on and between between electronic and nuclear motions and the idea that molecules passing through the transition state through to the products cannot pass back through to the reactants, this would suggest that there would be deviation between experimental and calculated values using Transition State theory.''

== EXERCISE 2: F - H - H 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?'' ====
Looking at the PES of F + H2 this indicated the reaction is exothermic because there is a decrease in the energy going from the reactants to products. This can be seen along the reaction path for this reaction where there is a dip in energy on the path to the formation of the products. This makes intuitive sense as the driving force of the reaction will be the formation of a strong F-H bond, in place of a weaker H-H bond. Additional evidence that the reaction is exothermic is that when a mep calculation was run with RAB = 1.812 and RBC = 0.74 with zero momentum for each case, the reaction proceeded and the products formed with no additional input of energy.

For the H + HF reaction, '''the reaction is surmised to be endothermic due to as there is no decrease in energy on forming the products. This is in line with intuition considering the strength of the H-F bond. The very high electronegativity of fluorine would make it so that significant energy would be required to cause this reaction to proceed. '''

==== ''Locate the approximate position of the transition state.'' ====
{| class="wikitable"
!Reaction
!rAB
!rBC
!Activation Energy/ Kj/mol
|-
|F + H2
|1.8108757
|0.7448759
|0.2673
|-
|H + HF
|0.7448756
|1.8108759
|30.251
|}
The positions of the transition states were located through experimentation of the initial AB and BC distances, until the forces acting along AB and BC were zero. The reasoning behind this comes from the notion that

<math>F = - \frac {\partial V}{\partial r} </math>

and seeing as we are at a saddle point, <math> \frac {\partial V}{\partial r} = 0 </math> and hence

<math> F = 0 </math>

 

where F is the force

[[File:Aja4117Surface Plot f + h2.png]]
[[File:Aja4117Surface Plot h + HF.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.'' ====

The set of reaction conditions that were used were:

Atom A: F

Atom B: H

Atom C: H

Distance AB: 2.25

Distance BC: 0.74

Momentum AB = -0.8

Momentum BC = 0.1

These conditions yielded a set of successful reactions, and post reaction and upon formation of F-H, there is transfer of energy in the form of kinetic energy to both the Hydrogen that is travelling away from the reaction complex and also kinetic energy in the bond vibration of F-H. This can be visualised in the Momenta vs Time graph below. Monitoring the emission spectrum of the product and observing electromagnetic radiation in the IR region would allow identification of the product.

[[File:Aja4117Surface Plot FH2 momenta vs time.png]]

For rBC = 0.74, with a momentum pAB = -0.5, and several values of pBC in the range -3 to 3 were explored. For pBC = -3, the reaction did proceed and there was significant transfer of kinetic energy to the product F-H. What is interesting to note is that the F-H bond forms twice. On the first formation, the bond breaks due to what is assumed to be, too high a vibrational energy. This leads to the reformation of the H-H bond, upon which there is still significant vibrational energy and combined with the electronegativity of the F atom causes the H-H bond to break and the F-H bond to form for the second time. The F-H bond is now stable, although still containing significant vibrational energy. For pBC = 3, the same phenomena occurs as for pBC = -3. 

==== ''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.'' ====
A transition state for an attractive potential energy surface occurs early in the reaction coordinate and involves high translational kinetic energy (evidenced by the random motion for the reactants) and results in a vibrationally excited product. This translational energy runs up the side of the valley and falls from side to side, showing that there is significant vibrational motion. We can therefore conclude that reactions with attractive potential energy surfaces proceed more efficiently if the energy is in relative translational motion.

In contrast, reactions with repulsive potential surfaces experience a late transition state and are expected to proceed with greater efficiency if the excess energy is present as vibrations. The energy is in the vibration of the reactant molecule and the motion causes the trajectory to move from side to side up the valley as it approaches the transition state. This energy is enough to tip the reactant molecules through the transition state and into the products, it is therefore apparent that the vibrational energy on the repulsive potential energy surface is key to the reactivity and formation of the products.

MRD:aja4117

2019-05-16T15:10:29Z

Aja4117: /* Locate the approximate position of the transition state. */

= [[Molecular reaction dynamics|'''Molecular reaction dynamics''']] =

== 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?''''' ===
The transition state is a saddle point which is defined as the point which is both a minimum along one vector and a maximum along an orthogonal vector. Along the reaction path, the transition state is a maximum and along the second vector, which is perpendicular to our initial vector it is a minimum. This can be visualised in the diagram below.

For a co linear system, as given for the H + H2 system there are two orthogonal degrees of freedom with the coordinates given as follows:

'''Z = r2 + r1'''
 

<math>\frac {\partial V}{\partial Z} = 0 </math>, <math>\frac {\partial^2 V}{\partial Z^2}>0 </math>

 

'''S = r2 - r1'''

 

<math>\frac {\partial V}{\partial S} = 0 </math>, <math>\frac {\partial^2 V}{\partial S^2}<0 </math>

 

Note that these are the orthogonal vectors.

The saddle point has the configuration in which r1 = r2 =r0 and this will be the point at which these two lines intersect on the PES. This is the transition state. S is the tangent to the reaction pathway at the maximum whilst Z bisects the reaction pathway at the same point, but will be seen as a minimum.

[[File:Aja4117Question 1.png]]

=== ''Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'' ===
The best estimate was AB = 0.9077425 and BC = 0.9077425. There are only two lines when looking at internuclear distances vs time as seen below. This indicates that two lines are at the same distance and one is covering the other. 

To locate the transition state, is to begin the reaction trajectories close to the transition state. Due to the transition state being a global maxima, the trajectories will either form the reactants or products in order to find the lowest energy configuration. Whether the product or reactants are formed is directly influenced by the trajectories. If the trajectories are closer to the reactants then the reactants will form and if they are closer to the products then the products will form.

[[File:Aja4117INT for HH.png]]

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

===== Trajectories from r1 = rts+δ, r2 = rts =====
Along the mep, because the atoms are moving infinitely slowly and the momenta and the velocities always reset to zero, there is no oscillation and hence no vibration. This is seen along the mep where there is no wavy lines. In contrast, when looking at the dynamics, there is oscillation as post collision there is transfer of kinetic energy that causes the molecule to vibrate. 

When considering the inter nuclear distance vs time, A-C and A-B distances increase rapidly over the first 250 steps after which the rate of increase of the distance begins to decrease. As the number of steps increases the rate of increase of distance begins to plateau as the gradient flattens. This can be rationalised when looking at the animation where post collision, the molecules initially rapidly move away from one another but the speed at which they move away from each other falls off with time. When considering the dynamic situation, A-C and A-B distances increase at a constant gradient over the course of the plot. The reasoning is that in the dynamic situation, the velocities and momenta are not reset to zero at each step.

[[File:Aja4117(dynamics type)Mep vs dynamics distance v time 1.png]]
[[File:Aja4117Distance vs time 1.png]]

For the same reason as stated as above, the momenta and velocities are reset to zero at each point this means that the momentum vs time plot is a straight line for the mep . However, when running the dynamic situation we see an initial sharp increase in the momentum followed by a constant oscillation of momentum that is due to the momentum being retained within the vibrational modes. For A-B, there is a sharp increase in the momentum which plateaus at the same instance as B-C begins to undergo constant oscillation. This is because A-B refers to the momentum of the system A-B. Once both reach constant momentum the combined system now has a set defined level of momentum with no way to lose it.

[[File:Aja4117(dynamics type)Mep vs dynamics momenta vs time1.png]]
[[File:Aja4117dynamics mom vs time1.png]]

===== Trajectories from r1 = rts, r2 = rts +δ =====
The result of switching the distances is that

<nowiki> </nowiki>Take note of the final values of the positions '''r1'''(t) '''r2'''(t) and '''p1'''(t) '''p2'''(t) for your trajectory for large enough t.

== Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed. ==

== What do you observe? STILL TO DO ==

=== Reactive and unreactive trajectories ===
{| class="wikitable" border="1"
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics
!Plots of trajectory
|-
| -1.25 || -2.5 ||-99.018||Yes||Reaction proceeds with minimal vibration||[[File:-1.25, -2.5.png]]
|
|-
| -1.5 || -2.0 ||-100.456||No||P2 does not have enough momentum to reach P1||[[File:-1.5, -2.0.png]]
|
|-
| -1.5 || -2.5 ||-98.956||Yes||Transfer of vibrational energy||[[File:-1.5, -2.5.png]]
|
|-
| -2.5 || -5.0 || -84.956
||No||Product forms but contains too much vibrational energy so dissociates||[[File:-2.5, -5.0.png ]]
|
|-
| -2.5 || -5.2 || -83.416
||Yes||Reaction does occur but recrosses the transition state barrier||[[File:-2.5,5.2.png]]
|
|}
The fact that the last experiment recrosses the transition state barrier is evidence of the breakdown of Transition State Theory, which is presented below. this states that once products are formed, the system cannot travel back through the transition state to the reactants.

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

There are two basic assumptions for Transition State Theory, the two most basic are the separation of the electronic and nuclear motions which is the same as the Born-Oppenheimer approximation from quantum mechanics. The second basic assumption is that the reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution.

There are however, additional assumptions and they are as follows:
# Molecular systems that have crossed the transition state in the direction of the products cannot turn around and reform reactants.
# In the transition state, motion along the reaction coordinate may be separated from the other motions and be treated classically as a translation.
# Even in the absence of an equilibrium between the reactant and product molecules, the transition states that are become products are distributed among their states according to the Maxwell-Boltzmann laws.

''Given that Transition State theory neglects quantum mechanical contributions such as influence on and between between electronic and nuclear motions and the idea that molecules passing through the transition state through to the products cannot pass back through to the reactants, this would suggest that there would be deviation between experimental and calculated values using Transition State theory.''

== EXERCISE 2: F - H - H 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?'' ====
Looking at the PES of F + H2 this indicated the reaction is exothermic because there is a decrease in the energy going from the reactants to products. This can be seen along the reaction path for this reaction where there is a dip in energy on the path to the formation of the products. This makes intuitive sense as the driving force of the reaction will be the formation of a strong F-H bond, in place of a weaker H-H bond. Additional evidence that the reaction is exothermic is that when a mep calculation was run with RAB = 1.812 and RBC = 0.74 with zero momentum for each case, the reaction proceeded and the products formed with no additional input of energy.

For the H + HF reaction, '''the reaction is surmised to be endothermic due to as there is no decrease in energy on forming the products. This is in line with intuition considering the strength of the H-F bond. The very high electronegativity of fluorine would make it so that significant energy would be required to cause this reaction to proceed. '''

==== ''Locate the approximate position of the transition state.'' ====
{| class="wikitable"
!Reaction
!rAB
!rBC
!Activation Energy/ Kj/mol
|-
|F + H2
|1.8108757
|0.7448759
|0.2673
|-
|H + HF
|0.7448756
|1.8108759
|30.251
|}
The positions of the transition states were located through experimentation of the initial AB and BC distances, until the forces acting along AB and BC were zero. The reasoning behind this comes from the notion that

<math>F = - \frac {\partial V}{\partial r} </math>

and seeing as we are at a saddle point, <math> \frac {\partial V}{\partial r} = 0 </math> and hence

<math> F = 0 <math/>

 

where F is the force

[[File:Aja4117Surface Plot f + h2.png]]
[[File:Aja4117Surface Plot h + HF.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.'' ====

The set of reaction conditions that were used were:

Atom A: F

Atom B: H

Atom C: H

Distance AB: 2.25

Distance BC: 0.74

Momentum AB = -0.8

Momentum BC = 0.1

These conditions yielded a set of successful reactions, and post reaction and upon formation of F-H, there is transfer of energy in the form of kinetic energy to both the Hydrogen that is travelling away from the reaction complex and also kinetic energy in the bond vibration of F-H. This can be visualised in the Momenta vs Time graph below. Monitoring the emission spectrum of the product and observing electromagnetic radiation in the IR region would allow identification of the product.

[[File:Aja4117Surface Plot FH2 momenta vs time.png]]

For rBC = 0.74, with a momentum pAB = -0.5, and several values of pBC in the range -3 to 3 were explored. For pBC = -3, the reaction did proceed and there was significant transfer of kinetic energy to the product F-H. What is interesting to note is that the F-H bond forms twice. On the first formation, the bond breaks due to what is assumed to be, too high a vibrational energy. This leads to the reformation of the H-H bond, upon which there is still significant vibrational energy and combined with the electronegativity of the F atom causes the H-H bond to break and the F-H bond to form for the second time. The F-H bond is now stable, although still containing significant vibrational energy. For pBC = 3, the same phenomena occurs as for pBC = -3. 

==== ''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.'' ====
A transition state for an attractive potential energy surface occurs early in the reaction coordinate and involves high translational kinetic energy (evidenced by the random motion for the reactants) and results in a vibrationally excited product. This translational energy runs up the side of the valley and falls from side to side, showing that there is significant vibrational motion. We can therefore conclude that reactions with attractive potential energy surfaces proceed more efficiently if the energy is in relative translational motion.

In contrast, reactions with repulsive potential surfaces experience a late transition state and are expected to proceed with greater efficiency if the excess energy is present as vibrations. The energy is in the vibration of the reactant molecule and the motion causes the trajectory to move from side to side up the valley as it approaches the transition state. This energy is enough to tip the reactant molecules through the transition state and into the products, it is therefore apparent that the vibrational energy on the repulsive potential energy surface is key to the reactivity and formation of the products.

MRD:aja4117

2019-05-16T15:08:57Z

Aja4117: /* Locate the approximate position of the transition state. */

= [[Molecular reaction dynamics|'''Molecular reaction dynamics''']] =

== 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?''''' ===
The transition state is a saddle point which is defined as the point which is both a minimum along one vector and a maximum along an orthogonal vector. Along the reaction path, the transition state is a maximum and along the second vector, which is perpendicular to our initial vector it is a minimum. This can be visualised in the diagram below.

For a co linear system, as given for the H + H2 system there are two orthogonal degrees of freedom with the coordinates given as follows:

'''Z = r2 + r1'''
 

<math>\frac {\partial V}{\partial Z} = 0 </math>, <math>\frac {\partial^2 V}{\partial Z^2}>0 </math>

 

'''S = r2 - r1'''

 

<math>\frac {\partial V}{\partial S} = 0 </math>, <math>\frac {\partial^2 V}{\partial S^2}<0 </math>

 

Note that these are the orthogonal vectors.

The saddle point has the configuration in which r1 = r2 =r0 and this will be the point at which these two lines intersect on the PES. This is the transition state. S is the tangent to the reaction pathway at the maximum whilst Z bisects the reaction pathway at the same point, but will be seen as a minimum.

[[File:Aja4117Question 1.png]]

=== ''Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'' ===
The best estimate was AB = 0.9077425 and BC = 0.9077425. There are only two lines when looking at internuclear distances vs time as seen below. This indicates that two lines are at the same distance and one is covering the other. 

To locate the transition state, is to begin the reaction trajectories close to the transition state. Due to the transition state being a global maxima, the trajectories will either form the reactants or products in order to find the lowest energy configuration. Whether the product or reactants are formed is directly influenced by the trajectories. If the trajectories are closer to the reactants then the reactants will form and if they are closer to the products then the products will form.

[[File:Aja4117INT for HH.png]]

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

===== Trajectories from r1 = rts+δ, r2 = rts =====
Along the mep, because the atoms are moving infinitely slowly and the momenta and the velocities always reset to zero, there is no oscillation and hence no vibration. This is seen along the mep where there is no wavy lines. In contrast, when looking at the dynamics, there is oscillation as post collision there is transfer of kinetic energy that causes the molecule to vibrate. 

When considering the inter nuclear distance vs time, A-C and A-B distances increase rapidly over the first 250 steps after which the rate of increase of the distance begins to decrease. As the number of steps increases the rate of increase of distance begins to plateau as the gradient flattens. This can be rationalised when looking at the animation where post collision, the molecules initially rapidly move away from one another but the speed at which they move away from each other falls off with time. When considering the dynamic situation, A-C and A-B distances increase at a constant gradient over the course of the plot. The reasoning is that in the dynamic situation, the velocities and momenta are not reset to zero at each step.

[[File:Aja4117(dynamics type)Mep vs dynamics distance v time 1.png]]
[[File:Aja4117Distance vs time 1.png]]

For the same reason as stated as above, the momenta and velocities are reset to zero at each point this means that the momentum vs time plot is a straight line for the mep . However, when running the dynamic situation we see an initial sharp increase in the momentum followed by a constant oscillation of momentum that is due to the momentum being retained within the vibrational modes. For A-B, there is a sharp increase in the momentum which plateaus at the same instance as B-C begins to undergo constant oscillation. This is because A-B refers to the momentum of the system A-B. Once both reach constant momentum the combined system now has a set defined level of momentum with no way to lose it.

[[File:Aja4117(dynamics type)Mep vs dynamics momenta vs time1.png]]
[[File:Aja4117dynamics mom vs time1.png]]

===== Trajectories from r1 = rts, r2 = rts +δ =====
The result of switching the distances is that

<nowiki> </nowiki>Take note of the final values of the positions '''r1'''(t) '''r2'''(t) and '''p1'''(t) '''p2'''(t) for your trajectory for large enough t.

== Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed. ==

== What do you observe? STILL TO DO ==

=== Reactive and unreactive trajectories ===
{| class="wikitable" border="1"
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics
!Plots of trajectory
|-
| -1.25 || -2.5 ||-99.018||Yes||Reaction proceeds with minimal vibration||[[File:-1.25, -2.5.png]]
|
|-
| -1.5 || -2.0 ||-100.456||No||P2 does not have enough momentum to reach P1||[[File:-1.5, -2.0.png]]
|
|-
| -1.5 || -2.5 ||-98.956||Yes||Transfer of vibrational energy||[[File:-1.5, -2.5.png]]
|
|-
| -2.5 || -5.0 || -84.956
||No||Product forms but contains too much vibrational energy so dissociates||[[File:-2.5, -5.0.png ]]
|
|-
| -2.5 || -5.2 || -83.416
||Yes||Reaction does occur but recrosses the transition state barrier||[[File:-2.5,5.2.png]]
|
|}
The fact that the last experiment recrosses the transition state barrier is evidence of the breakdown of Transition State Theory, which is presented below. this states that once products are formed, the system cannot travel back through the transition state to the reactants.

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

There are two basic assumptions for Transition State Theory, the two most basic are the separation of the electronic and nuclear motions which is the same as the Born-Oppenheimer approximation from quantum mechanics. The second basic assumption is that the reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution.

There are however, additional assumptions and they are as follows:
# Molecular systems that have crossed the transition state in the direction of the products cannot turn around and reform reactants.
# In the transition state, motion along the reaction coordinate may be separated from the other motions and be treated classically as a translation.
# Even in the absence of an equilibrium between the reactant and product molecules, the transition states that are become products are distributed among their states according to the Maxwell-Boltzmann laws.

''Given that Transition State theory neglects quantum mechanical contributions such as influence on and between between electronic and nuclear motions and the idea that molecules passing through the transition state through to the products cannot pass back through to the reactants, this would suggest that there would be deviation between experimental and calculated values using Transition State theory.''

== EXERCISE 2: F - H - H 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?'' ====
Looking at the PES of F + H2 this indicated the reaction is exothermic because there is a decrease in the energy going from the reactants to products. This can be seen along the reaction path for this reaction where there is a dip in energy on the path to the formation of the products. This makes intuitive sense as the driving force of the reaction will be the formation of a strong F-H bond, in place of a weaker H-H bond. Additional evidence that the reaction is exothermic is that when a mep calculation was run with RAB = 1.812 and RBC = 0.74 with zero momentum for each case, the reaction proceeded and the products formed with no additional input of energy.

For the H + HF reaction, '''the reaction is surmised to be endothermic due to as there is no decrease in energy on forming the products. This is in line with intuition considering the strength of the H-F bond. The very high electronegativity of fluorine would make it so that significant energy would be required to cause this reaction to proceed. '''

==== ''Locate the approximate position of the transition state.'' ====
{| class="wikitable"
!Reaction
!rAB
!rBC
!Activation Energy/ Kj/mol
|-
|F + H2
|1.8108757
|0.7448759
|0.2673
|-
|H + HF
|0.7448756
|1.8108759
|30.251
|}
The positions of the transition states were located through experimentation of the initial AB and BC distances, until the forces acting along AB and BC were zero. The reasoning behind this comes from the notion that

<math>F =\frac {\partial V}{\partial r} = 0 </math>

 

where F is the force

[[File:Aja4117Surface Plot f + h2.png]]
[[File:Aja4117Surface Plot h + HF.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.'' ====

The set of reaction conditions that were used were:

Atom A: F

Atom B: H

Atom C: H

Distance AB: 2.25

Distance BC: 0.74

Momentum AB = -0.8

Momentum BC = 0.1

These conditions yielded a set of successful reactions, and post reaction and upon formation of F-H, there is transfer of energy in the form of kinetic energy to both the Hydrogen that is travelling away from the reaction complex and also kinetic energy in the bond vibration of F-H. This can be visualised in the Momenta vs Time graph below. Monitoring the emission spectrum of the product and observing electromagnetic radiation in the IR region would allow identification of the product.

[[File:Aja4117Surface Plot FH2 momenta vs time.png]]

For rBC = 0.74, with a momentum pAB = -0.5, and several values of pBC in the range -3 to 3 were explored. For pBC = -3, the reaction did proceed and there was significant transfer of kinetic energy to the product F-H. What is interesting to note is that the F-H bond forms twice. On the first formation, the bond breaks due to what is assumed to be, too high a vibrational energy. This leads to the reformation of the H-H bond, upon which there is still significant vibrational energy and combined with the electronegativity of the F atom causes the H-H bond to break and the F-H bond to form for the second time. The F-H bond is now stable, although still containing significant vibrational energy. For pBC = 3, the same phenomena occurs as for pBC = -3. 

==== ''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.'' ====
A transition state for an attractive potential energy surface occurs early in the reaction coordinate and involves high translational kinetic energy (evidenced by the random motion for the reactants) and results in a vibrationally excited product. This translational energy runs up the side of the valley and falls from side to side, showing that there is significant vibrational motion. We can therefore conclude that reactions with attractive potential energy surfaces proceed more efficiently if the energy is in relative translational motion.

In contrast, reactions with repulsive potential surfaces experience a late transition state and are expected to proceed with greater efficiency if the excess energy is present as vibrations. The energy is in the vibration of the reactant molecule and the motion causes the trajectory to move from side to side up the valley as it approaches the transition state. This energy is enough to tip the reactant molecules through the transition state and into the products, it is therefore apparent that the vibrational energy on the repulsive potential energy surface is key to the reactivity and formation of the products.

MRD:aja4117

2019-05-16T15:08:36Z

Aja4117: /* Locate the approximate position of the transition state. */

= [[Molecular reaction dynamics|'''Molecular reaction dynamics''']] =

== 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?''''' ===
The transition state is a saddle point which is defined as the point which is both a minimum along one vector and a maximum along an orthogonal vector. Along the reaction path, the transition state is a maximum and along the second vector, which is perpendicular to our initial vector it is a minimum. This can be visualised in the diagram below.

For a co linear system, as given for the H + H2 system there are two orthogonal degrees of freedom with the coordinates given as follows:

'''Z = r2 + r1'''
 

<math>\frac {\partial V}{\partial Z} = 0 </math>, <math>\frac {\partial^2 V}{\partial Z^2}>0 </math>

 

'''S = r2 - r1'''

 

<math>\frac {\partial V}{\partial S} = 0 </math>, <math>\frac {\partial^2 V}{\partial S^2}<0 </math>

 

Note that these are the orthogonal vectors.

The saddle point has the configuration in which r1 = r2 =r0 and this will be the point at which these two lines intersect on the PES. This is the transition state. S is the tangent to the reaction pathway at the maximum whilst Z bisects the reaction pathway at the same point, but will be seen as a minimum.

[[File:Aja4117Question 1.png]]

=== ''Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'' ===
The best estimate was AB = 0.9077425 and BC = 0.9077425. There are only two lines when looking at internuclear distances vs time as seen below. This indicates that two lines are at the same distance and one is covering the other. 

To locate the transition state, is to begin the reaction trajectories close to the transition state. Due to the transition state being a global maxima, the trajectories will either form the reactants or products in order to find the lowest energy configuration. Whether the product or reactants are formed is directly influenced by the trajectories. If the trajectories are closer to the reactants then the reactants will form and if they are closer to the products then the products will form.

[[File:Aja4117INT for HH.png]]

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

===== Trajectories from r1 = rts+δ, r2 = rts =====
Along the mep, because the atoms are moving infinitely slowly and the momenta and the velocities always reset to zero, there is no oscillation and hence no vibration. This is seen along the mep where there is no wavy lines. In contrast, when looking at the dynamics, there is oscillation as post collision there is transfer of kinetic energy that causes the molecule to vibrate. 

When considering the inter nuclear distance vs time, A-C and A-B distances increase rapidly over the first 250 steps after which the rate of increase of the distance begins to decrease. As the number of steps increases the rate of increase of distance begins to plateau as the gradient flattens. This can be rationalised when looking at the animation where post collision, the molecules initially rapidly move away from one another but the speed at which they move away from each other falls off with time. When considering the dynamic situation, A-C and A-B distances increase at a constant gradient over the course of the plot. The reasoning is that in the dynamic situation, the velocities and momenta are not reset to zero at each step.

[[File:Aja4117(dynamics type)Mep vs dynamics distance v time 1.png]]
[[File:Aja4117Distance vs time 1.png]]

For the same reason as stated as above, the momenta and velocities are reset to zero at each point this means that the momentum vs time plot is a straight line for the mep . However, when running the dynamic situation we see an initial sharp increase in the momentum followed by a constant oscillation of momentum that is due to the momentum being retained within the vibrational modes. For A-B, there is a sharp increase in the momentum which plateaus at the same instance as B-C begins to undergo constant oscillation. This is because A-B refers to the momentum of the system A-B. Once both reach constant momentum the combined system now has a set defined level of momentum with no way to lose it.

[[File:Aja4117(dynamics type)Mep vs dynamics momenta vs time1.png]]
[[File:Aja4117dynamics mom vs time1.png]]

===== Trajectories from r1 = rts, r2 = rts +δ =====
The result of switching the distances is that

<nowiki> </nowiki>Take note of the final values of the positions '''r1'''(t) '''r2'''(t) and '''p1'''(t) '''p2'''(t) for your trajectory for large enough t.

== Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed. ==

== What do you observe? STILL TO DO ==

=== Reactive and unreactive trajectories ===
{| class="wikitable" border="1"
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics
!Plots of trajectory
|-
| -1.25 || -2.5 ||-99.018||Yes||Reaction proceeds with minimal vibration||[[File:-1.25, -2.5.png]]
|
|-
| -1.5 || -2.0 ||-100.456||No||P2 does not have enough momentum to reach P1||[[File:-1.5, -2.0.png]]
|
|-
| -1.5 || -2.5 ||-98.956||Yes||Transfer of vibrational energy||[[File:-1.5, -2.5.png]]
|
|-
| -2.5 || -5.0 || -84.956
||No||Product forms but contains too much vibrational energy so dissociates||[[File:-2.5, -5.0.png ]]
|
|-
| -2.5 || -5.2 || -83.416
||Yes||Reaction does occur but recrosses the transition state barrier||[[File:-2.5,5.2.png]]
|
|}
The fact that the last experiment recrosses the transition state barrier is evidence of the breakdown of Transition State Theory, which is presented below. this states that once products are formed, the system cannot travel back through the transition state to the reactants.

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

There are two basic assumptions for Transition State Theory, the two most basic are the separation of the electronic and nuclear motions which is the same as the Born-Oppenheimer approximation from quantum mechanics. The second basic assumption is that the reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution.

There are however, additional assumptions and they are as follows:
# Molecular systems that have crossed the transition state in the direction of the products cannot turn around and reform reactants.
# In the transition state, motion along the reaction coordinate may be separated from the other motions and be treated classically as a translation.
# Even in the absence of an equilibrium between the reactant and product molecules, the transition states that are become products are distributed among their states according to the Maxwell-Boltzmann laws.

''Given that Transition State theory neglects quantum mechanical contributions such as influence on and between between electronic and nuclear motions and the idea that molecules passing through the transition state through to the products cannot pass back through to the reactants, this would suggest that there would be deviation between experimental and calculated values using Transition State theory.''

== EXERCISE 2: F - H - H 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?'' ====
Looking at the PES of F + H2 this indicated the reaction is exothermic because there is a decrease in the energy going from the reactants to products. This can be seen along the reaction path for this reaction where there is a dip in energy on the path to the formation of the products. This makes intuitive sense as the driving force of the reaction will be the formation of a strong F-H bond, in place of a weaker H-H bond. Additional evidence that the reaction is exothermic is that when a mep calculation was run with RAB = 1.812 and RBC = 0.74 with zero momentum for each case, the reaction proceeded and the products formed with no additional input of energy.

For the H + HF reaction, '''the reaction is surmised to be endothermic due to as there is no decrease in energy on forming the products. This is in line with intuition considering the strength of the H-F bond. The very high electronegativity of fluorine would make it so that significant energy would be required to cause this reaction to proceed. '''

==== ''Locate the approximate position of the transition state.'' ====
{| class="wikitable"
!Reaction
!rAB
!rBC
!Activation Energy/ Kj/mol
|-
|F + H2
|1.8108757
|0.7448759
|0.2673
|-
|H + HF
|0.7448756
|1.8108759
|30.251
|}
The positions of the transition states were located through experimentation of the initial AB and BC distances, until the forces acting along AB and BC were zero. The reasoning behind this comes from the notion that

F = <math>\frac {\partial V}{\partial r} = 0 </math>

 

where F is the force

[[File:Aja4117Surface Plot f + h2.png]]
[[File:Aja4117Surface Plot h + HF.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.'' ====

The set of reaction conditions that were used were:

Atom A: F

Atom B: H

Atom C: H

Distance AB: 2.25

Distance BC: 0.74

Momentum AB = -0.8

Momentum BC = 0.1

These conditions yielded a set of successful reactions, and post reaction and upon formation of F-H, there is transfer of energy in the form of kinetic energy to both the Hydrogen that is travelling away from the reaction complex and also kinetic energy in the bond vibration of F-H. This can be visualised in the Momenta vs Time graph below. Monitoring the emission spectrum of the product and observing electromagnetic radiation in the IR region would allow identification of the product.

[[File:Aja4117Surface Plot FH2 momenta vs time.png]]

For rBC = 0.74, with a momentum pAB = -0.5, and several values of pBC in the range -3 to 3 were explored. For pBC = -3, the reaction did proceed and there was significant transfer of kinetic energy to the product F-H. What is interesting to note is that the F-H bond forms twice. On the first formation, the bond breaks due to what is assumed to be, too high a vibrational energy. This leads to the reformation of the H-H bond, upon which there is still significant vibrational energy and combined with the electronegativity of the F atom causes the H-H bond to break and the F-H bond to form for the second time. The F-H bond is now stable, although still containing significant vibrational energy. For pBC = 3, the same phenomena occurs as for pBC = -3. 

==== ''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.'' ====
A transition state for an attractive potential energy surface occurs early in the reaction coordinate and involves high translational kinetic energy (evidenced by the random motion for the reactants) and results in a vibrationally excited product. This translational energy runs up the side of the valley and falls from side to side, showing that there is significant vibrational motion. We can therefore conclude that reactions with attractive potential energy surfaces proceed more efficiently if the energy is in relative translational motion.

In contrast, reactions with repulsive potential surfaces experience a late transition state and are expected to proceed with greater efficiency if the excess energy is present as vibrations. The energy is in the vibration of the reactant molecule and the motion causes the trajectory to move from side to side up the valley as it approaches the transition state. This energy is enough to tip the reactant molecules through the transition state and into the products, it is therefore apparent that the vibrational energy on the repulsive potential energy surface is key to the reactivity and formation of the products.

MRD:aja4117

2019-05-16T15:08:22Z

Aja4117: /* Locate the approximate position of the transition state. */

= [[Molecular reaction dynamics|'''Molecular reaction dynamics''']] =

== 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?''''' ===
The transition state is a saddle point which is defined as the point which is both a minimum along one vector and a maximum along an orthogonal vector. Along the reaction path, the transition state is a maximum and along the second vector, which is perpendicular to our initial vector it is a minimum. This can be visualised in the diagram below.

For a co linear system, as given for the H + H2 system there are two orthogonal degrees of freedom with the coordinates given as follows:

'''Z = r2 + r1'''
 

<math>\frac {\partial V}{\partial Z} = 0 </math>, <math>\frac {\partial^2 V}{\partial Z^2}>0 </math>

 

'''S = r2 - r1'''

 

<math>\frac {\partial V}{\partial S} = 0 </math>, <math>\frac {\partial^2 V}{\partial S^2}<0 </math>

 

Note that these are the orthogonal vectors.

The saddle point has the configuration in which r1 = r2 =r0 and this will be the point at which these two lines intersect on the PES. This is the transition state. S is the tangent to the reaction pathway at the maximum whilst Z bisects the reaction pathway at the same point, but will be seen as a minimum.

[[File:Aja4117Question 1.png]]

=== ''Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'' ===
The best estimate was AB = 0.9077425 and BC = 0.9077425. There are only two lines when looking at internuclear distances vs time as seen below. This indicates that two lines are at the same distance and one is covering the other. 

To locate the transition state, is to begin the reaction trajectories close to the transition state. Due to the transition state being a global maxima, the trajectories will either form the reactants or products in order to find the lowest energy configuration. Whether the product or reactants are formed is directly influenced by the trajectories. If the trajectories are closer to the reactants then the reactants will form and if they are closer to the products then the products will form.

[[File:Aja4117INT for HH.png]]

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

===== Trajectories from r1 = rts+δ, r2 = rts =====
Along the mep, because the atoms are moving infinitely slowly and the momenta and the velocities always reset to zero, there is no oscillation and hence no vibration. This is seen along the mep where there is no wavy lines. In contrast, when looking at the dynamics, there is oscillation as post collision there is transfer of kinetic energy that causes the molecule to vibrate. 

When considering the inter nuclear distance vs time, A-C and A-B distances increase rapidly over the first 250 steps after which the rate of increase of the distance begins to decrease. As the number of steps increases the rate of increase of distance begins to plateau as the gradient flattens. This can be rationalised when looking at the animation where post collision, the molecules initially rapidly move away from one another but the speed at which they move away from each other falls off with time. When considering the dynamic situation, A-C and A-B distances increase at a constant gradient over the course of the plot. The reasoning is that in the dynamic situation, the velocities and momenta are not reset to zero at each step.

[[File:Aja4117(dynamics type)Mep vs dynamics distance v time 1.png]]
[[File:Aja4117Distance vs time 1.png]]

For the same reason as stated as above, the momenta and velocities are reset to zero at each point this means that the momentum vs time plot is a straight line for the mep . However, when running the dynamic situation we see an initial sharp increase in the momentum followed by a constant oscillation of momentum that is due to the momentum being retained within the vibrational modes. For A-B, there is a sharp increase in the momentum which plateaus at the same instance as B-C begins to undergo constant oscillation. This is because A-B refers to the momentum of the system A-B. Once both reach constant momentum the combined system now has a set defined level of momentum with no way to lose it.

[[File:Aja4117(dynamics type)Mep vs dynamics momenta vs time1.png]]
[[File:Aja4117dynamics mom vs time1.png]]

===== Trajectories from r1 = rts, r2 = rts +δ =====
The result of switching the distances is that

<nowiki> </nowiki>Take note of the final values of the positions '''r1'''(t) '''r2'''(t) and '''p1'''(t) '''p2'''(t) for your trajectory for large enough t.

== Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed. ==

== What do you observe? STILL TO DO ==

=== Reactive and unreactive trajectories ===
{| class="wikitable" border="1"
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics
!Plots of trajectory
|-
| -1.25 || -2.5 ||-99.018||Yes||Reaction proceeds with minimal vibration||[[File:-1.25, -2.5.png]]
|
|-
| -1.5 || -2.0 ||-100.456||No||P2 does not have enough momentum to reach P1||[[File:-1.5, -2.0.png]]
|
|-
| -1.5 || -2.5 ||-98.956||Yes||Transfer of vibrational energy||[[File:-1.5, -2.5.png]]
|
|-
| -2.5 || -5.0 || -84.956
||No||Product forms but contains too much vibrational energy so dissociates||[[File:-2.5, -5.0.png ]]
|
|-
| -2.5 || -5.2 || -83.416
||Yes||Reaction does occur but recrosses the transition state barrier||[[File:-2.5,5.2.png]]
|
|}
The fact that the last experiment recrosses the transition state barrier is evidence of the breakdown of Transition State Theory, which is presented below. this states that once products are formed, the system cannot travel back through the transition state to the reactants.

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

There are two basic assumptions for Transition State Theory, the two most basic are the separation of the electronic and nuclear motions which is the same as the Born-Oppenheimer approximation from quantum mechanics. The second basic assumption is that the reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution.

There are however, additional assumptions and they are as follows:
# Molecular systems that have crossed the transition state in the direction of the products cannot turn around and reform reactants.
# In the transition state, motion along the reaction coordinate may be separated from the other motions and be treated classically as a translation.
# Even in the absence of an equilibrium between the reactant and product molecules, the transition states that are become products are distributed among their states according to the Maxwell-Boltzmann laws.

''Given that Transition State theory neglects quantum mechanical contributions such as influence on and between between electronic and nuclear motions and the idea that molecules passing through the transition state through to the products cannot pass back through to the reactants, this would suggest that there would be deviation between experimental and calculated values using Transition State theory.''

== EXERCISE 2: F - H - H 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?'' ====
Looking at the PES of F + H2 this indicated the reaction is exothermic because there is a decrease in the energy going from the reactants to products. This can be seen along the reaction path for this reaction where there is a dip in energy on the path to the formation of the products. This makes intuitive sense as the driving force of the reaction will be the formation of a strong F-H bond, in place of a weaker H-H bond. Additional evidence that the reaction is exothermic is that when a mep calculation was run with RAB = 1.812 and RBC = 0.74 with zero momentum for each case, the reaction proceeded and the products formed with no additional input of energy.

For the H + HF reaction, '''the reaction is surmised to be endothermic due to as there is no decrease in energy on forming the products. This is in line with intuition considering the strength of the H-F bond. The very high electronegativity of fluorine would make it so that significant energy would be required to cause this reaction to proceed. '''

==== ''Locate the approximate position of the transition state.'' ====
{| class="wikitable"
!Reaction
!rAB
!rBC
!Activation Energy/ Kj/mol
|-
|F + H2
|1.8108757
|0.7448759
|0.2673
|-
|H + HF
|0.7448756
|1.8108759
|30.251
|}
The positions of the transition states were located through experimentation of the initial AB and BC distances, until the forces acting along AB and BC were zero. The reasoning behind this comes from the notion that

F = </math>\frac {\partial V}{\partial r} = 0 </math>

 

where F is the force

[[File:Aja4117Surface Plot f + h2.png]]
[[File:Aja4117Surface Plot h + HF.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.'' ====

The set of reaction conditions that were used were:

Atom A: F

Atom B: H

Atom C: H

Distance AB: 2.25

Distance BC: 0.74

Momentum AB = -0.8

Momentum BC = 0.1

These conditions yielded a set of successful reactions, and post reaction and upon formation of F-H, there is transfer of energy in the form of kinetic energy to both the Hydrogen that is travelling away from the reaction complex and also kinetic energy in the bond vibration of F-H. This can be visualised in the Momenta vs Time graph below. Monitoring the emission spectrum of the product and observing electromagnetic radiation in the IR region would allow identification of the product.

[[File:Aja4117Surface Plot FH2 momenta vs time.png]]

For rBC = 0.74, with a momentum pAB = -0.5, and several values of pBC in the range -3 to 3 were explored. For pBC = -3, the reaction did proceed and there was significant transfer of kinetic energy to the product F-H. What is interesting to note is that the F-H bond forms twice. On the first formation, the bond breaks due to what is assumed to be, too high a vibrational energy. This leads to the reformation of the H-H bond, upon which there is still significant vibrational energy and combined with the electronegativity of the F atom causes the H-H bond to break and the F-H bond to form for the second time. The F-H bond is now stable, although still containing significant vibrational energy. For pBC = 3, the same phenomena occurs as for pBC = -3. 

==== ''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.'' ====
A transition state for an attractive potential energy surface occurs early in the reaction coordinate and involves high translational kinetic energy (evidenced by the random motion for the reactants) and results in a vibrationally excited product. This translational energy runs up the side of the valley and falls from side to side, showing that there is significant vibrational motion. We can therefore conclude that reactions with attractive potential energy surfaces proceed more efficiently if the energy is in relative translational motion.

In contrast, reactions with repulsive potential surfaces experience a late transition state and are expected to proceed with greater efficiency if the excess energy is present as vibrations. The energy is in the vibration of the reactant molecule and the motion causes the trajectory to move from side to side up the valley as it approaches the transition state. This energy is enough to tip the reactant molecules through the transition state and into the products, it is therefore apparent that the vibrational energy on the repulsive potential energy surface is key to the reactivity and formation of the products.

MRD:aja4117

2019-05-16T15:08:01Z

Aja4117: /* EXERCISE 2: F - H - H system */

= [[Molecular reaction dynamics|'''Molecular reaction dynamics''']] =

== 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?''''' ===
The transition state is a saddle point which is defined as the point which is both a minimum along one vector and a maximum along an orthogonal vector. Along the reaction path, the transition state is a maximum and along the second vector, which is perpendicular to our initial vector it is a minimum. This can be visualised in the diagram below.

For a co linear system, as given for the H + H2 system there are two orthogonal degrees of freedom with the coordinates given as follows:

'''Z = r2 + r1'''
 

<math>\frac {\partial V}{\partial Z} = 0 </math>, <math>\frac {\partial^2 V}{\partial Z^2}>0 </math>

 

'''S = r2 - r1'''

 

<math>\frac {\partial V}{\partial S} = 0 </math>, <math>\frac {\partial^2 V}{\partial S^2}<0 </math>

 

Note that these are the orthogonal vectors.

The saddle point has the configuration in which r1 = r2 =r0 and this will be the point at which these two lines intersect on the PES. This is the transition state. S is the tangent to the reaction pathway at the maximum whilst Z bisects the reaction pathway at the same point, but will be seen as a minimum.

[[File:Aja4117Question 1.png]]

=== ''Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'' ===
The best estimate was AB = 0.9077425 and BC = 0.9077425. There are only two lines when looking at internuclear distances vs time as seen below. This indicates that two lines are at the same distance and one is covering the other. 

To locate the transition state, is to begin the reaction trajectories close to the transition state. Due to the transition state being a global maxima, the trajectories will either form the reactants or products in order to find the lowest energy configuration. Whether the product or reactants are formed is directly influenced by the trajectories. If the trajectories are closer to the reactants then the reactants will form and if they are closer to the products then the products will form.

[[File:Aja4117INT for HH.png]]

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

===== Trajectories from r1 = rts+δ, r2 = rts =====
Along the mep, because the atoms are moving infinitely slowly and the momenta and the velocities always reset to zero, there is no oscillation and hence no vibration. This is seen along the mep where there is no wavy lines. In contrast, when looking at the dynamics, there is oscillation as post collision there is transfer of kinetic energy that causes the molecule to vibrate. 

When considering the inter nuclear distance vs time, A-C and A-B distances increase rapidly over the first 250 steps after which the rate of increase of the distance begins to decrease. As the number of steps increases the rate of increase of distance begins to plateau as the gradient flattens. This can be rationalised when looking at the animation where post collision, the molecules initially rapidly move away from one another but the speed at which they move away from each other falls off with time. When considering the dynamic situation, A-C and A-B distances increase at a constant gradient over the course of the plot. The reasoning is that in the dynamic situation, the velocities and momenta are not reset to zero at each step.

[[File:Aja4117(dynamics type)Mep vs dynamics distance v time 1.png]]
[[File:Aja4117Distance vs time 1.png]]

For the same reason as stated as above, the momenta and velocities are reset to zero at each point this means that the momentum vs time plot is a straight line for the mep . However, when running the dynamic situation we see an initial sharp increase in the momentum followed by a constant oscillation of momentum that is due to the momentum being retained within the vibrational modes. For A-B, there is a sharp increase in the momentum which plateaus at the same instance as B-C begins to undergo constant oscillation. This is because A-B refers to the momentum of the system A-B. Once both reach constant momentum the combined system now has a set defined level of momentum with no way to lose it.

[[File:Aja4117(dynamics type)Mep vs dynamics momenta vs time1.png]]
[[File:Aja4117dynamics mom vs time1.png]]

===== Trajectories from r1 = rts, r2 = rts +δ =====
The result of switching the distances is that

<nowiki> </nowiki>Take note of the final values of the positions '''r1'''(t) '''r2'''(t) and '''p1'''(t) '''p2'''(t) for your trajectory for large enough t.

== Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed. ==

== What do you observe? STILL TO DO ==

=== Reactive and unreactive trajectories ===
{| class="wikitable" border="1"
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics
!Plots of trajectory
|-
| -1.25 || -2.5 ||-99.018||Yes||Reaction proceeds with minimal vibration||[[File:-1.25, -2.5.png]]
|
|-
| -1.5 || -2.0 ||-100.456||No||P2 does not have enough momentum to reach P1||[[File:-1.5, -2.0.png]]
|
|-
| -1.5 || -2.5 ||-98.956||Yes||Transfer of vibrational energy||[[File:-1.5, -2.5.png]]
|
|-
| -2.5 || -5.0 || -84.956
||No||Product forms but contains too much vibrational energy so dissociates||[[File:-2.5, -5.0.png ]]
|
|-
| -2.5 || -5.2 || -83.416
||Yes||Reaction does occur but recrosses the transition state barrier||[[File:-2.5,5.2.png]]
|
|}
The fact that the last experiment recrosses the transition state barrier is evidence of the breakdown of Transition State Theory, which is presented below. this states that once products are formed, the system cannot travel back through the transition state to the reactants.

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

There are two basic assumptions for Transition State Theory, the two most basic are the separation of the electronic and nuclear motions which is the same as the Born-Oppenheimer approximation from quantum mechanics. The second basic assumption is that the reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution.

There are however, additional assumptions and they are as follows:
# Molecular systems that have crossed the transition state in the direction of the products cannot turn around and reform reactants.
# In the transition state, motion along the reaction coordinate may be separated from the other motions and be treated classically as a translation.
# Even in the absence of an equilibrium between the reactant and product molecules, the transition states that are become products are distributed among their states according to the Maxwell-Boltzmann laws.

''Given that Transition State theory neglects quantum mechanical contributions such as influence on and between between electronic and nuclear motions and the idea that molecules passing through the transition state through to the products cannot pass back through to the reactants, this would suggest that there would be deviation between experimental and calculated values using Transition State theory.''

== EXERCISE 2: F - H - H 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?'' ====
Looking at the PES of F + H2 this indicated the reaction is exothermic because there is a decrease in the energy going from the reactants to products. This can be seen along the reaction path for this reaction where there is a dip in energy on the path to the formation of the products. This makes intuitive sense as the driving force of the reaction will be the formation of a strong F-H bond, in place of a weaker H-H bond. Additional evidence that the reaction is exothermic is that when a mep calculation was run with RAB = 1.812 and RBC = 0.74 with zero momentum for each case, the reaction proceeded and the products formed with no additional input of energy.

For the H + HF reaction, '''the reaction is surmised to be endothermic due to as there is no decrease in energy on forming the products. This is in line with intuition considering the strength of the H-F bond. The very high electronegativity of fluorine would make it so that significant energy would be required to cause this reaction to proceed. '''

==== ''Locate the approximate position of the transition state.'' ====
{| class="wikitable"
!Reaction
!rAB
!rBC
!Activation Energy/ Kj/mol
|-
|F + H2
|1.8108757
|0.7448759
|0.2673
|-
|H + HF
|0.7448756
|1.8108759
|30.251
|}
The positions of the transition states were located through experimentation of the initial AB and BC distances, until the forces acting along AB and BC were zero. The reasoning behind this comes from the notion that

</math> F = \frac {\partial V}{\partial r} = 0 </math>

 

where F is the force

[[File:Aja4117Surface Plot f + h2.png]]
[[File:Aja4117Surface Plot h + HF.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.'' ====

The set of reaction conditions that were used were:

Atom A: F

Atom B: H

Atom C: H

Distance AB: 2.25

Distance BC: 0.74

Momentum AB = -0.8

Momentum BC = 0.1

These conditions yielded a set of successful reactions, and post reaction and upon formation of F-H, there is transfer of energy in the form of kinetic energy to both the Hydrogen that is travelling away from the reaction complex and also kinetic energy in the bond vibration of F-H. This can be visualised in the Momenta vs Time graph below. Monitoring the emission spectrum of the product and observing electromagnetic radiation in the IR region would allow identification of the product.

[[File:Aja4117Surface Plot FH2 momenta vs time.png]]

For rBC = 0.74, with a momentum pAB = -0.5, and several values of pBC in the range -3 to 3 were explored. For pBC = -3, the reaction did proceed and there was significant transfer of kinetic energy to the product F-H. What is interesting to note is that the F-H bond forms twice. On the first formation, the bond breaks due to what is assumed to be, too high a vibrational energy. This leads to the reformation of the H-H bond, upon which there is still significant vibrational energy and combined with the electronegativity of the F atom causes the H-H bond to break and the F-H bond to form for the second time. The F-H bond is now stable, although still containing significant vibrational energy. For pBC = 3, the same phenomena occurs as for pBC = -3. 

==== ''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.'' ====
A transition state for an attractive potential energy surface occurs early in the reaction coordinate and involves high translational kinetic energy (evidenced by the random motion for the reactants) and results in a vibrationally excited product. This translational energy runs up the side of the valley and falls from side to side, showing that there is significant vibrational motion. We can therefore conclude that reactions with attractive potential energy surfaces proceed more efficiently if the energy is in relative translational motion.

In contrast, reactions with repulsive potential surfaces experience a late transition state and are expected to proceed with greater efficiency if the excess energy is present as vibrations. The energy is in the vibration of the reactant molecule and the motion causes the trajectory to move from side to side up the valley as it approaches the transition state. This energy is enough to tip the reactant molecules through the transition state and into the products, it is therefore apparent that the vibrational energy on the repulsive potential energy surface is key to the reactivity and formation of the products.

MRD:aja4117

2019-05-16T15:07:15Z

Aja4117: /* Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? */

= [[Molecular reaction dynamics|'''Molecular reaction dynamics''']] =

== 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?''''' ===
The transition state is a saddle point which is defined as the point which is both a minimum along one vector and a maximum along an orthogonal vector. Along the reaction path, the transition state is a maximum and along the second vector, which is perpendicular to our initial vector it is a minimum. This can be visualised in the diagram below.

For a co linear system, as given for the H + H2 system there are two orthogonal degrees of freedom with the coordinates given as follows:

'''Z = r2 + r1'''
 

<math>\frac {\partial V}{\partial Z} = 0 </math>, <math>\frac {\partial^2 V}{\partial Z^2}>0 </math>

 

'''S = r2 - r1'''

 

<math>\frac {\partial V}{\partial S} = 0 </math>, <math>\frac {\partial^2 V}{\partial S^2}<0 </math>

 

Note that these are the orthogonal vectors.

The saddle point has the configuration in which r1 = r2 =r0 and this will be the point at which these two lines intersect on the PES. This is the transition state. S is the tangent to the reaction pathway at the maximum whilst Z bisects the reaction pathway at the same point, but will be seen as a minimum.

[[File:Aja4117Question 1.png]]

=== ''Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'' ===
The best estimate was AB = 0.9077425 and BC = 0.9077425. There are only two lines when looking at internuclear distances vs time as seen below. This indicates that two lines are at the same distance and one is covering the other. 

To locate the transition state, is to begin the reaction trajectories close to the transition state. Due to the transition state being a global maxima, the trajectories will either form the reactants or products in order to find the lowest energy configuration. Whether the product or reactants are formed is directly influenced by the trajectories. If the trajectories are closer to the reactants then the reactants will form and if they are closer to the products then the products will form.

[[File:Aja4117INT for HH.png]]

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

===== Trajectories from r1 = rts+δ, r2 = rts =====
Along the mep, because the atoms are moving infinitely slowly and the momenta and the velocities always reset to zero, there is no oscillation and hence no vibration. This is seen along the mep where there is no wavy lines. In contrast, when looking at the dynamics, there is oscillation as post collision there is transfer of kinetic energy that causes the molecule to vibrate. 

When considering the inter nuclear distance vs time, A-C and A-B distances increase rapidly over the first 250 steps after which the rate of increase of the distance begins to decrease. As the number of steps increases the rate of increase of distance begins to plateau as the gradient flattens. This can be rationalised when looking at the animation where post collision, the molecules initially rapidly move away from one another but the speed at which they move away from each other falls off with time. When considering the dynamic situation, A-C and A-B distances increase at a constant gradient over the course of the plot. The reasoning is that in the dynamic situation, the velocities and momenta are not reset to zero at each step.

[[File:Aja4117(dynamics type)Mep vs dynamics distance v time 1.png]]
[[File:Aja4117Distance vs time 1.png]]

For the same reason as stated as above, the momenta and velocities are reset to zero at each point this means that the momentum vs time plot is a straight line for the mep . However, when running the dynamic situation we see an initial sharp increase in the momentum followed by a constant oscillation of momentum that is due to the momentum being retained within the vibrational modes. For A-B, there is a sharp increase in the momentum which plateaus at the same instance as B-C begins to undergo constant oscillation. This is because A-B refers to the momentum of the system A-B. Once both reach constant momentum the combined system now has a set defined level of momentum with no way to lose it.

[[File:Aja4117(dynamics type)Mep vs dynamics momenta vs time1.png]]
[[File:Aja4117dynamics mom vs time1.png]]

===== Trajectories from r1 = rts, r2 = rts +δ =====
The result of switching the distances is that

<nowiki> </nowiki>Take note of the final values of the positions '''r1'''(t) '''r2'''(t) and '''p1'''(t) '''p2'''(t) for your trajectory for large enough t.

== Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed. ==

== What do you observe? STILL TO DO ==

=== Reactive and unreactive trajectories ===
{| class="wikitable" border="1"
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics
!Plots of trajectory
|-
| -1.25 || -2.5 ||-99.018||Yes||Reaction proceeds with minimal vibration||[[File:-1.25, -2.5.png]]
|
|-
| -1.5 || -2.0 ||-100.456||No||P2 does not have enough momentum to reach P1||[[File:-1.5, -2.0.png]]
|
|-
| -1.5 || -2.5 ||-98.956||Yes||Transfer of vibrational energy||[[File:-1.5, -2.5.png]]
|
|-
| -2.5 || -5.0 || -84.956
||No||Product forms but contains too much vibrational energy so dissociates||[[File:-2.5, -5.0.png ]]
|
|-
| -2.5 || -5.2 || -83.416
||Yes||Reaction does occur but recrosses the transition state barrier||[[File:-2.5,5.2.png]]
|
|}
The fact that the last experiment recrosses the transition state barrier is evidence of the breakdown of Transition State Theory, which is presented below. this states that once products are formed, the system cannot travel back through the transition state to the reactants.

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

There are two basic assumptions for Transition State Theory, the two most basic are the separation of the electronic and nuclear motions which is the same as the Born-Oppenheimer approximation from quantum mechanics. The second basic assumption is that the reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution.

There are however, additional assumptions and they are as follows:
# Molecular systems that have crossed the transition state in the direction of the products cannot turn around and reform reactants.
# In the transition state, motion along the reaction coordinate may be separated from the other motions and be treated classically as a translation.
# Even in the absence of an equilibrium between the reactant and product molecules, the transition states that are become products are distributed among their states according to the Maxwell-Boltzmann laws.

''Given that Transition State theory neglects quantum mechanical contributions such as influence on and between between electronic and nuclear motions and the idea that molecules passing through the transition state through to the products cannot pass back through to the reactants, this would suggest that there would be deviation between experimental and calculated values using Transition State theory.''

== EXERCISE 2: F - H - H 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?'' ====
Looking at the PES of F + H2 this indicated the reaction is exothermic because there is a decrease in the energy going from the reactants to products. This can be seen along the reaction path for this reaction where there is a dip in energy on the path to the formation of the products. This makes intuitive sense as the driving force of the reaction will be the formation of a strong F-H bond, in place of a weaker H-H bond. Additional evidence that the reaction is exothermic is that when a mep calculation was run with RAB = 1.812 and RBC = 0.74 with zero momentum for each case, the reaction proceeded and the products formed with no additional input of energy.

For the H + HF reaction, '''the reaction is surmised to be endothermic due to as there is no decrease in energy on forming the products. This is in line with intuition considering the strength of the H-F bond. The very high electronegativity of fluorine would make it so that significant energy would be required to cause this reaction to proceed. '''

==== ''Locate the approximate position of the transition state.'' ====
{| class="wikitable"
!Reaction
!rAB
!rBC
!Activation Energy/ Kj/mol
|-
|F + H2
|1.8108757
|0.7448759
|0.2673
|-
|H + HF
|0.7448756
|1.8108759
|30.251
|}
The positions of the transition states were located through experimentation of the initial AB and BC distances, until the forces acting along AB and BC were zero. The reasoning behind this comes from the notion that

F = <math>\frac {\partial V}{\partial r} = 0

 

where F is the force

[[File:Aja4117Surface Plot f + h2.png]]
[[File:Aja4117Surface Plot h + HF.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.'' ====

The set of reaction conditions that were used were:

Atom A: F

Atom B: H

Atom C: H

Distance AB: 2.25

Distance BC: 0.74

Momentum AB = -0.8

Momentum BC = 0.1

These conditions yielded a set of successful reactions, and post reaction and upon formation of F-H, there is transfer of energy in the form of kinetic energy to both the Hydrogen that is travelling away from the reaction complex and also kinetic energy in the bond vibration of F-H. This can be visualised in the Momenta vs Time graph below. Monitoring the emission spectrum of the product and observing electromagnetic radiation in the IR region would allow identification of the product.

[[File:Aja4117Surface Plot FH2 momenta vs time.png]]

For rBC = 0.74, with a momentum pAB = -0.5, and several values of pBC in the range -3 to 3 were explored. For pBC = -3, the reaction did proceed and there was significant transfer of kinetic energy to the product F-H. What is interesting to note is that the F-H bond forms twice. On the first formation, the bond breaks due to what is assumed to be, too high a vibrational energy. This leads to the reformation of the H-H bond, upon which there is still significant vibrational energy and combined with the electronegativity of the F atom causes the H-H bond to break and the F-H bond to form for the second time. The F-H bond is now stable, although still containing significant vibrational energy. For pBC = 3, the same phenomena occurs as for pBC = -3. 

==== ''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.'' ====
A transition state for an attractive potential energy surface occurs early in the reaction coordinate and involves high translational kinetic energy (evidenced by the random motion for the reactants) and results in a vibrationally excited product. This translational energy runs up the side of the valley and falls from side to side, showing that there is significant vibrational motion. We can therefore conclude that reactions with attractive potential energy surfaces proceed more efficiently if the energy is in relative translational motion.

In contrast, reactions with repulsive potential surfaces experience a late transition state and are expected to proceed with greater efficiency if the excess energy is present as vibrations. The energy is in the vibration of the reactant molecule and the motion causes the trajectory to move from side to side up the valley as it approaches the transition state. This energy is enough to tip the reactant molecules through the transition state and into the products, it is therefore apparent that the vibrational energy on the repulsive potential energy surface is key to the reactivity and formation of the products.

MRD:aja4117

2019-05-16T15:07:01Z

Aja4117: /* EXERCISE 2: F - H - H system */

= [[Molecular reaction dynamics|'''Molecular reaction dynamics''']] =

== 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?''''' ===
The transition state is a saddle point which is defined as the point which is both a minimum along one vector and a maximum along an orthogonal vector. Along the reaction path, the transition state is a maximum and along the second vector, which is perpendicular to our initial vector it is a minimum. This can be visualised in the diagram below.

For a co linear system, as given for the H + H2 system there are two orthogonal degrees of freedom with the coordinates given as follows:

'''Z = r2 + r1'''
 

<math>\frac {\partial V}{\partial Z} = 0 </math>, <math>\frac {\partial^2 V}{\partial Z^2}>0 </math>

 

'''S = r2 - r1'''

 

<math>\frac {\partial V}{\partial S} = 0 </math>, <math>\frac {\partial^2 V}{\partial S^2}<0 </math>

 

Note that these are the orthogonal vectors.

The saddle point has the configuration in which r1 = r2 =r0 and this will be the point at which these two lines intersect on the PES. This is the transition state. S is the tangent to the reaction pathway at the maximum whilst Z bisects the reaction pathway at the same point, but will be seen as a minimum.

[[File:Aja4117Question 1.png]]

=== ''Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'' ===
The best estimate was AB = 0.9077425 and BC = 0.9077425. There are only two lines when looking at internuclear distances vs time as seen below. This indicates that two lines are at the same distance and one is covering the other. 

To locate the transition state, is to begin the reaction trajectories close to the transition state. Due to the transition state being a global maxima, the trajectories will either form the reactants or products in order to find the lowest energy configuration. Whether the product or reactants are formed is directly influenced by the trajectories. If the trajectories are closer to the reactants then the reactants will form and if they are closer to the products then the products will form.

[[File:Aja4117INT for HH.png]]

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

===== Trajectories from r1 = rts+δ, r2 = rts =====
Along the mep, because the atoms are moving infinitely slowly and the momenta and the velocities always reset to zero, there is no oscillation and hence no vibration. This is seen along the mep where there is no wavy lines. In contrast, when looking at the dynamics, there is oscillation as post collision there is transfer of kinetic energy that causes the molecule to vibrate. 

When considering the inter nuclear distance vs time, A-C and A-B distances increase rapidly over the first 250 steps after which the rate of increase of the distance begins to decrease. As the number of steps increases the rate of increase of distance begins to plateau as the gradient flattens. This can be rationalised when looking at the animation where post collision, the molecules initially rapidly move away from one another but the speed at which they move away from each other falls off with time. When considering the dynamic situation, A-C and A-B distances increase at a constant gradient over the course of the plot. The reasoning is that in the dynamic situation, the velocities and momenta are not reset to zero at each step.

[[File:Aja4117(dynamics type)Mep vs dynamics distance v time 1.png]]
[[File:Aja4117Distance vs time 1.png]]

For the same reason as stated as above, the momenta and velocities are reset to zero at each point this means that the momentum vs time plot is a straight line for the mep . However, when running the dynamic situation we see an initial sharp increase in the momentum followed by a constant oscillation of momentum that is due to the momentum being retained within the vibrational modes. For A-B, there is a sharp increase in the momentum which plateaus at the same instance as B-C begins to undergo constant oscillation. This is because A-B refers to the momentum of the system A-B. Once both reach constant momentum the combined system now has a set defined level of momentum with no way to lose it.

[[File:Aja4117(dynamics type)Mep vs dynamics momenta vs time1.png]]
[[File:Aja4117dynamics mom vs time1.png]]

===== Trajectories from r1 = rts, r2 = rts +δ =====
The result of switching the distances is that

<nowiki> </nowiki>Take note of the final values of the positions '''r1'''(t) '''r2'''(t) and '''p1'''(t) '''p2'''(t) for your trajectory for large enough t.

== Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed. ==

== What do you observe? STILL TO DO ==

=== Reactive and unreactive trajectories ===
{| class="wikitable" border="1"
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics
!Plots of trajectory
|-
| -1.25 || -2.5 ||-99.018||Yes||Reaction proceeds with minimal vibration||[[File:-1.25, -2.5.png]]
|
|-
| -1.5 || -2.0 ||-100.456||No||P2 does not have enough momentum to reach P1||[[File:-1.5, -2.0.png]]
|
|-
| -1.5 || -2.5 ||-98.956||Yes||Transfer of vibrational energy||[[File:-1.5, -2.5.png]]
|
|-
| -2.5 || -5.0 || -84.956
||No||Product forms but contains too much vibrational energy so dissociates||[[File:-2.5, -5.0.png ]]
|
|-
| -2.5 || -5.2 || -83.416
||Yes||Reaction does occur but recrosses the transition state barrier||[[File:-2.5,5.2.png]]
|
|}
The fact that the last experiment recrosses the transition state barrier is evidence of the breakdown of Transition State Theory, which is presented below. this states that once products are formed, the system cannot travel back through the transition state to the reactants.

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

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

There are two basic assumptions for Transition State Theory, the two most basic are the separation of the electronic and nuclear motions which is the same as the Born-Oppenheimer approximation from quantum mechanics. The second basic assumption is that the reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution.

There are however, additional assumptions and they are as follows:
# Molecular systems that have crossed the transition state in the direction of the products cannot turn around and reform reactants.
# In the transition state, motion along the reaction coordinate may be separated from the other motions and be treated classically as a translation.
# Even in the absence of an equilibrium between the reactant and product molecules, the transition states that are become products are distributed among their states according to the Maxwell-Boltzmann laws.

''Given that Transition State theory neglects quantum mechanical contributions such as influence on and between between electronic and nuclear motions and the idea that molecules passing through the transition state through to the products cannot pass back through to the reactants, this would suggest that there would be deviation between experimental and calculated values using Transition State theory.''

== EXERCISE 2: F - H - H 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?'' ====
Looking at the PES of F + H2 this indicated the reaction is exothermic because there is a decrease in the energy going from the reactants to products. This can be seen along the reaction path for this reaction where there is a dip in energy on the path to the formation of the products. This makes intuitive sense as the driving force of the reaction will be the formation of a strong F-H bond, in place of a weaker H-H bond. Additional evidence that the reaction is exothermic is that when a mep calculation was run with RAB = 1.812 and RBC = 0.74 with zero momentum for each case, the reaction proceeded and the products formed with no additional input of energy.

For the H + HF reaction, '''the reaction is surmised to be endothermic due to as there is no decrease in energy on forming the products. This is in line with intuition considering the strength of the H-F bond. The very high electronegativity of fluorine would make it so that significant energy would be required to cause this reaction to proceed. '''

==== ''Locate the approximate position of the transition state.'' ====
{| class="wikitable"
!Reaction
!rAB
!rBC
!Activation Energy/ Kj/mol
|-
|F + H2
|1.8108757
|0.7448759
|0.2673
|-
|H + HF
|0.7448756
|1.8108759
|30.251
|}
The positions of the transition states were located through experimentation of the initial AB and BC distances, until the forces acting along AB and BC were zero. The reasoning behind this comes from the notion that

F = <math>\frac {\partial V}{\partial r} = 0

 

where F is the force

[[File:Aja4117Surface Plot f + h2.png]]
[[File:Aja4117Surface Plot h + HF.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.'' ====

The set of reaction conditions that were used were:

Atom A: F

Atom B: H

Atom C: H

Distance AB: 2.25

Distance BC: 0.74

Momentum AB = -0.8

Momentum BC = 0.1

These conditions yielded a set of successful reactions, and post reaction and upon formation of F-H, there is transfer of energy in the form of kinetic energy to both the Hydrogen that is travelling away from the reaction complex and also kinetic energy in the bond vibration of F-H. This can be visualised in the Momenta vs Time graph below. Monitoring the emission spectrum of the product and observing electromagnetic radiation in the IR region would allow identification of the product.

[[File:Aja4117Surface Plot FH2 momenta vs time.png]]

For rBC = 0.74, with a momentum pAB = -0.5, and several values of pBC in the range -3 to 3 were explored. For pBC = -3, the reaction did proceed and there was significant transfer of kinetic energy to the product F-H. What is interesting to note is that the F-H bond forms twice. On the first formation, the bond breaks due to what is assumed to be, too high a vibrational energy. This leads to the reformation of the H-H bond, upon which there is still significant vibrational energy and combined with the electronegativity of the F atom causes the H-H bond to break and the F-H bond to form for the second time. The F-H bond is now stable, although still containing significant vibrational energy. For pBC = 3, the same phenomena occurs as for pBC = -3. 

==== ''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.'' ====
A transition state for an attractive potential energy surface occurs early in the reaction coordinate and involves high translational kinetic energy (evidenced by the random motion for the reactants) and results in a vibrationally excited product. This translational energy runs up the side of the valley and falls from side to side, showing that there is significant vibrational motion. We can therefore conclude that reactions with attractive potential energy surfaces proceed more efficiently if the energy is in relative translational motion.

In contrast, reactions with repulsive potential surfaces experience a late transition state and are expected to proceed with greater efficiency if the excess energy is present as vibrations. The energy is in the vibration of the reactant molecule and the motion causes the trajectory to move from side to side up the valley as it approaches the transition state. This energy is enough to tip the reactant molecules through the transition state and into the products, it is therefore apparent that the vibrational energy on the repulsive potential energy surface is key to the reactivity and formation of the products.

MRD:aja4117

2019-05-16T15:06:04Z

Aja4117: /* F + H2 */

= [[Molecular reaction dynamics|'''Molecular reaction dynamics''']] =

== 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?''''' ===
The transition state is a saddle point which is defined as the point which is both a minimum along one vector and a maximum along an orthogonal vector. Along the reaction path, the transition state is a maximum and along the second vector, which is perpendicular to our initial vector it is a minimum. This can be visualised in the diagram below.

For a co linear system, as given for the H + H2 system there are two orthogonal degrees of freedom with the coordinates given as follows:

'''Z = r2 + r1'''
 

<math>\frac {\partial V}{\partial Z} = 0 </math>, <math>\frac {\partial^2 V}{\partial Z^2}>0 </math>

 

'''S = r2 - r1'''

 

<math>\frac {\partial V}{\partial S} = 0 </math>, <math>\frac {\partial^2 V}{\partial S^2}<0 </math>

 

Note that these are the orthogonal vectors.

The saddle point has the configuration in which r1 = r2 =r0 and this will be the point at which these two lines intersect on the PES. This is the transition state. S is the tangent to the reaction pathway at the maximum whilst Z bisects the reaction pathway at the same point, but will be seen as a minimum.

[[File:Aja4117Question 1.png]]

=== ''Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'' ===
The best estimate was AB = 0.9077425 and BC = 0.9077425. There are only two lines when looking at internuclear distances vs time as seen below. This indicates that two lines are at the same distance and one is covering the other. 

To locate the transition state, is to begin the reaction trajectories close to the transition state. Due to the transition state being a global maxima, the trajectories will either form the reactants or products in order to find the lowest energy configuration. Whether the product or reactants are formed is directly influenced by the trajectories. If the trajectories are closer to the reactants then the reactants will form and if they are closer to the products then the products will form.

[[File:Aja4117INT for HH.png]]

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

===== Trajectories from r1 = rts+δ, r2 = rts =====
Along the mep, because the atoms are moving infinitely slowly and the momenta and the velocities always reset to zero, there is no oscillation and hence no vibration. This is seen along the mep where there is no wavy lines. In contrast, when looking at the dynamics, there is oscillation as post collision there is transfer of kinetic energy that causes the molecule to vibrate. 

When considering the inter nuclear distance vs time, A-C and A-B distances increase rapidly over the first 250 steps after which the rate of increase of the distance begins to decrease. As the number of steps increases the rate of increase of distance begins to plateau as the gradient flattens. This can be rationalised when looking at the animation where post collision, the molecules initially rapidly move away from one another but the speed at which they move away from each other falls off with time. When considering the dynamic situation, A-C and A-B distances increase at a constant gradient over the course of the plot. The reasoning is that in the dynamic situation, the velocities and momenta are not reset to zero at each step.

[[File:Aja4117(dynamics type)Mep vs dynamics distance v time 1.png]]
[[File:Aja4117Distance vs time 1.png]]

For the same reason as stated as above, the momenta and velocities are reset to zero at each point this means that the momentum vs time plot is a straight line for the mep . However, when running the dynamic situation we see an initial sharp increase in the momentum followed by a constant oscillation of momentum that is due to the momentum being retained within the vibrational modes. For A-B, there is a sharp increase in the momentum which plateaus at the same instance as B-C begins to undergo constant oscillation. This is because A-B refers to the momentum of the system A-B. Once both reach constant momentum the combined system now has a set defined level of momentum with no way to lose it.

[[File:Aja4117(dynamics type)Mep vs dynamics momenta vs time1.png]]
[[File:Aja4117dynamics mom vs time1.png]]

===== Trajectories from r1 = rts, r2 = rts +δ =====
The result of switching the distances is that

<nowiki> </nowiki>Take note of the final values of the positions '''r1'''(t) '''r2'''(t) and '''p1'''(t) '''p2'''(t) for your trajectory for large enough t.

== Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed. ==

== What do you observe? STILL TO DO ==

=== Reactive and unreactive trajectories ===
{| class="wikitable" border="1"
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics
!Plots of trajectory
|-
| -1.25 || -2.5 ||-99.018||Yes||Reaction proceeds with minimal vibration||[[File:-1.25, -2.5.png]]
|
|-
| -1.5 || -2.0 ||-100.456||No||P2 does not have enough momentum to reach P1||[[File:-1.5, -2.0.png]]
|
|-
| -1.5 || -2.5 ||-98.956||Yes||Transfer of vibrational energy||[[File:-1.5, -2.5.png]]
|
|-
| -2.5 || -5.0 || -84.956
||No||Product forms but contains too much vibrational energy so dissociates||[[File:-2.5, -5.0.png ]]
|
|-
| -2.5 || -5.2 || -83.416
||Yes||Reaction does occur but recrosses the transition state barrier||[[File:-2.5,5.2.png]]
|
|}
The fact that the last experiment recrosses the transition state barrier is evidence of the breakdown of Transition State Theory, which is presented below. this states that once products are formed, the system cannot travel back through the transition state to the reactants.

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

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

There are two basic assumptions for Transition State Theory, the two most basic are the separation of the electronic and nuclear motions which is the same as the Born-Oppenheimer approximation from quantum mechanics. The second basic assumption is that the reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution.

There are however, additional assumptions and they are as follows:
# Molecular systems that have crossed the transition state in the direction of the products cannot turn around and reform reactants.
# In the transition state, motion along the reaction coordinate may be separated from the other motions and be treated classically as a translation.
# Even in the absence of an equilibrium between the reactant and product molecules, the transition states that are become products are distributed among their states according to the Maxwell-Boltzmann laws.

''Given that Transition State theory neglects quantum mechanical contributions such as influence on and between between electronic and nuclear motions and the idea that molecules passing through the transition state through to the products cannot pass back through to the reactants, this would suggest that there would be deviation between experimental and calculated values using Transition State theory.''

== EXERCISE 2: F - H - H 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?'' ====
Looking at the PES of F + H2 this indicated the reaction is exothermic because there is a decrease in the energy going from the reactants to products. This can be seen along the reaction path for this reaction where there is a dip in energy on the path to the formation of the products. This makes intuitive sense as the driving force of the reaction will be the formation of a strong F-H bond, in place of a weaker H-H bond. Additional evidence that the reaction is exothermic is that when a mep calculation was run with RAB = 1.812 and RBC = 0.74 with zero momentum for each case, the reaction proceeded and the products formed with no additional input of energy.

For the H + HF reaction, '''the reaction is surmised to be endothermic due to as there is no decrease in energy on forming the products. This is in line with intuition considering the strength of the H-F bond. The very high electronegativity of fluorine would make it so that significant energy would be required to cause this reaction to proceed. '''

==== ''Locate the approximate position of the transition state.'' ====
{| class="wikitable"
!Reaction
!rAB
!rBC
!Activation Energy/ Kj/mol
|-
|F + H2
|1.8108757
|0.7448759
|0.2673
|-
|H + HF
|0.7448756
|1.8108759
|30.251
|}
The positions of the transition states were located through experimentation of the initial AB and BC distances, until the forces acting along AB and BC were zero. The reasoning behind this comes from the notion that

<math>F = \frac {\partial V}{\partial r} = 0

 

where F is the force

[[File:Aja4117Surface Plot f + h2.png]]
[[File:Aja4117Surface Plot h + HF.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.'' ====

The set of reaction conditions that were used were:

Atom A: F

Atom B: H

Atom C: H

Distance AB: 2.25

Distance BC: 0.74

Momentum AB = -0.8

Momentum BC = 0.1

These conditions yielded a set of successful reactions, and post reaction and upon formation of F-H, there is transfer of energy in the form of kinetic energy to both the Hydrogen that is travelling away from the reaction complex and also kinetic energy in the bond vibration of F-H. This can be visualised in the Momenta vs Time graph below. Monitoring the emission spectrum of the product and observing electromagnetic radiation in the IR region would allow identification of the product.

[[File:Aja4117Surface Plot FH2 momenta vs time.png]]

For rBC = 0.74, with a momentum pAB = -0.5, and several values of pBC in the range -3 to 3 were explored. For pBC = -3, the reaction did proceed and there was significant transfer of kinetic energy to the product F-H. What is interesting to note is that the F-H bond forms twice. On the first formation, the bond breaks due to what is assumed to be, too high a vibrational energy. This leads to the reformation of the H-H bond, upon which there is still significant vibrational energy and combined with the electronegativity of the F atom causes the H-H bond to break and the F-H bond to form for the second time. The F-H bond is now stable, although still containing significant vibrational energy. For pBC = 3, the same phenomena occurs as for pBC = -3. 

==== ''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.'' ====
A transition state for an attractive potential energy surface occurs early in the reaction coordinate and involves high translational kinetic energy (evidenced by the random motion for the reactants) and results in a vibrationally excited product. This translational energy runs up the side of the valley and falls from side to side, showing that there is significant vibrational motion. We can therefore conclude that reactions with attractive potential energy surfaces proceed more efficiently if the energy is in relative translational motion.

In contrast, reactions with repulsive potential surfaces experience a late transition state and are expected to proceed with greater efficiency if the excess energy is present as vibrations. The energy is in the vibration of the reactant molecule and the motion causes the trajectory to move from side to side up the valley as it approaches the transition state. This energy is enough to tip the reactant molecules through the transition state and into the products, it is therefore apparent that the vibrational energy on the repulsive potential energy surface is key to the reactivity and formation of the products.

MRD:aja4117

2019-05-16T15:02:07Z

Aja4117: /* Locate the approximate position of the transition state. */

= [[Molecular reaction dynamics|'''Molecular reaction dynamics''']] =

== 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?''''' ===
The transition state is a saddle point which is defined as the point which is both a minimum along one vector and a maximum along an orthogonal vector. Along the reaction path, the transition state is a maximum and along the second vector, which is perpendicular to our initial vector it is a minimum. This can be visualised in the diagram below.

For a co linear system, as given for the H + H2 system there are two orthogonal degrees of freedom with the coordinates given as follows:

'''Z = r2 + r1'''
 

<math>\frac {\partial V}{\partial Z} = 0 </math>, <math>\frac {\partial^2 V}{\partial Z^2}>0 </math>

 

'''S = r2 - r1'''

 

<math>\frac {\partial V}{\partial S} = 0 </math>, <math>\frac {\partial^2 V}{\partial S^2}<0 </math>

 

Note that these are the orthogonal vectors.

The saddle point has the configuration in which r1 = r2 =r0 and this will be the point at which these two lines intersect on the PES. This is the transition state. S is the tangent to the reaction pathway at the maximum whilst Z bisects the reaction pathway at the same point, but will be seen as a minimum.

[[File:Aja4117Question 1.png]]

=== ''Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'' ===
The best estimate was AB = 0.9077425 and BC = 0.9077425. There are only two lines when looking at internuclear distances vs time as seen below. This indicates that two lines are at the same distance and one is covering the other. 

To locate the transition state, is to begin the reaction trajectories close to the transition state. Due to the transition state being a global maxima, the trajectories will either form the reactants or products in order to find the lowest energy configuration. Whether the product or reactants are formed is directly influenced by the trajectories. If the trajectories are closer to the reactants then the reactants will form and if they are closer to the products then the products will form.

[[File:Aja4117INT for HH.png]]

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

===== Trajectories from r1 = rts+δ, r2 = rts =====
Along the mep, because the atoms are moving infinitely slowly and the momenta and the velocities always reset to zero, there is no oscillation and hence no vibration. This is seen along the mep where there is no wavy lines. In contrast, when looking at the dynamics, there is oscillation as post collision there is transfer of kinetic energy that causes the molecule to vibrate. 

When considering the inter nuclear distance vs time, A-C and A-B distances increase rapidly over the first 250 steps after which the rate of increase of the distance begins to decrease. As the number of steps increases the rate of increase of distance begins to plateau as the gradient flattens. This can be rationalised when looking at the animation where post collision, the molecules initially rapidly move away from one another but the speed at which they move away from each other falls off with time. When considering the dynamic situation, A-C and A-B distances increase at a constant gradient over the course of the plot. The reasoning is that in the dynamic situation, the velocities and momenta are not reset to zero at each step.

[[File:Aja4117(dynamics type)Mep vs dynamics distance v time 1.png]]
[[File:Aja4117Distance vs time 1.png]]

For the same reason as stated as above, the momenta and velocities are reset to zero at each point this means that the momentum vs time plot is a straight line for the mep . However, when running the dynamic situation we see an initial sharp increase in the momentum followed by a constant oscillation of momentum that is due to the momentum being retained within the vibrational modes. For A-B, there is a sharp increase in the momentum which plateaus at the same instance as B-C begins to undergo constant oscillation. This is because A-B refers to the momentum of the system A-B. Once both reach constant momentum the combined system now has a set defined level of momentum with no way to lose it.

[[File:Aja4117(dynamics type)Mep vs dynamics momenta vs time1.png]]
[[File:Aja4117dynamics mom vs time1.png]]

===== Trajectories from r1 = rts, r2 = rts +δ =====
The result of switching the distances is that

<nowiki> </nowiki>Take note of the final values of the positions '''r1'''(t) '''r2'''(t) and '''p1'''(t) '''p2'''(t) for your trajectory for large enough t.

== Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed. ==

== What do you observe? STILL TO DO ==

=== Reactive and unreactive trajectories ===
{| class="wikitable" border="1"
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics
!Plots of trajectory
|-
| -1.25 || -2.5 ||-99.018||Yes||Reaction proceeds with minimal vibration||[[File:-1.25, -2.5.png]]
|
|-
| -1.5 || -2.0 ||-100.456||No||P2 does not have enough momentum to reach P1||[[File:-1.5, -2.0.png]]
|
|-
| -1.5 || -2.5 ||-98.956||Yes||Transfer of vibrational energy||[[File:-1.5, -2.5.png]]
|
|-
| -2.5 || -5.0 || -84.956
||No||Product forms but contains too much vibrational energy so dissociates||[[File:-2.5, -5.0.png ]]
|
|-
| -2.5 || -5.2 || -83.416
||Yes||Reaction does occur but recrosses the transition state barrier||[[File:-2.5,5.2.png]]
|
|}
The fact that the last experiment recrosses the transition state barrier is evidence of the breakdown of Transition State Theory, which is presented below. this states that once products are formed, the system cannot travel back through the transition state to the reactants.

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

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

There are two basic assumptions for Transition State Theory, the two most basic are the separation of the electronic and nuclear motions which is the same as the Born-Oppenheimer approximation from quantum mechanics. The second basic assumption is that the reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution.

There are however, additional assumptions and they are as follows:
# Molecular systems that have crossed the transition state in the direction of the products cannot turn around and reform reactants.
# In the transition state, motion along the reaction coordinate may be separated from the other motions and be treated classically as a translation.
# Even in the absence of an equilibrium between the reactant and product molecules, the transition states that are become products are distributed among their states according to the Maxwell-Boltzmann laws.

''Given that Transition State theory neglects quantum mechanical contributions such as influence on and between between electronic and nuclear motions and the idea that molecules passing through the transition state through to the products cannot pass back through to the reactants, this would suggest that there would be deviation between experimental and calculated values using Transition State theory.''

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

=== ''F + H2'' ===

==== ''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?'' ====
Looking at the PES of F + H2 this indicated the reaction is exothermic because there is a decrease in the energy going from the reactants to products. This can be seen along the reaction path for this reaction where there is a dip in energy on the path to the formation of the products. This makes intuitive sense as the driving force of the reaction will be the formation of a strong F-H bond, in place of a weaker H-H bond. Additional evidence that the reaction is exothermic is that when a mep calculation was run with RAB = 1.812 and RBC = 0.74 with zero momentum for each case, the reaction proceeded and the products formed with no additional input of energy.

For the H + HF reaction, '''the reaction is surmised to be endothermic due to as there is no decrease in energy on forming the products. This is in line with intuition considering the strength of the H-F bond. The very high electronegativity of fluorine would make it so that significant energy would be required to cause this reaction to proceed. '''

==== ''Locate the approximate position of the transition state.'' ====
{| class="wikitable"
!Reaction
!rAB
!rBC
!Activation Energy/ Kj/mol
|-
|F + H2
|1.8108757
|0.7448759
|0.2673
|-
|H + HF
|0.7448756
|1.8108759
|30.251
|}

[[File:Aja4117Surface Plot f + h2.png]]
[[File:Aja4117Surface Plot h + HF.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.'' ====

The set of reaction conditions that were used were:

Atom A: F

Atom B: H

Atom C: H

Distance AB: 2.25

Distance BC: 0.74

Momentum AB = -0.8

Momentum BC = 0.1

These conditions yielded a set of successful reactions, and post reaction and upon formation of F-H, there is transfer of energy in the form of kinetic energy to both the Hydrogen that is travelling away from the reaction complex and also kinetic energy in the bond vibration of F-H. This can be visualised in the Momenta vs Time graph below. Monitoring the emission spectrum of the product and observing electromagnetic radiation in the IR region would allow identification of the product.

[[File:Aja4117Surface Plot FH2 momenta vs time.png]]

For rBC = 0.74, with a momentum pAB = -0.5, and several values of pBC in the range -3 to 3 were explored. For pBC = -3, the reaction did proceed and there was significant transfer of kinetic energy to the product F-H. What is interesting to note is that the F-H bond forms twice. On the first formation, the bond breaks due to what is assumed to be, too high a vibrational energy. This leads to the reformation of the H-H bond, upon which there is still significant vibrational energy and combined with the electronegativity of the F atom causes the H-H bond to break and the F-H bond to form for the second time. The F-H bond is now stable, although still containing significant vibrational energy. For pBC = 3, the same phenomena occurs as for pBC = -3. 

==== ''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.'' ====
A transition state for an attractive potential energy surface occurs early in the reaction coordinate and involves high translational kinetic energy (evidenced by the random motion for the reactants) and results in a vibrationally excited product. This translational energy runs up the side of the valley and falls from side to side, showing that there is significant vibrational motion. We can therefore conclude that reactions with attractive potential energy surfaces proceed more efficiently if the energy is in relative translational motion.

In contrast, reactions with repulsive potential surfaces experience a late transition state and are expected to proceed with greater efficiency if the excess energy is present as vibrations. The energy is in the vibration of the reactant molecule and the motion causes the trajectory to move from side to side up the valley as it approaches the transition state. This energy is enough to tip the reactant molecules through the transition state and into the products, it is therefore apparent that the vibrational energy on the repulsive potential energy surface is key to the reactivity and formation of the products.

MRD:aja4117

2019-05-16T15:01:17Z

Aja4117: /* Locate the approximate position of the transition state. */

= [[Molecular reaction dynamics|'''Molecular reaction dynamics''']] =

== 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?''''' ===
The transition state is a saddle point which is defined as the point which is both a minimum along one vector and a maximum along an orthogonal vector. Along the reaction path, the transition state is a maximum and along the second vector, which is perpendicular to our initial vector it is a minimum. This can be visualised in the diagram below.

For a co linear system, as given for the H + H2 system there are two orthogonal degrees of freedom with the coordinates given as follows:

'''Z = r2 + r1'''
 

<math>\frac {\partial V}{\partial Z} = 0 </math>, <math>\frac {\partial^2 V}{\partial Z^2}>0 </math>

 

'''S = r2 - r1'''

 

<math>\frac {\partial V}{\partial S} = 0 </math>, <math>\frac {\partial^2 V}{\partial S^2}<0 </math>

 

Note that these are the orthogonal vectors.

The saddle point has the configuration in which r1 = r2 =r0 and this will be the point at which these two lines intersect on the PES. This is the transition state. S is the tangent to the reaction pathway at the maximum whilst Z bisects the reaction pathway at the same point, but will be seen as a minimum.

[[File:Aja4117Question 1.png]]

=== ''Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'' ===
The best estimate was AB = 0.9077425 and BC = 0.9077425. There are only two lines when looking at internuclear distances vs time as seen below. This indicates that two lines are at the same distance and one is covering the other. 

To locate the transition state, is to begin the reaction trajectories close to the transition state. Due to the transition state being a global maxima, the trajectories will either form the reactants or products in order to find the lowest energy configuration. Whether the product or reactants are formed is directly influenced by the trajectories. If the trajectories are closer to the reactants then the reactants will form and if they are closer to the products then the products will form.

[[File:Aja4117INT for HH.png]]

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

===== Trajectories from r1 = rts+δ, r2 = rts =====
Along the mep, because the atoms are moving infinitely slowly and the momenta and the velocities always reset to zero, there is no oscillation and hence no vibration. This is seen along the mep where there is no wavy lines. In contrast, when looking at the dynamics, there is oscillation as post collision there is transfer of kinetic energy that causes the molecule to vibrate. 

When considering the inter nuclear distance vs time, A-C and A-B distances increase rapidly over the first 250 steps after which the rate of increase of the distance begins to decrease. As the number of steps increases the rate of increase of distance begins to plateau as the gradient flattens. This can be rationalised when looking at the animation where post collision, the molecules initially rapidly move away from one another but the speed at which they move away from each other falls off with time. When considering the dynamic situation, A-C and A-B distances increase at a constant gradient over the course of the plot. The reasoning is that in the dynamic situation, the velocities and momenta are not reset to zero at each step.

[[File:Aja4117(dynamics type)Mep vs dynamics distance v time 1.png]]
[[File:Aja4117Distance vs time 1.png]]

For the same reason as stated as above, the momenta and velocities are reset to zero at each point this means that the momentum vs time plot is a straight line for the mep . However, when running the dynamic situation we see an initial sharp increase in the momentum followed by a constant oscillation of momentum that is due to the momentum being retained within the vibrational modes. For A-B, there is a sharp increase in the momentum which plateaus at the same instance as B-C begins to undergo constant oscillation. This is because A-B refers to the momentum of the system A-B. Once both reach constant momentum the combined system now has a set defined level of momentum with no way to lose it.

[[File:Aja4117(dynamics type)Mep vs dynamics momenta vs time1.png]]
[[File:Aja4117dynamics mom vs time1.png]]

===== Trajectories from r1 = rts, r2 = rts +δ =====
The result of switching the distances is that

<nowiki> </nowiki>Take note of the final values of the positions '''r1'''(t) '''r2'''(t) and '''p1'''(t) '''p2'''(t) for your trajectory for large enough t.

== Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed. ==

== What do you observe? STILL TO DO ==

=== Reactive and unreactive trajectories ===
{| class="wikitable" border="1"
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics
!Plots of trajectory
|-
| -1.25 || -2.5 ||-99.018||Yes||Reaction proceeds with minimal vibration||[[File:-1.25, -2.5.png]]
|
|-
| -1.5 || -2.0 ||-100.456||No||P2 does not have enough momentum to reach P1||[[File:-1.5, -2.0.png]]
|
|-
| -1.5 || -2.5 ||-98.956||Yes||Transfer of vibrational energy||[[File:-1.5, -2.5.png]]
|
|-
| -2.5 || -5.0 || -84.956
||No||Product forms but contains too much vibrational energy so dissociates||[[File:-2.5, -5.0.png ]]
|
|-
| -2.5 || -5.2 || -83.416
||Yes||Reaction does occur but recrosses the transition state barrier||[[File:-2.5,5.2.png]]
|
|}
The fact that the last experiment recrosses the transition state barrier is evidence of the breakdown of Transition State Theory, which is presented below. this states that once products are formed, the system cannot travel back through the transition state to the reactants.

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

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

There are two basic assumptions for Transition State Theory, the two most basic are the separation of the electronic and nuclear motions which is the same as the Born-Oppenheimer approximation from quantum mechanics. The second basic assumption is that the reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution.

There are however, additional assumptions and they are as follows:
# Molecular systems that have crossed the transition state in the direction of the products cannot turn around and reform reactants.
# In the transition state, motion along the reaction coordinate may be separated from the other motions and be treated classically as a translation.
# Even in the absence of an equilibrium between the reactant and product molecules, the transition states that are become products are distributed among their states according to the Maxwell-Boltzmann laws.

''Given that Transition State theory neglects quantum mechanical contributions such as influence on and between between electronic and nuclear motions and the idea that molecules passing through the transition state through to the products cannot pass back through to the reactants, this would suggest that there would be deviation between experimental and calculated values using Transition State theory.''

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

=== ''F + H2'' ===

==== ''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?'' ====
Looking at the PES of F + H2 this indicated the reaction is exothermic because there is a decrease in the energy going from the reactants to products. This can be seen along the reaction path for this reaction where there is a dip in energy on the path to the formation of the products. This makes intuitive sense as the driving force of the reaction will be the formation of a strong F-H bond, in place of a weaker H-H bond. Additional evidence that the reaction is exothermic is that when a mep calculation was run with RAB = 1.812 and RBC = 0.74 with zero momentum for each case, the reaction proceeded and the products formed with no additional input of energy.

For the H + HF reaction, '''the reaction is surmised to be endothermic due to as there is no decrease in energy on forming the products. This is in line with intuition considering the strength of the H-F bond. The very high electronegativity of fluorine would make it so that significant energy would be required to cause this reaction to proceed. '''

==== ''Locate the approximate position of the transition state.'' ====
{| class="wikitable"
!Reaction
!rAB
!rBC
!Activation Energy Kj/mol
|-
|F + H2
|1.8108757
|0.7448759
|
|-
|H + HF
|0.7448756
|1.8108759
|
|}

[[File:Aja4117Surface Plot f + h2.png]]
[[File:Aja4117Surface Plot h + HF.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.'' ====

The set of reaction conditions that were used were:

Atom A: F

Atom B: H

Atom C: H

Distance AB: 2.25

Distance BC: 0.74

Momentum AB = -0.8

Momentum BC = 0.1

These conditions yielded a set of successful reactions, and post reaction and upon formation of F-H, there is transfer of energy in the form of kinetic energy to both the Hydrogen that is travelling away from the reaction complex and also kinetic energy in the bond vibration of F-H. This can be visualised in the Momenta vs Time graph below. Monitoring the emission spectrum of the product and observing electromagnetic radiation in the IR region would allow identification of the product.

[[File:Aja4117Surface Plot FH2 momenta vs time.png]]

For rBC = 0.74, with a momentum pAB = -0.5, and several values of pBC in the range -3 to 3 were explored. For pBC = -3, the reaction did proceed and there was significant transfer of kinetic energy to the product F-H. What is interesting to note is that the F-H bond forms twice. On the first formation, the bond breaks due to what is assumed to be, too high a vibrational energy. This leads to the reformation of the H-H bond, upon which there is still significant vibrational energy and combined with the electronegativity of the F atom causes the H-H bond to break and the F-H bond to form for the second time. The F-H bond is now stable, although still containing significant vibrational energy. For pBC = 3, the same phenomena occurs as for pBC = -3. 

==== ''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.'' ====
A transition state for an attractive potential energy surface occurs early in the reaction coordinate and involves high translational kinetic energy (evidenced by the random motion for the reactants) and results in a vibrationally excited product. This translational energy runs up the side of the valley and falls from side to side, showing that there is significant vibrational motion. We can therefore conclude that reactions with attractive potential energy surfaces proceed more efficiently if the energy is in relative translational motion.

In contrast, reactions with repulsive potential surfaces experience a late transition state and are expected to proceed with greater efficiency if the excess energy is present as vibrations. The energy is in the vibration of the reactant molecule and the motion causes the trajectory to move from side to side up the valley as it approaches the transition state. This energy is enough to tip the reactant molecules through the transition state and into the products, it is therefore apparent that the vibrational energy on the repulsive potential energy surface is key to the reactivity and formation of the products.

MRD:aja4117

2019-05-16T15:00:31Z

Aja4117: /* What do you observe? STILL TO DO */

= [[Molecular reaction dynamics|'''Molecular reaction dynamics''']] =

== 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?''''' ===
The transition state is a saddle point which is defined as the point which is both a minimum along one vector and a maximum along an orthogonal vector. Along the reaction path, the transition state is a maximum and along the second vector, which is perpendicular to our initial vector it is a minimum. This can be visualised in the diagram below.

For a co linear system, as given for the H + H2 system there are two orthogonal degrees of freedom with the coordinates given as follows:

'''Z = r2 + r1'''
 

<math>\frac {\partial V}{\partial Z} = 0 </math>, <math>\frac {\partial^2 V}{\partial Z^2}>0 </math>

 

'''S = r2 - r1'''

 

<math>\frac {\partial V}{\partial S} = 0 </math>, <math>\frac {\partial^2 V}{\partial S^2}<0 </math>

 

Note that these are the orthogonal vectors.

The saddle point has the configuration in which r1 = r2 =r0 and this will be the point at which these two lines intersect on the PES. This is the transition state. S is the tangent to the reaction pathway at the maximum whilst Z bisects the reaction pathway at the same point, but will be seen as a minimum.

[[File:Aja4117Question 1.png]]

=== ''Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'' ===
The best estimate was AB = 0.9077425 and BC = 0.9077425. There are only two lines when looking at internuclear distances vs time as seen below. This indicates that two lines are at the same distance and one is covering the other. 

To locate the transition state, is to begin the reaction trajectories close to the transition state. Due to the transition state being a global maxima, the trajectories will either form the reactants or products in order to find the lowest energy configuration. Whether the product or reactants are formed is directly influenced by the trajectories. If the trajectories are closer to the reactants then the reactants will form and if they are closer to the products then the products will form.

[[File:Aja4117INT for HH.png]]

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

===== Trajectories from r1 = rts+δ, r2 = rts =====
Along the mep, because the atoms are moving infinitely slowly and the momenta and the velocities always reset to zero, there is no oscillation and hence no vibration. This is seen along the mep where there is no wavy lines. In contrast, when looking at the dynamics, there is oscillation as post collision there is transfer of kinetic energy that causes the molecule to vibrate. 

When considering the inter nuclear distance vs time, A-C and A-B distances increase rapidly over the first 250 steps after which the rate of increase of the distance begins to decrease. As the number of steps increases the rate of increase of distance begins to plateau as the gradient flattens. This can be rationalised when looking at the animation where post collision, the molecules initially rapidly move away from one another but the speed at which they move away from each other falls off with time. When considering the dynamic situation, A-C and A-B distances increase at a constant gradient over the course of the plot. The reasoning is that in the dynamic situation, the velocities and momenta are not reset to zero at each step.

[[File:Aja4117(dynamics type)Mep vs dynamics distance v time 1.png]]
[[File:Aja4117Distance vs time 1.png]]

For the same reason as stated as above, the momenta and velocities are reset to zero at each point this means that the momentum vs time plot is a straight line for the mep . However, when running the dynamic situation we see an initial sharp increase in the momentum followed by a constant oscillation of momentum that is due to the momentum being retained within the vibrational modes. For A-B, there is a sharp increase in the momentum which plateaus at the same instance as B-C begins to undergo constant oscillation. This is because A-B refers to the momentum of the system A-B. Once both reach constant momentum the combined system now has a set defined level of momentum with no way to lose it.

[[File:Aja4117(dynamics type)Mep vs dynamics momenta vs time1.png]]
[[File:Aja4117dynamics mom vs time1.png]]

===== Trajectories from r1 = rts, r2 = rts +δ =====
The result of switching the distances is that

<nowiki> </nowiki>Take note of the final values of the positions '''r1'''(t) '''r2'''(t) and '''p1'''(t) '''p2'''(t) for your trajectory for large enough t.

== Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed. ==

== What do you observe? STILL TO DO ==

=== Reactive and unreactive trajectories ===
{| class="wikitable" border="1"
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics
!Plots of trajectory
|-
| -1.25 || -2.5 ||-99.018||Yes||Reaction proceeds with minimal vibration||[[File:-1.25, -2.5.png]]
|
|-
| -1.5 || -2.0 ||-100.456||No||P2 does not have enough momentum to reach P1||[[File:-1.5, -2.0.png]]
|
|-
| -1.5 || -2.5 ||-98.956||Yes||Transfer of vibrational energy||[[File:-1.5, -2.5.png]]
|
|-
| -2.5 || -5.0 || -84.956
||No||Product forms but contains too much vibrational energy so dissociates||[[File:-2.5, -5.0.png ]]
|
|-
| -2.5 || -5.2 || -83.416
||Yes||Reaction does occur but recrosses the transition state barrier||[[File:-2.5,5.2.png]]
|
|}
The fact that the last experiment recrosses the transition state barrier is evidence of the breakdown of Transition State Theory, which is presented below. this states that once products are formed, the system cannot travel back through the transition state to the reactants.

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

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

There are two basic assumptions for Transition State Theory, the two most basic are the separation of the electronic and nuclear motions which is the same as the Born-Oppenheimer approximation from quantum mechanics. The second basic assumption is that the reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution.

There are however, additional assumptions and they are as follows:
# Molecular systems that have crossed the transition state in the direction of the products cannot turn around and reform reactants.
# In the transition state, motion along the reaction coordinate may be separated from the other motions and be treated classically as a translation.
# Even in the absence of an equilibrium between the reactant and product molecules, the transition states that are become products are distributed among their states according to the Maxwell-Boltzmann laws.

''Given that Transition State theory neglects quantum mechanical contributions such as influence on and between between electronic and nuclear motions and the idea that molecules passing through the transition state through to the products cannot pass back through to the reactants, this would suggest that there would be deviation between experimental and calculated values using Transition State theory.''

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

=== ''F + H2'' ===

==== ''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?'' ====
Looking at the PES of F + H2 this indicated the reaction is exothermic because there is a decrease in the energy going from the reactants to products. This can be seen along the reaction path for this reaction where there is a dip in energy on the path to the formation of the products. This makes intuitive sense as the driving force of the reaction will be the formation of a strong F-H bond, in place of a weaker H-H bond. Additional evidence that the reaction is exothermic is that when a mep calculation was run with RAB = 1.812 and RBC = 0.74 with zero momentum for each case, the reaction proceeded and the products formed with no additional input of energy.

For the H + HF reaction, '''the reaction is surmised to be endothermic due to as there is no decrease in energy on forming the products. This is in line with intuition considering the strength of the H-F bond. The very high electronegativity of fluorine would make it so that significant energy would be required to cause this reaction to proceed. '''

==== ''Locate the approximate position of the transition state.'' ====
{| class="wikitable"
!Reaction
!rAB
!rBC
!Activation Energy Kcal/mol
|-
|F + H2
|1.8108757
|0.7448759
|
|-
|H + HF
|0.7448756
|1.8108759
|
|}

[[File:Aja4117Surface Plot f + h2.png]]
[[File:Aja4117Surface Plot h + HF.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.'' ====

The set of reaction conditions that were used were:

Atom A: F

Atom B: H

Atom C: H

Distance AB: 2.25

Distance BC: 0.74

Momentum AB = -0.8

Momentum BC = 0.1

These conditions yielded a set of successful reactions, and post reaction and upon formation of F-H, there is transfer of energy in the form of kinetic energy to both the Hydrogen that is travelling away from the reaction complex and also kinetic energy in the bond vibration of F-H. This can be visualised in the Momenta vs Time graph below. Monitoring the emission spectrum of the product and observing electromagnetic radiation in the IR region would allow identification of the product.

[[File:Aja4117Surface Plot FH2 momenta vs time.png]]

For rBC = 0.74, with a momentum pAB = -0.5, and several values of pBC in the range -3 to 3 were explored. For pBC = -3, the reaction did proceed and there was significant transfer of kinetic energy to the product F-H. What is interesting to note is that the F-H bond forms twice. On the first formation, the bond breaks due to what is assumed to be, too high a vibrational energy. This leads to the reformation of the H-H bond, upon which there is still significant vibrational energy and combined with the electronegativity of the F atom causes the H-H bond to break and the F-H bond to form for the second time. The F-H bond is now stable, although still containing significant vibrational energy. For pBC = 3, the same phenomena occurs as for pBC = -3. 

==== ''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.'' ====
A transition state for an attractive potential energy surface occurs early in the reaction coordinate and involves high translational kinetic energy (evidenced by the random motion for the reactants) and results in a vibrationally excited product. This translational energy runs up the side of the valley and falls from side to side, showing that there is significant vibrational motion. We can therefore conclude that reactions with attractive potential energy surfaces proceed more efficiently if the energy is in relative translational motion.

In contrast, reactions with repulsive potential surfaces experience a late transition state and are expected to proceed with greater efficiency if the excess energy is present as vibrations. The energy is in the vibration of the reactant molecule and the motion causes the trajectory to move from side to side up the valley as it approaches the transition state. This energy is enough to tip the reactant molecules through the transition state and into the products, it is therefore apparent that the vibrational energy on the repulsive potential energy surface is key to the reactivity and formation of the products.

MRD:aja4117

2019-05-16T14:57:56Z

Aja4117: /* What do you observe? STILL TO DO */

= [[Molecular reaction dynamics|'''Molecular reaction dynamics''']] =

== 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?''''' ===
The transition state is a saddle point which is defined as the point which is both a minimum along one vector and a maximum along an orthogonal vector. Along the reaction path, the transition state is a maximum and along the second vector, which is perpendicular to our initial vector it is a minimum. This can be visualised in the diagram below.

For a co linear system, as given for the H + H2 system there are two orthogonal degrees of freedom with the coordinates given as follows:

'''Z = r2 + r1'''
 

<math>\frac {\partial V}{\partial Z} = 0 </math>, <math>\frac {\partial^2 V}{\partial Z^2}>0 </math>

 

'''S = r2 - r1'''

 

<math>\frac {\partial V}{\partial S} = 0 </math>, <math>\frac {\partial^2 V}{\partial S^2}<0 </math>

 

Note that these are the orthogonal vectors.

The saddle point has the configuration in which r1 = r2 =r0 and this will be the point at which these two lines intersect on the PES. This is the transition state. S is the tangent to the reaction pathway at the maximum whilst Z bisects the reaction pathway at the same point, but will be seen as a minimum.

[[File:Aja4117Question 1.png]]

=== ''Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'' ===
The best estimate was AB = 0.9077425 and BC = 0.9077425. There are only two lines when looking at internuclear distances vs time as seen below. This indicates that two lines are at the same distance and one is covering the other. 

To locate the transition state, is to begin the reaction trajectories close to the transition state. Due to the transition state being a global maxima, the trajectories will either form the reactants or products in order to find the lowest energy configuration. Whether the product or reactants are formed is directly influenced by the trajectories. If the trajectories are closer to the reactants then the reactants will form and if they are closer to the products then the products will form.

[[File:Aja4117INT for HH.png]]

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

===== Trajectories from r1 = rts+δ, r2 = rts =====
Along the mep, because the atoms are moving infinitely slowly and the momenta and the velocities always reset to zero, there is no oscillation and hence no vibration. This is seen along the mep where there is no wavy lines. In contrast, when looking at the dynamics, there is oscillation as post collision there is transfer of kinetic energy that causes the molecule to vibrate. 

When considering the inter nuclear distance vs time, A-C and A-B distances increase rapidly over the first 250 steps after which the rate of increase of the distance begins to decrease. As the number of steps increases the rate of increase of distance begins to plateau as the gradient flattens. This can be rationalised when looking at the animation where post collision, the molecules initially rapidly move away from one another but the speed at which they move away from each other falls off with time. When considering the dynamic situation, A-C and A-B distances increase at a constant gradient over the course of the plot. The reasoning is that in the dynamic situation, the velocities and momenta are not reset to zero at each step.

[[File:Aja4117(dynamics type)Mep vs dynamics distance v time 1.png]]
[[File:Aja4117Distance vs time 1.png]]

For the same reason as stated as above, the momenta and velocities are reset to zero at each point this means that the momentum vs time plot is a straight line for the mep . However, when running the dynamic situation we see an initial sharp increase in the momentum followed by a constant oscillation of momentum that is due to the momentum being retained within the vibrational modes. For A-B, there is a sharp increase in the momentum which plateaus at the same instance as B-C begins to undergo constant oscillation. This is because A-B refers to the momentum of the system A-B. Once both reach constant momentum the combined system now has a set defined level of momentum with no way to lose it.

[[File:Aja4117(dynamics type)Mep vs dynamics momenta vs time1.png]]
[[File:Aja4117dynamics mom vs time1.png]]

===== Trajectories from r1 = rts, r2 = rts +δ =====
The result of switching the distances is that

<nowiki> </nowiki>Take note of the final values of the positions '''r1'''(t) '''r2'''(t) and '''p1'''(t) '''p2'''(t) for your trajectory for large enough t.

== Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed. ==

== What do you observe? STILL TO DO ==

=== Reactive and unreactive trajectories ===
{| class="wikitable" border="1"
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics
!Plots of trajectory
|-
| -1.25 || -2.5 ||-99.018||Yes||Reaction proceeds with minimal vibration
|
|-
| -1.5 || -2.0 ||-100.456||No||P2 does not have enough momentum to reach P1
|
|-
| -1.5 || -2.5 ||-98.956||Yes||Transfer of vibrational energy
|
|-
| -2.5 || -5.0 || -84.956
||No||Product forms but contains too much vibrational energy so dissociates
|
|-
| -2.5 || -5.2 || -83.416
||Yes||Reaction does occur but recrosses the transition state barrier
|
|}
The fact that the last experiment recrosses the transition state barrier is evidence of the breakdown of Transition State Theory, which is presented below. this states that once products are formed, the system cannot travel back through the transition state to the reactants.

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

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

There are two basic assumptions for Transition State Theory, the two most basic are the separation of the electronic and nuclear motions which is the same as the Born-Oppenheimer approximation from quantum mechanics. The second basic assumption is that the reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution.

There are however, additional assumptions and they are as follows:
# Molecular systems that have crossed the transition state in the direction of the products cannot turn around and reform reactants.
# In the transition state, motion along the reaction coordinate may be separated from the other motions and be treated classically as a translation.
# Even in the absence of an equilibrium between the reactant and product molecules, the transition states that are become products are distributed among their states according to the Maxwell-Boltzmann laws.

''Given that Transition State theory neglects quantum mechanical contributions such as influence on and between between electronic and nuclear motions and the idea that molecules passing through the transition state through to the products cannot pass back through to the reactants, this would suggest that there would be deviation between experimental and calculated values using Transition State theory.''

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

=== ''F + H2'' ===

==== ''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?'' ====
Looking at the PES of F + H2 this indicated the reaction is exothermic because there is a decrease in the energy going from the reactants to products. This can be seen along the reaction path for this reaction where there is a dip in energy on the path to the formation of the products. This makes intuitive sense as the driving force of the reaction will be the formation of a strong F-H bond, in place of a weaker H-H bond. Additional evidence that the reaction is exothermic is that when a mep calculation was run with RAB = 1.812 and RBC = 0.74 with zero momentum for each case, the reaction proceeded and the products formed with no additional input of energy.

For the H + HF reaction, '''the reaction is surmised to be endothermic due to as there is no decrease in energy on forming the products. This is in line with intuition considering the strength of the H-F bond. The very high electronegativity of fluorine would make it so that significant energy would be required to cause this reaction to proceed. '''

==== ''Locate the approximate position of the transition state.'' ====
{| class="wikitable"
!Reaction
!rAB
!rBC
!Activation Energy Kcal/mol
|-
|F + H2
|1.8108757
|0.7448759
|
|-
|H + HF
|0.7448756
|1.8108759
|
|}

[[File:Aja4117Surface Plot f + h2.png]]
[[File:Aja4117Surface Plot h + HF.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.'' ====

The set of reaction conditions that were used were:

Atom A: F

Atom B: H

Atom C: H

Distance AB: 2.25

Distance BC: 0.74

Momentum AB = -0.8

Momentum BC = 0.1

These conditions yielded a set of successful reactions, and post reaction and upon formation of F-H, there is transfer of energy in the form of kinetic energy to both the Hydrogen that is travelling away from the reaction complex and also kinetic energy in the bond vibration of F-H. This can be visualised in the Momenta vs Time graph below. Monitoring the emission spectrum of the product and observing electromagnetic radiation in the IR region would allow identification of the product.

[[File:Aja4117Surface Plot FH2 momenta vs time.png]]

For rBC = 0.74, with a momentum pAB = -0.5, and several values of pBC in the range -3 to 3 were explored. For pBC = -3, the reaction did proceed and there was significant transfer of kinetic energy to the product F-H. What is interesting to note is that the F-H bond forms twice. On the first formation, the bond breaks due to what is assumed to be, too high a vibrational energy. This leads to the reformation of the H-H bond, upon which there is still significant vibrational energy and combined with the electronegativity of the F atom causes the H-H bond to break and the F-H bond to form for the second time. The F-H bond is now stable, although still containing significant vibrational energy. For pBC = 3, the same phenomena occurs as for pBC = -3. 

==== ''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.'' ====
A transition state for an attractive potential energy surface occurs early in the reaction coordinate and involves high translational kinetic energy (evidenced by the random motion for the reactants) and results in a vibrationally excited product. This translational energy runs up the side of the valley and falls from side to side, showing that there is significant vibrational motion. We can therefore conclude that reactions with attractive potential energy surfaces proceed more efficiently if the energy is in relative translational motion.

In contrast, reactions with repulsive potential surfaces experience a late transition state and are expected to proceed with greater efficiency if the excess energy is present as vibrations. The energy is in the vibration of the reactant molecule and the motion causes the trajectory to move from side to side up the valley as it approaches the transition state. This energy is enough to tip the reactant molecules through the transition state and into the products, it is therefore apparent that the vibrational energy on the repulsive potential energy surface is key to the reactivity and formation of the products.

File:-2.5,5.2.png

2019-05-16T14:57:32Z

Aja4117:

File:-2.5, -5.0.png

2019-05-16T14:57:23Z

Aja4117: Aja4117 uploaded a new version of File:-2.5, -5.0.png

File:-1.5, -2.5.png

2019-05-16T14:57:11Z

Aja4117: Aja4117 uploaded a new version of File:-1.5, -2.5.png

File:-1.5, -2.0.png

2019-05-16T14:57:02Z

Aja4117: Aja4117 uploaded a new version of File:-1.5, -2.0.png

File:-1.25, -2.5.png

2019-05-16T14:56:49Z

Aja4117: Aja4117 uploaded a new version of File:-1.25, -2.5.png

MRD:aja4117

2019-05-16T14:43:09Z

Aja4117: /* EXERCISE 1: H + H2 system */

= [[Molecular reaction dynamics|'''Molecular reaction dynamics''']] =

== 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?''''' ===
The transition state is a saddle point which is defined as the point which is both a minimum along one vector and a maximum along an orthogonal vector. Along the reaction path, the transition state is a maximum and along the second vector, which is perpendicular to our initial vector it is a minimum. This can be visualised in the diagram below.

For a co linear system, as given for the H + H2 system there are two orthogonal degrees of freedom with the coordinates given as follows:

'''Z = r2 + r1'''
 

<math>\frac {\partial V}{\partial Z} = 0 </math>, <math>\frac {\partial^2 V}{\partial Z^2}>0 </math>

 

'''S = r2 - r1'''

 

<math>\frac {\partial V}{\partial S} = 0 </math>, <math>\frac {\partial^2 V}{\partial S^2}<0 </math>

 

Note that these are the orthogonal vectors.

The saddle point has the configuration in which r1 = r2 =r0 and this will be the point at which these two lines intersect on the PES. This is the transition state. S is the tangent to the reaction pathway at the maximum whilst Z bisects the reaction pathway at the same point, but will be seen as a minimum.

[[File:Aja4117Question 1.png]]

=== ''Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'' ===
The best estimate was AB = 0.9077425 and BC = 0.9077425. There are only two lines when looking at internuclear distances vs time as seen below. This indicates that two lines are at the same distance and one is covering the other. 

To locate the transition state, is to begin the reaction trajectories close to the transition state. Due to the transition state being a global maxima, the trajectories will either form the reactants or products in order to find the lowest energy configuration. Whether the product or reactants are formed is directly influenced by the trajectories. If the trajectories are closer to the reactants then the reactants will form and if they are closer to the products then the products will form.

[[File:Aja4117INT for HH.png]]

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

===== Trajectories from r1 = rts+δ, r2 = rts =====
Along the mep, because the atoms are moving infinitely slowly and the momenta and the velocities always reset to zero, there is no oscillation and hence no vibration. This is seen along the mep where there is no wavy lines. In contrast, when looking at the dynamics, there is oscillation as post collision there is transfer of kinetic energy that causes the molecule to vibrate. 

When considering the inter nuclear distance vs time, A-C and A-B distances increase rapidly over the first 250 steps after which the rate of increase of the distance begins to decrease. As the number of steps increases the rate of increase of distance begins to plateau as the gradient flattens. This can be rationalised when looking at the animation where post collision, the molecules initially rapidly move away from one another but the speed at which they move away from each other falls off with time. When considering the dynamic situation, A-C and A-B distances increase at a constant gradient over the course of the plot. The reasoning is that in the dynamic situation, the velocities and momenta are not reset to zero at each step.

[[File:Aja4117(dynamics type)Mep vs dynamics distance v time 1.png]]
[[File:Aja4117Distance vs time 1.png]]

For the same reason as stated as above, the momenta and velocities are reset to zero at each point this means that the momentum vs time plot is a straight line for the mep . However, when running the dynamic situation we see an initial sharp increase in the momentum followed by a constant oscillation of momentum that is due to the momentum being retained within the vibrational modes. For A-B, there is a sharp increase in the momentum which plateaus at the same instance as B-C begins to undergo constant oscillation. This is because A-B refers to the momentum of the system A-B. Once both reach constant momentum the combined system now has a set defined level of momentum with no way to lose it.

[[File:Aja4117(dynamics type)Mep vs dynamics momenta vs time1.png]]
[[File:Aja4117dynamics mom vs time1.png]]

===== Trajectories from r1 = rts, r2 = rts +δ =====
The result of switching the distances is that

<nowiki> </nowiki>Take note of the final values of the positions '''r1'''(t) '''r2'''(t) and '''p1'''(t) '''p2'''(t) for your trajectory for large enough t.

== Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed. ==

== What do you observe? STILL TO DO ==

=== Reactive and unreactive trajectories ===
{| class="wikitable" border="1"
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 ||-99.018||Yes||Reaction proceeds with minimal vibration||
|-
| -1.5 || -2.0 ||-100.456||No||P2 does not have enough momentum to reach P1||
|-
| -1.5 || -2.5 ||-98.956||Yes||Transfer of vibrational energy||
|-
| -2.5 || -5.0 || -84.956
||No||Product forms but contains too much vibrational energy so dissociates||
|-
| -2.5 || -5.2 || -83.416
||No||Product forms but contains too much vibrational energy so dissociates
||
|}

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

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

There are two basic assumptions for Transition State Theory, the two most basic are the separation of the electronic and nuclear motions which is the same as the Born-Oppenheimer approximation from quantum mechanics. The second basic assumption is that the reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution.

There are however, additional assumptions and they are as follows:
# Molecular systems that have crossed the transition state in the direction of the products cannot turn around and reform reactants.
# In the transition state, motion along the reaction coordinate may be separated from the other motions and be treated classically as a translation.
# even in the absence of an equilibrium between the reactant and product molecules, the transition states that are become products are distributed among their states according to the Maxwell-Boltzmann laws.

''Given that Transition State theory neglects quantum mechanical contributions such as influence on and between between electronic and nuclear motions and the idea that molecules passing through the transition state through to the products cannot pass back through to the reactants, this would suggest that there would be deviation between experimental and calculated values using Transition State theory.''

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

=== ''F + H2'' ===

==== ''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?'' ====
Looking at the PES of F + H2 this indicated the reaction is exothermic because there is a decrease in the energy going from the reactants to products. This can be seen along the reaction path for this reaction where there is a dip in energy on the path to the formation of the products. This makes intuitive sense as the driving force of the reaction will be the formation of a strong F-H bond, in place of a weaker H-H bond. Additional evidence that the reaction is exothermic is that when a mep calculation was run with RAB = 1.812 and RBC = 0.74 with zero momentum for each case, the reaction proceeded and the products formed with no additional input of energy.

For the H + HF reaction, '''the reaction is surmised to be endothermic due to as there is no decrease in energy on forming the products. This is in line with intuition considering the strength of the H-F bond. The very high electronegativity of fluorine would make it so that significant energy would be required to cause this reaction to proceed. '''

==== ''Locate the approximate position of the transition state.'' ====
{| class="wikitable"
!Reaction
!rAB
!rBC
!Activation Energy Kcal/mol
|-
|F + H2
|1.8108757
|0.7448759
|
|-
|H + HF
|0.7448756
|1.8108759
|
|}

[[File:Aja4117Surface Plot f + h2.png]]
[[File:Aja4117Surface Plot h + HF.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.'' ====

The set of reaction conditions that were used were:

Atom A: F

Atom B: H

Atom C: H

Distance AB: 2.25

Distance BC: 0.74

Momentum AB = -0.8

Momentum BC = 0.1

These conditions yielded a set of successful reactions, and post reaction and upon formation of F-H, there is transfer of energy in the form of kinetic energy to both the Hydrogen that is travelling away from the reaction complex and also kinetic energy in the bond vibration of F-H. This can be visualised in the Momenta vs Time graph below. Monitoring the emission spectrum of the product and observing electromagnetic radiation in the IR region would allow identification of the product.

[[File:Aja4117Surface Plot FH2 momenta vs time.png]]

For rBC = 0.74, with a momentum pAB = -0.5, and several values of pBC in the range -3 to 3 were explored. For pBC = -3, the reaction did proceed and there was significant transfer of kinetic energy to the product F-H. What is interesting to note is that the F-H bond forms twice. On the first formation, the bond breaks due to what is assumed to be, too high a vibrational energy. This leads to the reformation of the H-H bond, upon which there is still significant vibrational energy and combined with the electronegativity of the F atom causes the H-H bond to break and the F-H bond to form for the second time. The F-H bond is now stable, although still containing significant vibrational energy. For pBC = 3, the same phenomena occurs as for pBC = -3. 

==== ''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.'' ====
A transition state for an attractive potential energy surface occurs early in the reaction coordinate and involves high translational kinetic energy (evidenced by the random motion for the reactants) and results in a vibrationally excited product. This translational energy runs up the side of the valley and falls from side to side, showing that there is significant vibrational motion. We can therefore conclude that reactions with attractive potential energy surfaces proceed more efficiently if the energy is in relative translational motion.

In contrast, reactions with repulsive potential surfaces experience a late transition state and are expected to proceed with greater efficiency if the excess energy is present as vibrations. The energy is in the vibration of the reactant molecule and the motion causes the trajectory to move from side to side up the valley as it approaches the transition state. This energy is enough to tip the reactant molecules through the transition state and into the products, it is therefore apparent that the vibrational energy on the repulsive potential energy surface is key to the reactivity and formation of the products.

MRD:aja4117

2019-05-16T14:42:46Z

Aja4117: /* EXERCISE 1: H + H2 system */

= [[Molecular reaction dynamics|'''Molecular reaction dynamics''']] =

== 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?''''' ===
The transition state is a saddle point which is defined as the point which is both a minimum along one vector and a maximum along an orthogonal vector. Along the reaction path, the transition state is a maximum and along the second vector, which is perpendicular to our initial vector it is a minimum. This can be visualised in the diagram below.

For a co linear system, as given for the H + H2 system there are two orthogonal degrees of freedom with the coordinates given as follows:

'''Z = r2 + r1'''
 

<math>\frac {\partial V}{\partial Z} = 0 </math>, <math>\frac {\partial^2 V}{\partial Z^2}>0 </math>
 

'''S = r2 - r1'''

 
<math>\frac {\partial V}{\partial S} = 0 </math>, <math>\frac {\partial^2 V}{\partial S^2}<0 </math>
 
Note that these are the orthogonal vectors.

The saddle point has the configuration in which r1 = r2 =r0 and this will be the point at which these two lines intersect on the PES. This is the transition state. S is the tangent to the reaction pathway at the maximum whilst Z bisects the reaction pathway at the same point, but will be seen as a minimum.

[[File:Aja4117Question 1.png]]

=== ''Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'' ===
The best estimate was AB = 0.9077425 and BC = 0.9077425. There are only two lines when looking at internuclear distances vs time as seen below. This indicates that two lines are at the same distance and one is covering the other. 

To locate the transition state, is to begin the reaction trajectories close to the transition state. Due to the transition state being a global maxima, the trajectories will either form the reactants or products in order to find the lowest energy configuration. Whether the product or reactants are formed is directly influenced by the trajectories. If the trajectories are closer to the reactants then the reactants will form and if they are closer to the products then the products will form.

[[File:Aja4117INT for HH.png]]

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

===== Trajectories from r1 = rts+δ, r2 = rts =====
Along the mep, because the atoms are moving infinitely slowly and the momenta and the velocities always reset to zero, there is no oscillation and hence no vibration. This is seen along the mep where there is no wavy lines. In contrast, when looking at the dynamics, there is oscillation as post collision there is transfer of kinetic energy that causes the molecule to vibrate. 

When considering the inter nuclear distance vs time, A-C and A-B distances increase rapidly over the first 250 steps after which the rate of increase of the distance begins to decrease. As the number of steps increases the rate of increase of distance begins to plateau as the gradient flattens. This can be rationalised when looking at the animation where post collision, the molecules initially rapidly move away from one another but the speed at which they move away from each other falls off with time. When considering the dynamic situation, A-C and A-B distances increase at a constant gradient over the course of the plot. The reasoning is that in the dynamic situation, the velocities and momenta are not reset to zero at each step.

[[File:Aja4117(dynamics type)Mep vs dynamics distance v time 1.png]]
[[File:Aja4117Distance vs time 1.png]]

For the same reason as stated as above, the momenta and velocities are reset to zero at each point this means that the momentum vs time plot is a straight line for the mep . However, when running the dynamic situation we see an initial sharp increase in the momentum followed by a constant oscillation of momentum that is due to the momentum being retained within the vibrational modes. For A-B, there is a sharp increase in the momentum which plateaus at the same instance as B-C begins to undergo constant oscillation. This is because A-B refers to the momentum of the system A-B. Once both reach constant momentum the combined system now has a set defined level of momentum with no way to lose it.

[[File:Aja4117(dynamics type)Mep vs dynamics momenta vs time1.png]]
[[File:Aja4117dynamics mom vs time1.png]]

===== Trajectories from r1 = rts, r2 = rts +δ =====
The result of switching the distances is that

<nowiki> </nowiki>Take note of the final values of the positions '''r1'''(t) '''r2'''(t) and '''p1'''(t) '''p2'''(t) for your trajectory for large enough t.

== Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed. ==

== What do you observe? STILL TO DO ==

=== Reactive and unreactive trajectories ===
{| class="wikitable" border="1"
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 ||-99.018||Yes||Reaction proceeds with minimal vibration||
|-
| -1.5 || -2.0 ||-100.456||No||P2 does not have enough momentum to reach P1||
|-
| -1.5 || -2.5 ||-98.956||Yes||Transfer of vibrational energy||
|-
| -2.5 || -5.0 || -84.956
||No||Product forms but contains too much vibrational energy so dissociates||
|-
| -2.5 || -5.2 || -83.416
||No||Product forms but contains too much vibrational energy so dissociates
||
|}

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

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

There are two basic assumptions for Transition State Theory, the two most basic are the separation of the electronic and nuclear motions which is the same as the Born-Oppenheimer approximation from quantum mechanics. The second basic assumption is that the reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution.

There are however, additional assumptions and they are as follows:
# Molecular systems that have crossed the transition state in the direction of the products cannot turn around and reform reactants.
# In the transition state, motion along the reaction coordinate may be separated from the other motions and be treated classically as a translation.
# even in the absence of an equilibrium between the reactant and product molecules, the transition states that are become products are distributed among their states according to the Maxwell-Boltzmann laws.

''Given that Transition State theory neglects quantum mechanical contributions such as influence on and between between electronic and nuclear motions and the idea that molecules passing through the transition state through to the products cannot pass back through to the reactants, this would suggest that there would be deviation between experimental and calculated values using Transition State theory.''

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

=== ''F + H2'' ===

==== ''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?'' ====
Looking at the PES of F + H2 this indicated the reaction is exothermic because there is a decrease in the energy going from the reactants to products. This can be seen along the reaction path for this reaction where there is a dip in energy on the path to the formation of the products. This makes intuitive sense as the driving force of the reaction will be the formation of a strong F-H bond, in place of a weaker H-H bond. Additional evidence that the reaction is exothermic is that when a mep calculation was run with RAB = 1.812 and RBC = 0.74 with zero momentum for each case, the reaction proceeded and the products formed with no additional input of energy.

For the H + HF reaction, '''the reaction is surmised to be endothermic due to as there is no decrease in energy on forming the products. This is in line with intuition considering the strength of the H-F bond. The very high electronegativity of fluorine would make it so that significant energy would be required to cause this reaction to proceed. '''

==== ''Locate the approximate position of the transition state.'' ====
{| class="wikitable"
!Reaction
!rAB
!rBC
!Activation Energy Kcal/mol
|-
|F + H2
|1.8108757
|0.7448759
|
|-
|H + HF
|0.7448756
|1.8108759
|
|}

[[File:Aja4117Surface Plot f + h2.png]]
[[File:Aja4117Surface Plot h + HF.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.'' ====

The set of reaction conditions that were used were:

Atom A: F

Atom B: H

Atom C: H

Distance AB: 2.25

Distance BC: 0.74

Momentum AB = -0.8

Momentum BC = 0.1

These conditions yielded a set of successful reactions, and post reaction and upon formation of F-H, there is transfer of energy in the form of kinetic energy to both the Hydrogen that is travelling away from the reaction complex and also kinetic energy in the bond vibration of F-H. This can be visualised in the Momenta vs Time graph below. Monitoring the emission spectrum of the product and observing electromagnetic radiation in the IR region would allow identification of the product.

[[File:Aja4117Surface Plot FH2 momenta vs time.png]]

For rBC = 0.74, with a momentum pAB = -0.5, and several values of pBC in the range -3 to 3 were explored. For pBC = -3, the reaction did proceed and there was significant transfer of kinetic energy to the product F-H. What is interesting to note is that the F-H bond forms twice. On the first formation, the bond breaks due to what is assumed to be, too high a vibrational energy. This leads to the reformation of the H-H bond, upon which there is still significant vibrational energy and combined with the electronegativity of the F atom causes the H-H bond to break and the F-H bond to form for the second time. The F-H bond is now stable, although still containing significant vibrational energy. For pBC = 3, the same phenomena occurs as for pBC = -3. 

==== ''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.'' ====
A transition state for an attractive potential energy surface occurs early in the reaction coordinate and involves high translational kinetic energy (evidenced by the random motion for the reactants) and results in a vibrationally excited product. This translational energy runs up the side of the valley and falls from side to side, showing that there is significant vibrational motion. We can therefore conclude that reactions with attractive potential energy surfaces proceed more efficiently if the energy is in relative translational motion.

In contrast, reactions with repulsive potential surfaces experience a late transition state and are expected to proceed with greater efficiency if the excess energy is present as vibrations. The energy is in the vibration of the reactant molecule and the motion causes the trajectory to move from side to side up the valley as it approaches the transition state. This energy is enough to tip the reactant molecules through the transition state and into the products, it is therefore apparent that the vibrational energy on the repulsive potential energy surface is key to the reactivity and formation of the products.

MRD:aja4117

2019-05-16T14:42:13Z

Aja4117: /* EXERCISE 1: H + H2 system */

= [[Molecular reaction dynamics|'''Molecular reaction dynamics''']] =

== 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?''''' ===
The transition state is a saddle point which is defined as the point which is both a minimum along one vector and a maximum along an orthogonal vector. Along the reaction path, the transition state is a maximum and along the second vector, which is perpendicular to our initial vector it is a minimum. This can be visualised in the diagram below.

For a co linear system, as given for the H + H2 system there are two orthogonal degrees of freedom with the coordinates given as follows:

'''Z = r2 + r1'''

<math>\frac {\partial V}{\partial Z} = 0 </math>, <math>\frac {\partial^2 V}{\partial Z^2}>0 </math>

'''S = r2 - r1'''

<math>\frac {\partial V}{\partial S} = 0 </math>, <math>\frac {\partial^2 V}{\partial S^2}<0 </math>

Note that these are the orthogonal vectors.

The saddle point has the configuration in which r1 = r2 =r0 and this will be the point at which these two lines intersect on the PES. This is the transition state. S is the tangent to the reaction pathway at the maximum whilst Z bisects the reaction pathway at the same point, but will be seen as a minimum.

[[File:Aja4117Question 1.png]]

=== ''Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'' ===
The best estimate was AB = 0.9077425 and BC = 0.9077425. There are only two lines when looking at internuclear distances vs time as seen below. This indicates that two lines are at the same distance and one is covering the other. 

To locate the transition state, is to begin the reaction trajectories close to the transition state. Due to the transition state being a global maxima, the trajectories will either form the reactants or products in order to find the lowest energy configuration. Whether the product or reactants are formed is directly influenced by the trajectories. If the trajectories are closer to the reactants then the reactants will form and if they are closer to the products then the products will form.

[[File:Aja4117INT for HH.png]]

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

===== Trajectories from r1 = rts+δ, r2 = rts =====
Along the mep, because the atoms are moving infinitely slowly and the momenta and the velocities always reset to zero, there is no oscillation and hence no vibration. This is seen along the mep where there is no wavy lines. In contrast, when looking at the dynamics, there is oscillation as post collision there is transfer of kinetic energy that causes the molecule to vibrate. 

When considering the inter nuclear distance vs time, A-C and A-B distances increase rapidly over the first 250 steps after which the rate of increase of the distance begins to decrease. As the number of steps increases the rate of increase of distance begins to plateau as the gradient flattens. This can be rationalised when looking at the animation where post collision, the molecules initially rapidly move away from one another but the speed at which they move away from each other falls off with time. When considering the dynamic situation, A-C and A-B distances increase at a constant gradient over the course of the plot. The reasoning is that in the dynamic situation, the velocities and momenta are not reset to zero at each step.

[[File:Aja4117(dynamics type)Mep vs dynamics distance v time 1.png]]
[[File:Aja4117Distance vs time 1.png]]

For the same reason as stated as above, the momenta and velocities are reset to zero at each point this means that the momentum vs time plot is a straight line for the mep . However, when running the dynamic situation we see an initial sharp increase in the momentum followed by a constant oscillation of momentum that is due to the momentum being retained within the vibrational modes. For A-B, there is a sharp increase in the momentum which plateaus at the same instance as B-C begins to undergo constant oscillation. This is because A-B refers to the momentum of the system A-B. Once both reach constant momentum the combined system now has a set defined level of momentum with no way to lose it.

[[File:Aja4117(dynamics type)Mep vs dynamics momenta vs time1.png]]
[[File:Aja4117dynamics mom vs time1.png]]

===== Trajectories from r1 = rts, r2 = rts +δ =====
The result of switching the distances is that

<nowiki> </nowiki>Take note of the final values of the positions '''r1'''(t) '''r2'''(t) and '''p1'''(t) '''p2'''(t) for your trajectory for large enough t.

== Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed. ==

== What do you observe? STILL TO DO ==

=== Reactive and unreactive trajectories ===
{| class="wikitable" border="1"
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 ||-99.018||Yes||Reaction proceeds with minimal vibration||
|-
| -1.5 || -2.0 ||-100.456||No||P2 does not have enough momentum to reach P1||
|-
| -1.5 || -2.5 ||-98.956||Yes||Transfer of vibrational energy||
|-
| -2.5 || -5.0 || -84.956
||No||Product forms but contains too much vibrational energy so dissociates||
|-
| -2.5 || -5.2 || -83.416
||No||Product forms but contains too much vibrational energy so dissociates
||
|}

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

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

There are two basic assumptions for Transition State Theory, the two most basic are the separation of the electronic and nuclear motions which is the same as the Born-Oppenheimer approximation from quantum mechanics. The second basic assumption is that the reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution.

There are however, additional assumptions and they are as follows:
# Molecular systems that have crossed the transition state in the direction of the products cannot turn around and reform reactants.
# In the transition state, motion along the reaction coordinate may be separated from the other motions and be treated classically as a translation.
# even in the absence of an equilibrium between the reactant and product molecules, the transition states that are become products are distributed among their states according to the Maxwell-Boltzmann laws.

''Given that Transition State theory neglects quantum mechanical contributions such as influence on and between between electronic and nuclear motions and the idea that molecules passing through the transition state through to the products cannot pass back through to the reactants, this would suggest that there would be deviation between experimental and calculated values using Transition State theory.''

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

=== ''F + H2'' ===

==== ''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?'' ====
Looking at the PES of F + H2 this indicated the reaction is exothermic because there is a decrease in the energy going from the reactants to products. This can be seen along the reaction path for this reaction where there is a dip in energy on the path to the formation of the products. This makes intuitive sense as the driving force of the reaction will be the formation of a strong F-H bond, in place of a weaker H-H bond. Additional evidence that the reaction is exothermic is that when a mep calculation was run with RAB = 1.812 and RBC = 0.74 with zero momentum for each case, the reaction proceeded and the products formed with no additional input of energy.

For the H + HF reaction, '''the reaction is surmised to be endothermic due to as there is no decrease in energy on forming the products. This is in line with intuition considering the strength of the H-F bond. The very high electronegativity of fluorine would make it so that significant energy would be required to cause this reaction to proceed. '''

==== ''Locate the approximate position of the transition state.'' ====
{| class="wikitable"
!Reaction
!rAB
!rBC
!Activation Energy Kcal/mol
|-
|F + H2
|1.8108757
|0.7448759
|
|-
|H + HF
|0.7448756
|1.8108759
|
|}

[[File:Aja4117Surface Plot f + h2.png]]
[[File:Aja4117Surface Plot h + HF.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.'' ====

The set of reaction conditions that were used were:

Atom A: F

Atom B: H

Atom C: H

Distance AB: 2.25

Distance BC: 0.74

Momentum AB = -0.8

Momentum BC = 0.1

These conditions yielded a set of successful reactions, and post reaction and upon formation of F-H, there is transfer of energy in the form of kinetic energy to both the Hydrogen that is travelling away from the reaction complex and also kinetic energy in the bond vibration of F-H. This can be visualised in the Momenta vs Time graph below. Monitoring the emission spectrum of the product and observing electromagnetic radiation in the IR region would allow identification of the product.

[[File:Aja4117Surface Plot FH2 momenta vs time.png]]

For rBC = 0.74, with a momentum pAB = -0.5, and several values of pBC in the range -3 to 3 were explored. For pBC = -3, the reaction did proceed and there was significant transfer of kinetic energy to the product F-H. What is interesting to note is that the F-H bond forms twice. On the first formation, the bond breaks due to what is assumed to be, too high a vibrational energy. This leads to the reformation of the H-H bond, upon which there is still significant vibrational energy and combined with the electronegativity of the F atom causes the H-H bond to break and the F-H bond to form for the second time. The F-H bond is now stable, although still containing significant vibrational energy. For pBC = 3, the same phenomena occurs as for pBC = -3. 

==== ''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.'' ====
A transition state for an attractive potential energy surface occurs early in the reaction coordinate and involves high translational kinetic energy (evidenced by the random motion for the reactants) and results in a vibrationally excited product. This translational energy runs up the side of the valley and falls from side to side, showing that there is significant vibrational motion. We can therefore conclude that reactions with attractive potential energy surfaces proceed more efficiently if the energy is in relative translational motion.

In contrast, reactions with repulsive potential surfaces experience a late transition state and are expected to proceed with greater efficiency if the excess energy is present as vibrations. The energy is in the vibration of the reactant molecule and the motion causes the trajectory to move from side to side up the valley as it approaches the transition state. This energy is enough to tip the reactant molecules through the transition state and into the products, it is therefore apparent that the vibrational energy on the repulsive potential energy surface is key to the reactivity and formation of the products.

File:Aja4117INT for HH.png

2019-05-16T14:41:34Z

Aja4117:

MRD:aja4117

2019-05-16T14:39:48Z

Aja4117: /* EXERCISE 1: H + H2 system */

= [[Molecular reaction dynamics|'''Molecular reaction dynamics''']] =

== 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?''''' ===
The transition state is a saddle point which is defined as the point which is both a minimum along one vector and a maximum along an orthogonal vector. Along the reaction path, the transition state is a maximum and along the second vector, which is perpendicular to our initial vector it is a minimum. This can be visualised in the diagram below.

For a co linear system, as given for the H + H2 system there are two orthogonal degrees of freedom with the coordinates given as follows:

'''Z = r2 + r1'''

<math>\frac {\partial V}{\partial Z} = 0 </math>, <math>\frac {\partial^2 V}{\partial Z^2}>0 </math>

'''S = r2 - r1'''

<math>\frac {\partial V}{\partial S} = 0 </math>, <math>\frac {\partial^2 V}{\partial S^2}<0 </math>

Note that these are the orthogonal vectors.

The saddle point has the configuration in which r1 = r2 =r0 and this will be the point at which these two lines intersect on the PES. This is the transition state. S is the tangent to the reaction pathway at the maximum whilst Z bisects the reaction pathway at the same point, but will be seen as a minimum.

[[File:Aja4117Question 1.png]]

=== ''Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'' ===
The best estimate was AB = 0.9077425 and BC = 0.9077425. There are only two lines when looking at internuclear distances vs time as seen below. This indicates that two lines are at the same distance and one is covering the other. 

To locate the transition state, is to begin the reaction trajectories close to the transition state. Due to the transition state being a global maxima, the trajectories will either form the reactants or products in order to find the lowest energy configuration. Whether the product or reactants are formed is directly influenced by the trajectories. If the trajectories are closer to the reactants then the reactants will form and if they are closer to the products then the products will form.

[[File:Aja4117q2.PNG]]

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

===== Trajectories from r1 = rts+δ, r2 = rts =====
Along the mep, because the atoms are moving infinitely slowly and the momenta and the velocities always reset to zero, there is no oscillation and hence no vibration. This is seen along the mep where there is no wavy lines. In contrast, when looking at the dynamics, there is oscillation as post collision there is transfer of kinetic energy that causes the molecule to vibrate. 

When considering the inter nuclear distance vs time, A-C and A-B distances increase rapidly over the first 250 steps after which the rate of increase of the distance begins to decrease. As the number of steps increases the rate of increase of distance begins to plateau as the gradient flattens. This can be rationalised when looking at the animation where post collision, the molecules initially rapidly move away from one another but the speed at which they move away from each other falls off with time. When considering the dynamic situation, A-C and A-B distances increase at a constant gradient over the course of the plot. The reasoning is that in the dynamic situation, the velocities and momenta are not reset to zero at each step.

[[File:Aja4117(dynamics type)Mep vs dynamics distance v time 1.png]]
[[File:Aja4117Distance vs time 1.png]]

For the same reason as stated as above, the momenta and velocities are reset to zero at each point this means that the momentum vs time plot is a straight line for the mep . However, when running the dynamic situation we see an initial sharp increase in the momentum followed by a constant oscillation of momentum that is due to the momentum being retained within the vibrational modes. For A-B, there is a sharp increase in the momentum which plateaus at the same instance as B-C begins to undergo constant oscillation. This is because A-B refers to the momentum of the system A-B. Once both reach constant momentum the combined system now has a set defined level of momentum with no way to lose it.

[[File:Aja4117(dynamics type)Mep vs dynamics momenta vs time1.png]]
[[File:Aja4117dynamics mom vs time1.png]]

===== Trajectories from r1 = rts, r2 = rts +δ =====
The result of switching the distances is that

<nowiki> </nowiki>Take note of the final values of the positions '''r1'''(t) '''r2'''(t) and '''p1'''(t) '''p2'''(t) for your trajectory for large enough t.

== Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed. ==

== What do you observe? STILL TO DO ==

=== Reactive and unreactive trajectories ===
{| class="wikitable" border="1"
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 ||-99.018||Yes||Reaction proceeds with minimal vibration||
|-
| -1.5 || -2.0 ||-100.456||No||P2 does not have enough momentum to reach P1||
|-
| -1.5 || -2.5 ||-98.956||Yes||Transfer of vibrational energy||
|-
| -2.5 || -5.0 || -84.956
||No||Product forms but contains too much vibrational energy so dissociates||
|-
| -2.5 || -5.2 || -83.416
||No||Product forms but contains too much vibrational energy so dissociates
||
|}

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

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

There are two basic assumptions for Transition State Theory, the two most basic are the separation of the electronic and nuclear motions which is the same as the Born-Oppenheimer approximation from quantum mechanics. The second basic assumption is that the reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution.

There are however, additional assumptions and they are as follows:
# Molecular systems that have crossed the transition state in the direction of the products cannot turn around and reform reactants.
# In the transition state, motion along the reaction coordinate may be separated from the other motions and be treated classically as a translation.
# even in the absence of an equilibrium between the reactant and product molecules, the transition states that are become products are distributed among their states according to the Maxwell-Boltzmann laws.

''Given that Transition State theory neglects quantum mechanical contributions such as influence on and between between electronic and nuclear motions and the idea that molecules passing through the transition state through to the products cannot pass back through to the reactants, this would suggest that there would be deviation between experimental and calculated values using Transition State theory.''

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

=== ''F + H2'' ===

==== ''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?'' ====
Looking at the PES of F + H2 this indicated the reaction is exothermic because there is a decrease in the energy going from the reactants to products. This can be seen along the reaction path for this reaction where there is a dip in energy on the path to the formation of the products. This makes intuitive sense as the driving force of the reaction will be the formation of a strong F-H bond, in place of a weaker H-H bond. Additional evidence that the reaction is exothermic is that when a mep calculation was run with RAB = 1.812 and RBC = 0.74 with zero momentum for each case, the reaction proceeded and the products formed with no additional input of energy.

For the H + HF reaction, '''the reaction is surmised to be endothermic due to as there is no decrease in energy on forming the products. This is in line with intuition considering the strength of the H-F bond. The very high electronegativity of fluorine would make it so that significant energy would be required to cause this reaction to proceed. '''

==== ''Locate the approximate position of the transition state.'' ====
{| class="wikitable"
!Reaction
!rAB
!rBC
!Activation Energy Kcal/mol
|-
|F + H2
|1.8108757
|0.7448759
|
|-
|H + HF
|0.7448756
|1.8108759
|
|}

[[File:Aja4117Surface Plot f + h2.png]]
[[File:Aja4117Surface Plot h + HF.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.'' ====

The set of reaction conditions that were used were:

Atom A: F

Atom B: H

Atom C: H

Distance AB: 2.25

Distance BC: 0.74

Momentum AB = -0.8

Momentum BC = 0.1

These conditions yielded a set of successful reactions, and post reaction and upon formation of F-H, there is transfer of energy in the form of kinetic energy to both the Hydrogen that is travelling away from the reaction complex and also kinetic energy in the bond vibration of F-H. This can be visualised in the Momenta vs Time graph below. Monitoring the emission spectrum of the product and observing electromagnetic radiation in the IR region would allow identification of the product.

[[File:Aja4117Surface Plot FH2 momenta vs time.png]]

For rBC = 0.74, with a momentum pAB = -0.5, and several values of pBC in the range -3 to 3 were explored. For pBC = -3, the reaction did proceed and there was significant transfer of kinetic energy to the product F-H. What is interesting to note is that the F-H bond forms twice. On the first formation, the bond breaks due to what is assumed to be, too high a vibrational energy. This leads to the reformation of the H-H bond, upon which there is still significant vibrational energy and combined with the electronegativity of the F atom causes the H-H bond to break and the F-H bond to form for the second time. The F-H bond is now stable, although still containing significant vibrational energy. For pBC = 3, the same phenomena occurs as for pBC = -3. 

==== ''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.'' ====
A transition state for an attractive potential energy surface occurs early in the reaction coordinate and involves high translational kinetic energy (evidenced by the random motion for the reactants) and results in a vibrationally excited product. This translational energy runs up the side of the valley and falls from side to side, showing that there is significant vibrational motion. We can therefore conclude that reactions with attractive potential energy surfaces proceed more efficiently if the energy is in relative translational motion.

In contrast, reactions with repulsive potential surfaces experience a late transition state and are expected to proceed with greater efficiency if the excess energy is present as vibrations. The energy is in the vibration of the reactant molecule and the motion causes the trajectory to move from side to side up the valley as it approaches the transition state. This energy is enough to tip the reactant molecules through the transition state and into the products, it is therefore apparent that the vibrational energy on the repulsive potential energy surface is key to the reactivity and formation of the products.

MRD:aja4117

2019-05-16T14:39:27Z

Aja4117: /* EXERCISE 1: H + H2 system */

= [[Molecular reaction dynamics|'''Molecular reaction dynamics''']] =

== 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?''''' ===
The transition state is a saddle point which is defined as the point which is both a minimum along one vector and a maximum along an orthogonal vector. Along the reaction path, the transition state is a maximum and along the second vector, which is perpendicular to our initial vector it is a minimum. This can be visualised in the diagram below.

For a co linear system, as given for the H + H2 system there are two orthogonal degrees of freedom with the coordinates given as follows:

'''Z = r2 + r1'''

<math>\frac {\partial V}{\partial Z} = 0 </math>, <math>\frac {\partial^2 V}{\partial Z^2}>0 </math>

'''S = r2 - r1'''

<math>\frac {\partial V}{\partial S} = 0 </math>, <math>\frac {\partial^2 V}{\partial S^2}<0 </math>

Note that these are the orthogonal vectors.

The saddle point has the configuration in which r1 = r2 =r0 and this will be the point at which these two lines intersect on the PES. This is the transition state. S is the tangent to the reaction pathway at the maximum whilst Z bisects the reaction pathway at the same point, but will be seen as a minimum.

[[File:Aja4117Question 1.png]]

<math>\frac {\partial V}{\partial r} = 0 </math>

<math>\frac {\partial^2 V}{\partial r_{Y}^2}<0 </math>

=== ''Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'' ===
The best estimate was AB = 0.9077425 and BC = 0.9077425. There are only two lines when looking at internuclear distances vs time as seen below. This indicates that two lines are at the same distance and one is covering the other. 

To locate the transition state, is to begin the reaction trajectories close to the transition state. Due to the transition state being a global maxima, the trajectories will either form the reactants or products in order to find the lowest energy configuration. Whether the product or reactants are formed is directly influenced by the trajectories. If the trajectories are closer to the reactants then the reactants will form and if they are closer to the products then the products will form.

[[File:Aja4117q2.PNG]]

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

===== Trajectories from r1 = rts+δ, r2 = rts =====
Along the mep, because the atoms are moving infinitely slowly and the momenta and the velocities always reset to zero, there is no oscillation and hence no vibration. This is seen along the mep where there is no wavy lines. In contrast, when looking at the dynamics, there is oscillation as post collision there is transfer of kinetic energy that causes the molecule to vibrate. 

When considering the inter nuclear distance vs time, A-C and A-B distances increase rapidly over the first 250 steps after which the rate of increase of the distance begins to decrease. As the number of steps increases the rate of increase of distance begins to plateau as the gradient flattens. This can be rationalised when looking at the animation where post collision, the molecules initially rapidly move away from one another but the speed at which they move away from each other falls off with time. When considering the dynamic situation, A-C and A-B distances increase at a constant gradient over the course of the plot. The reasoning is that in the dynamic situation, the velocities and momenta are not reset to zero at each step.

[[File:Aja4117(dynamics type)Mep vs dynamics distance v time 1.png]]
[[File:Aja4117Distance vs time 1.png]]

For the same reason as stated as above, the momenta and velocities are reset to zero at each point this means that the momentum vs time plot is a straight line for the mep . However, when running the dynamic situation we see an initial sharp increase in the momentum followed by a constant oscillation of momentum that is due to the momentum being retained within the vibrational modes. For A-B, there is a sharp increase in the momentum which plateaus at the same instance as B-C begins to undergo constant oscillation. This is because A-B refers to the momentum of the system A-B. Once both reach constant momentum the combined system now has a set defined level of momentum with no way to lose it.

[[File:Aja4117(dynamics type)Mep vs dynamics momenta vs time1.png]]
[[File:Aja4117dynamics mom vs time1.png]]

===== Trajectories from r1 = rts, r2 = rts +δ =====
The result of switching the distances is that

<nowiki> </nowiki>Take note of the final values of the positions '''r1'''(t) '''r2'''(t) and '''p1'''(t) '''p2'''(t) for your trajectory for large enough t.

== Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed. ==

== What do you observe? STILL TO DO ==

=== Reactive and unreactive trajectories ===
{| class="wikitable" border="1"
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 ||-99.018||Yes||Reaction proceeds with minimal vibration||
|-
| -1.5 || -2.0 ||-100.456||No||P2 does not have enough momentum to reach P1||
|-
| -1.5 || -2.5 ||-98.956||Yes||Transfer of vibrational energy||
|-
| -2.5 || -5.0 || -84.956
||No||Product forms but contains too much vibrational energy so dissociates||
|-
| -2.5 || -5.2 || -83.416
||No||Product forms but contains too much vibrational energy so dissociates
||
|}

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

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

There are two basic assumptions for Transition State Theory, the two most basic are the separation of the electronic and nuclear motions which is the same as the Born-Oppenheimer approximation from quantum mechanics. The second basic assumption is that the reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution.

There are however, additional assumptions and they are as follows:
# Molecular systems that have crossed the transition state in the direction of the products cannot turn around and reform reactants.
# In the transition state, motion along the reaction coordinate may be separated from the other motions and be treated classically as a translation.
# even in the absence of an equilibrium between the reactant and product molecules, the transition states that are become products are distributed among their states according to the Maxwell-Boltzmann laws.

''Given that Transition State theory neglects quantum mechanical contributions such as influence on and between between electronic and nuclear motions and the idea that molecules passing through the transition state through to the products cannot pass back through to the reactants, this would suggest that there would be deviation between experimental and calculated values using Transition State theory.''

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

=== ''F + H2'' ===

==== ''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?'' ====
Looking at the PES of F + H2 this indicated the reaction is exothermic because there is a decrease in the energy going from the reactants to products. This can be seen along the reaction path for this reaction where there is a dip in energy on the path to the formation of the products. This makes intuitive sense as the driving force of the reaction will be the formation of a strong F-H bond, in place of a weaker H-H bond. Additional evidence that the reaction is exothermic is that when a mep calculation was run with RAB = 1.812 and RBC = 0.74 with zero momentum for each case, the reaction proceeded and the products formed with no additional input of energy.

For the H + HF reaction, '''the reaction is surmised to be endothermic due to as there is no decrease in energy on forming the products. This is in line with intuition considering the strength of the H-F bond. The very high electronegativity of fluorine would make it so that significant energy would be required to cause this reaction to proceed. '''

==== ''Locate the approximate position of the transition state.'' ====
{| class="wikitable"
!Reaction
!rAB
!rBC
!Activation Energy Kcal/mol
|-
|F + H2
|1.8108757
|0.7448759
|
|-
|H + HF
|0.7448756
|1.8108759
|
|}

[[File:Aja4117Surface Plot f + h2.png]]
[[File:Aja4117Surface Plot h + HF.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.'' ====

The set of reaction conditions that were used were:

Atom A: F

Atom B: H

Atom C: H

Distance AB: 2.25

Distance BC: 0.74

Momentum AB = -0.8

Momentum BC = 0.1

These conditions yielded a set of successful reactions, and post reaction and upon formation of F-H, there is transfer of energy in the form of kinetic energy to both the Hydrogen that is travelling away from the reaction complex and also kinetic energy in the bond vibration of F-H. This can be visualised in the Momenta vs Time graph below. Monitoring the emission spectrum of the product and observing electromagnetic radiation in the IR region would allow identification of the product.

[[File:Aja4117Surface Plot FH2 momenta vs time.png]]

For rBC = 0.74, with a momentum pAB = -0.5, and several values of pBC in the range -3 to 3 were explored. For pBC = -3, the reaction did proceed and there was significant transfer of kinetic energy to the product F-H. What is interesting to note is that the F-H bond forms twice. On the first formation, the bond breaks due to what is assumed to be, too high a vibrational energy. This leads to the reformation of the H-H bond, upon which there is still significant vibrational energy and combined with the electronegativity of the F atom causes the H-H bond to break and the F-H bond to form for the second time. The F-H bond is now stable, although still containing significant vibrational energy. For pBC = 3, the same phenomena occurs as for pBC = -3. 

==== ''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.'' ====
A transition state for an attractive potential energy surface occurs early in the reaction coordinate and involves high translational kinetic energy (evidenced by the random motion for the reactants) and results in a vibrationally excited product. This translational energy runs up the side of the valley and falls from side to side, showing that there is significant vibrational motion. We can therefore conclude that reactions with attractive potential energy surfaces proceed more efficiently if the energy is in relative translational motion.

In contrast, reactions with repulsive potential surfaces experience a late transition state and are expected to proceed with greater efficiency if the excess energy is present as vibrations. The energy is in the vibration of the reactant molecule and the motion causes the trajectory to move from side to side up the valley as it approaches the transition state. This energy is enough to tip the reactant molecules through the transition state and into the products, it is therefore apparent that the vibrational energy on the repulsive potential energy surface is key to the reactivity and formation of the products.

MRD:aja4117

2019-05-16T14:38:44Z

Aja4117: /* 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? */

= [[Molecular reaction dynamics|'''Molecular reaction dynamics''']] =

== 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?''''' ===
The transition state is a saddle point which is defined as the point which is both a minimum along one vector and a maximum along an orthogonal vector. Along the reaction path, the transition state is a maximum and along the second vector, which is perpendicular to our initial vector it is a minimum. This can be visualised in the diagram below.

For a co linear system, as given for the H + H2 system there are two orthogonal degrees of freedom with the coordinates given as follows:

'''Z = r2 + r1'''

<math>\frac {\partial V}{\partial Z} = 0 </math>, <math>\frac {\partial^2 V}{\partial Z^2}>0 </math>

'''S = r2 - r1'''

<math>\frac {\partial V}{\partial S} = 0 </math>, <math>\frac {\partial^2 V}{\partial S^2}<0 </math>

Note that these are the orthogonal vectors.

The saddle point has the configuration in which r1 = r2 =r0 and this will be the point at which these two lines intersect on the PES. This is the transition state. S is the tangent to the reaction pathway at the maximum whilst Z bisects the reaction pathway at the same point, but will be seen as a minimum.

[[File:Aja4117Question 1.png]]

<math>\frac {\partial V}{\partial r} = 0 </math>

<math>\frac {\partial^2 V}{\partial r_{Y}^2}<0 </math>

=== ''Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'' ===
The best estimate was AB = 0.9077425 and BC = 0.9077425. There are only two lines when looking at internuclear distances vs time as seen below. This indicates that two lines are at the same distance and one is covering the other. 

To locate the transition state, is to begin the reaction trajectories close to the transition state. Due to the transition state being a global maxima, the trajectories will either form the reactants or products in order to find the lowest energy configuration. Whether the product or reactants are formed is directly influenced by the trajectories. If the trajectories are closer to the reactants then the reactants will form and if they are closer to the products then the products will form.

[[File:Aja4117q2.PNG]]

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

===== Trajectories from r1 = rts+δ, r2 = rts =====
Along the mep, because the atoms are moving infinitely slowly and the momenta and the velocities always reset to zero, there is no oscillation and hence no vibration. This is seen along the mep where there is no wavy lines. In contrast, when looking at the dynamics, there is oscillation as post collision there is transfer of kinetic energy that causes the molecule to vibrate. 

When considering the inter nuclear distance vs time, A-C and A-B distances increase rapidly over the first 250 steps after which the rate of increase of the distance begins to decrease. As the number of steps increases the rate of increase of distance begins to plateau as the gradient flattens. This can be rationalised when looking at the animation where post collision, the molecules initially rapidly move away from one another but the speed at which they move away from each other falls off with time. When considering the dynamic situation, A-C and A-B distances increase at a constant gradient over the course of the plot. The reasoning is that in the dynamic situation, the velocities and momenta are not reset to zero at each step.

[[File:Aja4117(dynamics type)Mep vs dynamics distance v time 1.png]]
[[File:Aja4117Distance vs time 1.png]]

For the same reason as stated as above, the momenta and velocities are reset to zero at each point this means that the momentum vs time plot is a straight line for the mep . However, when running the dynamic situation we see an initial sharp increase in the momentum followed by a constant oscillation of momentum that is due to the momentum being retained within the vibrational modes. For A-B, there is a sharp increase in the momentum which plateaus at the same instance as B-C begins to undergo constant oscillation. This is because A-B refers to the momentum of the system A-B. Once both reach constant momentum the combined system now has a set defined level of momentum with no way to lose it.

[[File:Aja4117(dynamics type)Mep vs dynamics momenta vs time1.png]]
[[File:Aja4117dynamics mom vs time1.png]]

===== Trajectories from r1 = rts, r2 = rts +δ =====
The result of switching the distances is that

<nowiki> </nowiki>Take note of the final values of the positions '''r1'''(t) '''r2'''(t) and '''p1'''(t) '''p2'''(t) for your trajectory for large enough t.

== Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed. ==

== What do you observe? STILL TO DO ==

=== Reactive and unreactive trajectories ===
{| class="wikitable" border="1"
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 ||-99.018||Yes||Reaction proceeds with minimal vibration||
|-
| -1.5 || -2.0 ||-100.456||No||P2 does not have enough momentum to reach P1||
|-
| -1.5 || -2.5 ||-98.956||Yes||Transfer of vibrational energy||
|-
| -2.5 || -5.0 || -84.956
||No||Product forms but contains too much vibrational energy so dissociates||
|-
| -2.5 || -5.2 || -83.416
||No||Product forms but contains too much vibrational energy so dissociates
||
|}

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

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

There are two basic assumptions for Transition State Theory, the two most basic are the separation of the electronic and nuclear motions which is the same as the Born-Oppenheimer approximation from quantum mechanics. The second basic assumption is that the reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution.

There are however, additional assumptions and they are as follows:
# Molecular systems that have crossed the transition state in the direction of the products cannot turn around and reform reactants.
# In the transition state, motion along the reaction coordinate may be separated from the other motions and be treated classically as a translation.
# even in the absence of an equilibrium between the reactant and product molecules, the transition states that are become products are distributed among their states according to the Maxwell-Boltzmann laws.

''Given that Transition State theory neglects quantum mechanical contributions such as influence on and between between electronic and nuclear motions and the idea that molecules passing through the transition state through to the products cannot pass back through to the reactants, this would suggest that there would be deviation between experimental and calculated values using Transition State theory.''

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

=== ''F + H2'' ===

==== ''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?'' ====
Looking at the PES of F + H2 this indicated the reaction is exothermic because there is a decrease in the energy going from the reactants to products. This can be seen along the reaction path for this reaction where there is a dip in energy on the path to the formation of the products. This makes intuitive sense as the driving force of the reaction will be the formation of a strong F-H bond, in place of a weaker H-H bond. Additional evidence that the reaction is exothermic is that when a mep calculation was run with RAB = 1.812 and RBC = 0.74 with zero momentum for each case, the reaction proceeded and the products formed with no additional input of energy.

For the H + HF reaction, '''the reaction is surmised to be endothermic due to as there is no decrease in energy on forming the products. This is in line with intuition considering the strength of the H-F bond. The very high electronegativity of fluorine would make it so that significant energy would be required to cause this reaction to proceed. '''

==== ''Locate the approximate position of the transition state.'' ====
{| class="wikitable"
!Reaction
!rAB
!rBC
!Activation Energy Kcal/mol
|-
|F + H2
|1.8108757
|0.7448759
|
|-
|H + HF
|0.7448756
|1.8108759
|
|}

[[File:Aja4117Surface Plot f + h2.png]]
[[File:Aja4117Surface Plot h + HF.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.'' ====

The set of reaction conditions that were used were:

Atom A: F

Atom B: H

Atom C: H

Distance AB: 2.25

Distance BC: 0.74

Momentum AB = -0.8

Momentum BC = 0.1

These conditions yielded a set of successful reactions, and post reaction and upon formation of F-H, there is transfer of energy in the form of kinetic energy to both the Hydrogen that is travelling away from the reaction complex and also kinetic energy in the bond vibration of F-H. This can be visualised in the Momenta vs Time graph below. Monitoring the emission spectrum of the product and observing electromagnetic radiation in the IR region would allow identification of the product.

[[File:Aja4117Surface Plot FH2 momenta vs time.png]]

For rBC = 0.74, with a momentum pAB = -0.5, and several values of pBC in the range -3 to 3 were explored. For pBC = -3, the reaction did proceed and there was significant transfer of kinetic energy to the product F-H. What is interesting to note is that the F-H bond forms twice. On the first formation, the bond breaks due to what is assumed to be, too high a vibrational energy. This leads to the reformation of the H-H bond, upon which there is still significant vibrational energy and combined with the electronegativity of the F atom causes the H-H bond to break and the F-H bond to form for the second time. The F-H bond is now stable, although still containing significant vibrational energy. For pBC = 3, the same phenomena occurs as for pBC = -3. 

==== ''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.'' ====
A transition state for an attractive potential energy surface occurs early in the reaction coordinate and involves high translational kinetic energy (evidenced by the random motion for the reactants) and results in a vibrationally excited product. This translational energy runs up the side of the valley and falls from side to side, showing that there is significant vibrational motion. We can therefore conclude that reactions with attractive potential energy surfaces proceed more efficiently if the energy is in relative translational motion.

In contrast, reactions with repulsive potential surfaces experience a late transition state and are expected to proceed with greater efficiency if the excess energy is present as vibrations. The energy is in the vibration of the reactant molecule and the motion causes the trajectory to move from side to side up the valley as it approaches the transition state. This energy is enough to tip the reactant molecules through the transition state and into the products, it is therefore apparent that the vibrational energy on the repulsive potential energy surface is key to the reactivity and formation of the products.

MRD:aja4117

2019-05-16T14:38:13Z

Aja4117: /* EXERCISE 1: H + H2 system */

= [[Molecular reaction dynamics|'''Molecular reaction dynamics''']] =

== 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?''''' ===
The transition state is a saddle point which is defined as the point which is both a minimum along one vector and a maximum along an orthogonal vector. Along the reaction path, the transition state is a maximum and along the second vector, which is perpendicular to our initial vector it is a minimum. This can be visualised in the diagram below.

For a co linear system, as given for the H + H2 system there are two orthogonal degrees of freedom with the coordinates given as follows:

'''Z = r2 + r1'''

<math>\frac {\partial V}{\partial Z} = 0 </math>, <math>\frac {\partial^2 V}{\partial Z_{Y}^2}>0 </math>

'''S = r2 - r1'''

<math>\frac {\partial V}{\partial S} = 0 </math>, <math>\frac {\partial^2 V}{\partial S_{Y}^2}<0 </math>

Note that these are the orthogonal vectors.

The saddle point has the configuration in which r1 = r2 =r0 and this will be the point at which these two lines intersect on the PES. This is the transition state. S is the tangent to the reaction pathway at the maximum whilst Z bisects the reaction pathway at the same point, but will be seen as a minimum.

[[File:Aja4117Question 1.png]]

<math>\frac {\partial V}{\partial r} = 0 </math>

<math>\frac {\partial^2 V}{\partial r_{Y}^2}<0 </math>

=== ''Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'' ===
The best estimate was AB = 0.9077425 and BC = 0.9077425. There are only two lines when looking at internuclear distances vs time as seen below. This indicates that two lines are at the same distance and one is covering the other. 

To locate the transition state, is to begin the reaction trajectories close to the transition state. Due to the transition state being a global maxima, the trajectories will either form the reactants or products in order to find the lowest energy configuration. Whether the product or reactants are formed is directly influenced by the trajectories. If the trajectories are closer to the reactants then the reactants will form and if they are closer to the products then the products will form.

[[File:Aja4117q2.PNG]]

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

===== Trajectories from r1 = rts+δ, r2 = rts =====
Along the mep, because the atoms are moving infinitely slowly and the momenta and the velocities always reset to zero, there is no oscillation and hence no vibration. This is seen along the mep where there is no wavy lines. In contrast, when looking at the dynamics, there is oscillation as post collision there is transfer of kinetic energy that causes the molecule to vibrate. 

When considering the inter nuclear distance vs time, A-C and A-B distances increase rapidly over the first 250 steps after which the rate of increase of the distance begins to decrease. As the number of steps increases the rate of increase of distance begins to plateau as the gradient flattens. This can be rationalised when looking at the animation where post collision, the molecules initially rapidly move away from one another but the speed at which they move away from each other falls off with time. When considering the dynamic situation, A-C and A-B distances increase at a constant gradient over the course of the plot. The reasoning is that in the dynamic situation, the velocities and momenta are not reset to zero at each step.

[[File:Aja4117(dynamics type)Mep vs dynamics distance v time 1.png]]
[[File:Aja4117Distance vs time 1.png]]

For the same reason as stated as above, the momenta and velocities are reset to zero at each point this means that the momentum vs time plot is a straight line for the mep . However, when running the dynamic situation we see an initial sharp increase in the momentum followed by a constant oscillation of momentum that is due to the momentum being retained within the vibrational modes. For A-B, there is a sharp increase in the momentum which plateaus at the same instance as B-C begins to undergo constant oscillation. This is because A-B refers to the momentum of the system A-B. Once both reach constant momentum the combined system now has a set defined level of momentum with no way to lose it.

[[File:Aja4117(dynamics type)Mep vs dynamics momenta vs time1.png]]
[[File:Aja4117dynamics mom vs time1.png]]

===== Trajectories from r1 = rts, r2 = rts +δ =====
The result of switching the distances is that

<nowiki> </nowiki>Take note of the final values of the positions '''r1'''(t) '''r2'''(t) and '''p1'''(t) '''p2'''(t) for your trajectory for large enough t.

== Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed. ==

== What do you observe? STILL TO DO ==

=== Reactive and unreactive trajectories ===
{| class="wikitable" border="1"
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 ||-99.018||Yes||Reaction proceeds with minimal vibration||
|-
| -1.5 || -2.0 ||-100.456||No||P2 does not have enough momentum to reach P1||
|-
| -1.5 || -2.5 ||-98.956||Yes||Transfer of vibrational energy||
|-
| -2.5 || -5.0 || -84.956
||No||Product forms but contains too much vibrational energy so dissociates||
|-
| -2.5 || -5.2 || -83.416
||No||Product forms but contains too much vibrational energy so dissociates
||
|}

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

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

There are two basic assumptions for Transition State Theory, the two most basic are the separation of the electronic and nuclear motions which is the same as the Born-Oppenheimer approximation from quantum mechanics. The second basic assumption is that the reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution.

There are however, additional assumptions and they are as follows:
# Molecular systems that have crossed the transition state in the direction of the products cannot turn around and reform reactants.
# In the transition state, motion along the reaction coordinate may be separated from the other motions and be treated classically as a translation.
# even in the absence of an equilibrium between the reactant and product molecules, the transition states that are become products are distributed among their states according to the Maxwell-Boltzmann laws.

''Given that Transition State theory neglects quantum mechanical contributions such as influence on and between between electronic and nuclear motions and the idea that molecules passing through the transition state through to the products cannot pass back through to the reactants, this would suggest that there would be deviation between experimental and calculated values using Transition State theory.''

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

=== ''F + H2'' ===

==== ''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?'' ====
Looking at the PES of F + H2 this indicated the reaction is exothermic because there is a decrease in the energy going from the reactants to products. This can be seen along the reaction path for this reaction where there is a dip in energy on the path to the formation of the products. This makes intuitive sense as the driving force of the reaction will be the formation of a strong F-H bond, in place of a weaker H-H bond. Additional evidence that the reaction is exothermic is that when a mep calculation was run with RAB = 1.812 and RBC = 0.74 with zero momentum for each case, the reaction proceeded and the products formed with no additional input of energy.

For the H + HF reaction, '''the reaction is surmised to be endothermic due to as there is no decrease in energy on forming the products. This is in line with intuition considering the strength of the H-F bond. The very high electronegativity of fluorine would make it so that significant energy would be required to cause this reaction to proceed. '''

==== ''Locate the approximate position of the transition state.'' ====
{| class="wikitable"
!Reaction
!rAB
!rBC
!Activation Energy Kcal/mol
|-
|F + H2
|1.8108757
|0.7448759
|
|-
|H + HF
|0.7448756
|1.8108759
|
|}

[[File:Aja4117Surface Plot f + h2.png]]
[[File:Aja4117Surface Plot h + HF.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.'' ====

The set of reaction conditions that were used were:

Atom A: F

Atom B: H

Atom C: H

Distance AB: 2.25

Distance BC: 0.74

Momentum AB = -0.8

Momentum BC = 0.1

These conditions yielded a set of successful reactions, and post reaction and upon formation of F-H, there is transfer of energy in the form of kinetic energy to both the Hydrogen that is travelling away from the reaction complex and also kinetic energy in the bond vibration of F-H. This can be visualised in the Momenta vs Time graph below. Monitoring the emission spectrum of the product and observing electromagnetic radiation in the IR region would allow identification of the product.

[[File:Aja4117Surface Plot FH2 momenta vs time.png]]

For rBC = 0.74, with a momentum pAB = -0.5, and several values of pBC in the range -3 to 3 were explored. For pBC = -3, the reaction did proceed and there was significant transfer of kinetic energy to the product F-H. What is interesting to note is that the F-H bond forms twice. On the first formation, the bond breaks due to what is assumed to be, too high a vibrational energy. This leads to the reformation of the H-H bond, upon which there is still significant vibrational energy and combined with the electronegativity of the F atom causes the H-H bond to break and the F-H bond to form for the second time. The F-H bond is now stable, although still containing significant vibrational energy. For pBC = 3, the same phenomena occurs as for pBC = -3. 

==== ''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.'' ====
A transition state for an attractive potential energy surface occurs early in the reaction coordinate and involves high translational kinetic energy (evidenced by the random motion for the reactants) and results in a vibrationally excited product. This translational energy runs up the side of the valley and falls from side to side, showing that there is significant vibrational motion. We can therefore conclude that reactions with attractive potential energy surfaces proceed more efficiently if the energy is in relative translational motion.

In contrast, reactions with repulsive potential surfaces experience a late transition state and are expected to proceed with greater efficiency if the excess energy is present as vibrations. The energy is in the vibration of the reactant molecule and the motion causes the trajectory to move from side to side up the valley as it approaches the transition state. This energy is enough to tip the reactant molecules through the transition state and into the products, it is therefore apparent that the vibrational energy on the repulsive potential energy surface is key to the reactivity and formation of the products.

MRD:aja4117

2019-05-16T14:34:32Z

Aja4117: /* EXERCISE 1: H + H2 system */

= [[Molecular reaction dynamics|'''Molecular reaction dynamics''']] =

== 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?''''' ===
The transition state is a saddle point which is defined as the point which is both a minimum along one vector and a maximum along an orthogonal vector. Along the reaction path, the transition state is a maximum and along the second vector, which is perpendicular to our initial vector it is a minimum. This can be visualised in the diagram below.

To locate the transition state, is to begin the reaction trajectories close to the transition state. Due to the transition state being a global maxima, the trajectories will either form the reactants or products in order to find the lowest energy configuration. Whether the product or reactants are formed is directly influenced by the trajectories. If the trajectories are closer to the reactants then the reactants will form and if they are closer to the products then the products will form.

For a co linear system, as given for the H + H2 system there are two orthogonal degrees of freedom with the coordinates are given as follows:

'''Z = r2 + r1'''

'''S = r2 - r1'''

The saddle point has the configuration in which r1 = r2 =r0 and this will be the point at which these two lines intersect on the PES. This is also the transition state. S is the tangent to the reaction pathway at the maximum whilst Z bisects the reaction pathway at the same point, but will be seen as a minimum.

In addition, the reaction potential path U(s) is a maximum and thus its partial derivative is zero and its second partial derivative is negative. Verification of the transition state can be done when considering that the partial derivative of U with respect to Z, it will be zero and its second partial derivative will be positive.

[[File:Aja4117Question 1.png]]

Additionally, a transition state is a stationary point, who's first derivative is zero with respect to position.

<math>\frac {\partial V}{\partial r} = 0 </math>

As the transition state corresponds to the ma

<math>\frac {\partial^2 V}{\partial r_{Y}^2}<0 </math>

=== ''Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'' ===
The best estimate was AB = 0.9077425 and BC = 0.9077425. There are only two lines when looking at internuclear distances vs time as seen below. This indicates that two lines are at the same distance and one is covering the other. 

[[File:Aja4117q2.PNG]]

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

===== Trajectories from r1 = rts+δ, r2 = rts =====
Along the mep, because the atoms are moving infinitely slowly and the momenta and the velocities always reset to zero, there is no oscillation and hence no vibration. This is seen along the mep where there is no wavy lines. In contrast, when looking at the dynamics, there is oscillation as post collision there is transfer of kinetic energy that causes the molecule to vibrate. 

When considering the inter nuclear distance vs time, A-C and A-B distances increase rapidly over the first 250 steps after which the rate of increase of the distance begins to decrease. As the number of steps increases the rate of increase of distance begins to plateau as the gradient flattens. This can be rationalised when looking at the animation where post collision, the molecules initially rapidly move away from one another but the speed at which they move away from each other falls off with time. When considering the dynamic situation, A-C and A-B distances increase at a constant gradient over the course of the plot. The reasoning is that in the dynamic situation, the velocities and momenta are not reset to zero at each step.

[[File:Aja4117(dynamics type)Mep vs dynamics distance v time 1.png]]
[[File:Aja4117Distance vs time 1.png]]

For the same reason as stated as above, the momenta and velocities are reset to zero at each point this means that the momentum vs time plot is a straight line for the mep . However, when running the dynamic situation we see an initial sharp increase in the momentum followed by a constant oscillation of momentum that is due to the momentum being retained within the vibrational modes. For A-B, there is a sharp increase in the momentum which plateaus at the same instance as B-C begins to undergo constant oscillation. This is because A-B refers to the momentum of the system A-B. Once both reach constant momentum the combined system now has a set defined level of momentum with no way to lose it.

[[File:Aja4117(dynamics type)Mep vs dynamics momenta vs time1.png]]
[[File:Aja4117dynamics mom vs time1.png]]

===== Trajectories from r1 = rts, r2 = rts +δ =====
The result of switching the distances is that

<nowiki> </nowiki>Take note of the final values of the positions '''r1'''(t) '''r2'''(t) and '''p1'''(t) '''p2'''(t) for your trajectory for large enough t.

== Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed. ==

== What do you observe? STILL TO DO ==

=== Reactive and unreactive trajectories ===
{| class="wikitable" border="1"
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 ||-99.018||Yes||Reaction proceeds with minimal vibration||
|-
| -1.5 || -2.0 ||-100.456||No||P2 does not have enough momentum to reach P1||
|-
| -1.5 || -2.5 ||-98.956||Yes||Transfer of vibrational energy||
|-
| -2.5 || -5.0 || -84.956
||No||Product forms but contains too much vibrational energy so dissociates||
|-
| -2.5 || -5.2 || -83.416
||No||Product forms but contains too much vibrational energy so dissociates
||
|}

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

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

There are two basic assumptions for Transition State Theory, the two most basic are the separation of the electronic and nuclear motions which is the same as the Born-Oppenheimer approximation from quantum mechanics. The second basic assumption is that the reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution.

There are however, additional assumptions and they are as follows:
# Molecular systems that have crossed the transition state in the direction of the products cannot turn around and reform reactants.
# In the transition state, motion along the reaction coordinate may be separated from the other motions and be treated classically as a translation.
# even in the absence of an equilibrium between the reactant and product molecules, the transition states that are become products are distributed among their states according to the Maxwell-Boltzmann laws.

''Given that Transition State theory neglects quantum mechanical contributions such as influence on and between between electronic and nuclear motions and the idea that molecules passing through the transition state through to the products cannot pass back through to the reactants, this would suggest that there would be deviation between experimental and calculated values using Transition State theory.''

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

=== ''F + H2'' ===

==== ''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?'' ====
Looking at the PES of F + H2 this indicated the reaction is exothermic because there is a decrease in the energy going from the reactants to products. This can be seen along the reaction path for this reaction where there is a dip in energy on the path to the formation of the products. This makes intuitive sense as the driving force of the reaction will be the formation of a strong F-H bond, in place of a weaker H-H bond. Additional evidence that the reaction is exothermic is that when a mep calculation was run with RAB = 1.812 and RBC = 0.74 with zero momentum for each case, the reaction proceeded and the products formed with no additional input of energy.

For the H + HF reaction, '''the reaction is surmised to be endothermic due to as there is no decrease in energy on forming the products. This is in line with intuition considering the strength of the H-F bond. The very high electronegativity of fluorine would make it so that significant energy would be required to cause this reaction to proceed. '''

==== ''Locate the approximate position of the transition state.'' ====
{| class="wikitable"
!Reaction
!rAB
!rBC
!Activation Energy Kcal/mol
|-
|F + H2
|1.8108757
|0.7448759
|
|-
|H + HF
|0.7448756
|1.8108759
|
|}

[[File:Aja4117Surface Plot f + h2.png]]
[[File:Aja4117Surface Plot h + HF.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.'' ====

The set of reaction conditions that were used were:

Atom A: F

Atom B: H

Atom C: H

Distance AB: 2.25

Distance BC: 0.74

Momentum AB = -0.8

Momentum BC = 0.1

These conditions yielded a set of successful reactions, and post reaction and upon formation of F-H, there is transfer of energy in the form of kinetic energy to both the Hydrogen that is travelling away from the reaction complex and also kinetic energy in the bond vibration of F-H. This can be visualised in the Momenta vs Time graph below. Monitoring the emission spectrum of the product and observing electromagnetic radiation in the IR region would allow identification of the product.

[[File:Aja4117Surface Plot FH2 momenta vs time.png]]

For rBC = 0.74, with a momentum pAB = -0.5, and several values of pBC in the range -3 to 3 were explored. For pBC = -3, the reaction did proceed and there was significant transfer of kinetic energy to the product F-H. What is interesting to note is that the F-H bond forms twice. On the first formation, the bond breaks due to what is assumed to be, too high a vibrational energy. This leads to the reformation of the H-H bond, upon which there is still significant vibrational energy and combined with the electronegativity of the F atom causes the H-H bond to break and the F-H bond to form for the second time. The F-H bond is now stable, although still containing significant vibrational energy. For pBC = 3, the same phenomena occurs as for pBC = -3. 

==== ''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.'' ====
A transition state for an attractive potential energy surface occurs early in the reaction coordinate and involves high translational kinetic energy (evidenced by the random motion for the reactants) and results in a vibrationally excited product. This translational energy runs up the side of the valley and falls from side to side, showing that there is significant vibrational motion. We can therefore conclude that reactions with attractive potential energy surfaces proceed more efficiently if the energy is in relative translational motion.

In contrast, reactions with repulsive potential surfaces experience a late transition state and are expected to proceed with greater efficiency if the excess energy is present as vibrations. The energy is in the vibration of the reactant molecule and the motion causes the trajectory to move from side to side up the valley as it approaches the transition state. This energy is enough to tip the reactant molecules through the transition state and into the products, it is therefore apparent that the vibrational energy on the repulsive potential energy surface is key to the reactivity and formation of the products.

MRD:aja4117

2019-05-16T14:26:59Z

Aja4117: /* EXERCISE 1: H + H2 system */

= [[Molecular reaction dynamics|'''Molecular reaction dynamics''']] =

== 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?''''' ===
The transition state is defined as the point at which the partial derivative of the potential energy with respect to position is zero. Intuitively this is the '''maximum''' along the '''''minimum energy'' path'''.

To locate the transition state, is to begin the reaction trajectories close to the transition state. Due to the transition state being a global maxima, the trajectories will either form the reactants or products in order to find the lowest energy configuration. Whether the product or reactants are formed is directly influenced by the trajectories. If the trajectories are closer to the reactants then the reactants will form and if they are closer to the products then the products will form.

For a co linear system, as given for the H + H2 system there are two orthogonal degrees of freedom with the coordinates are given as follows:

'''Z = r2 + r1'''

'''S = r2 - r1'''

The saddle point has the configuration in which r1 = r2 =r0 and this will be the point at which these two lines intersect on the PES. This is also the transition state. S is the tangent to the reaction pathway at the maximum whilst Z bisects the reaction pathway at the same point, but will be seen as a minimum.

In addition, the reaction potential path U(s) is a maximum and thus its partial derivative is zero and its second partial derivative is negative. Verification of the transition state can be done when considering that the partial derivative of U with respect to Z, it will be zero and its second partial derivative will be positive.

[[File:Aja4117Question 1.png]]

Additionally, a transition state is a stationary point, whos first derivative is zero with respect to position.

<math>\frac {\partial V}{\partial r} = 0 </math>

As the transitoion state corresponds to the ma

<math>\frac {\partial^2 V}{\partial r_{Y}^2}<0 </math>

=== ''Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'' ===
The best estimate was AB = 0.9077425 and BC = 0.9077425. There are only two lines when looking at internuclear distances vs time as seen below. This indicates that two lines are at the same distance and one is covering the other. 

[[File:Aja4117q2.PNG]]

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

===== Trajectories from r1 = rts+δ, r2 = rts =====
Along the mep, because the atoms are moving infinitely slowly and the momenta and the velocities always reset to zero, there is no oscillation and hence no vibration. This is seen along the mep where there is no wavy lines. In contrast, when looking at the dynamics, there is oscillation as post collision there is transfer of kinetic energy that causes the molecule to vibrate. 

When considering the inter nuclear distance vs time, A-C and A-B distances increase rapidly over the first 250 steps after which the rate of increase of the distance begins to decrease. As the number of steps increases the rate of increase of distance begins to plateau as the gradient flattens. This can be rationalised when looking at the animation where post collision, the molecules initially rapidly move away from one another but the speed at which they move away from each other falls off with time. When considering the dynamic situation, A-C and A-B distances increase at a constant gradient over the course of the plot. The reasoning is that in the dynamic situation, the velocities and momenta are not reset to zero at each step.

[[File:Aja4117(dynamics type)Mep vs dynamics distance v time 1.png]]
[[File:Aja4117Distance vs time 1.png]]

For the same reason as stated as above, the momenta and velocities are reset to zero at each point this means that the momentum vs time plot is a straight line for the mep . However, when running the dynamic situation we see an initial sharp increase in the momentum followed by a constant oscillation of momentum that is due to the momentum being retained within the vibrational modes. For A-B, there is a sharp increase in the momentum which plateaus at the same instance as B-C begins to undergo constant oscillation. This is because A-B refers to the momentum of the system A-B. Once both reach constant momentum the combined system now has a set defined level of momentum with no way to lose it.

[[File:Aja4117(dynamics type)Mep vs dynamics momenta vs time1.png]]
[[File:Aja4117dynamics mom vs time1.png]]

===== Trajectories from r1 = rts, r2 = rts +δ =====
The result of switching the distances is that

<nowiki> </nowiki>Take note of the final values of the positions '''r1'''(t) '''r2'''(t) and '''p1'''(t) '''p2'''(t) for your trajectory for large enough t.

== Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed. ==

== What do you observe? STILL TO DO ==

=== Reactive and unreactive trajectories ===
{| class="wikitable" border="1"
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 ||-99.018||Yes||Reaction proceeds with minimal vibration||
|-
| -1.5 || -2.0 ||-100.456||No||P2 does not have enough momentum to reach P1||
|-
| -1.5 || -2.5 ||-98.956||Yes||Transfer of vibrational energy||
|-
| -2.5 || -5.0 || -84.956
||No||Product forms but contains too much vibrational energy so dissociates||
|-
| -2.5 || -5.2 || -83.416
||No||Product forms but contains too much vibrational energy so dissociates
||
|}

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

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

There are two basic assumptions for Transition State Theory, the two most basic are the separation of the electronic and nuclear motions which is the same as the Born-Oppenheimer approximation from quantum mechanics. The second basic assumption is that the reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution.

There are however, additional assumptions and they are as follows:
# Molecular systems that have crossed the transition state in the direction of the products cannot turn around and reform reactants.
# In the transition state, motion along the reaction coordinate may be separated from the other motions and be treated classically as a translation.
# even in the absence of an equilibrium between the reactant and product molecules, the transition states that are become products are distributed among their states according to the Maxwell-Boltzmann laws.

''Given that Transition State theory neglects quantum mechanical contributions such as influence on and between between electronic and nuclear motions and the idea that molecules passing through the transition state through to the products cannot pass back through to the reactants, this would suggest that there would be deviation between experimental and calculated values using Transition State theory.''

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

=== ''F + H2'' ===

==== ''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?'' ====
Looking at the PES of F + H2 this indicated the reaction is exothermic because there is a decrease in the energy going from the reactants to products. This can be seen along the reaction path for this reaction where there is a dip in energy on the path to the formation of the products. This makes intuitive sense as the driving force of the reaction will be the formation of a strong F-H bond, in place of a weaker H-H bond. Additional evidence that the reaction is exothermic is that when a mep calculation was run with RAB = 1.812 and RBC = 0.74 with zero momentum for each case, the reaction proceeded and the products formed with no additional input of energy.

For the H + HF reaction, '''the reaction is surmised to be endothermic due to as there is no decrease in energy on forming the products. This is in line with intuition considering the strength of the H-F bond. The very high electronegativity of fluorine would make it so that significant energy would be required to cause this reaction to proceed. '''

==== ''Locate the approximate position of the transition state.'' ====
{| class="wikitable"
!Reaction
!rAB
!rBC
!Activation Energy Kcal/mol
|-
|F + H2
|1.8108757
|0.7448759
|
|-
|H + HF
|0.7448756
|1.8108759
|
|}

[[File:Aja4117Surface Plot f + h2.png]]
[[File:Aja4117Surface Plot h + HF.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.'' ====

The set of reaction conditions that were used were:

Atom A: F

Atom B: H

Atom C: H

Distance AB: 2.25

Distance BC: 0.74

Momentum AB = -0.8

Momentum BC = 0.1

These conditions yielded a set of successful reactions, and post reaction and upon formation of F-H, there is transfer of energy in the form of kinetic energy to both the Hydrogen that is travelling away from the reaction complex and also kinetic energy in the bond vibration of F-H. This can be visualised in the Momenta vs Time graph below. Monitoring the emission spectrum of the product and observing electromagnetic radiation in the IR region would allow identification of the product.

[[File:Aja4117Surface Plot FH2 momenta vs time.png]]

For rBC = 0.74, with a momentum pAB = -0.5, and several values of pBC in the range -3 to 3 were explored. For pBC = -3, the reaction did proceed and there was significant transfer of kinetic energy to the product F-H. What is interesting to note is that the F-H bond forms twice. On the first formation, the bond breaks due to what is assumed to be, too high a vibrational energy. This leads to the reformation of the H-H bond, upon which there is still significant vibrational energy and combined with the electronegativity of the F atom causes the H-H bond to break and the F-H bond to form for the second time. The F-H bond is now stable, although still containing significant vibrational energy. For pBC = 3, the same phenomena occurs as for pBC = -3. 

==== ''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.'' ====
A transition state for an attractive potential energy surface occurs early in the reaction coordinate and involves high translational kinetic energy (evidenced by the random motion for the reactants) and results in a vibrationally excited product. This translational energy runs up the side of the valley and falls from side to side, showing that there is significant vibrational motion. We can therefore conclude that reactions with attractive potential energy surfaces proceed more efficiently if the energy is in relative translational motion.

In contrast, reactions with repulsive potential surfaces experience a late transition state and are expected to proceed with greater efficiency if the excess energy is present as vibrations. The energy is in the vibration of the reactant molecule and the motion causes the trajectory to move from side to side up the valley as it approaches the transition state. This energy is enough to tip the reactant molecules through the transition state and into the products, it is therefore apparent that the vibrational energy on the repulsive potential energy surface is key to the reactivity and formation of the products.

MRD:aja4117

2019-05-16T14:19:41Z

Aja4117: /* EXERCISE 1: H + H2 system */

= [[Molecular reaction dynamics|'''Molecular reaction dynamics''']] =

== 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?''''' ===
The transition state is defined as the point at which the partial derivative of the potential energy with respect to position is zero. Intuitively this is the '''maximum''' along the '''''minimum energy'' path'''.

To locate the transition state, is to begin the reaction trajectories close to the transition state. Due to the transition state being a global maxima, the trajectories will either form the reactants or products in order to find the lowest energy configuration. Whether the product or reactants are formed is directly influenced by the trajectories. If the trajectories are closer to the reactants then the reactants will form and if they are closer to the products then the products will form.

For a co linear system, as given for the H + H2 system there are two orthogonal degrees of freedom with the coordinates are given as follows:

'''Z = r2 + r1'''

'''S = r2 - r1'''

The saddle point has the configuration in which r1 = r2 =r0 and this will be the point at which these two lines intersect on the PES. This is also the transition state. S is the tangent to the reaction pathway at the maximum whilst Z bisects the reaction pathway at the same point, but will be seen as a minimum.

In addition, the reaction potential path U(s) is a maximum and thus its partial derivative is zero and its second partial derivative is negative. Verification of the transition state can be done when considering that the partial derivative of U with respect to Z, it will be zero and its second partial derivative will be positive.

[[File:Aja4117Question 1.png]]

Additionally, a transition state is a stationary point, whos first derivative is zero with respect to position.

<math>\frac {\partial V}{\partial r} =\frac {\partial V}{\partial r} = 0 </math>

=== ''Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'' ===
The best estimate was AB = 0.9077425 and BC = 0.9077425. There are only two lines when looking at internuclear distances vs time as seen below. This indicates that two lines are at the same distance and one is covering the other. 

[[File:Aja4117q2.PNG]]

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

===== Trajectories from r1 = rts+δ, r2 = rts =====
Along the mep, because the atoms are moving infinitely slowly and the momenta and the velocities always reset to zero, there is no oscillation and hence no vibration. This is seen along the mep where there is no wavy lines. In contrast, when looking at the dynamics, there is oscillation as post collision there is transfer of kinetic energy that causes the molecule to vibrate. 

When considering the inter nuclear distance vs time, A-C and A-B distances increase rapidly over the first 250 steps after which the rate of increase of the distance begins to decrease. As the number of steps increases the rate of increase of distance begins to plateau as the gradient flattens. This can be rationalised when looking at the animation where post collision, the molecules initially rapidly move away from one another but the speed at which they move away from each other falls off with time. When considering the dynamic situation, A-C and A-B distances increase at a constant gradient over the course of the plot. The reasoning is that in the dynamic situation, the velocities and momenta are not reset to zero at each step.

[[File:Aja4117(dynamics type)Mep vs dynamics distance v time 1.png]]
[[File:Aja4117Distance vs time 1.png]]

For the same reason as stated as above, the momenta and velocities are reset to zero at each point this means that the momentum vs time plot is a straight line for the mep . However, when running the dynamic situation we see an initial sharp increase in the momentum followed by a constant oscillation of momentum that is due to the momentum being retained within the vibrational modes. For A-B, there is a sharp increase in the momentum which plateaus at the same instance as B-C begins to undergo constant oscillation. This is because A-B refers to the momentum of the system A-B. Once both reach constant momentum the combined system now has a set defined level of momentum with no way to lose it.

[[File:Aja4117(dynamics type)Mep vs dynamics momenta vs time1.png]]
[[File:Aja4117dynamics mom vs time1.png]]

===== Trajectories from r1 = rts, r2 = rts +δ =====
The result of switching the distances is that

<nowiki> </nowiki>Take note of the final values of the positions '''r1'''(t) '''r2'''(t) and '''p1'''(t) '''p2'''(t) for your trajectory for large enough t.

== Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed. ==

== What do you observe? STILL TO DO ==

=== Reactive and unreactive trajectories ===
{| class="wikitable" border="1"
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 ||-99.018||Yes||Reaction proceeds with minimal vibration||
|-
| -1.5 || -2.0 ||-100.456||No||P2 does not have enough momentum to reach P1||
|-
| -1.5 || -2.5 ||-98.956||Yes||Transfer of vibrational energy||
|-
| -2.5 || -5.0 || -84.956
||No||Product forms but contains too much vibrational energy so dissociates||
|-
| -2.5 || -5.2 || -83.416
||No||Product forms but contains too much vibrational energy so dissociates
||
|}

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

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

There are two basic assumptions for Transition State Theory, the two most basic are the separation of the electronic and nuclear motions which is the same as the Born-Oppenheimer approximation from quantum mechanics. The second basic assumption is that the reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution.

There are however, additional assumptions and they are as follows:
# Molecular systems that have crossed the transition state in the direction of the products cannot turn around and reform reactants.
# In the transition state, motion along the reaction coordinate may be separated from the other motions and be treated classically as a translation.
# even in the absence of an equilibrium between the reactant and product molecules, the transition states that are become products are distributed among their states according to the Maxwell-Boltzmann laws.

''Given that Transition State theory neglects quantum mechanical contributions such as influence on and between between electronic and nuclear motions and the idea that molecules passing through the transition state through to the products cannot pass back through to the reactants, this would suggest that there would be deviation between experimental and calculated values using Transition State theory.''

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

=== ''F + H2'' ===

==== ''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?'' ====
Looking at the PES of F + H2 this indicated the reaction is exothermic because there is a decrease in the energy going from the reactants to products. This can be seen along the reaction path for this reaction where there is a dip in energy on the path to the formation of the products. This makes intuitive sense as the driving force of the reaction will be the formation of a strong F-H bond, in place of a weaker H-H bond. Additional evidence that the reaction is exothermic is that when a mep calculation was run with RAB = 1.812 and RBC = 0.74 with zero momentum for each case, the reaction proceeded and the products formed with no additional input of energy.

For the H + HF reaction, '''the reaction is surmised to be endothermic due to as there is no decrease in energy on forming the products. This is in line with intuition considering the strength of the H-F bond. The very high electronegativity of fluorine would make it so that significant energy would be required to cause this reaction to proceed. '''

==== ''Locate the approximate position of the transition state.'' ====
{| class="wikitable"
!Reaction
!rAB
!rBC
!Activation Energy Kcal/mol
|-
|F + H2
|1.8108757
|0.7448759
|
|-
|H + HF
|0.7448756
|1.8108759
|
|}

[[File:Aja4117Surface Plot f + h2.png]]
[[File:Aja4117Surface Plot h + HF.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.'' ====

The set of reaction conditions that were used were:

Atom A: F

Atom B: H

Atom C: H

Distance AB: 2.25

Distance BC: 0.74

Momentum AB = -0.8

Momentum BC = 0.1

These conditions yielded a set of successful reactions, and post reaction and upon formation of F-H, there is transfer of energy in the form of kinetic energy to both the Hydrogen that is travelling away from the reaction complex and also kinetic energy in the bond vibration of F-H. This can be visualised in the Momenta vs Time graph below. Monitoring the emission spectrum of the product and observing electromagnetic radiation in the IR region would allow identification of the product.

[[File:Aja4117Surface Plot FH2 momenta vs time.png]]

For rBC = 0.74, with a momentum pAB = -0.5, and several values of pBC in the range -3 to 3 were explored. For pBC = -3, the reaction did proceed and there was significant transfer of kinetic energy to the product F-H. What is interesting to note is that the F-H bond forms twice. On the first formation, the bond breaks due to what is assumed to be, too high a vibrational energy. This leads to the reformation of the H-H bond, upon which there is still significant vibrational energy and combined with the electronegativity of the F atom causes the H-H bond to break and the F-H bond to form for the second time. The F-H bond is now stable, although still containing significant vibrational energy. For pBC = 3, the same phenomena occurs as for pBC = -3. 

==== ''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.'' ====
A transition state for an attractive potential energy surface occurs early in the reaction coordinate and involves high translational kinetic energy (evidenced by the random motion for the reactants) and results in a vibrationally excited product. This translational energy runs up the side of the valley and falls from side to side, showing that there is significant vibrational motion. We can therefore conclude that reactions with attractive potential energy surfaces proceed more efficiently if the energy is in relative translational motion.

In contrast, reactions with repulsive potential surfaces experience a late transition state and are expected to proceed with greater efficiency if the excess energy is present as vibrations. The energy is in the vibration of the reactant molecule and the motion causes the trajectory to move from side to side up the valley as it approaches the transition state. This energy is enough to tip the reactant molecules through the transition state and into the products, it is therefore apparent that the vibrational energy on the repulsive potential energy surface is key to the reactivity and formation of the products.

MRD:aja4117

2019-05-16T14:19:08Z

Aja4117: /* EXERCISE 1: H + H2 system */

= [[Molecular reaction dynamics|'''Molecular reaction dynamics''']] =

== 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?''''' ===
The transition state is defined as the point at which the partial derivative of the potential energy with respect to position is zero. Intuitively this is the '''maximum''' along the '''''minimum energy'' path'''.

To locate the transition state, is to begin the reaction trajectories close to the transition state. Due to the transition state being a global maxima, the trajectories will either form the reactants or products in order to find the lowest energy configuration. Whether the product or reactants are formed is directly influenced by the trajectories. If the trajectories are closer to the reactants then the reactants will form and if they are closer to the products then the products will form.

For a co linear system, as given for the H + H2 system there are two orthogonal degrees of freedom with the coordinates are given as follows:

'''Z = r2 + r1'''

'''S = r2 - r1'''

The saddle point has the configuration in which r1 = r2 =r0 and this will be the point at which these two lines intersect on the PES. This is also the transition state. S is the tangent to the reaction pathway at the maximum whilst Z bisects the reaction pathway at the same point, but will be seen as a minimum.

In addition, the reaction potential path U(s) is a maximum and thus its partial derivative is zero and its second partial derivative is negative. Verification of the transition state can be done when considering that the partial derivative of U with respect to Z, it will be zero and its second partial derivative will be positive.

[[File:Aja4117Question 1.png]]

Additionally, a transition state is a stationary point, whos first derivative is zero with respect to position.

=== ''Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'' ===
The best estimate was AB = 0.9077425 and BC = 0.9077425. There are only two lines when looking at internuclear distances vs time as seen below. This indicates that two lines are at the same distance and one is covering the other. 

[[File:Aja4117q2.PNG]]

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

===== Trajectories from r1 = rts+δ, r2 = rts =====
Along the mep, because the atoms are moving infinitely slowly and the momenta and the velocities always reset to zero, there is no oscillation and hence no vibration. This is seen along the mep where there is no wavy lines. In contrast, when looking at the dynamics, there is oscillation as post collision there is transfer of kinetic energy that causes the molecule to vibrate. 

When considering the inter nuclear distance vs time, A-C and A-B distances increase rapidly over the first 250 steps after which the rate of increase of the distance begins to decrease. As the number of steps increases the rate of increase of distance begins to plateau as the gradient flattens. This can be rationalised when looking at the animation where post collision, the molecules initially rapidly move away from one another but the speed at which they move away from each other falls off with time. When considering the dynamic situation, A-C and A-B distances increase at a constant gradient over the course of the plot. The reasoning is that in the dynamic situation, the velocities and momenta are not reset to zero at each step.

[[File:Aja4117(dynamics type)Mep vs dynamics distance v time 1.png]]
[[File:Aja4117Distance vs time 1.png]]

For the same reason as stated as above, the momenta and velocities are reset to zero at each point this means that the momentum vs time plot is a straight line for the mep . However, when running the dynamic situation we see an initial sharp increase in the momentum followed by a constant oscillation of momentum that is due to the momentum being retained within the vibrational modes. For A-B, there is a sharp increase in the momentum which plateaus at the same instance as B-C begins to undergo constant oscillation. This is because A-B refers to the momentum of the system A-B. Once both reach constant momentum the combined system now has a set defined level of momentum with no way to lose it.

[[File:Aja4117(dynamics type)Mep vs dynamics momenta vs time1.png]]
[[File:Aja4117dynamics mom vs time1.png]]

===== Trajectories from r1 = rts, r2 = rts +δ =====
The result of switching the distances is that

<nowiki> </nowiki>Take note of the final values of the positions '''r1'''(t) '''r2'''(t) and '''p1'''(t) '''p2'''(t) for your trajectory for large enough t.

== Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed. ==

== What do you observe? STILL TO DO ==

=== Reactive and unreactive trajectories ===
{| class="wikitable" border="1"
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 ||-99.018||Yes||Reaction proceeds with minimal vibration||
|-
| -1.5 || -2.0 ||-100.456||No||P2 does not have enough momentum to reach P1||
|-
| -1.5 || -2.5 ||-98.956||Yes||Transfer of vibrational energy||
|-
| -2.5 || -5.0 || -84.956
||No||Product forms but contains too much vibrational energy so dissociates||
|-
| -2.5 || -5.2 || -83.416
||No||Product forms but contains too much vibrational energy so dissociates
||
|}

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

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

There are two basic assumptions for Transition State Theory, the two most basic are the separation of the electronic and nuclear motions which is the same as the Born-Oppenheimer approximation from quantum mechanics. The second basic assumption is that the reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution.

There are however, additional assumptions and they are as follows:
# Molecular systems that have crossed the transition state in the direction of the products cannot turn around and reform reactants.
# In the transition state, motion along the reaction coordinate may be separated from the other motions and be treated classically as a translation.
# even in the absence of an equilibrium between the reactant and product molecules, the transition states that are become products are distributed among their states according to the Maxwell-Boltzmann laws.

''Given that Transition State theory neglects quantum mechanical contributions such as influence on and between between electronic and nuclear motions and the idea that molecules passing through the transition state through to the products cannot pass back through to the reactants, this would suggest that there would be deviation between experimental and calculated values using Transition State theory.''

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

=== ''F + H2'' ===

==== ''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?'' ====
Looking at the PES of F + H2 this indicated the reaction is exothermic because there is a decrease in the energy going from the reactants to products. This can be seen along the reaction path for this reaction where there is a dip in energy on the path to the formation of the products. This makes intuitive sense as the driving force of the reaction will be the formation of a strong F-H bond, in place of a weaker H-H bond. Additional evidence that the reaction is exothermic is that when a mep calculation was run with RAB = 1.812 and RBC = 0.74 with zero momentum for each case, the reaction proceeded and the products formed with no additional input of energy.

For the H + HF reaction, '''the reaction is surmised to be endothermic due to as there is no decrease in energy on forming the products. This is in line with intuition considering the strength of the H-F bond. The very high electronegativity of fluorine would make it so that significant energy would be required to cause this reaction to proceed. '''

==== ''Locate the approximate position of the transition state.'' ====
{| class="wikitable"
!Reaction
!rAB
!rBC
!Activation Energy Kcal/mol
|-
|F + H2
|1.8108757
|0.7448759
|
|-
|H + HF
|0.7448756
|1.8108759
|
|}

[[File:Aja4117Surface Plot f + h2.png]]
[[File:Aja4117Surface Plot h + HF.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.'' ====

The set of reaction conditions that were used were:

Atom A: F

Atom B: H

Atom C: H

Distance AB: 2.25

Distance BC: 0.74

Momentum AB = -0.8

Momentum BC = 0.1

These conditions yielded a set of successful reactions, and post reaction and upon formation of F-H, there is transfer of energy in the form of kinetic energy to both the Hydrogen that is travelling away from the reaction complex and also kinetic energy in the bond vibration of F-H. This can be visualised in the Momenta vs Time graph below. Monitoring the emission spectrum of the product and observing electromagnetic radiation in the IR region would allow identification of the product.

[[File:Aja4117Surface Plot FH2 momenta vs time.png]]

For rBC = 0.74, with a momentum pAB = -0.5, and several values of pBC in the range -3 to 3 were explored. For pBC = -3, the reaction did proceed and there was significant transfer of kinetic energy to the product F-H. What is interesting to note is that the F-H bond forms twice. On the first formation, the bond breaks due to what is assumed to be, too high a vibrational energy. This leads to the reformation of the H-H bond, upon which there is still significant vibrational energy and combined with the electronegativity of the F atom causes the H-H bond to break and the F-H bond to form for the second time. The F-H bond is now stable, although still containing significant vibrational energy. For pBC = 3, the same phenomena occurs as for pBC = -3. 

==== ''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.'' ====
A transition state for an attractive potential energy surface occurs early in the reaction coordinate and involves high translational kinetic energy (evidenced by the random motion for the reactants) and results in a vibrationally excited product. This translational energy runs up the side of the valley and falls from side to side, showing that there is significant vibrational motion. We can therefore conclude that reactions with attractive potential energy surfaces proceed more efficiently if the energy is in relative translational motion.

In contrast, reactions with repulsive potential surfaces experience a late transition state and are expected to proceed with greater efficiency if the excess energy is present as vibrations. The energy is in the vibration of the reactant molecule and the motion causes the trajectory to move from side to side up the valley as it approaches the transition state. This energy is enough to tip the reactant molecules through the transition state and into the products, it is therefore apparent that the vibrational energy on the repulsive potential energy surface is key to the reactivity and formation of the products.

MRD:aja4117

2019-05-16T14:17:40Z

Aja4117:

= [[Molecular reaction dynamics|'''Molecular reaction dynamics''']] =

== 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?''''' ===
The transition state is defined as the point at which the partial derivative of the potential energy with respect to position is zero. Intuitively this is the '''maximum''' along the '''''minimum energy'' path'''.

To locate the transition state, is to begin the reaction trajectories close to the transition state. Due to the transition state being a global maxima, the trajectories will either form the reactants or products in order to find the lowest energy configuration. Whether the product or reactants are formed is directly influenced by the trajectories. If the trajectories are closer to the reactants then the reactants will form and if they are closer to the products then the products will form.

For a co linear system, as given for the H + H2 system there are two orthogonal degrees of freedom with the coordinates are given as follows:

'''Z = r2 + r1'''

'''S = r2 - r1'''

The saddle point has the configuration in which r1 = r2 =r0 and this will be the point at which these two lines intersect on the PES. This is also the transition state. S is the tangent to the reaction pathway at the maximum whilst Z bisects the reaction pathway at the same point, but will be seen as a minimum.

In addition, the reaction potential path U(s) is a maximum and thus its partial derivative is zero and its second partial derivative is negative. Verification of the transition state can be done when considering that the partial derivative of U with respect to Z, it will be zero and its second partial derivative will be positive.

[[File:Aja4117Question 1.png]]

=== ''Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'' ===
The best estimate was AB = 0.9077425 and BC = 0.9077425. There are only two lines when looking at internuclear distances vs time as seen below. This indicates that two lines are at the same distance and one is covering the other. 

[[File:Aja4117q2.PNG]]

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

===== Trajectories from r1 = rts+δ, r2 = rts =====
Along the mep, because the atoms are moving infinitely slowly and the momenta and the velocities always reset to zero, there is no oscillation and hence no vibration. This is seen along the mep where there is no wavy lines. In contrast, when looking at the dynamics, there is oscillation as post collision there is transfer of kinetic energy that causes the molecule to vibrate. 

When considering the inter nuclear distance vs time, A-C and A-B distances increase rapidly over the first 250 steps after which the rate of increase of the distance begins to decrease. As the number of steps increases the rate of increase of distance begins to plateau as the gradient flattens. This can be rationalised when looking at the animation where post collision, the molecules initially rapidly move away from one another but the speed at which they move away from each other falls off with time. When considering the dynamic situation, A-C and A-B distances increase at a constant gradient over the course of the plot. The reasoning is that in the dynamic situation, the velocities and momenta are not reset to zero at each step.

[[File:Aja4117(dynamics type)Mep vs dynamics distance v time 1.png]]
[[File:Aja4117Distance vs time 1.png]]

For the same reason as stated as above, the momenta and velocities are reset to zero at each point this means that the momentum vs time plot is a straight line for the mep . However, when running the dynamic situation we see an initial sharp increase in the momentum followed by a constant oscillation of momentum that is due to the momentum being retained within the vibrational modes. For A-B, there is a sharp increase in the momentum which plateaus at the same instance as B-C begins to undergo constant oscillation. This is because A-B refers to the momentum of the system A-B. Once both reach constant momentum the combined system now has a set defined level of momentum with no way to lose it.

[[File:Aja4117(dynamics type)Mep vs dynamics momenta vs time1.png]]
[[File:Aja4117dynamics mom vs time1.png]]

===== Trajectories from r1 = rts, r2 = rts +δ =====
The result of switching the distances is that

<nowiki> </nowiki>Take note of the final values of the positions '''r1'''(t) '''r2'''(t) and '''p1'''(t) '''p2'''(t) for your trajectory for large enough t.

== Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed. ==

== What do you observe? STILL TO DO ==

=== Reactive and unreactive trajectories ===
{| class="wikitable" border="1"
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 ||-99.018||Yes||Reaction proceeds with minimal vibration||
|-
| -1.5 || -2.0 ||-100.456||No||P2 does not have enough momentum to reach P1||
|-
| -1.5 || -2.5 ||-98.956||Yes||Transfer of vibrational energy||
|-
| -2.5 || -5.0 || -84.956
||No||Product forms but contains too much vibrational energy so dissociates||
|-
| -2.5 || -5.2 || -83.416
||No||Product forms but contains too much vibrational energy so dissociates
||
|}

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

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

There are two basic assumptions for Transition State Theory, the two most basic are the separation of the electronic and nuclear motions which is the same as the Born-Oppenheimer approximation from quantum mechanics. The second basic assumption is that the reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution.

There are however, additional assumptions and they are as follows:
# Molecular systems that have crossed the transition state in the direction of the products cannot turn around and reform reactants.
# In the transition state, motion along the reaction coordinate may be separated from the other motions and be treated classically as a translation.
# even in the absence of an equilibrium between the reactant and product molecules, the transition states that are become products are distributed among their states according to the Maxwell-Boltzmann laws.

''Given that Transition State theory neglects quantum mechanical contributions such as influence on and between between electronic and nuclear motions and the idea that molecules passing through the transition state through to the products cannot pass back through to the reactants, this would suggest that there would be deviation between experimental and calculated values using Transition State theory.''

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

=== ''F + H2'' ===

==== ''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?'' ====
Looking at the PES of F + H2 this indicated the reaction is exothermic because there is a decrease in the energy going from the reactants to products. This can be seen along the reaction path for this reaction where there is a dip in energy on the path to the formation of the products. This makes intuitive sense as the driving force of the reaction will be the formation of a strong F-H bond, in place of a weaker H-H bond. Additional evidence that the reaction is exothermic is that when a mep calculation was run with RAB = 1.812 and RBC = 0.74 with zero momentum for each case, the reaction proceeded and the products formed with no additional input of energy.

For the H + HF reaction, '''the reaction is surmised to be endothermic due to as there is no decrease in energy on forming the products. This is in line with intuition considering the strength of the H-F bond. The very high electronegativity of fluorine would make it so that significant energy would be required to cause this reaction to proceed. '''

==== ''Locate the approximate position of the transition state.'' ====
{| class="wikitable"
!Reaction
!rAB
!rBC
!Activation Energy Kcal/mol
|-
|F + H2
|1.8108757
|0.7448759
|
|-
|H + HF
|0.7448756
|1.8108759
|
|}

[[File:Aja4117Surface Plot f + h2.png]]
[[File:Aja4117Surface Plot h + HF.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.'' ====

The set of reaction conditions that were used were:

Atom A: F

Atom B: H

Atom C: H

Distance AB: 2.25

Distance BC: 0.74

Momentum AB = -0.8

Momentum BC = 0.1

These conditions yielded a set of successful reactions, and post reaction and upon formation of F-H, there is transfer of energy in the form of kinetic energy to both the Hydrogen that is travelling away from the reaction complex and also kinetic energy in the bond vibration of F-H. This can be visualised in the Momenta vs Time graph below. Monitoring the emission spectrum of the product and observing electromagnetic radiation in the IR region would allow identification of the product.

[[File:Aja4117Surface Plot FH2 momenta vs time.png]]

For rBC = 0.74, with a momentum pAB = -0.5, and several values of pBC in the range -3 to 3 were explored. For pBC = -3, the reaction did proceed and there was significant transfer of kinetic energy to the product F-H. What is interesting to note is that the F-H bond forms twice. On the first formation, the bond breaks due to what is assumed to be, too high a vibrational energy. This leads to the reformation of the H-H bond, upon which there is still significant vibrational energy and combined with the electronegativity of the F atom causes the H-H bond to break and the F-H bond to form for the second time. The F-H bond is now stable, although still containing significant vibrational energy. For pBC = 3, the same phenomena occurs as for pBC = -3. 

==== ''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.'' ====
A transition state for an attractive potential energy surface occurs early in the reaction coordinate and involves high translational kinetic energy (evidenced by the random motion for the reactants) and results in a vibrationally excited product. This translational energy runs up the side of the valley and falls from side to side, showing that there is significant vibrational motion. We can therefore conclude that reactions with attractive potential energy surfaces proceed more efficiently if the energy is in relative translational motion.

In contrast, reactions with repulsive potential surfaces experience a late transition state and are expected to proceed with greater efficiency if the excess energy is present as vibrations. The energy is in the vibration of the reactant molecule and the motion causes the trajectory to move from side to side up the valley as it approaches the transition state. This energy is enough to tip the reactant molecules through the transition state and into the products, it is therefore apparent that the vibrational energy on the repulsive potential energy surface is key to the reactivity and formation of the products.

MRD:aja4117

2019-05-16T14:15:54Z

Aja4117: /* EXERCISE 1: H + H2 system */

= [[Molecular reaction dynamics|'''Molecular reaction dynamics''']] =

== 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?''''' ===
The transition state is defined as the point at which the partial derivative of the potential energy with respect to position is zero. Intuitively this is the '''maximum''' along the '''''minimum energy'' path'''.

To locate the transition state, is to begin the reaction trajectories close to the transition state. Due to the transition state being a global maxima, the trajectories will either form the reactants or products in order to find the lowest energy configuration. Whether the product or reactants are formed is directly influenced by the trajectories. If the trajectories are closer to the reactants then the reactants will form and if they are closer to the products then the products will form.

For a co linear system, as given for the H + H2 system there are two orthogonal degrees of freedom with the coordinates are given as follows:

'''Z = r2 + r1'''

'''S = r2 - r1'''

The saddle point has the configuration in which r1 = r2 =r0 and this will be the point at which these two lines intersect on the PES. This is also the transition state. S is the tangent to the reaction pathway at the maximum whilst Z bisects the reaction pathway at the same point, but will be seen as a minimum.

In addition, the reaction potential path U(s) is a maximum and thus its partial derivative is zero and its second partial derivative is negative. Verification of the transition state can be done when considering that the partial derivative of U with respect to Z, it will be zero and its second partial derivative will be positive.

[[File:Aja4117Question 1.png]]

=== ''Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'' ===
The best estimate was AB = 0.9077425 and BC = 0.9077425. There are only two lines when looking at internuclear distances vs time as seen below. This indicates that two lines are at the same distance and one is covering the other. 

[[File:Aja4117q2.PNG]]

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

===== Trajectories from r1 = rts+δ, r2 = rts =====
Along the mep, because the atoms are moving infinitely slowly and the momenta and the velocities always reset to zero, there is no oscillation and hence no vibration. This is seen along the mep where there is no wavy lines. In contrast, when looking at the dynamics, there is oscillation as post collision there is transfer of kinetic energy that causes the molecule to vibrate. 

When considering the inter nuclear distance vs time, A-C and A-B distances increase rapidly over the first 250 steps after which the rate of increase of the distance begins to decrease. As the number of steps increases the rate of increase of distance begins to plateau as the gradient flattens. This can be rationalised when looking at the animation where post collision, the molecules initially rapidly move away from one another but the speed at which they move away from each other falls off with time. When considering the dynamic situation, A-C and A-B distances increase at a constant gradient over the course of the plot. The reasoning is that in the dynamic situation, the velocities and momenta are not reset to zero at each step.

[[File:Aja4117(dynamics type)Mep vs dynamics distance v time 1.png]]
[[File:Aja4117Distance vs time 1.png]]

For the same reason as stated as above, the momenta and velocities are reset to zero at each point this means that the momentum vs time plot is a straight line for the mep . However, when running the dynamic situation we see an initial sharp increase in the momentum followed by a constant oscillation of momentum that is due to the momentum being retained within the vibrational modes. For A-B, there is a sharp increase in the momentum which plateaus at the same instance as B-C begins to undergo constant oscillation. This is because A-B refers to the momentum of the system A-B. Once both reach constant momentum the combined system now has a set defined level of momentum with no way to lose it.

[[File:Aja4117(dynamics type)Mep vs dynamics momenta vs time1.png]]
[[File:Aja4117dynamics mom vs time1.png]]

===== Trajectories from r1 = rts, r2 = rts +δ =====
The result of switching the distances is that

<nowiki> </nowiki>Take note of the final values of the positions '''r1'''(t) '''r2'''(t) and '''p1'''(t) '''p2'''(t) for your trajectory for large enough t.

== Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed. ==

== What do you observe? STILL TO DO ==

=== Reactive and unreactive trajectories ===
{| class="wikitable" border="1"
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 ||-99.018||Yes||Reaction proceeds with minimal vibration||
|-
| -1.5 || -2.0 ||-100.456||No||P2 does not have enough momentum to reach P1||
|-
| -1.5 || -2.5 ||-98.956||Yes||Transfer of vibrational energy||
|-
| -2.5 || -5.0 || -84.956
||No||Product forms but contains too much vibrational energy so dissociates||
|-
| -2.5 || -5.2 || -83.416
||No||Product forms but contains too much vibrational energy so dissociates
||
|}

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

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

There are two basic assumptions for Transition State Theory, the two most basic are the separation of the electronic and nuclear motions which is the same as the Born-Oppenheimer approximation from quantum mechanics. The second basic assumption is that the reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution.

There are however, additional assumptions and they are as follows:
# Molecular systems that have crossed the transition state in the direction of the products cannot turn around and reform reactants.
# In the transition state, motion along the reaction coordinate may be separated from the other motions and be treated classically as a translation.
# even in the absence of an equilibrium between the reactant and product molecules, the transition states that are become products are distributed among their states according to the Maxwell-Boltzmann laws.

''Given that Transition State theory neglects quantum mechanical contributions such as influence on and between between electronic and nuclear motions and the idea that molecules passing through the transition state through to the products cannot pass back through to the reactants, this would suggest that there would be deviation between experimental and calculated values using Transition State theory.''

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

=== ''F + H2'' ===

==== ''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?'' ====
Looking at the PES of F + H2 this indicated the reaction is exothermic because there is a decrease in the energy going from the reactants to products. This can be seen along the reaction path for this reaction where there is a dip in energy on the path to the formation of the products. This makes intuitive sense as the driving force of the reaction will be the formation of a strong F-H bond, in place of a weaker H-H bond. Additional evidence that the reaction is exothermic is that when a mep calculation was run with RAB = 1.812 and RBC = 0.74 with zero momentum for each case, the reaction proceeded and the products formed with no additional input of energy.

For the H + HF reaction, '''the reaction is surmised to be endothermic due to as there is no decrease in energy on forming the products. This is in line with intuition considering the strength of the H-F bond. The very high electronegativity of fluorine would make it so that significant energy would be required to cause this reaction to proceed. '''

==== ''Locate the approximate position of the transition state.'' ====
{| class="wikitable"
!Reaction
!rAB
!rBC
!Activation Energy Kcal/mol
|-
|F + H2
|1.8108757
|0.7448759
|
|-
|H + HF
|0.7448756
|1.8108759
|
|}

==== ''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.'' ====

The set of reaction conditions that were used were:

Atom A: F

Atom B: H

Atom C: H

Distance AB: 2.25

Distance BC: 0.74

Momentum AB = -0.8

Momentum BC = 0.1

These conditions yielded a set of successful reactions, and post reaction and upon formation of F-H, there is transfer of energy in the form of kinetic energy to both the Hydrogen that is travelling away from the reaction complex and also kinetic energy in the bond vibration of F-H. This can be visualised in the Momenta vs Time graph below. Monitoring the emission spectrum of the product and observing electromagnetic radiation in the IR region would allow identification of the product.

[[File:Aja4117Surface Plot FH2 momenta vs time.png]]

For rBC = 0.74, with a momentum pAB = -0.5, and several values of pBC in the range -3 to 3 were explored. For pBC = -3, the reaction did proceed and there was significant transfer of kinetic energy to the product F-H. What is interesting to note is that the F-H bond forms twice. On the first formation, the bond breaks due to what is assumed to be, too high a vibrational energy. This leads to the reformation of the H-H bond, upon which there is still significant vibrational energy and combined with the electronegativity of the F atom causes the H-H bond to break and the F-H bond to form for the second time. The F-H bond is now stable, although still containing significant vibrational energy. For pBC = 3, the same phenomena occurs as for pBC = -3. 

==== ''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.'' ====
A transition state for an attractive potential energy surface occurs early in the reaction coordinate and involves high translational kinetic energy (evidenced by the random motion for the reactants) and results in a vibrationally excited product. This translational energy runs up the side of the valley and falls from side to side, showing that there is significant vibrational motion. We can therefore conclude that reactions with attractive potential energy surfaces proceed more efficiently if the energy is in relative translational motion.

In contrast, reactions with repulsive potential surfaces experience a late transition state and are expected to proceed with greater efficiency if the excess energy is present as vibrations. The energy is in the vibration of the reactant molecule and the motion causes the trajectory to move from side to side up the valley as it approaches the transition state. This energy is enough to tip the reactant molecules through the transition state and into the products, it is therefore apparent that the vibrational energy on the repulsive potential energy surface is key to the reactivity and formation of the products.

File:Aja4117Surface Plot h + HF.png

2019-05-16T14:15:36Z

Aja4117:

File:Aja4117Surface Plot f + h2.png

2019-05-16T14:15:06Z

Aja4117:

MRD:aja4117

2019-05-16T06:45:22Z

Aja4117: /* EXERCISE 2: F - H - H system */

= [[Molecular reaction dynamics|'''Molecular reaction dynamics''']] =

== 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?''''' ===
The transition state is defined as the point at which the partial derivative of the potential energy with respect to position is zero. Intuitively this is the '''maximum''' along the '''''minimum energy'' path'''.

To locate the transition state, is to begin the reaction trajectories close to the transition state. Due to the transition state being a global maxima, the trajectories will either form the reactants or products in order to find the lowest energy configuration. Whether the product or reactants are formed is directly influenced by the trajectories. If the trajectories are closer to the reactants then the reactants will form and if they are closer to the products then the products will form.

For a co linear system, as given for the H + H2 system there are two orthogonal degrees of freedom with the coordinates are given as follows:

'''Z = r2 + r1'''

'''S = r2 - r1'''

The saddle point has the configuration in which r1 = r2 =r0 and this will be the point at which these two lines intersect on the PES. This is also the transition state. S is the tangent to the reaction pathway at the maximum whilst Z bisects the reaction pathway at the same point, but will be seen as a minimum.

In addition, the reaction potential path U(s) is a maximum and thus its partial derivative is zero and its second partial derivative is negative. Verification of the transition state can be done when considering that the partial derivative of U with respect to Z, it will be zero and its second partial derivative will be positive.

[[File:Aja4117Question 1.png]]

=== ''Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'' ===
The best estimate was AB = 0.9077425 and BC = 0.9077425. There are only two lines when looking at internuclear distances vs time as seen below. This indicates that two lines are at the same distance and one is covering the other. 

[[File:Aja4117q2.PNG]]

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

===== Trajectories from r1 = rts+δ, r2 = rts =====
Along the mep, because the atoms are moving infinitely slowly and the momenta and the velocities always reset to zero, there is no oscillation and hence no vibration. This is seen along the mep where there is no wavy lines. In contrast, when looking at the dynamics, there is oscillation as post collision there is transfer of kinetic energy that causes the molecule to vibrate. 

When considering the inter nuclear distance vs time, A-C and A-B distances increase rapidly over the first 250 steps after which the rate of increase of the distance begins to decrease. As the number of steps increases the rate of increase of distance begins to plateau as the gradient flattens. This can be rationalised when looking at the animation where post collision, the molecules initially rapidly move away from one another but the speed at which they move away from each other falls off with time. When considering the dynamic situation, A-C and A-B distances increase at a constant gradient over the course of the plot. The reasoning is that in the dynamic situation, the velocities and momenta are not reset to zero at each step.

[[File:Aja4117(dynamics type)Mep vs dynamics distance v time 1.png]]
[[File:Aja4117Distance vs time 1.png]]

For the same reason as stated as above, the momenta and velocities are reset to zero at each point this means that the momentum vs time plot is a straight line for the mep . However, when running the dynamic situation we see an initial sharp increase in the momentum followed by a constant oscillation of momentum that is due to the momentum being retained within the vibrational modes. For A-B, there is a sharp increase in the momentum which plateaus at the same instance as B-C begins to undergo constant oscillation. This is because A-B refers to the momentum of the system A-B. Once both reach constant momentum the combined system now has a set defined level of momentum with no way to lose it.

[[File:Aja4117(dynamics type)Mep vs dynamics momenta vs time1.png]]
[[File:Aja4117dynamics mom vs time1.png]]

===== Trajectories from r1 = rts, r2 = rts +δ =====
The result of switching the distances is that

<nowiki> </nowiki>Take note of the final values of the positions '''r1'''(t) '''r2'''(t) and '''p1'''(t) '''p2'''(t) for your trajectory for large enough t.

== Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed. ==

== What do you observe? STILL TO DO ==

=== Reactive and unreactive trajectories ===
{| class="wikitable" border="1"
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 ||-99.018||Yes||Reaction proceeds with minimal vibration||
|-
| -1.5 || -2.0 ||-100.456||No||P2 does not have enough momentum to reach P1||
|-
| -1.5 || -2.5 ||-98.956||Yes||Transfer of vibrational energy||
|-
| -2.5 || -5.0 || -84.956
||No||Product forms but contains too much vibrational energy so dissociates||
|-
| -2.5 || -5.2 || -83.416
||No||Product forms but contains too much vibrational energy so dissociates
||
|}

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

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

There are two basic assumptions for Transition State Theory, the two most basic are the separation of the electronic and nuclear motions which is the same as the Born-Oppenheimer approximation from quantum mechanics. The second basic assumption is that the reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution.

There are however, additional assumptions and they are as follows:
# Molecular systems that have crossed the transition state in the direction of the products cannot turn around and reform reactants.
# In the transition state, motion along the reaction coordinate may be separated from the other motions and be treated classically as a translation.
# even in the absence of an equilibrium between the reactant and product molecules, the transition states that are become products are distributed among their states according to the Maxwell-Boltzmann laws.

''Given that Transition State theory neglects quantum mechanical contributions such as influence on and between between electronic and nuclear motions and the idea that molecules passing through the transition state through to the products cannot pass back through to the reactants, this would suggest that there would be deviation between experimental and calculated values using Transition State theory.''

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

=== ''F + H2'' ===

==== ''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?'' ====
Looking at the PES of F + H2 this indicated the reaction is exothermic because there is a decrease in the energy going from the reactants to products. This can be seen along the reaction path for this reaction where there is a dip in energy on the path to the formation of the products. This makes intuitive sense as the driving force of the reaction will be the formation of a strong F-H bond, in place of a weaker H-H bond. Additional evidence that the reaction is exothermic is that when a mep calculation was run with RAB = 1.812 and RBC = 0.74 with zero momentum for each case, the reaction proceeded and the products formed with no additional input of energy.

For the H + HF reaction, '''the reaction is surmised to be endothermic due to as there is no decrease in energy on forming the products. This is in line with intuition considering the strength of the H-F bond. The very high electronegativity of fluorine would make it so that significant energy would be required to cause this reaction to proceed. '''

==== ''Locate the approximate position of the transition state.'' ====
For the reaction of F + H2 the approximate position of the transition state was reported to be 

==== ''Report the activation energy for both reactions.'' ====
 

==== ''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.'' ====

The set of reaction conditions that were used were:

Atom A: F

Atom B: H

Atom C: H

Distance AB: 2.25

Distance BC: 0.74

Momentum AB = -0.8

Momentum BC = 0.1

These conditions yielded a set of successful reactions, and post reaction and upon formation of F-H, there is transfer of energy in the form of kinetic energy to both the Hydrogen that is travelling away from the reaction complex and also kinetic energy in the bond vibration of F-H. This can be visualised in the Momenta vs Time graph below. '''HOW COULD THIS BE IDENTIFIED??'''

[[File:Aja4117Surface Plot FH2 momenta vs time.png]]

For rBC = 0.74, with a momentum pAB = -0.5, and several values of pBC in the range -3 to 3 were explored. For pBC = -3, the reaction did proceed and there was significant transfer of kinetic energy to the product F-H. What is interesting to note is that the F-H bond forms twice. On the first formation, the bond breaks due to what is assumed to be, too high a vibrational energy. This leads to the reformation of the H-H bond, upon which there is still significant vibrational energy and combined with the electronegativity of the F atom causes the H-H bond to break and the F-H bond to form for the second time. The F-H bond is now stable, although still containing significant vibrational energy. For pBC = 3, the same phenomena occurs as for pBC = -3. 

==== ''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.'' ====
A transition state for an attractive potential energy surface occurs early in the reaction coordinate and involves high translational kinetic energy (evidenced by the random motion for the reactants) and results in a vibrationally excited product. This translational energy runs up the side of the valley and falls from side to side, showing that there is significant vibrational motion. We can therefore conclude that reactions with attractive potential energy surfaces proceed more efficiently if the energy is in relative translational motion.

In contrast, reactions with repulsive potential surfaces experience a late transition state and are expected to proceed with greater efficiency if the excess energy is present as vibrations. The energy is in the vibration of the reactant molecule and the motion causes the trajectory to move from side to side up the valley as it approaches the transition state. This energy is enough to tip the reactant molecules through the transition state and into the products, it is therefore apparent that the vibrational energy on the repulsive potential energy surface is key to the reactivity and formation of the products.

MRD:aja4117

2019-05-15T23:05:53Z

Aja4117: /* 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. */

= [[Molecular reaction dynamics|'''Molecular reaction dynamics''']] =

== 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?''''' ===
The transition state is defined as the point at which the partial derivative of the potential energy with respect to position is zero. Intuitively this is the '''maximum''' along the '''''minimum energy'' path'''.

To locate the transition state, is to begin the reaction trajectories close to the transition state. Due to the transition state being a global maxima, the trajectories will either form the reactants or products in order to find the lowest energy configuration. Whether the product or reactants are formed is directly influenced by the trajectories. If the trajectories are closer to the reactants then the reactants will form and if they are closer to the products then the products will form.

For a co linear system, as given for the H + H2 system there are two orthogonal degrees of freedom with the coordinates are given as follows:

'''Z = r2 + r1'''

'''S = r2 - r1'''

The saddle point has the configuration in which r1 = r2 =r0 and this will be the point at which these two lines intersect on the PES. This is also the transition state. S is the tangent to the reaction pathway at the maximum whilst Z bisects the reaction pathway at the same point, but will be seen as a minimum.

In addition, the reaction potential path U(s) is a maximum and thus its partial derivative is zero and its second partial derivative is negative. Verification of the transition state can be done when considering that the partial derivative of U with respect to Z, it will be zero and its second partial derivative will be positive.

[[File:Aja4117Question 1.png]]

=== ''Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'' ===
The best estimate was AB = 0.9077425 and BC = 0.9077425. There are only two lines when looking at internuclear distances vs time as seen below. This indicates that two lines are at the same distance and one is covering the other. 

[[File:Aja4117q2.PNG]]

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

===== Trajectories from r1 = rts+δ, r2 = rts =====
Along the mep, because the atoms are moving infinitely slowly and the momenta and the velocities always reset to zero, there is no oscillation and hence no vibration. This is seen along the mep where there is no wavy lines. In contrast, when looking at the dynamics, there is oscillation as post collision there is transfer of kinetic energy that causes the molecule to vibrate. 

When considering the inter nuclear distance vs time, A-C and A-B distances increase rapidly over the first 250 steps after which the rate of increase of the distance begins to decrease. As the number of steps increases the rate of increase of distance begins to plateau as the gradient flattens. This can be rationalised when looking at the animation where post collision, the molecules initially rapidly move away from one another but the speed at which they move away from each other falls off with time. When considering the dynamic situation, A-C and A-B distances increase at a constant gradient over the course of the plot. The reasoning is that in the dynamic situation, the velocities and momenta are not reset to zero at each step.

[[File:Aja4117(dynamics type)Mep vs dynamics distance v time 1.png]]
[[File:Aja4117Distance vs time 1.png]]

For the same reason as stated as above, the momenta and velocities are reset to zero at each point this means that the momentum vs time plot is a straight line for the mep . However, when running the dynamic situation we see an initial sharp increase in the momentum followed by a constant oscillation of momentum that is due to the momentum being retained within the vibrational modes. For A-B, there is a sharp increase in the momentum which plateaus at the same instance as B-C begins to undergo constant oscillation. This is because A-B refers to the momentum of the system A-B. Once both reach constant momentum the combined system now has a set defined level of momentum with no way to lose it.

[[File:Aja4117(dynamics type)Mep vs dynamics momenta vs time1.png]]
[[File:Aja4117dynamics mom vs time1.png]]

===== Trajectories from r1 = rts, r2 = rts +δ =====
The result of switching the distances is that

<nowiki> </nowiki>Take note of the final values of the positions '''r1'''(t) '''r2'''(t) and '''p1'''(t) '''p2'''(t) for your trajectory for large enough t.

== Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed. ==

== What do you observe? STILL TO DO ==

=== Reactive and unreactive trajectories ===
{| class="wikitable" border="1"
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 ||-99.018||Yes||Reaction proceeds with minimal vibration||
|-
| -1.5 || -2.0 ||-100.456||No||P2 does not have enough momentum to reach P1||
|-
| -1.5 || -2.5 ||-98.956||Yes||Transfer of vibrational energy||
|-
| -2.5 || -5.0 || -84.956
||No||Product forms but contains too much vibrational energy so dissociates||
|-
| -2.5 || -5.2 || -83.416
||No||Product forms but contains too much vibrational energy so dissociates
||
|}

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

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

There are two basic assumptions for Transition State Theory, the two most basic are the separation of the electronic and nuclear motions which is the same as the Born-Oppenheimer approximation from quantum mechanics. The second basic assumption is that the reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution.

There are however, additional assumptions and they are as follows:
# Molecular systems that have crossed the transition state in the direction of the products cannot turn around and reform reactants.
# In the transition state, motion along the reaction coordinate may be separated from the other motions and be treated classically as a translation.
# even in the absence of an equilibrium between the reactant and product molecules, the transition states that are become products are distributed among their states according to the Maxwell-Boltzmann laws.

''Given that Transition State theory neglects quantum mechanical contributions such as influence on and between between electronic and nuclear motions and the idea that molecules passing through the transition state through to the products cannot pass back through to the reactants, this would suggest that there would be deviation between experimental and calculated values using Transition State theory.''

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

=== ''F + H2'' ===

==== ''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?'' ====
Looking at the PES of F + H2 this indicated the reaction is exothermic because there is a decrease in the energy going from the reactants to products. This can be seen along the reaction path for this reaction where there is a dip in energy on the path to the formation of the products. This makes intuitive sense as the driving force of the reaction will be the formation of a strong F-H bond, in place of a weaker H-H bond.

For the H + HF reaction, '''the reaction is surmised to be endothermic due to as there is no decrease in energy on forming the products. This is in line with intuition considering the strength of the H-F bond. The very high electronegativity of fluorine would make it so that significant energy would be required to cause this reaction to proceed. '''

==== ''Locate the approximate position of the transition state.'' ====
For the reaction of F + H2 the approximate position of the transition state was reported to be 

==== ''Report the activation energy for both reactions.'' ====
 

==== ''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.'' ====

The set of reaction conditions that were used were:

Atom A: F

Atom B: H

Atom C: H

Distance AB: 2.25

Distance BC: 0.74

Momentum AB = -0.8

Momentum BC = 0.1

These conditions yielded a set of successful reactions, and post reaction and upon formation of F-H, there is transfer of energy in the form of kinetic energy to both the Hydrogen that is travelling away from the reaction complex and also kinetic energy in the bond vibration of F-H. This can be visualised in the Momenta vs Time graph below. '''HOW COULD THIS BE IDENTIFIED??'''

[[File:Aja4117Surface Plot FH2 momenta vs time.png]]

For rBC = 0.74, with a momentum pAB = -0.5, and several values of pBC in the range -3 to 3 were explored. For pBC = -3, the reaction did proceed and there was significant transfer of kinetic energy to the product F-H. What is interesting to note is that the F-H bond forms twice. On the first formation, the bond breaks due to what is assumed to be, too high a vibrational energy. This leads to the reformation of the H-H bond, upon which there is still significant vibrational energy and combined with the electronegativity of the F atom causes the H-H bond to break and the F-H bond to form for the second time. The F-H bond is now stable, although still containing significant vibrational energy. For pBC = 3, the same phenomena occurs as for pBC = -3. 

==== ''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.'' ====
A transition state for an attractive potential energy surface occurs early in the reaction coordinate and involves high translational kinetic energy (evidenced by the random motion for the reactants) and results in a vibrationally excited product. This translational energy runs up the side of the valley and falls from side to side, showing that there is significant vibrational motion. We can therefore conclude that reactions with attractive potential energy surfaces proceed more efficiently if the energy is in relative translational motion.

In contrast, reactions with repulsive potential surfaces experience a late transition state and are expected to proceed with greater efficiency if the excess energy is present as vibrations. The energy is in the vibration of the reactant molecule and the motion causes the trajectory to move from side to side up the valley as it approaches the transition state. This energy is enough to tip the reactant molecules through the transition state and into the products, it is therefore apparent that the vibrational energy on the repulsive potential energy surface is key to the reactivity and formation of the products.

MRD:aja4117

2019-05-15T22:39:44Z

Aja4117: /* EXERCISE 2: F - H - H system */

= [[Molecular reaction dynamics|'''Molecular reaction dynamics''']] =

== 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?''''' ===
The transition state is defined as the point at which the partial derivative of the potential energy with respect to position is zero. Intuitively this is the '''maximum''' along the '''''minimum energy'' path'''.

To locate the transition state, is to begin the reaction trajectories close to the transition state. Due to the transition state being a global maxima, the trajectories will either form the reactants or products in order to find the lowest energy configuration. Whether the product or reactants are formed is directly influenced by the trajectories. If the trajectories are closer to the reactants then the reactants will form and if they are closer to the products then the products will form.

For a co linear system, as given for the H + H2 system there are two orthogonal degrees of freedom with the coordinates are given as follows:

'''Z = r2 + r1'''

'''S = r2 - r1'''

The saddle point has the configuration in which r1 = r2 =r0 and this will be the point at which these two lines intersect on the PES. This is also the transition state. S is the tangent to the reaction pathway at the maximum whilst Z bisects the reaction pathway at the same point, but will be seen as a minimum.

In addition, the reaction potential path U(s) is a maximum and thus its partial derivative is zero and its second partial derivative is negative. Verification of the transition state can be done when considering that the partial derivative of U with respect to Z, it will be zero and its second partial derivative will be positive.

[[File:Aja4117Question 1.png]]

=== ''Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'' ===
The best estimate was AB = 0.9077425 and BC = 0.9077425. There are only two lines when looking at internuclear distances vs time as seen below. This indicates that two lines are at the same distance and one is covering the other. 

[[File:Aja4117q2.PNG]]

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

===== Trajectories from r1 = rts+δ, r2 = rts =====
Along the mep, because the atoms are moving infinitely slowly and the momenta and the velocities always reset to zero, there is no oscillation and hence no vibration. This is seen along the mep where there is no wavy lines. In contrast, when looking at the dynamics, there is oscillation as post collision there is transfer of kinetic energy that causes the molecule to vibrate. 

When considering the inter nuclear distance vs time, A-C and A-B distances increase rapidly over the first 250 steps after which the rate of increase of the distance begins to decrease. As the number of steps increases the rate of increase of distance begins to plateau as the gradient flattens. This can be rationalised when looking at the animation where post collision, the molecules initially rapidly move away from one another but the speed at which they move away from each other falls off with time. When considering the dynamic situation, A-C and A-B distances increase at a constant gradient over the course of the plot. The reasoning is that in the dynamic situation, the velocities and momenta are not reset to zero at each step.

[[File:Aja4117(dynamics type)Mep vs dynamics distance v time 1.png]]
[[File:Aja4117Distance vs time 1.png]]

For the same reason as stated as above, the momenta and velocities are reset to zero at each point this means that the momentum vs time plot is a straight line for the mep . However, when running the dynamic situation we see an initial sharp increase in the momentum followed by a constant oscillation of momentum that is due to the momentum being retained within the vibrational modes. For A-B, there is a sharp increase in the momentum which plateaus at the same instance as B-C begins to undergo constant oscillation. This is because A-B refers to the momentum of the system A-B. Once both reach constant momentum the combined system now has a set defined level of momentum with no way to lose it.

[[File:Aja4117(dynamics type)Mep vs dynamics momenta vs time1.png]]
[[File:Aja4117dynamics mom vs time1.png]]

===== Trajectories from r1 = rts, r2 = rts +δ =====
The result of switching the distances is that

<nowiki> </nowiki>Take note of the final values of the positions '''r1'''(t) '''r2'''(t) and '''p1'''(t) '''p2'''(t) for your trajectory for large enough t.

== Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed. ==

== What do you observe? STILL TO DO ==

=== Reactive and unreactive trajectories ===
{| class="wikitable" border="1"
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 ||-99.018||Yes||Reaction proceeds with minimal vibration||
|-
| -1.5 || -2.0 ||-100.456||No||P2 does not have enough momentum to reach P1||
|-
| -1.5 || -2.5 ||-98.956||Yes||Transfer of vibrational energy||
|-
| -2.5 || -5.0 || -84.956
||No||Product forms but contains too much vibrational energy so dissociates||
|-
| -2.5 || -5.2 || -83.416
||No||Product forms but contains too much vibrational energy so dissociates
||
|}

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

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

There are two basic assumptions for Transition State Theory, the two most basic are the separation of the electronic and nuclear motions which is the same as the Born-Oppenheimer approximation from quantum mechanics. The second basic assumption is that the reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution.

There are however, additional assumptions and they are as follows:
# Molecular systems that have crossed the transition state in the direction of the products cannot turn around and reform reactants.
# In the transition state, motion along the reaction coordinate may be separated from the other motions and be treated classically as a translation.
# even in the absence of an equilibrium between the reactant and product molecules, the transition states that are become products are distributed among their states according to the Maxwell-Boltzmann laws.

''Given that Transition State theory neglects quantum mechanical contributions such as influence on and between between electronic and nuclear motions and the idea that molecules passing through the transition state through to the products cannot pass back through to the reactants, this would suggest that there would be deviation between experimental and calculated values using Transition State theory.''

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

=== ''F + H2'' ===

==== ''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?'' ====
Looking at the PES of F + H2 this indicated the reaction is exothermic because there is a decrease in the energy going from the reactants to products. This can be seen along the reaction path for this reaction where there is a dip in energy on the path to the formation of the products. This makes intuitive sense as the driving force of the reaction will be the formation of a strong F-H bond, in place of a weaker H-H bond.

For the H + HF reaction, '''the reaction is surmised to be endothermic due to as there is no decrease in energy on forming the products. This is in line with intuition considering the strength of the H-F bond. The very high electronegativity of fluorine would make it so that significant energy would be required to cause this reaction to proceed. '''

==== ''Locate the approximate position of the transition state.'' ====
For the reaction of F + H2 the approximate position of the transition state was reported to be 

==== ''Report the activation energy for both reactions.'' ====
 

==== ''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.'' ====

The set of reaction conditions that were used were:

Atom A: F

Atom B: H

Atom C: H

Distance AB: 2.25

Distance BC: 0.74

Momentum AB = -0.8

Momentum BC = 0.1

These conditions yielded a set of successful reactions, and post reaction and upon formation of F-H, there is transfer of energy in the form of kinetic energy to both the Hydrogen that is travelling away from the reaction complex and also kinetic energy in the bond vibration of F-H. This can be visualised in the Momenta vs Time graph below. '''HOW COULD THIS BE IDENTIFIED??'''

[[File:Aja4117Surface Plot FH2 momenta vs time.png]]

For rBC = 0.74, with a momentum pAB = -0.5, and several values of pBC in the range -3 to 3 were explored. For pBC = -3, the reaction did proceed and there was significant transfer of kinetic energy to the product F-H. What is interesting to note is that the F-H bond forms twice. On the first formation, the bond breaks due to what is assumed to be, too high a vibrational energy. This leads to the reformation of the H-H bond, upon which there is still significant vibrational energy and combined with the electronegativity of the F atom causes the H-H bond to break and the F-H bond to form for the second time. The F-H bond is now stable, although still containing significant vibrational energy. For pBC = 3, the same phenomena occurs as for pBC = -3. 

==== ''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.'' ====

=== ''H + HF '' ===

MRD:aja4117

2019-05-15T22:13:48Z

Aja4117:

= [[Molecular reaction dynamics|'''Molecular reaction dynamics''']] =

== 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?''''' ===
The transition state is defined as the point at which the partial derivative of the potential energy with respect to position is zero. Intuitively this is the '''maximum''' along the '''''minimum energy'' path'''.

To locate the transition state, is to begin the reaction trajectories close to the transition state. Due to the transition state being a global maxima, the trajectories will either form the reactants or products in order to find the lowest energy configuration. Whether the product or reactants are formed is directly influenced by the trajectories. If the trajectories are closer to the reactants then the reactants will form and if they are closer to the products then the products will form.

For a co linear system, as given for the H + H2 system there are two orthogonal degrees of freedom with the coordinates are given as follows:

'''Z = r2 + r1'''

'''S = r2 - r1'''

The saddle point has the configuration in which r1 = r2 =r0 and this will be the point at which these two lines intersect on the PES. This is also the transition state. S is the tangent to the reaction pathway at the maximum whilst Z bisects the reaction pathway at the same point, but will be seen as a minimum.

In addition, the reaction potential path U(s) is a maximum and thus its partial derivative is zero and its second partial derivative is negative. Verification of the transition state can be done when considering that the partial derivative of U with respect to Z, it will be zero and its second partial derivative will be positive.

[[File:Aja4117Question 1.png]]

=== ''Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'' ===
The best estimate was AB = 0.9077425 and BC = 0.9077425. There are only two lines when looking at internuclear distances vs time as seen below. This indicates that two lines are at the same distance and one is covering the other. 

[[File:Aja4117q2.PNG]]

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

===== Trajectories from r1 = rts+δ, r2 = rts =====
Along the mep, because the atoms are moving infinitely slowly and the momenta and the velocities always reset to zero, there is no oscillation and hence no vibration. This is seen along the mep where there is no wavy lines. In contrast, when looking at the dynamics, there is oscillation as post collision there is transfer of kinetic energy that causes the molecule to vibrate. 

When considering the inter nuclear distance vs time, A-C and A-B distances increase rapidly over the first 250 steps after which the rate of increase of the distance begins to decrease. As the number of steps increases the rate of increase of distance begins to plateau as the gradient flattens. This can be rationalised when looking at the animation where post collision, the molecules initially rapidly move away from one another but the speed at which they move away from each other falls off with time. When considering the dynamic situation, A-C and A-B distances increase at a constant gradient over the course of the plot. The reasoning is that in the dynamic situation, the velocities and momenta are not reset to zero at each step.

[[File:Aja4117(dynamics type)Mep vs dynamics distance v time 1.png]]
[[File:Aja4117Distance vs time 1.png]]

For the same reason as stated as above, the momenta and velocities are reset to zero at each point this means that the momentum vs time plot is a straight line for the mep . However, when running the dynamic situation we see an initial sharp increase in the momentum followed by a constant oscillation of momentum that is due to the momentum being retained within the vibrational modes. For A-B, there is a sharp increase in the momentum which plateaus at the same instance as B-C begins to undergo constant oscillation. This is because A-B refers to the momentum of the system A-B. Once both reach constant momentum the combined system now has a set defined level of momentum with no way to lose it.

[[File:Aja4117(dynamics type)Mep vs dynamics momenta vs time1.png]]
[[File:Aja4117dynamics mom vs time1.png]]

===== Trajectories from r1 = rts, r2 = rts +δ =====
The result of switching the distances is that

<nowiki> </nowiki>Take note of the final values of the positions '''r1'''(t) '''r2'''(t) and '''p1'''(t) '''p2'''(t) for your trajectory for large enough t.

== Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed. ==

== What do you observe? STILL TO DO ==

=== Reactive and unreactive trajectories ===
{| class="wikitable" border="1"
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 ||-99.018||Yes||Reaction proceeds with minimal vibration||
|-
| -1.5 || -2.0 ||-100.456||No||P2 does not have enough momentum to reach P1||
|-
| -1.5 || -2.5 ||-98.956||Yes||Transfer of vibrational energy||
|-
| -2.5 || -5.0 || -84.956
||No||Product forms but contains too much vibrational energy so dissociates||
|-
| -2.5 || -5.2 || -83.416
||No||Product forms but contains too much vibrational energy so dissociates
||
|}

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

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

There are two basic assumptions for Transition State Theory, the two most basic are the separation of the electronic and nuclear motions which is the same as the Born-Oppenheimer approximation from quantum mechanics. The second basic assumption is that the reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution.

There are however, additional assumptions and they are as follows:
# Molecular systems that have crossed the transition state in the direction of the products cannot turn around and reform reactants.
# In the transition state, motion along the reaction coordinate may be separated from the other motions and be treated classically as a translation.
# even in the absence of an equilibrium between the reactant and product molecules, the transition states that are become products are distributed among their states according to the Maxwell-Boltzmann laws.

''Given that Transition State theory neglects quantum mechanical contributions such as influence on and between between electronic and nuclear motions and the idea that molecules passing through the transition state through to the products cannot pass back through to the reactants, this would suggest that there would be deviation between experimental and calculated values using Transition State theory.''

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

=== ''F + H2'' ===

==== ''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?'' ====
Looking at the PES of F + H2 this indicated the reaction is exothermic because there is a decrease in the energy going from the reactants to products. This can be seen along the reaction path for this reaction where there is a dip in energy on the path to the formation of the products. This makes intuitive sense as the driving force of the reaction will be the formation of a strong F-H bond, in place of a weaker H-H bond.

For the H + HF reaction, '''the reaction is surmised to be endothermic due to as there is no decrease in energy on forming the products. This is in line with intuition considering the strength of the H-F bond. The very high electronegativity of fluorine would make it so that significant energy would be required to cause this reaction to proceed. '''

==== ''Locate the approximate position of the transition state.'' ====
For the reaction of F + H2 the approximate position of the transition state was reported to be 

==== ''Report the activation energy for both reactions.'' ====
 

==== ''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.'' ====

The set of reaction conditions that were used were:

Atom A: F

Atom B: H

Atom C: H

Distance AB: 2.25

Distance BC: 0.74

Momentum AB = -0.8

Momentum BC = 0.1

These conditions yielded a set of successful reactions, and post reaction and upon formation of F-H, there is transfer of energy in the form of kinetic energy to both the Hydrogen that is travelling away from the reaction complex and also kinetic energy in the bond vibration of F-H. This can be visualised in the Momenta vs Time graph below.

[[File:Aja4117Surface Plot FH2 momenta vs time.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.'' ====

=== ''H + HF '' ===

MRD:aja4117

2019-05-15T22:13:02Z

Aja4117: /* EXERCISE 2: F - H - H syste */

= [[Molecular reaction dynamics|'''Molecular reaction dynamics''']] =

== 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?''''' ===
The transition state is defined as the point at which the partial derivative of the potential energy with respect to position is zero. Intuitively this is the '''maximum''' along the '''''minimum energy'' path'''.

To locate the transition state, is to begin the reaction trajectories close to the transition state. Due to the transition state being a global maxima, the trajectories will either form the reactants or products in order to find the lowest energy configuration. Whether the product or reactants are formed is directly influenced by the trajectories. If the trajectories are closer to the reactants then the reactants will form and if they are closer to the products then the products will form.

For a co linear system, as given for the H + H2 system there are two orthogonal degrees of freedom with the coordinates are given as follows:

'''Z = r2 + r1'''

'''S = r2 - r1'''

The saddle point has the configuration in which r1 = r2 =r0 and this will be the point at which these two lines intersect on the PES. This is also the transition state. S is the tangent to the reaction pathway at the maximum whilst Z bisects the reaction pathway at the same point, but will be seen as a minimum.

In addition, the reaction potential path U(s) is a maximum and thus its partial derivative is zero and its second partial derivative is negative. Verification of the transition state can be done when considering that the partial derivative of U with respect to Z, it will be zero and its second partial derivative will be positive.

[[File:Aja4117Question 1.png]]

=== ''Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'' ===
The best estimate was AB = 0.9077425 and BC = 0.9077425. There are only two lines when looking at internuclear distances vs time as seen below. This indicates that two lines are at the same distance and one is covering the other. 

[[File:Aja4117q2.PNG]]

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

===== Trajectories from r1 = rts+δ, r2 = rts =====
Along the mep, because the atoms are moving infinitely slowly and the momenta and the velocities always reset to zero, there is no oscillation and hence no vibration. This is seen along the mep where there is no wavy lines. In contrast, when looking at the dynamics, there is oscillation as post collision there is transfer of kinetic energy that causes the molecule to vibrate. 

When considering the inter nuclear distance vs time, A-C and A-B distances increase rapidly over the first 250 steps after which the rate of increase of the distance begins to decrease. As the number of steps increases the rate of increase of distance begins to plateau as the gradient flattens. This can be rationalised when looking at the animation where post collision, the molecules initially rapidly move away from one another but the speed at which they move away from each other falls off with time. When considering the dynamic situation, A-C and A-B distances increase at a constant gradient over the course of the plot. The reasoning is that in the dynamic situation, the velocities and momenta are not reset to zero at each step.

[[File:Aja4117(dynamics type)Mep vs dynamics distance v time 1.png]]
[[File:Aja4117Distance vs time 1.png]]

For the same reason as stated as above, the momenta and velocities are reset to zero at each point this means that the momentum vs time plot is a straight line for the mep . However, when running the dynamic situation we see an initial sharp increase in the momentum followed by a constant oscillation of momentum that is due to the momentum being retained within the vibrational modes. For A-B, there is a sharp increase in the momentum which plateaus at the same instance as B-C begins to undergo constant oscillation. This is because A-B refers to the momentum of the system A-B. Once both reach constant momentum the combined system now has a set defined level of momentum with no way to lose it.

[[File:Aja4117(dynamics type)Mep vs dynamics momenta vs time1.png]]
[[File:Aja4117dynamics mom vs time1.png]]

===== Trajectories from r1 = rts, r2 = rts +δ =====
The result of switching the distances is that

<nowiki> </nowiki>Take note of the final values of the positions '''r1'''(t) '''r2'''(t) and '''p1'''(t) '''p2'''(t) for your trajectory for large enough t.

== Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed. ==

== What do you observe? STILL TO DO ==

=== Reactive and unreactive trajectories ===
{| class="wikitable" border="1"
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 ||-99.018||Yes||Reaction proceeds with minimal vibration||
|-
| -1.5 || -2.0 ||-100.456||No||P2 does not have enough momentum to reach P1||
|-
| -1.5 || -2.5 ||-98.956||Yes||Transfer of vibrational energy||
|-
| -2.5 || -5.0 || -84.956
||No||Product forms but contains too much vibrational energy so dissociates||
|-
| -2.5 || -5.2 || -83.416
||No||Product forms but contains too much vibrational energy so dissociates
||
|}

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

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

There are two basic assumptions for Transition State Theory, the two most basic are the separation of the electronic and nuclear motions which is the same as the Born-Oppenheimer approximation from quantum mechanics. The second basic assumption is that the reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution.

There are however, additional assumptions and they are as follows:
# Molecular systems that have crossed the transition state in the direction of the products cannot turn around and reform reactants.
# In the transition state, motion along the reaction coordinate may be separated from the other motions and be treated classically as a translation.
# even in the absence of an equilibrium between the reactant and product molecules, the transition states that are become products are distributed among their states according to the Maxwell-Boltzmann laws.

''Given that Transition State theory neglects quantum mechanical contributions such as influence on and between between electronic and nuclear motions and the idea that molecules passing through the transition state through to the products cannot pass back through to the reactants, this would suggest that there would be deviation between experimental and calculated values using Transition State theory.''

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

=== ''F + H2'' ===

==== ''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?'' ====
Looking at the PES of F + H2 this indicated the reaction is exothermic because there is a decrease in the energy going from the reactants to products. This can be seen along the reaction path for this reaction where there is a dip in energy on the path to the formation of the products. This makes intuitive sense as the driving force of the reaction will be the formation of a strong F-H bond, in place of a weaker H-H bond.

For the H + HF reaction, '''the reaction is surmised to be endothermic due to as there is no decrease in energy on forming the products. This is in line with intuition considering the strength of the H-F bond. The very high electronegativity of fluorine would make it so that significant energy would be required to cause this reaction to proceed. '''

==== ''Locate the approximate position of the transition state.'' ====
For the reaction of F + H2 the approximate position of the transition state was reported to be 

==== ''Report the activation energy for both reactions.'' ====
 

==== ''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.'' ====

The set of reaction conditions that were used were:

Atom A: F

Atom B: H

Atom C: H

Distance AB: 2.25

Distance BC: 0.74

Momentum AB = -0.8

Momentum BC = 0.1

These conditions yielded a set of successful reactions, and post reaction and upon formation of F-H, there is transfer of energy in the form of kinetic energy to both the Hydrogen that is travelling away from the reaction complex and also kinetic energy in the bond vibration of F-H. This can be visualised in the Momenta vs Time graph below.
==== ''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.'' ====

=== ''H + HF '' ===

File:Aja4117Surface Plot FH2 momenta vs time.png

2019-05-15T22:12:36Z

Aja4117:

MRD:aja4117

2019-05-15T21:09:55Z

Aja4117: /* State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? */

MRD:aja4117

2019-05-14T17:47:24Z

Aja4117:

MRD:aja4117

2019-05-14T17:05:57Z

Aja4117:

MRD:aja4117

2019-05-14T16:28:49Z

Aja4117: /* Trajectories from r1 = rts+δ, r2 = rts */