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MRD:mlw115RXNDy

2018-05-22T16:37:23Z

Mlw115: /* Reaction Dynamics */

= Reaction Dynamics =

= Exercise 1: H2 + H system=

==Dynamics from the transition state region ==

 Question 1. What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

Where a reaction is modeled on a potential energy surface the lowest energy path between the reactants and products is seen as a "valley floor" of the surface. The transition structure is a saddle point along the minimum energy ridge.

The gradient of the potential energy surface is zero at the minimum, along the direction of the reaction path and so the first derivative with respect to the inter-nuclear distance is zero. For the saddle point the gradient of the potential energy surface is zero orthogonal to the direction of the reaction pathway.

δV(ri)/δri = 0

Second derivatives
δV2(ri)/δri2 > 0 Minima
δV2(ri)/δri2 < 0 Maxima

where V is the potential energy and r related to the nuclear positions.

The minimum pathway and saddle point can be distinguished using the second derivatives with respect to atom positions. The second derivative of a maximum point is negative, whereas the second derivative of a minimum is positive. As the saddle point is a maxima along the minimum path the second derivative is negative. The second derivative of the minimum energy path in the direction of the reaction will be positive.

==Estimating Transition State Position ==

 Question 2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. 

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.908||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot 1 !! Surface Plot 2!! Internuclear distance vs Time
|-
|[[File:Mlw115_H2H_saddle.png|300px]]||[[File:mlw115_H2H_TS.png|300px]]||[[File:Mlw115_H2H_TS_INDvT.png|300px]]|
|}

The best guess of inter nuclear distances at the transition state was found using trial and error, to find the positions that would result in the minimum trajectory. As the system is symmetric the distances would be equal. The two surface plots are included to demonstrate the "reaction path". The internuclear distance vs time plot demonstrates no change in any bond lengths at the transition state so the overall momenta is 0 and potential energy is at the maximum. This agrees with that the transition state is a stationary point.

== Reaction Path ==
===MEP vs Dynamics===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.909||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot (MEP) !! Surface Plot (Dynamics
|-
|[[File:Mlw115_H2H_MEP_surfaceplot.png|300px]]||[[File:mlw115_H2H_MEP_surfaceplot_dynamics.png|300px]]
|}

 Question 3. Comment on how the mep and the trajectory you just calculated differ.

It can be seen in the MEP plot that the trajectory follows the minimum energy valley floor without any obvious deviations or oscillation. This does not account for the inter-molecular vibrations between the molecules,. The MEP trajectory is corresponding to infinitely slow motion, it does not accurately represent the atomic motion of the reaction. The trajectory oscillates in the dynamics plot, also in much less steps the AB bond distance increases considerably more. This plot is more realistic as it accounts for the motion of atoms during the reaction.

{| class="wikitable" border="1"
|+ Final Geometry
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| 2.486||1.167
|}

=== Reversed conditions ===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| -2.486||-1.167
|}

[[File:mlw115_H2H_dynamics_reversed_momenta.png|300px]]

From taking the previous runs final conditions and setting these as the initial conditions and reversing the momenta values the reaction seems to run in reverse, reaching an end exactly at the transition state.

===Determining Reactivity===

Question 4. Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2.0
|}

{| class="wikitable" border=1
! pAB !! pBC!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -1.25 || -2.5 || -99.018|| Yes || [[File:Mlw115_Surface_Plot_1.25_2.5.png|300px]] || Initially the HAHB distance changes very little as the vibrational energy is quite low, as the HC approaches it can be seen that the BC distance decreases linearly. The molecules pass over the transition state, after this point the HAHB distance decreases and the HBHC distance oscillates slightly, with a higher vibrational energy than the HAHB molecule. This overall process represents the HAHB bond breaking and forming a HBHC molecule.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mlw115_Surface_Plot_1.5_2.0.png|300px]] || The molecule approaches the maxima with some vibrational energy, shown by oscillations in the initial path, the energy of the system is not high enough to pass over the transition state energy maxima so the HC "rebounds" and the HBHC distance begins to increase. The HAHB molecule oscillations increase.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mlw115_Surface_Plot_1.5_2.5.png|300px]] || The reaction proceeds similarly to the first case with HC approaching and forming a bond with HB as there is sufficient energy to proceed over the transition state maxima. The vibrational energy throughout is slightly higher than the first case, demonstrated by higher oscillations in atom distances.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mlw115_Surface_Plot_2.5_5.png|300px]] || In this case the energy is high enough and the atoms are able to pass over the transition state energy maxima, however the vibrational energy of the HBHC system formed seems to be too high to form the molecule and the HC atom rebounds and the distance begins to decrease and so the HAHB is reformed but with considerably higher oscillations.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mlw115_Surface_Plot_2.5_5.2.png|300px]] || The HC approaches the oscillating HAHB system, the HB atom rebounds between the two atoms, as it can be seen on the surface plot the system passes over the transition state maxima, back over to the other side and back again as this occurs. Finally the HBHC bond is formed with a high vibrationally energy
|}

===Transition State Theory===

 Question 5 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? 

Transition state theory is used to understand how elementary chemical reactions take place assuming a quasi-equilibirum between reactanting complexes and those at the transition state. The rate of reaction can be understood by inspecting the complexes at the maximum saddle point of the minimum energy pathway. The reactants are in equilibrium with the activated transition state complexes and can be converted to the products.

This theory is useful to theoretically understand a chemical reaction however it operates on a series of assumptions which limit the theory.
Main Assumptions:
Equilibrium is close in the systems
Only one reaction pathway occurs
There is negligable barrier recrossing

For instance the atoms are assumed to behave classically, ignoring tunelling. Tunneling would be more prominent in reactions with low activation barriers. Also the theory assumes that the saddle point with the lowest energy will be passed over to reach the products, however this is not the case when higher vibrational energies will be populated.

Clearly a key issue seen in the data obtained above is that there is considerable barrier recrossing. Therefore the transition state theory would fail in this case. The TST would assume that all collisions of an energy above the barrier of the transition state would result in a successful reaction. Clearly this is not the case and the experimental rate would be much lower than the TST would predict.

<ref>https://goldbook.iupac.org/html/T/T06470.html</ref> <ref>https://pubs.acs.org/doi/pdf/10.1021/j100238a002</ref>

=Exercise 2: H-H-F System=

==PES Inspection==
Question 6. 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?

For the forward reaction of H2 + F --> HF + H is exothermic, as shown in the surface plot below the energy of the reactants H2 is higher than the energy of the product HF. As the energy absorbed in breaking the H-H bond in the first case is less than the energy released forming the H-F bond which means that in total energy is released which results in an exothermic reaction.

{| class="wikitable" border=1
! Surface Plot H2+F->HF+H !! Surface Plot HF+H -> H2+F
|-
|[[File:mlw115_Surface_Plot_HHF_labelled.png|300px]]||[[File:mlw115_Surface_Plot_HHF_H2Prod.png|300px]]|
|}

{| class="wikitable" border=1
! Bond!! Strength (kJ/mol)
|-
|H-F|| 565
|-
|H-H|| 432
|}

==Determining Transition State ==

Question 7. Locate the approximate position of the transition state.

A = HA B = HB and C = F

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! Surface Plot !! Internuclear distance vs Time
|-
|[[File:Mlw115_Surface_Plot_TS_HHF.png|400px]]||[[File:Mlw115 INDvT HHF TS.png|350px]]|
|}

== Calculating Activation Energy ==
Question 8. Report the activation energy for both reactions.

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.74||0
|-
| BC|| 1.8130/1.8135||0
|}

{| class="wikitable" border=1
! BC= 1.8135!! BC= 1.8135 Zoomed!! BC= 1.8130
|-
|[[File:Mlw115_HHF_EvT_1.8135.png|300px]] ‎||[[File:Mlw115_Plot_1.813_EvT_zoomT0_Resized.png|300px]]||[[File:Mlw115_HHF_EvT_1.813.png|320px]]
|}

{| class="wikitable" border=1
|+Activation energy HF+H->H2+F
|-
! !! Energy (Kcal/mol)
|-
|Initial energy || -103.752
|-
|Final energy|| -133.89
|-
|EA|| 30.138
|}

{| class="wikitable" border=1
|+Activation energy H2+F->HF+H
|-
! !! Energy (Kcal/mol)
|-
|Initial energy|| -103.742
|-
|Final energy|| -103.752
|-
|EA|| 0.01
|}

==Reaction Dynamics==
=== F+ H2 ===
{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.74||0
|-
| BC|| 2.3||-2.7
|}

{| class="wikitable" border=1
!Surface Plot!! Internuclear momentum vs time
|-
|[[File:Mlw_Surface_Plot_HHF_Reactive.png]]|| [[File:Mlw_INMvT_HHF_Reactive.png]]
|}

{| class="wikitable" border=1
!Energy!!
|-
|Kinetic|| 3.837
|-
|Potential|| -103.886

|-
|Total|| -100.049
|}

Question 9. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?

As energy is always conserved and can only be transformed from different forms to another. It can be seen from the animation that as the H-H (AB) bond is broken and the H-F (BC) bond is formed the vibrational energy (and so the kinetic energy) considerably increases for H-F. The internuclear momentum vs time graph demonstrates this, as the momentum is much higher for the H-F atom. As the vibrational energy begins to dampen the energy will be released as heat so experimentally an increase in temperature could confirm the reaction had occurred. Spectroscopy can be used to determine the vibrational energy of the bonds formed.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -3 || -0.5 || -100.339|| No || [[File:Mlw115_Surface_Plot_HHF_-3_0.5.png|300px]] || H2 approaches the fluorine atom slowly but when a collision is made the H rebounds straight back into the other H and they continue their path oscillating together in the opposite direction. This demonstrates recrossing, as the activation barrier is passed over but the products are not formed
|-

| 3 || -0.5 || -93.254 || No || [[File:Mlw115_Surface_Plot_HHF_3_0.5.png|300px]] || This is similar to the previous case however the hydrogen rebounds twice into flourine but on the second rebounds remains returns back to oscillating with the original H atom
|-
| 2 || -0.5|| -98.753|| No || [[File: Mlw115_Surface_Plot_HHF_2_0.5.png|300px]] || The HF molecule oscillates within the energy well at high vibrational energy after collision. The system remains in the products side after the reaction however the molecule remains in a high energetic state
|-
| 1.5|| -0.5|| 103.754|| No || [[File: Mlw115_Surface_Plot_HHF_1.5_0.5.png|300px]] || Similarly to the previous situation the fluorine slowly approaches the hydrogen molecule with a high vibrational energy and a high vibration energy molecule of HF is formed
|-
|0.1||-0.8|| 103.364|| Yes|| [[File: Mlw115_Surface_Plot_HHF_0.1_0.8.png|300px]] || Despite having lower total energy than the previous cases the reaction goes to completion, increasing the momentum of the fluorine atom has a considerably more pronounced effect on the success of the reaction compared to any changes in the vibraitonal momentum of the oscillating hydrogen molecule.
|}

=== HF+F ===

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 2
|-
|BC || 0.74
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -5 || 0.2 || -103.364 || No || [[File:Mlw115_Surface_Plot_HFH_5_0.2.png|300px]] || Approaching H rebounds and reaction doesn't complete

|}

A reactive pathway was found by trial and error around the reversed conditions from the reverse reaction.

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 5.0||-1.5
|-
| BC|| 1.2||1.0
|}

[[File: Mlw115_Surface_Plot_HFH_-1.5_1.png| 400px]]

==Polanyi's Empirical Rules ==

Question 10. Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.

Polanyi's rules state that in reactions with an early transition state (most closely resembling the reactants) translational energy most effectively promotes a successful reaction. Conversely where the transition state is late (resembling products closely) vibrational energy, specifically of the reactants most effectively promotes a successful reaction. <ref>The Journal of Chemical Physics 138, 234104 (2013); https://doi.org/10.1063/1.4810007</ref>

This is demonstrated in the above, where the forward reaction H2+F-> HF+H has an early transition state, as the activation barrier is so low and the transition state closely resembles the reactants. In this case a high translational energy of the incoming atom (fluorine) is more effective in resulting in a successful reaction and increasing the vibrational energy of the H2 molecule has little impact. This is illustrated nicely in the table above.

In the reverse reaction HF + H -> H2+F the transition state is late, as the activated complex much more closely resembles the product. A high vibrational energy of the HF molecule results in a successful reaction whereas increasing the translational energy of the approaching H atom has little impact.

=References=
<references/>

MRD:mlw115RXNDy

2018-05-22T16:22:37Z

Mlw115: /* Reaction Dynamics */

= Reaction Dynamics =

= Exercise 1: H2 + H system=

==Dynamics from the transition state region ==

 Question 1. What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

Where a reaction is modeled on a potential energy surface the lowest energy path between the reactants and products is seen as a "valley floor" of the surface. The transition structure is a saddle point along the minimum energy ridge.

The gradient of the potential energy surface is zero at the minimum, along the direction of the reaction path and so the first derivative with respect to the inter-nuclear distance is zero. For the saddle point the gradient of the potential energy surface is zero orthogonal to the direction of the reaction pathway.

δV(ri)/δri = 0

Second derivatives
δV2(ri)/δri2 > 0 Minima
δV2(ri)/δri2 < 0 Maxima

where V is the potential energy and r related to the nuclear positions.

The minimum pathway and saddle point can be distinguished using the second derivatives with respect to atom positions. The second derivative of a maximum point is negative, whereas the second derivative of a minimum is positive. As the saddle point is a maxima along the minimum path the second derivative is negative. The second derivative of the minimum energy path in the direction of the reaction will be positive.

==Estimating Transition State Position ==

 Question 2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. 

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.908||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot 1 !! Surface Plot 2!! Internuclear distance vs Time
|-
|[[File:Mlw115_H2H_saddle.png|300px]]||[[File:mlw115_H2H_TS.png|300px]]||[[File:Mlw115_H2H_TS_INDvT.png|300px]]|
|}

The best guess of inter nuclear distances at the transition state was found using trial and error, to find the positions that would result in the minimum trajectory. As the system is symmetric the distances would be equal. The two surface plots are included to demonstrate the "reaction path". The internuclear distance vs time plot demonstrates no change in any bond lengths at the transition state so the overall momenta is 0 and potential energy is at the maximum. This agrees with that the transition state is a stationary point.

== Reaction Path ==
===MEP vs Dynamics===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.909||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot (MEP) !! Surface Plot (Dynamics
|-
|[[File:Mlw115_H2H_MEP_surfaceplot.png|300px]]||[[File:mlw115_H2H_MEP_surfaceplot_dynamics.png|300px]]
|}

 Question 3. Comment on how the mep and the trajectory you just calculated differ.

It can be seen in the MEP plot that the trajectory follows the minimum energy valley floor without any obvious deviations or oscillation. This does not account for the inter-molecular vibrations between the molecules,. The MEP trajectory is corresponding to infinitely slow motion, it does not accurately represent the atomic motion of the reaction. The trajectory oscillates in the dynamics plot, also in much less steps the AB bond distance increases considerably more. This plot is more realistic as it accounts for the motion of atoms during the reaction.

{| class="wikitable" border="1"
|+ Final Geometry
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| 2.486||1.167
|}

=== Reversed conditions ===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| -2.486||-1.167
|}

[[File:mlw115_H2H_dynamics_reversed_momenta.png|300px]]

From taking the previous runs final conditions and setting these as the initial conditions and reversing the momenta values the reaction seems to run in reverse, reaching an end exactly at the transition state.

===Determining Reactivity===

Question 4. Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2.0
|}

{| class="wikitable" border=1
! pAB !! pBC!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -1.25 || -2.5 || -99.018|| Yes || [[File:Mlw115_Surface_Plot_1.25_2.5.png|300px]] || Initially the HAHB distance changes very little as the vibrational energy is quite low, as the HC approaches it can be seen that the BC distance decreases linearly. The molecules pass over the transition state, after this point the HAHB distance decreases and the HBHC distance oscillates slightly, with a higher vibrational energy than the HAHB molecule. This overall process represents the HAHB bond breaking and forming a HBHC molecule.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mlw115_Surface_Plot_1.5_2.0.png|300px]] || The molecule approaches the maxima with some vibrational energy, shown by oscillations in the initial path, the energy of the system is not high enough to pass over the transition state energy maxima so the HC "rebounds" and the HBHC distance begins to increase. The HAHB molecule oscillations increase.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mlw115_Surface_Plot_1.5_2.5.png|300px]] || The reaction proceeds similarly to the first case with HC approaching and forming a bond with HB as there is sufficient energy to proceed over the transition state maxima. The vibrational energy throughout is slightly higher than the first case, demonstrated by higher oscillations in atom distances.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mlw115_Surface_Plot_2.5_5.png|300px]] || In this case the energy is high enough and the atoms are able to pass over the transition state energy maxima, however the vibrational energy of the HBHC system formed seems to be too high to form the molecule and the HC atom rebounds and the distance begins to decrease and so the HAHB is reformed but with considerably higher oscillations.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mlw115_Surface_Plot_2.5_5.2.png|300px]] || The HC approaches the oscillating HAHB system, the HB atom rebounds between the two atoms, as it can be seen on the surface plot the system passes over the transition state maxima, back over to the other side and back again as this occurs. Finally the HBHC bond is formed with a high vibrationally energy
|}

===Transition State Theory===

 Question 5 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? 

Transition state theory is used to understand how elementary chemical reactions take place assuming a quasi-equilibirum between reactanting complexes and those at the transition state. The rate of reaction can be understood by inspecting the complexes at the maximum saddle point of the minimum energy pathway. The reactants are in equilibrium with the activated transition state complexes and can be converted to the products.

This theory is useful to theoretically understand a chemical reaction however it operates on a series of assumptions which limit the theory.
Main Assumptions:
Equilibrium is close in the systems
Only one reaction pathway occurs
There is negligable barrier recrossing

For instance the atoms are assumed to behave classically, ignoring tunelling. Tunneling would be more prominent in reactions with low activation barriers. Also the theory assumes that the saddle point with the lowest energy will be passed over to reach the products, however this is not the case when higher vibrational energies will be populated.

Clearly a key issue seen in the data obtained above is that there is considerable barrier recrossing. Therefore the transition state theory would fail in this case. The TST would assume that all collisions of an energy above the barrier of the transition state would result in a successful reaction. Clearly this is not the case and the experimental rate would be much lower than the TST would predict.

<ref>https://goldbook.iupac.org/html/T/T06470.html</ref> <ref>https://pubs.acs.org/doi/pdf/10.1021/j100238a002</ref>

=Exercise 2: H-H-F System=

==PES Inspection==
Question 6. 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?

For the forward reaction of H2 + F --> HF + H is exothermic, as shown in the surface plot below the energy of the reactants H2 is higher than the energy of the product HF. As the energy absorbed in breaking the H-H bond in the first case is less than the energy released forming the H-F bond which means that in total energy is released which results in an exothermic reaction.

{| class="wikitable" border=1
! Surface Plot H2+F->HF+H !! Surface Plot HF+H -> H2+F
|-
|[[File:mlw115_Surface_Plot_HHF_labelled.png|300px]]||[[File:mlw115_Surface_Plot_HHF_H2Prod.png|300px]]|
|}

{| class="wikitable" border=1
! Bond!! Strength (kJ/mol)
|-
|H-F|| 565
|-
|H-H|| 432
|}

==Determining Transition State ==

Question 7. Locate the approximate position of the transition state.

A = HA B = HB and C = F

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! Surface Plot !! Internuclear distance vs Time
|-
|[[File:Mlw115_Surface_Plot_TS_HHF.png|400px]]||[[File:Mlw115 INDvT HHF TS.png|350px]]|
|}

== Calculating Activation Energy ==
Question 8. Report the activation energy for both reactions.

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.74||0
|-
| BC|| 1.8130/1.8135||0
|}

{| class="wikitable" border=1
! BC= 1.8135!! BC= 1.8135 Zoomed!! BC= 1.8130
|-
|[[File:Mlw115_HHF_EvT_1.8135.png|300px]] ‎||[[File:Mlw115_Plot_1.813_EvT_zoomT0_Resized.png|300px]]||[[File:Mlw115_HHF_EvT_1.813.png|320px]]
|}

{| class="wikitable" border=1
|+Activation energy HF+H->H2+F
|-
! !! Energy (Kcal/mol)
|-
|Initial energy || -103.752
|-
|Final energy|| -133.89
|-
|EA|| 30.138
|}

{| class="wikitable" border=1
|+Activation energy H2+F->HF+H
|-
! !! Energy (Kcal/mol)
|-
|Initial energy|| -103.742
|-
|Final energy|| -103.752
|-
|EA|| 0.01
|}

==Reaction Dynamics==
=== F+ H2 ===
{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.74||0
|-
| BC|| 2.3||-2.7
|}

{| class="wikitable" border=1
!Surface Plot!! Internuclear momentum vs time
|-
|[[File:Mlw_Surface_Plot_HHF_Reactive.png]]|| [[File:Mlw_INMvT_HHF_Reactive.png]]
|}

{| class="wikitable" border=1
!Energy!!
|-
|Kinetic|| 3.837
|-
|Potential|| -103.886

|-
|Total|| -100.049
|}

Question 9. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?

As energy is always conserved and can only be transformed from different forms to another. It can be seen from the animation that as the H-H (AB) bond is broken and the H-F (BC) bond is formed the vibrational energy (and so the kinetic energy) considerably increases for H-F. The internuclear momentum vs time graph demonstrates this, as the momentum is much higher for the H-F atom. As the vibrational energy begins to dampen the energy will be released as heat so experimentally an increase in temperature could confirm the reaction had occurred. Spectroscopy can be used to determine the vibrational energy of the bonds formed.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -3 || -0.5 || -100.339|| No || [[File:Mlw115_Surface_Plot_HHF_-3_0.5.png|300px]] || H2 approaches the fluorine atom slowly but when a collision is made the H rebounds straight back into the other H and they continue their path oscillating together in the opposite direction
|-

| 3 || -0.5 || -93.254 || No || [[File:Mlw115_Surface_Plot_HHF_3_0.5.png|300px]] || This is similar to the previous case however the hydrogen rebounds twice into flourine but on the second rebounds remains returns back to oscillating with the original H atom
|-
| 2 || -0.5|| -98.753|| No || [[File: Mlw115_Surface_Plot_HHF_2_0.5.png|300px]] || The HF molecule oscillates within the energy well at high vibrational energy after collision
|-
| 1.5|| -0.5|| 103.754|| No || [[File: Mlw115_Surface_Plot_HHF_1.5_0.5.png|300px]] || Similarly to the previous situation
|-
|0.1||-0.8|| 103.364|| Yes|| [[File: Mlw115_Surface_Plot_HHF_0.1_0.8.png|300px]] || Despite having lower total energy than the previous cases the reaction goes to completion as the momentum between the reacting atoms is sufficient
|}

=== HF+F ===

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 2
|-
|BC || 0.74
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -5 || 0.2 || -103.364 || No || [[File:Mlw115_Surface_Plot_HFH_5_0.2.png|300px]] || Approaching H rebounds and reaction doesn't complete

|}

A reactive pathway was found by trial and error around the reversed conditions from the reverse reaction.

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 5.0||-1.5
|-
| BC|| 1.2||1.0
|}

[[File: Mlw115_Surface_Plot_HFH_-1.5_1.png| 400px]]

==Polanyi's Empirical Rules ==

Question 10. Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.

Polanyi's rules state that in reactions with an early transition state (most closely resembling the reactants) translational energy most effectively promotes a successful reaction. Conversely where the transition state is late (resembling products closely) vibrational energy, specifically of the reactants most effectively promotes a successful reaction. <ref>The Journal of Chemical Physics 138, 234104 (2013); https://doi.org/10.1063/1.4810007</ref>

This is demonstrated in the above, where the forward reaction H2+F-> HF+H has an early transition state, as the activation barrier is so low and the transition state closely resembles the reactants. In this case a high translational energy of the incoming atom (fluorine) is more effective in resulting in a successful reaction and increasing the vibrational energy of the H2 molecule has little impact. This is illustrated nicely in the table above.

In the reverse reaction HF + H -> H2+F the transition state is late, as the activated complex much more closely resembles the product. A high vibrational energy of the HF molecule results in a successful reaction whereas increasing the translational energy of the approaching H atom has little impact.

=References=
<references/>

MRD:mlw115RXNDy

2018-05-22T16:18:33Z

Mlw115: /* Reaction Dynamics */

= Reaction Dynamics =

= Exercise 1: H2 + H system=

==Dynamics from the transition state region ==

 Question 1. What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

Where a reaction is modeled on a potential energy surface the lowest energy path between the reactants and products is seen as a "valley floor" of the surface. The transition structure is a saddle point along the minimum energy ridge.

The gradient of the potential energy surface is zero at the minimum, along the direction of the reaction path and so the first derivative with respect to the inter-nuclear distance is zero. For the saddle point the gradient of the potential energy surface is zero orthogonal to the direction of the reaction pathway.

δV(ri)/δri = 0

Second derivatives
δV2(ri)/δri2 > 0 Minima
δV2(ri)/δri2 < 0 Maxima

where V is the potential energy and r related to the nuclear positions.

The minimum pathway and saddle point can be distinguished using the second derivatives with respect to atom positions. The second derivative of a maximum point is negative, whereas the second derivative of a minimum is positive. As the saddle point is a maxima along the minimum path the second derivative is negative. The second derivative of the minimum energy path in the direction of the reaction will be positive.

==Estimating Transition State Position ==

 Question 2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. 

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.908||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot 1 !! Surface Plot 2!! Internuclear distance vs Time
|-
|[[File:Mlw115_H2H_saddle.png|300px]]||[[File:mlw115_H2H_TS.png|300px]]||[[File:Mlw115_H2H_TS_INDvT.png|300px]]|
|}

The best guess of inter nuclear distances at the transition state was found using trial and error, to find the positions that would result in the minimum trajectory. As the system is symmetric the distances would be equal. The two surface plots are included to demonstrate the "reaction path". The internuclear distance vs time plot demonstrates no change in any bond lengths at the transition state so the overall momenta is 0 and potential energy is at the maximum. This agrees with that the transition state is a stationary point.

== Reaction Path ==
===MEP vs Dynamics===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.909||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot (MEP) !! Surface Plot (Dynamics
|-
|[[File:Mlw115_H2H_MEP_surfaceplot.png|300px]]||[[File:mlw115_H2H_MEP_surfaceplot_dynamics.png|300px]]
|}

 Question 3. Comment on how the mep and the trajectory you just calculated differ.

It can be seen in the MEP plot that the trajectory follows the minimum energy valley floor without any obvious deviations or oscillation. This does not account for the inter-molecular vibrations between the molecules,. The MEP trajectory is corresponding to infinitely slow motion, it does not accurately represent the atomic motion of the reaction. The trajectory oscillates in the dynamics plot, also in much less steps the AB bond distance increases considerably more. This plot is more realistic as it accounts for the motion of atoms during the reaction.

{| class="wikitable" border="1"
|+ Final Geometry
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| 2.486||1.167
|}

=== Reversed conditions ===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| -2.486||-1.167
|}

[[File:mlw115_H2H_dynamics_reversed_momenta.png|300px]]

From taking the previous runs final conditions and setting these as the initial conditions and reversing the momenta values the reaction seems to run in reverse, reaching an end exactly at the transition state.

===Determining Reactivity===

Question 4. Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2.0
|}

{| class="wikitable" border=1
! pAB !! pBC!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -1.25 || -2.5 || -99.018|| Yes || [[File:Mlw115_Surface_Plot_1.25_2.5.png|300px]] || Initially the HAHB distance changes very little as the vibrational energy is quite low, as the HC approaches it can be seen that the BC distance decreases linearly. The molecules pass over the transition state, after this point the HAHB distance decreases and the HBHC distance oscillates slightly, with a higher vibrational energy than the HAHB molecule. This overall process represents the HAHB bond breaking and forming a HBHC molecule.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mlw115_Surface_Plot_1.5_2.0.png|300px]] || The molecule approaches the maxima with some vibrational energy, shown by oscillations in the initial path, the energy of the system is not high enough to pass over the transition state energy maxima so the HC "rebounds" and the HBHC distance begins to increase. The HAHB molecule oscillations increase.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mlw115_Surface_Plot_1.5_2.5.png|300px]] || The reaction proceeds similarly to the first case with HC approaching and forming a bond with HB as there is sufficient energy to proceed over the transition state maxima. The vibrational energy throughout is slightly higher than the first case, demonstrated by higher oscillations in atom distances.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mlw115_Surface_Plot_2.5_5.png|300px]] || In this case the energy is high enough and the atoms are able to pass over the transition state energy maxima, however the vibrational energy of the HBHC system formed seems to be too high to form the molecule and the HC atom rebounds and the distance begins to decrease and so the HAHB is reformed but with considerably higher oscillations.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mlw115_Surface_Plot_2.5_5.2.png|300px]] || The HC approaches the oscillating HAHB system, the HB atom rebounds between the two atoms, as it can be seen on the surface plot the system passes over the transition state maxima, back over to the other side and back again as this occurs. Finally the HBHC bond is formed with a high vibrationally energy
|}

===Transition State Theory===

 Question 5 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? 

Transition state theory is used to understand how elementary chemical reactions take place assuming a quasi-equilibirum between reactanting complexes and those at the transition state. The rate of reaction can be understood by inspecting the complexes at the maximum saddle point of the minimum energy pathway. The reactants are in equilibrium with the activated transition state complexes and can be converted to the products.

This theory is useful to theoretically understand a chemical reaction however it operates on a series of assumptions which limit the theory.
Main Assumptions:
Equilibrium is close in the systems
Only one reaction pathway occurs
There is negligable barrier recrossing

For instance the atoms are assumed to behave classically, ignoring tunelling. Tunneling would be more prominent in reactions with low activation barriers. Also the theory assumes that the saddle point with the lowest energy will be passed over to reach the products, however this is not the case when higher vibrational energies will be populated.

Clearly a key issue seen in the data obtained above is that there is considerable barrier recrossing. Therefore the transition state theory would fail in this case. The TST would assume that all collisions of an energy above the barrier of the transition state would result in a successful reaction. Clearly this is not the case and the experimental rate would be much lower than the TST would predict.

<ref>https://goldbook.iupac.org/html/T/T06470.html</ref> <ref>https://pubs.acs.org/doi/pdf/10.1021/j100238a002</ref>

=Exercise 2: H-H-F System=

==PES Inspection==
Question 6. 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?

For the forward reaction of H2 + F --> HF + H is exothermic, as shown in the surface plot below the energy of the reactants H2 is higher than the energy of the product HF. As the energy absorbed in breaking the H-H bond in the first case is less than the energy released forming the H-F bond which means that in total energy is released which results in an exothermic reaction.

{| class="wikitable" border=1
! Surface Plot H2+F->HF+H !! Surface Plot HF+H -> H2+F
|-
|[[File:mlw115_Surface_Plot_HHF_labelled.png|300px]]||[[File:mlw115_Surface_Plot_HHF_H2Prod.png|300px]]|
|}

{| class="wikitable" border=1
! Bond!! Strength (kJ/mol)
|-
|H-F|| 565
|-
|H-H|| 432
|}

==Determining Transition State ==

Question 7. Locate the approximate position of the transition state.

A = HA B = HB and C = F

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! Surface Plot !! Internuclear distance vs Time
|-
|[[File:Mlw115_Surface_Plot_TS_HHF.png|400px]]||[[File:Mlw115 INDvT HHF TS.png|350px]]|
|}

== Calculating Activation Energy ==
Question 8. Report the activation energy for both reactions.

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.74||0
|-
| BC|| 1.8130/1.8135||0
|}

{| class="wikitable" border=1
! BC= 1.8135!! BC= 1.8135 Zoomed!! BC= 1.8130
|-
|[[File:Mlw115_HHF_EvT_1.8135.png|300px]] ‎||[[File:Mlw115_Plot_1.813_EvT_zoomT0_Resized.png|300px]]||[[File:Mlw115_HHF_EvT_1.813.png|320px]]
|}

{| class="wikitable" border=1
|+Activation energy HF+H->H2+F
|-
! !! Energy (Kcal/mol)
|-
|Initial energy || -103.752
|-
|Final energy|| -133.89
|-
|EA|| 30.138
|}

{| class="wikitable" border=1
|+Activation energy H2+F->HF+H
|-
! !! Energy (Kcal/mol)
|-
|Initial energy|| -103.742
|-
|Final energy|| -103.752
|-
|EA|| 0.01
|}

==Reaction Dynamics==
=== F+ H2 ===
{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.74||0
|-
| BC|| 2.3||-2.7
|}

{| class="wikitable" border=1
!Surface Plot!! Internuclear momentum vs time
|-
|[[File:Mlw_Surface_Plot_HHF_Reactive.png]]|| [[File:Mlw_INMvT_HHF_Reactive.png]]
|}

{| class="wikitable" border=1
!Energy!!
|-
|Kinetic|| 3.837
|-
|Potential|| -103.886

|-
|Total|| -100.049
|}

Question 9. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?

It can be seen from the animation that as the H-H (AB) bond is broken and the H-F (BC) bond is formed the vibrational energy (and so the kinetic energy) considerably increases for H-F. The internuclear momentum vs time graph demonstrates this, as the momentum is much higher for the H-F atom. This vibrational energy will be released as heat so experimentally an increase in temperature could confirm the reaction had occurred.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -3 || -0.5 || -100.339|| No || [[File:Mlw115_Surface_Plot_HHF_-3_0.5.png|300px]] || H2 approaches the fluorine atom slowly but when a collision is made the H rebounds straight back into the other H and they continue their path oscillating together in the opposite direction
|-

| 3 || -0.5 || -93.254 || No || [[File:Mlw115_Surface_Plot_HHF_3_0.5.png|300px]] || This is similar to the previous case however the hydrogen rebounds twice into flourine but on the second rebounds remains returns back to oscillating with the original H atom
|-
| 2 || -0.5|| -98.753|| No || [[File: Mlw115_Surface_Plot_HHF_2_0.5.png|300px]] || The HF molecule oscillates within the energy well at high vibrational energy after collision
|-
| 1.5|| -0.5|| 103.754|| No || [[File: Mlw115_Surface_Plot_HHF_1.5_0.5.png|300px]] || Similarly to the previous situation
|-
|0.1||-0.8|| 103.364|| Yes|| [[File: Mlw115_Surface_Plot_HHF_0.1_0.8.png|300px]] || Despite having lower total energy than the previous cases the reaction goes to completion as the momentum between the reacting atoms is sufficient
|}

=== HF+F ===

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 2
|-
|BC || 0.74
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -5 || 0.2 || -103.364 || No || [[File:Mlw115_Surface_Plot_HFH_5_0.2.png|300px]] || Approaching H rebounds and reaction doesn't complete

|}

A reactive pathway was found by trial and error around the reversed conditions from the reverse reaction.

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 5.0||-1.5
|-
| BC|| 1.2||1.0
|}

[[File: Mlw115_Surface_Plot_HFH_-1.5_1.png| 400px]]

==Polanyi's Empirical Rules ==

Question 10. Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.

Polanyi's rules state that in reactions with an early transition state (most closely resembling the reactants) translational energy most effectively promotes a successful reaction. Conversely where the transition state is late (resembling products closely) vibrational energy, specifically of the reactants most effectively promotes a successful reaction. <ref>The Journal of Chemical Physics 138, 234104 (2013); https://doi.org/10.1063/1.4810007</ref>

This is demonstrated in the above, where the forward reaction H2+F-> HF+H has an early transition state, as the activation barrier is so low and the transition state closely resembles the reactants. In this case a high translational energy of the incoming atom (fluorine) is more effective in resulting in a successful reaction and increasing the vibrational energy of the H2 molecule has little impact. This is illustrated nicely in the table above.

In the reverse reaction HF + H -> H2+F the transition state is late, as the activated complex much more closely resembles the product. A high vibrational energy of the HF molecule results in a successful reaction whereas increasing the translational energy of the approaching H atom has little impact.

=References=
<references/>

MRD:mlw115RXNDy

2018-05-22T16:15:26Z

Mlw115: /* Calculating Activation Energy */

= Reaction Dynamics =

= Exercise 1: H2 + H system=

==Dynamics from the transition state region ==

 Question 1. What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

Where a reaction is modeled on a potential energy surface the lowest energy path between the reactants and products is seen as a "valley floor" of the surface. The transition structure is a saddle point along the minimum energy ridge.

The gradient of the potential energy surface is zero at the minimum, along the direction of the reaction path and so the first derivative with respect to the inter-nuclear distance is zero. For the saddle point the gradient of the potential energy surface is zero orthogonal to the direction of the reaction pathway.

δV(ri)/δri = 0

Second derivatives
δV2(ri)/δri2 > 0 Minima
δV2(ri)/δri2 < 0 Maxima

where V is the potential energy and r related to the nuclear positions.

The minimum pathway and saddle point can be distinguished using the second derivatives with respect to atom positions. The second derivative of a maximum point is negative, whereas the second derivative of a minimum is positive. As the saddle point is a maxima along the minimum path the second derivative is negative. The second derivative of the minimum energy path in the direction of the reaction will be positive.

==Estimating Transition State Position ==

 Question 2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. 

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.908||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot 1 !! Surface Plot 2!! Internuclear distance vs Time
|-
|[[File:Mlw115_H2H_saddle.png|300px]]||[[File:mlw115_H2H_TS.png|300px]]||[[File:Mlw115_H2H_TS_INDvT.png|300px]]|
|}

The best guess of inter nuclear distances at the transition state was found using trial and error, to find the positions that would result in the minimum trajectory. As the system is symmetric the distances would be equal. The two surface plots are included to demonstrate the "reaction path". The internuclear distance vs time plot demonstrates no change in any bond lengths at the transition state so the overall momenta is 0 and potential energy is at the maximum. This agrees with that the transition state is a stationary point.

== Reaction Path ==
===MEP vs Dynamics===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.909||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot (MEP) !! Surface Plot (Dynamics
|-
|[[File:Mlw115_H2H_MEP_surfaceplot.png|300px]]||[[File:mlw115_H2H_MEP_surfaceplot_dynamics.png|300px]]
|}

 Question 3. Comment on how the mep and the trajectory you just calculated differ.

It can be seen in the MEP plot that the trajectory follows the minimum energy valley floor without any obvious deviations or oscillation. This does not account for the inter-molecular vibrations between the molecules,. The MEP trajectory is corresponding to infinitely slow motion, it does not accurately represent the atomic motion of the reaction. The trajectory oscillates in the dynamics plot, also in much less steps the AB bond distance increases considerably more. This plot is more realistic as it accounts for the motion of atoms during the reaction.

{| class="wikitable" border="1"
|+ Final Geometry
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| 2.486||1.167
|}

=== Reversed conditions ===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| -2.486||-1.167
|}

[[File:mlw115_H2H_dynamics_reversed_momenta.png|300px]]

From taking the previous runs final conditions and setting these as the initial conditions and reversing the momenta values the reaction seems to run in reverse, reaching an end exactly at the transition state.

===Determining Reactivity===

Question 4. Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2.0
|}

{| class="wikitable" border=1
! pAB !! pBC!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -1.25 || -2.5 || -99.018|| Yes || [[File:Mlw115_Surface_Plot_1.25_2.5.png|300px]] || Initially the HAHB distance changes very little as the vibrational energy is quite low, as the HC approaches it can be seen that the BC distance decreases linearly. The molecules pass over the transition state, after this point the HAHB distance decreases and the HBHC distance oscillates slightly, with a higher vibrational energy than the HAHB molecule. This overall process represents the HAHB bond breaking and forming a HBHC molecule.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mlw115_Surface_Plot_1.5_2.0.png|300px]] || The molecule approaches the maxima with some vibrational energy, shown by oscillations in the initial path, the energy of the system is not high enough to pass over the transition state energy maxima so the HC "rebounds" and the HBHC distance begins to increase. The HAHB molecule oscillations increase.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mlw115_Surface_Plot_1.5_2.5.png|300px]] || The reaction proceeds similarly to the first case with HC approaching and forming a bond with HB as there is sufficient energy to proceed over the transition state maxima. The vibrational energy throughout is slightly higher than the first case, demonstrated by higher oscillations in atom distances.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mlw115_Surface_Plot_2.5_5.png|300px]] || In this case the energy is high enough and the atoms are able to pass over the transition state energy maxima, however the vibrational energy of the HBHC system formed seems to be too high to form the molecule and the HC atom rebounds and the distance begins to decrease and so the HAHB is reformed but with considerably higher oscillations.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mlw115_Surface_Plot_2.5_5.2.png|300px]] || The HC approaches the oscillating HAHB system, the HB atom rebounds between the two atoms, as it can be seen on the surface plot the system passes over the transition state maxima, back over to the other side and back again as this occurs. Finally the HBHC bond is formed with a high vibrationally energy
|}

===Transition State Theory===

 Question 5 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? 

Transition state theory is used to understand how elementary chemical reactions take place assuming a quasi-equilibirum between reactanting complexes and those at the transition state. The rate of reaction can be understood by inspecting the complexes at the maximum saddle point of the minimum energy pathway. The reactants are in equilibrium with the activated transition state complexes and can be converted to the products.

This theory is useful to theoretically understand a chemical reaction however it operates on a series of assumptions which limit the theory.
Main Assumptions:
Equilibrium is close in the systems
Only one reaction pathway occurs
There is negligable barrier recrossing

For instance the atoms are assumed to behave classically, ignoring tunelling. Tunneling would be more prominent in reactions with low activation barriers. Also the theory assumes that the saddle point with the lowest energy will be passed over to reach the products, however this is not the case when higher vibrational energies will be populated.

Clearly a key issue seen in the data obtained above is that there is considerable barrier recrossing. Therefore the transition state theory would fail in this case. The TST would assume that all collisions of an energy above the barrier of the transition state would result in a successful reaction. Clearly this is not the case and the experimental rate would be much lower than the TST would predict.

<ref>https://goldbook.iupac.org/html/T/T06470.html</ref> <ref>https://pubs.acs.org/doi/pdf/10.1021/j100238a002</ref>

=Exercise 2: H-H-F System=

==PES Inspection==
Question 6. 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?

For the forward reaction of H2 + F --> HF + H is exothermic, as shown in the surface plot below the energy of the reactants H2 is higher than the energy of the product HF. As the energy absorbed in breaking the H-H bond in the first case is less than the energy released forming the H-F bond which means that in total energy is released which results in an exothermic reaction.

{| class="wikitable" border=1
! Surface Plot H2+F->HF+H !! Surface Plot HF+H -> H2+F
|-
|[[File:mlw115_Surface_Plot_HHF_labelled.png|300px]]||[[File:mlw115_Surface_Plot_HHF_H2Prod.png|300px]]|
|}

{| class="wikitable" border=1
! Bond!! Strength (kJ/mol)
|-
|H-F|| 565
|-
|H-H|| 432
|}

==Determining Transition State ==

Question 7. Locate the approximate position of the transition state.

A = HA B = HB and C = F

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! Surface Plot !! Internuclear distance vs Time
|-
|[[File:Mlw115_Surface_Plot_TS_HHF.png|400px]]||[[File:Mlw115 INDvT HHF TS.png|350px]]|
|}

== Calculating Activation Energy ==
Question 8. Report the activation energy for both reactions.

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.74||0
|-
| BC|| 1.8130/1.8135||0
|}

{| class="wikitable" border=1
! BC= 1.8135!! BC= 1.8135 Zoomed!! BC= 1.8130
|-
|[[File:Mlw115_HHF_EvT_1.8135.png|300px]] ‎||[[File:Mlw115_Plot_1.813_EvT_zoomT0_Resized.png|300px]]||[[File:Mlw115_HHF_EvT_1.813.png|320px]]
|}

{| class="wikitable" border=1
|+Activation energy HF+H->H2+F
|-
! !! Energy (Kcal/mol)
|-
|Initial energy || -103.752
|-
|Final energy|| -133.89
|-
|EA|| 30.138
|}

{| class="wikitable" border=1
|+Activation energy H2+F->HF+H
|-
! !! Energy (Kcal/mol)
|-
|Initial energy|| -103.742
|-
|Final energy|| -103.752
|-
|EA|| 0.01
|}

==Reaction Dynamics==
=== F+ H2 ===
{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.74||0
|-
| BC|| 2.3||-2.7
|}

{| class="wikitable" border=1
!Surface Plot!! Internuclear momentum vs time
|-
|[[File:Mlw_Surface_Plot_HHF_Reactive.png]]|| [[File:Mlw_INMvT_HHF_Reactive.png]]
|}

{| class="wikitable" border=1
!Energy!!
|-
|Kinetic|| 3.837
|-
|Potential|| -103.886

|-
|Total|| -100.049
|}

Question 9. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?

It can be seen from the animation that as the H-H (AB) bond is broken and the H-F (BC) bond is formed the vibrational energy (and so the kinetic energy) considerably increases for H-F. The internuclear momentum vs time graph demonstrates this, as the momentum is much higher for the H-F atom. This vibrational energy will be released as heat so experimentally an increase in temperature could confirm the reaction had occurred.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -3 || -0.5 || -100.339|| No || [[File:Mlw115_Surface_Plot_HHF_-3_0.5.png|300px]] || H2 approaches the fluorine atom slowly but when a collision is made the H rebounds straight back into the other H and they continue their path oscillating together in the opposite direction
|-

| 3 || -0.5 || -93.254 || No || [[File:Mlw115_Surface_Plot_HHF_3_0.5.png|300px]] || This is similar to the previous case however the hydrogen rebounds twice into flourine but on the second rebounds remains returns back to oscillating with the original H atom
|-
| 2 || -0.5|| -98.753|| No || [[File: Mlw115_Surface_Plot_HHF_2_0.5.png|300px]] || The HF molecule oscillates within the energy well at high vibrational energy after collision
|-
| 1.5|| -0.5|| 103.754|| No || [[File: Mlw115_Surface_Plot_HHF_1.5_0.5.png|300px]] || Similarly to the previous situation
|-
|0.1||-0.8|| 103.364|| Yes|| [[File: Mlw115_Surface_Plot_HHF_0.1_0.8.png|300px]] || Despite having lower total energy than the previous cases the reaction goes to completion as the momentum between the reacting atoms is sufficient
|}

Despite the H2 energy being higher than the activation energy if the BC atoms do not collide with a high enough energy the reaction will return back over the peak and the original reactants will be produced.

=== HF+F ===

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 2
|-
|BC || 0.74
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -5 || 0.2 || -103.364 || No || [[File:Mlw115_Surface_Plot_HFH_5_0.2.png|300px]] || Approaching H rebounds and reaction doesn't complete

|}

A reactive pathway was found by trial and error around the reversed conditions from the reverse reaction.

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 5.0||-1.5
|-
| BC|| 1.2||1.0
|}

[[File: Mlw115_Surface_Plot_HFH_-1.5_1.png| 400px]]

==Polanyi's Empirical Rules ==

Question 10. Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.

Polanyi's rules state that in reactions with an early transition state (most closely resembling the reactants) translational energy most effectively promotes a successful reaction. Conversely where the transition state is late (resembling products closely) vibrational energy, specifically of the reactants most effectively promotes a successful reaction. <ref>The Journal of Chemical Physics 138, 234104 (2013); https://doi.org/10.1063/1.4810007</ref>

This is demonstrated in the above, where the forward reaction H2+F-> HF+H has an early transition state, as the activation barrier is so low and the transition state closely resembles the reactants. In this case a high translational energy of the incoming atom (fluorine) is more effective in resulting in a successful reaction and increasing the vibrational energy of the H2 molecule has little impact. This is illustrated nicely in the table above.

In the reverse reaction HF + H -> H2+F the transition state is late, as the activated complex much more closely resembles the product. A high vibrational energy of the HF molecule results in a successful reaction whereas increasing the translational energy of the approaching H atom has little impact.

=References=
<references/>

MRD:mlw115RXNDy

2018-05-22T16:15:01Z

Mlw115: /* Calculating Activation Energy */

= Reaction Dynamics =

= Exercise 1: H2 + H system=

==Dynamics from the transition state region ==

 Question 1. What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

Where a reaction is modeled on a potential energy surface the lowest energy path between the reactants and products is seen as a "valley floor" of the surface. The transition structure is a saddle point along the minimum energy ridge.

The gradient of the potential energy surface is zero at the minimum, along the direction of the reaction path and so the first derivative with respect to the inter-nuclear distance is zero. For the saddle point the gradient of the potential energy surface is zero orthogonal to the direction of the reaction pathway.

δV(ri)/δri = 0

Second derivatives
δV2(ri)/δri2 > 0 Minima
δV2(ri)/δri2 < 0 Maxima

where V is the potential energy and r related to the nuclear positions.

The minimum pathway and saddle point can be distinguished using the second derivatives with respect to atom positions. The second derivative of a maximum point is negative, whereas the second derivative of a minimum is positive. As the saddle point is a maxima along the minimum path the second derivative is negative. The second derivative of the minimum energy path in the direction of the reaction will be positive.

==Estimating Transition State Position ==

 Question 2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. 

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.908||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot 1 !! Surface Plot 2!! Internuclear distance vs Time
|-
|[[File:Mlw115_H2H_saddle.png|300px]]||[[File:mlw115_H2H_TS.png|300px]]||[[File:Mlw115_H2H_TS_INDvT.png|300px]]|
|}

The best guess of inter nuclear distances at the transition state was found using trial and error, to find the positions that would result in the minimum trajectory. As the system is symmetric the distances would be equal. The two surface plots are included to demonstrate the "reaction path". The internuclear distance vs time plot demonstrates no change in any bond lengths at the transition state so the overall momenta is 0 and potential energy is at the maximum. This agrees with that the transition state is a stationary point.

== Reaction Path ==
===MEP vs Dynamics===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.909||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot (MEP) !! Surface Plot (Dynamics
|-
|[[File:Mlw115_H2H_MEP_surfaceplot.png|300px]]||[[File:mlw115_H2H_MEP_surfaceplot_dynamics.png|300px]]
|}

 Question 3. Comment on how the mep and the trajectory you just calculated differ.

It can be seen in the MEP plot that the trajectory follows the minimum energy valley floor without any obvious deviations or oscillation. This does not account for the inter-molecular vibrations between the molecules,. The MEP trajectory is corresponding to infinitely slow motion, it does not accurately represent the atomic motion of the reaction. The trajectory oscillates in the dynamics plot, also in much less steps the AB bond distance increases considerably more. This plot is more realistic as it accounts for the motion of atoms during the reaction.

{| class="wikitable" border="1"
|+ Final Geometry
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| 2.486||1.167
|}

=== Reversed conditions ===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| -2.486||-1.167
|}

[[File:mlw115_H2H_dynamics_reversed_momenta.png|300px]]

From taking the previous runs final conditions and setting these as the initial conditions and reversing the momenta values the reaction seems to run in reverse, reaching an end exactly at the transition state.

===Determining Reactivity===

Question 4. Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2.0
|}

{| class="wikitable" border=1
! pAB !! pBC!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -1.25 || -2.5 || -99.018|| Yes || [[File:Mlw115_Surface_Plot_1.25_2.5.png|300px]] || Initially the HAHB distance changes very little as the vibrational energy is quite low, as the HC approaches it can be seen that the BC distance decreases linearly. The molecules pass over the transition state, after this point the HAHB distance decreases and the HBHC distance oscillates slightly, with a higher vibrational energy than the HAHB molecule. This overall process represents the HAHB bond breaking and forming a HBHC molecule.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mlw115_Surface_Plot_1.5_2.0.png|300px]] || The molecule approaches the maxima with some vibrational energy, shown by oscillations in the initial path, the energy of the system is not high enough to pass over the transition state energy maxima so the HC "rebounds" and the HBHC distance begins to increase. The HAHB molecule oscillations increase.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mlw115_Surface_Plot_1.5_2.5.png|300px]] || The reaction proceeds similarly to the first case with HC approaching and forming a bond with HB as there is sufficient energy to proceed over the transition state maxima. The vibrational energy throughout is slightly higher than the first case, demonstrated by higher oscillations in atom distances.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mlw115_Surface_Plot_2.5_5.png|300px]] || In this case the energy is high enough and the atoms are able to pass over the transition state energy maxima, however the vibrational energy of the HBHC system formed seems to be too high to form the molecule and the HC atom rebounds and the distance begins to decrease and so the HAHB is reformed but with considerably higher oscillations.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mlw115_Surface_Plot_2.5_5.2.png|300px]] || The HC approaches the oscillating HAHB system, the HB atom rebounds between the two atoms, as it can be seen on the surface plot the system passes over the transition state maxima, back over to the other side and back again as this occurs. Finally the HBHC bond is formed with a high vibrationally energy
|}

===Transition State Theory===

 Question 5 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? 

Transition state theory is used to understand how elementary chemical reactions take place assuming a quasi-equilibirum between reactanting complexes and those at the transition state. The rate of reaction can be understood by inspecting the complexes at the maximum saddle point of the minimum energy pathway. The reactants are in equilibrium with the activated transition state complexes and can be converted to the products.

This theory is useful to theoretically understand a chemical reaction however it operates on a series of assumptions which limit the theory.
Main Assumptions:
Equilibrium is close in the systems
Only one reaction pathway occurs
There is negligable barrier recrossing

For instance the atoms are assumed to behave classically, ignoring tunelling. Tunneling would be more prominent in reactions with low activation barriers. Also the theory assumes that the saddle point with the lowest energy will be passed over to reach the products, however this is not the case when higher vibrational energies will be populated.

Clearly a key issue seen in the data obtained above is that there is considerable barrier recrossing. Therefore the transition state theory would fail in this case. The TST would assume that all collisions of an energy above the barrier of the transition state would result in a successful reaction. Clearly this is not the case and the experimental rate would be much lower than the TST would predict.

<ref>https://goldbook.iupac.org/html/T/T06470.html</ref> <ref>https://pubs.acs.org/doi/pdf/10.1021/j100238a002</ref>

=Exercise 2: H-H-F System=

==PES Inspection==
Question 6. 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?

For the forward reaction of H2 + F --> HF + H is exothermic, as shown in the surface plot below the energy of the reactants H2 is higher than the energy of the product HF. As the energy absorbed in breaking the H-H bond in the first case is less than the energy released forming the H-F bond which means that in total energy is released which results in an exothermic reaction.

{| class="wikitable" border=1
! Surface Plot H2+F->HF+H !! Surface Plot HF+H -> H2+F
|-
|[[File:mlw115_Surface_Plot_HHF_labelled.png|300px]]||[[File:mlw115_Surface_Plot_HHF_H2Prod.png|300px]]|
|}

{| class="wikitable" border=1
! Bond!! Strength (kJ/mol)
|-
|H-F|| 565
|-
|H-H|| 432
|}

==Determining Transition State ==

Question 7. Locate the approximate position of the transition state.

A = HA B = HB and C = F

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! Surface Plot !! Internuclear distance vs Time
|-
|[[File:Mlw115_Surface_Plot_TS_HHF.png|400px]]||[[File:Mlw115 INDvT HHF TS.png|350px]]|
|}

== Calculating Activation Energy ==
Question 8. Report the activation energy for both reactions.

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.74||0
|-
| BC|| 1.8130/1.8135||0
|}

{| class="wikitable" border=1
! BC= 1.8135!! BC= 1.8135 Zoomed!! BC= 1.8130
|-
|[[File:Mlw115_HHF_EvT_1.8135.png|300px]] ‎||[[File:Mlw115_Plot_1.813_EvT_zoomT0_Resized.png|300px]]||[[File:Mlw115_HHF_EvT_1.813.png|320px]]
|}

pAB = pBC =0

distAB = 0.74

{| class="wikitable" border=1
|+Activation energy HF+H->H2+F
|-
! !! Energy (Kcal/mol)
|-
|Initial energy || -103.752
|-
|Final energy|| -133.89
|-
|EA|| 30.138
|}

{| class="wikitable" border=1
|+Activation energy H2+F->HF+H
|-
! !! Energy (Kcal/mol)
|-
|Initial energy|| -103.742
|-
|Final energy|| -103.752
|-
|EA|| 0.01
|}

==Reaction Dynamics==
=== F+ H2 ===
{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.74||0
|-
| BC|| 2.3||-2.7
|}

{| class="wikitable" border=1
!Surface Plot!! Internuclear momentum vs time
|-
|[[File:Mlw_Surface_Plot_HHF_Reactive.png]]|| [[File:Mlw_INMvT_HHF_Reactive.png]]
|}

{| class="wikitable" border=1
!Energy!!
|-
|Kinetic|| 3.837
|-
|Potential|| -103.886

|-
|Total|| -100.049
|}

Question 9. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?

It can be seen from the animation that as the H-H (AB) bond is broken and the H-F (BC) bond is formed the vibrational energy (and so the kinetic energy) considerably increases for H-F. The internuclear momentum vs time graph demonstrates this, as the momentum is much higher for the H-F atom. This vibrational energy will be released as heat so experimentally an increase in temperature could confirm the reaction had occurred.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -3 || -0.5 || -100.339|| No || [[File:Mlw115_Surface_Plot_HHF_-3_0.5.png|300px]] || H2 approaches the fluorine atom slowly but when a collision is made the H rebounds straight back into the other H and they continue their path oscillating together in the opposite direction
|-

| 3 || -0.5 || -93.254 || No || [[File:Mlw115_Surface_Plot_HHF_3_0.5.png|300px]] || This is similar to the previous case however the hydrogen rebounds twice into flourine but on the second rebounds remains returns back to oscillating with the original H atom
|-
| 2 || -0.5|| -98.753|| No || [[File: Mlw115_Surface_Plot_HHF_2_0.5.png|300px]] || The HF molecule oscillates within the energy well at high vibrational energy after collision
|-
| 1.5|| -0.5|| 103.754|| No || [[File: Mlw115_Surface_Plot_HHF_1.5_0.5.png|300px]] || Similarly to the previous situation
|-
|0.1||-0.8|| 103.364|| Yes|| [[File: Mlw115_Surface_Plot_HHF_0.1_0.8.png|300px]] || Despite having lower total energy than the previous cases the reaction goes to completion as the momentum between the reacting atoms is sufficient
|}

Despite the H2 energy being higher than the activation energy if the BC atoms do not collide with a high enough energy the reaction will return back over the peak and the original reactants will be produced.

=== HF+F ===

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 2
|-
|BC || 0.74
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -5 || 0.2 || -103.364 || No || [[File:Mlw115_Surface_Plot_HFH_5_0.2.png|300px]] || Approaching H rebounds and reaction doesn't complete

|}

A reactive pathway was found by trial and error around the reversed conditions from the reverse reaction.

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 5.0||-1.5
|-
| BC|| 1.2||1.0
|}

[[File: Mlw115_Surface_Plot_HFH_-1.5_1.png| 400px]]

==Polanyi's Empirical Rules ==

Question 10. Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.

Polanyi's rules state that in reactions with an early transition state (most closely resembling the reactants) translational energy most effectively promotes a successful reaction. Conversely where the transition state is late (resembling products closely) vibrational energy, specifically of the reactants most effectively promotes a successful reaction. <ref>The Journal of Chemical Physics 138, 234104 (2013); https://doi.org/10.1063/1.4810007</ref>

This is demonstrated in the above, where the forward reaction H2+F-> HF+H has an early transition state, as the activation barrier is so low and the transition state closely resembles the reactants. In this case a high translational energy of the incoming atom (fluorine) is more effective in resulting in a successful reaction and increasing the vibrational energy of the H2 molecule has little impact. This is illustrated nicely in the table above.

In the reverse reaction HF + H -> H2+F the transition state is late, as the activated complex much more closely resembles the product. A high vibrational energy of the HF molecule results in a successful reaction whereas increasing the translational energy of the approaching H atom has little impact.

=References=
<references/>

MRD:mlw115RXNDy

2018-05-22T16:12:22Z

Mlw115: /* PES Inspection */

= Reaction Dynamics =

= Exercise 1: H2 + H system=

==Dynamics from the transition state region ==

 Question 1. What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

Where a reaction is modeled on a potential energy surface the lowest energy path between the reactants and products is seen as a "valley floor" of the surface. The transition structure is a saddle point along the minimum energy ridge.

The gradient of the potential energy surface is zero at the minimum, along the direction of the reaction path and so the first derivative with respect to the inter-nuclear distance is zero. For the saddle point the gradient of the potential energy surface is zero orthogonal to the direction of the reaction pathway.

δV(ri)/δri = 0

Second derivatives
δV2(ri)/δri2 > 0 Minima
δV2(ri)/δri2 < 0 Maxima

where V is the potential energy and r related to the nuclear positions.

The minimum pathway and saddle point can be distinguished using the second derivatives with respect to atom positions. The second derivative of a maximum point is negative, whereas the second derivative of a minimum is positive. As the saddle point is a maxima along the minimum path the second derivative is negative. The second derivative of the minimum energy path in the direction of the reaction will be positive.

==Estimating Transition State Position ==

 Question 2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. 

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.908||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot 1 !! Surface Plot 2!! Internuclear distance vs Time
|-
|[[File:Mlw115_H2H_saddle.png|300px]]||[[File:mlw115_H2H_TS.png|300px]]||[[File:Mlw115_H2H_TS_INDvT.png|300px]]|
|}

The best guess of inter nuclear distances at the transition state was found using trial and error, to find the positions that would result in the minimum trajectory. As the system is symmetric the distances would be equal. The two surface plots are included to demonstrate the "reaction path". The internuclear distance vs time plot demonstrates no change in any bond lengths at the transition state so the overall momenta is 0 and potential energy is at the maximum. This agrees with that the transition state is a stationary point.

== Reaction Path ==
===MEP vs Dynamics===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.909||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot (MEP) !! Surface Plot (Dynamics
|-
|[[File:Mlw115_H2H_MEP_surfaceplot.png|300px]]||[[File:mlw115_H2H_MEP_surfaceplot_dynamics.png|300px]]
|}

 Question 3. Comment on how the mep and the trajectory you just calculated differ.

It can be seen in the MEP plot that the trajectory follows the minimum energy valley floor without any obvious deviations or oscillation. This does not account for the inter-molecular vibrations between the molecules,. The MEP trajectory is corresponding to infinitely slow motion, it does not accurately represent the atomic motion of the reaction. The trajectory oscillates in the dynamics plot, also in much less steps the AB bond distance increases considerably more. This plot is more realistic as it accounts for the motion of atoms during the reaction.

{| class="wikitable" border="1"
|+ Final Geometry
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| 2.486||1.167
|}

=== Reversed conditions ===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| -2.486||-1.167
|}

[[File:mlw115_H2H_dynamics_reversed_momenta.png|300px]]

From taking the previous runs final conditions and setting these as the initial conditions and reversing the momenta values the reaction seems to run in reverse, reaching an end exactly at the transition state.

===Determining Reactivity===

Question 4. Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2.0
|}

{| class="wikitable" border=1
! pAB !! pBC!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -1.25 || -2.5 || -99.018|| Yes || [[File:Mlw115_Surface_Plot_1.25_2.5.png|300px]] || Initially the HAHB distance changes very little as the vibrational energy is quite low, as the HC approaches it can be seen that the BC distance decreases linearly. The molecules pass over the transition state, after this point the HAHB distance decreases and the HBHC distance oscillates slightly, with a higher vibrational energy than the HAHB molecule. This overall process represents the HAHB bond breaking and forming a HBHC molecule.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mlw115_Surface_Plot_1.5_2.0.png|300px]] || The molecule approaches the maxima with some vibrational energy, shown by oscillations in the initial path, the energy of the system is not high enough to pass over the transition state energy maxima so the HC "rebounds" and the HBHC distance begins to increase. The HAHB molecule oscillations increase.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mlw115_Surface_Plot_1.5_2.5.png|300px]] || The reaction proceeds similarly to the first case with HC approaching and forming a bond with HB as there is sufficient energy to proceed over the transition state maxima. The vibrational energy throughout is slightly higher than the first case, demonstrated by higher oscillations in atom distances.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mlw115_Surface_Plot_2.5_5.png|300px]] || In this case the energy is high enough and the atoms are able to pass over the transition state energy maxima, however the vibrational energy of the HBHC system formed seems to be too high to form the molecule and the HC atom rebounds and the distance begins to decrease and so the HAHB is reformed but with considerably higher oscillations.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mlw115_Surface_Plot_2.5_5.2.png|300px]] || The HC approaches the oscillating HAHB system, the HB atom rebounds between the two atoms, as it can be seen on the surface plot the system passes over the transition state maxima, back over to the other side and back again as this occurs. Finally the HBHC bond is formed with a high vibrationally energy
|}

===Transition State Theory===

 Question 5 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? 

Transition state theory is used to understand how elementary chemical reactions take place assuming a quasi-equilibirum between reactanting complexes and those at the transition state. The rate of reaction can be understood by inspecting the complexes at the maximum saddle point of the minimum energy pathway. The reactants are in equilibrium with the activated transition state complexes and can be converted to the products.

This theory is useful to theoretically understand a chemical reaction however it operates on a series of assumptions which limit the theory.
Main Assumptions:
Equilibrium is close in the systems
Only one reaction pathway occurs
There is negligable barrier recrossing

For instance the atoms are assumed to behave classically, ignoring tunelling. Tunneling would be more prominent in reactions with low activation barriers. Also the theory assumes that the saddle point with the lowest energy will be passed over to reach the products, however this is not the case when higher vibrational energies will be populated.

Clearly a key issue seen in the data obtained above is that there is considerable barrier recrossing. Therefore the transition state theory would fail in this case. The TST would assume that all collisions of an energy above the barrier of the transition state would result in a successful reaction. Clearly this is not the case and the experimental rate would be much lower than the TST would predict.

<ref>https://goldbook.iupac.org/html/T/T06470.html</ref> <ref>https://pubs.acs.org/doi/pdf/10.1021/j100238a002</ref>

=Exercise 2: H-H-F System=

==PES Inspection==
Question 6. 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?

For the forward reaction of H2 + F --> HF + H is exothermic, as shown in the surface plot below the energy of the reactants H2 is higher than the energy of the product HF. As the energy absorbed in breaking the H-H bond in the first case is less than the energy released forming the H-F bond which means that in total energy is released which results in an exothermic reaction.

{| class="wikitable" border=1
! Surface Plot H2+F->HF+H !! Surface Plot HF+H -> H2+F
|-
|[[File:mlw115_Surface_Plot_HHF_labelled.png|300px]]||[[File:mlw115_Surface_Plot_HHF_H2Prod.png|300px]]|
|}

{| class="wikitable" border=1
! Bond!! Strength (kJ/mol)
|-
|H-F|| 565
|-
|H-H|| 432
|}

==Determining Transition State ==

Question 7. Locate the approximate position of the transition state.

A = HA B = HB and C = F

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! Surface Plot !! Internuclear distance vs Time
|-
|[[File:Mlw115_Surface_Plot_TS_HHF.png|400px]]||[[File:Mlw115 INDvT HHF TS.png|350px]]|
|}

== Calculating Activation Energy ==
Question 8. Report the activation energy for both reactions.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! BC= 1.8135!! BC= 1.8135 Zoomed!! BC= 1.813
|-
|[[File:Mlw115_HHF_EvT_1.8135.png|300px]] ‎||[[File:Mlw115_Plot_1.813_EvT_zoomT0_Resized.png|300px]]||[[File:Mlw115_HHF_EvT_1.813.png|320px]]
|}

pAB = pBC =0

distAB = 0.74

{| class="wikitable" border=1
|+Activation energy HF+H->H2+F
|-
! !! Energy (Kcal/mol)
|-
|Initial energy || -103.752
|-
|Final energy|| -133.89
|-
|EA|| 30.138
|}

{| class="wikitable" border=1
|+Activation energy H2+F->HF+H
|-
! !! Energy (Kcal/mol)
|-
|Initial energy|| -103.742
|-
|Final energy|| -103.752
|-
|EA|| 0.01
|}

==Reaction Dynamics==
=== F+ H2 ===
{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.74||0
|-
| BC|| 2.3||-2.7
|}

{| class="wikitable" border=1
!Surface Plot!! Internuclear momentum vs time
|-
|[[File:Mlw_Surface_Plot_HHF_Reactive.png]]|| [[File:Mlw_INMvT_HHF_Reactive.png]]
|}

{| class="wikitable" border=1
!Energy!!
|-
|Kinetic|| 3.837
|-
|Potential|| -103.886

|-
|Total|| -100.049
|}

Question 9. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?

It can be seen from the animation that as the H-H (AB) bond is broken and the H-F (BC) bond is formed the vibrational energy (and so the kinetic energy) considerably increases for H-F. The internuclear momentum vs time graph demonstrates this, as the momentum is much higher for the H-F atom. This vibrational energy will be released as heat so experimentally an increase in temperature could confirm the reaction had occurred.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -3 || -0.5 || -100.339|| No || [[File:Mlw115_Surface_Plot_HHF_-3_0.5.png|300px]] || H2 approaches the fluorine atom slowly but when a collision is made the H rebounds straight back into the other H and they continue their path oscillating together in the opposite direction
|-

| 3 || -0.5 || -93.254 || No || [[File:Mlw115_Surface_Plot_HHF_3_0.5.png|300px]] || This is similar to the previous case however the hydrogen rebounds twice into flourine but on the second rebounds remains returns back to oscillating with the original H atom
|-
| 2 || -0.5|| -98.753|| No || [[File: Mlw115_Surface_Plot_HHF_2_0.5.png|300px]] || The HF molecule oscillates within the energy well at high vibrational energy after collision
|-
| 1.5|| -0.5|| 103.754|| No || [[File: Mlw115_Surface_Plot_HHF_1.5_0.5.png|300px]] || Similarly to the previous situation
|-
|0.1||-0.8|| 103.364|| Yes|| [[File: Mlw115_Surface_Plot_HHF_0.1_0.8.png|300px]] || Despite having lower total energy than the previous cases the reaction goes to completion as the momentum between the reacting atoms is sufficient
|}

Despite the H2 energy being higher than the activation energy if the BC atoms do not collide with a high enough energy the reaction will return back over the peak and the original reactants will be produced.

=== HF+F ===

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 2
|-
|BC || 0.74
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -5 || 0.2 || -103.364 || No || [[File:Mlw115_Surface_Plot_HFH_5_0.2.png|300px]] || Approaching H rebounds and reaction doesn't complete

|}

A reactive pathway was found by trial and error around the reversed conditions from the reverse reaction.

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 5.0||-1.5
|-
| BC|| 1.2||1.0
|}

[[File: Mlw115_Surface_Plot_HFH_-1.5_1.png| 400px]]

==Polanyi's Empirical Rules ==

Question 10. Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.

Polanyi's rules state that in reactions with an early transition state (most closely resembling the reactants) translational energy most effectively promotes a successful reaction. Conversely where the transition state is late (resembling products closely) vibrational energy, specifically of the reactants most effectively promotes a successful reaction. <ref>The Journal of Chemical Physics 138, 234104 (2013); https://doi.org/10.1063/1.4810007</ref>

This is demonstrated in the above, where the forward reaction H2+F-> HF+H has an early transition state, as the activation barrier is so low and the transition state closely resembles the reactants. In this case a high translational energy of the incoming atom (fluorine) is more effective in resulting in a successful reaction and increasing the vibrational energy of the H2 molecule has little impact. This is illustrated nicely in the table above.

In the reverse reaction HF + H -> H2+F the transition state is late, as the activated complex much more closely resembles the product. A high vibrational energy of the HF molecule results in a successful reaction whereas increasing the translational energy of the approaching H atom has little impact.

=References=
<references/>

MRD:mlw115RXNDy

2018-05-22T16:08:13Z

Mlw115: /* Transition State Theory */

= Reaction Dynamics =

= Exercise 1: H2 + H system=

==Dynamics from the transition state region ==

 Question 1. What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

Where a reaction is modeled on a potential energy surface the lowest energy path between the reactants and products is seen as a "valley floor" of the surface. The transition structure is a saddle point along the minimum energy ridge.

The gradient of the potential energy surface is zero at the minimum, along the direction of the reaction path and so the first derivative with respect to the inter-nuclear distance is zero. For the saddle point the gradient of the potential energy surface is zero orthogonal to the direction of the reaction pathway.

δV(ri)/δri = 0

Second derivatives
δV2(ri)/δri2 > 0 Minima
δV2(ri)/δri2 < 0 Maxima

where V is the potential energy and r related to the nuclear positions.

The minimum pathway and saddle point can be distinguished using the second derivatives with respect to atom positions. The second derivative of a maximum point is negative, whereas the second derivative of a minimum is positive. As the saddle point is a maxima along the minimum path the second derivative is negative. The second derivative of the minimum energy path in the direction of the reaction will be positive.

==Estimating Transition State Position ==

 Question 2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. 

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.908||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot 1 !! Surface Plot 2!! Internuclear distance vs Time
|-
|[[File:Mlw115_H2H_saddle.png|300px]]||[[File:mlw115_H2H_TS.png|300px]]||[[File:Mlw115_H2H_TS_INDvT.png|300px]]|
|}

The best guess of inter nuclear distances at the transition state was found using trial and error, to find the positions that would result in the minimum trajectory. As the system is symmetric the distances would be equal. The two surface plots are included to demonstrate the "reaction path". The internuclear distance vs time plot demonstrates no change in any bond lengths at the transition state so the overall momenta is 0 and potential energy is at the maximum. This agrees with that the transition state is a stationary point.

== Reaction Path ==
===MEP vs Dynamics===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.909||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot (MEP) !! Surface Plot (Dynamics
|-
|[[File:Mlw115_H2H_MEP_surfaceplot.png|300px]]||[[File:mlw115_H2H_MEP_surfaceplot_dynamics.png|300px]]
|}

 Question 3. Comment on how the mep and the trajectory you just calculated differ.

It can be seen in the MEP plot that the trajectory follows the minimum energy valley floor without any obvious deviations or oscillation. This does not account for the inter-molecular vibrations between the molecules,. The MEP trajectory is corresponding to infinitely slow motion, it does not accurately represent the atomic motion of the reaction. The trajectory oscillates in the dynamics plot, also in much less steps the AB bond distance increases considerably more. This plot is more realistic as it accounts for the motion of atoms during the reaction.

{| class="wikitable" border="1"
|+ Final Geometry
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| 2.486||1.167
|}

=== Reversed conditions ===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| -2.486||-1.167
|}

[[File:mlw115_H2H_dynamics_reversed_momenta.png|300px]]

From taking the previous runs final conditions and setting these as the initial conditions and reversing the momenta values the reaction seems to run in reverse, reaching an end exactly at the transition state.

===Determining Reactivity===

Question 4. Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2.0
|}

{| class="wikitable" border=1
! pAB !! pBC!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -1.25 || -2.5 || -99.018|| Yes || [[File:Mlw115_Surface_Plot_1.25_2.5.png|300px]] || Initially the HAHB distance changes very little as the vibrational energy is quite low, as the HC approaches it can be seen that the BC distance decreases linearly. The molecules pass over the transition state, after this point the HAHB distance decreases and the HBHC distance oscillates slightly, with a higher vibrational energy than the HAHB molecule. This overall process represents the HAHB bond breaking and forming a HBHC molecule.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mlw115_Surface_Plot_1.5_2.0.png|300px]] || The molecule approaches the maxima with some vibrational energy, shown by oscillations in the initial path, the energy of the system is not high enough to pass over the transition state energy maxima so the HC "rebounds" and the HBHC distance begins to increase. The HAHB molecule oscillations increase.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mlw115_Surface_Plot_1.5_2.5.png|300px]] || The reaction proceeds similarly to the first case with HC approaching and forming a bond with HB as there is sufficient energy to proceed over the transition state maxima. The vibrational energy throughout is slightly higher than the first case, demonstrated by higher oscillations in atom distances.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mlw115_Surface_Plot_2.5_5.png|300px]] || In this case the energy is high enough and the atoms are able to pass over the transition state energy maxima, however the vibrational energy of the HBHC system formed seems to be too high to form the molecule and the HC atom rebounds and the distance begins to decrease and so the HAHB is reformed but with considerably higher oscillations.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mlw115_Surface_Plot_2.5_5.2.png|300px]] || The HC approaches the oscillating HAHB system, the HB atom rebounds between the two atoms, as it can be seen on the surface plot the system passes over the transition state maxima, back over to the other side and back again as this occurs. Finally the HBHC bond is formed with a high vibrationally energy
|}

===Transition State Theory===

 Question 5 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? 

Transition state theory is used to understand how elementary chemical reactions take place assuming a quasi-equilibirum between reactanting complexes and those at the transition state. The rate of reaction can be understood by inspecting the complexes at the maximum saddle point of the minimum energy pathway. The reactants are in equilibrium with the activated transition state complexes and can be converted to the products.

This theory is useful to theoretically understand a chemical reaction however it operates on a series of assumptions which limit the theory.
Main Assumptions:
Equilibrium is close in the systems
Only one reaction pathway occurs
There is negligable barrier recrossing

For instance the atoms are assumed to behave classically, ignoring tunelling. Tunneling would be more prominent in reactions with low activation barriers. Also the theory assumes that the saddle point with the lowest energy will be passed over to reach the products, however this is not the case when higher vibrational energies will be populated.

Clearly a key issue seen in the data obtained above is that there is considerable barrier recrossing. Therefore the transition state theory would fail in this case. The TST would assume that all collisions of an energy above the barrier of the transition state would result in a successful reaction. Clearly this is not the case and the experimental rate would be much lower than the TST would predict.

<ref>https://goldbook.iupac.org/html/T/T06470.html</ref> <ref>https://pubs.acs.org/doi/pdf/10.1021/j100238a002</ref>

=Exercise 2: H-H-F System=

==PES Inspection==
Question 6. 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?

For the forward reaction of H2 + F --> HF + H is exothermic, as shown in the surface plot below the energy of the reactants H2 is higher than the energy of the product HF. As the energy absorbed in breaking the H-H bond in the first case is less than the energy released forming the H-F bond which means that in total energy is released which results in an exothermic reaction.

{| class="wikitable" border=1
! Surface Plot H2+F->HF+H !! Surface Plot HF+H -> H2+F
|-
|[[File:mlw115_Surface_Plot_HHF_labelled.png|300px]]||[[File:mlw115_Surface_Plot_HHF_H2Prod.png|300px]]|
|}

==Determining Transition State ==

Question 7. Locate the approximate position of the transition state.

A = HA B = HB and C = F

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! Surface Plot !! Internuclear distance vs Time
|-
|[[File:Mlw115_Surface_Plot_TS_HHF.png|400px]]||[[File:Mlw115 INDvT HHF TS.png|350px]]|
|}

== Calculating Activation Energy ==
Question 8. Report the activation energy for both reactions.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! BC= 1.8135!! BC= 1.8135 Zoomed!! BC= 1.813
|-
|[[File:Mlw115_HHF_EvT_1.8135.png|300px]] ‎||[[File:Mlw115_Plot_1.813_EvT_zoomT0_Resized.png|300px]]||[[File:Mlw115_HHF_EvT_1.813.png|320px]]
|}

pAB = pBC =0

distAB = 0.74

{| class="wikitable" border=1
|+Activation energy HF+H->H2+F
|-
! !! Energy (Kcal/mol)
|-
|Initial energy || -103.752
|-
|Final energy|| -133.89
|-
|EA|| 30.138
|}

{| class="wikitable" border=1
|+Activation energy H2+F->HF+H
|-
! !! Energy (Kcal/mol)
|-
|Initial energy|| -103.742
|-
|Final energy|| -103.752
|-
|EA|| 0.01
|}

==Reaction Dynamics==
=== F+ H2 ===
{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.74||0
|-
| BC|| 2.3||-2.7
|}

{| class="wikitable" border=1
!Surface Plot!! Internuclear momentum vs time
|-
|[[File:Mlw_Surface_Plot_HHF_Reactive.png]]|| [[File:Mlw_INMvT_HHF_Reactive.png]]
|}

{| class="wikitable" border=1
!Energy!!
|-
|Kinetic|| 3.837
|-
|Potential|| -103.886

|-
|Total|| -100.049
|}

Question 9. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?

It can be seen from the animation that as the H-H (AB) bond is broken and the H-F (BC) bond is formed the vibrational energy (and so the kinetic energy) considerably increases for H-F. The internuclear momentum vs time graph demonstrates this, as the momentum is much higher for the H-F atom. This vibrational energy will be released as heat so experimentally an increase in temperature could confirm the reaction had occurred.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -3 || -0.5 || -100.339|| No || [[File:Mlw115_Surface_Plot_HHF_-3_0.5.png|300px]] || H2 approaches the fluorine atom slowly but when a collision is made the H rebounds straight back into the other H and they continue their path oscillating together in the opposite direction
|-

| 3 || -0.5 || -93.254 || No || [[File:Mlw115_Surface_Plot_HHF_3_0.5.png|300px]] || This is similar to the previous case however the hydrogen rebounds twice into flourine but on the second rebounds remains returns back to oscillating with the original H atom
|-
| 2 || -0.5|| -98.753|| No || [[File: Mlw115_Surface_Plot_HHF_2_0.5.png|300px]] || The HF molecule oscillates within the energy well at high vibrational energy after collision
|-
| 1.5|| -0.5|| 103.754|| No || [[File: Mlw115_Surface_Plot_HHF_1.5_0.5.png|300px]] || Similarly to the previous situation
|-
|0.1||-0.8|| 103.364|| Yes|| [[File: Mlw115_Surface_Plot_HHF_0.1_0.8.png|300px]] || Despite having lower total energy than the previous cases the reaction goes to completion as the momentum between the reacting atoms is sufficient
|}

Despite the H2 energy being higher than the activation energy if the BC atoms do not collide with a high enough energy the reaction will return back over the peak and the original reactants will be produced.

=== HF+F ===

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 2
|-
|BC || 0.74
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -5 || 0.2 || -103.364 || No || [[File:Mlw115_Surface_Plot_HFH_5_0.2.png|300px]] || Approaching H rebounds and reaction doesn't complete

|}

A reactive pathway was found by trial and error around the reversed conditions from the reverse reaction.

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 5.0||-1.5
|-
| BC|| 1.2||1.0
|}

[[File: Mlw115_Surface_Plot_HFH_-1.5_1.png| 400px]]

==Polanyi's Empirical Rules ==

Question 10. Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.

Polanyi's rules state that in reactions with an early transition state (most closely resembling the reactants) translational energy most effectively promotes a successful reaction. Conversely where the transition state is late (resembling products closely) vibrational energy, specifically of the reactants most effectively promotes a successful reaction. <ref>The Journal of Chemical Physics 138, 234104 (2013); https://doi.org/10.1063/1.4810007</ref>

This is demonstrated in the above, where the forward reaction H2+F-> HF+H has an early transition state, as the activation barrier is so low and the transition state closely resembles the reactants. In this case a high translational energy of the incoming atom (fluorine) is more effective in resulting in a successful reaction and increasing the vibrational energy of the H2 molecule has little impact. This is illustrated nicely in the table above.

In the reverse reaction HF + H -> H2+F the transition state is late, as the activated complex much more closely resembles the product. A high vibrational energy of the HF molecule results in a successful reaction whereas increasing the translational energy of the approaching H atom has little impact.

=References=
<references/>

MRD:mlw115RXNDy

2018-05-22T16:01:27Z

Mlw115: /* Transition State Theory */

= Reaction Dynamics =

= Exercise 1: H2 + H system=

==Dynamics from the transition state region ==

 Question 1. What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

Where a reaction is modeled on a potential energy surface the lowest energy path between the reactants and products is seen as a "valley floor" of the surface. The transition structure is a saddle point along the minimum energy ridge.

The gradient of the potential energy surface is zero at the minimum, along the direction of the reaction path and so the first derivative with respect to the inter-nuclear distance is zero. For the saddle point the gradient of the potential energy surface is zero orthogonal to the direction of the reaction pathway.

δV(ri)/δri = 0

Second derivatives
δV2(ri)/δri2 > 0 Minima
δV2(ri)/δri2 < 0 Maxima

where V is the potential energy and r related to the nuclear positions.

The minimum pathway and saddle point can be distinguished using the second derivatives with respect to atom positions. The second derivative of a maximum point is negative, whereas the second derivative of a minimum is positive. As the saddle point is a maxima along the minimum path the second derivative is negative. The second derivative of the minimum energy path in the direction of the reaction will be positive.

==Estimating Transition State Position ==

 Question 2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. 

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.908||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot 1 !! Surface Plot 2!! Internuclear distance vs Time
|-
|[[File:Mlw115_H2H_saddle.png|300px]]||[[File:mlw115_H2H_TS.png|300px]]||[[File:Mlw115_H2H_TS_INDvT.png|300px]]|
|}

The best guess of inter nuclear distances at the transition state was found using trial and error, to find the positions that would result in the minimum trajectory. As the system is symmetric the distances would be equal. The two surface plots are included to demonstrate the "reaction path". The internuclear distance vs time plot demonstrates no change in any bond lengths at the transition state so the overall momenta is 0 and potential energy is at the maximum. This agrees with that the transition state is a stationary point.

== Reaction Path ==
===MEP vs Dynamics===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.909||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot (MEP) !! Surface Plot (Dynamics
|-
|[[File:Mlw115_H2H_MEP_surfaceplot.png|300px]]||[[File:mlw115_H2H_MEP_surfaceplot_dynamics.png|300px]]
|}

 Question 3. Comment on how the mep and the trajectory you just calculated differ.

It can be seen in the MEP plot that the trajectory follows the minimum energy valley floor without any obvious deviations or oscillation. This does not account for the inter-molecular vibrations between the molecules,. The MEP trajectory is corresponding to infinitely slow motion, it does not accurately represent the atomic motion of the reaction. The trajectory oscillates in the dynamics plot, also in much less steps the AB bond distance increases considerably more. This plot is more realistic as it accounts for the motion of atoms during the reaction.

{| class="wikitable" border="1"
|+ Final Geometry
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| 2.486||1.167
|}

=== Reversed conditions ===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| -2.486||-1.167
|}

[[File:mlw115_H2H_dynamics_reversed_momenta.png|300px]]

From taking the previous runs final conditions and setting these as the initial conditions and reversing the momenta values the reaction seems to run in reverse, reaching an end exactly at the transition state.

===Determining Reactivity===

Question 4. Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2.0
|}

{| class="wikitable" border=1
! pAB !! pBC!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -1.25 || -2.5 || -99.018|| Yes || [[File:Mlw115_Surface_Plot_1.25_2.5.png|300px]] || Initially the HAHB distance changes very little as the vibrational energy is quite low, as the HC approaches it can be seen that the BC distance decreases linearly. The molecules pass over the transition state, after this point the HAHB distance decreases and the HBHC distance oscillates slightly, with a higher vibrational energy than the HAHB molecule. This overall process represents the HAHB bond breaking and forming a HBHC molecule.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mlw115_Surface_Plot_1.5_2.0.png|300px]] || The molecule approaches the maxima with some vibrational energy, shown by oscillations in the initial path, the energy of the system is not high enough to pass over the transition state energy maxima so the HC "rebounds" and the HBHC distance begins to increase. The HAHB molecule oscillations increase.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mlw115_Surface_Plot_1.5_2.5.png|300px]] || The reaction proceeds similarly to the first case with HC approaching and forming a bond with HB as there is sufficient energy to proceed over the transition state maxima. The vibrational energy throughout is slightly higher than the first case, demonstrated by higher oscillations in atom distances.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mlw115_Surface_Plot_2.5_5.png|300px]] || In this case the energy is high enough and the atoms are able to pass over the transition state energy maxima, however the vibrational energy of the HBHC system formed seems to be too high to form the molecule and the HC atom rebounds and the distance begins to decrease and so the HAHB is reformed but with considerably higher oscillations.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mlw115_Surface_Plot_2.5_5.2.png|300px]] || The HC approaches the oscillating HAHB system, the HB atom rebounds between the two atoms, as it can be seen on the surface plot the system passes over the transition state maxima, back over to the other side and back again as this occurs. Finally the HBHC bond is formed with a high vibrationally energy
|}

===Transition State Theory===

 Question 5 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? 

Transition state theory is used to understand how elementary chemical reactions take place assuming a quasi-equilibirum between reactanting complexes and those at the transition state. The rate of reaction can be understood by inspecting the complexes at the maximum saddle point of the minimum energy pathway. The reactants are in equilibrium with the activated transition state complexes and can be converted to the products.

The theory assumes there is a surface in space that divides the space into a region of reactants and products and that once the reactants cross over to the products region there is no recrossing. Conventional TST has a dividing surface at the saddle point. The theory calculates a one-way rate constant at equlibrium <ref>https://pubs.acs.org/doi/pdf/10.1021/jp953748q</ref>

This theory is useful to theoretically understand a chemical reaction however it operates on a series of assumptions which limit the theory. For instance the atoms are assumed to behave classically, ignoring tunelling. Tunneling would be more prominent in reactions with low activation barriers. Also the theory assumes that the saddle point with the lowest energy will be passed over to reach the products, however this is not the case when higher vibrational energies will be populated.

<ref>https://goldbook.iupac.org/html/T/T06470.html</ref> <ref>https://pubs.acs.org/doi/pdf/10.1021/j100238a002</ref>

=Exercise 2: H-H-F System=

==PES Inspection==
Question 6. 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?

For the forward reaction of H2 + F --> HF + H is exothermic, as shown in the surface plot below the energy of the reactants H2 is higher than the energy of the product HF. As the energy absorbed in breaking the H-H bond in the first case is less than the energy released forming the H-F bond which means that in total energy is released which results in an exothermic reaction.

{| class="wikitable" border=1
! Surface Plot H2+F->HF+H !! Surface Plot HF+H -> H2+F
|-
|[[File:mlw115_Surface_Plot_HHF_labelled.png|300px]]||[[File:mlw115_Surface_Plot_HHF_H2Prod.png|300px]]|
|}

==Determining Transition State ==

Question 7. Locate the approximate position of the transition state.

A = HA B = HB and C = F

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! Surface Plot !! Internuclear distance vs Time
|-
|[[File:Mlw115_Surface_Plot_TS_HHF.png|400px]]||[[File:Mlw115 INDvT HHF TS.png|350px]]|
|}

== Calculating Activation Energy ==
Question 8. Report the activation energy for both reactions.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! BC= 1.8135!! BC= 1.8135 Zoomed!! BC= 1.813
|-
|[[File:Mlw115_HHF_EvT_1.8135.png|300px]] ‎||[[File:Mlw115_Plot_1.813_EvT_zoomT0_Resized.png|300px]]||[[File:Mlw115_HHF_EvT_1.813.png|320px]]
|}

pAB = pBC =0

distAB = 0.74

{| class="wikitable" border=1
|+Activation energy HF+H->H2+F
|-
! !! Energy (Kcal/mol)
|-
|Initial energy || -103.752
|-
|Final energy|| -133.89
|-
|EA|| 30.138
|}

{| class="wikitable" border=1
|+Activation energy H2+F->HF+H
|-
! !! Energy (Kcal/mol)
|-
|Initial energy|| -103.742
|-
|Final energy|| -103.752
|-
|EA|| 0.01
|}

==Reaction Dynamics==
=== F+ H2 ===
{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.74||0
|-
| BC|| 2.3||-2.7
|}

{| class="wikitable" border=1
!Surface Plot!! Internuclear momentum vs time
|-
|[[File:Mlw_Surface_Plot_HHF_Reactive.png]]|| [[File:Mlw_INMvT_HHF_Reactive.png]]
|}

{| class="wikitable" border=1
!Energy!!
|-
|Kinetic|| 3.837
|-
|Potential|| -103.886

|-
|Total|| -100.049
|}

Question 9. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?

It can be seen from the animation that as the H-H (AB) bond is broken and the H-F (BC) bond is formed the vibrational energy (and so the kinetic energy) considerably increases for H-F. The internuclear momentum vs time graph demonstrates this, as the momentum is much higher for the H-F atom. This vibrational energy will be released as heat so experimentally an increase in temperature could confirm the reaction had occurred.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -3 || -0.5 || -100.339|| No || [[File:Mlw115_Surface_Plot_HHF_-3_0.5.png|300px]] || H2 approaches the fluorine atom slowly but when a collision is made the H rebounds straight back into the other H and they continue their path oscillating together in the opposite direction
|-

| 3 || -0.5 || -93.254 || No || [[File:Mlw115_Surface_Plot_HHF_3_0.5.png|300px]] || This is similar to the previous case however the hydrogen rebounds twice into flourine but on the second rebounds remains returns back to oscillating with the original H atom
|-
| 2 || -0.5|| -98.753|| No || [[File: Mlw115_Surface_Plot_HHF_2_0.5.png|300px]] || The HF molecule oscillates within the energy well at high vibrational energy after collision
|-
| 1.5|| -0.5|| 103.754|| No || [[File: Mlw115_Surface_Plot_HHF_1.5_0.5.png|300px]] || Similarly to the previous situation
|-
|0.1||-0.8|| 103.364|| Yes|| [[File: Mlw115_Surface_Plot_HHF_0.1_0.8.png|300px]] || Despite having lower total energy than the previous cases the reaction goes to completion as the momentum between the reacting atoms is sufficient
|}

Despite the H2 energy being higher than the activation energy if the BC atoms do not collide with a high enough energy the reaction will return back over the peak and the original reactants will be produced.

=== HF+F ===

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 2
|-
|BC || 0.74
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -5 || 0.2 || -103.364 || No || [[File:Mlw115_Surface_Plot_HFH_5_0.2.png|300px]] || Approaching H rebounds and reaction doesn't complete

|}

A reactive pathway was found by trial and error around the reversed conditions from the reverse reaction.

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 5.0||-1.5
|-
| BC|| 1.2||1.0
|}

[[File: Mlw115_Surface_Plot_HFH_-1.5_1.png| 400px]]

==Polanyi's Empirical Rules ==

Question 10. Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.

Polanyi's rules state that in reactions with an early transition state (most closely resembling the reactants) translational energy most effectively promotes a successful reaction. Conversely where the transition state is late (resembling products closely) vibrational energy, specifically of the reactants most effectively promotes a successful reaction. <ref>The Journal of Chemical Physics 138, 234104 (2013); https://doi.org/10.1063/1.4810007</ref>

This is demonstrated in the above, where the forward reaction H2+F-> HF+H has an early transition state, as the activation barrier is so low and the transition state closely resembles the reactants. In this case a high translational energy of the incoming atom (fluorine) is more effective in resulting in a successful reaction and increasing the vibrational energy of the H2 molecule has little impact. This is illustrated nicely in the table above.

In the reverse reaction HF + H -> H2+F the transition state is late, as the activated complex much more closely resembles the product. A high vibrational energy of the HF molecule results in a successful reaction whereas increasing the translational energy of the approaching H atom has little impact.

=References=
<references/>

MRD:mlw115RXNDy

2018-05-22T15:48:18Z

Mlw115: /* Polanyi's Empirical Rules */

= Reaction Dynamics =

= Exercise 1: H2 + H system=

==Dynamics from the transition state region ==

 Question 1. What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

Where a reaction is modeled on a potential energy surface the lowest energy path between the reactants and products is seen as a "valley floor" of the surface. The transition structure is a saddle point along the minimum energy ridge.

The gradient of the potential energy surface is zero at the minimum, along the direction of the reaction path and so the first derivative with respect to the inter-nuclear distance is zero. For the saddle point the gradient of the potential energy surface is zero orthogonal to the direction of the reaction pathway.

δV(ri)/δri = 0

Second derivatives
δV2(ri)/δri2 > 0 Minima
δV2(ri)/δri2 < 0 Maxima

where V is the potential energy and r related to the nuclear positions.

The minimum pathway and saddle point can be distinguished using the second derivatives with respect to atom positions. The second derivative of a maximum point is negative, whereas the second derivative of a minimum is positive. As the saddle point is a maxima along the minimum path the second derivative is negative. The second derivative of the minimum energy path in the direction of the reaction will be positive.

==Estimating Transition State Position ==

 Question 2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. 

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.908||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot 1 !! Surface Plot 2!! Internuclear distance vs Time
|-
|[[File:Mlw115_H2H_saddle.png|300px]]||[[File:mlw115_H2H_TS.png|300px]]||[[File:Mlw115_H2H_TS_INDvT.png|300px]]|
|}

The best guess of inter nuclear distances at the transition state was found using trial and error, to find the positions that would result in the minimum trajectory. As the system is symmetric the distances would be equal. The two surface plots are included to demonstrate the "reaction path". The internuclear distance vs time plot demonstrates no change in any bond lengths at the transition state so the overall momenta is 0 and potential energy is at the maximum. This agrees with that the transition state is a stationary point.

== Reaction Path ==
===MEP vs Dynamics===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.909||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot (MEP) !! Surface Plot (Dynamics
|-
|[[File:Mlw115_H2H_MEP_surfaceplot.png|300px]]||[[File:mlw115_H2H_MEP_surfaceplot_dynamics.png|300px]]
|}

 Question 3. Comment on how the mep and the trajectory you just calculated differ.

It can be seen in the MEP plot that the trajectory follows the minimum energy valley floor without any obvious deviations or oscillation. This does not account for the inter-molecular vibrations between the molecules,. The MEP trajectory is corresponding to infinitely slow motion, it does not accurately represent the atomic motion of the reaction. The trajectory oscillates in the dynamics plot, also in much less steps the AB bond distance increases considerably more. This plot is more realistic as it accounts for the motion of atoms during the reaction.

{| class="wikitable" border="1"
|+ Final Geometry
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| 2.486||1.167
|}

=== Reversed conditions ===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| -2.486||-1.167
|}

[[File:mlw115_H2H_dynamics_reversed_momenta.png|300px]]

From taking the previous runs final conditions and setting these as the initial conditions and reversing the momenta values the reaction seems to run in reverse, reaching an end exactly at the transition state.

===Determining Reactivity===

Question 4. Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2.0
|}

{| class="wikitable" border=1
! pAB !! pBC!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -1.25 || -2.5 || -99.018|| Yes || [[File:Mlw115_Surface_Plot_1.25_2.5.png|300px]] || Initially the HAHB distance changes very little as the vibrational energy is quite low, as the HC approaches it can be seen that the BC distance decreases linearly. The molecules pass over the transition state, after this point the HAHB distance decreases and the HBHC distance oscillates slightly, with a higher vibrational energy than the HAHB molecule. This overall process represents the HAHB bond breaking and forming a HBHC molecule.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mlw115_Surface_Plot_1.5_2.0.png|300px]] || The molecule approaches the maxima with some vibrational energy, shown by oscillations in the initial path, the energy of the system is not high enough to pass over the transition state energy maxima so the HC "rebounds" and the HBHC distance begins to increase. The HAHB molecule oscillations increase.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mlw115_Surface_Plot_1.5_2.5.png|300px]] || The reaction proceeds similarly to the first case with HC approaching and forming a bond with HB as there is sufficient energy to proceed over the transition state maxima. The vibrational energy throughout is slightly higher than the first case, demonstrated by higher oscillations in atom distances.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mlw115_Surface_Plot_2.5_5.png|300px]] || In this case the energy is high enough and the atoms are able to pass over the transition state energy maxima, however the vibrational energy of the HBHC system formed seems to be too high to form the molecule and the HC atom rebounds and the distance begins to decrease and so the HAHB is reformed but with considerably higher oscillations.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mlw115_Surface_Plot_2.5_5.2.png|300px]] || The HC approaches the oscillating HAHB system, the HB atom rebounds between the two atoms, as it can be seen on the surface plot the system passes over the transition state maxima, back over to the other side and back again as this occurs. Finally the HBHC bond is formed with a high vibrationally energy
|}

===Transition State Theory===

 Question 5 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? 

Transition state theory is used to understand how elementary chemical reactions take place assuming a quasi-equilibirum between reactanting complexes and those at the transition state. The rate of reaction can be understood by inspecting the complexes at the maximum saddle point of the minimum energy pathway. The reactants are in equilibrium with the activated transition state complexes and can be converted to the products.

This theory is useful to theoretically understand a chemical reaction however it operates on a series of assumptions which limit the theory. For instance the atoms are assumed to behave classically, ignoring tunelling. Tunneling would be more prominent in reactions with low activation barriers. Also the theory assumes that the saddle point with the lowest energy will be passed over to reach the products, however this is not the case when higher vibrational energies will be populated.

<ref>https://goldbook.iupac.org/html/T/T06470.html</ref> <ref>https://pubs.acs.org/doi/pdf/10.1021/j100238a002</ref>

=Exercise 2: H-H-F System=

==PES Inspection==
Question 6. 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?

For the forward reaction of H2 + F --> HF + H is exothermic, as shown in the surface plot below the energy of the reactants H2 is higher than the energy of the product HF. As the energy absorbed in breaking the H-H bond in the first case is less than the energy released forming the H-F bond which means that in total energy is released which results in an exothermic reaction.

{| class="wikitable" border=1
! Surface Plot H2+F->HF+H !! Surface Plot HF+H -> H2+F
|-
|[[File:mlw115_Surface_Plot_HHF_labelled.png|300px]]||[[File:mlw115_Surface_Plot_HHF_H2Prod.png|300px]]|
|}

==Determining Transition State ==

Question 7. Locate the approximate position of the transition state.

A = HA B = HB and C = F

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! Surface Plot !! Internuclear distance vs Time
|-
|[[File:Mlw115_Surface_Plot_TS_HHF.png|400px]]||[[File:Mlw115 INDvT HHF TS.png|350px]]|
|}

== Calculating Activation Energy ==
Question 8. Report the activation energy for both reactions.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! BC= 1.8135!! BC= 1.8135 Zoomed!! BC= 1.813
|-
|[[File:Mlw115_HHF_EvT_1.8135.png|300px]] ‎||[[File:Mlw115_Plot_1.813_EvT_zoomT0_Resized.png|300px]]||[[File:Mlw115_HHF_EvT_1.813.png|320px]]
|}

pAB = pBC =0

distAB = 0.74

{| class="wikitable" border=1
|+Activation energy HF+H->H2+F
|-
! !! Energy (Kcal/mol)
|-
|Initial energy || -103.752
|-
|Final energy|| -133.89
|-
|EA|| 30.138
|}

{| class="wikitable" border=1
|+Activation energy H2+F->HF+H
|-
! !! Energy (Kcal/mol)
|-
|Initial energy|| -103.742
|-
|Final energy|| -103.752
|-
|EA|| 0.01
|}

==Reaction Dynamics==
=== F+ H2 ===
{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.74||0
|-
| BC|| 2.3||-2.7
|}

{| class="wikitable" border=1
!Surface Plot!! Internuclear momentum vs time
|-
|[[File:Mlw_Surface_Plot_HHF_Reactive.png]]|| [[File:Mlw_INMvT_HHF_Reactive.png]]
|}

{| class="wikitable" border=1
!Energy!!
|-
|Kinetic|| 3.837
|-
|Potential|| -103.886

|-
|Total|| -100.049
|}

Question 9. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?

It can be seen from the animation that as the H-H (AB) bond is broken and the H-F (BC) bond is formed the vibrational energy (and so the kinetic energy) considerably increases for H-F. The internuclear momentum vs time graph demonstrates this, as the momentum is much higher for the H-F atom. This vibrational energy will be released as heat so experimentally an increase in temperature could confirm the reaction had occurred.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -3 || -0.5 || -100.339|| No || [[File:Mlw115_Surface_Plot_HHF_-3_0.5.png|300px]] || H2 approaches the fluorine atom slowly but when a collision is made the H rebounds straight back into the other H and they continue their path oscillating together in the opposite direction
|-

| 3 || -0.5 || -93.254 || No || [[File:Mlw115_Surface_Plot_HHF_3_0.5.png|300px]] || This is similar to the previous case however the hydrogen rebounds twice into flourine but on the second rebounds remains returns back to oscillating with the original H atom
|-
| 2 || -0.5|| -98.753|| No || [[File: Mlw115_Surface_Plot_HHF_2_0.5.png|300px]] || The HF molecule oscillates within the energy well at high vibrational energy after collision
|-
| 1.5|| -0.5|| 103.754|| No || [[File: Mlw115_Surface_Plot_HHF_1.5_0.5.png|300px]] || Similarly to the previous situation
|-
|0.1||-0.8|| 103.364|| Yes|| [[File: Mlw115_Surface_Plot_HHF_0.1_0.8.png|300px]] || Despite having lower total energy than the previous cases the reaction goes to completion as the momentum between the reacting atoms is sufficient
|}

Despite the H2 energy being higher than the activation energy if the BC atoms do not collide with a high enough energy the reaction will return back over the peak and the original reactants will be produced.

=== HF+F ===

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 2
|-
|BC || 0.74
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -5 || 0.2 || -103.364 || No || [[File:Mlw115_Surface_Plot_HFH_5_0.2.png|300px]] || Approaching H rebounds and reaction doesn't complete

|}

A reactive pathway was found by trial and error around the reversed conditions from the reverse reaction.

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 5.0||-1.5
|-
| BC|| 1.2||1.0
|}

[[File: Mlw115_Surface_Plot_HFH_-1.5_1.png| 400px]]

==Polanyi's Empirical Rules ==

Question 10. Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.

Polanyi's rules state that in reactions with an early transition state (most closely resembling the reactants) translational energy most effectively promotes a successful reaction. Conversely where the transition state is late (resembling products closely) vibrational energy, specifically of the reactants most effectively promotes a successful reaction. <ref>The Journal of Chemical Physics 138, 234104 (2013); https://doi.org/10.1063/1.4810007</ref>

This is demonstrated in the above, where the forward reaction H2+F-> HF+H has an early transition state, as the activation barrier is so low and the transition state closely resembles the reactants. In this case a high translational energy of the incoming atom (fluorine) is more effective in resulting in a successful reaction and increasing the vibrational energy of the H2 molecule has little impact. This is illustrated nicely in the table above.

In the reverse reaction HF + H -> H2+F the transition state is late, as the activated complex much more closely resembles the product. A high vibrational energy of the HF molecule results in a successful reaction whereas increasing the translational energy of the approaching H atom has little impact.

=References=
<references/>

MRD:mlw115RXNDy

2018-05-22T15:40:02Z

Mlw115: /* Polanyi's Empirical Rules */

= Reaction Dynamics =

= Exercise 1: H2 + H system=

==Dynamics from the transition state region ==

 Question 1. What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

Where a reaction is modeled on a potential energy surface the lowest energy path between the reactants and products is seen as a "valley floor" of the surface. The transition structure is a saddle point along the minimum energy ridge.

The gradient of the potential energy surface is zero at the minimum, along the direction of the reaction path and so the first derivative with respect to the inter-nuclear distance is zero. For the saddle point the gradient of the potential energy surface is zero orthogonal to the direction of the reaction pathway.

δV(ri)/δri = 0

Second derivatives
δV2(ri)/δri2 > 0 Minima
δV2(ri)/δri2 < 0 Maxima

where V is the potential energy and r related to the nuclear positions.

The minimum pathway and saddle point can be distinguished using the second derivatives with respect to atom positions. The second derivative of a maximum point is negative, whereas the second derivative of a minimum is positive. As the saddle point is a maxima along the minimum path the second derivative is negative. The second derivative of the minimum energy path in the direction of the reaction will be positive.

==Estimating Transition State Position ==

 Question 2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. 

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.908||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot 1 !! Surface Plot 2!! Internuclear distance vs Time
|-
|[[File:Mlw115_H2H_saddle.png|300px]]||[[File:mlw115_H2H_TS.png|300px]]||[[File:Mlw115_H2H_TS_INDvT.png|300px]]|
|}

The best guess of inter nuclear distances at the transition state was found using trial and error, to find the positions that would result in the minimum trajectory. As the system is symmetric the distances would be equal. The two surface plots are included to demonstrate the "reaction path". The internuclear distance vs time plot demonstrates no change in any bond lengths at the transition state so the overall momenta is 0 and potential energy is at the maximum. This agrees with that the transition state is a stationary point.

== Reaction Path ==
===MEP vs Dynamics===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.909||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot (MEP) !! Surface Plot (Dynamics
|-
|[[File:Mlw115_H2H_MEP_surfaceplot.png|300px]]||[[File:mlw115_H2H_MEP_surfaceplot_dynamics.png|300px]]
|}

 Question 3. Comment on how the mep and the trajectory you just calculated differ.

It can be seen in the MEP plot that the trajectory follows the minimum energy valley floor without any obvious deviations or oscillation. This does not account for the inter-molecular vibrations between the molecules,. The MEP trajectory is corresponding to infinitely slow motion, it does not accurately represent the atomic motion of the reaction. The trajectory oscillates in the dynamics plot, also in much less steps the AB bond distance increases considerably more. This plot is more realistic as it accounts for the motion of atoms during the reaction.

{| class="wikitable" border="1"
|+ Final Geometry
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| 2.486||1.167
|}

=== Reversed conditions ===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| -2.486||-1.167
|}

[[File:mlw115_H2H_dynamics_reversed_momenta.png|300px]]

From taking the previous runs final conditions and setting these as the initial conditions and reversing the momenta values the reaction seems to run in reverse, reaching an end exactly at the transition state.

===Determining Reactivity===

Question 4. Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2.0
|}

{| class="wikitable" border=1
! pAB !! pBC!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -1.25 || -2.5 || -99.018|| Yes || [[File:Mlw115_Surface_Plot_1.25_2.5.png|300px]] || Initially the HAHB distance changes very little as the vibrational energy is quite low, as the HC approaches it can be seen that the BC distance decreases linearly. The molecules pass over the transition state, after this point the HAHB distance decreases and the HBHC distance oscillates slightly, with a higher vibrational energy than the HAHB molecule. This overall process represents the HAHB bond breaking and forming a HBHC molecule.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mlw115_Surface_Plot_1.5_2.0.png|300px]] || The molecule approaches the maxima with some vibrational energy, shown by oscillations in the initial path, the energy of the system is not high enough to pass over the transition state energy maxima so the HC "rebounds" and the HBHC distance begins to increase. The HAHB molecule oscillations increase.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mlw115_Surface_Plot_1.5_2.5.png|300px]] || The reaction proceeds similarly to the first case with HC approaching and forming a bond with HB as there is sufficient energy to proceed over the transition state maxima. The vibrational energy throughout is slightly higher than the first case, demonstrated by higher oscillations in atom distances.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mlw115_Surface_Plot_2.5_5.png|300px]] || In this case the energy is high enough and the atoms are able to pass over the transition state energy maxima, however the vibrational energy of the HBHC system formed seems to be too high to form the molecule and the HC atom rebounds and the distance begins to decrease and so the HAHB is reformed but with considerably higher oscillations.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mlw115_Surface_Plot_2.5_5.2.png|300px]] || The HC approaches the oscillating HAHB system, the HB atom rebounds between the two atoms, as it can be seen on the surface plot the system passes over the transition state maxima, back over to the other side and back again as this occurs. Finally the HBHC bond is formed with a high vibrationally energy
|}

===Transition State Theory===

 Question 5 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? 

Transition state theory is used to understand how elementary chemical reactions take place assuming a quasi-equilibirum between reactanting complexes and those at the transition state. The rate of reaction can be understood by inspecting the complexes at the maximum saddle point of the minimum energy pathway. The reactants are in equilibrium with the activated transition state complexes and can be converted to the products.

This theory is useful to theoretically understand a chemical reaction however it operates on a series of assumptions which limit the theory. For instance the atoms are assumed to behave classically, ignoring tunelling. Tunneling would be more prominent in reactions with low activation barriers. Also the theory assumes that the saddle point with the lowest energy will be passed over to reach the products, however this is not the case when higher vibrational energies will be populated.

<ref>https://goldbook.iupac.org/html/T/T06470.html</ref> <ref>https://pubs.acs.org/doi/pdf/10.1021/j100238a002</ref>

=Exercise 2: H-H-F System=

==PES Inspection==
Question 6. 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?

For the forward reaction of H2 + F --> HF + H is exothermic, as shown in the surface plot below the energy of the reactants H2 is higher than the energy of the product HF. As the energy absorbed in breaking the H-H bond in the first case is less than the energy released forming the H-F bond which means that in total energy is released which results in an exothermic reaction.

{| class="wikitable" border=1
! Surface Plot H2+F->HF+H !! Surface Plot HF+H -> H2+F
|-
|[[File:mlw115_Surface_Plot_HHF_labelled.png|300px]]||[[File:mlw115_Surface_Plot_HHF_H2Prod.png|300px]]|
|}

==Determining Transition State ==

Question 7. Locate the approximate position of the transition state.

A = HA B = HB and C = F

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! Surface Plot !! Internuclear distance vs Time
|-
|[[File:Mlw115_Surface_Plot_TS_HHF.png|400px]]||[[File:Mlw115 INDvT HHF TS.png|350px]]|
|}

== Calculating Activation Energy ==
Question 8. Report the activation energy for both reactions.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! BC= 1.8135!! BC= 1.8135 Zoomed!! BC= 1.813
|-
|[[File:Mlw115_HHF_EvT_1.8135.png|300px]] ‎||[[File:Mlw115_Plot_1.813_EvT_zoomT0_Resized.png|300px]]||[[File:Mlw115_HHF_EvT_1.813.png|320px]]
|}

pAB = pBC =0

distAB = 0.74

{| class="wikitable" border=1
|+Activation energy HF+H->H2+F
|-
! !! Energy (Kcal/mol)
|-
|Initial energy || -103.752
|-
|Final energy|| -133.89
|-
|EA|| 30.138
|}

{| class="wikitable" border=1
|+Activation energy H2+F->HF+H
|-
! !! Energy (Kcal/mol)
|-
|Initial energy|| -103.742
|-
|Final energy|| -103.752
|-
|EA|| 0.01
|}

==Reaction Dynamics==
=== F+ H2 ===
{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.74||0
|-
| BC|| 2.3||-2.7
|}

{| class="wikitable" border=1
!Surface Plot!! Internuclear momentum vs time
|-
|[[File:Mlw_Surface_Plot_HHF_Reactive.png]]|| [[File:Mlw_INMvT_HHF_Reactive.png]]
|}

{| class="wikitable" border=1
!Energy!!
|-
|Kinetic|| 3.837
|-
|Potential|| -103.886

|-
|Total|| -100.049
|}

Question 9. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?

It can be seen from the animation that as the H-H (AB) bond is broken and the H-F (BC) bond is formed the vibrational energy (and so the kinetic energy) considerably increases for H-F. The internuclear momentum vs time graph demonstrates this, as the momentum is much higher for the H-F atom. This vibrational energy will be released as heat so experimentally an increase in temperature could confirm the reaction had occurred.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -3 || -0.5 || -100.339|| No || [[File:Mlw115_Surface_Plot_HHF_-3_0.5.png|300px]] || H2 approaches the fluorine atom slowly but when a collision is made the H rebounds straight back into the other H and they continue their path oscillating together in the opposite direction
|-

| 3 || -0.5 || -93.254 || No || [[File:Mlw115_Surface_Plot_HHF_3_0.5.png|300px]] || This is similar to the previous case however the hydrogen rebounds twice into flourine but on the second rebounds remains returns back to oscillating with the original H atom
|-
| 2 || -0.5|| -98.753|| No || [[File: Mlw115_Surface_Plot_HHF_2_0.5.png|300px]] || The HF molecule oscillates within the energy well at high vibrational energy after collision
|-
| 1.5|| -0.5|| 103.754|| No || [[File: Mlw115_Surface_Plot_HHF_1.5_0.5.png|300px]] || Similarly to the previous situation
|-
|0.1||-0.8|| 103.364|| Yes|| [[File: Mlw115_Surface_Plot_HHF_0.1_0.8.png|300px]] || Despite having lower total energy than the previous cases the reaction goes to completion as the momentum between the reacting atoms is sufficient
|}

Despite the H2 energy being higher than the activation energy if the BC atoms do not collide with a high enough energy the reaction will return back over the peak and the original reactants will be produced.

=== HF+F ===

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 2
|-
|BC || 0.74
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -5 || 0.2 || -103.364 || No || [[File:Mlw115_Surface_Plot_HFH_5_0.2.png|300px]] || Approaching H rebounds and reaction doesn't complete

|}

A reactive pathway was found by trial and error around the reversed conditions from the reverse reaction.

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 5.0||-1.5
|-
| BC|| 1.2||1.0
|}

[[File: Mlw115_Surface_Plot_HFH_-1.5_1.png| 400px]]

==Polanyi's Empirical Rules ==

Question 10. Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.

Polanyi's rules state that in reactions with an early transition state (most closely resembling the reactants) translational energy most effectively promotes a successful reaction. Conversely where the transition state is late (resembling products closely) vibrational energy, specifically of the reactants most effectively promotes a successful reaction. <ref>The Journal of Chemical Physics 138, 234104 (2013); https://doi.org/10.1063/1.4810007</ref>

This is demonstrated in the above, where the forward reaction H2+F-> HF+H has an early transition state, as the activation barrier is so low and the transition state closely resembles the reactants. In this case a high translational energy

=References=
<references/>

MRD:mlw115RXNDy

2018-05-22T15:35:37Z

Mlw115: /* Polanyi's Empirical Rules */

= Reaction Dynamics =

= Exercise 1: H2 + H system=

==Dynamics from the transition state region ==

 Question 1. What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

Where a reaction is modeled on a potential energy surface the lowest energy path between the reactants and products is seen as a "valley floor" of the surface. The transition structure is a saddle point along the minimum energy ridge.

The gradient of the potential energy surface is zero at the minimum, along the direction of the reaction path and so the first derivative with respect to the inter-nuclear distance is zero. For the saddle point the gradient of the potential energy surface is zero orthogonal to the direction of the reaction pathway.

δV(ri)/δri = 0

Second derivatives
δV2(ri)/δri2 > 0 Minima
δV2(ri)/δri2 < 0 Maxima

where V is the potential energy and r related to the nuclear positions.

The minimum pathway and saddle point can be distinguished using the second derivatives with respect to atom positions. The second derivative of a maximum point is negative, whereas the second derivative of a minimum is positive. As the saddle point is a maxima along the minimum path the second derivative is negative. The second derivative of the minimum energy path in the direction of the reaction will be positive.

==Estimating Transition State Position ==

 Question 2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. 

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.908||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot 1 !! Surface Plot 2!! Internuclear distance vs Time
|-
|[[File:Mlw115_H2H_saddle.png|300px]]||[[File:mlw115_H2H_TS.png|300px]]||[[File:Mlw115_H2H_TS_INDvT.png|300px]]|
|}

The best guess of inter nuclear distances at the transition state was found using trial and error, to find the positions that would result in the minimum trajectory. As the system is symmetric the distances would be equal. The two surface plots are included to demonstrate the "reaction path". The internuclear distance vs time plot demonstrates no change in any bond lengths at the transition state so the overall momenta is 0 and potential energy is at the maximum. This agrees with that the transition state is a stationary point.

== Reaction Path ==
===MEP vs Dynamics===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.909||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot (MEP) !! Surface Plot (Dynamics
|-
|[[File:Mlw115_H2H_MEP_surfaceplot.png|300px]]||[[File:mlw115_H2H_MEP_surfaceplot_dynamics.png|300px]]
|}

 Question 3. Comment on how the mep and the trajectory you just calculated differ.

It can be seen in the MEP plot that the trajectory follows the minimum energy valley floor without any obvious deviations or oscillation. This does not account for the inter-molecular vibrations between the molecules,. The MEP trajectory is corresponding to infinitely slow motion, it does not accurately represent the atomic motion of the reaction. The trajectory oscillates in the dynamics plot, also in much less steps the AB bond distance increases considerably more. This plot is more realistic as it accounts for the motion of atoms during the reaction.

{| class="wikitable" border="1"
|+ Final Geometry
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| 2.486||1.167
|}

=== Reversed conditions ===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| -2.486||-1.167
|}

[[File:mlw115_H2H_dynamics_reversed_momenta.png|300px]]

From taking the previous runs final conditions and setting these as the initial conditions and reversing the momenta values the reaction seems to run in reverse, reaching an end exactly at the transition state.

===Determining Reactivity===

Question 4. Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2.0
|}

{| class="wikitable" border=1
! pAB !! pBC!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -1.25 || -2.5 || -99.018|| Yes || [[File:Mlw115_Surface_Plot_1.25_2.5.png|300px]] || Initially the HAHB distance changes very little as the vibrational energy is quite low, as the HC approaches it can be seen that the BC distance decreases linearly. The molecules pass over the transition state, after this point the HAHB distance decreases and the HBHC distance oscillates slightly, with a higher vibrational energy than the HAHB molecule. This overall process represents the HAHB bond breaking and forming a HBHC molecule.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mlw115_Surface_Plot_1.5_2.0.png|300px]] || The molecule approaches the maxima with some vibrational energy, shown by oscillations in the initial path, the energy of the system is not high enough to pass over the transition state energy maxima so the HC "rebounds" and the HBHC distance begins to increase. The HAHB molecule oscillations increase.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mlw115_Surface_Plot_1.5_2.5.png|300px]] || The reaction proceeds similarly to the first case with HC approaching and forming a bond with HB as there is sufficient energy to proceed over the transition state maxima. The vibrational energy throughout is slightly higher than the first case, demonstrated by higher oscillations in atom distances.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mlw115_Surface_Plot_2.5_5.png|300px]] || In this case the energy is high enough and the atoms are able to pass over the transition state energy maxima, however the vibrational energy of the HBHC system formed seems to be too high to form the molecule and the HC atom rebounds and the distance begins to decrease and so the HAHB is reformed but with considerably higher oscillations.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mlw115_Surface_Plot_2.5_5.2.png|300px]] || The HC approaches the oscillating HAHB system, the HB atom rebounds between the two atoms, as it can be seen on the surface plot the system passes over the transition state maxima, back over to the other side and back again as this occurs. Finally the HBHC bond is formed with a high vibrationally energy
|}

===Transition State Theory===

 Question 5 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? 

Transition state theory is used to understand how elementary chemical reactions take place assuming a quasi-equilibirum between reactanting complexes and those at the transition state. The rate of reaction can be understood by inspecting the complexes at the maximum saddle point of the minimum energy pathway. The reactants are in equilibrium with the activated transition state complexes and can be converted to the products.

This theory is useful to theoretically understand a chemical reaction however it operates on a series of assumptions which limit the theory. For instance the atoms are assumed to behave classically, ignoring tunelling. Tunneling would be more prominent in reactions with low activation barriers. Also the theory assumes that the saddle point with the lowest energy will be passed over to reach the products, however this is not the case when higher vibrational energies will be populated.

<ref>https://goldbook.iupac.org/html/T/T06470.html</ref> <ref>https://pubs.acs.org/doi/pdf/10.1021/j100238a002</ref>

=Exercise 2: H-H-F System=

==PES Inspection==
Question 6. 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?

For the forward reaction of H2 + F --> HF + H is exothermic, as shown in the surface plot below the energy of the reactants H2 is higher than the energy of the product HF. As the energy absorbed in breaking the H-H bond in the first case is less than the energy released forming the H-F bond which means that in total energy is released which results in an exothermic reaction.

{| class="wikitable" border=1
! Surface Plot H2+F->HF+H !! Surface Plot HF+H -> H2+F
|-
|[[File:mlw115_Surface_Plot_HHF_labelled.png|300px]]||[[File:mlw115_Surface_Plot_HHF_H2Prod.png|300px]]|
|}

==Determining Transition State ==

Question 7. Locate the approximate position of the transition state.

A = HA B = HB and C = F

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! Surface Plot !! Internuclear distance vs Time
|-
|[[File:Mlw115_Surface_Plot_TS_HHF.png|400px]]||[[File:Mlw115 INDvT HHF TS.png|350px]]|
|}

== Calculating Activation Energy ==
Question 8. Report the activation energy for both reactions.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! BC= 1.8135!! BC= 1.8135 Zoomed!! BC= 1.813
|-
|[[File:Mlw115_HHF_EvT_1.8135.png|300px]] ‎||[[File:Mlw115_Plot_1.813_EvT_zoomT0_Resized.png|300px]]||[[File:Mlw115_HHF_EvT_1.813.png|320px]]
|}

pAB = pBC =0

distAB = 0.74

{| class="wikitable" border=1
|+Activation energy HF+H->H2+F
|-
! !! Energy (Kcal/mol)
|-
|Initial energy || -103.752
|-
|Final energy|| -133.89
|-
|EA|| 30.138
|}

{| class="wikitable" border=1
|+Activation energy H2+F->HF+H
|-
! !! Energy (Kcal/mol)
|-
|Initial energy|| -103.742
|-
|Final energy|| -103.752
|-
|EA|| 0.01
|}

==Reaction Dynamics==
=== F+ H2 ===
{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.74||0
|-
| BC|| 2.3||-2.7
|}

{| class="wikitable" border=1
!Surface Plot!! Internuclear momentum vs time
|-
|[[File:Mlw_Surface_Plot_HHF_Reactive.png]]|| [[File:Mlw_INMvT_HHF_Reactive.png]]
|}

{| class="wikitable" border=1
!Energy!!
|-
|Kinetic|| 3.837
|-
|Potential|| -103.886

|-
|Total|| -100.049
|}

Question 9. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?

It can be seen from the animation that as the H-H (AB) bond is broken and the H-F (BC) bond is formed the vibrational energy (and so the kinetic energy) considerably increases for H-F. The internuclear momentum vs time graph demonstrates this, as the momentum is much higher for the H-F atom. This vibrational energy will be released as heat so experimentally an increase in temperature could confirm the reaction had occurred.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -3 || -0.5 || -100.339|| No || [[File:Mlw115_Surface_Plot_HHF_-3_0.5.png|300px]] || H2 approaches the fluorine atom slowly but when a collision is made the H rebounds straight back into the other H and they continue their path oscillating together in the opposite direction
|-

| 3 || -0.5 || -93.254 || No || [[File:Mlw115_Surface_Plot_HHF_3_0.5.png|300px]] || This is similar to the previous case however the hydrogen rebounds twice into flourine but on the second rebounds remains returns back to oscillating with the original H atom
|-
| 2 || -0.5|| -98.753|| No || [[File: Mlw115_Surface_Plot_HHF_2_0.5.png|300px]] || The HF molecule oscillates within the energy well at high vibrational energy after collision
|-
| 1.5|| -0.5|| 103.754|| No || [[File: Mlw115_Surface_Plot_HHF_1.5_0.5.png|300px]] || Similarly to the previous situation
|-
|0.1||-0.8|| 103.364|| Yes|| [[File: Mlw115_Surface_Plot_HHF_0.1_0.8.png|300px]] || Despite having lower total energy than the previous cases the reaction goes to completion as the momentum between the reacting atoms is sufficient
|}

Despite the H2 energy being higher than the activation energy if the BC atoms do not collide with a high enough energy the reaction will return back over the peak and the original reactants will be produced.

=== HF+F ===

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 2
|-
|BC || 0.74
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -5 || 0.2 || -103.364 || No || [[File:Mlw115_Surface_Plot_HFH_5_0.2.png|300px]] || Approaching H rebounds and reaction doesn't complete

|}

A reactive pathway was found by trial and error around the reversed conditions from the reverse reaction.

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 5.0||-1.5
|-
| BC|| 1.2||1.0
|}

[[File: Mlw115_Surface_Plot_HFH_-1.5_1.png| 400px]]

==Polanyi's Empirical Rules ==

Question 10. Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.

Polanyi's rules state that in reactions with an early transition state (most closely resembling the reactants) translational energy most effectively promotes a successful reaction. Conversely where the transition state is late (resembling products closely) vibrational energy, specifically of the reactants most effectively promotes a successful reaction. <ref>The Journal of Chemical Physics 138, 234104 (2013); https://doi.org/10.1063/1.4810007</ref>

This is demonstrated in the above, where

=References=
<references/>

MRD:mlw115RXNDy

2018-05-22T15:34:12Z

Mlw115: /* Transition State Theory */

= Reaction Dynamics =

= Exercise 1: H2 + H system=

==Dynamics from the transition state region ==

 Question 1. What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

Where a reaction is modeled on a potential energy surface the lowest energy path between the reactants and products is seen as a "valley floor" of the surface. The transition structure is a saddle point along the minimum energy ridge.

The gradient of the potential energy surface is zero at the minimum, along the direction of the reaction path and so the first derivative with respect to the inter-nuclear distance is zero. For the saddle point the gradient of the potential energy surface is zero orthogonal to the direction of the reaction pathway.

δV(ri)/δri = 0

Second derivatives
δV2(ri)/δri2 > 0 Minima
δV2(ri)/δri2 < 0 Maxima

where V is the potential energy and r related to the nuclear positions.

The minimum pathway and saddle point can be distinguished using the second derivatives with respect to atom positions. The second derivative of a maximum point is negative, whereas the second derivative of a minimum is positive. As the saddle point is a maxima along the minimum path the second derivative is negative. The second derivative of the minimum energy path in the direction of the reaction will be positive.

==Estimating Transition State Position ==

 Question 2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. 

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.908||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot 1 !! Surface Plot 2!! Internuclear distance vs Time
|-
|[[File:Mlw115_H2H_saddle.png|300px]]||[[File:mlw115_H2H_TS.png|300px]]||[[File:Mlw115_H2H_TS_INDvT.png|300px]]|
|}

The best guess of inter nuclear distances at the transition state was found using trial and error, to find the positions that would result in the minimum trajectory. As the system is symmetric the distances would be equal. The two surface plots are included to demonstrate the "reaction path". The internuclear distance vs time plot demonstrates no change in any bond lengths at the transition state so the overall momenta is 0 and potential energy is at the maximum. This agrees with that the transition state is a stationary point.

== Reaction Path ==
===MEP vs Dynamics===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.909||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot (MEP) !! Surface Plot (Dynamics
|-
|[[File:Mlw115_H2H_MEP_surfaceplot.png|300px]]||[[File:mlw115_H2H_MEP_surfaceplot_dynamics.png|300px]]
|}

 Question 3. Comment on how the mep and the trajectory you just calculated differ.

It can be seen in the MEP plot that the trajectory follows the minimum energy valley floor without any obvious deviations or oscillation. This does not account for the inter-molecular vibrations between the molecules,. The MEP trajectory is corresponding to infinitely slow motion, it does not accurately represent the atomic motion of the reaction. The trajectory oscillates in the dynamics plot, also in much less steps the AB bond distance increases considerably more. This plot is more realistic as it accounts for the motion of atoms during the reaction.

{| class="wikitable" border="1"
|+ Final Geometry
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| 2.486||1.167
|}

=== Reversed conditions ===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| -2.486||-1.167
|}

[[File:mlw115_H2H_dynamics_reversed_momenta.png|300px]]

From taking the previous runs final conditions and setting these as the initial conditions and reversing the momenta values the reaction seems to run in reverse, reaching an end exactly at the transition state.

===Determining Reactivity===

Question 4. Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2.0
|}

{| class="wikitable" border=1
! pAB !! pBC!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -1.25 || -2.5 || -99.018|| Yes || [[File:Mlw115_Surface_Plot_1.25_2.5.png|300px]] || Initially the HAHB distance changes very little as the vibrational energy is quite low, as the HC approaches it can be seen that the BC distance decreases linearly. The molecules pass over the transition state, after this point the HAHB distance decreases and the HBHC distance oscillates slightly, with a higher vibrational energy than the HAHB molecule. This overall process represents the HAHB bond breaking and forming a HBHC molecule.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mlw115_Surface_Plot_1.5_2.0.png|300px]] || The molecule approaches the maxima with some vibrational energy, shown by oscillations in the initial path, the energy of the system is not high enough to pass over the transition state energy maxima so the HC "rebounds" and the HBHC distance begins to increase. The HAHB molecule oscillations increase.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mlw115_Surface_Plot_1.5_2.5.png|300px]] || The reaction proceeds similarly to the first case with HC approaching and forming a bond with HB as there is sufficient energy to proceed over the transition state maxima. The vibrational energy throughout is slightly higher than the first case, demonstrated by higher oscillations in atom distances.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mlw115_Surface_Plot_2.5_5.png|300px]] || In this case the energy is high enough and the atoms are able to pass over the transition state energy maxima, however the vibrational energy of the HBHC system formed seems to be too high to form the molecule and the HC atom rebounds and the distance begins to decrease and so the HAHB is reformed but with considerably higher oscillations.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mlw115_Surface_Plot_2.5_5.2.png|300px]] || The HC approaches the oscillating HAHB system, the HB atom rebounds between the two atoms, as it can be seen on the surface plot the system passes over the transition state maxima, back over to the other side and back again as this occurs. Finally the HBHC bond is formed with a high vibrationally energy
|}

===Transition State Theory===

 Question 5 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? 

Transition state theory is used to understand how elementary chemical reactions take place assuming a quasi-equilibirum between reactanting complexes and those at the transition state. The rate of reaction can be understood by inspecting the complexes at the maximum saddle point of the minimum energy pathway. The reactants are in equilibrium with the activated transition state complexes and can be converted to the products.

This theory is useful to theoretically understand a chemical reaction however it operates on a series of assumptions which limit the theory. For instance the atoms are assumed to behave classically, ignoring tunelling. Tunneling would be more prominent in reactions with low activation barriers. Also the theory assumes that the saddle point with the lowest energy will be passed over to reach the products, however this is not the case when higher vibrational energies will be populated.

<ref>https://goldbook.iupac.org/html/T/T06470.html</ref> <ref>https://pubs.acs.org/doi/pdf/10.1021/j100238a002</ref>

=Exercise 2: H-H-F System=

==PES Inspection==
Question 6. 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?

For the forward reaction of H2 + F --> HF + H is exothermic, as shown in the surface plot below the energy of the reactants H2 is higher than the energy of the product HF. As the energy absorbed in breaking the H-H bond in the first case is less than the energy released forming the H-F bond which means that in total energy is released which results in an exothermic reaction.

{| class="wikitable" border=1
! Surface Plot H2+F->HF+H !! Surface Plot HF+H -> H2+F
|-
|[[File:mlw115_Surface_Plot_HHF_labelled.png|300px]]||[[File:mlw115_Surface_Plot_HHF_H2Prod.png|300px]]|
|}

==Determining Transition State ==

Question 7. Locate the approximate position of the transition state.

A = HA B = HB and C = F

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! Surface Plot !! Internuclear distance vs Time
|-
|[[File:Mlw115_Surface_Plot_TS_HHF.png|400px]]||[[File:Mlw115 INDvT HHF TS.png|350px]]|
|}

== Calculating Activation Energy ==
Question 8. Report the activation energy for both reactions.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! BC= 1.8135!! BC= 1.8135 Zoomed!! BC= 1.813
|-
|[[File:Mlw115_HHF_EvT_1.8135.png|300px]] ‎||[[File:Mlw115_Plot_1.813_EvT_zoomT0_Resized.png|300px]]||[[File:Mlw115_HHF_EvT_1.813.png|320px]]
|}

pAB = pBC =0

distAB = 0.74

{| class="wikitable" border=1
|+Activation energy HF+H->H2+F
|-
! !! Energy (Kcal/mol)
|-
|Initial energy || -103.752
|-
|Final energy|| -133.89
|-
|EA|| 30.138
|}

{| class="wikitable" border=1
|+Activation energy H2+F->HF+H
|-
! !! Energy (Kcal/mol)
|-
|Initial energy|| -103.742
|-
|Final energy|| -103.752
|-
|EA|| 0.01
|}

==Reaction Dynamics==
=== F+ H2 ===
{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.74||0
|-
| BC|| 2.3||-2.7
|}

{| class="wikitable" border=1
!Surface Plot!! Internuclear momentum vs time
|-
|[[File:Mlw_Surface_Plot_HHF_Reactive.png]]|| [[File:Mlw_INMvT_HHF_Reactive.png]]
|}

{| class="wikitable" border=1
!Energy!!
|-
|Kinetic|| 3.837
|-
|Potential|| -103.886

|-
|Total|| -100.049
|}

Question 9. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?

It can be seen from the animation that as the H-H (AB) bond is broken and the H-F (BC) bond is formed the vibrational energy (and so the kinetic energy) considerably increases for H-F. The internuclear momentum vs time graph demonstrates this, as the momentum is much higher for the H-F atom. This vibrational energy will be released as heat so experimentally an increase in temperature could confirm the reaction had occurred.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -3 || -0.5 || -100.339|| No || [[File:Mlw115_Surface_Plot_HHF_-3_0.5.png|300px]] || H2 approaches the fluorine atom slowly but when a collision is made the H rebounds straight back into the other H and they continue their path oscillating together in the opposite direction
|-

| 3 || -0.5 || -93.254 || No || [[File:Mlw115_Surface_Plot_HHF_3_0.5.png|300px]] || This is similar to the previous case however the hydrogen rebounds twice into flourine but on the second rebounds remains returns back to oscillating with the original H atom
|-
| 2 || -0.5|| -98.753|| No || [[File: Mlw115_Surface_Plot_HHF_2_0.5.png|300px]] || The HF molecule oscillates within the energy well at high vibrational energy after collision
|-
| 1.5|| -0.5|| 103.754|| No || [[File: Mlw115_Surface_Plot_HHF_1.5_0.5.png|300px]] || Similarly to the previous situation
|-
|0.1||-0.8|| 103.364|| Yes|| [[File: Mlw115_Surface_Plot_HHF_0.1_0.8.png|300px]] || Despite having lower total energy than the previous cases the reaction goes to completion as the momentum between the reacting atoms is sufficient
|}

Despite the H2 energy being higher than the activation energy if the BC atoms do not collide with a high enough energy the reaction will return back over the peak and the original reactants will be produced.

=== HF+F ===

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 2
|-
|BC || 0.74
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -5 || 0.2 || -103.364 || No || [[File:Mlw115_Surface_Plot_HFH_5_0.2.png|300px]] || Approaching H rebounds and reaction doesn't complete

|}

A reactive pathway was found by trial and error around the reversed conditions from the reverse reaction.

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 5.0||-1.5
|-
| BC|| 1.2||1.0
|}

[[File: Mlw115_Surface_Plot_HFH_-1.5_1.png| 400px]]

==Polanyi's Empirical Rules ==

Question 10. Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.

Polanyi's rules state that in reactions with an early transition state (most closely resembling the reactants) translational energy most effectively promotes a successful reaction. Conversely where the transition state is late (resembling products closely) vibrational energy, specifically of the reactants most effectively promotes a successful reaction. <ref>The Journal of Chemical Physics 138, 234104 (2013); https://doi.org/10.1063/1.4810007</ref>

=References=
<references/>

MRD:mlw115RXNDy

2018-05-22T15:26:57Z

Mlw115: /* Estimating Transition State Position */

= Reaction Dynamics =

= Exercise 1: H2 + H system=

==Dynamics from the transition state region ==

 Question 1. What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

Where a reaction is modeled on a potential energy surface the lowest energy path between the reactants and products is seen as a "valley floor" of the surface. The transition structure is a saddle point along the minimum energy ridge.

The gradient of the potential energy surface is zero at the minimum, along the direction of the reaction path and so the first derivative with respect to the inter-nuclear distance is zero. For the saddle point the gradient of the potential energy surface is zero orthogonal to the direction of the reaction pathway.

δV(ri)/δri = 0

Second derivatives
δV2(ri)/δri2 > 0 Minima
δV2(ri)/δri2 < 0 Maxima

where V is the potential energy and r related to the nuclear positions.

The minimum pathway and saddle point can be distinguished using the second derivatives with respect to atom positions. The second derivative of a maximum point is negative, whereas the second derivative of a minimum is positive. As the saddle point is a maxima along the minimum path the second derivative is negative. The second derivative of the minimum energy path in the direction of the reaction will be positive.

==Estimating Transition State Position ==

 Question 2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. 

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.908||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot 1 !! Surface Plot 2!! Internuclear distance vs Time
|-
|[[File:Mlw115_H2H_saddle.png|300px]]||[[File:mlw115_H2H_TS.png|300px]]||[[File:Mlw115_H2H_TS_INDvT.png|300px]]|
|}

The best guess of inter nuclear distances at the transition state was found using trial and error, to find the positions that would result in the minimum trajectory. As the system is symmetric the distances would be equal. The two surface plots are included to demonstrate the "reaction path". The internuclear distance vs time plot demonstrates no change in any bond lengths at the transition state so the overall momenta is 0 and potential energy is at the maximum. This agrees with that the transition state is a stationary point.

== Reaction Path ==
===MEP vs Dynamics===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.909||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot (MEP) !! Surface Plot (Dynamics
|-
|[[File:Mlw115_H2H_MEP_surfaceplot.png|300px]]||[[File:mlw115_H2H_MEP_surfaceplot_dynamics.png|300px]]
|}

 Question 3. Comment on how the mep and the trajectory you just calculated differ.

It can be seen in the MEP plot that the trajectory follows the minimum energy valley floor without any obvious deviations or oscillation. This does not account for the inter-molecular vibrations between the molecules,. The MEP trajectory is corresponding to infinitely slow motion, it does not accurately represent the atomic motion of the reaction. The trajectory oscillates in the dynamics plot, also in much less steps the AB bond distance increases considerably more. This plot is more realistic as it accounts for the motion of atoms during the reaction.

{| class="wikitable" border="1"
|+ Final Geometry
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| 2.486||1.167
|}

=== Reversed conditions ===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| -2.486||-1.167
|}

[[File:mlw115_H2H_dynamics_reversed_momenta.png|300px]]

From taking the previous runs final conditions and setting these as the initial conditions and reversing the momenta values the reaction seems to run in reverse, reaching an end exactly at the transition state.

===Determining Reactivity===

Question 4. Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2.0
|}

{| class="wikitable" border=1
! pAB !! pBC!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -1.25 || -2.5 || -99.018|| Yes || [[File:Mlw115_Surface_Plot_1.25_2.5.png|300px]] || Initially the HAHB distance changes very little as the vibrational energy is quite low, as the HC approaches it can be seen that the BC distance decreases linearly. The molecules pass over the transition state, after this point the HAHB distance decreases and the HBHC distance oscillates slightly, with a higher vibrational energy than the HAHB molecule. This overall process represents the HAHB bond breaking and forming a HBHC molecule.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mlw115_Surface_Plot_1.5_2.0.png|300px]] || The molecule approaches the maxima with some vibrational energy, shown by oscillations in the initial path, the energy of the system is not high enough to pass over the transition state energy maxima so the HC "rebounds" and the HBHC distance begins to increase. The HAHB molecule oscillations increase.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mlw115_Surface_Plot_1.5_2.5.png|300px]] || The reaction proceeds similarly to the first case with HC approaching and forming a bond with HB as there is sufficient energy to proceed over the transition state maxima. The vibrational energy throughout is slightly higher than the first case, demonstrated by higher oscillations in atom distances.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mlw115_Surface_Plot_2.5_5.png|300px]] || In this case the energy is high enough and the atoms are able to pass over the transition state energy maxima, however the vibrational energy of the HBHC system formed seems to be too high to form the molecule and the HC atom rebounds and the distance begins to decrease and so the HAHB is reformed but with considerably higher oscillations.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mlw115_Surface_Plot_2.5_5.2.png|300px]] || The HC approaches the oscillating HAHB system, the HB atom rebounds between the two atoms, as it can be seen on the surface plot the system passes over the transition state maxima, back over to the other side and back again as this occurs. Finally the HBHC bond is formed with a high vibrationally energy
|}

===Transition State Theory===

 Question 5 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? 

Transition state theory is used to understand how elementary chemical reactions take place assuming a quasi-equilibirum between reactanting complexes and those at the transition state. The rate of reaction can be understood by inspecting the complexes at the maximum saddle point of the minimum energy pathway. The reactants are in equilibrium with the activated transition state complexes and can be converted to the products.

This theory is useful to theoretically understand a chemical reaction however it operates on a series of assumptions which limit the theory. For instance the atoms are assumed to behave classically, ignoring tunelling. Tunneling would be more prominent in reactions with low activation barriers. Also the theory assumes that the saddle point with the lowest energy will be passed over to reach the products, however this is not the case when higher vibrational energies will be populated.

<ref>https://goldbook.iupac.org/html/T/T06470.html</ref> <ref>https://pubs.acs.org/doi/pdf/10.1021/j100238a002</ref>

=Exercise 2: H-H-F System=

==PES Inspection==
Question 6. 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?

For the forward reaction of H2 + F --> HF + H is exothermic, as shown in the surface plot below the energy of the reactants H2 is higher than the energy of the product HF. As the energy absorbed in breaking the H-H bond in the first case is less than the energy released forming the H-F bond which means that in total energy is released which results in an exothermic reaction.

{| class="wikitable" border=1
! Surface Plot H2+F->HF+H !! Surface Plot HF+H -> H2+F
|-
|[[File:mlw115_Surface_Plot_HHF_labelled.png|300px]]||[[File:mlw115_Surface_Plot_HHF_H2Prod.png|300px]]|
|}

==Determining Transition State ==

Question 7. Locate the approximate position of the transition state.

A = HA B = HB and C = F

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! Surface Plot !! Internuclear distance vs Time
|-
|[[File:Mlw115_Surface_Plot_TS_HHF.png|400px]]||[[File:Mlw115 INDvT HHF TS.png|350px]]|
|}

== Calculating Activation Energy ==
Question 8. Report the activation energy for both reactions.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! BC= 1.8135!! BC= 1.8135 Zoomed!! BC= 1.813
|-
|[[File:Mlw115_HHF_EvT_1.8135.png|300px]] ‎||[[File:Mlw115_Plot_1.813_EvT_zoomT0_Resized.png|300px]]||[[File:Mlw115_HHF_EvT_1.813.png|320px]]
|}

pAB = pBC =0

distAB = 0.74

{| class="wikitable" border=1
|+Activation energy HF+H->H2+F
|-
! !! Energy (Kcal/mol)
|-
|Initial energy || -103.752
|-
|Final energy|| -133.89
|-
|EA|| 30.138
|}

{| class="wikitable" border=1
|+Activation energy H2+F->HF+H
|-
! !! Energy (Kcal/mol)
|-
|Initial energy|| -103.742
|-
|Final energy|| -103.752
|-
|EA|| 0.01
|}

==Reaction Dynamics==
=== F+ H2 ===
{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.74||0
|-
| BC|| 2.3||-2.7
|}

{| class="wikitable" border=1
!Surface Plot!! Internuclear momentum vs time
|-
|[[File:Mlw_Surface_Plot_HHF_Reactive.png]]|| [[File:Mlw_INMvT_HHF_Reactive.png]]
|}

{| class="wikitable" border=1
!Energy!!
|-
|Kinetic|| 3.837
|-
|Potential|| -103.886

|-
|Total|| -100.049
|}

Question 9. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?

It can be seen from the animation that as the H-H (AB) bond is broken and the H-F (BC) bond is formed the vibrational energy (and so the kinetic energy) considerably increases for H-F. The internuclear momentum vs time graph demonstrates this, as the momentum is much higher for the H-F atom. This vibrational energy will be released as heat so experimentally an increase in temperature could confirm the reaction had occurred.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -3 || -0.5 || -100.339|| No || [[File:Mlw115_Surface_Plot_HHF_-3_0.5.png|300px]] || H2 approaches the fluorine atom slowly but when a collision is made the H rebounds straight back into the other H and they continue their path oscillating together in the opposite direction
|-

| 3 || -0.5 || -93.254 || No || [[File:Mlw115_Surface_Plot_HHF_3_0.5.png|300px]] || This is similar to the previous case however the hydrogen rebounds twice into flourine but on the second rebounds remains returns back to oscillating with the original H atom
|-
| 2 || -0.5|| -98.753|| No || [[File: Mlw115_Surface_Plot_HHF_2_0.5.png|300px]] || The HF molecule oscillates within the energy well at high vibrational energy after collision
|-
| 1.5|| -0.5|| 103.754|| No || [[File: Mlw115_Surface_Plot_HHF_1.5_0.5.png|300px]] || Similarly to the previous situation
|-
|0.1||-0.8|| 103.364|| Yes|| [[File: Mlw115_Surface_Plot_HHF_0.1_0.8.png|300px]] || Despite having lower total energy than the previous cases the reaction goes to completion as the momentum between the reacting atoms is sufficient
|}

Despite the H2 energy being higher than the activation energy if the BC atoms do not collide with a high enough energy the reaction will return back over the peak and the original reactants will be produced.

=== HF+F ===

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 2
|-
|BC || 0.74
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -5 || 0.2 || -103.364 || No || [[File:Mlw115_Surface_Plot_HFH_5_0.2.png|300px]] || Approaching H rebounds and reaction doesn't complete

|}

A reactive pathway was found by trial and error around the reversed conditions from the reverse reaction.

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 5.0||-1.5
|-
| BC|| 1.2||1.0
|}

[[File: Mlw115_Surface_Plot_HFH_-1.5_1.png| 400px]]

==Polanyi's Empirical Rules ==

Question 10. Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.

Polanyi's rules state that in reactions with an early transition state (most closely resembling the reactants) translational energy most effectively promotes a successful reaction. Conversely where the transition state is late (resembling products closely) vibrational energy, specifically of the reactants most effectively promotes a successful reaction. <ref>The Journal of Chemical Physics 138, 234104 (2013); https://doi.org/10.1063/1.4810007</ref>

=References=
<references/>

MRD:mlw115RXNDy

2018-05-22T15:23:53Z

Mlw115: /* Introduction */

= Reaction Dynamics =

= Exercise 1: H2 + H system=

==Dynamics from the transition state region ==

 Question 1. What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

Where a reaction is modeled on a potential energy surface the lowest energy path between the reactants and products is seen as a "valley floor" of the surface. The transition structure is a saddle point along the minimum energy ridge.

The gradient of the potential energy surface is zero at the minimum, along the direction of the reaction path and so the first derivative with respect to the inter-nuclear distance is zero. For the saddle point the gradient of the potential energy surface is zero orthogonal to the direction of the reaction pathway.

δV(ri)/δri = 0

Second derivatives
δV2(ri)/δri2 > 0 Minima
δV2(ri)/δri2 < 0 Maxima

where V is the potential energy and r related to the nuclear positions.

The minimum pathway and saddle point can be distinguished using the second derivatives with respect to atom positions. The second derivative of a maximum point is negative, whereas the second derivative of a minimum is positive. As the saddle point is a maxima along the minimum path the second derivative is negative. The second derivative of the minimum energy path in the direction of the reaction will be positive.

==Estimating Transition State Position ==

 Question 2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. 

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.908||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot 1 !! Surface Plot 2!! Internuclear distance vs Time
|-
|[[File:Mlw115_H2H_saddle.png|300px]]||[[File:mlw115_H2H_TS.png|300px]]||[[File:Mlw115_H2H_TS_INDvT.png|300px]]|
|}

The best guess of inter nuclear distances at the transition state was found using trial and error, to find the positions that would result in the minimum trajectory. As the system is symmetric the distances would be equal. The two surface plots are included to demonstrate the "reaction path". The internuclear distance vs time plot demonstrates no change in any bond lengths at the transition state so the overall momenta is 0 and potential energy is at the maximum.

== Reaction Path ==
===MEP vs Dynamics===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.909||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot (MEP) !! Surface Plot (Dynamics
|-
|[[File:Mlw115_H2H_MEP_surfaceplot.png|300px]]||[[File:mlw115_H2H_MEP_surfaceplot_dynamics.png|300px]]
|}

 Question 3. Comment on how the mep and the trajectory you just calculated differ.

It can be seen in the MEP plot that the trajectory follows the minimum energy valley floor without any obvious deviations or oscillation. This does not account for the inter-molecular vibrations between the molecules,. The MEP trajectory is corresponding to infinitely slow motion, it does not accurately represent the atomic motion of the reaction. The trajectory oscillates in the dynamics plot, also in much less steps the AB bond distance increases considerably more. This plot is more realistic as it accounts for the motion of atoms during the reaction.

{| class="wikitable" border="1"
|+ Final Geometry
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| 2.486||1.167
|}

=== Reversed conditions ===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| -2.486||-1.167
|}

[[File:mlw115_H2H_dynamics_reversed_momenta.png|300px]]

From taking the previous runs final conditions and setting these as the initial conditions and reversing the momenta values the reaction seems to run in reverse, reaching an end exactly at the transition state.

===Determining Reactivity===

Question 4. Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2.0
|}

{| class="wikitable" border=1
! pAB !! pBC!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -1.25 || -2.5 || -99.018|| Yes || [[File:Mlw115_Surface_Plot_1.25_2.5.png|300px]] || Initially the HAHB distance changes very little as the vibrational energy is quite low, as the HC approaches it can be seen that the BC distance decreases linearly. The molecules pass over the transition state, after this point the HAHB distance decreases and the HBHC distance oscillates slightly, with a higher vibrational energy than the HAHB molecule. This overall process represents the HAHB bond breaking and forming a HBHC molecule.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mlw115_Surface_Plot_1.5_2.0.png|300px]] || The molecule approaches the maxima with some vibrational energy, shown by oscillations in the initial path, the energy of the system is not high enough to pass over the transition state energy maxima so the HC "rebounds" and the HBHC distance begins to increase. The HAHB molecule oscillations increase.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mlw115_Surface_Plot_1.5_2.5.png|300px]] || The reaction proceeds similarly to the first case with HC approaching and forming a bond with HB as there is sufficient energy to proceed over the transition state maxima. The vibrational energy throughout is slightly higher than the first case, demonstrated by higher oscillations in atom distances.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mlw115_Surface_Plot_2.5_5.png|300px]] || In this case the energy is high enough and the atoms are able to pass over the transition state energy maxima, however the vibrational energy of the HBHC system formed seems to be too high to form the molecule and the HC atom rebounds and the distance begins to decrease and so the HAHB is reformed but with considerably higher oscillations.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mlw115_Surface_Plot_2.5_5.2.png|300px]] || The HC approaches the oscillating HAHB system, the HB atom rebounds between the two atoms, as it can be seen on the surface plot the system passes over the transition state maxima, back over to the other side and back again as this occurs. Finally the HBHC bond is formed with a high vibrationally energy
|}

===Transition State Theory===

 Question 5 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? 

Transition state theory is used to understand how elementary chemical reactions take place assuming a quasi-equilibirum between reactanting complexes and those at the transition state. The rate of reaction can be understood by inspecting the complexes at the maximum saddle point of the minimum energy pathway. The reactants are in equilibrium with the activated transition state complexes and can be converted to the products.

This theory is useful to theoretically understand a chemical reaction however it operates on a series of assumptions which limit the theory. For instance the atoms are assumed to behave classically, ignoring tunelling. Tunneling would be more prominent in reactions with low activation barriers. Also the theory assumes that the saddle point with the lowest energy will be passed over to reach the products, however this is not the case when higher vibrational energies will be populated.

<ref>https://goldbook.iupac.org/html/T/T06470.html</ref> <ref>https://pubs.acs.org/doi/pdf/10.1021/j100238a002</ref>

=Exercise 2: H-H-F System=

==PES Inspection==
Question 6. 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?

For the forward reaction of H2 + F --> HF + H is exothermic, as shown in the surface plot below the energy of the reactants H2 is higher than the energy of the product HF. As the energy absorbed in breaking the H-H bond in the first case is less than the energy released forming the H-F bond which means that in total energy is released which results in an exothermic reaction.

{| class="wikitable" border=1
! Surface Plot H2+F->HF+H !! Surface Plot HF+H -> H2+F
|-
|[[File:mlw115_Surface_Plot_HHF_labelled.png|300px]]||[[File:mlw115_Surface_Plot_HHF_H2Prod.png|300px]]|
|}

==Determining Transition State ==

Question 7. Locate the approximate position of the transition state.

A = HA B = HB and C = F

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! Surface Plot !! Internuclear distance vs Time
|-
|[[File:Mlw115_Surface_Plot_TS_HHF.png|400px]]||[[File:Mlw115 INDvT HHF TS.png|350px]]|
|}

== Calculating Activation Energy ==
Question 8. Report the activation energy for both reactions.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! BC= 1.8135!! BC= 1.8135 Zoomed!! BC= 1.813
|-
|[[File:Mlw115_HHF_EvT_1.8135.png|300px]] ‎||[[File:Mlw115_Plot_1.813_EvT_zoomT0_Resized.png|300px]]||[[File:Mlw115_HHF_EvT_1.813.png|320px]]
|}

pAB = pBC =0

distAB = 0.74

{| class="wikitable" border=1
|+Activation energy HF+H->H2+F
|-
! !! Energy (Kcal/mol)
|-
|Initial energy || -103.752
|-
|Final energy|| -133.89
|-
|EA|| 30.138
|}

{| class="wikitable" border=1
|+Activation energy H2+F->HF+H
|-
! !! Energy (Kcal/mol)
|-
|Initial energy|| -103.742
|-
|Final energy|| -103.752
|-
|EA|| 0.01
|}

==Reaction Dynamics==
=== F+ H2 ===
{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.74||0
|-
| BC|| 2.3||-2.7
|}

{| class="wikitable" border=1
!Surface Plot!! Internuclear momentum vs time
|-
|[[File:Mlw_Surface_Plot_HHF_Reactive.png]]|| [[File:Mlw_INMvT_HHF_Reactive.png]]
|}

{| class="wikitable" border=1
!Energy!!
|-
|Kinetic|| 3.837
|-
|Potential|| -103.886

|-
|Total|| -100.049
|}

Question 9. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?

It can be seen from the animation that as the H-H (AB) bond is broken and the H-F (BC) bond is formed the vibrational energy (and so the kinetic energy) considerably increases for H-F. The internuclear momentum vs time graph demonstrates this, as the momentum is much higher for the H-F atom. This vibrational energy will be released as heat so experimentally an increase in temperature could confirm the reaction had occurred.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -3 || -0.5 || -100.339|| No || [[File:Mlw115_Surface_Plot_HHF_-3_0.5.png|300px]] || H2 approaches the fluorine atom slowly but when a collision is made the H rebounds straight back into the other H and they continue their path oscillating together in the opposite direction
|-

| 3 || -0.5 || -93.254 || No || [[File:Mlw115_Surface_Plot_HHF_3_0.5.png|300px]] || This is similar to the previous case however the hydrogen rebounds twice into flourine but on the second rebounds remains returns back to oscillating with the original H atom
|-
| 2 || -0.5|| -98.753|| No || [[File: Mlw115_Surface_Plot_HHF_2_0.5.png|300px]] || The HF molecule oscillates within the energy well at high vibrational energy after collision
|-
| 1.5|| -0.5|| 103.754|| No || [[File: Mlw115_Surface_Plot_HHF_1.5_0.5.png|300px]] || Similarly to the previous situation
|-
|0.1||-0.8|| 103.364|| Yes|| [[File: Mlw115_Surface_Plot_HHF_0.1_0.8.png|300px]] || Despite having lower total energy than the previous cases the reaction goes to completion as the momentum between the reacting atoms is sufficient
|}

Despite the H2 energy being higher than the activation energy if the BC atoms do not collide with a high enough energy the reaction will return back over the peak and the original reactants will be produced.

=== HF+F ===

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 2
|-
|BC || 0.74
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -5 || 0.2 || -103.364 || No || [[File:Mlw115_Surface_Plot_HFH_5_0.2.png|300px]] || Approaching H rebounds and reaction doesn't complete

|}

A reactive pathway was found by trial and error around the reversed conditions from the reverse reaction.

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 5.0||-1.5
|-
| BC|| 1.2||1.0
|}

[[File: Mlw115_Surface_Plot_HFH_-1.5_1.png| 400px]]

==Polanyi's Empirical Rules ==

Question 10. Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.

Polanyi's rules state that in reactions with an early transition state (most closely resembling the reactants) translational energy most effectively promotes a successful reaction. Conversely where the transition state is late (resembling products closely) vibrational energy, specifically of the reactants most effectively promotes a successful reaction. <ref>The Journal of Chemical Physics 138, 234104 (2013); https://doi.org/10.1063/1.4810007</ref>

=References=
<references/>

MRD:mlw115RXNDy

2018-05-22T15:20:02Z

Mlw115: /* Transition State Theory */

= Reaction Dynamics =

==Introduction==

The reactivity of atom-diatom systems are studied in order to gain insight on the internal degrees of freedom of the system. Particularly vibrational and translational energy are considered with respect to overcoming the transition state energy barrier.

The atoms are considered to behave classically and the interactions are expressed as potential energy surfaces.

= Exercise 1: H2 + H system=

==Dynamics from the transition state region ==

 Question 1. What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

Where a reaction is modeled on a potential energy surface the lowest energy path between the reactants and products is seen as a "valley floor" of the surface. The transition structure is a saddle point along the minimum energy ridge.

The gradient of the potential energy surface is zero at the minimum, along the direction of the reaction path and so the first derivative with respect to the inter-nuclear distance is zero. For the saddle point the gradient of the potential energy surface is zero orthogonal to the direction of the reaction pathway.

δV(ri)/δri = 0

Second derivatives
δV2(ri)/δri2 > 0 Minima
δV2(ri)/δri2 < 0 Maxima

where V is the potential energy and r related to the nuclear positions.

The minimum pathway and saddle point can be distinguished using the second derivatives with respect to atom positions. The second derivative of a maximum point is negative, whereas the second derivative of a minimum is positive. As the saddle point is a maxima along the minimum path the second derivative is negative. The second derivative of the minimum energy path in the direction of the reaction will be positive.

==Estimating Transition State Position ==

 Question 2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. 

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.908||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot 1 !! Surface Plot 2!! Internuclear distance vs Time
|-
|[[File:Mlw115_H2H_saddle.png|300px]]||[[File:mlw115_H2H_TS.png|300px]]||[[File:Mlw115_H2H_TS_INDvT.png|300px]]|
|}

The best guess of inter nuclear distances at the transition state was found using trial and error, to find the positions that would result in the minimum trajectory. As the system is symmetric the distances would be equal. The two surface plots are included to demonstrate the "reaction path". The internuclear distance vs time plot demonstrates no change in any bond lengths at the transition state so the overall momenta is 0 and potential energy is at the maximum.

== Reaction Path ==
===MEP vs Dynamics===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.909||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot (MEP) !! Surface Plot (Dynamics
|-
|[[File:Mlw115_H2H_MEP_surfaceplot.png|300px]]||[[File:mlw115_H2H_MEP_surfaceplot_dynamics.png|300px]]
|}

 Question 3. Comment on how the mep and the trajectory you just calculated differ.

It can be seen in the MEP plot that the trajectory follows the minimum energy valley floor without any obvious deviations or oscillation. This does not account for the inter-molecular vibrations between the molecules,. The MEP trajectory is corresponding to infinitely slow motion, it does not accurately represent the atomic motion of the reaction. The trajectory oscillates in the dynamics plot, also in much less steps the AB bond distance increases considerably more. This plot is more realistic as it accounts for the motion of atoms during the reaction.

{| class="wikitable" border="1"
|+ Final Geometry
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| 2.486||1.167
|}

=== Reversed conditions ===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| -2.486||-1.167
|}

[[File:mlw115_H2H_dynamics_reversed_momenta.png|300px]]

From taking the previous runs final conditions and setting these as the initial conditions and reversing the momenta values the reaction seems to run in reverse, reaching an end exactly at the transition state.

===Determining Reactivity===

Question 4. Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2.0
|}

{| class="wikitable" border=1
! pAB !! pBC!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -1.25 || -2.5 || -99.018|| Yes || [[File:Mlw115_Surface_Plot_1.25_2.5.png|300px]] || Initially the HAHB distance changes very little as the vibrational energy is quite low, as the HC approaches it can be seen that the BC distance decreases linearly. The molecules pass over the transition state, after this point the HAHB distance decreases and the HBHC distance oscillates slightly, with a higher vibrational energy than the HAHB molecule. This overall process represents the HAHB bond breaking and forming a HBHC molecule.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mlw115_Surface_Plot_1.5_2.0.png|300px]] || The molecule approaches the maxima with some vibrational energy, shown by oscillations in the initial path, the energy of the system is not high enough to pass over the transition state energy maxima so the HC "rebounds" and the HBHC distance begins to increase. The HAHB molecule oscillations increase.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mlw115_Surface_Plot_1.5_2.5.png|300px]] || The reaction proceeds similarly to the first case with HC approaching and forming a bond with HB as there is sufficient energy to proceed over the transition state maxima. The vibrational energy throughout is slightly higher than the first case, demonstrated by higher oscillations in atom distances.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mlw115_Surface_Plot_2.5_5.png|300px]] || In this case the energy is high enough and the atoms are able to pass over the transition state energy maxima, however the vibrational energy of the HBHC system formed seems to be too high to form the molecule and the HC atom rebounds and the distance begins to decrease and so the HAHB is reformed but with considerably higher oscillations.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mlw115_Surface_Plot_2.5_5.2.png|300px]] || The HC approaches the oscillating HAHB system, the HB atom rebounds between the two atoms, as it can be seen on the surface plot the system passes over the transition state maxima, back over to the other side and back again as this occurs. Finally the HBHC bond is formed with a high vibrationally energy
|}

===Transition State Theory===

 Question 5 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? 

Transition state theory is used to understand how elementary chemical reactions take place assuming a quasi-equilibirum between reactanting complexes and those at the transition state. The rate of reaction can be understood by inspecting the complexes at the maximum saddle point of the minimum energy pathway. The reactants are in equilibrium with the activated transition state complexes and can be converted to the products.

This theory is useful to theoretically understand a chemical reaction however it operates on a series of assumptions which limit the theory. For instance the atoms are assumed to behave classically, ignoring tunelling. Tunneling would be more prominent in reactions with low activation barriers. Also the theory assumes that the saddle point with the lowest energy will be passed over to reach the products, however this is not the case when higher vibrational energies will be populated.

<ref>https://goldbook.iupac.org/html/T/T06470.html</ref> <ref>https://pubs.acs.org/doi/pdf/10.1021/j100238a002</ref>

=Exercise 2: H-H-F System=

==PES Inspection==
Question 6. 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?

For the forward reaction of H2 + F --> HF + H is exothermic, as shown in the surface plot below the energy of the reactants H2 is higher than the energy of the product HF. As the energy absorbed in breaking the H-H bond in the first case is less than the energy released forming the H-F bond which means that in total energy is released which results in an exothermic reaction.

{| class="wikitable" border=1
! Surface Plot H2+F->HF+H !! Surface Plot HF+H -> H2+F
|-
|[[File:mlw115_Surface_Plot_HHF_labelled.png|300px]]||[[File:mlw115_Surface_Plot_HHF_H2Prod.png|300px]]|
|}

==Determining Transition State ==

Question 7. Locate the approximate position of the transition state.

A = HA B = HB and C = F

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! Surface Plot !! Internuclear distance vs Time
|-
|[[File:Mlw115_Surface_Plot_TS_HHF.png|400px]]||[[File:Mlw115 INDvT HHF TS.png|350px]]|
|}

== Calculating Activation Energy ==
Question 8. Report the activation energy for both reactions.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! BC= 1.8135!! BC= 1.8135 Zoomed!! BC= 1.813
|-
|[[File:Mlw115_HHF_EvT_1.8135.png|300px]] ‎||[[File:Mlw115_Plot_1.813_EvT_zoomT0_Resized.png|300px]]||[[File:Mlw115_HHF_EvT_1.813.png|320px]]
|}

pAB = pBC =0

distAB = 0.74

{| class="wikitable" border=1
|+Activation energy HF+H->H2+F
|-
! !! Energy (Kcal/mol)
|-
|Initial energy || -103.752
|-
|Final energy|| -133.89
|-
|EA|| 30.138
|}

{| class="wikitable" border=1
|+Activation energy H2+F->HF+H
|-
! !! Energy (Kcal/mol)
|-
|Initial energy|| -103.742
|-
|Final energy|| -103.752
|-
|EA|| 0.01
|}

==Reaction Dynamics==
=== F+ H2 ===
{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.74||0
|-
| BC|| 2.3||-2.7
|}

{| class="wikitable" border=1
!Surface Plot!! Internuclear momentum vs time
|-
|[[File:Mlw_Surface_Plot_HHF_Reactive.png]]|| [[File:Mlw_INMvT_HHF_Reactive.png]]
|}

{| class="wikitable" border=1
!Energy!!
|-
|Kinetic|| 3.837
|-
|Potential|| -103.886

|-
|Total|| -100.049
|}

Question 9. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?

It can be seen from the animation that as the H-H (AB) bond is broken and the H-F (BC) bond is formed the vibrational energy (and so the kinetic energy) considerably increases for H-F. The internuclear momentum vs time graph demonstrates this, as the momentum is much higher for the H-F atom. This vibrational energy will be released as heat so experimentally an increase in temperature could confirm the reaction had occurred.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -3 || -0.5 || -100.339|| No || [[File:Mlw115_Surface_Plot_HHF_-3_0.5.png|300px]] || H2 approaches the fluorine atom slowly but when a collision is made the H rebounds straight back into the other H and they continue their path oscillating together in the opposite direction
|-

| 3 || -0.5 || -93.254 || No || [[File:Mlw115_Surface_Plot_HHF_3_0.5.png|300px]] || This is similar to the previous case however the hydrogen rebounds twice into flourine but on the second rebounds remains returns back to oscillating with the original H atom
|-
| 2 || -0.5|| -98.753|| No || [[File: Mlw115_Surface_Plot_HHF_2_0.5.png|300px]] || The HF molecule oscillates within the energy well at high vibrational energy after collision
|-
| 1.5|| -0.5|| 103.754|| No || [[File: Mlw115_Surface_Plot_HHF_1.5_0.5.png|300px]] || Similarly to the previous situation
|-
|0.1||-0.8|| 103.364|| Yes|| [[File: Mlw115_Surface_Plot_HHF_0.1_0.8.png|300px]] || Despite having lower total energy than the previous cases the reaction goes to completion as the momentum between the reacting atoms is sufficient
|}

Despite the H2 energy being higher than the activation energy if the BC atoms do not collide with a high enough energy the reaction will return back over the peak and the original reactants will be produced.

=== HF+F ===

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 2
|-
|BC || 0.74
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -5 || 0.2 || -103.364 || No || [[File:Mlw115_Surface_Plot_HFH_5_0.2.png|300px]] || Approaching H rebounds and reaction doesn't complete

|}

A reactive pathway was found by trial and error around the reversed conditions from the reverse reaction.

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 5.0||-1.5
|-
| BC|| 1.2||1.0
|}

[[File: Mlw115_Surface_Plot_HFH_-1.5_1.png| 400px]]

==Polanyi's Empirical Rules ==

Question 10. Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.

Polanyi's rules state that in reactions with an early transition state (most closely resembling the reactants) translational energy most effectively promotes a successful reaction. Conversely where the transition state is late (resembling products closely) vibrational energy, specifically of the reactants most effectively promotes a successful reaction. <ref>The Journal of Chemical Physics 138, 234104 (2013); https://doi.org/10.1063/1.4810007</ref>

=References=
<references/>

MRD:mlw115RXNDy

2018-05-22T15:16:00Z

Mlw115: /* Reversed starting positions */

= Reaction Dynamics =

==Introduction==

The reactivity of atom-diatom systems are studied in order to gain insight on the internal degrees of freedom of the system. Particularly vibrational and translational energy are considered with respect to overcoming the transition state energy barrier.

The atoms are considered to behave classically and the interactions are expressed as potential energy surfaces.

= Exercise 1: H2 + H system=

==Dynamics from the transition state region ==

 Question 1. What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

Where a reaction is modeled on a potential energy surface the lowest energy path between the reactants and products is seen as a "valley floor" of the surface. The transition structure is a saddle point along the minimum energy ridge.

The gradient of the potential energy surface is zero at the minimum, along the direction of the reaction path and so the first derivative with respect to the inter-nuclear distance is zero. For the saddle point the gradient of the potential energy surface is zero orthogonal to the direction of the reaction pathway.

δV(ri)/δri = 0

Second derivatives
δV2(ri)/δri2 > 0 Minima
δV2(ri)/δri2 < 0 Maxima

where V is the potential energy and r related to the nuclear positions.

The minimum pathway and saddle point can be distinguished using the second derivatives with respect to atom positions. The second derivative of a maximum point is negative, whereas the second derivative of a minimum is positive. As the saddle point is a maxima along the minimum path the second derivative is negative. The second derivative of the minimum energy path in the direction of the reaction will be positive.

==Estimating Transition State Position ==

 Question 2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. 

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.908||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot 1 !! Surface Plot 2!! Internuclear distance vs Time
|-
|[[File:Mlw115_H2H_saddle.png|300px]]||[[File:mlw115_H2H_TS.png|300px]]||[[File:Mlw115_H2H_TS_INDvT.png|300px]]|
|}

The best guess of inter nuclear distances at the transition state was found using trial and error, to find the positions that would result in the minimum trajectory. As the system is symmetric the distances would be equal. The two surface plots are included to demonstrate the "reaction path". The internuclear distance vs time plot demonstrates no change in any bond lengths at the transition state so the overall momenta is 0 and potential energy is at the maximum.

== Reaction Path ==
===MEP vs Dynamics===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.909||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot (MEP) !! Surface Plot (Dynamics
|-
|[[File:Mlw115_H2H_MEP_surfaceplot.png|300px]]||[[File:mlw115_H2H_MEP_surfaceplot_dynamics.png|300px]]
|}

 Question 3. Comment on how the mep and the trajectory you just calculated differ.

It can be seen in the MEP plot that the trajectory follows the minimum energy valley floor without any obvious deviations or oscillation. This does not account for the inter-molecular vibrations between the molecules,. The MEP trajectory is corresponding to infinitely slow motion, it does not accurately represent the atomic motion of the reaction. The trajectory oscillates in the dynamics plot, also in much less steps the AB bond distance increases considerably more. This plot is more realistic as it accounts for the motion of atoms during the reaction.

{| class="wikitable" border="1"
|+ Final Geometry
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| 2.486||1.167
|}

=== Reversed conditions ===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| -2.486||-1.167
|}

[[File:mlw115_H2H_dynamics_reversed_momenta.png|300px]]

From taking the previous runs final conditions and setting these as the initial conditions and reversing the momenta values the reaction seems to run in reverse, reaching an end exactly at the transition state.

===Determining Reactivity===

Question 4. Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2.0
|}

{| class="wikitable" border=1
! pAB !! pBC!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -1.25 || -2.5 || -99.018|| Yes || [[File:Mlw115_Surface_Plot_1.25_2.5.png|300px]] || Initially the HAHB distance changes very little as the vibrational energy is quite low, as the HC approaches it can be seen that the BC distance decreases linearly. The molecules pass over the transition state, after this point the HAHB distance decreases and the HBHC distance oscillates slightly, with a higher vibrational energy than the HAHB molecule. This overall process represents the HAHB bond breaking and forming a HBHC molecule.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mlw115_Surface_Plot_1.5_2.0.png|300px]] || The molecule approaches the maxima with some vibrational energy, shown by oscillations in the initial path, the energy of the system is not high enough to pass over the transition state energy maxima so the HC "rebounds" and the HBHC distance begins to increase. The HAHB molecule oscillations increase.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mlw115_Surface_Plot_1.5_2.5.png|300px]] || The reaction proceeds similarly to the first case with HC approaching and forming a bond with HB as there is sufficient energy to proceed over the transition state maxima. The vibrational energy throughout is slightly higher than the first case, demonstrated by higher oscillations in atom distances.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mlw115_Surface_Plot_2.5_5.png|300px]] || In this case the energy is high enough and the atoms are able to pass over the transition state energy maxima, however the vibrational energy of the HBHC system formed seems to be too high to form the molecule and the HC atom rebounds and the distance begins to decrease and so the HAHB is reformed but with considerably higher oscillations.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mlw115_Surface_Plot_2.5_5.2.png|300px]] || The HC approaches the oscillating HAHB system, the HB atom rebounds between the two atoms, as it can be seen on the surface plot the system passes over the transition state maxima, back over to the other side and back again as this occurs. Finally the HBHC bond is formed with a high vibrationally energy
|}

===Transition State Theory===

 Question 5 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? 

Transition state theory is used to understand how elementary chemical reactions take place assuming a quasi-equilibirum between reactanting complexes and those at the transition state. The rate of reaction can be understood by inspecting the complexes at the maximum saddle point of the minimum energy pathway. The reactants are in equilibrium with the activated transition state complexes and can be converted to the products.

This theory is useful to theoretically understand a chemical reaction however it operates on a series of assumptions which limit the theory. For instance the atoms are assumed to behave classically, ignoring tunelling. Tunneling would be more prominent in reactions with low activation barriers. Also the theory assumes that the saddle point with the lowest energy will be passed over to reach the products, however this is not the case when higher vibrational energies will be populated.

<ref>https://goldbook.iupac.org/html/T/T06470.html</ref>

=Exercise 2: H-H-F System=

==PES Inspection==
Question 6. 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?

For the forward reaction of H2 + F --> HF + H is exothermic, as shown in the surface plot below the energy of the reactants H2 is higher than the energy of the product HF. As the energy absorbed in breaking the H-H bond in the first case is less than the energy released forming the H-F bond which means that in total energy is released which results in an exothermic reaction.

{| class="wikitable" border=1
! Surface Plot H2+F->HF+H !! Surface Plot HF+H -> H2+F
|-
|[[File:mlw115_Surface_Plot_HHF_labelled.png|300px]]||[[File:mlw115_Surface_Plot_HHF_H2Prod.png|300px]]|
|}

==Determining Transition State ==

Question 7. Locate the approximate position of the transition state.

A = HA B = HB and C = F

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! Surface Plot !! Internuclear distance vs Time
|-
|[[File:Mlw115_Surface_Plot_TS_HHF.png|400px]]||[[File:Mlw115 INDvT HHF TS.png|350px]]|
|}

== Calculating Activation Energy ==
Question 8. Report the activation energy for both reactions.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! BC= 1.8135!! BC= 1.8135 Zoomed!! BC= 1.813
|-
|[[File:Mlw115_HHF_EvT_1.8135.png|300px]] ‎||[[File:Mlw115_Plot_1.813_EvT_zoomT0_Resized.png|300px]]||[[File:Mlw115_HHF_EvT_1.813.png|320px]]
|}

pAB = pBC =0

distAB = 0.74

{| class="wikitable" border=1
|+Activation energy HF+H->H2+F
|-
! !! Energy (Kcal/mol)
|-
|Initial energy || -103.752
|-
|Final energy|| -133.89
|-
|EA|| 30.138
|}

{| class="wikitable" border=1
|+Activation energy H2+F->HF+H
|-
! !! Energy (Kcal/mol)
|-
|Initial energy|| -103.742
|-
|Final energy|| -103.752
|-
|EA|| 0.01
|}

==Reaction Dynamics==
=== F+ H2 ===
{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.74||0
|-
| BC|| 2.3||-2.7
|}

{| class="wikitable" border=1
!Surface Plot!! Internuclear momentum vs time
|-
|[[File:Mlw_Surface_Plot_HHF_Reactive.png]]|| [[File:Mlw_INMvT_HHF_Reactive.png]]
|}

{| class="wikitable" border=1
!Energy!!
|-
|Kinetic|| 3.837
|-
|Potential|| -103.886

|-
|Total|| -100.049
|}

Question 9. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?

It can be seen from the animation that as the H-H (AB) bond is broken and the H-F (BC) bond is formed the vibrational energy (and so the kinetic energy) considerably increases for H-F. The internuclear momentum vs time graph demonstrates this, as the momentum is much higher for the H-F atom. This vibrational energy will be released as heat so experimentally an increase in temperature could confirm the reaction had occurred.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -3 || -0.5 || -100.339|| No || [[File:Mlw115_Surface_Plot_HHF_-3_0.5.png|300px]] || H2 approaches the fluorine atom slowly but when a collision is made the H rebounds straight back into the other H and they continue their path oscillating together in the opposite direction
|-

| 3 || -0.5 || -93.254 || No || [[File:Mlw115_Surface_Plot_HHF_3_0.5.png|300px]] || This is similar to the previous case however the hydrogen rebounds twice into flourine but on the second rebounds remains returns back to oscillating with the original H atom
|-
| 2 || -0.5|| -98.753|| No || [[File: Mlw115_Surface_Plot_HHF_2_0.5.png|300px]] || The HF molecule oscillates within the energy well at high vibrational energy after collision
|-
| 1.5|| -0.5|| 103.754|| No || [[File: Mlw115_Surface_Plot_HHF_1.5_0.5.png|300px]] || Similarly to the previous situation
|-
|0.1||-0.8|| 103.364|| Yes|| [[File: Mlw115_Surface_Plot_HHF_0.1_0.8.png|300px]] || Despite having lower total energy than the previous cases the reaction goes to completion as the momentum between the reacting atoms is sufficient
|}

Despite the H2 energy being higher than the activation energy if the BC atoms do not collide with a high enough energy the reaction will return back over the peak and the original reactants will be produced.

=== HF+F ===

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 2
|-
|BC || 0.74
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -5 || 0.2 || -103.364 || No || [[File:Mlw115_Surface_Plot_HFH_5_0.2.png|300px]] || Approaching H rebounds and reaction doesn't complete

|}

A reactive pathway was found by trial and error around the reversed conditions from the reverse reaction.

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 5.0||-1.5
|-
| BC|| 1.2||1.0
|}

[[File: Mlw115_Surface_Plot_HFH_-1.5_1.png| 400px]]

==Polanyi's Empirical Rules ==

Question 10. Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.

Polanyi's rules state that in reactions with an early transition state (most closely resembling the reactants) translational energy most effectively promotes a successful reaction. Conversely where the transition state is late (resembling products closely) vibrational energy, specifically of the reactants most effectively promotes a successful reaction. <ref>The Journal of Chemical Physics 138, 234104 (2013); https://doi.org/10.1063/1.4810007</ref>

=References=
<references/>

MRD:mlw115RXNDy

2018-05-22T15:15:11Z

Mlw115: /* MEP vs Dynamics */

= Reaction Dynamics =

==Introduction==

The reactivity of atom-diatom systems are studied in order to gain insight on the internal degrees of freedom of the system. Particularly vibrational and translational energy are considered with respect to overcoming the transition state energy barrier.

The atoms are considered to behave classically and the interactions are expressed as potential energy surfaces.

= Exercise 1: H2 + H system=

==Dynamics from the transition state region ==

 Question 1. What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

Where a reaction is modeled on a potential energy surface the lowest energy path between the reactants and products is seen as a "valley floor" of the surface. The transition structure is a saddle point along the minimum energy ridge.

The gradient of the potential energy surface is zero at the minimum, along the direction of the reaction path and so the first derivative with respect to the inter-nuclear distance is zero. For the saddle point the gradient of the potential energy surface is zero orthogonal to the direction of the reaction pathway.

δV(ri)/δri = 0

Second derivatives
δV2(ri)/δri2 > 0 Minima
δV2(ri)/δri2 < 0 Maxima

where V is the potential energy and r related to the nuclear positions.

The minimum pathway and saddle point can be distinguished using the second derivatives with respect to atom positions. The second derivative of a maximum point is negative, whereas the second derivative of a minimum is positive. As the saddle point is a maxima along the minimum path the second derivative is negative. The second derivative of the minimum energy path in the direction of the reaction will be positive.

==Estimating Transition State Position ==

 Question 2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. 

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.908||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot 1 !! Surface Plot 2!! Internuclear distance vs Time
|-
|[[File:Mlw115_H2H_saddle.png|300px]]||[[File:mlw115_H2H_TS.png|300px]]||[[File:Mlw115_H2H_TS_INDvT.png|300px]]|
|}

The best guess of inter nuclear distances at the transition state was found using trial and error, to find the positions that would result in the minimum trajectory. As the system is symmetric the distances would be equal. The two surface plots are included to demonstrate the "reaction path". The internuclear distance vs time plot demonstrates no change in any bond lengths at the transition state so the overall momenta is 0 and potential energy is at the maximum.

== Reaction Path ==
===MEP vs Dynamics===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.909||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot (MEP) !! Surface Plot (Dynamics
|-
|[[File:Mlw115_H2H_MEP_surfaceplot.png|300px]]||[[File:mlw115_H2H_MEP_surfaceplot_dynamics.png|300px]]
|}

 Question 3. Comment on how the mep and the trajectory you just calculated differ.

It can be seen in the MEP plot that the trajectory follows the minimum energy valley floor without any obvious deviations or oscillation. This does not account for the inter-molecular vibrations between the molecules,. The MEP trajectory is corresponding to infinitely slow motion, it does not accurately represent the atomic motion of the reaction. The trajectory oscillates in the dynamics plot, also in much less steps the AB bond distance increases considerably more. This plot is more realistic as it accounts for the motion of atoms during the reaction.

{| class="wikitable" border="1"
|+ Final Geometry
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| 2.486||1.167
|}

=== Reversed starting positions ===

What would change if we used the initial conditions r1 = rts and r2 = rts+0.01 instead? ****** DO THIS ******

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| -2.486||-1.167
|}

[[File:mlw115_H2H_dynamics_reversed_momenta.png|300px]]

From taking the previous runs final conditions and setting these as the initial conditions and reversing the momenta values the reaction seems to run in reverse, reaching an end exactly at the transition state.

===Determining Reactivity===

Question 4. Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2.0
|}

{| class="wikitable" border=1
! pAB !! pBC!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -1.25 || -2.5 || -99.018|| Yes || [[File:Mlw115_Surface_Plot_1.25_2.5.png|300px]] || Initially the HAHB distance changes very little as the vibrational energy is quite low, as the HC approaches it can be seen that the BC distance decreases linearly. The molecules pass over the transition state, after this point the HAHB distance decreases and the HBHC distance oscillates slightly, with a higher vibrational energy than the HAHB molecule. This overall process represents the HAHB bond breaking and forming a HBHC molecule.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mlw115_Surface_Plot_1.5_2.0.png|300px]] || The molecule approaches the maxima with some vibrational energy, shown by oscillations in the initial path, the energy of the system is not high enough to pass over the transition state energy maxima so the HC "rebounds" and the HBHC distance begins to increase. The HAHB molecule oscillations increase.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mlw115_Surface_Plot_1.5_2.5.png|300px]] || The reaction proceeds similarly to the first case with HC approaching and forming a bond with HB as there is sufficient energy to proceed over the transition state maxima. The vibrational energy throughout is slightly higher than the first case, demonstrated by higher oscillations in atom distances.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mlw115_Surface_Plot_2.5_5.png|300px]] || In this case the energy is high enough and the atoms are able to pass over the transition state energy maxima, however the vibrational energy of the HBHC system formed seems to be too high to form the molecule and the HC atom rebounds and the distance begins to decrease and so the HAHB is reformed but with considerably higher oscillations.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mlw115_Surface_Plot_2.5_5.2.png|300px]] || The HC approaches the oscillating HAHB system, the HB atom rebounds between the two atoms, as it can be seen on the surface plot the system passes over the transition state maxima, back over to the other side and back again as this occurs. Finally the HBHC bond is formed with a high vibrationally energy
|}

===Transition State Theory===

 Question 5 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? 

Transition state theory is used to understand how elementary chemical reactions take place assuming a quasi-equilibirum between reactanting complexes and those at the transition state. The rate of reaction can be understood by inspecting the complexes at the maximum saddle point of the minimum energy pathway. The reactants are in equilibrium with the activated transition state complexes and can be converted to the products.

This theory is useful to theoretically understand a chemical reaction however it operates on a series of assumptions which limit the theory. For instance the atoms are assumed to behave classically, ignoring tunelling. Tunneling would be more prominent in reactions with low activation barriers. Also the theory assumes that the saddle point with the lowest energy will be passed over to reach the products, however this is not the case when higher vibrational energies will be populated.

<ref>https://goldbook.iupac.org/html/T/T06470.html</ref>

=Exercise 2: H-H-F System=

==PES Inspection==
Question 6. 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?

For the forward reaction of H2 + F --> HF + H is exothermic, as shown in the surface plot below the energy of the reactants H2 is higher than the energy of the product HF. As the energy absorbed in breaking the H-H bond in the first case is less than the energy released forming the H-F bond which means that in total energy is released which results in an exothermic reaction.

{| class="wikitable" border=1
! Surface Plot H2+F->HF+H !! Surface Plot HF+H -> H2+F
|-
|[[File:mlw115_Surface_Plot_HHF_labelled.png|300px]]||[[File:mlw115_Surface_Plot_HHF_H2Prod.png|300px]]|
|}

==Determining Transition State ==

Question 7. Locate the approximate position of the transition state.

A = HA B = HB and C = F

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! Surface Plot !! Internuclear distance vs Time
|-
|[[File:Mlw115_Surface_Plot_TS_HHF.png|400px]]||[[File:Mlw115 INDvT HHF TS.png|350px]]|
|}

== Calculating Activation Energy ==
Question 8. Report the activation energy for both reactions.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! BC= 1.8135!! BC= 1.8135 Zoomed!! BC= 1.813
|-
|[[File:Mlw115_HHF_EvT_1.8135.png|300px]] ‎||[[File:Mlw115_Plot_1.813_EvT_zoomT0_Resized.png|300px]]||[[File:Mlw115_HHF_EvT_1.813.png|320px]]
|}

pAB = pBC =0

distAB = 0.74

{| class="wikitable" border=1
|+Activation energy HF+H->H2+F
|-
! !! Energy (Kcal/mol)
|-
|Initial energy || -103.752
|-
|Final energy|| -133.89
|-
|EA|| 30.138
|}

{| class="wikitable" border=1
|+Activation energy H2+F->HF+H
|-
! !! Energy (Kcal/mol)
|-
|Initial energy|| -103.742
|-
|Final energy|| -103.752
|-
|EA|| 0.01
|}

==Reaction Dynamics==
=== F+ H2 ===
{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.74||0
|-
| BC|| 2.3||-2.7
|}

{| class="wikitable" border=1
!Surface Plot!! Internuclear momentum vs time
|-
|[[File:Mlw_Surface_Plot_HHF_Reactive.png]]|| [[File:Mlw_INMvT_HHF_Reactive.png]]
|}

{| class="wikitable" border=1
!Energy!!
|-
|Kinetic|| 3.837
|-
|Potential|| -103.886

|-
|Total|| -100.049
|}

Question 9. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?

It can be seen from the animation that as the H-H (AB) bond is broken and the H-F (BC) bond is formed the vibrational energy (and so the kinetic energy) considerably increases for H-F. The internuclear momentum vs time graph demonstrates this, as the momentum is much higher for the H-F atom. This vibrational energy will be released as heat so experimentally an increase in temperature could confirm the reaction had occurred.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -3 || -0.5 || -100.339|| No || [[File:Mlw115_Surface_Plot_HHF_-3_0.5.png|300px]] || H2 approaches the fluorine atom slowly but when a collision is made the H rebounds straight back into the other H and they continue their path oscillating together in the opposite direction
|-

| 3 || -0.5 || -93.254 || No || [[File:Mlw115_Surface_Plot_HHF_3_0.5.png|300px]] || This is similar to the previous case however the hydrogen rebounds twice into flourine but on the second rebounds remains returns back to oscillating with the original H atom
|-
| 2 || -0.5|| -98.753|| No || [[File: Mlw115_Surface_Plot_HHF_2_0.5.png|300px]] || The HF molecule oscillates within the energy well at high vibrational energy after collision
|-
| 1.5|| -0.5|| 103.754|| No || [[File: Mlw115_Surface_Plot_HHF_1.5_0.5.png|300px]] || Similarly to the previous situation
|-
|0.1||-0.8|| 103.364|| Yes|| [[File: Mlw115_Surface_Plot_HHF_0.1_0.8.png|300px]] || Despite having lower total energy than the previous cases the reaction goes to completion as the momentum between the reacting atoms is sufficient
|}

Despite the H2 energy being higher than the activation energy if the BC atoms do not collide with a high enough energy the reaction will return back over the peak and the original reactants will be produced.

=== HF+F ===

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 2
|-
|BC || 0.74
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -5 || 0.2 || -103.364 || No || [[File:Mlw115_Surface_Plot_HFH_5_0.2.png|300px]] || Approaching H rebounds and reaction doesn't complete

|}

A reactive pathway was found by trial and error around the reversed conditions from the reverse reaction.

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 5.0||-1.5
|-
| BC|| 1.2||1.0
|}

[[File: Mlw115_Surface_Plot_HFH_-1.5_1.png| 400px]]

==Polanyi's Empirical Rules ==

Question 10. Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.

Polanyi's rules state that in reactions with an early transition state (most closely resembling the reactants) translational energy most effectively promotes a successful reaction. Conversely where the transition state is late (resembling products closely) vibrational energy, specifically of the reactants most effectively promotes a successful reaction. <ref>The Journal of Chemical Physics 138, 234104 (2013); https://doi.org/10.1063/1.4810007</ref>

=References=
<references/>

MRD:mlw115RXNDy

2018-05-22T15:02:26Z

Mlw115: /* HF+F */

= Reaction Dynamics =

==Introduction==

The reactivity of atom-diatom systems are studied in order to gain insight on the internal degrees of freedom of the system. Particularly vibrational and translational energy are considered with respect to overcoming the transition state energy barrier.

The atoms are considered to behave classically and the interactions are expressed as potential energy surfaces.

= Exercise 1: H2 + H system=

==Dynamics from the transition state region ==

 Question 1. What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

Where a reaction is modeled on a potential energy surface the lowest energy path between the reactants and products is seen as a "valley floor" of the surface. The transition structure is a saddle point along the minimum energy ridge.

The gradient of the potential energy surface is zero at the minimum, along the direction of the reaction path and so the first derivative with respect to the inter-nuclear distance is zero. For the saddle point the gradient of the potential energy surface is zero orthogonal to the direction of the reaction pathway.

δV(ri)/δri = 0

Second derivatives
δV2(ri)/δri2 > 0 Minima
δV2(ri)/δri2 < 0 Maxima

where V is the potential energy and r related to the nuclear positions.

The minimum pathway and saddle point can be distinguished using the second derivatives with respect to atom positions. The second derivative of a maximum point is negative, whereas the second derivative of a minimum is positive. As the saddle point is a maxima along the minimum path the second derivative is negative. The second derivative of the minimum energy path in the direction of the reaction will be positive.

==Estimating Transition State Position ==

 Question 2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. 

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.908||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot 1 !! Surface Plot 2!! Internuclear distance vs Time
|-
|[[File:Mlw115_H2H_saddle.png|300px]]||[[File:mlw115_H2H_TS.png|300px]]||[[File:Mlw115_H2H_TS_INDvT.png|300px]]|
|}

The best guess of inter nuclear distances at the transition state was found using trial and error, to find the positions that would result in the minimum trajectory. As the system is symmetric the distances would be equal. The two surface plots are included to demonstrate the "reaction path". The internuclear distance vs time plot demonstrates no change in any bond lengths at the transition state so the overall momenta is 0 and potential energy is at the maximum.

== Reaction Path ==
===MEP vs Dynamics===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.909||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot (MEP) !! Surface Plot (Dynamics
|-
|[[File:Mlw115_H2H_MEP_surfaceplot.png|300px]]||[[File:mlw115_H2H_MEP_surfaceplot_dynamics.png|300px]]
|}

 Question 3. Comment on how the mep and the trajectory you just calculated differ.

It can be seen in the MEP plot that the trajectory follows the minimum energy valley floor without any obvious deviations or oscillation. The MEP trajectory is corresponding to infinitely slow motion, it does not accurately represent the atomic motion of the reaction. The trajectory oscillates in the dynamics plot, also in much less steps the AB bond distance increases considerably more. This plot is more realistic as it accounts for the motion of atoms during the reaction.

{| class="wikitable" border="1"
|+ Final Geometry
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| 2.486||1.167
|}

=== Reversed starting positions ===

What would change if we used the initial conditions r1 = rts and r2 = rts+0.01 instead? ****** DO THIS ******

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| -2.486||-1.167
|}

[[File:mlw115_H2H_dynamics_reversed_momenta.png|300px]]

From taking the previous runs final conditions and setting these as the initial conditions and reversing the momenta values the reaction seems to run in reverse, reaching an end exactly at the transition state.

===Determining Reactivity===

Question 4. Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2.0
|}

{| class="wikitable" border=1
! pAB !! pBC!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -1.25 || -2.5 || -99.018|| Yes || [[File:Mlw115_Surface_Plot_1.25_2.5.png|300px]] || Initially the HAHB distance changes very little as the vibrational energy is quite low, as the HC approaches it can be seen that the BC distance decreases linearly. The molecules pass over the transition state, after this point the HAHB distance decreases and the HBHC distance oscillates slightly, with a higher vibrational energy than the HAHB molecule. This overall process represents the HAHB bond breaking and forming a HBHC molecule.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mlw115_Surface_Plot_1.5_2.0.png|300px]] || The molecule approaches the maxima with some vibrational energy, shown by oscillations in the initial path, the energy of the system is not high enough to pass over the transition state energy maxima so the HC "rebounds" and the HBHC distance begins to increase. The HAHB molecule oscillations increase.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mlw115_Surface_Plot_1.5_2.5.png|300px]] || The reaction proceeds similarly to the first case with HC approaching and forming a bond with HB as there is sufficient energy to proceed over the transition state maxima. The vibrational energy throughout is slightly higher than the first case, demonstrated by higher oscillations in atom distances.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mlw115_Surface_Plot_2.5_5.png|300px]] || In this case the energy is high enough and the atoms are able to pass over the transition state energy maxima, however the vibrational energy of the HBHC system formed seems to be too high to form the molecule and the HC atom rebounds and the distance begins to decrease and so the HAHB is reformed but with considerably higher oscillations.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mlw115_Surface_Plot_2.5_5.2.png|300px]] || The HC approaches the oscillating HAHB system, the HB atom rebounds between the two atoms, as it can be seen on the surface plot the system passes over the transition state maxima, back over to the other side and back again as this occurs. Finally the HBHC bond is formed with a high vibrationally energy
|}

===Transition State Theory===

 Question 5 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? 

Transition state theory is used to understand how elementary chemical reactions take place assuming a quasi-equilibirum between reactanting complexes and those at the transition state. The rate of reaction can be understood by inspecting the complexes at the maximum saddle point of the minimum energy pathway. The reactants are in equilibrium with the activated transition state complexes and can be converted to the products.

This theory is useful to theoretically understand a chemical reaction however it operates on a series of assumptions which limit the theory. For instance the atoms are assumed to behave classically, ignoring tunelling. Tunneling would be more prominent in reactions with low activation barriers. Also the theory assumes that the saddle point with the lowest energy will be passed over to reach the products, however this is not the case when higher vibrational energies will be populated.

<ref>https://goldbook.iupac.org/html/T/T06470.html</ref>

=Exercise 2: H-H-F System=

==PES Inspection==
Question 6. 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?

For the forward reaction of H2 + F --> HF + H is exothermic, as shown in the surface plot below the energy of the reactants H2 is higher than the energy of the product HF. As the energy absorbed in breaking the H-H bond in the first case is less than the energy released forming the H-F bond which means that in total energy is released which results in an exothermic reaction.

{| class="wikitable" border=1
! Surface Plot H2+F->HF+H !! Surface Plot HF+H -> H2+F
|-
|[[File:mlw115_Surface_Plot_HHF_labelled.png|300px]]||[[File:mlw115_Surface_Plot_HHF_H2Prod.png|300px]]|
|}

==Determining Transition State ==

Question 7. Locate the approximate position of the transition state.

A = HA B = HB and C = F

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! Surface Plot !! Internuclear distance vs Time
|-
|[[File:Mlw115_Surface_Plot_TS_HHF.png|400px]]||[[File:Mlw115 INDvT HHF TS.png|350px]]|
|}

== Calculating Activation Energy ==
Question 8. Report the activation energy for both reactions.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! BC= 1.8135!! BC= 1.8135 Zoomed!! BC= 1.813
|-
|[[File:Mlw115_HHF_EvT_1.8135.png|300px]] ‎||[[File:Mlw115_Plot_1.813_EvT_zoomT0_Resized.png|300px]]||[[File:Mlw115_HHF_EvT_1.813.png|320px]]
|}

pAB = pBC =0

distAB = 0.74

{| class="wikitable" border=1
|+Activation energy HF+H->H2+F
|-
! !! Energy (Kcal/mol)
|-
|Initial energy || -103.752
|-
|Final energy|| -133.89
|-
|EA|| 30.138
|}

{| class="wikitable" border=1
|+Activation energy H2+F->HF+H
|-
! !! Energy (Kcal/mol)
|-
|Initial energy|| -103.742
|-
|Final energy|| -103.752
|-
|EA|| 0.01
|}

==Reaction Dynamics==
=== F+ H2 ===
{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.74||0
|-
| BC|| 2.3||-2.7
|}

{| class="wikitable" border=1
!Surface Plot!! Internuclear momentum vs time
|-
|[[File:Mlw_Surface_Plot_HHF_Reactive.png]]|| [[File:Mlw_INMvT_HHF_Reactive.png]]
|}

{| class="wikitable" border=1
!Energy!!
|-
|Kinetic|| 3.837
|-
|Potential|| -103.886

|-
|Total|| -100.049
|}

Question 9. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?

It can be seen from the animation that as the H-H (AB) bond is broken and the H-F (BC) bond is formed the vibrational energy (and so the kinetic energy) considerably increases for H-F. The internuclear momentum vs time graph demonstrates this, as the momentum is much higher for the H-F atom. This vibrational energy will be released as heat so experimentally an increase in temperature could confirm the reaction had occurred.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -3 || -0.5 || -100.339|| No || [[File:Mlw115_Surface_Plot_HHF_-3_0.5.png|300px]] || H2 approaches the fluorine atom slowly but when a collision is made the H rebounds straight back into the other H and they continue their path oscillating together in the opposite direction
|-

| 3 || -0.5 || -93.254 || No || [[File:Mlw115_Surface_Plot_HHF_3_0.5.png|300px]] || This is similar to the previous case however the hydrogen rebounds twice into flourine but on the second rebounds remains returns back to oscillating with the original H atom
|-
| 2 || -0.5|| -98.753|| No || [[File: Mlw115_Surface_Plot_HHF_2_0.5.png|300px]] || The HF molecule oscillates within the energy well at high vibrational energy after collision
|-
| 1.5|| -0.5|| 103.754|| No || [[File: Mlw115_Surface_Plot_HHF_1.5_0.5.png|300px]] || Similarly to the previous situation
|-
|0.1||-0.8|| 103.364|| Yes|| [[File: Mlw115_Surface_Plot_HHF_0.1_0.8.png|300px]] || Despite having lower total energy than the previous cases the reaction goes to completion as the momentum between the reacting atoms is sufficient
|}

Despite the H2 energy being higher than the activation energy if the BC atoms do not collide with a high enough energy the reaction will return back over the peak and the original reactants will be produced.

=== HF+F ===

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 2
|-
|BC || 0.74
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -5 || 0.2 || -103.364 || No || [[File:Mlw115_Surface_Plot_HFH_5_0.2.png|300px]] || Approaching H rebounds and reaction doesn't complete

|}

A reactive pathway was found by trial and error around the reversed conditions from the reverse reaction.

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 5.0||-1.5
|-
| BC|| 1.2||1.0
|}

[[File: Mlw115_Surface_Plot_HFH_-1.5_1.png| 400px]]

==Polanyi's Empirical Rules ==

Question 10. Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.

Polanyi's rules state that in reactions with an early transition state (most closely resembling the reactants) translational energy most effectively promotes a successful reaction. Conversely where the transition state is late (resembling products closely) vibrational energy, specifically of the reactants most effectively promotes a successful reaction. <ref>The Journal of Chemical Physics 138, 234104 (2013); https://doi.org/10.1063/1.4810007</ref>

=References=
<references/>

MRD:mlw115RXNDy

2018-05-22T15:00:26Z

Mlw115: /* Reaction Dynamics */

= Reaction Dynamics =

==Introduction==

The reactivity of atom-diatom systems are studied in order to gain insight on the internal degrees of freedom of the system. Particularly vibrational and translational energy are considered with respect to overcoming the transition state energy barrier.

The atoms are considered to behave classically and the interactions are expressed as potential energy surfaces.

= Exercise 1: H2 + H system=

==Dynamics from the transition state region ==

 Question 1. What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

Where a reaction is modeled on a potential energy surface the lowest energy path between the reactants and products is seen as a "valley floor" of the surface. The transition structure is a saddle point along the minimum energy ridge.

The gradient of the potential energy surface is zero at the minimum, along the direction of the reaction path and so the first derivative with respect to the inter-nuclear distance is zero. For the saddle point the gradient of the potential energy surface is zero orthogonal to the direction of the reaction pathway.

δV(ri)/δri = 0

Second derivatives
δV2(ri)/δri2 > 0 Minima
δV2(ri)/δri2 < 0 Maxima

where V is the potential energy and r related to the nuclear positions.

The minimum pathway and saddle point can be distinguished using the second derivatives with respect to atom positions. The second derivative of a maximum point is negative, whereas the second derivative of a minimum is positive. As the saddle point is a maxima along the minimum path the second derivative is negative. The second derivative of the minimum energy path in the direction of the reaction will be positive.

==Estimating Transition State Position ==

 Question 2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. 

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.908||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot 1 !! Surface Plot 2!! Internuclear distance vs Time
|-
|[[File:Mlw115_H2H_saddle.png|300px]]||[[File:mlw115_H2H_TS.png|300px]]||[[File:Mlw115_H2H_TS_INDvT.png|300px]]|
|}

The best guess of inter nuclear distances at the transition state was found using trial and error, to find the positions that would result in the minimum trajectory. As the system is symmetric the distances would be equal. The two surface plots are included to demonstrate the "reaction path". The internuclear distance vs time plot demonstrates no change in any bond lengths at the transition state so the overall momenta is 0 and potential energy is at the maximum.

== Reaction Path ==
===MEP vs Dynamics===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.909||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot (MEP) !! Surface Plot (Dynamics
|-
|[[File:Mlw115_H2H_MEP_surfaceplot.png|300px]]||[[File:mlw115_H2H_MEP_surfaceplot_dynamics.png|300px]]
|}

 Question 3. Comment on how the mep and the trajectory you just calculated differ.

It can be seen in the MEP plot that the trajectory follows the minimum energy valley floor without any obvious deviations or oscillation. The MEP trajectory is corresponding to infinitely slow motion, it does not accurately represent the atomic motion of the reaction. The trajectory oscillates in the dynamics plot, also in much less steps the AB bond distance increases considerably more. This plot is more realistic as it accounts for the motion of atoms during the reaction.

{| class="wikitable" border="1"
|+ Final Geometry
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| 2.486||1.167
|}

=== Reversed starting positions ===

What would change if we used the initial conditions r1 = rts and r2 = rts+0.01 instead? ****** DO THIS ******

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| -2.486||-1.167
|}

[[File:mlw115_H2H_dynamics_reversed_momenta.png|300px]]

From taking the previous runs final conditions and setting these as the initial conditions and reversing the momenta values the reaction seems to run in reverse, reaching an end exactly at the transition state.

===Determining Reactivity===

Question 4. Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2.0
|}

{| class="wikitable" border=1
! pAB !! pBC!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -1.25 || -2.5 || -99.018|| Yes || [[File:Mlw115_Surface_Plot_1.25_2.5.png|300px]] || Initially the HAHB distance changes very little as the vibrational energy is quite low, as the HC approaches it can be seen that the BC distance decreases linearly. The molecules pass over the transition state, after this point the HAHB distance decreases and the HBHC distance oscillates slightly, with a higher vibrational energy than the HAHB molecule. This overall process represents the HAHB bond breaking and forming a HBHC molecule.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mlw115_Surface_Plot_1.5_2.0.png|300px]] || The molecule approaches the maxima with some vibrational energy, shown by oscillations in the initial path, the energy of the system is not high enough to pass over the transition state energy maxima so the HC "rebounds" and the HBHC distance begins to increase. The HAHB molecule oscillations increase.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mlw115_Surface_Plot_1.5_2.5.png|300px]] || The reaction proceeds similarly to the first case with HC approaching and forming a bond with HB as there is sufficient energy to proceed over the transition state maxima. The vibrational energy throughout is slightly higher than the first case, demonstrated by higher oscillations in atom distances.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mlw115_Surface_Plot_2.5_5.png|300px]] || In this case the energy is high enough and the atoms are able to pass over the transition state energy maxima, however the vibrational energy of the HBHC system formed seems to be too high to form the molecule and the HC atom rebounds and the distance begins to decrease and so the HAHB is reformed but with considerably higher oscillations.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mlw115_Surface_Plot_2.5_5.2.png|300px]] || The HC approaches the oscillating HAHB system, the HB atom rebounds between the two atoms, as it can be seen on the surface plot the system passes over the transition state maxima, back over to the other side and back again as this occurs. Finally the HBHC bond is formed with a high vibrationally energy
|}

===Transition State Theory===

 Question 5 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? 

Transition state theory is used to understand how elementary chemical reactions take place assuming a quasi-equilibirum between reactanting complexes and those at the transition state. The rate of reaction can be understood by inspecting the complexes at the maximum saddle point of the minimum energy pathway. The reactants are in equilibrium with the activated transition state complexes and can be converted to the products.

This theory is useful to theoretically understand a chemical reaction however it operates on a series of assumptions which limit the theory. For instance the atoms are assumed to behave classically, ignoring tunelling. Tunneling would be more prominent in reactions with low activation barriers. Also the theory assumes that the saddle point with the lowest energy will be passed over to reach the products, however this is not the case when higher vibrational energies will be populated.

<ref>https://goldbook.iupac.org/html/T/T06470.html</ref>

=Exercise 2: H-H-F System=

==PES Inspection==
Question 6. 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?

For the forward reaction of H2 + F --> HF + H is exothermic, as shown in the surface plot below the energy of the reactants H2 is higher than the energy of the product HF. As the energy absorbed in breaking the H-H bond in the first case is less than the energy released forming the H-F bond which means that in total energy is released which results in an exothermic reaction.

{| class="wikitable" border=1
! Surface Plot H2+F->HF+H !! Surface Plot HF+H -> H2+F
|-
|[[File:mlw115_Surface_Plot_HHF_labelled.png|300px]]||[[File:mlw115_Surface_Plot_HHF_H2Prod.png|300px]]|
|}

==Determining Transition State ==

Question 7. Locate the approximate position of the transition state.

A = HA B = HB and C = F

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! Surface Plot !! Internuclear distance vs Time
|-
|[[File:Mlw115_Surface_Plot_TS_HHF.png|400px]]||[[File:Mlw115 INDvT HHF TS.png|350px]]|
|}

== Calculating Activation Energy ==
Question 8. Report the activation energy for both reactions.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! BC= 1.8135!! BC= 1.8135 Zoomed!! BC= 1.813
|-
|[[File:Mlw115_HHF_EvT_1.8135.png|300px]] ‎||[[File:Mlw115_Plot_1.813_EvT_zoomT0_Resized.png|300px]]||[[File:Mlw115_HHF_EvT_1.813.png|320px]]
|}

pAB = pBC =0

distAB = 0.74

{| class="wikitable" border=1
|+Activation energy HF+H->H2+F
|-
! !! Energy (Kcal/mol)
|-
|Initial energy || -103.752
|-
|Final energy|| -133.89
|-
|EA|| 30.138
|}

{| class="wikitable" border=1
|+Activation energy H2+F->HF+H
|-
! !! Energy (Kcal/mol)
|-
|Initial energy|| -103.742
|-
|Final energy|| -103.752
|-
|EA|| 0.01
|}

==Reaction Dynamics==
=== F+ H2 ===
{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.74||0
|-
| BC|| 2.3||-2.7
|}

{| class="wikitable" border=1
!Surface Plot!! Internuclear momentum vs time
|-
|[[File:Mlw_Surface_Plot_HHF_Reactive.png]]|| [[File:Mlw_INMvT_HHF_Reactive.png]]
|}

{| class="wikitable" border=1
!Energy!!
|-
|Kinetic|| 3.837
|-
|Potential|| -103.886

|-
|Total|| -100.049
|}

Question 9. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?

It can be seen from the animation that as the H-H (AB) bond is broken and the H-F (BC) bond is formed the vibrational energy (and so the kinetic energy) considerably increases for H-F. The internuclear momentum vs time graph demonstrates this, as the momentum is much higher for the H-F atom. This vibrational energy will be released as heat so experimentally an increase in temperature could confirm the reaction had occurred.

=== HF+F ===

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -3 || -0.5 || -100.339|| No || [[File:Mlw115_Surface_Plot_HHF_-3_0.5.png|300px]] || H2 approaches the fluorine atom slowly but when a collision is made the H rebounds straight back into the other H and they continue their path oscillating together in the opposite direction
|-

| 3 || -0.5 || -93.254 || No || [[File:Mlw115_Surface_Plot_HHF_3_0.5.png|300px]] || This is similar to the previous case however the hydrogen rebounds twice into flourine but on the second rebounds remains returns back to oscillating with the original H atom
|-
| 2 || -0.5|| -98.753|| No || [[File: Mlw115_Surface_Plot_HHF_2_0.5.png|300px]] || The HF molecule oscillates within the energy well at high vibrational energy after collision
|-
| 1.5|| -0.5|| 103.754|| No || [[File: Mlw115_Surface_Plot_HHF_1.5_0.5.png|300px]] || Similarly to the previous situation
|-
|0.1||-0.8|| 103.364|| Yes|| [[File: Mlw115_Surface_Plot_HHF_0.1_0.8.png|300px]] || Despite having lower total energy than the previous cases the reaction goes to completion as the momentum between the reacting atoms is sufficient
|}

Despite the H2 energy being higher than the activation energy if the BC atoms do not collide with a high enough energy the reaction will return back over the peak and the original reactants will be produced.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 2
|-
|BC || 0.74
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -5 || 0.2 || -103.364 || No || [[File:Mlw115_Surface_Plot_HFH_5_0.2.png|300px]] || Approaching H rebounds and reaction doesn't complete
|-
| 2.21||0.37||
|}

A reactive pathway was found by trial and error around the reversed conditions from the reverse reaction.

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 5.0||-1.5
|-
| BC|| 1.2||1.0
|}

[[File: Mlw115_Surface_Plot_HFH_-1.5_1.png| 400px]]

==Polanyi's Empirical Rules ==

Question 10. Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.

Polanyi's rules state that in reactions with an early transition state (most closely resembling the reactants) translational energy most effectively promotes a successful reaction. Conversely where the transition state is late (resembling products closely) vibrational energy, specifically of the reactants most effectively promotes a successful reaction. <ref>The Journal of Chemical Physics 138, 234104 (2013); https://doi.org/10.1063/1.4810007</ref>

=References=
<references/>

MRD:mlw115RXNDy

2018-05-22T14:59:30Z

Mlw115: /* Reaction Dynamics */

= Reaction Dynamics =

==Introduction==

The reactivity of atom-diatom systems are studied in order to gain insight on the internal degrees of freedom of the system. Particularly vibrational and translational energy are considered with respect to overcoming the transition state energy barrier.

The atoms are considered to behave classically and the interactions are expressed as potential energy surfaces.

= Exercise 1: H2 + H system=

==Dynamics from the transition state region ==

 Question 1. What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

Where a reaction is modeled on a potential energy surface the lowest energy path between the reactants and products is seen as a "valley floor" of the surface. The transition structure is a saddle point along the minimum energy ridge.

The gradient of the potential energy surface is zero at the minimum, along the direction of the reaction path and so the first derivative with respect to the inter-nuclear distance is zero. For the saddle point the gradient of the potential energy surface is zero orthogonal to the direction of the reaction pathway.

δV(ri)/δri = 0

Second derivatives
δV2(ri)/δri2 > 0 Minima
δV2(ri)/δri2 < 0 Maxima

where V is the potential energy and r related to the nuclear positions.

The minimum pathway and saddle point can be distinguished using the second derivatives with respect to atom positions. The second derivative of a maximum point is negative, whereas the second derivative of a minimum is positive. As the saddle point is a maxima along the minimum path the second derivative is negative. The second derivative of the minimum energy path in the direction of the reaction will be positive.

==Estimating Transition State Position ==

 Question 2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. 

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.908||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot 1 !! Surface Plot 2!! Internuclear distance vs Time
|-
|[[File:Mlw115_H2H_saddle.png|300px]]||[[File:mlw115_H2H_TS.png|300px]]||[[File:Mlw115_H2H_TS_INDvT.png|300px]]|
|}

The best guess of inter nuclear distances at the transition state was found using trial and error, to find the positions that would result in the minimum trajectory. As the system is symmetric the distances would be equal. The two surface plots are included to demonstrate the "reaction path". The internuclear distance vs time plot demonstrates no change in any bond lengths at the transition state so the overall momenta is 0 and potential energy is at the maximum.

== Reaction Path ==
===MEP vs Dynamics===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.909||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot (MEP) !! Surface Plot (Dynamics
|-
|[[File:Mlw115_H2H_MEP_surfaceplot.png|300px]]||[[File:mlw115_H2H_MEP_surfaceplot_dynamics.png|300px]]
|}

 Question 3. Comment on how the mep and the trajectory you just calculated differ.

It can be seen in the MEP plot that the trajectory follows the minimum energy valley floor without any obvious deviations or oscillation. The MEP trajectory is corresponding to infinitely slow motion, it does not accurately represent the atomic motion of the reaction. The trajectory oscillates in the dynamics plot, also in much less steps the AB bond distance increases considerably more. This plot is more realistic as it accounts for the motion of atoms during the reaction.

{| class="wikitable" border="1"
|+ Final Geometry
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| 2.486||1.167
|}

=== Reversed starting positions ===

What would change if we used the initial conditions r1 = rts and r2 = rts+0.01 instead? ****** DO THIS ******

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| -2.486||-1.167
|}

[[File:mlw115_H2H_dynamics_reversed_momenta.png|300px]]

From taking the previous runs final conditions and setting these as the initial conditions and reversing the momenta values the reaction seems to run in reverse, reaching an end exactly at the transition state.

===Determining Reactivity===

Question 4. Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2.0
|}

{| class="wikitable" border=1
! pAB !! pBC!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -1.25 || -2.5 || -99.018|| Yes || [[File:Mlw115_Surface_Plot_1.25_2.5.png|300px]] || Initially the HAHB distance changes very little as the vibrational energy is quite low, as the HC approaches it can be seen that the BC distance decreases linearly. The molecules pass over the transition state, after this point the HAHB distance decreases and the HBHC distance oscillates slightly, with a higher vibrational energy than the HAHB molecule. This overall process represents the HAHB bond breaking and forming a HBHC molecule.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mlw115_Surface_Plot_1.5_2.0.png|300px]] || The molecule approaches the maxima with some vibrational energy, shown by oscillations in the initial path, the energy of the system is not high enough to pass over the transition state energy maxima so the HC "rebounds" and the HBHC distance begins to increase. The HAHB molecule oscillations increase.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mlw115_Surface_Plot_1.5_2.5.png|300px]] || The reaction proceeds similarly to the first case with HC approaching and forming a bond with HB as there is sufficient energy to proceed over the transition state maxima. The vibrational energy throughout is slightly higher than the first case, demonstrated by higher oscillations in atom distances.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mlw115_Surface_Plot_2.5_5.png|300px]] || In this case the energy is high enough and the atoms are able to pass over the transition state energy maxima, however the vibrational energy of the HBHC system formed seems to be too high to form the molecule and the HC atom rebounds and the distance begins to decrease and so the HAHB is reformed but with considerably higher oscillations.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mlw115_Surface_Plot_2.5_5.2.png|300px]] || The HC approaches the oscillating HAHB system, the HB atom rebounds between the two atoms, as it can be seen on the surface plot the system passes over the transition state maxima, back over to the other side and back again as this occurs. Finally the HBHC bond is formed with a high vibrationally energy
|}

===Transition State Theory===

 Question 5 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? 

Transition state theory is used to understand how elementary chemical reactions take place assuming a quasi-equilibirum between reactanting complexes and those at the transition state. The rate of reaction can be understood by inspecting the complexes at the maximum saddle point of the minimum energy pathway. The reactants are in equilibrium with the activated transition state complexes and can be converted to the products.

This theory is useful to theoretically understand a chemical reaction however it operates on a series of assumptions which limit the theory. For instance the atoms are assumed to behave classically, ignoring tunelling. Tunneling would be more prominent in reactions with low activation barriers. Also the theory assumes that the saddle point with the lowest energy will be passed over to reach the products, however this is not the case when higher vibrational energies will be populated.

<ref>https://goldbook.iupac.org/html/T/T06470.html</ref>

=Exercise 2: H-H-F System=

==PES Inspection==
Question 6. 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?

For the forward reaction of H2 + F --> HF + H is exothermic, as shown in the surface plot below the energy of the reactants H2 is higher than the energy of the product HF. As the energy absorbed in breaking the H-H bond in the first case is less than the energy released forming the H-F bond which means that in total energy is released which results in an exothermic reaction.

{| class="wikitable" border=1
! Surface Plot H2+F->HF+H !! Surface Plot HF+H -> H2+F
|-
|[[File:mlw115_Surface_Plot_HHF_labelled.png|300px]]||[[File:mlw115_Surface_Plot_HHF_H2Prod.png|300px]]|
|}

==Determining Transition State ==

Question 7. Locate the approximate position of the transition state.

A = HA B = HB and C = F

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! Surface Plot !! Internuclear distance vs Time
|-
|[[File:Mlw115_Surface_Plot_TS_HHF.png|400px]]||[[File:Mlw115 INDvT HHF TS.png|350px]]|
|}

== Calculating Activation Energy ==
Question 8. Report the activation energy for both reactions.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! BC= 1.8135!! BC= 1.8135 Zoomed!! BC= 1.813
|-
|[[File:Mlw115_HHF_EvT_1.8135.png|300px]] ‎||[[File:Mlw115_Plot_1.813_EvT_zoomT0_Resized.png|300px]]||[[File:Mlw115_HHF_EvT_1.813.png|320px]]
|}

pAB = pBC =0

distAB = 0.74

{| class="wikitable" border=1
|+Activation energy HF+H->H2+F
|-
! !! Energy (Kcal/mol)
|-
|Initial energy || -103.752
|-
|Final energy|| -133.89
|-
|EA|| 30.138
|}

{| class="wikitable" border=1
|+Activation energy H2+F->HF+H
|-
! !! Energy (Kcal/mol)
|-
|Initial energy|| -103.742
|-
|Final energy|| -103.752
|-
|EA|| 0.01
|}

==Reaction Dynamics==

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.74||0
|-
| BC|| 2.3||-2.7
|}

{| class="wikitable" border=1
!Surface Plot!! Internuclear momentum vs time
|-
|[[File:Mlw_Surface_Plot_HHF_Reactive.png]]|| [[File:Mlw_INMvT_HHF_Reactive.png]]
|}

{| class="wikitable" border=1
!Energy!!
|-
|Kinetic|| 3.837
|-
|Potential|| -103.886

|-
|Total|| -100.049
|}

Question 9. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?

It can be seen from the animation that as the H-H (AB) bond is broken and the H-F (BC) bond is formed the vibrational energy (and so the kinetic energy) considerably increases for H-F. The internuclear momentum vs time graph demonstrates this, as the momentum is much higher for the H-F atom. This vibrational energy will be released as heat so experimentally an increase in temperature could confirm the reaction had occurred.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -3 || -0.5 || -100.339|| No || [[File:Mlw115_Surface_Plot_HHF_-3_0.5.png|300px]] || H2 approaches the fluorine atom slowly but when a collision is made the H rebounds straight back into the other H and they continue their path oscillating together in the opposite direction
|-

| 3 || -0.5 || -93.254 || No || [[File:Mlw115_Surface_Plot_HHF_3_0.5.png|300px]] || This is similar to the previous case however the hydrogen rebounds twice into flourine but on the second rebounds remains returns back to oscillating with the original H atom
|-
| 2 || -0.5|| -98.753|| No || [[File: Mlw115_Surface_Plot_HHF_2_0.5.png|300px]] || The HF molecule oscillates within the energy well at high vibrational energy after collision
|-
| 1.5|| -0.5|| 103.754|| No || [[File: Mlw115_Surface_Plot_HHF_1.5_0.5.png|300px]] || Similarly to the previous situation
|-
|0.1||-0.8|| 103.364|| Yes|| [[File: Mlw115_Surface_Plot_HHF_0.1_0.8.png|300px]] || Despite having lower total energy than the previous cases the reaction goes to completion as the momentum between the reacting atoms is sufficient
|}

Despite the H2 energy being higher than the activation energy if the BC atoms do not collide with a high enough energy the reaction will return back over the peak and the original reactants will be produced.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 2
|-
|BC || 0.74
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -5 || 0.2 || -103.364 || No || [[File:Mlw115_Surface_Plot_HFH_5_0.2.png|300px]] || Approaching H rebounds and reaction doesn't complete
|-
| 2.21||0.37||
|}

A reactive pathway was found by trial and error around the reversed conditions from the reverse reaction.

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 5.0||-1.5
|-
| BC|| 1.2||1.0
|}

[[File: Mlw115_Surface_Plot_HFH_-1.5_1.png| 400px]]

==Polanyi's Empirical Rules ==

Question 10. Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.

Polanyi's rules state that in reactions with an early transition state (most closely resembling the reactants) translational energy most effectively promotes a successful reaction. Conversely where the transition state is late (resembling products closely) vibrational energy, specifically of the reactants most effectively promotes a successful reaction. <ref>The Journal of Chemical Physics 138, 234104 (2013); https://doi.org/10.1063/1.4810007</ref>

=References=
<references/>

MRD:mlw115RXNDy

2018-05-22T14:58:25Z

Mlw115: /* Reaction Dynamics */

= Reaction Dynamics =

==Introduction==

The reactivity of atom-diatom systems are studied in order to gain insight on the internal degrees of freedom of the system. Particularly vibrational and translational energy are considered with respect to overcoming the transition state energy barrier.

The atoms are considered to behave classically and the interactions are expressed as potential energy surfaces.

= Exercise 1: H2 + H system=

==Dynamics from the transition state region ==

 Question 1. What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

Where a reaction is modeled on a potential energy surface the lowest energy path between the reactants and products is seen as a "valley floor" of the surface. The transition structure is a saddle point along the minimum energy ridge.

The gradient of the potential energy surface is zero at the minimum, along the direction of the reaction path and so the first derivative with respect to the inter-nuclear distance is zero. For the saddle point the gradient of the potential energy surface is zero orthogonal to the direction of the reaction pathway.

δV(ri)/δri = 0

Second derivatives
δV2(ri)/δri2 > 0 Minima
δV2(ri)/δri2 < 0 Maxima

where V is the potential energy and r related to the nuclear positions.

The minimum pathway and saddle point can be distinguished using the second derivatives with respect to atom positions. The second derivative of a maximum point is negative, whereas the second derivative of a minimum is positive. As the saddle point is a maxima along the minimum path the second derivative is negative. The second derivative of the minimum energy path in the direction of the reaction will be positive.

==Estimating Transition State Position ==

 Question 2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. 

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.908||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot 1 !! Surface Plot 2!! Internuclear distance vs Time
|-
|[[File:Mlw115_H2H_saddle.png|300px]]||[[File:mlw115_H2H_TS.png|300px]]||[[File:Mlw115_H2H_TS_INDvT.png|300px]]|
|}

The best guess of inter nuclear distances at the transition state was found using trial and error, to find the positions that would result in the minimum trajectory. As the system is symmetric the distances would be equal. The two surface plots are included to demonstrate the "reaction path". The internuclear distance vs time plot demonstrates no change in any bond lengths at the transition state so the overall momenta is 0 and potential energy is at the maximum.

== Reaction Path ==
===MEP vs Dynamics===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.909||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot (MEP) !! Surface Plot (Dynamics
|-
|[[File:Mlw115_H2H_MEP_surfaceplot.png|300px]]||[[File:mlw115_H2H_MEP_surfaceplot_dynamics.png|300px]]
|}

 Question 3. Comment on how the mep and the trajectory you just calculated differ.

It can be seen in the MEP plot that the trajectory follows the minimum energy valley floor without any obvious deviations or oscillation. The MEP trajectory is corresponding to infinitely slow motion, it does not accurately represent the atomic motion of the reaction. The trajectory oscillates in the dynamics plot, also in much less steps the AB bond distance increases considerably more. This plot is more realistic as it accounts for the motion of atoms during the reaction.

{| class="wikitable" border="1"
|+ Final Geometry
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| 2.486||1.167
|}

=== Reversed starting positions ===

What would change if we used the initial conditions r1 = rts and r2 = rts+0.01 instead? ****** DO THIS ******

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| -2.486||-1.167
|}

[[File:mlw115_H2H_dynamics_reversed_momenta.png|300px]]

From taking the previous runs final conditions and setting these as the initial conditions and reversing the momenta values the reaction seems to run in reverse, reaching an end exactly at the transition state.

===Determining Reactivity===

Question 4. Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2.0
|}

{| class="wikitable" border=1
! pAB !! pBC!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -1.25 || -2.5 || -99.018|| Yes || [[File:Mlw115_Surface_Plot_1.25_2.5.png|300px]] || Initially the HAHB distance changes very little as the vibrational energy is quite low, as the HC approaches it can be seen that the BC distance decreases linearly. The molecules pass over the transition state, after this point the HAHB distance decreases and the HBHC distance oscillates slightly, with a higher vibrational energy than the HAHB molecule. This overall process represents the HAHB bond breaking and forming a HBHC molecule.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mlw115_Surface_Plot_1.5_2.0.png|300px]] || The molecule approaches the maxima with some vibrational energy, shown by oscillations in the initial path, the energy of the system is not high enough to pass over the transition state energy maxima so the HC "rebounds" and the HBHC distance begins to increase. The HAHB molecule oscillations increase.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mlw115_Surface_Plot_1.5_2.5.png|300px]] || The reaction proceeds similarly to the first case with HC approaching and forming a bond with HB as there is sufficient energy to proceed over the transition state maxima. The vibrational energy throughout is slightly higher than the first case, demonstrated by higher oscillations in atom distances.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mlw115_Surface_Plot_2.5_5.png|300px]] || In this case the energy is high enough and the atoms are able to pass over the transition state energy maxima, however the vibrational energy of the HBHC system formed seems to be too high to form the molecule and the HC atom rebounds and the distance begins to decrease and so the HAHB is reformed but with considerably higher oscillations.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mlw115_Surface_Plot_2.5_5.2.png|300px]] || The HC approaches the oscillating HAHB system, the HB atom rebounds between the two atoms, as it can be seen on the surface plot the system passes over the transition state maxima, back over to the other side and back again as this occurs. Finally the HBHC bond is formed with a high vibrationally energy
|}

===Transition State Theory===

 Question 5 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? 

Transition state theory is used to understand how elementary chemical reactions take place assuming a quasi-equilibirum between reactanting complexes and those at the transition state. The rate of reaction can be understood by inspecting the complexes at the maximum saddle point of the minimum energy pathway. The reactants are in equilibrium with the activated transition state complexes and can be converted to the products.

This theory is useful to theoretically understand a chemical reaction however it operates on a series of assumptions which limit the theory. For instance the atoms are assumed to behave classically, ignoring tunelling. Tunneling would be more prominent in reactions with low activation barriers. Also the theory assumes that the saddle point with the lowest energy will be passed over to reach the products, however this is not the case when higher vibrational energies will be populated.

<ref>https://goldbook.iupac.org/html/T/T06470.html</ref>

=Exercise 2: H-H-F System=

==PES Inspection==
Question 6. 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?

For the forward reaction of H2 + F --> HF + H is exothermic, as shown in the surface plot below the energy of the reactants H2 is higher than the energy of the product HF. As the energy absorbed in breaking the H-H bond in the first case is less than the energy released forming the H-F bond which means that in total energy is released which results in an exothermic reaction.

{| class="wikitable" border=1
! Surface Plot H2+F->HF+H !! Surface Plot HF+H -> H2+F
|-
|[[File:mlw115_Surface_Plot_HHF_labelled.png|300px]]||[[File:mlw115_Surface_Plot_HHF_H2Prod.png|300px]]|
|}

==Determining Transition State ==

Question 7. Locate the approximate position of the transition state.

A = HA B = HB and C = F

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! Surface Plot !! Internuclear distance vs Time
|-
|[[File:Mlw115_Surface_Plot_TS_HHF.png|400px]]||[[File:Mlw115 INDvT HHF TS.png|350px]]|
|}

== Calculating Activation Energy ==
Question 8. Report the activation energy for both reactions.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! BC= 1.8135!! BC= 1.8135 Zoomed!! BC= 1.813
|-
|[[File:Mlw115_HHF_EvT_1.8135.png|300px]] ‎||[[File:Mlw115_Plot_1.813_EvT_zoomT0_Resized.png|300px]]||[[File:Mlw115_HHF_EvT_1.813.png|320px]]
|}

pAB = pBC =0

distAB = 0.74

{| class="wikitable" border=1
|+Activation energy HF+H->H2+F
|-
! !! Energy (Kcal/mol)
|-
|Initial energy || -103.752
|-
|Final energy|| -133.89
|-
|EA|| 30.138
|}

{| class="wikitable" border=1
|+Activation energy H2+F->HF+H
|-
! !! Energy (Kcal/mol)
|-
|Initial energy|| -103.742
|-
|Final energy|| -103.752
|-
|EA|| 0.01
|}

==Reaction Dynamics==

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.74||0
|-
| BC|| 2.3||-2.7
|}

{| class="wikitable" border=1
!Surface Plot!! Internuclear momentum vs time
|-
|[[File:Mlw_Surface_Plot_HHF_Reactive.png]]|| [[File:Mlw_INMvT_HHF_Reactive.png]]
|}

{| class="wikitable" border=1
!Energy!!
|-
|Kinetic|| 3.837
|-
|Potential|| -103.886

|-
|Total|| -100.049
|}

Question 9. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?

It can be seen from the animation that as the H-H (AB) bond is broken and the H-F (BC) bond is formed the vibrational energy (and so the kinetic energy) considerably increases for H-F. The internuclear momentum vs time graph demonstrates this, as the momentum is much higher for the H-F atom. This vibrational energy will be released as heat so experimentally an increase in temperature could confirm the reaction had occurred.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -3 || -0.5 || -100.339|| No || [[File:Mlw115_Surface_Plot_HHF_-3_0.5.png|300px]] || H2 approaches the fluorine atom slowly but when a collision is made the H rebounds straight back into the other H and they continue their path oscillating together in the opposite direction
|-

| 3 || -0.5 || -93.254 || No || [[File:Mlw115_Surface_Plot_HHF_3_0.5.png|300px]] || This is similar to the previous case however the hydrogen rebounds twice into flourine but on the second rebounds remains returns back to oscillating with the original H atom
|-
| 2 || -0.5|| -98.753|| No || [[File: Mlw115_Surface_Plot_HHF_2_0.5.png|300px]] || The HF molecule oscillates within the energy well at high vibrational energy after collision
|-
| 1.5|| -0.5|| 103.754|| No || [[File: Mlw115_Surface_Plot_HHF_1.5_0.5.png|300px]] || Similarly to the previous situation
|-
|0.1||-0.8|| 103.364|| Yes|| [[File: Mlw115_Surface_Plot_HHF_0.1_0.8.png|300px]] || Despite having lower total energy than the previous cases the reaction goes to completion as the momentum between the reacting atoms is sufficient
|}

Despite the H2 energy being higher than the activation energy if the BC atoms do not collide with a high enough energy the reaction will return back over the peak and the original reactants will be produced.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 2
|-
|BC || 0.74
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -5 || 0.2 || -103.364 || No || [[File:Mlw115_Surface_Plot_HFH_5_0.2.png|300px]] || Approaching H rebounds and reaction doesn't complete
|-
| 2.21||0.37||
|}

A reactive pathway was found by trial and error around the reversed conditions from the reverse reaction.

AB D 5.0 M -1.5
BC D 1.2 M 1.0

[[File: Mlw115_Surface_Plot_HFH_-1.5_1.png| 400px]]

===Polanyi's Empirical Rules ===

Question 10. Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.

Polanyi's rules state that in reactions with an early transition state (most closely resembling the reactants) translational energy most effectively promotes a successful reaction. Conversely where the transition state is late (resembling products closely) vibrational energy, specifically of the reactants most effectively promotes a successful reaction. <ref>The Journal of Chemical Physics 138, 234104 (2013); https://doi.org/10.1063/1.4810007</ref>

=References=
<references/>

MRD:mlw115RXNDy

2018-05-22T14:56:06Z

Mlw115: /* Polanyi's Empirical Rules */

= Reaction Dynamics =

==Introduction==

The reactivity of atom-diatom systems are studied in order to gain insight on the internal degrees of freedom of the system. Particularly vibrational and translational energy are considered with respect to overcoming the transition state energy barrier.

The atoms are considered to behave classically and the interactions are expressed as potential energy surfaces.

= Exercise 1: H2 + H system=

==Dynamics from the transition state region ==

 Question 1. What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

Where a reaction is modeled on a potential energy surface the lowest energy path between the reactants and products is seen as a "valley floor" of the surface. The transition structure is a saddle point along the minimum energy ridge.

The gradient of the potential energy surface is zero at the minimum, along the direction of the reaction path and so the first derivative with respect to the inter-nuclear distance is zero. For the saddle point the gradient of the potential energy surface is zero orthogonal to the direction of the reaction pathway.

δV(ri)/δri = 0

Second derivatives
δV2(ri)/δri2 > 0 Minima
δV2(ri)/δri2 < 0 Maxima

where V is the potential energy and r related to the nuclear positions.

The minimum pathway and saddle point can be distinguished using the second derivatives with respect to atom positions. The second derivative of a maximum point is negative, whereas the second derivative of a minimum is positive. As the saddle point is a maxima along the minimum path the second derivative is negative. The second derivative of the minimum energy path in the direction of the reaction will be positive.

==Estimating Transition State Position ==

 Question 2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. 

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.908||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot 1 !! Surface Plot 2!! Internuclear distance vs Time
|-
|[[File:Mlw115_H2H_saddle.png|300px]]||[[File:mlw115_H2H_TS.png|300px]]||[[File:Mlw115_H2H_TS_INDvT.png|300px]]|
|}

The best guess of inter nuclear distances at the transition state was found using trial and error, to find the positions that would result in the minimum trajectory. As the system is symmetric the distances would be equal. The two surface plots are included to demonstrate the "reaction path". The internuclear distance vs time plot demonstrates no change in any bond lengths at the transition state so the overall momenta is 0 and potential energy is at the maximum.

== Reaction Path ==
===MEP vs Dynamics===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.909||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot (MEP) !! Surface Plot (Dynamics
|-
|[[File:Mlw115_H2H_MEP_surfaceplot.png|300px]]||[[File:mlw115_H2H_MEP_surfaceplot_dynamics.png|300px]]
|}

 Question 3. Comment on how the mep and the trajectory you just calculated differ.

It can be seen in the MEP plot that the trajectory follows the minimum energy valley floor without any obvious deviations or oscillation. The MEP trajectory is corresponding to infinitely slow motion, it does not accurately represent the atomic motion of the reaction. The trajectory oscillates in the dynamics plot, also in much less steps the AB bond distance increases considerably more. This plot is more realistic as it accounts for the motion of atoms during the reaction.

{| class="wikitable" border="1"
|+ Final Geometry
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| 2.486||1.167
|}

=== Reversed starting positions ===

What would change if we used the initial conditions r1 = rts and r2 = rts+0.01 instead? ****** DO THIS ******

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| -2.486||-1.167
|}

[[File:mlw115_H2H_dynamics_reversed_momenta.png|300px]]

From taking the previous runs final conditions and setting these as the initial conditions and reversing the momenta values the reaction seems to run in reverse, reaching an end exactly at the transition state.

===Determining Reactivity===

Question 4. Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2.0
|}

{| class="wikitable" border=1
! pAB !! pBC!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -1.25 || -2.5 || -99.018|| Yes || [[File:Mlw115_Surface_Plot_1.25_2.5.png|300px]] || Initially the HAHB distance changes very little as the vibrational energy is quite low, as the HC approaches it can be seen that the BC distance decreases linearly. The molecules pass over the transition state, after this point the HAHB distance decreases and the HBHC distance oscillates slightly, with a higher vibrational energy than the HAHB molecule. This overall process represents the HAHB bond breaking and forming a HBHC molecule.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mlw115_Surface_Plot_1.5_2.0.png|300px]] || The molecule approaches the maxima with some vibrational energy, shown by oscillations in the initial path, the energy of the system is not high enough to pass over the transition state energy maxima so the HC "rebounds" and the HBHC distance begins to increase. The HAHB molecule oscillations increase.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mlw115_Surface_Plot_1.5_2.5.png|300px]] || The reaction proceeds similarly to the first case with HC approaching and forming a bond with HB as there is sufficient energy to proceed over the transition state maxima. The vibrational energy throughout is slightly higher than the first case, demonstrated by higher oscillations in atom distances.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mlw115_Surface_Plot_2.5_5.png|300px]] || In this case the energy is high enough and the atoms are able to pass over the transition state energy maxima, however the vibrational energy of the HBHC system formed seems to be too high to form the molecule and the HC atom rebounds and the distance begins to decrease and so the HAHB is reformed but with considerably higher oscillations.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mlw115_Surface_Plot_2.5_5.2.png|300px]] || The HC approaches the oscillating HAHB system, the HB atom rebounds between the two atoms, as it can be seen on the surface plot the system passes over the transition state maxima, back over to the other side and back again as this occurs. Finally the HBHC bond is formed with a high vibrationally energy
|}

===Transition State Theory===

 Question 5 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? 

Transition state theory is used to understand how elementary chemical reactions take place assuming a quasi-equilibirum between reactanting complexes and those at the transition state. The rate of reaction can be understood by inspecting the complexes at the maximum saddle point of the minimum energy pathway. The reactants are in equilibrium with the activated transition state complexes and can be converted to the products.

This theory is useful to theoretically understand a chemical reaction however it operates on a series of assumptions which limit the theory. For instance the atoms are assumed to behave classically, ignoring tunelling. Tunneling would be more prominent in reactions with low activation barriers. Also the theory assumes that the saddle point with the lowest energy will be passed over to reach the products, however this is not the case when higher vibrational energies will be populated.

<ref>https://goldbook.iupac.org/html/T/T06470.html</ref>

=Exercise 2: H-H-F System=

==PES Inspection==
Question 6. 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?

For the forward reaction of H2 + F --> HF + H is exothermic, as shown in the surface plot below the energy of the reactants H2 is higher than the energy of the product HF. As the energy absorbed in breaking the H-H bond in the first case is less than the energy released forming the H-F bond which means that in total energy is released which results in an exothermic reaction.

{| class="wikitable" border=1
! Surface Plot H2+F->HF+H !! Surface Plot HF+H -> H2+F
|-
|[[File:mlw115_Surface_Plot_HHF_labelled.png|300px]]||[[File:mlw115_Surface_Plot_HHF_H2Prod.png|300px]]|
|}

==Determining Transition State ==

Question 7. Locate the approximate position of the transition state.

A = HA B = HB and C = F

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! Surface Plot !! Internuclear distance vs Time
|-
|[[File:Mlw115_Surface_Plot_TS_HHF.png|400px]]||[[File:Mlw115 INDvT HHF TS.png|350px]]|
|}

== Calculating Activation Energy ==
Question 8. Report the activation energy for both reactions.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! BC= 1.8135!! BC= 1.8135 Zoomed!! BC= 1.813
|-
|[[File:Mlw115_HHF_EvT_1.8135.png|300px]] ‎||[[File:Mlw115_Plot_1.813_EvT_zoomT0_Resized.png|300px]]||[[File:Mlw115_HHF_EvT_1.813.png|320px]]
|}

pAB = pBC =0

distAB = 0.74

{| class="wikitable" border=1
|+Activation energy HF+H->H2+F
|-
! !! Energy (Kcal/mol)
|-
|Initial energy || -103.752
|-
|Final energy|| -133.89
|-
|EA|| 30.138
|}

{| class="wikitable" border=1
|+Activation energy H2+F->HF+H
|-
! !! Energy (Kcal/mol)
|-
|Initial energy|| -103.742
|-
|Final energy|| -103.752
|-
|EA|| 0.01
|}

==Reaction Dynamics==

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.74||0
|-
| BC|| 2.3||-2.7
|}

{| class="wikitable" border=1
!Surface Plot!! Internuclear momentum vs time
|-
|[[File:Mlw_Surface_Plot_HHF_Reactive.png]]|| [[File:Mlw_INMvT_HHF_Reactive.png]]
|}

{| class="wikitable" border=1
!Energy!!
|-
|Kinetic|| 3.837
|-
|Potential|| -103.886

|-
|Total|| -100.049
|}

Question 9. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?

It can be seen from the animation that as the H-H (AB) bond is broken and the H-F (BC) bond is formed the vibrational energy (and so the kinetic energy) considerably increases for H-F. The internuclear momentum vs time graph demonstrates this, as the momentum is much higher for the H-F atom. This vibrational energy will be released as heat so experimentally an increase in temperature could confirm the reaction had occurred.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -3 || -0.5 || -100.339|| No || [[File:Mlw115_Surface_Plot_HHF_-3_0.5.png|300px]] || H2 approaches the fluorine atom slowly but when a collision is made the H rebounds straight back into the other H and they continue their path oscillating together in the opposite direction
|-

| 3 || -0.5 || -93.254 || No || [[File:Mlw115_Surface_Plot_HHF_3_0.5.png|300px]] || This is similar to the previous case however the hydrogen rebounds twice into flourine but on the second rebounds remains returns back to oscillating with the original H atom
|-
| 2 || -0.5|| -98.753|| No || [[File: Mlw115_Surface_Plot_HHF_2_0.5.png|300px]] || The HF molecule oscillates within the energy well at high vibrational energy after collision
|-
| 1.5|| -0.5|| 103.754|| No || [[File: Mlw115_Surface_Plot_HHF_1.5_0.5.png|300px]] || Similarly to the previous situation
|-
|0.1||-0.8|| 103.364|| Yes|| [[File: Mlw115_Surface_Plot_HHF_0.1_0.8.png|300px]] || Despite having lower total energy than the previous cases the reaction goes to completion as the momentum between the reacting atoms is sufficient
|}

Despite the H2 energy being higher than the activation energy if the BC atoms do not collide with a high enough energy the reaction will return back over the peak and the original reactants will be produced.

====H-F + H====

rAB= 2 rBC = 0.74
{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -5 || 0.2 || -103.364 || No || [[File:Mlw115_Surface_Plot_HFH_5_0.2.png|300px]] || Approaching H rebounds and reaction doesn't complete
|-
| 2.21||0.37||
|}

Reversal of conditions:
AB D 5.07 M -2.22
BC D 1.19 M 0.37

Random Guessing:
AB D 5.0 M -1.5
BC D 1.2 M 1.0

[[File: Mlw115_Surface_Plot_HFH_-1.5_1.png| 400px]]

===Polanyi's Empirical Rules ===

Question 10. Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.

Polanyi's rules state that in reactions with an early transition state (most closely resembling the reactants) translational energy most effectively promotes a successful reaction. Conversely where the transition state is late (resembling products closely) vibrational energy, specifically of the reactants most effectively promotes a successful reaction. <ref>The Journal of Chemical Physics 138, 234104 (2013); https://doi.org/10.1063/1.4810007</ref>

=References=
<references/>

MRD:mlw115RXNDy

2018-05-22T14:54:18Z

Mlw115: /* Polanyi's Empirical Rules */

= Reaction Dynamics =

==Introduction==

The reactivity of atom-diatom systems are studied in order to gain insight on the internal degrees of freedom of the system. Particularly vibrational and translational energy are considered with respect to overcoming the transition state energy barrier.

The atoms are considered to behave classically and the interactions are expressed as potential energy surfaces.

= Exercise 1: H2 + H system=

==Dynamics from the transition state region ==

 Question 1. What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

Where a reaction is modeled on a potential energy surface the lowest energy path between the reactants and products is seen as a "valley floor" of the surface. The transition structure is a saddle point along the minimum energy ridge.

The gradient of the potential energy surface is zero at the minimum, along the direction of the reaction path and so the first derivative with respect to the inter-nuclear distance is zero. For the saddle point the gradient of the potential energy surface is zero orthogonal to the direction of the reaction pathway.

δV(ri)/δri = 0

Second derivatives
δV2(ri)/δri2 > 0 Minima
δV2(ri)/δri2 < 0 Maxima

where V is the potential energy and r related to the nuclear positions.

The minimum pathway and saddle point can be distinguished using the second derivatives with respect to atom positions. The second derivative of a maximum point is negative, whereas the second derivative of a minimum is positive. As the saddle point is a maxima along the minimum path the second derivative is negative. The second derivative of the minimum energy path in the direction of the reaction will be positive.

==Estimating Transition State Position ==

 Question 2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. 

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.908||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot 1 !! Surface Plot 2!! Internuclear distance vs Time
|-
|[[File:Mlw115_H2H_saddle.png|300px]]||[[File:mlw115_H2H_TS.png|300px]]||[[File:Mlw115_H2H_TS_INDvT.png|300px]]|
|}

The best guess of inter nuclear distances at the transition state was found using trial and error, to find the positions that would result in the minimum trajectory. As the system is symmetric the distances would be equal. The two surface plots are included to demonstrate the "reaction path". The internuclear distance vs time plot demonstrates no change in any bond lengths at the transition state so the overall momenta is 0 and potential energy is at the maximum.

== Reaction Path ==
===MEP vs Dynamics===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.909||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot (MEP) !! Surface Plot (Dynamics
|-
|[[File:Mlw115_H2H_MEP_surfaceplot.png|300px]]||[[File:mlw115_H2H_MEP_surfaceplot_dynamics.png|300px]]
|}

 Question 3. Comment on how the mep and the trajectory you just calculated differ.

It can be seen in the MEP plot that the trajectory follows the minimum energy valley floor without any obvious deviations or oscillation. The MEP trajectory is corresponding to infinitely slow motion, it does not accurately represent the atomic motion of the reaction. The trajectory oscillates in the dynamics plot, also in much less steps the AB bond distance increases considerably more. This plot is more realistic as it accounts for the motion of atoms during the reaction.

{| class="wikitable" border="1"
|+ Final Geometry
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| 2.486||1.167
|}

=== Reversed starting positions ===

What would change if we used the initial conditions r1 = rts and r2 = rts+0.01 instead? ****** DO THIS ******

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| -2.486||-1.167
|}

[[File:mlw115_H2H_dynamics_reversed_momenta.png|300px]]

From taking the previous runs final conditions and setting these as the initial conditions and reversing the momenta values the reaction seems to run in reverse, reaching an end exactly at the transition state.

===Determining Reactivity===

Question 4. Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2.0
|}

{| class="wikitable" border=1
! pAB !! pBC!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -1.25 || -2.5 || -99.018|| Yes || [[File:Mlw115_Surface_Plot_1.25_2.5.png|300px]] || Initially the HAHB distance changes very little as the vibrational energy is quite low, as the HC approaches it can be seen that the BC distance decreases linearly. The molecules pass over the transition state, after this point the HAHB distance decreases and the HBHC distance oscillates slightly, with a higher vibrational energy than the HAHB molecule. This overall process represents the HAHB bond breaking and forming a HBHC molecule.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mlw115_Surface_Plot_1.5_2.0.png|300px]] || The molecule approaches the maxima with some vibrational energy, shown by oscillations in the initial path, the energy of the system is not high enough to pass over the transition state energy maxima so the HC "rebounds" and the HBHC distance begins to increase. The HAHB molecule oscillations increase.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mlw115_Surface_Plot_1.5_2.5.png|300px]] || The reaction proceeds similarly to the first case with HC approaching and forming a bond with HB as there is sufficient energy to proceed over the transition state maxima. The vibrational energy throughout is slightly higher than the first case, demonstrated by higher oscillations in atom distances.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mlw115_Surface_Plot_2.5_5.png|300px]] || In this case the energy is high enough and the atoms are able to pass over the transition state energy maxima, however the vibrational energy of the HBHC system formed seems to be too high to form the molecule and the HC atom rebounds and the distance begins to decrease and so the HAHB is reformed but with considerably higher oscillations.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mlw115_Surface_Plot_2.5_5.2.png|300px]] || The HC approaches the oscillating HAHB system, the HB atom rebounds between the two atoms, as it can be seen on the surface plot the system passes over the transition state maxima, back over to the other side and back again as this occurs. Finally the HBHC bond is formed with a high vibrationally energy
|}

===Transition State Theory===

 Question 5 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? 

Transition state theory is used to understand how elementary chemical reactions take place assuming a quasi-equilibirum between reactanting complexes and those at the transition state. The rate of reaction can be understood by inspecting the complexes at the maximum saddle point of the minimum energy pathway. The reactants are in equilibrium with the activated transition state complexes and can be converted to the products.

This theory is useful to theoretically understand a chemical reaction however it operates on a series of assumptions which limit the theory. For instance the atoms are assumed to behave classically, ignoring tunelling. Tunneling would be more prominent in reactions with low activation barriers. Also the theory assumes that the saddle point with the lowest energy will be passed over to reach the products, however this is not the case when higher vibrational energies will be populated.

<ref>https://goldbook.iupac.org/html/T/T06470.html</ref>

=Exercise 2: H-H-F System=

==PES Inspection==
Question 6. 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?

For the forward reaction of H2 + F --> HF + H is exothermic, as shown in the surface plot below the energy of the reactants H2 is higher than the energy of the product HF. As the energy absorbed in breaking the H-H bond in the first case is less than the energy released forming the H-F bond which means that in total energy is released which results in an exothermic reaction.

{| class="wikitable" border=1
! Surface Plot H2+F->HF+H !! Surface Plot HF+H -> H2+F
|-
|[[File:mlw115_Surface_Plot_HHF_labelled.png|300px]]||[[File:mlw115_Surface_Plot_HHF_H2Prod.png|300px]]|
|}

==Determining Transition State ==

Question 7. Locate the approximate position of the transition state.

A = HA B = HB and C = F

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! Surface Plot !! Internuclear distance vs Time
|-
|[[File:Mlw115_Surface_Plot_TS_HHF.png|400px]]||[[File:Mlw115 INDvT HHF TS.png|350px]]|
|}

== Calculating Activation Energy ==
Question 8. Report the activation energy for both reactions.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! BC= 1.8135!! BC= 1.8135 Zoomed!! BC= 1.813
|-
|[[File:Mlw115_HHF_EvT_1.8135.png|300px]] ‎||[[File:Mlw115_Plot_1.813_EvT_zoomT0_Resized.png|300px]]||[[File:Mlw115_HHF_EvT_1.813.png|320px]]
|}

pAB = pBC =0

distAB = 0.74

{| class="wikitable" border=1
|+Activation energy HF+H->H2+F
|-
! !! Energy (Kcal/mol)
|-
|Initial energy || -103.752
|-
|Final energy|| -133.89
|-
|EA|| 30.138
|}

{| class="wikitable" border=1
|+Activation energy H2+F->HF+H
|-
! !! Energy (Kcal/mol)
|-
|Initial energy|| -103.742
|-
|Final energy|| -103.752
|-
|EA|| 0.01
|}

==Reaction Dynamics==

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.74||0
|-
| BC|| 2.3||-2.7
|}

{| class="wikitable" border=1
!Surface Plot!! Internuclear momentum vs time
|-
|[[File:Mlw_Surface_Plot_HHF_Reactive.png]]|| [[File:Mlw_INMvT_HHF_Reactive.png]]
|}

{| class="wikitable" border=1
!Energy!!
|-
|Kinetic|| 3.837
|-
|Potential|| -103.886

|-
|Total|| -100.049
|}

Question 9. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?

It can be seen from the animation that as the H-H (AB) bond is broken and the H-F (BC) bond is formed the vibrational energy (and so the kinetic energy) considerably increases for H-F. The internuclear momentum vs time graph demonstrates this, as the momentum is much higher for the H-F atom. This vibrational energy will be released as heat so experimentally an increase in temperature could confirm the reaction had occurred.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -3 || -0.5 || -100.339|| No || [[File:Mlw115_Surface_Plot_HHF_-3_0.5.png|300px]] || H2 approaches the fluorine atom slowly but when a collision is made the H rebounds straight back into the other H and they continue their path oscillating together in the opposite direction
|-

| 3 || -0.5 || -93.254 || No || [[File:Mlw115_Surface_Plot_HHF_3_0.5.png|300px]] || This is similar to the previous case however the hydrogen rebounds twice into flourine but on the second rebounds remains returns back to oscillating with the original H atom
|-
| 2 || -0.5|| -98.753|| No || [[File: Mlw115_Surface_Plot_HHF_2_0.5.png|300px]] || The HF molecule oscillates within the energy well at high vibrational energy after collision
|-
| 1.5|| -0.5|| 103.754|| No || [[File: Mlw115_Surface_Plot_HHF_1.5_0.5.png|300px]] || Similarly to the previous situation
|-
|0.1||-0.8|| 103.364|| Yes|| [[File: Mlw115_Surface_Plot_HHF_0.1_0.8.png|300px]] || Despite having lower total energy than the previous cases the reaction goes to completion as the momentum between the reacting atoms is sufficient
|}

Despite the H2 energy being higher than the activation energy if the BC atoms do not collide with a high enough energy the reaction will return back over the peak and the original reactants will be produced.

===Polanyi's Empirical Rules ===

Question 10. Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.

Polanyi's rules state that in reactions with an early transition state (most closely resembling the reactants) translational energy most effectively promotes a successful reaction. Conversely where the transition state is late (resembling products closely) vibrational energy, specifically of the reactants most effectively promotes a successful reaction. <ref>The Journal of Chemical Physics 138, 234104 (2013); https://doi.org/10.1063/1.4810007</ref>

====H-F + H====
rAB= 2 rBC = 0.74
{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -5 || 0.2 || -103.364 || No || [[File:Mlw115_Surface_Plot_HFH_5_0.2.png|300px]] || Approaching H rebounds and reaction doesn't complete
|-
| 2.21||0.37||
|}

Reversal of conditions:
AB D 5.07 M -2.22
BC D 1.19 M 0.37

Random Guessing:
AB D 5.0 M -1.5
BC D 1.2 M 1.0

[[File: Mlw115_Surface_Plot_HFH_-1.5_1.png| 400px]]

=References=
<references/>

MRD:mlw115RXNDy

2018-05-22T14:38:23Z

Mlw115: /* Polanyi's Empirical Rules */

= Reaction Dynamics =

==Introduction==

The reactivity of atom-diatom systems are studied in order to gain insight on the internal degrees of freedom of the system. Particularly vibrational and translational energy are considered with respect to overcoming the transition state energy barrier.

The atoms are considered to behave classically and the interactions are expressed as potential energy surfaces.

= Exercise 1: H2 + H system=

==Dynamics from the transition state region ==

 Question 1. What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

Where a reaction is modeled on a potential energy surface the lowest energy path between the reactants and products is seen as a "valley floor" of the surface. The transition structure is a saddle point along the minimum energy ridge.

The gradient of the potential energy surface is zero at the minimum, along the direction of the reaction path and so the first derivative with respect to the inter-nuclear distance is zero. For the saddle point the gradient of the potential energy surface is zero orthogonal to the direction of the reaction pathway.

δV(ri)/δri = 0

Second derivatives
δV2(ri)/δri2 > 0 Minima
δV2(ri)/δri2 < 0 Maxima

where V is the potential energy and r related to the nuclear positions.

The minimum pathway and saddle point can be distinguished using the second derivatives with respect to atom positions. The second derivative of a maximum point is negative, whereas the second derivative of a minimum is positive. As the saddle point is a maxima along the minimum path the second derivative is negative. The second derivative of the minimum energy path in the direction of the reaction will be positive.

==Estimating Transition State Position ==

 Question 2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. 

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.908||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot 1 !! Surface Plot 2!! Internuclear distance vs Time
|-
|[[File:Mlw115_H2H_saddle.png|300px]]||[[File:mlw115_H2H_TS.png|300px]]||[[File:Mlw115_H2H_TS_INDvT.png|300px]]|
|}

The best guess of inter nuclear distances at the transition state was found using trial and error, to find the positions that would result in the minimum trajectory. As the system is symmetric the distances would be equal. The two surface plots are included to demonstrate the "reaction path". The internuclear distance vs time plot demonstrates no change in any bond lengths at the transition state so the overall momenta is 0 and potential energy is at the maximum.

== Reaction Path ==
===MEP vs Dynamics===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.909||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot (MEP) !! Surface Plot (Dynamics
|-
|[[File:Mlw115_H2H_MEP_surfaceplot.png|300px]]||[[File:mlw115_H2H_MEP_surfaceplot_dynamics.png|300px]]
|}

 Question 3. Comment on how the mep and the trajectory you just calculated differ.

It can be seen in the MEP plot that the trajectory follows the minimum energy valley floor without any obvious deviations or oscillation. The MEP trajectory is corresponding to infinitely slow motion, it does not accurately represent the atomic motion of the reaction. The trajectory oscillates in the dynamics plot, also in much less steps the AB bond distance increases considerably more. This plot is more realistic as it accounts for the motion of atoms during the reaction.

{| class="wikitable" border="1"
|+ Final Geometry
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| 2.486||1.167
|}

=== Reversed starting positions ===

What would change if we used the initial conditions r1 = rts and r2 = rts+0.01 instead? ****** DO THIS ******

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| -2.486||-1.167
|}

[[File:mlw115_H2H_dynamics_reversed_momenta.png|300px]]

From taking the previous runs final conditions and setting these as the initial conditions and reversing the momenta values the reaction seems to run in reverse, reaching an end exactly at the transition state.

===Determining Reactivity===

Question 4. Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2.0
|}

{| class="wikitable" border=1
! pAB !! pBC!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -1.25 || -2.5 || -99.018|| Yes || [[File:Mlw115_Surface_Plot_1.25_2.5.png|300px]] || Initially the HAHB distance changes very little as the vibrational energy is quite low, as the HC approaches it can be seen that the BC distance decreases linearly. The molecules pass over the transition state, after this point the HAHB distance decreases and the HBHC distance oscillates slightly, with a higher vibrational energy than the HAHB molecule. This overall process represents the HAHB bond breaking and forming a HBHC molecule.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mlw115_Surface_Plot_1.5_2.0.png|300px]] || The molecule approaches the maxima with some vibrational energy, shown by oscillations in the initial path, the energy of the system is not high enough to pass over the transition state energy maxima so the HC "rebounds" and the HBHC distance begins to increase. The HAHB molecule oscillations increase.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mlw115_Surface_Plot_1.5_2.5.png|300px]] || The reaction proceeds similarly to the first case with HC approaching and forming a bond with HB as there is sufficient energy to proceed over the transition state maxima. The vibrational energy throughout is slightly higher than the first case, demonstrated by higher oscillations in atom distances.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mlw115_Surface_Plot_2.5_5.png|300px]] || In this case the energy is high enough and the atoms are able to pass over the transition state energy maxima, however the vibrational energy of the HBHC system formed seems to be too high to form the molecule and the HC atom rebounds and the distance begins to decrease and so the HAHB is reformed but with considerably higher oscillations.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mlw115_Surface_Plot_2.5_5.2.png|300px]] || The HC approaches the oscillating HAHB system, the HB atom rebounds between the two atoms, as it can be seen on the surface plot the system passes over the transition state maxima, back over to the other side and back again as this occurs. Finally the HBHC bond is formed with a high vibrationally energy
|}

===Transition State Theory===

 Question 5 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? 

Transition state theory is used to understand how elementary chemical reactions take place assuming a quasi-equilibirum between reactanting complexes and those at the transition state. The rate of reaction can be understood by inspecting the complexes at the maximum saddle point of the minimum energy pathway. The reactants are in equilibrium with the activated transition state complexes and can be converted to the products.

This theory is useful to theoretically understand a chemical reaction however it operates on a series of assumptions which limit the theory. For instance the atoms are assumed to behave classically, ignoring tunelling. Tunneling would be more prominent in reactions with low activation barriers. Also the theory assumes that the saddle point with the lowest energy will be passed over to reach the products, however this is not the case when higher vibrational energies will be populated.

<ref>https://goldbook.iupac.org/html/T/T06470.html</ref>

=Exercise 2: H-H-F System=

==PES Inspection==
Question 6. 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?

For the forward reaction of H2 + F --> HF + H is exothermic, as shown in the surface plot below the energy of the reactants H2 is higher than the energy of the product HF. As the energy absorbed in breaking the H-H bond in the first case is less than the energy released forming the H-F bond which means that in total energy is released which results in an exothermic reaction.

{| class="wikitable" border=1
! Surface Plot H2+F->HF+H !! Surface Plot HF+H -> H2+F
|-
|[[File:mlw115_Surface_Plot_HHF_labelled.png|300px]]||[[File:mlw115_Surface_Plot_HHF_H2Prod.png|300px]]|
|}

==Determining Transition State ==

Question 7. Locate the approximate position of the transition state.

A = HA B = HB and C = F

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! Surface Plot !! Internuclear distance vs Time
|-
|[[File:Mlw115_Surface_Plot_TS_HHF.png|400px]]||[[File:Mlw115 INDvT HHF TS.png|350px]]|
|}

== Calculating Activation Energy ==
Question 8. Report the activation energy for both reactions.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! BC= 1.8135!! BC= 1.8135 Zoomed!! BC= 1.813
|-
|[[File:Mlw115_HHF_EvT_1.8135.png|300px]] ‎||[[File:Mlw115_Plot_1.813_EvT_zoomT0_Resized.png|300px]]||[[File:Mlw115_HHF_EvT_1.813.png|320px]]
|}

pAB = pBC =0

distAB = 0.74

{| class="wikitable" border=1
|+Activation energy HF+H->H2+F
|-
! !! Energy (Kcal/mol)
|-
|Initial energy || -103.752
|-
|Final energy|| -133.89
|-
|EA|| 30.138
|}

{| class="wikitable" border=1
|+Activation energy H2+F->HF+H
|-
! !! Energy (Kcal/mol)
|-
|Initial energy|| -103.742
|-
|Final energy|| -103.752
|-
|EA|| 0.01
|}

==Reaction Dynamics==

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.74||0
|-
| BC|| 2.3||-2.7
|}

{| class="wikitable" border=1
!Surface Plot!! Internuclear momentum vs time
|-
|[[File:Mlw_Surface_Plot_HHF_Reactive.png]]|| [[File:Mlw_INMvT_HHF_Reactive.png]]
|}

{| class="wikitable" border=1
!Energy!!
|-
|Kinetic|| 3.837
|-
|Potential|| -103.886

|-
|Total|| -100.049
|}

Question 9. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?

It can be seen from the animation that as the H-H (AB) bond is broken and the H-F (BC) bond is formed the vibrational energy (and so the kinetic energy) considerably increases for H-F. The internuclear momentum vs time graph demonstrates this, as the momentum is much higher for the H-F atom. This vibrational energy will be released as heat so experimentally an increase in temperature could confirm the reaction had occurred.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -3 || -0.5 || -100.339|| No || [[File:Mlw115_Surface_Plot_HHF_-3_0.5.png|300px]] || H2 approaches the fluorine atom slowly but when a collision is made the H rebounds straight back into the other H and they continue their path oscillating together in the opposite direction
|-

| 3 || -0.5 || -93.254 || No || [[File:Mlw115_Surface_Plot_HHF_3_0.5.png|300px]] || This is similar to the previous case however the hydrogen rebounds twice into flourine but on the second rebounds remains returns back to oscillating with the original H atom
|-
| 2 || -0.5|| -98.753|| No || [[File: Mlw115_Surface_Plot_HHF_2_0.5.png|300px]] || The HF molecule oscillates within the energy well at high vibrational energy after collision
|-
| 1.5|| -0.5|| 103.754|| No || [[File: Mlw115_Surface_Plot_HHF_1.5_0.5.png|300px]] || Similarly to the previous situation
|-
|0.1||-0.8|| 103.364|| Yes|| [[File: Mlw115_Surface_Plot_HHF_0.1_0.8.png|300px]] || Despite having lower total energy than the previous cases the reaction goes to completion as the momentum between the reacting atoms is sufficient
|}

Despite the H2 energy being higher than the activation energy if the BC atoms do not collide with a high enough energy the reaction will return back over the peak and the original reactants will be produced.

===Polanyi's Empirical Rules ===

Question 10Discuss 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.

Polanyi's rules state that in reactions with an early transition state (most closely resembling the reactants) translational energy most effectively promotes a successful reaction. Conversely where the transition state is late (resembling products closely) vibrational energy, specifically of the reactants most effectively promotes a successful reaction. <ref>The Journal of Chemical Physics 138, 234104 (2013); https://doi.org/10.1063/1.4810007</ref>

====H-F + H====
rAB= 2 rBC = 0.74
{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -5 || 0.2 || -103.364 || No || [[File:Mlw115_Surface_Plot_HFH_5_0.2.png|300px]] || Approaching H rebounds and reaction doesn't complete
|-
| 2.21||0.37||
|}

Reversal of conditions:
AB D 5.07 M -2.22
BC D 1.19 M 0.37

Random Guessing:
AB D 5.0 M -1.5
BC D 1.2 M 1.0

[[File: Mlw115_Surface_Plot_HFH_-1.5_1.png| 400px]]

=References=
<references/>

MRD:mlw115RXNDy

2018-05-22T14:37:41Z

Mlw115: /* Reaction Dynamics */

= Reaction Dynamics =

==Introduction==

The reactivity of atom-diatom systems are studied in order to gain insight on the internal degrees of freedom of the system. Particularly vibrational and translational energy are considered with respect to overcoming the transition state energy barrier.

The atoms are considered to behave classically and the interactions are expressed as potential energy surfaces.

= Exercise 1: H2 + H system=

==Dynamics from the transition state region ==

 Question 1. What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

Where a reaction is modeled on a potential energy surface the lowest energy path between the reactants and products is seen as a "valley floor" of the surface. The transition structure is a saddle point along the minimum energy ridge.

The gradient of the potential energy surface is zero at the minimum, along the direction of the reaction path and so the first derivative with respect to the inter-nuclear distance is zero. For the saddle point the gradient of the potential energy surface is zero orthogonal to the direction of the reaction pathway.

δV(ri)/δri = 0

Second derivatives
δV2(ri)/δri2 > 0 Minima
δV2(ri)/δri2 < 0 Maxima

where V is the potential energy and r related to the nuclear positions.

The minimum pathway and saddle point can be distinguished using the second derivatives with respect to atom positions. The second derivative of a maximum point is negative, whereas the second derivative of a minimum is positive. As the saddle point is a maxima along the minimum path the second derivative is negative. The second derivative of the minimum energy path in the direction of the reaction will be positive.

==Estimating Transition State Position ==

 Question 2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. 

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.908||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot 1 !! Surface Plot 2!! Internuclear distance vs Time
|-
|[[File:Mlw115_H2H_saddle.png|300px]]||[[File:mlw115_H2H_TS.png|300px]]||[[File:Mlw115_H2H_TS_INDvT.png|300px]]|
|}

The best guess of inter nuclear distances at the transition state was found using trial and error, to find the positions that would result in the minimum trajectory. As the system is symmetric the distances would be equal. The two surface plots are included to demonstrate the "reaction path". The internuclear distance vs time plot demonstrates no change in any bond lengths at the transition state so the overall momenta is 0 and potential energy is at the maximum.

== Reaction Path ==
===MEP vs Dynamics===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.909||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot (MEP) !! Surface Plot (Dynamics
|-
|[[File:Mlw115_H2H_MEP_surfaceplot.png|300px]]||[[File:mlw115_H2H_MEP_surfaceplot_dynamics.png|300px]]
|}

 Question 3. Comment on how the mep and the trajectory you just calculated differ.

It can be seen in the MEP plot that the trajectory follows the minimum energy valley floor without any obvious deviations or oscillation. The MEP trajectory is corresponding to infinitely slow motion, it does not accurately represent the atomic motion of the reaction. The trajectory oscillates in the dynamics plot, also in much less steps the AB bond distance increases considerably more. This plot is more realistic as it accounts for the motion of atoms during the reaction.

{| class="wikitable" border="1"
|+ Final Geometry
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| 2.486||1.167
|}

=== Reversed starting positions ===

What would change if we used the initial conditions r1 = rts and r2 = rts+0.01 instead? ****** DO THIS ******

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| -2.486||-1.167
|}

[[File:mlw115_H2H_dynamics_reversed_momenta.png|300px]]

From taking the previous runs final conditions and setting these as the initial conditions and reversing the momenta values the reaction seems to run in reverse, reaching an end exactly at the transition state.

===Determining Reactivity===

Question 4. Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2.0
|}

{| class="wikitable" border=1
! pAB !! pBC!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -1.25 || -2.5 || -99.018|| Yes || [[File:Mlw115_Surface_Plot_1.25_2.5.png|300px]] || Initially the HAHB distance changes very little as the vibrational energy is quite low, as the HC approaches it can be seen that the BC distance decreases linearly. The molecules pass over the transition state, after this point the HAHB distance decreases and the HBHC distance oscillates slightly, with a higher vibrational energy than the HAHB molecule. This overall process represents the HAHB bond breaking and forming a HBHC molecule.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mlw115_Surface_Plot_1.5_2.0.png|300px]] || The molecule approaches the maxima with some vibrational energy, shown by oscillations in the initial path, the energy of the system is not high enough to pass over the transition state energy maxima so the HC "rebounds" and the HBHC distance begins to increase. The HAHB molecule oscillations increase.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mlw115_Surface_Plot_1.5_2.5.png|300px]] || The reaction proceeds similarly to the first case with HC approaching and forming a bond with HB as there is sufficient energy to proceed over the transition state maxima. The vibrational energy throughout is slightly higher than the first case, demonstrated by higher oscillations in atom distances.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mlw115_Surface_Plot_2.5_5.png|300px]] || In this case the energy is high enough and the atoms are able to pass over the transition state energy maxima, however the vibrational energy of the HBHC system formed seems to be too high to form the molecule and the HC atom rebounds and the distance begins to decrease and so the HAHB is reformed but with considerably higher oscillations.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mlw115_Surface_Plot_2.5_5.2.png|300px]] || The HC approaches the oscillating HAHB system, the HB atom rebounds between the two atoms, as it can be seen on the surface plot the system passes over the transition state maxima, back over to the other side and back again as this occurs. Finally the HBHC bond is formed with a high vibrationally energy
|}

===Transition State Theory===

 Question 5 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? 

Transition state theory is used to understand how elementary chemical reactions take place assuming a quasi-equilibirum between reactanting complexes and those at the transition state. The rate of reaction can be understood by inspecting the complexes at the maximum saddle point of the minimum energy pathway. The reactants are in equilibrium with the activated transition state complexes and can be converted to the products.

This theory is useful to theoretically understand a chemical reaction however it operates on a series of assumptions which limit the theory. For instance the atoms are assumed to behave classically, ignoring tunelling. Tunneling would be more prominent in reactions with low activation barriers. Also the theory assumes that the saddle point with the lowest energy will be passed over to reach the products, however this is not the case when higher vibrational energies will be populated.

<ref>https://goldbook.iupac.org/html/T/T06470.html</ref>

=Exercise 2: H-H-F System=

==PES Inspection==
Question 6. 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?

For the forward reaction of H2 + F --> HF + H is exothermic, as shown in the surface plot below the energy of the reactants H2 is higher than the energy of the product HF. As the energy absorbed in breaking the H-H bond in the first case is less than the energy released forming the H-F bond which means that in total energy is released which results in an exothermic reaction.

{| class="wikitable" border=1
! Surface Plot H2+F->HF+H !! Surface Plot HF+H -> H2+F
|-
|[[File:mlw115_Surface_Plot_HHF_labelled.png|300px]]||[[File:mlw115_Surface_Plot_HHF_H2Prod.png|300px]]|
|}

==Determining Transition State ==

Question 7. Locate the approximate position of the transition state.

A = HA B = HB and C = F

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! Surface Plot !! Internuclear distance vs Time
|-
|[[File:Mlw115_Surface_Plot_TS_HHF.png|400px]]||[[File:Mlw115 INDvT HHF TS.png|350px]]|
|}

== Calculating Activation Energy ==
Question 8. Report the activation energy for both reactions.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! BC= 1.8135!! BC= 1.8135 Zoomed!! BC= 1.813
|-
|[[File:Mlw115_HHF_EvT_1.8135.png|300px]] ‎||[[File:Mlw115_Plot_1.813_EvT_zoomT0_Resized.png|300px]]||[[File:Mlw115_HHF_EvT_1.813.png|320px]]
|}

pAB = pBC =0

distAB = 0.74

{| class="wikitable" border=1
|+Activation energy HF+H->H2+F
|-
! !! Energy (Kcal/mol)
|-
|Initial energy || -103.752
|-
|Final energy|| -133.89
|-
|EA|| 30.138
|}

{| class="wikitable" border=1
|+Activation energy H2+F->HF+H
|-
! !! Energy (Kcal/mol)
|-
|Initial energy|| -103.742
|-
|Final energy|| -103.752
|-
|EA|| 0.01
|}

==Reaction Dynamics==

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.74||0
|-
| BC|| 2.3||-2.7
|}

{| class="wikitable" border=1
!Surface Plot!! Internuclear momentum vs time
|-
|[[File:Mlw_Surface_Plot_HHF_Reactive.png]]|| [[File:Mlw_INMvT_HHF_Reactive.png]]
|}

{| class="wikitable" border=1
!Energy!!
|-
|Kinetic|| 3.837
|-
|Potential|| -103.886

|-
|Total|| -100.049
|}

Question 9. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?

It can be seen from the animation that as the H-H (AB) bond is broken and the H-F (BC) bond is formed the vibrational energy (and so the kinetic energy) considerably increases for H-F. The internuclear momentum vs time graph demonstrates this, as the momentum is much higher for the H-F atom. This vibrational energy will be released as heat so experimentally an increase in temperature could confirm the reaction had occurred.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -3 || -0.5 || -100.339|| No || [[File:Mlw115_Surface_Plot_HHF_-3_0.5.png|300px]] || H2 approaches the fluorine atom slowly but when a collision is made the H rebounds straight back into the other H and they continue their path oscillating together in the opposite direction
|-

| 3 || -0.5 || -93.254 || No || [[File:Mlw115_Surface_Plot_HHF_3_0.5.png|300px]] || This is similar to the previous case however the hydrogen rebounds twice into flourine but on the second rebounds remains returns back to oscillating with the original H atom
|-
| 2 || -0.5|| -98.753|| No || [[File: Mlw115_Surface_Plot_HHF_2_0.5.png|300px]] || The HF molecule oscillates within the energy well at high vibrational energy after collision
|-
| 1.5|| -0.5|| 103.754|| No || [[File: Mlw115_Surface_Plot_HHF_1.5_0.5.png|300px]] || Similarly to the previous situation
|-
|0.1||-0.8|| 103.364|| Yes|| [[File: Mlw115_Surface_Plot_HHF_0.1_0.8.png|300px]] || Despite having lower total energy than the previous cases the reaction goes to completion as the momentum between the reacting atoms is sufficient
|}

Despite the H2 energy being higher than the activation energy if the BC atoms do not collide with a high enough energy the reaction will return back over the peak and the original reactants will be produced.

===Polanyi's Empirical Rules ===

Polanyi's rules state that in reactions with an early transition state (most closely resembling the reactants) translational energy most effectively promotes a successful reaction. Conversely where the transition state is late (resembling products closely) vibrational energy, specifically of the reactants most effectively promotes a successful reaction. <ref>The Journal of Chemical Physics 138, 234104 (2013); https://doi.org/10.1063/1.4810007</ref>

====H-F + H====
rAB= 2 rBC = 0.74
{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -5 || 0.2 || -103.364 || No || [[File:Mlw115_Surface_Plot_HFH_5_0.2.png|300px]] || Approaching H rebounds and reaction doesn't complete
|-
| 2.21||0.37||
|}

Reversal of conditions:
AB D 5.07 M -2.22
BC D 1.19 M 0.37

Random Guessing:
AB D 5.0 M -1.5
BC D 1.2 M 1.0

[[File: Mlw115_Surface_Plot_HFH_-1.5_1.png| 400px]]

=References=
<references/>

MRD:mlw115RXNDy

2018-05-22T14:36:56Z

Mlw115: /* Calculating Activation Energy */

= Reaction Dynamics =

==Introduction==

The reactivity of atom-diatom systems are studied in order to gain insight on the internal degrees of freedom of the system. Particularly vibrational and translational energy are considered with respect to overcoming the transition state energy barrier.

The atoms are considered to behave classically and the interactions are expressed as potential energy surfaces.

= Exercise 1: H2 + H system=

==Dynamics from the transition state region ==

 Question 1. What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

Where a reaction is modeled on a potential energy surface the lowest energy path between the reactants and products is seen as a "valley floor" of the surface. The transition structure is a saddle point along the minimum energy ridge.

The gradient of the potential energy surface is zero at the minimum, along the direction of the reaction path and so the first derivative with respect to the inter-nuclear distance is zero. For the saddle point the gradient of the potential energy surface is zero orthogonal to the direction of the reaction pathway.

δV(ri)/δri = 0

Second derivatives
δV2(ri)/δri2 > 0 Minima
δV2(ri)/δri2 < 0 Maxima

where V is the potential energy and r related to the nuclear positions.

The minimum pathway and saddle point can be distinguished using the second derivatives with respect to atom positions. The second derivative of a maximum point is negative, whereas the second derivative of a minimum is positive. As the saddle point is a maxima along the minimum path the second derivative is negative. The second derivative of the minimum energy path in the direction of the reaction will be positive.

==Estimating Transition State Position ==

 Question 2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. 

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.908||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot 1 !! Surface Plot 2!! Internuclear distance vs Time
|-
|[[File:Mlw115_H2H_saddle.png|300px]]||[[File:mlw115_H2H_TS.png|300px]]||[[File:Mlw115_H2H_TS_INDvT.png|300px]]|
|}

The best guess of inter nuclear distances at the transition state was found using trial and error, to find the positions that would result in the minimum trajectory. As the system is symmetric the distances would be equal. The two surface plots are included to demonstrate the "reaction path". The internuclear distance vs time plot demonstrates no change in any bond lengths at the transition state so the overall momenta is 0 and potential energy is at the maximum.

== Reaction Path ==
===MEP vs Dynamics===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.909||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot (MEP) !! Surface Plot (Dynamics
|-
|[[File:Mlw115_H2H_MEP_surfaceplot.png|300px]]||[[File:mlw115_H2H_MEP_surfaceplot_dynamics.png|300px]]
|}

 Question 3. Comment on how the mep and the trajectory you just calculated differ.

It can be seen in the MEP plot that the trajectory follows the minimum energy valley floor without any obvious deviations or oscillation. The MEP trajectory is corresponding to infinitely slow motion, it does not accurately represent the atomic motion of the reaction. The trajectory oscillates in the dynamics plot, also in much less steps the AB bond distance increases considerably more. This plot is more realistic as it accounts for the motion of atoms during the reaction.

{| class="wikitable" border="1"
|+ Final Geometry
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| 2.486||1.167
|}

=== Reversed starting positions ===

What would change if we used the initial conditions r1 = rts and r2 = rts+0.01 instead? ****** DO THIS ******

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| -2.486||-1.167
|}

[[File:mlw115_H2H_dynamics_reversed_momenta.png|300px]]

From taking the previous runs final conditions and setting these as the initial conditions and reversing the momenta values the reaction seems to run in reverse, reaching an end exactly at the transition state.

===Determining Reactivity===

Question 4. Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2.0
|}

{| class="wikitable" border=1
! pAB !! pBC!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -1.25 || -2.5 || -99.018|| Yes || [[File:Mlw115_Surface_Plot_1.25_2.5.png|300px]] || Initially the HAHB distance changes very little as the vibrational energy is quite low, as the HC approaches it can be seen that the BC distance decreases linearly. The molecules pass over the transition state, after this point the HAHB distance decreases and the HBHC distance oscillates slightly, with a higher vibrational energy than the HAHB molecule. This overall process represents the HAHB bond breaking and forming a HBHC molecule.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mlw115_Surface_Plot_1.5_2.0.png|300px]] || The molecule approaches the maxima with some vibrational energy, shown by oscillations in the initial path, the energy of the system is not high enough to pass over the transition state energy maxima so the HC "rebounds" and the HBHC distance begins to increase. The HAHB molecule oscillations increase.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mlw115_Surface_Plot_1.5_2.5.png|300px]] || The reaction proceeds similarly to the first case with HC approaching and forming a bond with HB as there is sufficient energy to proceed over the transition state maxima. The vibrational energy throughout is slightly higher than the first case, demonstrated by higher oscillations in atom distances.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mlw115_Surface_Plot_2.5_5.png|300px]] || In this case the energy is high enough and the atoms are able to pass over the transition state energy maxima, however the vibrational energy of the HBHC system formed seems to be too high to form the molecule and the HC atom rebounds and the distance begins to decrease and so the HAHB is reformed but with considerably higher oscillations.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mlw115_Surface_Plot_2.5_5.2.png|300px]] || The HC approaches the oscillating HAHB system, the HB atom rebounds between the two atoms, as it can be seen on the surface plot the system passes over the transition state maxima, back over to the other side and back again as this occurs. Finally the HBHC bond is formed with a high vibrationally energy
|}

===Transition State Theory===

 Question 5 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? 

Transition state theory is used to understand how elementary chemical reactions take place assuming a quasi-equilibirum between reactanting complexes and those at the transition state. The rate of reaction can be understood by inspecting the complexes at the maximum saddle point of the minimum energy pathway. The reactants are in equilibrium with the activated transition state complexes and can be converted to the products.

This theory is useful to theoretically understand a chemical reaction however it operates on a series of assumptions which limit the theory. For instance the atoms are assumed to behave classically, ignoring tunelling. Tunneling would be more prominent in reactions with low activation barriers. Also the theory assumes that the saddle point with the lowest energy will be passed over to reach the products, however this is not the case when higher vibrational energies will be populated.

<ref>https://goldbook.iupac.org/html/T/T06470.html</ref>

=Exercise 2: H-H-F System=

==PES Inspection==
Question 6. 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?

For the forward reaction of H2 + F --> HF + H is exothermic, as shown in the surface plot below the energy of the reactants H2 is higher than the energy of the product HF. As the energy absorbed in breaking the H-H bond in the first case is less than the energy released forming the H-F bond which means that in total energy is released which results in an exothermic reaction.

{| class="wikitable" border=1
! Surface Plot H2+F->HF+H !! Surface Plot HF+H -> H2+F
|-
|[[File:mlw115_Surface_Plot_HHF_labelled.png|300px]]||[[File:mlw115_Surface_Plot_HHF_H2Prod.png|300px]]|
|}

==Determining Transition State ==

Question 7. Locate the approximate position of the transition state.

A = HA B = HB and C = F

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! Surface Plot !! Internuclear distance vs Time
|-
|[[File:Mlw115_Surface_Plot_TS_HHF.png|400px]]||[[File:Mlw115 INDvT HHF TS.png|350px]]|
|}

== Calculating Activation Energy ==
Question 8. Report the activation energy for both reactions.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! BC= 1.8135!! BC= 1.8135 Zoomed!! BC= 1.813
|-
|[[File:Mlw115_HHF_EvT_1.8135.png|300px]] ‎||[[File:Mlw115_Plot_1.813_EvT_zoomT0_Resized.png|300px]]||[[File:Mlw115_HHF_EvT_1.813.png|320px]]
|}

pAB = pBC =0

distAB = 0.74

{| class="wikitable" border=1
|+Activation energy HF+H->H2+F
|-
! !! Energy (Kcal/mol)
|-
|Initial energy || -103.752
|-
|Final energy|| -133.89
|-
|EA|| 30.138
|}

{| class="wikitable" border=1
|+Activation energy H2+F->HF+H
|-
! !! Energy (Kcal/mol)
|-
|Initial energy|| -103.742
|-
|Final energy|| -103.752
|-
|EA|| 0.01
|}

==Reaction Dynamics==

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.74||0
|-
| BC|| 2.3||-2.7
|}

{| class="wikitable" border=1
!Surface Plot!! Internuclear momentum vs time
|-
|[[File:Mlw_Surface_Plot_HHF_Reactive.png]]|| [[File:Mlw_INMvT_HHF_Reactive.png]]
|}

{| class="wikitable" border=1
!Energy!!
|-
|Kinetic|| 3.837
|-
|Potential|| -103.886

|-
|Total|| -100.049
|}

It can be seen from the animation that as the H-H (AB) bond is broken and the H-F (BC) bond is formed the vibrational energy (and so the kinetic energy) considerably increases for H-F. The internuclear momentum vs time graph demonstrates this, as the momentum is much higher for the H-F atom. This vibrational energy will be released as heat so experimentally an increase in temperature could confirm the reaction had occurred.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -3 || -0.5 || -100.339|| No || [[File:Mlw115_Surface_Plot_HHF_-3_0.5.png|300px]] || H2 approaches the fluorine atom slowly but when a collision is made the H rebounds straight back into the other H and they continue their path oscillating together in the opposite direction
|-

| 3 || -0.5 || -93.254 || No || [[File:Mlw115_Surface_Plot_HHF_3_0.5.png|300px]] || This is similar to the previous case however the hydrogen rebounds twice into flourine but on the second rebounds remains returns back to oscillating with the original H atom
|-
| 2 || -0.5|| -98.753|| No || [[File: Mlw115_Surface_Plot_HHF_2_0.5.png|300px]] || The HF molecule oscillates within the energy well at high vibrational energy after collision
|-
| 1.5|| -0.5|| 103.754|| No || [[File: Mlw115_Surface_Plot_HHF_1.5_0.5.png|300px]] || Similarly to the previous situation
|-
|0.1||-0.8|| 103.364|| Yes|| [[File: Mlw115_Surface_Plot_HHF_0.1_0.8.png|300px]] || Despite having lower total energy than the previous cases the reaction goes to completion as the momentum between the reacting atoms is sufficient
|}

Despite the H2 energy being higher than the activation energy if the BC atoms do not collide with a high enough energy the reaction will return back over the peak and the original reactants will be produced.

===Polanyi's Empirical Rules ===

Polanyi's rules state that in reactions with an early transition state (most closely resembling the reactants) translational energy most effectively promotes a successful reaction. Conversely where the transition state is late (resembling products closely) vibrational energy, specifically of the reactants most effectively promotes a successful reaction. <ref>The Journal of Chemical Physics 138, 234104 (2013); https://doi.org/10.1063/1.4810007</ref>

====H-F + H====
rAB= 2 rBC = 0.74
{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -5 || 0.2 || -103.364 || No || [[File:Mlw115_Surface_Plot_HFH_5_0.2.png|300px]] || Approaching H rebounds and reaction doesn't complete
|-
| 2.21||0.37||
|}

Reversal of conditions:
AB D 5.07 M -2.22
BC D 1.19 M 0.37

Random Guessing:
AB D 5.0 M -1.5
BC D 1.2 M 1.0

[[File: Mlw115_Surface_Plot_HFH_-1.5_1.png| 400px]]

=References=
<references/>

MRD:mlw115RXNDy

2018-05-22T14:36:10Z

Mlw115: /* Determining Transition State */

= Reaction Dynamics =

==Introduction==

The reactivity of atom-diatom systems are studied in order to gain insight on the internal degrees of freedom of the system. Particularly vibrational and translational energy are considered with respect to overcoming the transition state energy barrier.

The atoms are considered to behave classically and the interactions are expressed as potential energy surfaces.

= Exercise 1: H2 + H system=

==Dynamics from the transition state region ==

 Question 1. What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

Where a reaction is modeled on a potential energy surface the lowest energy path between the reactants and products is seen as a "valley floor" of the surface. The transition structure is a saddle point along the minimum energy ridge.

The gradient of the potential energy surface is zero at the minimum, along the direction of the reaction path and so the first derivative with respect to the inter-nuclear distance is zero. For the saddle point the gradient of the potential energy surface is zero orthogonal to the direction of the reaction pathway.

δV(ri)/δri = 0

Second derivatives
δV2(ri)/δri2 > 0 Minima
δV2(ri)/δri2 < 0 Maxima

where V is the potential energy and r related to the nuclear positions.

The minimum pathway and saddle point can be distinguished using the second derivatives with respect to atom positions. The second derivative of a maximum point is negative, whereas the second derivative of a minimum is positive. As the saddle point is a maxima along the minimum path the second derivative is negative. The second derivative of the minimum energy path in the direction of the reaction will be positive.

==Estimating Transition State Position ==

 Question 2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. 

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.908||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot 1 !! Surface Plot 2!! Internuclear distance vs Time
|-
|[[File:Mlw115_H2H_saddle.png|300px]]||[[File:mlw115_H2H_TS.png|300px]]||[[File:Mlw115_H2H_TS_INDvT.png|300px]]|
|}

The best guess of inter nuclear distances at the transition state was found using trial and error, to find the positions that would result in the minimum trajectory. As the system is symmetric the distances would be equal. The two surface plots are included to demonstrate the "reaction path". The internuclear distance vs time plot demonstrates no change in any bond lengths at the transition state so the overall momenta is 0 and potential energy is at the maximum.

== Reaction Path ==
===MEP vs Dynamics===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.909||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot (MEP) !! Surface Plot (Dynamics
|-
|[[File:Mlw115_H2H_MEP_surfaceplot.png|300px]]||[[File:mlw115_H2H_MEP_surfaceplot_dynamics.png|300px]]
|}

 Question 3. Comment on how the mep and the trajectory you just calculated differ.

It can be seen in the MEP plot that the trajectory follows the minimum energy valley floor without any obvious deviations or oscillation. The MEP trajectory is corresponding to infinitely slow motion, it does not accurately represent the atomic motion of the reaction. The trajectory oscillates in the dynamics plot, also in much less steps the AB bond distance increases considerably more. This plot is more realistic as it accounts for the motion of atoms during the reaction.

{| class="wikitable" border="1"
|+ Final Geometry
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| 2.486||1.167
|}

=== Reversed starting positions ===

What would change if we used the initial conditions r1 = rts and r2 = rts+0.01 instead? ****** DO THIS ******

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| -2.486||-1.167
|}

[[File:mlw115_H2H_dynamics_reversed_momenta.png|300px]]

From taking the previous runs final conditions and setting these as the initial conditions and reversing the momenta values the reaction seems to run in reverse, reaching an end exactly at the transition state.

===Determining Reactivity===

Question 4. Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2.0
|}

{| class="wikitable" border=1
! pAB !! pBC!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -1.25 || -2.5 || -99.018|| Yes || [[File:Mlw115_Surface_Plot_1.25_2.5.png|300px]] || Initially the HAHB distance changes very little as the vibrational energy is quite low, as the HC approaches it can be seen that the BC distance decreases linearly. The molecules pass over the transition state, after this point the HAHB distance decreases and the HBHC distance oscillates slightly, with a higher vibrational energy than the HAHB molecule. This overall process represents the HAHB bond breaking and forming a HBHC molecule.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mlw115_Surface_Plot_1.5_2.0.png|300px]] || The molecule approaches the maxima with some vibrational energy, shown by oscillations in the initial path, the energy of the system is not high enough to pass over the transition state energy maxima so the HC "rebounds" and the HBHC distance begins to increase. The HAHB molecule oscillations increase.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mlw115_Surface_Plot_1.5_2.5.png|300px]] || The reaction proceeds similarly to the first case with HC approaching and forming a bond with HB as there is sufficient energy to proceed over the transition state maxima. The vibrational energy throughout is slightly higher than the first case, demonstrated by higher oscillations in atom distances.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mlw115_Surface_Plot_2.5_5.png|300px]] || In this case the energy is high enough and the atoms are able to pass over the transition state energy maxima, however the vibrational energy of the HBHC system formed seems to be too high to form the molecule and the HC atom rebounds and the distance begins to decrease and so the HAHB is reformed but with considerably higher oscillations.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mlw115_Surface_Plot_2.5_5.2.png|300px]] || The HC approaches the oscillating HAHB system, the HB atom rebounds between the two atoms, as it can be seen on the surface plot the system passes over the transition state maxima, back over to the other side and back again as this occurs. Finally the HBHC bond is formed with a high vibrationally energy
|}

===Transition State Theory===

 Question 5 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? 

Transition state theory is used to understand how elementary chemical reactions take place assuming a quasi-equilibirum between reactanting complexes and those at the transition state. The rate of reaction can be understood by inspecting the complexes at the maximum saddle point of the minimum energy pathway. The reactants are in equilibrium with the activated transition state complexes and can be converted to the products.

This theory is useful to theoretically understand a chemical reaction however it operates on a series of assumptions which limit the theory. For instance the atoms are assumed to behave classically, ignoring tunelling. Tunneling would be more prominent in reactions with low activation barriers. Also the theory assumes that the saddle point with the lowest energy will be passed over to reach the products, however this is not the case when higher vibrational energies will be populated.

<ref>https://goldbook.iupac.org/html/T/T06470.html</ref>

=Exercise 2: H-H-F System=

==PES Inspection==
Question 6. 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?

For the forward reaction of H2 + F --> HF + H is exothermic, as shown in the surface plot below the energy of the reactants H2 is higher than the energy of the product HF. As the energy absorbed in breaking the H-H bond in the first case is less than the energy released forming the H-F bond which means that in total energy is released which results in an exothermic reaction.

{| class="wikitable" border=1
! Surface Plot H2+F->HF+H !! Surface Plot HF+H -> H2+F
|-
|[[File:mlw115_Surface_Plot_HHF_labelled.png|300px]]||[[File:mlw115_Surface_Plot_HHF_H2Prod.png|300px]]|
|}

==Determining Transition State ==

Question 7. Locate the approximate position of the transition state.

A = HA B = HB and C = F

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! Surface Plot !! Internuclear distance vs Time
|-
|[[File:Mlw115_Surface_Plot_TS_HHF.png|400px]]||[[File:Mlw115 INDvT HHF TS.png|350px]]|
|}

== Calculating Activation Energy ==

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! BC= 1.8135!! BC= 1.8135 Zoomed!! BC= 1.813
|-
|[[File:Mlw115_HHF_EvT_1.8135.png|300px]] ‎||[[File:Mlw115_Plot_1.813_EvT_zoomT0_Resized.png|300px]]||[[File:Mlw115_HHF_EvT_1.813.png|320px]]
|}

pAB = pBC =0

distAB = 0.74

{| class="wikitable" border=1
|+Activation energy HF+H->H2+F
|-
! !! Energy (Kcal/mol)
|-
|Initial energy || -103.752
|-
|Final energy|| -133.89
|-
|EA|| 30.138
|}

{| class="wikitable" border=1
|+Activation energy H2+F->HF+H
|-
! !! Energy (Kcal/mol)
|-
|Initial energy|| -103.742
|-
|Final energy|| -103.752
|-
|EA|| 0.01
|}

==Reaction Dynamics==

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.74||0
|-
| BC|| 2.3||-2.7
|}

{| class="wikitable" border=1
!Surface Plot!! Internuclear momentum vs time
|-
|[[File:Mlw_Surface_Plot_HHF_Reactive.png]]|| [[File:Mlw_INMvT_HHF_Reactive.png]]
|}

{| class="wikitable" border=1
!Energy!!
|-
|Kinetic|| 3.837
|-
|Potential|| -103.886

|-
|Total|| -100.049
|}

It can be seen from the animation that as the H-H (AB) bond is broken and the H-F (BC) bond is formed the vibrational energy (and so the kinetic energy) considerably increases for H-F. The internuclear momentum vs time graph demonstrates this, as the momentum is much higher for the H-F atom. This vibrational energy will be released as heat so experimentally an increase in temperature could confirm the reaction had occurred.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -3 || -0.5 || -100.339|| No || [[File:Mlw115_Surface_Plot_HHF_-3_0.5.png|300px]] || H2 approaches the fluorine atom slowly but when a collision is made the H rebounds straight back into the other H and they continue their path oscillating together in the opposite direction
|-

| 3 || -0.5 || -93.254 || No || [[File:Mlw115_Surface_Plot_HHF_3_0.5.png|300px]] || This is similar to the previous case however the hydrogen rebounds twice into flourine but on the second rebounds remains returns back to oscillating with the original H atom
|-
| 2 || -0.5|| -98.753|| No || [[File: Mlw115_Surface_Plot_HHF_2_0.5.png|300px]] || The HF molecule oscillates within the energy well at high vibrational energy after collision
|-
| 1.5|| -0.5|| 103.754|| No || [[File: Mlw115_Surface_Plot_HHF_1.5_0.5.png|300px]] || Similarly to the previous situation
|-
|0.1||-0.8|| 103.364|| Yes|| [[File: Mlw115_Surface_Plot_HHF_0.1_0.8.png|300px]] || Despite having lower total energy than the previous cases the reaction goes to completion as the momentum between the reacting atoms is sufficient
|}

Despite the H2 energy being higher than the activation energy if the BC atoms do not collide with a high enough energy the reaction will return back over the peak and the original reactants will be produced.

===Polanyi's Empirical Rules ===

Polanyi's rules state that in reactions with an early transition state (most closely resembling the reactants) translational energy most effectively promotes a successful reaction. Conversely where the transition state is late (resembling products closely) vibrational energy, specifically of the reactants most effectively promotes a successful reaction. <ref>The Journal of Chemical Physics 138, 234104 (2013); https://doi.org/10.1063/1.4810007</ref>

====H-F + H====
rAB= 2 rBC = 0.74
{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -5 || 0.2 || -103.364 || No || [[File:Mlw115_Surface_Plot_HFH_5_0.2.png|300px]] || Approaching H rebounds and reaction doesn't complete
|-
| 2.21||0.37||
|}

Reversal of conditions:
AB D 5.07 M -2.22
BC D 1.19 M 0.37

Random Guessing:
AB D 5.0 M -1.5
BC D 1.2 M 1.0

[[File: Mlw115_Surface_Plot_HFH_-1.5_1.png| 400px]]

=References=
<references/>

MRD:mlw115RXNDy

2018-05-22T14:35:35Z

Mlw115: /* PES Inspection */

= Reaction Dynamics =

==Introduction==

The reactivity of atom-diatom systems are studied in order to gain insight on the internal degrees of freedom of the system. Particularly vibrational and translational energy are considered with respect to overcoming the transition state energy barrier.

The atoms are considered to behave classically and the interactions are expressed as potential energy surfaces.

= Exercise 1: H2 + H system=

==Dynamics from the transition state region ==

 Question 1. What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

Where a reaction is modeled on a potential energy surface the lowest energy path between the reactants and products is seen as a "valley floor" of the surface. The transition structure is a saddle point along the minimum energy ridge.

The gradient of the potential energy surface is zero at the minimum, along the direction of the reaction path and so the first derivative with respect to the inter-nuclear distance is zero. For the saddle point the gradient of the potential energy surface is zero orthogonal to the direction of the reaction pathway.

δV(ri)/δri = 0

Second derivatives
δV2(ri)/δri2 > 0 Minima
δV2(ri)/δri2 < 0 Maxima

where V is the potential energy and r related to the nuclear positions.

The minimum pathway and saddle point can be distinguished using the second derivatives with respect to atom positions. The second derivative of a maximum point is negative, whereas the second derivative of a minimum is positive. As the saddle point is a maxima along the minimum path the second derivative is negative. The second derivative of the minimum energy path in the direction of the reaction will be positive.

==Estimating Transition State Position ==

 Question 2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. 

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.908||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot 1 !! Surface Plot 2!! Internuclear distance vs Time
|-
|[[File:Mlw115_H2H_saddle.png|300px]]||[[File:mlw115_H2H_TS.png|300px]]||[[File:Mlw115_H2H_TS_INDvT.png|300px]]|
|}

The best guess of inter nuclear distances at the transition state was found using trial and error, to find the positions that would result in the minimum trajectory. As the system is symmetric the distances would be equal. The two surface plots are included to demonstrate the "reaction path". The internuclear distance vs time plot demonstrates no change in any bond lengths at the transition state so the overall momenta is 0 and potential energy is at the maximum.

== Reaction Path ==
===MEP vs Dynamics===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.909||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot (MEP) !! Surface Plot (Dynamics
|-
|[[File:Mlw115_H2H_MEP_surfaceplot.png|300px]]||[[File:mlw115_H2H_MEP_surfaceplot_dynamics.png|300px]]
|}

 Question 3. Comment on how the mep and the trajectory you just calculated differ.

It can be seen in the MEP plot that the trajectory follows the minimum energy valley floor without any obvious deviations or oscillation. The MEP trajectory is corresponding to infinitely slow motion, it does not accurately represent the atomic motion of the reaction. The trajectory oscillates in the dynamics plot, also in much less steps the AB bond distance increases considerably more. This plot is more realistic as it accounts for the motion of atoms during the reaction.

{| class="wikitable" border="1"
|+ Final Geometry
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| 2.486||1.167
|}

=== Reversed starting positions ===

What would change if we used the initial conditions r1 = rts and r2 = rts+0.01 instead? ****** DO THIS ******

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| -2.486||-1.167
|}

[[File:mlw115_H2H_dynamics_reversed_momenta.png|300px]]

From taking the previous runs final conditions and setting these as the initial conditions and reversing the momenta values the reaction seems to run in reverse, reaching an end exactly at the transition state.

===Determining Reactivity===

Question 4. Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2.0
|}

{| class="wikitable" border=1
! pAB !! pBC!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -1.25 || -2.5 || -99.018|| Yes || [[File:Mlw115_Surface_Plot_1.25_2.5.png|300px]] || Initially the HAHB distance changes very little as the vibrational energy is quite low, as the HC approaches it can be seen that the BC distance decreases linearly. The molecules pass over the transition state, after this point the HAHB distance decreases and the HBHC distance oscillates slightly, with a higher vibrational energy than the HAHB molecule. This overall process represents the HAHB bond breaking and forming a HBHC molecule.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mlw115_Surface_Plot_1.5_2.0.png|300px]] || The molecule approaches the maxima with some vibrational energy, shown by oscillations in the initial path, the energy of the system is not high enough to pass over the transition state energy maxima so the HC "rebounds" and the HBHC distance begins to increase. The HAHB molecule oscillations increase.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mlw115_Surface_Plot_1.5_2.5.png|300px]] || The reaction proceeds similarly to the first case with HC approaching and forming a bond with HB as there is sufficient energy to proceed over the transition state maxima. The vibrational energy throughout is slightly higher than the first case, demonstrated by higher oscillations in atom distances.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mlw115_Surface_Plot_2.5_5.png|300px]] || In this case the energy is high enough and the atoms are able to pass over the transition state energy maxima, however the vibrational energy of the HBHC system formed seems to be too high to form the molecule and the HC atom rebounds and the distance begins to decrease and so the HAHB is reformed but with considerably higher oscillations.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mlw115_Surface_Plot_2.5_5.2.png|300px]] || The HC approaches the oscillating HAHB system, the HB atom rebounds between the two atoms, as it can be seen on the surface plot the system passes over the transition state maxima, back over to the other side and back again as this occurs. Finally the HBHC bond is formed with a high vibrationally energy
|}

===Transition State Theory===

 Question 5 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? 

Transition state theory is used to understand how elementary chemical reactions take place assuming a quasi-equilibirum between reactanting complexes and those at the transition state. The rate of reaction can be understood by inspecting the complexes at the maximum saddle point of the minimum energy pathway. The reactants are in equilibrium with the activated transition state complexes and can be converted to the products.

This theory is useful to theoretically understand a chemical reaction however it operates on a series of assumptions which limit the theory. For instance the atoms are assumed to behave classically, ignoring tunelling. Tunneling would be more prominent in reactions with low activation barriers. Also the theory assumes that the saddle point with the lowest energy will be passed over to reach the products, however this is not the case when higher vibrational energies will be populated.

<ref>https://goldbook.iupac.org/html/T/T06470.html</ref>

=Exercise 2: H-H-F System=

==PES Inspection==
Question 6. 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?

For the forward reaction of H2 + F --> HF + H is exothermic, as shown in the surface plot below the energy of the reactants H2 is higher than the energy of the product HF. As the energy absorbed in breaking the H-H bond in the first case is less than the energy released forming the H-F bond which means that in total energy is released which results in an exothermic reaction.

{| class="wikitable" border=1
! Surface Plot H2+F->HF+H !! Surface Plot HF+H -> H2+F
|-
|[[File:mlw115_Surface_Plot_HHF_labelled.png|300px]]||[[File:mlw115_Surface_Plot_HHF_H2Prod.png|300px]]|
|}

==Determining Transition State ==

A = HA B = HB and C = F

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! Surface Plot !! Internuclear distance vs Time
|-
|[[File:Mlw115_Surface_Plot_TS_HHF.png|400px]]||[[File:Mlw115 INDvT HHF TS.png|350px]]|
|}

== Calculating Activation Energy ==

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! BC= 1.8135!! BC= 1.8135 Zoomed!! BC= 1.813
|-
|[[File:Mlw115_HHF_EvT_1.8135.png|300px]] ‎||[[File:Mlw115_Plot_1.813_EvT_zoomT0_Resized.png|300px]]||[[File:Mlw115_HHF_EvT_1.813.png|320px]]
|}

pAB = pBC =0

distAB = 0.74

{| class="wikitable" border=1
|+Activation energy HF+H->H2+F
|-
! !! Energy (Kcal/mol)
|-
|Initial energy || -103.752
|-
|Final energy|| -133.89
|-
|EA|| 30.138
|}

{| class="wikitable" border=1
|+Activation energy H2+F->HF+H
|-
! !! Energy (Kcal/mol)
|-
|Initial energy|| -103.742
|-
|Final energy|| -103.752
|-
|EA|| 0.01
|}

==Reaction Dynamics==

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.74||0
|-
| BC|| 2.3||-2.7
|}

{| class="wikitable" border=1
!Surface Plot!! Internuclear momentum vs time
|-
|[[File:Mlw_Surface_Plot_HHF_Reactive.png]]|| [[File:Mlw_INMvT_HHF_Reactive.png]]
|}

{| class="wikitable" border=1
!Energy!!
|-
|Kinetic|| 3.837
|-
|Potential|| -103.886

|-
|Total|| -100.049
|}

It can be seen from the animation that as the H-H (AB) bond is broken and the H-F (BC) bond is formed the vibrational energy (and so the kinetic energy) considerably increases for H-F. The internuclear momentum vs time graph demonstrates this, as the momentum is much higher for the H-F atom. This vibrational energy will be released as heat so experimentally an increase in temperature could confirm the reaction had occurred.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -3 || -0.5 || -100.339|| No || [[File:Mlw115_Surface_Plot_HHF_-3_0.5.png|300px]] || H2 approaches the fluorine atom slowly but when a collision is made the H rebounds straight back into the other H and they continue their path oscillating together in the opposite direction
|-

| 3 || -0.5 || -93.254 || No || [[File:Mlw115_Surface_Plot_HHF_3_0.5.png|300px]] || This is similar to the previous case however the hydrogen rebounds twice into flourine but on the second rebounds remains returns back to oscillating with the original H atom
|-
| 2 || -0.5|| -98.753|| No || [[File: Mlw115_Surface_Plot_HHF_2_0.5.png|300px]] || The HF molecule oscillates within the energy well at high vibrational energy after collision
|-
| 1.5|| -0.5|| 103.754|| No || [[File: Mlw115_Surface_Plot_HHF_1.5_0.5.png|300px]] || Similarly to the previous situation
|-
|0.1||-0.8|| 103.364|| Yes|| [[File: Mlw115_Surface_Plot_HHF_0.1_0.8.png|300px]] || Despite having lower total energy than the previous cases the reaction goes to completion as the momentum between the reacting atoms is sufficient
|}

Despite the H2 energy being higher than the activation energy if the BC atoms do not collide with a high enough energy the reaction will return back over the peak and the original reactants will be produced.

===Polanyi's Empirical Rules ===

Polanyi's rules state that in reactions with an early transition state (most closely resembling the reactants) translational energy most effectively promotes a successful reaction. Conversely where the transition state is late (resembling products closely) vibrational energy, specifically of the reactants most effectively promotes a successful reaction. <ref>The Journal of Chemical Physics 138, 234104 (2013); https://doi.org/10.1063/1.4810007</ref>

====H-F + H====
rAB= 2 rBC = 0.74
{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -5 || 0.2 || -103.364 || No || [[File:Mlw115_Surface_Plot_HFH_5_0.2.png|300px]] || Approaching H rebounds and reaction doesn't complete
|-
| 2.21||0.37||
|}

Reversal of conditions:
AB D 5.07 M -2.22
BC D 1.19 M 0.37

Random Guessing:
AB D 5.0 M -1.5
BC D 1.2 M 1.0

[[File: Mlw115_Surface_Plot_HFH_-1.5_1.png| 400px]]

=References=
<references/>

MRD:mlw115RXNDy

2018-05-22T14:34:42Z

Mlw115: /* Dynamics from the transition state region */

= Reaction Dynamics =

==Introduction==

The reactivity of atom-diatom systems are studied in order to gain insight on the internal degrees of freedom of the system. Particularly vibrational and translational energy are considered with respect to overcoming the transition state energy barrier.

The atoms are considered to behave classically and the interactions are expressed as potential energy surfaces.

= Exercise 1: H2 + H system=

==Dynamics from the transition state region ==

 Question 1. What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

Where a reaction is modeled on a potential energy surface the lowest energy path between the reactants and products is seen as a "valley floor" of the surface. The transition structure is a saddle point along the minimum energy ridge.

The gradient of the potential energy surface is zero at the minimum, along the direction of the reaction path and so the first derivative with respect to the inter-nuclear distance is zero. For the saddle point the gradient of the potential energy surface is zero orthogonal to the direction of the reaction pathway.

δV(ri)/δri = 0

Second derivatives
δV2(ri)/δri2 > 0 Minima
δV2(ri)/δri2 < 0 Maxima

where V is the potential energy and r related to the nuclear positions.

The minimum pathway and saddle point can be distinguished using the second derivatives with respect to atom positions. The second derivative of a maximum point is negative, whereas the second derivative of a minimum is positive. As the saddle point is a maxima along the minimum path the second derivative is negative. The second derivative of the minimum energy path in the direction of the reaction will be positive.

==Estimating Transition State Position ==

 Question 2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. 

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.908||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot 1 !! Surface Plot 2!! Internuclear distance vs Time
|-
|[[File:Mlw115_H2H_saddle.png|300px]]||[[File:mlw115_H2H_TS.png|300px]]||[[File:Mlw115_H2H_TS_INDvT.png|300px]]|
|}

The best guess of inter nuclear distances at the transition state was found using trial and error, to find the positions that would result in the minimum trajectory. As the system is symmetric the distances would be equal. The two surface plots are included to demonstrate the "reaction path". The internuclear distance vs time plot demonstrates no change in any bond lengths at the transition state so the overall momenta is 0 and potential energy is at the maximum.

== Reaction Path ==
===MEP vs Dynamics===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.909||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot (MEP) !! Surface Plot (Dynamics
|-
|[[File:Mlw115_H2H_MEP_surfaceplot.png|300px]]||[[File:mlw115_H2H_MEP_surfaceplot_dynamics.png|300px]]
|}

 Question 3. Comment on how the mep and the trajectory you just calculated differ.

It can be seen in the MEP plot that the trajectory follows the minimum energy valley floor without any obvious deviations or oscillation. The MEP trajectory is corresponding to infinitely slow motion, it does not accurately represent the atomic motion of the reaction. The trajectory oscillates in the dynamics plot, also in much less steps the AB bond distance increases considerably more. This plot is more realistic as it accounts for the motion of atoms during the reaction.

{| class="wikitable" border="1"
|+ Final Geometry
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| 2.486||1.167
|}

=== Reversed starting positions ===

What would change if we used the initial conditions r1 = rts and r2 = rts+0.01 instead? ****** DO THIS ******

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| -2.486||-1.167
|}

[[File:mlw115_H2H_dynamics_reversed_momenta.png|300px]]

From taking the previous runs final conditions and setting these as the initial conditions and reversing the momenta values the reaction seems to run in reverse, reaching an end exactly at the transition state.

===Determining Reactivity===

Question 4. Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2.0
|}

{| class="wikitable" border=1
! pAB !! pBC!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -1.25 || -2.5 || -99.018|| Yes || [[File:Mlw115_Surface_Plot_1.25_2.5.png|300px]] || Initially the HAHB distance changes very little as the vibrational energy is quite low, as the HC approaches it can be seen that the BC distance decreases linearly. The molecules pass over the transition state, after this point the HAHB distance decreases and the HBHC distance oscillates slightly, with a higher vibrational energy than the HAHB molecule. This overall process represents the HAHB bond breaking and forming a HBHC molecule.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mlw115_Surface_Plot_1.5_2.0.png|300px]] || The molecule approaches the maxima with some vibrational energy, shown by oscillations in the initial path, the energy of the system is not high enough to pass over the transition state energy maxima so the HC "rebounds" and the HBHC distance begins to increase. The HAHB molecule oscillations increase.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mlw115_Surface_Plot_1.5_2.5.png|300px]] || The reaction proceeds similarly to the first case with HC approaching and forming a bond with HB as there is sufficient energy to proceed over the transition state maxima. The vibrational energy throughout is slightly higher than the first case, demonstrated by higher oscillations in atom distances.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mlw115_Surface_Plot_2.5_5.png|300px]] || In this case the energy is high enough and the atoms are able to pass over the transition state energy maxima, however the vibrational energy of the HBHC system formed seems to be too high to form the molecule and the HC atom rebounds and the distance begins to decrease and so the HAHB is reformed but with considerably higher oscillations.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mlw115_Surface_Plot_2.5_5.2.png|300px]] || The HC approaches the oscillating HAHB system, the HB atom rebounds between the two atoms, as it can be seen on the surface plot the system passes over the transition state maxima, back over to the other side and back again as this occurs. Finally the HBHC bond is formed with a high vibrationally energy
|}

===Transition State Theory===

 Question 5 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? 

Transition state theory is used to understand how elementary chemical reactions take place assuming a quasi-equilibirum between reactanting complexes and those at the transition state. The rate of reaction can be understood by inspecting the complexes at the maximum saddle point of the minimum energy pathway. The reactants are in equilibrium with the activated transition state complexes and can be converted to the products.

This theory is useful to theoretically understand a chemical reaction however it operates on a series of assumptions which limit the theory. For instance the atoms are assumed to behave classically, ignoring tunelling. Tunneling would be more prominent in reactions with low activation barriers. Also the theory assumes that the saddle point with the lowest energy will be passed over to reach the products, however this is not the case when higher vibrational energies will be populated.

<ref>https://goldbook.iupac.org/html/T/T06470.html</ref>

=Exercise 2: H-H-F System=

==PES Inspection==

For the forward reaction of H2 + F --> HF + H is exothermic, as shown in the surface plot below the energy of the reactants H2 is higher than the energy of the product HF. As the energy absorbed in breaking the H-H bond in the first case is less than the energy released forming the H-F bond which means that in total energy is released which results in an exothermic reaction.

{| class="wikitable" border=1
! Surface Plot H2+F->HF+H !! Surface Plot HF+H -> H2+F
|-
|[[File:mlw115_Surface_Plot_HHF_labelled.png|300px]]||[[File:mlw115_Surface_Plot_HHF_H2Prod.png|300px]]|
|}

==Determining Transition State ==

A = HA B = HB and C = F

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! Surface Plot !! Internuclear distance vs Time
|-
|[[File:Mlw115_Surface_Plot_TS_HHF.png|400px]]||[[File:Mlw115 INDvT HHF TS.png|350px]]|
|}

== Calculating Activation Energy ==

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! BC= 1.8135!! BC= 1.8135 Zoomed!! BC= 1.813
|-
|[[File:Mlw115_HHF_EvT_1.8135.png|300px]] ‎||[[File:Mlw115_Plot_1.813_EvT_zoomT0_Resized.png|300px]]||[[File:Mlw115_HHF_EvT_1.813.png|320px]]
|}

pAB = pBC =0

distAB = 0.74

{| class="wikitable" border=1
|+Activation energy HF+H->H2+F
|-
! !! Energy (Kcal/mol)
|-
|Initial energy || -103.752
|-
|Final energy|| -133.89
|-
|EA|| 30.138
|}

{| class="wikitable" border=1
|+Activation energy H2+F->HF+H
|-
! !! Energy (Kcal/mol)
|-
|Initial energy|| -103.742
|-
|Final energy|| -103.752
|-
|EA|| 0.01
|}

==Reaction Dynamics==

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.74||0
|-
| BC|| 2.3||-2.7
|}

{| class="wikitable" border=1
!Surface Plot!! Internuclear momentum vs time
|-
|[[File:Mlw_Surface_Plot_HHF_Reactive.png]]|| [[File:Mlw_INMvT_HHF_Reactive.png]]
|}

{| class="wikitable" border=1
!Energy!!
|-
|Kinetic|| 3.837
|-
|Potential|| -103.886

|-
|Total|| -100.049
|}

It can be seen from the animation that as the H-H (AB) bond is broken and the H-F (BC) bond is formed the vibrational energy (and so the kinetic energy) considerably increases for H-F. The internuclear momentum vs time graph demonstrates this, as the momentum is much higher for the H-F atom. This vibrational energy will be released as heat so experimentally an increase in temperature could confirm the reaction had occurred.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -3 || -0.5 || -100.339|| No || [[File:Mlw115_Surface_Plot_HHF_-3_0.5.png|300px]] || H2 approaches the fluorine atom slowly but when a collision is made the H rebounds straight back into the other H and they continue their path oscillating together in the opposite direction
|-

| 3 || -0.5 || -93.254 || No || [[File:Mlw115_Surface_Plot_HHF_3_0.5.png|300px]] || This is similar to the previous case however the hydrogen rebounds twice into flourine but on the second rebounds remains returns back to oscillating with the original H atom
|-
| 2 || -0.5|| -98.753|| No || [[File: Mlw115_Surface_Plot_HHF_2_0.5.png|300px]] || The HF molecule oscillates within the energy well at high vibrational energy after collision
|-
| 1.5|| -0.5|| 103.754|| No || [[File: Mlw115_Surface_Plot_HHF_1.5_0.5.png|300px]] || Similarly to the previous situation
|-
|0.1||-0.8|| 103.364|| Yes|| [[File: Mlw115_Surface_Plot_HHF_0.1_0.8.png|300px]] || Despite having lower total energy than the previous cases the reaction goes to completion as the momentum between the reacting atoms is sufficient
|}

Despite the H2 energy being higher than the activation energy if the BC atoms do not collide with a high enough energy the reaction will return back over the peak and the original reactants will be produced.

===Polanyi's Empirical Rules ===

Polanyi's rules state that in reactions with an early transition state (most closely resembling the reactants) translational energy most effectively promotes a successful reaction. Conversely where the transition state is late (resembling products closely) vibrational energy, specifically of the reactants most effectively promotes a successful reaction. <ref>The Journal of Chemical Physics 138, 234104 (2013); https://doi.org/10.1063/1.4810007</ref>

====H-F + H====
rAB= 2 rBC = 0.74
{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -5 || 0.2 || -103.364 || No || [[File:Mlw115_Surface_Plot_HFH_5_0.2.png|300px]] || Approaching H rebounds and reaction doesn't complete
|-
| 2.21||0.37||
|}

Reversal of conditions:
AB D 5.07 M -2.22
BC D 1.19 M 0.37

Random Guessing:
AB D 5.0 M -1.5
BC D 1.2 M 1.0

[[File: Mlw115_Surface_Plot_HFH_-1.5_1.png| 400px]]

=References=
<references/>

MRD:mlw115RXNDy

2018-05-22T14:29:54Z

Mlw115: /* Transition State Theory */

= Reaction Dynamics =

==Introduction==

The reactivity of atom-diatom systems are studied in order to gain insight on the internal degrees of freedom of the system. Particularly vibrational and translational energy are considered with respect to overcoming the transition state energy barrier.

The atoms are considered to behave classically and the interactions are expressed as potential energy surfaces.

= Exercise 1: H2 + H system=

==Dynamics from the transition state region ==

 Question 1. What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

Where a reaction is modeled on a potential energy surface the lowest energy path between the reactants and products is seen as a "valley floor" of the surface. The transition structure is a saddle point along the minimum energy ridge.

The gradient of the potential energy surface is zero at the minima and so the first derivative with respect to the inter-nuclear distance is zero. This is also the case for the peak of the saddle point.

δV(ri)/δri = 0

However using second derivatives
δV2(ri)/δri2 > 0 Minima
δV2(ri)/δri2 < 0 Maxima

where V is the potential energy and r related to the nuclear positions.

The second derivative of a maximum point is negative, whereas the second derivative of a minimum is positive. As the saddle point is a maxima along the minimum path the second derivative is negative.

==Estimating Transition State Position ==

 Question 2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. 

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.908||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot 1 !! Surface Plot 2!! Internuclear distance vs Time
|-
|[[File:Mlw115_H2H_saddle.png|300px]]||[[File:mlw115_H2H_TS.png|300px]]||[[File:Mlw115_H2H_TS_INDvT.png|300px]]|
|}

The best guess of inter nuclear distances at the transition state was found using trial and error, to find the positions that would result in the minimum trajectory. As the system is symmetric the distances would be equal. The two surface plots are included to demonstrate the "reaction path". The internuclear distance vs time plot demonstrates no change in any bond lengths at the transition state so the overall momenta is 0 and potential energy is at the maximum.

== Reaction Path ==
===MEP vs Dynamics===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.909||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot (MEP) !! Surface Plot (Dynamics
|-
|[[File:Mlw115_H2H_MEP_surfaceplot.png|300px]]||[[File:mlw115_H2H_MEP_surfaceplot_dynamics.png|300px]]
|}

 Question 3. Comment on how the mep and the trajectory you just calculated differ.

It can be seen in the MEP plot that the trajectory follows the minimum energy valley floor without any obvious deviations or oscillation. The MEP trajectory is corresponding to infinitely slow motion, it does not accurately represent the atomic motion of the reaction. The trajectory oscillates in the dynamics plot, also in much less steps the AB bond distance increases considerably more. This plot is more realistic as it accounts for the motion of atoms during the reaction.

{| class="wikitable" border="1"
|+ Final Geometry
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| 2.486||1.167
|}

=== Reversed starting positions ===

What would change if we used the initial conditions r1 = rts and r2 = rts+0.01 instead? ****** DO THIS ******

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| -2.486||-1.167
|}

[[File:mlw115_H2H_dynamics_reversed_momenta.png|300px]]

From taking the previous runs final conditions and setting these as the initial conditions and reversing the momenta values the reaction seems to run in reverse, reaching an end exactly at the transition state.

===Determining Reactivity===

Question 4. Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2.0
|}

{| class="wikitable" border=1
! pAB !! pBC!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -1.25 || -2.5 || -99.018|| Yes || [[File:Mlw115_Surface_Plot_1.25_2.5.png|300px]] || Initially the HAHB distance changes very little as the vibrational energy is quite low, as the HC approaches it can be seen that the BC distance decreases linearly. The molecules pass over the transition state, after this point the HAHB distance decreases and the HBHC distance oscillates slightly, with a higher vibrational energy than the HAHB molecule. This overall process represents the HAHB bond breaking and forming a HBHC molecule.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mlw115_Surface_Plot_1.5_2.0.png|300px]] || The molecule approaches the maxima with some vibrational energy, shown by oscillations in the initial path, the energy of the system is not high enough to pass over the transition state energy maxima so the HC "rebounds" and the HBHC distance begins to increase. The HAHB molecule oscillations increase.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mlw115_Surface_Plot_1.5_2.5.png|300px]] || The reaction proceeds similarly to the first case with HC approaching and forming a bond with HB as there is sufficient energy to proceed over the transition state maxima. The vibrational energy throughout is slightly higher than the first case, demonstrated by higher oscillations in atom distances.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mlw115_Surface_Plot_2.5_5.png|300px]] || In this case the energy is high enough and the atoms are able to pass over the transition state energy maxima, however the vibrational energy of the HBHC system formed seems to be too high to form the molecule and the HC atom rebounds and the distance begins to decrease and so the HAHB is reformed but with considerably higher oscillations.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mlw115_Surface_Plot_2.5_5.2.png|300px]] || The HC approaches the oscillating HAHB system, the HB atom rebounds between the two atoms, as it can be seen on the surface plot the system passes over the transition state maxima, back over to the other side and back again as this occurs. Finally the HBHC bond is formed with a high vibrationally energy
|}

===Transition State Theory===

 Question 5 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? 

Transition state theory is used to understand how elementary chemical reactions take place assuming a quasi-equilibirum between reactanting complexes and those at the transition state. The rate of reaction can be understood by inspecting the complexes at the maximum saddle point of the minimum energy pathway. The reactants are in equilibrium with the activated transition state complexes and can be converted to the products.

This theory is useful to theoretically understand a chemical reaction however it operates on a series of assumptions which limit the theory. For instance the atoms are assumed to behave classically, ignoring tunelling. Tunneling would be more prominent in reactions with low activation barriers. Also the theory assumes that the saddle point with the lowest energy will be passed over to reach the products, however this is not the case when higher vibrational energies will be populated.

<ref>https://goldbook.iupac.org/html/T/T06470.html</ref>

=Exercise 2: H-H-F System=

==PES Inspection==

For the forward reaction of H2 + F --> HF + H is exothermic, as shown in the surface plot below the energy of the reactants H2 is higher than the energy of the product HF. As the energy absorbed in breaking the H-H bond in the first case is less than the energy released forming the H-F bond which means that in total energy is released which results in an exothermic reaction.

{| class="wikitable" border=1
! Surface Plot H2+F->HF+H !! Surface Plot HF+H -> H2+F
|-
|[[File:mlw115_Surface_Plot_HHF_labelled.png|300px]]||[[File:mlw115_Surface_Plot_HHF_H2Prod.png|300px]]|
|}

==Determining Transition State ==

A = HA B = HB and C = F

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! Surface Plot !! Internuclear distance vs Time
|-
|[[File:Mlw115_Surface_Plot_TS_HHF.png|400px]]||[[File:Mlw115 INDvT HHF TS.png|350px]]|
|}

== Calculating Activation Energy ==

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! BC= 1.8135!! BC= 1.8135 Zoomed!! BC= 1.813
|-
|[[File:Mlw115_HHF_EvT_1.8135.png|300px]] ‎||[[File:Mlw115_Plot_1.813_EvT_zoomT0_Resized.png|300px]]||[[File:Mlw115_HHF_EvT_1.813.png|320px]]
|}

pAB = pBC =0

distAB = 0.74

{| class="wikitable" border=1
|+Activation energy HF+H->H2+F
|-
! !! Energy (Kcal/mol)
|-
|Initial energy || -103.752
|-
|Final energy|| -133.89
|-
|EA|| 30.138
|}

{| class="wikitable" border=1
|+Activation energy H2+F->HF+H
|-
! !! Energy (Kcal/mol)
|-
|Initial energy|| -103.742
|-
|Final energy|| -103.752
|-
|EA|| 0.01
|}

==Reaction Dynamics==

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.74||0
|-
| BC|| 2.3||-2.7
|}

{| class="wikitable" border=1
!Surface Plot!! Internuclear momentum vs time
|-
|[[File:Mlw_Surface_Plot_HHF_Reactive.png]]|| [[File:Mlw_INMvT_HHF_Reactive.png]]
|}

{| class="wikitable" border=1
!Energy!!
|-
|Kinetic|| 3.837
|-
|Potential|| -103.886

|-
|Total|| -100.049
|}

It can be seen from the animation that as the H-H (AB) bond is broken and the H-F (BC) bond is formed the vibrational energy (and so the kinetic energy) considerably increases for H-F. The internuclear momentum vs time graph demonstrates this, as the momentum is much higher for the H-F atom. This vibrational energy will be released as heat so experimentally an increase in temperature could confirm the reaction had occurred.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -3 || -0.5 || -100.339|| No || [[File:Mlw115_Surface_Plot_HHF_-3_0.5.png|300px]] || H2 approaches the fluorine atom slowly but when a collision is made the H rebounds straight back into the other H and they continue their path oscillating together in the opposite direction
|-

| 3 || -0.5 || -93.254 || No || [[File:Mlw115_Surface_Plot_HHF_3_0.5.png|300px]] || This is similar to the previous case however the hydrogen rebounds twice into flourine but on the second rebounds remains returns back to oscillating with the original H atom
|-
| 2 || -0.5|| -98.753|| No || [[File: Mlw115_Surface_Plot_HHF_2_0.5.png|300px]] || The HF molecule oscillates within the energy well at high vibrational energy after collision
|-
| 1.5|| -0.5|| 103.754|| No || [[File: Mlw115_Surface_Plot_HHF_1.5_0.5.png|300px]] || Similarly to the previous situation
|-
|0.1||-0.8|| 103.364|| Yes|| [[File: Mlw115_Surface_Plot_HHF_0.1_0.8.png|300px]] || Despite having lower total energy than the previous cases the reaction goes to completion as the momentum between the reacting atoms is sufficient
|}

Despite the H2 energy being higher than the activation energy if the BC atoms do not collide with a high enough energy the reaction will return back over the peak and the original reactants will be produced.

===Polanyi's Empirical Rules ===

Polanyi's rules state that in reactions with an early transition state (most closely resembling the reactants) translational energy most effectively promotes a successful reaction. Conversely where the transition state is late (resembling products closely) vibrational energy, specifically of the reactants most effectively promotes a successful reaction. <ref>The Journal of Chemical Physics 138, 234104 (2013); https://doi.org/10.1063/1.4810007</ref>

====H-F + H====
rAB= 2 rBC = 0.74
{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -5 || 0.2 || -103.364 || No || [[File:Mlw115_Surface_Plot_HFH_5_0.2.png|300px]] || Approaching H rebounds and reaction doesn't complete
|-
| 2.21||0.37||
|}

Reversal of conditions:
AB D 5.07 M -2.22
BC D 1.19 M 0.37

Random Guessing:
AB D 5.0 M -1.5
BC D 1.2 M 1.0

[[File: Mlw115_Surface_Plot_HFH_-1.5_1.png| 400px]]

=References=
<references/>

MRD:mlw115RXNDy

2018-05-22T14:22:18Z

Mlw115: /* Transition State Theory */

= Reaction Dynamics =

==Introduction==

The reactivity of atom-diatom systems are studied in order to gain insight on the internal degrees of freedom of the system. Particularly vibrational and translational energy are considered with respect to overcoming the transition state energy barrier.

The atoms are considered to behave classically and the interactions are expressed as potential energy surfaces.

= Exercise 1: H2 + H system=

==Dynamics from the transition state region ==

 Question 1. What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

Where a reaction is modeled on a potential energy surface the lowest energy path between the reactants and products is seen as a "valley floor" of the surface. The transition structure is a saddle point along the minimum energy ridge.

The gradient of the potential energy surface is zero at the minima and so the first derivative with respect to the inter-nuclear distance is zero. This is also the case for the peak of the saddle point.

δV(ri)/δri = 0

However using second derivatives
δV2(ri)/δri2 > 0 Minima
δV2(ri)/δri2 < 0 Maxima

where V is the potential energy and r related to the nuclear positions.

The second derivative of a maximum point is negative, whereas the second derivative of a minimum is positive. As the saddle point is a maxima along the minimum path the second derivative is negative.

==Estimating Transition State Position ==

 Question 2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. 

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.908||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot 1 !! Surface Plot 2!! Internuclear distance vs Time
|-
|[[File:Mlw115_H2H_saddle.png|300px]]||[[File:mlw115_H2H_TS.png|300px]]||[[File:Mlw115_H2H_TS_INDvT.png|300px]]|
|}

The best guess of inter nuclear distances at the transition state was found using trial and error, to find the positions that would result in the minimum trajectory. As the system is symmetric the distances would be equal. The two surface plots are included to demonstrate the "reaction path". The internuclear distance vs time plot demonstrates no change in any bond lengths at the transition state so the overall momenta is 0 and potential energy is at the maximum.

== Reaction Path ==
===MEP vs Dynamics===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.909||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot (MEP) !! Surface Plot (Dynamics
|-
|[[File:Mlw115_H2H_MEP_surfaceplot.png|300px]]||[[File:mlw115_H2H_MEP_surfaceplot_dynamics.png|300px]]
|}

 Question 3. Comment on how the mep and the trajectory you just calculated differ.

It can be seen in the MEP plot that the trajectory follows the minimum energy valley floor without any obvious deviations or oscillation. The MEP trajectory is corresponding to infinitely slow motion, it does not accurately represent the atomic motion of the reaction. The trajectory oscillates in the dynamics plot, also in much less steps the AB bond distance increases considerably more. This plot is more realistic as it accounts for the motion of atoms during the reaction.

{| class="wikitable" border="1"
|+ Final Geometry
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| 2.486||1.167
|}

=== Reversed starting positions ===

What would change if we used the initial conditions r1 = rts and r2 = rts+0.01 instead? ****** DO THIS ******

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| -2.486||-1.167
|}

[[File:mlw115_H2H_dynamics_reversed_momenta.png|300px]]

From taking the previous runs final conditions and setting these as the initial conditions and reversing the momenta values the reaction seems to run in reverse, reaching an end exactly at the transition state.

===Determining Reactivity===

Question 4. Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2.0
|}

{| class="wikitable" border=1
! pAB !! pBC!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -1.25 || -2.5 || -99.018|| Yes || [[File:Mlw115_Surface_Plot_1.25_2.5.png|300px]] || Initially the HAHB distance changes very little as the vibrational energy is quite low, as the HC approaches it can be seen that the BC distance decreases linearly. The molecules pass over the transition state, after this point the HAHB distance decreases and the HBHC distance oscillates slightly, with a higher vibrational energy than the HAHB molecule. This overall process represents the HAHB bond breaking and forming a HBHC molecule.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mlw115_Surface_Plot_1.5_2.0.png|300px]] || The molecule approaches the maxima with some vibrational energy, shown by oscillations in the initial path, the energy of the system is not high enough to pass over the transition state energy maxima so the HC "rebounds" and the HBHC distance begins to increase. The HAHB molecule oscillations increase.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mlw115_Surface_Plot_1.5_2.5.png|300px]] || The reaction proceeds similarly to the first case with HC approaching and forming a bond with HB as there is sufficient energy to proceed over the transition state maxima. The vibrational energy throughout is slightly higher than the first case, demonstrated by higher oscillations in atom distances.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mlw115_Surface_Plot_2.5_5.png|300px]] || In this case the energy is high enough and the atoms are able to pass over the transition state energy maxima, however the vibrational energy of the HBHC system formed seems to be too high to form the molecule and the HC atom rebounds and the distance begins to decrease and so the HAHB is reformed but with considerably higher oscillations.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mlw115_Surface_Plot_2.5_5.2.png|300px]] || The HC approaches the oscillating HAHB system, the HB atom rebounds between the two atoms, as it can be seen on the surface plot the system passes over the transition state maxima, back over to the other side and back again as this occurs. Finally the HBHC bond is formed with a high vibrationally energy
|}

===Transition State Theory===

 Question 5 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? 

Transition state theory is used to understand how elementary chemical reactions take place assuming a quasi-equilibirum between reactanting complexes and those at the transition state. The rate of reaction can be understood by inspecting the complexes at the maximum saddle point of the minimum energy pathway. The reactants are in equilibrium with the activated transition state complexes and can be converted to the products.

This theory is useful to theoretically understand a chemical reaction however it operates on a series of assumptions which limit the theory. For instance the atoms are assumed to behave classically, ignoring tunelling

<ref>https://goldbook.iupac.org/html/T/T06470.html</ref>

=Exercise 2: H-H-F System=

==PES Inspection==

For the forward reaction of H2 + F --> HF + H is exothermic, as shown in the surface plot below the energy of the reactants H2 is higher than the energy of the product HF. As the energy absorbed in breaking the H-H bond in the first case is less than the energy released forming the H-F bond which means that in total energy is released which results in an exothermic reaction.

{| class="wikitable" border=1
! Surface Plot H2+F->HF+H !! Surface Plot HF+H -> H2+F
|-
|[[File:mlw115_Surface_Plot_HHF_labelled.png|300px]]||[[File:mlw115_Surface_Plot_HHF_H2Prod.png|300px]]|
|}

==Determining Transition State ==

A = HA B = HB and C = F

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! Surface Plot !! Internuclear distance vs Time
|-
|[[File:Mlw115_Surface_Plot_TS_HHF.png|400px]]||[[File:Mlw115 INDvT HHF TS.png|350px]]|
|}

== Calculating Activation Energy ==

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! BC= 1.8135!! BC= 1.8135 Zoomed!! BC= 1.813
|-
|[[File:Mlw115_HHF_EvT_1.8135.png|300px]] ‎||[[File:Mlw115_Plot_1.813_EvT_zoomT0_Resized.png|300px]]||[[File:Mlw115_HHF_EvT_1.813.png|320px]]
|}

pAB = pBC =0

distAB = 0.74

{| class="wikitable" border=1
|+Activation energy HF+H->H2+F
|-
! !! Energy (Kcal/mol)
|-
|Initial energy || -103.752
|-
|Final energy|| -133.89
|-
|EA|| 30.138
|}

{| class="wikitable" border=1
|+Activation energy H2+F->HF+H
|-
! !! Energy (Kcal/mol)
|-
|Initial energy|| -103.742
|-
|Final energy|| -103.752
|-
|EA|| 0.01
|}

==Reaction Dynamics==

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.74||0
|-
| BC|| 2.3||-2.7
|}

{| class="wikitable" border=1
!Surface Plot!! Internuclear momentum vs time
|-
|[[File:Mlw_Surface_Plot_HHF_Reactive.png]]|| [[File:Mlw_INMvT_HHF_Reactive.png]]
|}

{| class="wikitable" border=1
!Energy!!
|-
|Kinetic|| 3.837
|-
|Potential|| -103.886

|-
|Total|| -100.049
|}

It can be seen from the animation that as the H-H (AB) bond is broken and the H-F (BC) bond is formed the vibrational energy (and so the kinetic energy) considerably increases for H-F. The internuclear momentum vs time graph demonstrates this, as the momentum is much higher for the H-F atom. This vibrational energy will be released as heat so experimentally an increase in temperature could confirm the reaction had occurred.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -3 || -0.5 || -100.339|| No || [[File:Mlw115_Surface_Plot_HHF_-3_0.5.png|300px]] || H2 approaches the fluorine atom slowly but when a collision is made the H rebounds straight back into the other H and they continue their path oscillating together in the opposite direction
|-

| 3 || -0.5 || -93.254 || No || [[File:Mlw115_Surface_Plot_HHF_3_0.5.png|300px]] || This is similar to the previous case however the hydrogen rebounds twice into flourine but on the second rebounds remains returns back to oscillating with the original H atom
|-
| 2 || -0.5|| -98.753|| No || [[File: Mlw115_Surface_Plot_HHF_2_0.5.png|300px]] || The HF molecule oscillates within the energy well at high vibrational energy after collision
|-
| 1.5|| -0.5|| 103.754|| No || [[File: Mlw115_Surface_Plot_HHF_1.5_0.5.png|300px]] || Similarly to the previous situation
|-
|0.1||-0.8|| 103.364|| Yes|| [[File: Mlw115_Surface_Plot_HHF_0.1_0.8.png|300px]] || Despite having lower total energy than the previous cases the reaction goes to completion as the momentum between the reacting atoms is sufficient
|}

Despite the H2 energy being higher than the activation energy if the BC atoms do not collide with a high enough energy the reaction will return back over the peak and the original reactants will be produced.

===Polanyi's Empirical Rules ===

Polanyi's rules state that in reactions with an early transition state (most closely resembling the reactants) translational energy most effectively promotes a successful reaction. Conversely where the transition state is late (resembling products closely) vibrational energy, specifically of the reactants most effectively promotes a successful reaction. <ref>The Journal of Chemical Physics 138, 234104 (2013); https://doi.org/10.1063/1.4810007</ref>

====H-F + H====
rAB= 2 rBC = 0.74
{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -5 || 0.2 || -103.364 || No || [[File:Mlw115_Surface_Plot_HFH_5_0.2.png|300px]] || Approaching H rebounds and reaction doesn't complete
|-
| 2.21||0.37||
|}

Reversal of conditions:
AB D 5.07 M -2.22
BC D 1.19 M 0.37

Random Guessing:
AB D 5.0 M -1.5
BC D 1.2 M 1.0

[[File: Mlw115_Surface_Plot_HFH_-1.5_1.png| 400px]]

=References=
<references/>

MRD:mlw115RXNDy

2018-05-22T14:04:21Z

Mlw115: /* Determining Reactivity */

= Reaction Dynamics =

==Introduction==

The reactivity of atom-diatom systems are studied in order to gain insight on the internal degrees of freedom of the system. Particularly vibrational and translational energy are considered with respect to overcoming the transition state energy barrier.

The atoms are considered to behave classically and the interactions are expressed as potential energy surfaces.

= Exercise 1: H2 + H system=

==Dynamics from the transition state region ==

 Question 1. What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

Where a reaction is modeled on a potential energy surface the lowest energy path between the reactants and products is seen as a "valley floor" of the surface. The transition structure is a saddle point along the minimum energy ridge.

The gradient of the potential energy surface is zero at the minima and so the first derivative with respect to the inter-nuclear distance is zero. This is also the case for the peak of the saddle point.

δV(ri)/δri = 0

However using second derivatives
δV2(ri)/δri2 > 0 Minima
δV2(ri)/δri2 < 0 Maxima

where V is the potential energy and r related to the nuclear positions.

The second derivative of a maximum point is negative, whereas the second derivative of a minimum is positive. As the saddle point is a maxima along the minimum path the second derivative is negative.

==Estimating Transition State Position ==

 Question 2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. 

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.908||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot 1 !! Surface Plot 2!! Internuclear distance vs Time
|-
|[[File:Mlw115_H2H_saddle.png|300px]]||[[File:mlw115_H2H_TS.png|300px]]||[[File:Mlw115_H2H_TS_INDvT.png|300px]]|
|}

The best guess of inter nuclear distances at the transition state was found using trial and error, to find the positions that would result in the minimum trajectory. As the system is symmetric the distances would be equal. The two surface plots are included to demonstrate the "reaction path". The internuclear distance vs time plot demonstrates no change in any bond lengths at the transition state so the overall momenta is 0 and potential energy is at the maximum.

== Reaction Path ==
===MEP vs Dynamics===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.909||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot (MEP) !! Surface Plot (Dynamics
|-
|[[File:Mlw115_H2H_MEP_surfaceplot.png|300px]]||[[File:mlw115_H2H_MEP_surfaceplot_dynamics.png|300px]]
|}

 Question 3. Comment on how the mep and the trajectory you just calculated differ.

It can be seen in the MEP plot that the trajectory follows the minimum energy valley floor without any obvious deviations or oscillation. The MEP trajectory is corresponding to infinitely slow motion, it does not accurately represent the atomic motion of the reaction. The trajectory oscillates in the dynamics plot, also in much less steps the AB bond distance increases considerably more. This plot is more realistic as it accounts for the motion of atoms during the reaction.

{| class="wikitable" border="1"
|+ Final Geometry
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| 2.486||1.167
|}

=== Reversed starting positions ===

What would change if we used the initial conditions r1 = rts and r2 = rts+0.01 instead? ****** DO THIS ******

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| -2.486||-1.167
|}

[[File:mlw115_H2H_dynamics_reversed_momenta.png|300px]]

From taking the previous runs final conditions and setting these as the initial conditions and reversing the momenta values the reaction seems to run in reverse, reaching an end exactly at the transition state.

===Determining Reactivity===

Question 4. Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2.0
|}

{| class="wikitable" border=1
! pAB !! pBC!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -1.25 || -2.5 || -99.018|| Yes || [[File:Mlw115_Surface_Plot_1.25_2.5.png|300px]] || Initially the HAHB distance changes very little as the vibrational energy is quite low, as the HC approaches it can be seen that the BC distance decreases linearly. The molecules pass over the transition state, after this point the HAHB distance decreases and the HBHC distance oscillates slightly, with a higher vibrational energy than the HAHB molecule. This overall process represents the HAHB bond breaking and forming a HBHC molecule.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mlw115_Surface_Plot_1.5_2.0.png|300px]] || The molecule approaches the maxima with some vibrational energy, shown by oscillations in the initial path, the energy of the system is not high enough to pass over the transition state energy maxima so the HC "rebounds" and the HBHC distance begins to increase. The HAHB molecule oscillations increase.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mlw115_Surface_Plot_1.5_2.5.png|300px]] || The reaction proceeds similarly to the first case with HC approaching and forming a bond with HB as there is sufficient energy to proceed over the transition state maxima. The vibrational energy throughout is slightly higher than the first case, demonstrated by higher oscillations in atom distances.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mlw115_Surface_Plot_2.5_5.png|300px]] || In this case the energy is high enough and the atoms are able to pass over the transition state energy maxima, however the vibrational energy of the HBHC system formed seems to be too high to form the molecule and the HC atom rebounds and the distance begins to decrease and so the HAHB is reformed but with considerably higher oscillations.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mlw115_Surface_Plot_2.5_5.2.png|300px]] || The HC approaches the oscillating HAHB system, the HB atom rebounds between the two atoms, as it can be seen on the surface plot the system passes over the transition state maxima, back over to the other side and back again as this occurs. Finally the HBHC bond is formed with a high vibrationally energy
|}

===Transition State Theory===

TS state theory assumes that if the energy is high enough to cross the Transition state energy maxima then the reaction will be able to go to completion. However as demonstrated in the penultimate example the system can cross back over the maxima and the initial product will be reformed resulting in an incomplete reaction. This therefore will incorrectly predict the rate of reaction to be higher than experimentally measured as not all collisions above the necessary energy will result in a successful reaction.

=Exercise 2: H-H-F System=

==PES Inspection==

For the forward reaction of H2 + F --> HF + H is exothermic, as shown in the surface plot below the energy of the reactants H2 is higher than the energy of the product HF. As the energy absorbed in breaking the H-H bond in the first case is less than the energy released forming the H-F bond which means that in total energy is released which results in an exothermic reaction.

{| class="wikitable" border=1
! Surface Plot H2+F->HF+H !! Surface Plot HF+H -> H2+F
|-
|[[File:mlw115_Surface_Plot_HHF_labelled.png|300px]]||[[File:mlw115_Surface_Plot_HHF_H2Prod.png|300px]]|
|}

==Determining Transition State ==

A = HA B = HB and C = F

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! Surface Plot !! Internuclear distance vs Time
|-
|[[File:Mlw115_Surface_Plot_TS_HHF.png|400px]]||[[File:Mlw115 INDvT HHF TS.png|350px]]|
|}

== Calculating Activation Energy ==

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! BC= 1.8135!! BC= 1.8135 Zoomed!! BC= 1.813
|-
|[[File:Mlw115_HHF_EvT_1.8135.png|300px]] ‎||[[File:Mlw115_Plot_1.813_EvT_zoomT0_Resized.png|300px]]||[[File:Mlw115_HHF_EvT_1.813.png|320px]]
|}

pAB = pBC =0

distAB = 0.74

{| class="wikitable" border=1
|+Activation energy HF+H->H2+F
|-
! !! Energy (Kcal/mol)
|-
|Initial energy || -103.752
|-
|Final energy|| -133.89
|-
|EA|| 30.138
|}

{| class="wikitable" border=1
|+Activation energy H2+F->HF+H
|-
! !! Energy (Kcal/mol)
|-
|Initial energy|| -103.742
|-
|Final energy|| -103.752
|-
|EA|| 0.01
|}

==Reaction Dynamics==

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.74||0
|-
| BC|| 2.3||-2.7
|}

{| class="wikitable" border=1
!Surface Plot!! Internuclear momentum vs time
|-
|[[File:Mlw_Surface_Plot_HHF_Reactive.png]]|| [[File:Mlw_INMvT_HHF_Reactive.png]]
|}

{| class="wikitable" border=1
!Energy!!
|-
|Kinetic|| 3.837
|-
|Potential|| -103.886

|-
|Total|| -100.049
|}

It can be seen from the animation that as the H-H (AB) bond is broken and the H-F (BC) bond is formed the vibrational energy (and so the kinetic energy) considerably increases for H-F. The internuclear momentum vs time graph demonstrates this, as the momentum is much higher for the H-F atom. This vibrational energy will be released as heat so experimentally an increase in temperature could confirm the reaction had occurred.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -3 || -0.5 || -100.339|| No || [[File:Mlw115_Surface_Plot_HHF_-3_0.5.png|300px]] || H2 approaches the fluorine atom slowly but when a collision is made the H rebounds straight back into the other H and they continue their path oscillating together in the opposite direction
|-

| 3 || -0.5 || -93.254 || No || [[File:Mlw115_Surface_Plot_HHF_3_0.5.png|300px]] || This is similar to the previous case however the hydrogen rebounds twice into flourine but on the second rebounds remains returns back to oscillating with the original H atom
|-
| 2 || -0.5|| -98.753|| No || [[File: Mlw115_Surface_Plot_HHF_2_0.5.png|300px]] || The HF molecule oscillates within the energy well at high vibrational energy after collision
|-
| 1.5|| -0.5|| 103.754|| No || [[File: Mlw115_Surface_Plot_HHF_1.5_0.5.png|300px]] || Similarly to the previous situation
|-
|0.1||-0.8|| 103.364|| Yes|| [[File: Mlw115_Surface_Plot_HHF_0.1_0.8.png|300px]] || Despite having lower total energy than the previous cases the reaction goes to completion as the momentum between the reacting atoms is sufficient
|}

Despite the H2 energy being higher than the activation energy if the BC atoms do not collide with a high enough energy the reaction will return back over the peak and the original reactants will be produced.

===Polanyi's Empirical Rules ===

Polanyi's rules state that in reactions with an early transition state (most closely resembling the reactants) translational energy most effectively promotes a successful reaction. Conversely where the transition state is late (resembling products closely) vibrational energy, specifically of the reactants most effectively promotes a successful reaction. <ref>The Journal of Chemical Physics 138, 234104 (2013); https://doi.org/10.1063/1.4810007</ref>

====H-F + H====
rAB= 2 rBC = 0.74
{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -5 || 0.2 || -103.364 || No || [[File:Mlw115_Surface_Plot_HFH_5_0.2.png|300px]] || Approaching H rebounds and reaction doesn't complete
|-
| 2.21||0.37||
|}

Reversal of conditions:
AB D 5.07 M -2.22
BC D 1.19 M 0.37

Random Guessing:
AB D 5.0 M -1.5
BC D 1.2 M 1.0

[[File: Mlw115_Surface_Plot_HFH_-1.5_1.png| 400px]]

=References=
<references/>

MRD:mlw115RXNDy

2018-05-22T14:03:36Z

Mlw115: /* Determining Reactivity */

= Reaction Dynamics =

==Introduction==

The reactivity of atom-diatom systems are studied in order to gain insight on the internal degrees of freedom of the system. Particularly vibrational and translational energy are considered with respect to overcoming the transition state energy barrier.

The atoms are considered to behave classically and the interactions are expressed as potential energy surfaces.

= Exercise 1: H2 + H system=

==Dynamics from the transition state region ==

 Question 1. What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

Where a reaction is modeled on a potential energy surface the lowest energy path between the reactants and products is seen as a "valley floor" of the surface. The transition structure is a saddle point along the minimum energy ridge.

The gradient of the potential energy surface is zero at the minima and so the first derivative with respect to the inter-nuclear distance is zero. This is also the case for the peak of the saddle point.

δV(ri)/δri = 0

However using second derivatives
δV2(ri)/δri2 > 0 Minima
δV2(ri)/δri2 < 0 Maxima

where V is the potential energy and r related to the nuclear positions.

The second derivative of a maximum point is negative, whereas the second derivative of a minimum is positive. As the saddle point is a maxima along the minimum path the second derivative is negative.

==Estimating Transition State Position ==

 Question 2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. 

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.908||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot 1 !! Surface Plot 2!! Internuclear distance vs Time
|-
|[[File:Mlw115_H2H_saddle.png|300px]]||[[File:mlw115_H2H_TS.png|300px]]||[[File:Mlw115_H2H_TS_INDvT.png|300px]]|
|}

The best guess of inter nuclear distances at the transition state was found using trial and error, to find the positions that would result in the minimum trajectory. As the system is symmetric the distances would be equal. The two surface plots are included to demonstrate the "reaction path". The internuclear distance vs time plot demonstrates no change in any bond lengths at the transition state so the overall momenta is 0 and potential energy is at the maximum.

== Reaction Path ==
===MEP vs Dynamics===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.909||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot (MEP) !! Surface Plot (Dynamics
|-
|[[File:Mlw115_H2H_MEP_surfaceplot.png|300px]]||[[File:mlw115_H2H_MEP_surfaceplot_dynamics.png|300px]]
|}

 Question 3. Comment on how the mep and the trajectory you just calculated differ.

It can be seen in the MEP plot that the trajectory follows the minimum energy valley floor without any obvious deviations or oscillation. The MEP trajectory is corresponding to infinitely slow motion, it does not accurately represent the atomic motion of the reaction. The trajectory oscillates in the dynamics plot, also in much less steps the AB bond distance increases considerably more. This plot is more realistic as it accounts for the motion of atoms during the reaction.

{| class="wikitable" border="1"
|+ Final Geometry
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| 2.486||1.167
|}

=== Reversed starting positions ===

What would change if we used the initial conditions r1 = rts and r2 = rts+0.01 instead? ****** DO THIS ******

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| -2.486||-1.167
|}

[[File:mlw115_H2H_dynamics_reversed_momenta.png|300px]]

From taking the previous runs final conditions and setting these as the initial conditions and reversing the momenta values the reaction seems to run in reverse, reaching an end exactly at the transition state.

===Determining Reactivity===

 rAB = 0.74 

 rBC = 2.0 

Question 4. Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.

{| class="wikitable" border=1
! pAB !! pBC!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -1.25 || -2.5 || -99.018|| Yes || [[File:Mlw115_Surface_Plot_1.25_2.5.png|300px]] || Initially the HAHB distance changes very little as the vibrational energy is quite low, as the HC approaches it can be seen that the BC distance decreases linearly. The molecules pass over the transition state, after this point the HAHB distance decreases and the HBHC distance oscillates slightly, with a higher vibrational energy than the HAHB molecule. This overall process represents the HAHB bond breaking and forming a HBHC molecule.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mlw115_Surface_Plot_1.5_2.0.png|300px]] || The molecule approaches the maxima with some vibrational energy, shown by oscillations in the initial path, the energy of the system is not high enough to pass over the transition state energy maxima so the HC "rebounds" and the HBHC distance begins to increase. The HAHB molecule oscillations increase.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mlw115_Surface_Plot_1.5_2.5.png|300px]] || The reaction proceeds similarly to the first case with HC approaching and forming a bond with HB as there is sufficient energy to proceed over the transition state maxima. The vibrational energy throughout is slightly higher than the first case, demonstrated by higher oscillations in atom distances.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mlw115_Surface_Plot_2.5_5.png|300px]] || In this case the energy is high enough and the atoms are able to pass over the transition state energy maxima, however the vibrational energy of the HBHC system formed seems to be too high to form the molecule and the HC atom rebounds and the distance begins to decrease and so the HAHB is reformed but with considerably higher oscillations.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mlw115_Surface_Plot_2.5_5.2.png|300px]] || The HC approaches the oscillating HAHB system, the HB atom rebounds between the two atoms, as it can be seen on the surface plot the system passes over the transition state maxima, back over to the other side and back again as this occurs. Finally the HBHC bond is formed with a high vibrationally energy
|}

===Transition State Theory===

TS state theory assumes that if the energy is high enough to cross the Transition state energy maxima then the reaction will be able to go to completion. However as demonstrated in the penultimate example the system can cross back over the maxima and the initial product will be reformed resulting in an incomplete reaction. This therefore will incorrectly predict the rate of reaction to be higher than experimentally measured as not all collisions above the necessary energy will result in a successful reaction.

=Exercise 2: H-H-F System=

==PES Inspection==

For the forward reaction of H2 + F --> HF + H is exothermic, as shown in the surface plot below the energy of the reactants H2 is higher than the energy of the product HF. As the energy absorbed in breaking the H-H bond in the first case is less than the energy released forming the H-F bond which means that in total energy is released which results in an exothermic reaction.

{| class="wikitable" border=1
! Surface Plot H2+F->HF+H !! Surface Plot HF+H -> H2+F
|-
|[[File:mlw115_Surface_Plot_HHF_labelled.png|300px]]||[[File:mlw115_Surface_Plot_HHF_H2Prod.png|300px]]|
|}

==Determining Transition State ==

A = HA B = HB and C = F

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! Surface Plot !! Internuclear distance vs Time
|-
|[[File:Mlw115_Surface_Plot_TS_HHF.png|400px]]||[[File:Mlw115 INDvT HHF TS.png|350px]]|
|}

== Calculating Activation Energy ==

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! BC= 1.8135!! BC= 1.8135 Zoomed!! BC= 1.813
|-
|[[File:Mlw115_HHF_EvT_1.8135.png|300px]] ‎||[[File:Mlw115_Plot_1.813_EvT_zoomT0_Resized.png|300px]]||[[File:Mlw115_HHF_EvT_1.813.png|320px]]
|}

pAB = pBC =0

distAB = 0.74

{| class="wikitable" border=1
|+Activation energy HF+H->H2+F
|-
! !! Energy (Kcal/mol)
|-
|Initial energy || -103.752
|-
|Final energy|| -133.89
|-
|EA|| 30.138
|}

{| class="wikitable" border=1
|+Activation energy H2+F->HF+H
|-
! !! Energy (Kcal/mol)
|-
|Initial energy|| -103.742
|-
|Final energy|| -103.752
|-
|EA|| 0.01
|}

==Reaction Dynamics==

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.74||0
|-
| BC|| 2.3||-2.7
|}

{| class="wikitable" border=1
!Surface Plot!! Internuclear momentum vs time
|-
|[[File:Mlw_Surface_Plot_HHF_Reactive.png]]|| [[File:Mlw_INMvT_HHF_Reactive.png]]
|}

{| class="wikitable" border=1
!Energy!!
|-
|Kinetic|| 3.837
|-
|Potential|| -103.886

|-
|Total|| -100.049
|}

It can be seen from the animation that as the H-H (AB) bond is broken and the H-F (BC) bond is formed the vibrational energy (and so the kinetic energy) considerably increases for H-F. The internuclear momentum vs time graph demonstrates this, as the momentum is much higher for the H-F atom. This vibrational energy will be released as heat so experimentally an increase in temperature could confirm the reaction had occurred.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -3 || -0.5 || -100.339|| No || [[File:Mlw115_Surface_Plot_HHF_-3_0.5.png|300px]] || H2 approaches the fluorine atom slowly but when a collision is made the H rebounds straight back into the other H and they continue their path oscillating together in the opposite direction
|-

| 3 || -0.5 || -93.254 || No || [[File:Mlw115_Surface_Plot_HHF_3_0.5.png|300px]] || This is similar to the previous case however the hydrogen rebounds twice into flourine but on the second rebounds remains returns back to oscillating with the original H atom
|-
| 2 || -0.5|| -98.753|| No || [[File: Mlw115_Surface_Plot_HHF_2_0.5.png|300px]] || The HF molecule oscillates within the energy well at high vibrational energy after collision
|-
| 1.5|| -0.5|| 103.754|| No || [[File: Mlw115_Surface_Plot_HHF_1.5_0.5.png|300px]] || Similarly to the previous situation
|-
|0.1||-0.8|| 103.364|| Yes|| [[File: Mlw115_Surface_Plot_HHF_0.1_0.8.png|300px]] || Despite having lower total energy than the previous cases the reaction goes to completion as the momentum between the reacting atoms is sufficient
|}

Despite the H2 energy being higher than the activation energy if the BC atoms do not collide with a high enough energy the reaction will return back over the peak and the original reactants will be produced.

===Polanyi's Empirical Rules ===

Polanyi's rules state that in reactions with an early transition state (most closely resembling the reactants) translational energy most effectively promotes a successful reaction. Conversely where the transition state is late (resembling products closely) vibrational energy, specifically of the reactants most effectively promotes a successful reaction. <ref>The Journal of Chemical Physics 138, 234104 (2013); https://doi.org/10.1063/1.4810007</ref>

====H-F + H====
rAB= 2 rBC = 0.74
{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -5 || 0.2 || -103.364 || No || [[File:Mlw115_Surface_Plot_HFH_5_0.2.png|300px]] || Approaching H rebounds and reaction doesn't complete
|-
| 2.21||0.37||
|}

Reversal of conditions:
AB D 5.07 M -2.22
BC D 1.19 M 0.37

Random Guessing:
AB D 5.0 M -1.5
BC D 1.2 M 1.0

[[File: Mlw115_Surface_Plot_HFH_-1.5_1.png| 400px]]

=References=
<references/>

MRD:mlw115RXNDy

2018-05-22T14:03:23Z

Mlw115: /* Determining Reactivity */

= Reaction Dynamics =

==Introduction==

The reactivity of atom-diatom systems are studied in order to gain insight on the internal degrees of freedom of the system. Particularly vibrational and translational energy are considered with respect to overcoming the transition state energy barrier.

The atoms are considered to behave classically and the interactions are expressed as potential energy surfaces.

= Exercise 1: H2 + H system=

==Dynamics from the transition state region ==

 Question 1. What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

Where a reaction is modeled on a potential energy surface the lowest energy path between the reactants and products is seen as a "valley floor" of the surface. The transition structure is a saddle point along the minimum energy ridge.

The gradient of the potential energy surface is zero at the minima and so the first derivative with respect to the inter-nuclear distance is zero. This is also the case for the peak of the saddle point.

δV(ri)/δri = 0

However using second derivatives
δV2(ri)/δri2 > 0 Minima
δV2(ri)/δri2 < 0 Maxima

where V is the potential energy and r related to the nuclear positions.

The second derivative of a maximum point is negative, whereas the second derivative of a minimum is positive. As the saddle point is a maxima along the minimum path the second derivative is negative.

==Estimating Transition State Position ==

 Question 2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. 

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.908||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot 1 !! Surface Plot 2!! Internuclear distance vs Time
|-
|[[File:Mlw115_H2H_saddle.png|300px]]||[[File:mlw115_H2H_TS.png|300px]]||[[File:Mlw115_H2H_TS_INDvT.png|300px]]|
|}

The best guess of inter nuclear distances at the transition state was found using trial and error, to find the positions that would result in the minimum trajectory. As the system is symmetric the distances would be equal. The two surface plots are included to demonstrate the "reaction path". The internuclear distance vs time plot demonstrates no change in any bond lengths at the transition state so the overall momenta is 0 and potential energy is at the maximum.

== Reaction Path ==
===MEP vs Dynamics===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.909||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot (MEP) !! Surface Plot (Dynamics
|-
|[[File:Mlw115_H2H_MEP_surfaceplot.png|300px]]||[[File:mlw115_H2H_MEP_surfaceplot_dynamics.png|300px]]
|}

 Question 3. Comment on how the mep and the trajectory you just calculated differ.

It can be seen in the MEP plot that the trajectory follows the minimum energy valley floor without any obvious deviations or oscillation. The MEP trajectory is corresponding to infinitely slow motion, it does not accurately represent the atomic motion of the reaction. The trajectory oscillates in the dynamics plot, also in much less steps the AB bond distance increases considerably more. This plot is more realistic as it accounts for the motion of atoms during the reaction.

{| class="wikitable" border="1"
|+ Final Geometry
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| 2.486||1.167
|}

=== Reversed starting positions ===

What would change if we used the initial conditions r1 = rts and r2 = rts+0.01 instead? ****** DO THIS ******

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| -2.486||-1.167
|}

[[File:mlw115_H2H_dynamics_reversed_momenta.png|300px]]

From taking the previous runs final conditions and setting these as the initial conditions and reversing the momenta values the reaction seems to run in reverse, reaching an end exactly at the transition state.

===Determining Reactivity===

 rAB = 0.74 

 rBC = 2.0 
Question 4. Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.

{| class="wikitable" border=1
! pAB !! pBC!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -1.25 || -2.5 || -99.018|| Yes || [[File:Mlw115_Surface_Plot_1.25_2.5.png|300px]] || Initially the HAHB distance changes very little as the vibrational energy is quite low, as the HC approaches it can be seen that the BC distance decreases linearly. The molecules pass over the transition state, after this point the HAHB distance decreases and the HBHC distance oscillates slightly, with a higher vibrational energy than the HAHB molecule. This overall process represents the HAHB bond breaking and forming a HBHC molecule.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mlw115_Surface_Plot_1.5_2.0.png|300px]] || The molecule approaches the maxima with some vibrational energy, shown by oscillations in the initial path, the energy of the system is not high enough to pass over the transition state energy maxima so the HC "rebounds" and the HBHC distance begins to increase. The HAHB molecule oscillations increase.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mlw115_Surface_Plot_1.5_2.5.png|300px]] || The reaction proceeds similarly to the first case with HC approaching and forming a bond with HB as there is sufficient energy to proceed over the transition state maxima. The vibrational energy throughout is slightly higher than the first case, demonstrated by higher oscillations in atom distances.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mlw115_Surface_Plot_2.5_5.png|300px]] || In this case the energy is high enough and the atoms are able to pass over the transition state energy maxima, however the vibrational energy of the HBHC system formed seems to be too high to form the molecule and the HC atom rebounds and the distance begins to decrease and so the HAHB is reformed but with considerably higher oscillations.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mlw115_Surface_Plot_2.5_5.2.png|300px]] || The HC approaches the oscillating HAHB system, the HB atom rebounds between the two atoms, as it can be seen on the surface plot the system passes over the transition state maxima, back over to the other side and back again as this occurs. Finally the HBHC bond is formed with a high vibrationally energy
|}

===Transition State Theory===

TS state theory assumes that if the energy is high enough to cross the Transition state energy maxima then the reaction will be able to go to completion. However as demonstrated in the penultimate example the system can cross back over the maxima and the initial product will be reformed resulting in an incomplete reaction. This therefore will incorrectly predict the rate of reaction to be higher than experimentally measured as not all collisions above the necessary energy will result in a successful reaction.

=Exercise 2: H-H-F System=

==PES Inspection==

For the forward reaction of H2 + F --> HF + H is exothermic, as shown in the surface plot below the energy of the reactants H2 is higher than the energy of the product HF. As the energy absorbed in breaking the H-H bond in the first case is less than the energy released forming the H-F bond which means that in total energy is released which results in an exothermic reaction.

{| class="wikitable" border=1
! Surface Plot H2+F->HF+H !! Surface Plot HF+H -> H2+F
|-
|[[File:mlw115_Surface_Plot_HHF_labelled.png|300px]]||[[File:mlw115_Surface_Plot_HHF_H2Prod.png|300px]]|
|}

==Determining Transition State ==

A = HA B = HB and C = F

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! Surface Plot !! Internuclear distance vs Time
|-
|[[File:Mlw115_Surface_Plot_TS_HHF.png|400px]]||[[File:Mlw115 INDvT HHF TS.png|350px]]|
|}

== Calculating Activation Energy ==

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! BC= 1.8135!! BC= 1.8135 Zoomed!! BC= 1.813
|-
|[[File:Mlw115_HHF_EvT_1.8135.png|300px]] ‎||[[File:Mlw115_Plot_1.813_EvT_zoomT0_Resized.png|300px]]||[[File:Mlw115_HHF_EvT_1.813.png|320px]]
|}

pAB = pBC =0

distAB = 0.74

{| class="wikitable" border=1
|+Activation energy HF+H->H2+F
|-
! !! Energy (Kcal/mol)
|-
|Initial energy || -103.752
|-
|Final energy|| -133.89
|-
|EA|| 30.138
|}

{| class="wikitable" border=1
|+Activation energy H2+F->HF+H
|-
! !! Energy (Kcal/mol)
|-
|Initial energy|| -103.742
|-
|Final energy|| -103.752
|-
|EA|| 0.01
|}

==Reaction Dynamics==

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.74||0
|-
| BC|| 2.3||-2.7
|}

{| class="wikitable" border=1
!Surface Plot!! Internuclear momentum vs time
|-
|[[File:Mlw_Surface_Plot_HHF_Reactive.png]]|| [[File:Mlw_INMvT_HHF_Reactive.png]]
|}

{| class="wikitable" border=1
!Energy!!
|-
|Kinetic|| 3.837
|-
|Potential|| -103.886

|-
|Total|| -100.049
|}

It can be seen from the animation that as the H-H (AB) bond is broken and the H-F (BC) bond is formed the vibrational energy (and so the kinetic energy) considerably increases for H-F. The internuclear momentum vs time graph demonstrates this, as the momentum is much higher for the H-F atom. This vibrational energy will be released as heat so experimentally an increase in temperature could confirm the reaction had occurred.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -3 || -0.5 || -100.339|| No || [[File:Mlw115_Surface_Plot_HHF_-3_0.5.png|300px]] || H2 approaches the fluorine atom slowly but when a collision is made the H rebounds straight back into the other H and they continue their path oscillating together in the opposite direction
|-

| 3 || -0.5 || -93.254 || No || [[File:Mlw115_Surface_Plot_HHF_3_0.5.png|300px]] || This is similar to the previous case however the hydrogen rebounds twice into flourine but on the second rebounds remains returns back to oscillating with the original H atom
|-
| 2 || -0.5|| -98.753|| No || [[File: Mlw115_Surface_Plot_HHF_2_0.5.png|300px]] || The HF molecule oscillates within the energy well at high vibrational energy after collision
|-
| 1.5|| -0.5|| 103.754|| No || [[File: Mlw115_Surface_Plot_HHF_1.5_0.5.png|300px]] || Similarly to the previous situation
|-
|0.1||-0.8|| 103.364|| Yes|| [[File: Mlw115_Surface_Plot_HHF_0.1_0.8.png|300px]] || Despite having lower total energy than the previous cases the reaction goes to completion as the momentum between the reacting atoms is sufficient
|}

Despite the H2 energy being higher than the activation energy if the BC atoms do not collide with a high enough energy the reaction will return back over the peak and the original reactants will be produced.

===Polanyi's Empirical Rules ===

Polanyi's rules state that in reactions with an early transition state (most closely resembling the reactants) translational energy most effectively promotes a successful reaction. Conversely where the transition state is late (resembling products closely) vibrational energy, specifically of the reactants most effectively promotes a successful reaction. <ref>The Journal of Chemical Physics 138, 234104 (2013); https://doi.org/10.1063/1.4810007</ref>

====H-F + H====
rAB= 2 rBC = 0.74
{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -5 || 0.2 || -103.364 || No || [[File:Mlw115_Surface_Plot_HFH_5_0.2.png|300px]] || Approaching H rebounds and reaction doesn't complete
|-
| 2.21||0.37||
|}

Reversal of conditions:
AB D 5.07 M -2.22
BC D 1.19 M 0.37

Random Guessing:
AB D 5.0 M -1.5
BC D 1.2 M 1.0

[[File: Mlw115_Surface_Plot_HFH_-1.5_1.png| 400px]]

=References=
<references/>

MRD:mlw115RXNDy

2018-05-22T14:02:09Z

Mlw115: /* Reaction Path */

= Reaction Dynamics =

==Introduction==

The reactivity of atom-diatom systems are studied in order to gain insight on the internal degrees of freedom of the system. Particularly vibrational and translational energy are considered with respect to overcoming the transition state energy barrier.

The atoms are considered to behave classically and the interactions are expressed as potential energy surfaces.

= Exercise 1: H2 + H system=

==Dynamics from the transition state region ==

 Question 1. What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

Where a reaction is modeled on a potential energy surface the lowest energy path between the reactants and products is seen as a "valley floor" of the surface. The transition structure is a saddle point along the minimum energy ridge.

The gradient of the potential energy surface is zero at the minima and so the first derivative with respect to the inter-nuclear distance is zero. This is also the case for the peak of the saddle point.

δV(ri)/δri = 0

However using second derivatives
δV2(ri)/δri2 > 0 Minima
δV2(ri)/δri2 < 0 Maxima

where V is the potential energy and r related to the nuclear positions.

The second derivative of a maximum point is negative, whereas the second derivative of a minimum is positive. As the saddle point is a maxima along the minimum path the second derivative is negative.

==Estimating Transition State Position ==

 Question 2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. 

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.908||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot 1 !! Surface Plot 2!! Internuclear distance vs Time
|-
|[[File:Mlw115_H2H_saddle.png|300px]]||[[File:mlw115_H2H_TS.png|300px]]||[[File:Mlw115_H2H_TS_INDvT.png|300px]]|
|}

The best guess of inter nuclear distances at the transition state was found using trial and error, to find the positions that would result in the minimum trajectory. As the system is symmetric the distances would be equal. The two surface plots are included to demonstrate the "reaction path". The internuclear distance vs time plot demonstrates no change in any bond lengths at the transition state so the overall momenta is 0 and potential energy is at the maximum.

== Reaction Path ==
===MEP vs Dynamics===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.909||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot (MEP) !! Surface Plot (Dynamics
|-
|[[File:Mlw115_H2H_MEP_surfaceplot.png|300px]]||[[File:mlw115_H2H_MEP_surfaceplot_dynamics.png|300px]]
|}

 Question 3. Comment on how the mep and the trajectory you just calculated differ.

It can be seen in the MEP plot that the trajectory follows the minimum energy valley floor without any obvious deviations or oscillation. The MEP trajectory is corresponding to infinitely slow motion, it does not accurately represent the atomic motion of the reaction. The trajectory oscillates in the dynamics plot, also in much less steps the AB bond distance increases considerably more. This plot is more realistic as it accounts for the motion of atoms during the reaction.

{| class="wikitable" border="1"
|+ Final Geometry
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| 2.486||1.167
|}

=== Reversed starting positions ===

What would change if we used the initial conditions r1 = rts and r2 = rts+0.01 instead? ****** DO THIS ******

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| -2.486||-1.167
|}

[[File:mlw115_H2H_dynamics_reversed_momenta.png|300px]]

From taking the previous runs final conditions and setting these as the initial conditions and reversing the momenta values the reaction seems to run in reverse, reaching an end exactly at the transition state.

===Determining Reactivity===

 rAB = 0.74 

 rBC = 2.0 

{| class="wikitable" border=1
! pAB !! pBC!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -1.25 || -2.5 || -99.018|| Yes || [[File:Mlw115_Surface_Plot_1.25_2.5.png|300px]] || Initially the HAHB distance changes very little as the vibrational energy is quite low, as the HC approaches it can be seen that the BC distance decreases linearly. The molecules pass over the transition state, after this point the HAHB distance decreases and the HBHC distance oscillates slightly, with a higher vibrational energy than the HAHB molecule. This overall process represents the HAHB bond breaking and forming a HBHC molecule.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mlw115_Surface_Plot_1.5_2.0.png|300px]] || The molecule approaches the maxima with some vibrational energy, shown by oscillations in the initial path, the energy of the system is not high enough to pass over the transition state energy maxima so the HC "rebounds" and the HBHC distance begins to increase. The HAHB molecule oscillations increase.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mlw115_Surface_Plot_1.5_2.5.png|300px]] || The reaction proceeds similarly to the first case with HC approaching and forming a bond with HB as there is sufficient energy to proceed over the transition state maxima. The vibrational energy throughout is slightly higher than the first case, demonstrated by higher oscillations in atom distances.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mlw115_Surface_Plot_2.5_5.png|300px]] || In this case the energy is high enough and the atoms are able to pass over the transition state energy maxima, however the vibrational energy of the HBHC system formed seems to be too high to form the molecule and the HC atom rebounds and the distance begins to decrease and so the HAHB is reformed but with considerably higher oscillations.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mlw115_Surface_Plot_2.5_5.2.png|300px]] || The HC approaches the oscillating HAHB system, the HB atom rebounds between the two atoms, as it can be seen on the surface plot the system passes over the transition state maxima, back over to the other side and back again as this occurs. Finally the HBHC bond is formed with a high vibrationally energy
|}

===Transition State Theory===

TS state theory assumes that if the energy is high enough to cross the Transition state energy maxima then the reaction will be able to go to completion. However as demonstrated in the penultimate example the system can cross back over the maxima and the initial product will be reformed resulting in an incomplete reaction. This therefore will incorrectly predict the rate of reaction to be higher than experimentally measured as not all collisions above the necessary energy will result in a successful reaction.

=Exercise 2: H-H-F System=

==PES Inspection==

For the forward reaction of H2 + F --> HF + H is exothermic, as shown in the surface plot below the energy of the reactants H2 is higher than the energy of the product HF. As the energy absorbed in breaking the H-H bond in the first case is less than the energy released forming the H-F bond which means that in total energy is released which results in an exothermic reaction.

{| class="wikitable" border=1
! Surface Plot H2+F->HF+H !! Surface Plot HF+H -> H2+F
|-
|[[File:mlw115_Surface_Plot_HHF_labelled.png|300px]]||[[File:mlw115_Surface_Plot_HHF_H2Prod.png|300px]]|
|}

==Determining Transition State ==

A = HA B = HB and C = F

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! Surface Plot !! Internuclear distance vs Time
|-
|[[File:Mlw115_Surface_Plot_TS_HHF.png|400px]]||[[File:Mlw115 INDvT HHF TS.png|350px]]|
|}

== Calculating Activation Energy ==

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! BC= 1.8135!! BC= 1.8135 Zoomed!! BC= 1.813
|-
|[[File:Mlw115_HHF_EvT_1.8135.png|300px]] ‎||[[File:Mlw115_Plot_1.813_EvT_zoomT0_Resized.png|300px]]||[[File:Mlw115_HHF_EvT_1.813.png|320px]]
|}

pAB = pBC =0

distAB = 0.74

{| class="wikitable" border=1
|+Activation energy HF+H->H2+F
|-
! !! Energy (Kcal/mol)
|-
|Initial energy || -103.752
|-
|Final energy|| -133.89
|-
|EA|| 30.138
|}

{| class="wikitable" border=1
|+Activation energy H2+F->HF+H
|-
! !! Energy (Kcal/mol)
|-
|Initial energy|| -103.742
|-
|Final energy|| -103.752
|-
|EA|| 0.01
|}

==Reaction Dynamics==

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.74||0
|-
| BC|| 2.3||-2.7
|}

{| class="wikitable" border=1
!Surface Plot!! Internuclear momentum vs time
|-
|[[File:Mlw_Surface_Plot_HHF_Reactive.png]]|| [[File:Mlw_INMvT_HHF_Reactive.png]]
|}

{| class="wikitable" border=1
!Energy!!
|-
|Kinetic|| 3.837
|-
|Potential|| -103.886

|-
|Total|| -100.049
|}

It can be seen from the animation that as the H-H (AB) bond is broken and the H-F (BC) bond is formed the vibrational energy (and so the kinetic energy) considerably increases for H-F. The internuclear momentum vs time graph demonstrates this, as the momentum is much higher for the H-F atom. This vibrational energy will be released as heat so experimentally an increase in temperature could confirm the reaction had occurred.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -3 || -0.5 || -100.339|| No || [[File:Mlw115_Surface_Plot_HHF_-3_0.5.png|300px]] || H2 approaches the fluorine atom slowly but when a collision is made the H rebounds straight back into the other H and they continue their path oscillating together in the opposite direction
|-

| 3 || -0.5 || -93.254 || No || [[File:Mlw115_Surface_Plot_HHF_3_0.5.png|300px]] || This is similar to the previous case however the hydrogen rebounds twice into flourine but on the second rebounds remains returns back to oscillating with the original H atom
|-
| 2 || -0.5|| -98.753|| No || [[File: Mlw115_Surface_Plot_HHF_2_0.5.png|300px]] || The HF molecule oscillates within the energy well at high vibrational energy after collision
|-
| 1.5|| -0.5|| 103.754|| No || [[File: Mlw115_Surface_Plot_HHF_1.5_0.5.png|300px]] || Similarly to the previous situation
|-
|0.1||-0.8|| 103.364|| Yes|| [[File: Mlw115_Surface_Plot_HHF_0.1_0.8.png|300px]] || Despite having lower total energy than the previous cases the reaction goes to completion as the momentum between the reacting atoms is sufficient
|}

Despite the H2 energy being higher than the activation energy if the BC atoms do not collide with a high enough energy the reaction will return back over the peak and the original reactants will be produced.

===Polanyi's Empirical Rules ===

Polanyi's rules state that in reactions with an early transition state (most closely resembling the reactants) translational energy most effectively promotes a successful reaction. Conversely where the transition state is late (resembling products closely) vibrational energy, specifically of the reactants most effectively promotes a successful reaction. <ref>The Journal of Chemical Physics 138, 234104 (2013); https://doi.org/10.1063/1.4810007</ref>

====H-F + H====
rAB= 2 rBC = 0.74
{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -5 || 0.2 || -103.364 || No || [[File:Mlw115_Surface_Plot_HFH_5_0.2.png|300px]] || Approaching H rebounds and reaction doesn't complete
|-
| 2.21||0.37||
|}

Reversal of conditions:
AB D 5.07 M -2.22
BC D 1.19 M 0.37

Random Guessing:
AB D 5.0 M -1.5
BC D 1.2 M 1.0

[[File: Mlw115_Surface_Plot_HFH_-1.5_1.png| 400px]]

=References=
<references/>

MRD:mlw115RXNDy

2018-05-22T14:01:49Z

Mlw115: /* MEP vs Dynamics */

= Reaction Dynamics =

==Introduction==

The reactivity of atom-diatom systems are studied in order to gain insight on the internal degrees of freedom of the system. Particularly vibrational and translational energy are considered with respect to overcoming the transition state energy barrier.

The atoms are considered to behave classically and the interactions are expressed as potential energy surfaces.

= Exercise 1: H2 + H system=

==Dynamics from the transition state region ==

 Question 1. What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

Where a reaction is modeled on a potential energy surface the lowest energy path between the reactants and products is seen as a "valley floor" of the surface. The transition structure is a saddle point along the minimum energy ridge.

The gradient of the potential energy surface is zero at the minima and so the first derivative with respect to the inter-nuclear distance is zero. This is also the case for the peak of the saddle point.

δV(ri)/δri = 0

However using second derivatives
δV2(ri)/δri2 > 0 Minima
δV2(ri)/δri2 < 0 Maxima

where V is the potential energy and r related to the nuclear positions.

The second derivative of a maximum point is negative, whereas the second derivative of a minimum is positive. As the saddle point is a maxima along the minimum path the second derivative is negative.

==Estimating Transition State Position ==

 Question 2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. 

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.908||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot 1 !! Surface Plot 2!! Internuclear distance vs Time
|-
|[[File:Mlw115_H2H_saddle.png|300px]]||[[File:mlw115_H2H_TS.png|300px]]||[[File:Mlw115_H2H_TS_INDvT.png|300px]]|
|}

The best guess of inter nuclear distances at the transition state was found using trial and error, to find the positions that would result in the minimum trajectory. As the system is symmetric the distances would be equal. The two surface plots are included to demonstrate the "reaction path". The internuclear distance vs time plot demonstrates no change in any bond lengths at the transition state so the overall momenta is 0 and potential energy is at the maximum.

== Reaction Path ==
===MEP vs Dynamics===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.909||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot (MEP) !! Surface Plot (Dynamics
|-
|[[File:Mlw115_H2H_MEP_surfaceplot.png|300px]]||[[File:mlw115_H2H_MEP_surfaceplot_dynamics.png|300px]]
|}

 Question 3. Comment on how the mep and the trajectory you just calculated differ.

It can be seen in the MEP plot that the trajectory follows the minimum energy valley floor without any obvious deviations or oscillation. The MEP trajectory is corresponding to infinitely slow motion, it does not accurately represent the atomic motion of the reaction. The trajectory oscillates in the dynamics plot, also in much less steps the AB bond distance increases considerably more. This plot is more realistic as it accounts for the motion of atoms during the reaction.

{| class="wikitable" border="1"
|+ Final Geometry
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| 2.486||1.167
|}

=== Reversed starting positions ===

What would change if we used the initial conditions r1 = rts and r2 = rts+0.01 instead? ****** DO THIS ******

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| -2.486||-1.167
|}

[[File:mlw115_H2H_dynamics_reversed_momenta.png|300px]]

From taking the previous runs final conditions and setting these as the initial conditions and reversing the momenta values the reaction seems to run in reverse, reaching an end exactly at the transition state.

===Determining Reactivity===

 rAB = 0.74 

 rBC = 2.0 

{| class="wikitable" border=1
! pAB !! pBC!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -1.25 || -2.5 || -99.018|| Yes || [[File:Mlw115_Surface_Plot_1.25_2.5.png|300px]] || Initially the HAHB distance changes very little as the vibrational energy is quite low, as the HC approaches it can be seen that the BC distance decreases linearly. The molecules pass over the transition state, after this point the HAHB distance decreases and the HBHC distance oscillates slightly, with a higher vibrational energy than the HAHB molecule. This overall process represents the HAHB bond breaking and forming a HBHC molecule.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mlw115_Surface_Plot_1.5_2.0.png|300px]] || The molecule approaches the maxima with some vibrational energy, shown by oscillations in the initial path, the energy of the system is not high enough to pass over the transition state energy maxima so the HC "rebounds" and the HBHC distance begins to increase. The HAHB molecule oscillations increase.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mlw115_Surface_Plot_1.5_2.5.png|300px]] || The reaction proceeds similarly to the first case with HC approaching and forming a bond with HB as there is sufficient energy to proceed over the transition state maxima. The vibrational energy throughout is slightly higher than the first case, demonstrated by higher oscillations in atom distances.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mlw115_Surface_Plot_2.5_5.png|300px]] || In this case the energy is high enough and the atoms are able to pass over the transition state energy maxima, however the vibrational energy of the HBHC system formed seems to be too high to form the molecule and the HC atom rebounds and the distance begins to decrease and so the HAHB is reformed but with considerably higher oscillations.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mlw115_Surface_Plot_2.5_5.2.png|300px]] || The HC approaches the oscillating HAHB system, the HB atom rebounds between the two atoms, as it can be seen on the surface plot the system passes over the transition state maxima, back over to the other side and back again as this occurs. Finally the HBHC bond is formed with a high vibrationally energy
|}

===Transition State Theory===

TS state theory assumes that if the energy is high enough to cross the Transition state energy maxima then the reaction will be able to go to completion. However as demonstrated in the penultimate example the system can cross back over the maxima and the initial product will be reformed resulting in an incomplete reaction. This therefore will incorrectly predict the rate of reaction to be higher than experimentally measured as not all collisions above the necessary energy will result in a successful reaction.

=Exercise 2: H-H-F System=

==PES Inspection==

For the forward reaction of H2 + F --> HF + H is exothermic, as shown in the surface plot below the energy of the reactants H2 is higher than the energy of the product HF. As the energy absorbed in breaking the H-H bond in the first case is less than the energy released forming the H-F bond which means that in total energy is released which results in an exothermic reaction.

{| class="wikitable" border=1
! Surface Plot H2+F->HF+H !! Surface Plot HF+H -> H2+F
|-
|[[File:mlw115_Surface_Plot_HHF_labelled.png|300px]]||[[File:mlw115_Surface_Plot_HHF_H2Prod.png|300px]]|
|}

==Determining Transition State ==

A = HA B = HB and C = F

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! Surface Plot !! Internuclear distance vs Time
|-
|[[File:Mlw115_Surface_Plot_TS_HHF.png|400px]]||[[File:Mlw115 INDvT HHF TS.png|350px]]|
|}

== Calculating Activation Energy ==

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! BC= 1.8135!! BC= 1.8135 Zoomed!! BC= 1.813
|-
|[[File:Mlw115_HHF_EvT_1.8135.png|300px]] ‎||[[File:Mlw115_Plot_1.813_EvT_zoomT0_Resized.png|300px]]||[[File:Mlw115_HHF_EvT_1.813.png|320px]]
|}

pAB = pBC =0

distAB = 0.74

{| class="wikitable" border=1
|+Activation energy HF+H->H2+F
|-
! !! Energy (Kcal/mol)
|-
|Initial energy || -103.752
|-
|Final energy|| -133.89
|-
|EA|| 30.138
|}

{| class="wikitable" border=1
|+Activation energy H2+F->HF+H
|-
! !! Energy (Kcal/mol)
|-
|Initial energy|| -103.742
|-
|Final energy|| -103.752
|-
|EA|| 0.01
|}

==Reaction Dynamics==

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.74||0
|-
| BC|| 2.3||-2.7
|}

{| class="wikitable" border=1
!Surface Plot!! Internuclear momentum vs time
|-
|[[File:Mlw_Surface_Plot_HHF_Reactive.png]]|| [[File:Mlw_INMvT_HHF_Reactive.png]]
|}

{| class="wikitable" border=1
!Energy!!
|-
|Kinetic|| 3.837
|-
|Potential|| -103.886

|-
|Total|| -100.049
|}

It can be seen from the animation that as the H-H (AB) bond is broken and the H-F (BC) bond is formed the vibrational energy (and so the kinetic energy) considerably increases for H-F. The internuclear momentum vs time graph demonstrates this, as the momentum is much higher for the H-F atom. This vibrational energy will be released as heat so experimentally an increase in temperature could confirm the reaction had occurred.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -3 || -0.5 || -100.339|| No || [[File:Mlw115_Surface_Plot_HHF_-3_0.5.png|300px]] || H2 approaches the fluorine atom slowly but when a collision is made the H rebounds straight back into the other H and they continue their path oscillating together in the opposite direction
|-

| 3 || -0.5 || -93.254 || No || [[File:Mlw115_Surface_Plot_HHF_3_0.5.png|300px]] || This is similar to the previous case however the hydrogen rebounds twice into flourine but on the second rebounds remains returns back to oscillating with the original H atom
|-
| 2 || -0.5|| -98.753|| No || [[File: Mlw115_Surface_Plot_HHF_2_0.5.png|300px]] || The HF molecule oscillates within the energy well at high vibrational energy after collision
|-
| 1.5|| -0.5|| 103.754|| No || [[File: Mlw115_Surface_Plot_HHF_1.5_0.5.png|300px]] || Similarly to the previous situation
|-
|0.1||-0.8|| 103.364|| Yes|| [[File: Mlw115_Surface_Plot_HHF_0.1_0.8.png|300px]] || Despite having lower total energy than the previous cases the reaction goes to completion as the momentum between the reacting atoms is sufficient
|}

Despite the H2 energy being higher than the activation energy if the BC atoms do not collide with a high enough energy the reaction will return back over the peak and the original reactants will be produced.

===Polanyi's Empirical Rules ===

Polanyi's rules state that in reactions with an early transition state (most closely resembling the reactants) translational energy most effectively promotes a successful reaction. Conversely where the transition state is late (resembling products closely) vibrational energy, specifically of the reactants most effectively promotes a successful reaction. <ref>The Journal of Chemical Physics 138, 234104 (2013); https://doi.org/10.1063/1.4810007</ref>

====H-F + H====
rAB= 2 rBC = 0.74
{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -5 || 0.2 || -103.364 || No || [[File:Mlw115_Surface_Plot_HFH_5_0.2.png|300px]] || Approaching H rebounds and reaction doesn't complete
|-
| 2.21||0.37||
|}

Reversal of conditions:
AB D 5.07 M -2.22
BC D 1.19 M 0.37

Random Guessing:
AB D 5.0 M -1.5
BC D 1.2 M 1.0

[[File: Mlw115_Surface_Plot_HFH_-1.5_1.png| 400px]]

=References=
<references/>

MRD:mlw115RXNDy

2018-05-22T13:59:15Z

Mlw115: /* Estimating Transition State Position */

= Reaction Dynamics =

==Introduction==

The reactivity of atom-diatom systems are studied in order to gain insight on the internal degrees of freedom of the system. Particularly vibrational and translational energy are considered with respect to overcoming the transition state energy barrier.

The atoms are considered to behave classically and the interactions are expressed as potential energy surfaces.

= Exercise 1: H2 + H system=

==Dynamics from the transition state region ==

 Question 1. What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

Where a reaction is modeled on a potential energy surface the lowest energy path between the reactants and products is seen as a "valley floor" of the surface. The transition structure is a saddle point along the minimum energy ridge.

The gradient of the potential energy surface is zero at the minima and so the first derivative with respect to the inter-nuclear distance is zero. This is also the case for the peak of the saddle point.

δV(ri)/δri = 0

However using second derivatives
δV2(ri)/δri2 > 0 Minima
δV2(ri)/δri2 < 0 Maxima

where V is the potential energy and r related to the nuclear positions.

The second derivative of a maximum point is negative, whereas the second derivative of a minimum is positive. As the saddle point is a maxima along the minimum path the second derivative is negative.

==Estimating Transition State Position ==

 Question 2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. 

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.908||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot 1 !! Surface Plot 2!! Internuclear distance vs Time
|-
|[[File:Mlw115_H2H_saddle.png|300px]]||[[File:mlw115_H2H_TS.png|300px]]||[[File:Mlw115_H2H_TS_INDvT.png|300px]]|
|}

The best guess of inter nuclear distances at the transition state was found using trial and error, to find the positions that would result in the minimum trajectory. As the system is symmetric the distances would be equal. The two surface plots are included to demonstrate the "reaction path". The internuclear distance vs time plot demonstrates no change in any bond lengths at the transition state so the overall momenta is 0 and potential energy is at the maximum.

== Reaction Path ==
===MEP vs Dynamics===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.909||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot (MEP) !! Surface Plot (Dynamics
|-
|[[File:Mlw115_H2H_MEP_surfaceplot.png|300px]]||[[File:mlw115_H2H_MEP_surfaceplot_dynamics.png|300px]]
|}

It can be seen in this plot that the trajectory follows the minimum energy valley floor without any obvious deviations or oscillation.

The trajectory oscillates in this plot, also in much less steps the AB bond distance increases considerably more. This plot is more realistic as it accounts for the motion of atoms during the reaction.

{| class="wikitable" border="1"
|+ Final Geometry
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| 2.486||1.167
|}

=== Reversed starting positions ===

What would change if we used the initial conditions r1 = rts and r2 = rts+0.01 instead? ****** DO THIS ******

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| -2.486||-1.167
|}

[[File:mlw115_H2H_dynamics_reversed_momenta.png|300px]]

From taking the previous runs final conditions and setting these as the initial conditions and reversing the momenta values the reaction seems to run in reverse, reaching an end exactly at the transition state.

===Determining Reactivity===

 rAB = 0.74 

 rBC = 2.0 

{| class="wikitable" border=1
! pAB !! pBC!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -1.25 || -2.5 || -99.018|| Yes || [[File:Mlw115_Surface_Plot_1.25_2.5.png|300px]] || Initially the HAHB distance changes very little as the vibrational energy is quite low, as the HC approaches it can be seen that the BC distance decreases linearly. The molecules pass over the transition state, after this point the HAHB distance decreases and the HBHC distance oscillates slightly, with a higher vibrational energy than the HAHB molecule. This overall process represents the HAHB bond breaking and forming a HBHC molecule.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mlw115_Surface_Plot_1.5_2.0.png|300px]] || The molecule approaches the maxima with some vibrational energy, shown by oscillations in the initial path, the energy of the system is not high enough to pass over the transition state energy maxima so the HC "rebounds" and the HBHC distance begins to increase. The HAHB molecule oscillations increase.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mlw115_Surface_Plot_1.5_2.5.png|300px]] || The reaction proceeds similarly to the first case with HC approaching and forming a bond with HB as there is sufficient energy to proceed over the transition state maxima. The vibrational energy throughout is slightly higher than the first case, demonstrated by higher oscillations in atom distances.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mlw115_Surface_Plot_2.5_5.png|300px]] || In this case the energy is high enough and the atoms are able to pass over the transition state energy maxima, however the vibrational energy of the HBHC system formed seems to be too high to form the molecule and the HC atom rebounds and the distance begins to decrease and so the HAHB is reformed but with considerably higher oscillations.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mlw115_Surface_Plot_2.5_5.2.png|300px]] || The HC approaches the oscillating HAHB system, the HB atom rebounds between the two atoms, as it can be seen on the surface plot the system passes over the transition state maxima, back over to the other side and back again as this occurs. Finally the HBHC bond is formed with a high vibrationally energy
|}

===Transition State Theory===

TS state theory assumes that if the energy is high enough to cross the Transition state energy maxima then the reaction will be able to go to completion. However as demonstrated in the penultimate example the system can cross back over the maxima and the initial product will be reformed resulting in an incomplete reaction. This therefore will incorrectly predict the rate of reaction to be higher than experimentally measured as not all collisions above the necessary energy will result in a successful reaction.

=Exercise 2: H-H-F System=

==PES Inspection==

For the forward reaction of H2 + F --> HF + H is exothermic, as shown in the surface plot below the energy of the reactants H2 is higher than the energy of the product HF. As the energy absorbed in breaking the H-H bond in the first case is less than the energy released forming the H-F bond which means that in total energy is released which results in an exothermic reaction.

{| class="wikitable" border=1
! Surface Plot H2+F->HF+H !! Surface Plot HF+H -> H2+F
|-
|[[File:mlw115_Surface_Plot_HHF_labelled.png|300px]]||[[File:mlw115_Surface_Plot_HHF_H2Prod.png|300px]]|
|}

==Determining Transition State ==

A = HA B = HB and C = F

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! Surface Plot !! Internuclear distance vs Time
|-
|[[File:Mlw115_Surface_Plot_TS_HHF.png|400px]]||[[File:Mlw115 INDvT HHF TS.png|350px]]|
|}

== Calculating Activation Energy ==

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! BC= 1.8135!! BC= 1.8135 Zoomed!! BC= 1.813
|-
|[[File:Mlw115_HHF_EvT_1.8135.png|300px]] ‎||[[File:Mlw115_Plot_1.813_EvT_zoomT0_Resized.png|300px]]||[[File:Mlw115_HHF_EvT_1.813.png|320px]]
|}

pAB = pBC =0

distAB = 0.74

{| class="wikitable" border=1
|+Activation energy HF+H->H2+F
|-
! !! Energy (Kcal/mol)
|-
|Initial energy || -103.752
|-
|Final energy|| -133.89
|-
|EA|| 30.138
|}

{| class="wikitable" border=1
|+Activation energy H2+F->HF+H
|-
! !! Energy (Kcal/mol)
|-
|Initial energy|| -103.742
|-
|Final energy|| -103.752
|-
|EA|| 0.01
|}

==Reaction Dynamics==

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.74||0
|-
| BC|| 2.3||-2.7
|}

{| class="wikitable" border=1
!Surface Plot!! Internuclear momentum vs time
|-
|[[File:Mlw_Surface_Plot_HHF_Reactive.png]]|| [[File:Mlw_INMvT_HHF_Reactive.png]]
|}

{| class="wikitable" border=1
!Energy!!
|-
|Kinetic|| 3.837
|-
|Potential|| -103.886

|-
|Total|| -100.049
|}

It can be seen from the animation that as the H-H (AB) bond is broken and the H-F (BC) bond is formed the vibrational energy (and so the kinetic energy) considerably increases for H-F. The internuclear momentum vs time graph demonstrates this, as the momentum is much higher for the H-F atom. This vibrational energy will be released as heat so experimentally an increase in temperature could confirm the reaction had occurred.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -3 || -0.5 || -100.339|| No || [[File:Mlw115_Surface_Plot_HHF_-3_0.5.png|300px]] || H2 approaches the fluorine atom slowly but when a collision is made the H rebounds straight back into the other H and they continue their path oscillating together in the opposite direction
|-

| 3 || -0.5 || -93.254 || No || [[File:Mlw115_Surface_Plot_HHF_3_0.5.png|300px]] || This is similar to the previous case however the hydrogen rebounds twice into flourine but on the second rebounds remains returns back to oscillating with the original H atom
|-
| 2 || -0.5|| -98.753|| No || [[File: Mlw115_Surface_Plot_HHF_2_0.5.png|300px]] || The HF molecule oscillates within the energy well at high vibrational energy after collision
|-
| 1.5|| -0.5|| 103.754|| No || [[File: Mlw115_Surface_Plot_HHF_1.5_0.5.png|300px]] || Similarly to the previous situation
|-
|0.1||-0.8|| 103.364|| Yes|| [[File: Mlw115_Surface_Plot_HHF_0.1_0.8.png|300px]] || Despite having lower total energy than the previous cases the reaction goes to completion as the momentum between the reacting atoms is sufficient
|}

Despite the H2 energy being higher than the activation energy if the BC atoms do not collide with a high enough energy the reaction will return back over the peak and the original reactants will be produced.

===Polanyi's Empirical Rules ===

Polanyi's rules state that in reactions with an early transition state (most closely resembling the reactants) translational energy most effectively promotes a successful reaction. Conversely where the transition state is late (resembling products closely) vibrational energy, specifically of the reactants most effectively promotes a successful reaction. <ref>The Journal of Chemical Physics 138, 234104 (2013); https://doi.org/10.1063/1.4810007</ref>

====H-F + H====
rAB= 2 rBC = 0.74
{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -5 || 0.2 || -103.364 || No || [[File:Mlw115_Surface_Plot_HFH_5_0.2.png|300px]] || Approaching H rebounds and reaction doesn't complete
|-
| 2.21||0.37||
|}

Reversal of conditions:
AB D 5.07 M -2.22
BC D 1.19 M 0.37

Random Guessing:
AB D 5.0 M -1.5
BC D 1.2 M 1.0

[[File: Mlw115_Surface_Plot_HFH_-1.5_1.png| 400px]]

=References=
<references/>

MRD:mlw115RXNDy

2018-05-22T13:57:24Z

Mlw115: /* Dynamics from the transition state region */

= Reaction Dynamics =

==Introduction==

The reactivity of atom-diatom systems are studied in order to gain insight on the internal degrees of freedom of the system. Particularly vibrational and translational energy are considered with respect to overcoming the transition state energy barrier.

The atoms are considered to behave classically and the interactions are expressed as potential energy surfaces.

= Exercise 1: H2 + H system=

==Dynamics from the transition state region ==

 Question 1. What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

Where a reaction is modeled on a potential energy surface the lowest energy path between the reactants and products is seen as a "valley floor" of the surface. The transition structure is a saddle point along the minimum energy ridge.

The gradient of the potential energy surface is zero at the minima and so the first derivative with respect to the inter-nuclear distance is zero. This is also the case for the peak of the saddle point.

δV(ri)/δri = 0

However using second derivatives
δV2(ri)/δri2 > 0 Minima
δV2(ri)/δri2 < 0 Maxima

where V is the potential energy and r related to the nuclear positions.

The second derivative of a maximum point is negative, whereas the second derivative of a minimum is positive. As the saddle point is a maxima along the minimum path the second derivative is negative.

==Estimating Transition State Position ==

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.908||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot 1 !! Surface Plot 2!! Internuclear distance vs Time
|-
|[[File:Mlw115_H2H_saddle.png|300px]]||[[File:mlw115_H2H_TS.png|300px]]||[[File:Mlw115_H2H_TS_INDvT.png|300px]]|
|}

The best guess of inter nuclear distances at the transition state was found using trial and error, to find the positions that would result in the minimum trajectory. As the system is symmetric the distances would be equal. The two surface plots are included to demonstrate the "reaction path" which involves only slight oscillations of the molecules. The internuclear distance vs time plot demonstrates no change in any bond lengths at the transition state so the overall momenta is 0 and potential energy is at the maximum.

== Reaction Path ==
===MEP vs Dynamics===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.909||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot (MEP) !! Surface Plot (Dynamics
|-
|[[File:Mlw115_H2H_MEP_surfaceplot.png|300px]]||[[File:mlw115_H2H_MEP_surfaceplot_dynamics.png|300px]]
|}

It can be seen in this plot that the trajectory follows the minimum energy valley floor without any obvious deviations or oscillation.

The trajectory oscillates in this plot, also in much less steps the AB bond distance increases considerably more. This plot is more realistic as it accounts for the motion of atoms during the reaction.

{| class="wikitable" border="1"
|+ Final Geometry
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| 2.486||1.167
|}

=== Reversed starting positions ===

What would change if we used the initial conditions r1 = rts and r2 = rts+0.01 instead? ****** DO THIS ******

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| -2.486||-1.167
|}

[[File:mlw115_H2H_dynamics_reversed_momenta.png|300px]]

From taking the previous runs final conditions and setting these as the initial conditions and reversing the momenta values the reaction seems to run in reverse, reaching an end exactly at the transition state.

===Determining Reactivity===

 rAB = 0.74 

 rBC = 2.0 

{| class="wikitable" border=1
! pAB !! pBC!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -1.25 || -2.5 || -99.018|| Yes || [[File:Mlw115_Surface_Plot_1.25_2.5.png|300px]] || Initially the HAHB distance changes very little as the vibrational energy is quite low, as the HC approaches it can be seen that the BC distance decreases linearly. The molecules pass over the transition state, after this point the HAHB distance decreases and the HBHC distance oscillates slightly, with a higher vibrational energy than the HAHB molecule. This overall process represents the HAHB bond breaking and forming a HBHC molecule.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mlw115_Surface_Plot_1.5_2.0.png|300px]] || The molecule approaches the maxima with some vibrational energy, shown by oscillations in the initial path, the energy of the system is not high enough to pass over the transition state energy maxima so the HC "rebounds" and the HBHC distance begins to increase. The HAHB molecule oscillations increase.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mlw115_Surface_Plot_1.5_2.5.png|300px]] || The reaction proceeds similarly to the first case with HC approaching and forming a bond with HB as there is sufficient energy to proceed over the transition state maxima. The vibrational energy throughout is slightly higher than the first case, demonstrated by higher oscillations in atom distances.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mlw115_Surface_Plot_2.5_5.png|300px]] || In this case the energy is high enough and the atoms are able to pass over the transition state energy maxima, however the vibrational energy of the HBHC system formed seems to be too high to form the molecule and the HC atom rebounds and the distance begins to decrease and so the HAHB is reformed but with considerably higher oscillations.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mlw115_Surface_Plot_2.5_5.2.png|300px]] || The HC approaches the oscillating HAHB system, the HB atom rebounds between the two atoms, as it can be seen on the surface plot the system passes over the transition state maxima, back over to the other side and back again as this occurs. Finally the HBHC bond is formed with a high vibrationally energy
|}

===Transition State Theory===

TS state theory assumes that if the energy is high enough to cross the Transition state energy maxima then the reaction will be able to go to completion. However as demonstrated in the penultimate example the system can cross back over the maxima and the initial product will be reformed resulting in an incomplete reaction. This therefore will incorrectly predict the rate of reaction to be higher than experimentally measured as not all collisions above the necessary energy will result in a successful reaction.

=Exercise 2: H-H-F System=

==PES Inspection==

For the forward reaction of H2 + F --> HF + H is exothermic, as shown in the surface plot below the energy of the reactants H2 is higher than the energy of the product HF. As the energy absorbed in breaking the H-H bond in the first case is less than the energy released forming the H-F bond which means that in total energy is released which results in an exothermic reaction.

{| class="wikitable" border=1
! Surface Plot H2+F->HF+H !! Surface Plot HF+H -> H2+F
|-
|[[File:mlw115_Surface_Plot_HHF_labelled.png|300px]]||[[File:mlw115_Surface_Plot_HHF_H2Prod.png|300px]]|
|}

==Determining Transition State ==

A = HA B = HB and C = F

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! Surface Plot !! Internuclear distance vs Time
|-
|[[File:Mlw115_Surface_Plot_TS_HHF.png|400px]]||[[File:Mlw115 INDvT HHF TS.png|350px]]|
|}

== Calculating Activation Energy ==

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! BC= 1.8135!! BC= 1.8135 Zoomed!! BC= 1.813
|-
|[[File:Mlw115_HHF_EvT_1.8135.png|300px]] ‎||[[File:Mlw115_Plot_1.813_EvT_zoomT0_Resized.png|300px]]||[[File:Mlw115_HHF_EvT_1.813.png|320px]]
|}

pAB = pBC =0

distAB = 0.74

{| class="wikitable" border=1
|+Activation energy HF+H->H2+F
|-
! !! Energy (Kcal/mol)
|-
|Initial energy || -103.752
|-
|Final energy|| -133.89
|-
|EA|| 30.138
|}

{| class="wikitable" border=1
|+Activation energy H2+F->HF+H
|-
! !! Energy (Kcal/mol)
|-
|Initial energy|| -103.742
|-
|Final energy|| -103.752
|-
|EA|| 0.01
|}

==Reaction Dynamics==

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.74||0
|-
| BC|| 2.3||-2.7
|}

{| class="wikitable" border=1
!Surface Plot!! Internuclear momentum vs time
|-
|[[File:Mlw_Surface_Plot_HHF_Reactive.png]]|| [[File:Mlw_INMvT_HHF_Reactive.png]]
|}

{| class="wikitable" border=1
!Energy!!
|-
|Kinetic|| 3.837
|-
|Potential|| -103.886

|-
|Total|| -100.049
|}

It can be seen from the animation that as the H-H (AB) bond is broken and the H-F (BC) bond is formed the vibrational energy (and so the kinetic energy) considerably increases for H-F. The internuclear momentum vs time graph demonstrates this, as the momentum is much higher for the H-F atom. This vibrational energy will be released as heat so experimentally an increase in temperature could confirm the reaction had occurred.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -3 || -0.5 || -100.339|| No || [[File:Mlw115_Surface_Plot_HHF_-3_0.5.png|300px]] || H2 approaches the fluorine atom slowly but when a collision is made the H rebounds straight back into the other H and they continue their path oscillating together in the opposite direction
|-

| 3 || -0.5 || -93.254 || No || [[File:Mlw115_Surface_Plot_HHF_3_0.5.png|300px]] || This is similar to the previous case however the hydrogen rebounds twice into flourine but on the second rebounds remains returns back to oscillating with the original H atom
|-
| 2 || -0.5|| -98.753|| No || [[File: Mlw115_Surface_Plot_HHF_2_0.5.png|300px]] || The HF molecule oscillates within the energy well at high vibrational energy after collision
|-
| 1.5|| -0.5|| 103.754|| No || [[File: Mlw115_Surface_Plot_HHF_1.5_0.5.png|300px]] || Similarly to the previous situation
|-
|0.1||-0.8|| 103.364|| Yes|| [[File: Mlw115_Surface_Plot_HHF_0.1_0.8.png|300px]] || Despite having lower total energy than the previous cases the reaction goes to completion as the momentum between the reacting atoms is sufficient
|}

Despite the H2 energy being higher than the activation energy if the BC atoms do not collide with a high enough energy the reaction will return back over the peak and the original reactants will be produced.

===Polanyi's Empirical Rules ===

Polanyi's rules state that in reactions with an early transition state (most closely resembling the reactants) translational energy most effectively promotes a successful reaction. Conversely where the transition state is late (resembling products closely) vibrational energy, specifically of the reactants most effectively promotes a successful reaction. <ref>The Journal of Chemical Physics 138, 234104 (2013); https://doi.org/10.1063/1.4810007</ref>

====H-F + H====
rAB= 2 rBC = 0.74
{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -5 || 0.2 || -103.364 || No || [[File:Mlw115_Surface_Plot_HFH_5_0.2.png|300px]] || Approaching H rebounds and reaction doesn't complete
|-
| 2.21||0.37||
|}

Reversal of conditions:
AB D 5.07 M -2.22
BC D 1.19 M 0.37

Random Guessing:
AB D 5.0 M -1.5
BC D 1.2 M 1.0

[[File: Mlw115_Surface_Plot_HFH_-1.5_1.png| 400px]]

=References=
<references/>

MRD:mlw115RXNDy

2018-05-22T13:56:27Z

Mlw115: /* Dynamics from the transition state region */

= Reaction Dynamics =

==Introduction==

The reactivity of atom-diatom systems are studied in order to gain insight on the internal degrees of freedom of the system. Particularly vibrational and translational energy are considered with respect to overcoming the transition state energy barrier.

The atoms are considered to behave classically and the interactions are expressed as potential energy surfaces.

= Exercise 1: H2 + H system=

==Dynamics from the transition state region ==

 Question 1. What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

Where a reaction is modeled on a potential energy surface the lowest energy path between the reactants and products is seen as a "valley floor" of the surface. The transition structure is a saddle point along the minimum energy ridge.

The gradient of the potential energy surface is zero at the minima and so the first derivative with respect to the inter-nuclear distance is zero. This is also the case for the peak of the saddle point.

δV(ri)/δri = 0

However using second derivatives
δV2(ri)/δri2 > 0 Minima
δV2(ri)/δri2 < 0 Maxima

where V is the potential energy and r related to the nuclear positions.

The second derivative of a maximum point is negative, whereas the second derivative of a minimum is positive. As the saddle point is a maxima along the minimum path the second derivative is negative.

==Estimating Transition State Position ==

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.908||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot 1 !! Surface Plot 2!! Internuclear distance vs Time
|-
|[[File:Mlw115_H2H_saddle.png|300px]]||[[File:mlw115_H2H_TS.png|300px]]||[[File:Mlw115_H2H_TS_INDvT.png|300px]]|
|}

The best guess of inter nuclear distances at the transition state was found using trial and error, to find the positions that would result in the minimum trajectory. As the system is symmetric the distances would be equal. The two surface plots are included to demonstrate the "reaction path" which involves only slight oscillations of the molecules. The internuclear distance vs time plot demonstrates no change in any bond lengths at the transition state so the overall momenta is 0 and potential energy is at the maximum.

== Reaction Path ==
===MEP vs Dynamics===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.909||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot (MEP) !! Surface Plot (Dynamics
|-
|[[File:Mlw115_H2H_MEP_surfaceplot.png|300px]]||[[File:mlw115_H2H_MEP_surfaceplot_dynamics.png|300px]]
|}

It can be seen in this plot that the trajectory follows the minimum energy valley floor without any obvious deviations or oscillation.

The trajectory oscillates in this plot, also in much less steps the AB bond distance increases considerably more. This plot is more realistic as it accounts for the motion of atoms during the reaction.

{| class="wikitable" border="1"
|+ Final Geometry
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| 2.486||1.167
|}

=== Reversed starting positions ===

What would change if we used the initial conditions r1 = rts and r2 = rts+0.01 instead? ****** DO THIS ******

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| -2.486||-1.167
|}

[[File:mlw115_H2H_dynamics_reversed_momenta.png|300px]]

From taking the previous runs final conditions and setting these as the initial conditions and reversing the momenta values the reaction seems to run in reverse, reaching an end exactly at the transition state.

===Determining Reactivity===

 rAB = 0.74 

 rBC = 2.0 

{| class="wikitable" border=1
! pAB !! pBC!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -1.25 || -2.5 || -99.018|| Yes || [[File:Mlw115_Surface_Plot_1.25_2.5.png|300px]] || Initially the HAHB distance changes very little as the vibrational energy is quite low, as the HC approaches it can be seen that the BC distance decreases linearly. The molecules pass over the transition state, after this point the HAHB distance decreases and the HBHC distance oscillates slightly, with a higher vibrational energy than the HAHB molecule. This overall process represents the HAHB bond breaking and forming a HBHC molecule.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mlw115_Surface_Plot_1.5_2.0.png|300px]] || The molecule approaches the maxima with some vibrational energy, shown by oscillations in the initial path, the energy of the system is not high enough to pass over the transition state energy maxima so the HC "rebounds" and the HBHC distance begins to increase. The HAHB molecule oscillations increase.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mlw115_Surface_Plot_1.5_2.5.png|300px]] || The reaction proceeds similarly to the first case with HC approaching and forming a bond with HB as there is sufficient energy to proceed over the transition state maxima. The vibrational energy throughout is slightly higher than the first case, demonstrated by higher oscillations in atom distances.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mlw115_Surface_Plot_2.5_5.png|300px]] || In this case the energy is high enough and the atoms are able to pass over the transition state energy maxima, however the vibrational energy of the HBHC system formed seems to be too high to form the molecule and the HC atom rebounds and the distance begins to decrease and so the HAHB is reformed but with considerably higher oscillations.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mlw115_Surface_Plot_2.5_5.2.png|300px]] || The HC approaches the oscillating HAHB system, the HB atom rebounds between the two atoms, as it can be seen on the surface plot the system passes over the transition state maxima, back over to the other side and back again as this occurs. Finally the HBHC bond is formed with a high vibrationally energy
|}

===Transition State Theory===

TS state theory assumes that if the energy is high enough to cross the Transition state energy maxima then the reaction will be able to go to completion. However as demonstrated in the penultimate example the system can cross back over the maxima and the initial product will be reformed resulting in an incomplete reaction. This therefore will incorrectly predict the rate of reaction to be higher than experimentally measured as not all collisions above the necessary energy will result in a successful reaction.

=Exercise 2: H-H-F System=

==PES Inspection==

For the forward reaction of H2 + F --> HF + H is exothermic, as shown in the surface plot below the energy of the reactants H2 is higher than the energy of the product HF. As the energy absorbed in breaking the H-H bond in the first case is less than the energy released forming the H-F bond which means that in total energy is released which results in an exothermic reaction.

{| class="wikitable" border=1
! Surface Plot H2+F->HF+H !! Surface Plot HF+H -> H2+F
|-
|[[File:mlw115_Surface_Plot_HHF_labelled.png|300px]]||[[File:mlw115_Surface_Plot_HHF_H2Prod.png|300px]]|
|}

==Determining Transition State ==

A = HA B = HB and C = F

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! Surface Plot !! Internuclear distance vs Time
|-
|[[File:Mlw115_Surface_Plot_TS_HHF.png|400px]]||[[File:Mlw115 INDvT HHF TS.png|350px]]|
|}

== Calculating Activation Energy ==

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! BC= 1.8135!! BC= 1.8135 Zoomed!! BC= 1.813
|-
|[[File:Mlw115_HHF_EvT_1.8135.png|300px]] ‎||[[File:Mlw115_Plot_1.813_EvT_zoomT0_Resized.png|300px]]||[[File:Mlw115_HHF_EvT_1.813.png|320px]]
|}

pAB = pBC =0

distAB = 0.74

{| class="wikitable" border=1
|+Activation energy HF+H->H2+F
|-
! !! Energy (Kcal/mol)
|-
|Initial energy || -103.752
|-
|Final energy|| -133.89
|-
|EA|| 30.138
|}

{| class="wikitable" border=1
|+Activation energy H2+F->HF+H
|-
! !! Energy (Kcal/mol)
|-
|Initial energy|| -103.742
|-
|Final energy|| -103.752
|-
|EA|| 0.01
|}

==Reaction Dynamics==

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.74||0
|-
| BC|| 2.3||-2.7
|}

{| class="wikitable" border=1
!Surface Plot!! Internuclear momentum vs time
|-
|[[File:Mlw_Surface_Plot_HHF_Reactive.png]]|| [[File:Mlw_INMvT_HHF_Reactive.png]]
|}

{| class="wikitable" border=1
!Energy!!
|-
|Kinetic|| 3.837
|-
|Potential|| -103.886

|-
|Total|| -100.049
|}

It can be seen from the animation that as the H-H (AB) bond is broken and the H-F (BC) bond is formed the vibrational energy (and so the kinetic energy) considerably increases for H-F. The internuclear momentum vs time graph demonstrates this, as the momentum is much higher for the H-F atom. This vibrational energy will be released as heat so experimentally an increase in temperature could confirm the reaction had occurred.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -3 || -0.5 || -100.339|| No || [[File:Mlw115_Surface_Plot_HHF_-3_0.5.png|300px]] || H2 approaches the fluorine atom slowly but when a collision is made the H rebounds straight back into the other H and they continue their path oscillating together in the opposite direction
|-

| 3 || -0.5 || -93.254 || No || [[File:Mlw115_Surface_Plot_HHF_3_0.5.png|300px]] || This is similar to the previous case however the hydrogen rebounds twice into flourine but on the second rebounds remains returns back to oscillating with the original H atom
|-
| 2 || -0.5|| -98.753|| No || [[File: Mlw115_Surface_Plot_HHF_2_0.5.png|300px]] || The HF molecule oscillates within the energy well at high vibrational energy after collision
|-
| 1.5|| -0.5|| 103.754|| No || [[File: Mlw115_Surface_Plot_HHF_1.5_0.5.png|300px]] || Similarly to the previous situation
|-
|0.1||-0.8|| 103.364|| Yes|| [[File: Mlw115_Surface_Plot_HHF_0.1_0.8.png|300px]] || Despite having lower total energy than the previous cases the reaction goes to completion as the momentum between the reacting atoms is sufficient
|}

Despite the H2 energy being higher than the activation energy if the BC atoms do not collide with a high enough energy the reaction will return back over the peak and the original reactants will be produced.

===Polanyi's Empirical Rules ===

Polanyi's rules state that in reactions with an early transition state (most closely resembling the reactants) translational energy most effectively promotes a successful reaction. Conversely where the transition state is late (resembling products closely) vibrational energy, specifically of the reactants most effectively promotes a successful reaction. <ref>The Journal of Chemical Physics 138, 234104 (2013); https://doi.org/10.1063/1.4810007</ref>

====H-F + H====
rAB= 2 rBC = 0.74
{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -5 || 0.2 || -103.364 || No || [[File:Mlw115_Surface_Plot_HFH_5_0.2.png|300px]] || Approaching H rebounds and reaction doesn't complete
|-
| 2.21||0.37||
|}

Reversal of conditions:
AB D 5.07 M -2.22
BC D 1.19 M 0.37

Random Guessing:
AB D 5.0 M -1.5
BC D 1.2 M 1.0

[[File: Mlw115_Surface_Plot_HFH_-1.5_1.png| 400px]]

=References=
<references/>

MRD:mlw115RXNDy

2018-05-22T13:40:47Z

Mlw115: /* Reaction Dynamics */

= Reaction Dynamics =

==Introduction==

The reactivity of atom-diatom systems are studied in order to gain insight on the internal degrees of freedom of the system. Particularly vibrational and translational energy are considered with respect to overcoming the transition state energy barrier.

The atoms are considered to behave classically and the interactions are expressed as potential energy surfaces.

= Exercise 1: H2 + H system=

==Dynamics from the transition state region ==

Where a reaction is modeled on a potential energy surface the lowest energy path between the reactants and products is seen as a "valley floor" of the surface. The transition structure is a saddle point along the minimum energy ridge.

The gradient of the potential energy surface is zero at the minima and so the first derivative with respect to the is zero. This is also the case for the peak of the saddle point.

δV(ri)/δri = 0

However using second derivatives
δV2(ri)/δri2 > 0 Minima
δV2(ri)/δri2 < 0 Maxima

The second derivative of a maximum point is negative, whereas the second derivative of a minimum is positive. As the saddle point is a maxima along the minimum path the second derivative is negative

==Estimating Transition State Position ==

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.908||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot 1 !! Surface Plot 2!! Internuclear distance vs Time
|-
|[[File:Mlw115_H2H_saddle.png|300px]]||[[File:mlw115_H2H_TS.png|300px]]||[[File:Mlw115_H2H_TS_INDvT.png|300px]]|
|}

The best guess of inter nuclear distances at the transition state was found using trial and error, to find the positions that would result in the minimum trajectory. As the system is symmetric the distances would be equal. The two surface plots are included to demonstrate the "reaction path" which involves only slight oscillations of the molecules. The internuclear distance vs time plot demonstrates no change in any bond lengths at the transition state so the overall momenta is 0 and potential energy is at the maximum.

== Reaction Path ==
===MEP vs Dynamics===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.909||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot (MEP) !! Surface Plot (Dynamics
|-
|[[File:Mlw115_H2H_MEP_surfaceplot.png|300px]]||[[File:mlw115_H2H_MEP_surfaceplot_dynamics.png|300px]]
|}

It can be seen in this plot that the trajectory follows the minimum energy valley floor without any obvious deviations or oscillation.

The trajectory oscillates in this plot, also in much less steps the AB bond distance increases considerably more. This plot is more realistic as it accounts for the motion of atoms during the reaction.

{| class="wikitable" border="1"
|+ Final Geometry
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| 2.486||1.167
|}

=== Reversed starting positions ===

What would change if we used the initial conditions r1 = rts and r2 = rts+0.01 instead? ****** DO THIS ******

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| -2.486||-1.167
|}

[[File:mlw115_H2H_dynamics_reversed_momenta.png|300px]]

From taking the previous runs final conditions and setting these as the initial conditions and reversing the momenta values the reaction seems to run in reverse, reaching an end exactly at the transition state.

===Determining Reactivity===

 rAB = 0.74 

 rBC = 2.0 

{| class="wikitable" border=1
! pAB !! pBC!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -1.25 || -2.5 || -99.018|| Yes || [[File:Mlw115_Surface_Plot_1.25_2.5.png|300px]] || Initially the HAHB distance changes very little as the vibrational energy is quite low, as the HC approaches it can be seen that the BC distance decreases linearly. The molecules pass over the transition state, after this point the HAHB distance decreases and the HBHC distance oscillates slightly, with a higher vibrational energy than the HAHB molecule. This overall process represents the HAHB bond breaking and forming a HBHC molecule.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mlw115_Surface_Plot_1.5_2.0.png|300px]] || The molecule approaches the maxima with some vibrational energy, shown by oscillations in the initial path, the energy of the system is not high enough to pass over the transition state energy maxima so the HC "rebounds" and the HBHC distance begins to increase. The HAHB molecule oscillations increase.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mlw115_Surface_Plot_1.5_2.5.png|300px]] || The reaction proceeds similarly to the first case with HC approaching and forming a bond with HB as there is sufficient energy to proceed over the transition state maxima. The vibrational energy throughout is slightly higher than the first case, demonstrated by higher oscillations in atom distances.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mlw115_Surface_Plot_2.5_5.png|300px]] || In this case the energy is high enough and the atoms are able to pass over the transition state energy maxima, however the vibrational energy of the HBHC system formed seems to be too high to form the molecule and the HC atom rebounds and the distance begins to decrease and so the HAHB is reformed but with considerably higher oscillations.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mlw115_Surface_Plot_2.5_5.2.png|300px]] || The HC approaches the oscillating HAHB system, the HB atom rebounds between the two atoms, as it can be seen on the surface plot the system passes over the transition state maxima, back over to the other side and back again as this occurs. Finally the HBHC bond is formed with a high vibrationally energy
|}

===Transition State Theory===

TS state theory assumes that if the energy is high enough to cross the Transition state energy maxima then the reaction will be able to go to completion. However as demonstrated in the penultimate example the system can cross back over the maxima and the initial product will be reformed resulting in an incomplete reaction. This therefore will incorrectly predict the rate of reaction to be higher than experimentally measured as not all collisions above the necessary energy will result in a successful reaction.

=Exercise 2: H-H-F System=

==PES Inspection==

For the forward reaction of H2 + F --> HF + H is exothermic, as shown in the surface plot below the energy of the reactants H2 is higher than the energy of the product HF. As the energy absorbed in breaking the H-H bond in the first case is less than the energy released forming the H-F bond which means that in total energy is released which results in an exothermic reaction.

{| class="wikitable" border=1
! Surface Plot H2+F->HF+H !! Surface Plot HF+H -> H2+F
|-
|[[File:mlw115_Surface_Plot_HHF_labelled.png|300px]]||[[File:mlw115_Surface_Plot_HHF_H2Prod.png|300px]]|
|}

==Determining Transition State ==

A = HA B = HB and C = F

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! Surface Plot !! Internuclear distance vs Time
|-
|[[File:Mlw115_Surface_Plot_TS_HHF.png|400px]]||[[File:Mlw115 INDvT HHF TS.png|350px]]|
|}

== Calculating Activation Energy ==

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! BC= 1.8135!! BC= 1.8135 Zoomed!! BC= 1.813
|-
|[[File:Mlw115_HHF_EvT_1.8135.png|300px]] ‎||[[File:Mlw115_Plot_1.813_EvT_zoomT0_Resized.png|300px]]||[[File:Mlw115_HHF_EvT_1.813.png|320px]]
|}

pAB = pBC =0

distAB = 0.74

{| class="wikitable" border=1
|+Activation energy HF+H->H2+F
|-
! !! Energy (Kcal/mol)
|-
|Initial energy || -103.752
|-
|Final energy|| -133.89
|-
|EA|| 30.138
|}

{| class="wikitable" border=1
|+Activation energy H2+F->HF+H
|-
! !! Energy (Kcal/mol)
|-
|Initial energy|| -103.742
|-
|Final energy|| -103.752
|-
|EA|| 0.01
|}

==Reaction Dynamics==

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.74||0
|-
| BC|| 2.3||-2.7
|}

{| class="wikitable" border=1
!Surface Plot!! Internuclear momentum vs time
|-
|[[File:Mlw_Surface_Plot_HHF_Reactive.png]]|| [[File:Mlw_INMvT_HHF_Reactive.png]]
|}

{| class="wikitable" border=1
!Energy!!
|-
|Kinetic|| 3.837
|-
|Potential|| -103.886

|-
|Total|| -100.049
|}

It can be seen from the animation that as the H-H (AB) bond is broken and the H-F (BC) bond is formed the vibrational energy (and so the kinetic energy) considerably increases for H-F. The internuclear momentum vs time graph demonstrates this, as the momentum is much higher for the H-F atom. This vibrational energy will be released as heat so experimentally an increase in temperature could confirm the reaction had occurred.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -3 || -0.5 || -100.339|| No || [[File:Mlw115_Surface_Plot_HHF_-3_0.5.png|300px]] || H2 approaches the fluorine atom slowly but when a collision is made the H rebounds straight back into the other H and they continue their path oscillating together in the opposite direction
|-

| 3 || -0.5 || -93.254 || No || [[File:Mlw115_Surface_Plot_HHF_3_0.5.png|300px]] || This is similar to the previous case however the hydrogen rebounds twice into flourine but on the second rebounds remains returns back to oscillating with the original H atom
|-
| 2 || -0.5|| -98.753|| No || [[File: Mlw115_Surface_Plot_HHF_2_0.5.png|300px]] || The HF molecule oscillates within the energy well at high vibrational energy after collision
|-
| 1.5|| -0.5|| 103.754|| No || [[File: Mlw115_Surface_Plot_HHF_1.5_0.5.png|300px]] || Similarly to the previous situation
|-
|0.1||-0.8|| 103.364|| Yes|| [[File: Mlw115_Surface_Plot_HHF_0.1_0.8.png|300px]] || Despite having lower total energy than the previous cases the reaction goes to completion as the momentum between the reacting atoms is sufficient
|}

Despite the H2 energy being higher than the activation energy if the BC atoms do not collide with a high enough energy the reaction will return back over the peak and the original reactants will be produced.

===Polanyi's Empirical Rules ===

Polanyi's rules state that in reactions with an early transition state (most closely resembling the reactants) translational energy most effectively promotes a successful reaction. Conversely where the transition state is late (resembling products closely) vibrational energy, specifically of the reactants most effectively promotes a successful reaction. <ref>The Journal of Chemical Physics 138, 234104 (2013); https://doi.org/10.1063/1.4810007</ref>

====H-F + H====
rAB= 2 rBC = 0.74
{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -5 || 0.2 || -103.364 || No || [[File:Mlw115_Surface_Plot_HFH_5_0.2.png|300px]] || Approaching H rebounds and reaction doesn't complete
|-
| 2.21||0.37||
|}

Reversal of conditions:
AB D 5.07 M -2.22
BC D 1.19 M 0.37

Random Guessing:
AB D 5.0 M -1.5
BC D 1.2 M 1.0

[[File: Mlw115_Surface_Plot_HFH_-1.5_1.png| 400px]]

=References=
<references/>

MRD:mlw115RXNDy

2018-05-22T13:37:39Z

Mlw115: /* Calculating Activation Energy */

= Reaction Dynamics =

==Introduction==

The reactivity of atom-diatom systems are studied in order to gain insight on the internal degrees of freedom of the system. Particularly vibrational and translational energy are considered with respect to overcoming the transition state energy barrier.

The atoms are considered to behave classically and the interactions are expressed as potential energy surfaces.

= Exercise 1: H2 + H system=

==Dynamics from the transition state region ==

Where a reaction is modeled on a potential energy surface the lowest energy path between the reactants and products is seen as a "valley floor" of the surface. The transition structure is a saddle point along the minimum energy ridge.

The gradient of the potential energy surface is zero at the minima and so the first derivative with respect to the is zero. This is also the case for the peak of the saddle point.

δV(ri)/δri = 0

However using second derivatives
δV2(ri)/δri2 > 0 Minima
δV2(ri)/δri2 < 0 Maxima

The second derivative of a maximum point is negative, whereas the second derivative of a minimum is positive. As the saddle point is a maxima along the minimum path the second derivative is negative

==Estimating Transition State Position ==

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.908||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot 1 !! Surface Plot 2!! Internuclear distance vs Time
|-
|[[File:Mlw115_H2H_saddle.png|300px]]||[[File:mlw115_H2H_TS.png|300px]]||[[File:Mlw115_H2H_TS_INDvT.png|300px]]|
|}

The best guess of inter nuclear distances at the transition state was found using trial and error, to find the positions that would result in the minimum trajectory. As the system is symmetric the distances would be equal. The two surface plots are included to demonstrate the "reaction path" which involves only slight oscillations of the molecules. The internuclear distance vs time plot demonstrates no change in any bond lengths at the transition state so the overall momenta is 0 and potential energy is at the maximum.

== Reaction Path ==
===MEP vs Dynamics===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.909||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot (MEP) !! Surface Plot (Dynamics
|-
|[[File:Mlw115_H2H_MEP_surfaceplot.png|300px]]||[[File:mlw115_H2H_MEP_surfaceplot_dynamics.png|300px]]
|}

It can be seen in this plot that the trajectory follows the minimum energy valley floor without any obvious deviations or oscillation.

The trajectory oscillates in this plot, also in much less steps the AB bond distance increases considerably more. This plot is more realistic as it accounts for the motion of atoms during the reaction.

{| class="wikitable" border="1"
|+ Final Geometry
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| 2.486||1.167
|}

=== Reversed starting positions ===

What would change if we used the initial conditions r1 = rts and r2 = rts+0.01 instead? ****** DO THIS ******

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| -2.486||-1.167
|}

[[File:mlw115_H2H_dynamics_reversed_momenta.png|300px]]

From taking the previous runs final conditions and setting these as the initial conditions and reversing the momenta values the reaction seems to run in reverse, reaching an end exactly at the transition state.

===Determining Reactivity===

 rAB = 0.74 

 rBC = 2.0 

{| class="wikitable" border=1
! pAB !! pBC!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -1.25 || -2.5 || -99.018|| Yes || [[File:Mlw115_Surface_Plot_1.25_2.5.png|300px]] || Initially the HAHB distance changes very little as the vibrational energy is quite low, as the HC approaches it can be seen that the BC distance decreases linearly. The molecules pass over the transition state, after this point the HAHB distance decreases and the HBHC distance oscillates slightly, with a higher vibrational energy than the HAHB molecule. This overall process represents the HAHB bond breaking and forming a HBHC molecule.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mlw115_Surface_Plot_1.5_2.0.png|300px]] || The molecule approaches the maxima with some vibrational energy, shown by oscillations in the initial path, the energy of the system is not high enough to pass over the transition state energy maxima so the HC "rebounds" and the HBHC distance begins to increase. The HAHB molecule oscillations increase.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mlw115_Surface_Plot_1.5_2.5.png|300px]] || The reaction proceeds similarly to the first case with HC approaching and forming a bond with HB as there is sufficient energy to proceed over the transition state maxima. The vibrational energy throughout is slightly higher than the first case, demonstrated by higher oscillations in atom distances.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mlw115_Surface_Plot_2.5_5.png|300px]] || In this case the energy is high enough and the atoms are able to pass over the transition state energy maxima, however the vibrational energy of the HBHC system formed seems to be too high to form the molecule and the HC atom rebounds and the distance begins to decrease and so the HAHB is reformed but with considerably higher oscillations.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mlw115_Surface_Plot_2.5_5.2.png|300px]] || The HC approaches the oscillating HAHB system, the HB atom rebounds between the two atoms, as it can be seen on the surface plot the system passes over the transition state maxima, back over to the other side and back again as this occurs. Finally the HBHC bond is formed with a high vibrationally energy
|}

===Transition State Theory===

TS state theory assumes that if the energy is high enough to cross the Transition state energy maxima then the reaction will be able to go to completion. However as demonstrated in the penultimate example the system can cross back over the maxima and the initial product will be reformed resulting in an incomplete reaction. This therefore will incorrectly predict the rate of reaction to be higher than experimentally measured as not all collisions above the necessary energy will result in a successful reaction.

=Exercise 2: H-H-F System=

==PES Inspection==

For the forward reaction of H2 + F --> HF + H is exothermic, as shown in the surface plot below the energy of the reactants H2 is higher than the energy of the product HF. As the energy absorbed in breaking the H-H bond in the first case is less than the energy released forming the H-F bond which means that in total energy is released which results in an exothermic reaction.

{| class="wikitable" border=1
! Surface Plot H2+F->HF+H !! Surface Plot HF+H -> H2+F
|-
|[[File:mlw115_Surface_Plot_HHF_labelled.png|300px]]||[[File:mlw115_Surface_Plot_HHF_H2Prod.png|300px]]|
|}

==Determining Transition State ==

A = HA B = HB and C = F

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! Surface Plot !! Internuclear distance vs Time
|-
|[[File:Mlw115_Surface_Plot_TS_HHF.png|400px]]||[[File:Mlw115 INDvT HHF TS.png|350px]]|
|}

== Calculating Activation Energy ==

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! BC= 1.8135!! BC= 1.8135 Zoomed!! BC= 1.813
|-
|[[File:Mlw115_HHF_EvT_1.8135.png|300px]] ‎||[[File:Mlw115_Plot_1.813_EvT_zoomT0_Resized.png|300px]]||[[File:Mlw115_HHF_EvT_1.813.png|320px]]
|}

pAB = pBC =0

distAB = 0.74

{| class="wikitable" border=1
|+Activation energy HF+H->H2+F
|-
! !! Energy (Kcal/mol)
|-
|Initial energy || -103.752
|-
|Final energy|| -133.89
|-
|EA|| 30.138
|}

{| class="wikitable" border=1
|+Activation energy H2+F->HF+H
|-
! !! Energy (Kcal/mol)
|-
|Initial energy|| -103.742
|-
|Final energy|| -103.752
|-
|EA|| 0.01
|}

==Reaction Dynamics==

A=H B=H C=F

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.74||0
|-
| BC|| 2.3||-2.7
|}

[[File:Mlw_Surface_Plot_HHF_Reactive.png]]

{| class="wikitable" border=1
!Energy!!
|-
|Kinetic|| 3.837
|-
|Potential|| -103.886

|-
|Total|| -100.049
|}

Internuclear momentum vs Time

[[File:Mlw_INMvT_HHF_Reactive.png]]

It can be seen from the animation that as the H-H (AB) bond is broken and the H-F (BC) bond is formed the vibrational energy (and so the kinetic energy) considerably increases for H-F. The internuclear momentum vs time graph demonstrates this, as the momentum is much higher for the H-F atom. This vibrational energy will be released as heat so experimentally an increase in temperature could confirm the reaction had occurred.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -3 || -0.5 || -100.339|| No || [[File:Mlw115_Surface_Plot_HHF_-3_0.5.png|300px]] || H2 approaches the fluorine atom slowly but when a collision is made the H rebounds straight back into the other H and they continue their path oscillating together in the opposite direction
|-

| 3 || -0.5 || -93.254 || No || [[File:Mlw115_Surface_Plot_HHF_3_0.5.png|300px]] || This is similar to the previous case however the hydrogen rebounds twice into flourine but on the second rebounds remains returns back to oscillating with the original H atom
|-
| 2 || -0.5|| -98.753|| No || [[File: Mlw115_Surface_Plot_HHF_2_0.5.png|300px]] || The HF molecule oscillates within the energy well at high vibrational energy after collision
|-
| 1.5|| -0.5|| 103.754|| No || [[File: Mlw115_Surface_Plot_HHF_1.5_0.5.png|300px]] || Similarly to the previous situation
|-
|0.1||-0.8|| 103.364|| Yes|| [[File: Mlw115_Surface_Plot_HHF_0.1_0.8.png|300px]] || Despite having lower total energy than the previous cases the reaction goes to completion as the momentum between the reacting atoms is sufficient
|}

Despite the H2 energy being higher than the activation energy if the BC atoms do not collide with a high enough energy the reaction will return back over the peak and the original reactants will be produced.

===Polanyi's Empirical Rules ===

Polanyi's rules state that in reactions with an early transition state (most closely resembling the reactants) translational energy most effectively promotes a successful reaction. Conversely where the transition state is late (resembling products closely) vibrational energy, specifically of the reactants most effectively promotes a successful reaction. <ref>The Journal of Chemical Physics 138, 234104 (2013); https://doi.org/10.1063/1.4810007</ref>

====H-F + H====
rAB= 2 rBC = 0.74
{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -5 || 0.2 || -103.364 || No || [[File:Mlw115_Surface_Plot_HFH_5_0.2.png|300px]] || Approaching H rebounds and reaction doesn't complete
|-
| 2.21||0.37||
|}

Reversal of conditions:
AB D 5.07 M -2.22
BC D 1.19 M 0.37

Random Guessing:
AB D 5.0 M -1.5
BC D 1.2 M 1.0

[[File: Mlw115_Surface_Plot_HFH_-1.5_1.png| 400px]]

=References=
<references/>

MRD:mlw115RXNDy

2018-05-22T13:37:22Z

Mlw115: /* Calculating Activation Energy */

= Reaction Dynamics =

==Introduction==

The reactivity of atom-diatom systems are studied in order to gain insight on the internal degrees of freedom of the system. Particularly vibrational and translational energy are considered with respect to overcoming the transition state energy barrier.

The atoms are considered to behave classically and the interactions are expressed as potential energy surfaces.

= Exercise 1: H2 + H system=

==Dynamics from the transition state region ==

Where a reaction is modeled on a potential energy surface the lowest energy path between the reactants and products is seen as a "valley floor" of the surface. The transition structure is a saddle point along the minimum energy ridge.

The gradient of the potential energy surface is zero at the minima and so the first derivative with respect to the is zero. This is also the case for the peak of the saddle point.

δV(ri)/δri = 0

However using second derivatives
δV2(ri)/δri2 > 0 Minima
δV2(ri)/δri2 < 0 Maxima

The second derivative of a maximum point is negative, whereas the second derivative of a minimum is positive. As the saddle point is a maxima along the minimum path the second derivative is negative

==Estimating Transition State Position ==

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.908||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot 1 !! Surface Plot 2!! Internuclear distance vs Time
|-
|[[File:Mlw115_H2H_saddle.png|300px]]||[[File:mlw115_H2H_TS.png|300px]]||[[File:Mlw115_H2H_TS_INDvT.png|300px]]|
|}

The best guess of inter nuclear distances at the transition state was found using trial and error, to find the positions that would result in the minimum trajectory. As the system is symmetric the distances would be equal. The two surface plots are included to demonstrate the "reaction path" which involves only slight oscillations of the molecules. The internuclear distance vs time plot demonstrates no change in any bond lengths at the transition state so the overall momenta is 0 and potential energy is at the maximum.

== Reaction Path ==
===MEP vs Dynamics===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.909||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot (MEP) !! Surface Plot (Dynamics
|-
|[[File:Mlw115_H2H_MEP_surfaceplot.png|300px]]||[[File:mlw115_H2H_MEP_surfaceplot_dynamics.png|300px]]
|}

It can be seen in this plot that the trajectory follows the minimum energy valley floor without any obvious deviations or oscillation.

The trajectory oscillates in this plot, also in much less steps the AB bond distance increases considerably more. This plot is more realistic as it accounts for the motion of atoms during the reaction.

{| class="wikitable" border="1"
|+ Final Geometry
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| 2.486||1.167
|}

=== Reversed starting positions ===

What would change if we used the initial conditions r1 = rts and r2 = rts+0.01 instead? ****** DO THIS ******

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| -2.486||-1.167
|}

[[File:mlw115_H2H_dynamics_reversed_momenta.png|300px]]

From taking the previous runs final conditions and setting these as the initial conditions and reversing the momenta values the reaction seems to run in reverse, reaching an end exactly at the transition state.

===Determining Reactivity===

 rAB = 0.74 

 rBC = 2.0 

{| class="wikitable" border=1
! pAB !! pBC!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -1.25 || -2.5 || -99.018|| Yes || [[File:Mlw115_Surface_Plot_1.25_2.5.png|300px]] || Initially the HAHB distance changes very little as the vibrational energy is quite low, as the HC approaches it can be seen that the BC distance decreases linearly. The molecules pass over the transition state, after this point the HAHB distance decreases and the HBHC distance oscillates slightly, with a higher vibrational energy than the HAHB molecule. This overall process represents the HAHB bond breaking and forming a HBHC molecule.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mlw115_Surface_Plot_1.5_2.0.png|300px]] || The molecule approaches the maxima with some vibrational energy, shown by oscillations in the initial path, the energy of the system is not high enough to pass over the transition state energy maxima so the HC "rebounds" and the HBHC distance begins to increase. The HAHB molecule oscillations increase.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mlw115_Surface_Plot_1.5_2.5.png|300px]] || The reaction proceeds similarly to the first case with HC approaching and forming a bond with HB as there is sufficient energy to proceed over the transition state maxima. The vibrational energy throughout is slightly higher than the first case, demonstrated by higher oscillations in atom distances.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mlw115_Surface_Plot_2.5_5.png|300px]] || In this case the energy is high enough and the atoms are able to pass over the transition state energy maxima, however the vibrational energy of the HBHC system formed seems to be too high to form the molecule and the HC atom rebounds and the distance begins to decrease and so the HAHB is reformed but with considerably higher oscillations.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mlw115_Surface_Plot_2.5_5.2.png|300px]] || The HC approaches the oscillating HAHB system, the HB atom rebounds between the two atoms, as it can be seen on the surface plot the system passes over the transition state maxima, back over to the other side and back again as this occurs. Finally the HBHC bond is formed with a high vibrationally energy
|}

===Transition State Theory===

TS state theory assumes that if the energy is high enough to cross the Transition state energy maxima then the reaction will be able to go to completion. However as demonstrated in the penultimate example the system can cross back over the maxima and the initial product will be reformed resulting in an incomplete reaction. This therefore will incorrectly predict the rate of reaction to be higher than experimentally measured as not all collisions above the necessary energy will result in a successful reaction.

=Exercise 2: H-H-F System=

==PES Inspection==

For the forward reaction of H2 + F --> HF + H is exothermic, as shown in the surface plot below the energy of the reactants H2 is higher than the energy of the product HF. As the energy absorbed in breaking the H-H bond in the first case is less than the energy released forming the H-F bond which means that in total energy is released which results in an exothermic reaction.

{| class="wikitable" border=1
! Surface Plot H2+F->HF+H !! Surface Plot HF+H -> H2+F
|-
|[[File:mlw115_Surface_Plot_HHF_labelled.png|300px]]||[[File:mlw115_Surface_Plot_HHF_H2Prod.png|300px]]|
|}

==Determining Transition State ==

A = HA B = HB and C = F

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! Surface Plot !! Internuclear distance vs Time
|-
|[[File:Mlw115_Surface_Plot_TS_HHF.png|400px]]||[[File:Mlw115 INDvT HHF TS.png|350px]]|
|}

== Calculating Activation Energy ==

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! BC= 1.8135!! BC= 1.8135 Zoomed!! BC= 1.813
|-
|[[File:Mlw115_HHF_EvT_1.8135.png|300px]] ‎||[[File:Mlw115_Plot_1.813_EvT_zoomT0_Resized.png|300px]]||[[File:Mlw115_HHF_EvT_1.813.png|320px]]
|}

pAB = pBC =0

distAB = 0.74

{| class="wikitable" border=1
|Activation energy HF+H->H2+F
|-
! !! Energy (Kcal/mol)
|-
|Initial energy || -103.752
|-
|Final energy|| -133.89
|-
|EA|| 30.138
|}

{| class="wikitable" border=1
|+Activation energy H2+F->HF+H
|-
! !! Energy (Kcal/mol)
|-
|Initial energy|| -103.742
|-
|Final energy|| -103.752
|-
|EA|| 0.01
|}

==Reaction Dynamics==

A=H B=H C=F

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.74||0
|-
| BC|| 2.3||-2.7
|}

[[File:Mlw_Surface_Plot_HHF_Reactive.png]]

{| class="wikitable" border=1
!Energy!!
|-
|Kinetic|| 3.837
|-
|Potential|| -103.886

|-
|Total|| -100.049
|}

Internuclear momentum vs Time

[[File:Mlw_INMvT_HHF_Reactive.png]]

It can be seen from the animation that as the H-H (AB) bond is broken and the H-F (BC) bond is formed the vibrational energy (and so the kinetic energy) considerably increases for H-F. The internuclear momentum vs time graph demonstrates this, as the momentum is much higher for the H-F atom. This vibrational energy will be released as heat so experimentally an increase in temperature could confirm the reaction had occurred.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -3 || -0.5 || -100.339|| No || [[File:Mlw115_Surface_Plot_HHF_-3_0.5.png|300px]] || H2 approaches the fluorine atom slowly but when a collision is made the H rebounds straight back into the other H and they continue their path oscillating together in the opposite direction
|-

| 3 || -0.5 || -93.254 || No || [[File:Mlw115_Surface_Plot_HHF_3_0.5.png|300px]] || This is similar to the previous case however the hydrogen rebounds twice into flourine but on the second rebounds remains returns back to oscillating with the original H atom
|-
| 2 || -0.5|| -98.753|| No || [[File: Mlw115_Surface_Plot_HHF_2_0.5.png|300px]] || The HF molecule oscillates within the energy well at high vibrational energy after collision
|-
| 1.5|| -0.5|| 103.754|| No || [[File: Mlw115_Surface_Plot_HHF_1.5_0.5.png|300px]] || Similarly to the previous situation
|-
|0.1||-0.8|| 103.364|| Yes|| [[File: Mlw115_Surface_Plot_HHF_0.1_0.8.png|300px]] || Despite having lower total energy than the previous cases the reaction goes to completion as the momentum between the reacting atoms is sufficient
|}

Despite the H2 energy being higher than the activation energy if the BC atoms do not collide with a high enough energy the reaction will return back over the peak and the original reactants will be produced.

===Polanyi's Empirical Rules ===

Polanyi's rules state that in reactions with an early transition state (most closely resembling the reactants) translational energy most effectively promotes a successful reaction. Conversely where the transition state is late (resembling products closely) vibrational energy, specifically of the reactants most effectively promotes a successful reaction. <ref>The Journal of Chemical Physics 138, 234104 (2013); https://doi.org/10.1063/1.4810007</ref>

====H-F + H====
rAB= 2 rBC = 0.74
{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -5 || 0.2 || -103.364 || No || [[File:Mlw115_Surface_Plot_HFH_5_0.2.png|300px]] || Approaching H rebounds and reaction doesn't complete
|-
| 2.21||0.37||
|}

Reversal of conditions:
AB D 5.07 M -2.22
BC D 1.19 M 0.37

Random Guessing:
AB D 5.0 M -1.5
BC D 1.2 M 1.0

[[File: Mlw115_Surface_Plot_HFH_-1.5_1.png| 400px]]

=References=
<references/>

MRD:mlw115RXNDy

2018-05-22T13:35:41Z

Mlw115: /* Calculating Activation Energy */

= Reaction Dynamics =

==Introduction==

The reactivity of atom-diatom systems are studied in order to gain insight on the internal degrees of freedom of the system. Particularly vibrational and translational energy are considered with respect to overcoming the transition state energy barrier.

The atoms are considered to behave classically and the interactions are expressed as potential energy surfaces.

= Exercise 1: H2 + H system=

==Dynamics from the transition state region ==

Where a reaction is modeled on a potential energy surface the lowest energy path between the reactants and products is seen as a "valley floor" of the surface. The transition structure is a saddle point along the minimum energy ridge.

The gradient of the potential energy surface is zero at the minima and so the first derivative with respect to the is zero. This is also the case for the peak of the saddle point.

δV(ri)/δri = 0

However using second derivatives
δV2(ri)/δri2 > 0 Minima
δV2(ri)/δri2 < 0 Maxima

The second derivative of a maximum point is negative, whereas the second derivative of a minimum is positive. As the saddle point is a maxima along the minimum path the second derivative is negative

==Estimating Transition State Position ==

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.908||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot 1 !! Surface Plot 2!! Internuclear distance vs Time
|-
|[[File:Mlw115_H2H_saddle.png|300px]]||[[File:mlw115_H2H_TS.png|300px]]||[[File:Mlw115_H2H_TS_INDvT.png|300px]]|
|}

The best guess of inter nuclear distances at the transition state was found using trial and error, to find the positions that would result in the minimum trajectory. As the system is symmetric the distances would be equal. The two surface plots are included to demonstrate the "reaction path" which involves only slight oscillations of the molecules. The internuclear distance vs time plot demonstrates no change in any bond lengths at the transition state so the overall momenta is 0 and potential energy is at the maximum.

== Reaction Path ==
===MEP vs Dynamics===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.909||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot (MEP) !! Surface Plot (Dynamics
|-
|[[File:Mlw115_H2H_MEP_surfaceplot.png|300px]]||[[File:mlw115_H2H_MEP_surfaceplot_dynamics.png|300px]]
|}

It can be seen in this plot that the trajectory follows the minimum energy valley floor without any obvious deviations or oscillation.

The trajectory oscillates in this plot, also in much less steps the AB bond distance increases considerably more. This plot is more realistic as it accounts for the motion of atoms during the reaction.

{| class="wikitable" border="1"
|+ Final Geometry
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| 2.486||1.167
|}

=== Reversed starting positions ===

What would change if we used the initial conditions r1 = rts and r2 = rts+0.01 instead? ****** DO THIS ******

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| -2.486||-1.167
|}

[[File:mlw115_H2H_dynamics_reversed_momenta.png|300px]]

From taking the previous runs final conditions and setting these as the initial conditions and reversing the momenta values the reaction seems to run in reverse, reaching an end exactly at the transition state.

===Determining Reactivity===

 rAB = 0.74 

 rBC = 2.0 

{| class="wikitable" border=1
! pAB !! pBC!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -1.25 || -2.5 || -99.018|| Yes || [[File:Mlw115_Surface_Plot_1.25_2.5.png|300px]] || Initially the HAHB distance changes very little as the vibrational energy is quite low, as the HC approaches it can be seen that the BC distance decreases linearly. The molecules pass over the transition state, after this point the HAHB distance decreases and the HBHC distance oscillates slightly, with a higher vibrational energy than the HAHB molecule. This overall process represents the HAHB bond breaking and forming a HBHC molecule.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mlw115_Surface_Plot_1.5_2.0.png|300px]] || The molecule approaches the maxima with some vibrational energy, shown by oscillations in the initial path, the energy of the system is not high enough to pass over the transition state energy maxima so the HC "rebounds" and the HBHC distance begins to increase. The HAHB molecule oscillations increase.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mlw115_Surface_Plot_1.5_2.5.png|300px]] || The reaction proceeds similarly to the first case with HC approaching and forming a bond with HB as there is sufficient energy to proceed over the transition state maxima. The vibrational energy throughout is slightly higher than the first case, demonstrated by higher oscillations in atom distances.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mlw115_Surface_Plot_2.5_5.png|300px]] || In this case the energy is high enough and the atoms are able to pass over the transition state energy maxima, however the vibrational energy of the HBHC system formed seems to be too high to form the molecule and the HC atom rebounds and the distance begins to decrease and so the HAHB is reformed but with considerably higher oscillations.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mlw115_Surface_Plot_2.5_5.2.png|300px]] || The HC approaches the oscillating HAHB system, the HB atom rebounds between the two atoms, as it can be seen on the surface plot the system passes over the transition state maxima, back over to the other side and back again as this occurs. Finally the HBHC bond is formed with a high vibrationally energy
|}

===Transition State Theory===

TS state theory assumes that if the energy is high enough to cross the Transition state energy maxima then the reaction will be able to go to completion. However as demonstrated in the penultimate example the system can cross back over the maxima and the initial product will be reformed resulting in an incomplete reaction. This therefore will incorrectly predict the rate of reaction to be higher than experimentally measured as not all collisions above the necessary energy will result in a successful reaction.

=Exercise 2: H-H-F System=

==PES Inspection==

For the forward reaction of H2 + F --> HF + H is exothermic, as shown in the surface plot below the energy of the reactants H2 is higher than the energy of the product HF. As the energy absorbed in breaking the H-H bond in the first case is less than the energy released forming the H-F bond which means that in total energy is released which results in an exothermic reaction.

{| class="wikitable" border=1
! Surface Plot H2+F->HF+H !! Surface Plot HF+H -> H2+F
|-
|[[File:mlw115_Surface_Plot_HHF_labelled.png|300px]]||[[File:mlw115_Surface_Plot_HHF_H2Prod.png|300px]]|
|}

==Determining Transition State ==

A = HA B = HB and C = F

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! Surface Plot !! Internuclear distance vs Time
|-
|[[File:Mlw115_Surface_Plot_TS_HHF.png|400px]]||[[File:Mlw115 INDvT HHF TS.png|350px]]|
|}

== Calculating Activation Energy ==

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! BC= 1.8135!! BC= 1.8135 Zoomed!! BC= 1.813
|-
|[[File:Mlw115_HHF_EvT_1.8135.png|300px]] ‎||[[File:Mlw115_Plot_1.813_EvT_zoomT0_Resized.png|300px]]||[[File:Mlw115_HHF_EvT_1.813.png|320px]]
|}

pAB = pBC =0

distAB = 0.74

{| class="wikitable" border=1
|+1.813 Activation energy HF+H -> H2+F
|-
! !! Energy (Kcal/mol)
|-
|Initial energy || -103.752
|-
|Final energy|| -133.89
|-
|EA|| 30.138
|}

{| class="wikitable" border=1
|+1.8135 Activation energy H2+F->HF+H
|-
! !! Energy (Kcal/mol)
|-
|Initial energy|| -103.742
|-
|Final energy|| -103.752
|-
|EA|| 0.01
|}

==Reaction Dynamics==

A=H B=H C=F

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.74||0
|-
| BC|| 2.3||-2.7
|}

[[File:Mlw_Surface_Plot_HHF_Reactive.png]]

{| class="wikitable" border=1
!Energy!!
|-
|Kinetic|| 3.837
|-
|Potential|| -103.886

|-
|Total|| -100.049
|}

Internuclear momentum vs Time

[[File:Mlw_INMvT_HHF_Reactive.png]]

It can be seen from the animation that as the H-H (AB) bond is broken and the H-F (BC) bond is formed the vibrational energy (and so the kinetic energy) considerably increases for H-F. The internuclear momentum vs time graph demonstrates this, as the momentum is much higher for the H-F atom. This vibrational energy will be released as heat so experimentally an increase in temperature could confirm the reaction had occurred.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -3 || -0.5 || -100.339|| No || [[File:Mlw115_Surface_Plot_HHF_-3_0.5.png|300px]] || H2 approaches the fluorine atom slowly but when a collision is made the H rebounds straight back into the other H and they continue their path oscillating together in the opposite direction
|-

| 3 || -0.5 || -93.254 || No || [[File:Mlw115_Surface_Plot_HHF_3_0.5.png|300px]] || This is similar to the previous case however the hydrogen rebounds twice into flourine but on the second rebounds remains returns back to oscillating with the original H atom
|-
| 2 || -0.5|| -98.753|| No || [[File: Mlw115_Surface_Plot_HHF_2_0.5.png|300px]] || The HF molecule oscillates within the energy well at high vibrational energy after collision
|-
| 1.5|| -0.5|| 103.754|| No || [[File: Mlw115_Surface_Plot_HHF_1.5_0.5.png|300px]] || Similarly to the previous situation
|-
|0.1||-0.8|| 103.364|| Yes|| [[File: Mlw115_Surface_Plot_HHF_0.1_0.8.png|300px]] || Despite having lower total energy than the previous cases the reaction goes to completion as the momentum between the reacting atoms is sufficient
|}

Despite the H2 energy being higher than the activation energy if the BC atoms do not collide with a high enough energy the reaction will return back over the peak and the original reactants will be produced.

===Polanyi's Empirical Rules ===

Polanyi's rules state that in reactions with an early transition state (most closely resembling the reactants) translational energy most effectively promotes a successful reaction. Conversely where the transition state is late (resembling products closely) vibrational energy, specifically of the reactants most effectively promotes a successful reaction. <ref>The Journal of Chemical Physics 138, 234104 (2013); https://doi.org/10.1063/1.4810007</ref>

====H-F + H====
rAB= 2 rBC = 0.74
{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -5 || 0.2 || -103.364 || No || [[File:Mlw115_Surface_Plot_HFH_5_0.2.png|300px]] || Approaching H rebounds and reaction doesn't complete
|-
| 2.21||0.37||
|}

Reversal of conditions:
AB D 5.07 M -2.22
BC D 1.19 M 0.37

Random Guessing:
AB D 5.0 M -1.5
BC D 1.2 M 1.0

[[File: Mlw115_Surface_Plot_HFH_-1.5_1.png| 400px]]

=References=
<references/>

MRD:mlw115RXNDy

2018-05-22T13:31:39Z

Mlw115: /* Calculating Activation Energy */

= Reaction Dynamics =

==Introduction==

The reactivity of atom-diatom systems are studied in order to gain insight on the internal degrees of freedom of the system. Particularly vibrational and translational energy are considered with respect to overcoming the transition state energy barrier.

The atoms are considered to behave classically and the interactions are expressed as potential energy surfaces.

= Exercise 1: H2 + H system=

==Dynamics from the transition state region ==

Where a reaction is modeled on a potential energy surface the lowest energy path between the reactants and products is seen as a "valley floor" of the surface. The transition structure is a saddle point along the minimum energy ridge.

The gradient of the potential energy surface is zero at the minima and so the first derivative with respect to the is zero. This is also the case for the peak of the saddle point.

δV(ri)/δri = 0

However using second derivatives
δV2(ri)/δri2 > 0 Minima
δV2(ri)/δri2 < 0 Maxima

The second derivative of a maximum point is negative, whereas the second derivative of a minimum is positive. As the saddle point is a maxima along the minimum path the second derivative is negative

==Estimating Transition State Position ==

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.908||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot 1 !! Surface Plot 2!! Internuclear distance vs Time
|-
|[[File:Mlw115_H2H_saddle.png|300px]]||[[File:mlw115_H2H_TS.png|300px]]||[[File:Mlw115_H2H_TS_INDvT.png|300px]]|
|}

The best guess of inter nuclear distances at the transition state was found using trial and error, to find the positions that would result in the minimum trajectory. As the system is symmetric the distances would be equal. The two surface plots are included to demonstrate the "reaction path" which involves only slight oscillations of the molecules. The internuclear distance vs time plot demonstrates no change in any bond lengths at the transition state so the overall momenta is 0 and potential energy is at the maximum.

== Reaction Path ==
===MEP vs Dynamics===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.909||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot (MEP) !! Surface Plot (Dynamics
|-
|[[File:Mlw115_H2H_MEP_surfaceplot.png|300px]]||[[File:mlw115_H2H_MEP_surfaceplot_dynamics.png|300px]]
|}

It can be seen in this plot that the trajectory follows the minimum energy valley floor without any obvious deviations or oscillation.

The trajectory oscillates in this plot, also in much less steps the AB bond distance increases considerably more. This plot is more realistic as it accounts for the motion of atoms during the reaction.

{| class="wikitable" border="1"
|+ Final Geometry
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| 2.486||1.167
|}

=== Reversed starting positions ===

What would change if we used the initial conditions r1 = rts and r2 = rts+0.01 instead? ****** DO THIS ******

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| -2.486||-1.167
|}

[[File:mlw115_H2H_dynamics_reversed_momenta.png|300px]]

From taking the previous runs final conditions and setting these as the initial conditions and reversing the momenta values the reaction seems to run in reverse, reaching an end exactly at the transition state.

===Determining Reactivity===

 rAB = 0.74 

 rBC = 2.0 

{| class="wikitable" border=1
! pAB !! pBC!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -1.25 || -2.5 || -99.018|| Yes || [[File:Mlw115_Surface_Plot_1.25_2.5.png|300px]] || Initially the HAHB distance changes very little as the vibrational energy is quite low, as the HC approaches it can be seen that the BC distance decreases linearly. The molecules pass over the transition state, after this point the HAHB distance decreases and the HBHC distance oscillates slightly, with a higher vibrational energy than the HAHB molecule. This overall process represents the HAHB bond breaking and forming a HBHC molecule.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mlw115_Surface_Plot_1.5_2.0.png|300px]] || The molecule approaches the maxima with some vibrational energy, shown by oscillations in the initial path, the energy of the system is not high enough to pass over the transition state energy maxima so the HC "rebounds" and the HBHC distance begins to increase. The HAHB molecule oscillations increase.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mlw115_Surface_Plot_1.5_2.5.png|300px]] || The reaction proceeds similarly to the first case with HC approaching and forming a bond with HB as there is sufficient energy to proceed over the transition state maxima. The vibrational energy throughout is slightly higher than the first case, demonstrated by higher oscillations in atom distances.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mlw115_Surface_Plot_2.5_5.png|300px]] || In this case the energy is high enough and the atoms are able to pass over the transition state energy maxima, however the vibrational energy of the HBHC system formed seems to be too high to form the molecule and the HC atom rebounds and the distance begins to decrease and so the HAHB is reformed but with considerably higher oscillations.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mlw115_Surface_Plot_2.5_5.2.png|300px]] || The HC approaches the oscillating HAHB system, the HB atom rebounds between the two atoms, as it can be seen on the surface plot the system passes over the transition state maxima, back over to the other side and back again as this occurs. Finally the HBHC bond is formed with a high vibrationally energy
|}

===Transition State Theory===

TS state theory assumes that if the energy is high enough to cross the Transition state energy maxima then the reaction will be able to go to completion. However as demonstrated in the penultimate example the system can cross back over the maxima and the initial product will be reformed resulting in an incomplete reaction. This therefore will incorrectly predict the rate of reaction to be higher than experimentally measured as not all collisions above the necessary energy will result in a successful reaction.

=Exercise 2: H-H-F System=

==PES Inspection==

For the forward reaction of H2 + F --> HF + H is exothermic, as shown in the surface plot below the energy of the reactants H2 is higher than the energy of the product HF. As the energy absorbed in breaking the H-H bond in the first case is less than the energy released forming the H-F bond which means that in total energy is released which results in an exothermic reaction.

{| class="wikitable" border=1
! Surface Plot H2+F->HF+H !! Surface Plot HF+H -> H2+F
|-
|[[File:mlw115_Surface_Plot_HHF_labelled.png|300px]]||[[File:mlw115_Surface_Plot_HHF_H2Prod.png|300px]]|
|}

==Determining Transition State ==

A = HA B = HB and C = F

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! Surface Plot !! Internuclear distance vs Time
|-
|[[File:Mlw115_Surface_Plot_TS_HHF.png|400px]]||[[File:Mlw115 INDvT HHF TS.png|350px]]|
|}

== Calculating Activation Energy ==

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! BC= 1.8135!! BC= 1.8135 Zoomed!! BC= 1.813
|-
|[[File:Mlw115_HHF_EvT_1.8135.png|300px]] ‎||[[File:Mlw115_Plot_1.813_EvT_zoomT0_Resized.png|300px]]||[[File:Mlw115_HHF_EvT_1.813.png|320px]]
|}

pAB = pBC =0

distAB = 0.74

{| class="wikitable" border=1
|+1.813
|-
! !! Energy (Kcal/mol)
|-
|Top || -103.752
|-
|Bottom || -133.89
|-
|EA|| 30.138
|}

{| class="wikitable" border=1
|+1.8135
|-
! !! Energy (Kcal/mol)
|-
|Top of hump|| -103.772
|-
|Bottom of hump|| -103.742
|-
|EA|| 0.01
|}

==Reaction Dynamics==

A=H B=H C=F

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.74||0
|-
| BC|| 2.3||-2.7
|}

[[File:Mlw_Surface_Plot_HHF_Reactive.png]]

{| class="wikitable" border=1
!Energy!!
|-
|Kinetic|| 3.837
|-
|Potential|| -103.886

|-
|Total|| -100.049
|}

Internuclear momentum vs Time

[[File:Mlw_INMvT_HHF_Reactive.png]]

It can be seen from the animation that as the H-H (AB) bond is broken and the H-F (BC) bond is formed the vibrational energy (and so the kinetic energy) considerably increases for H-F. The internuclear momentum vs time graph demonstrates this, as the momentum is much higher for the H-F atom. This vibrational energy will be released as heat so experimentally an increase in temperature could confirm the reaction had occurred.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -3 || -0.5 || -100.339|| No || [[File:Mlw115_Surface_Plot_HHF_-3_0.5.png|300px]] || H2 approaches the fluorine atom slowly but when a collision is made the H rebounds straight back into the other H and they continue their path oscillating together in the opposite direction
|-

| 3 || -0.5 || -93.254 || No || [[File:Mlw115_Surface_Plot_HHF_3_0.5.png|300px]] || This is similar to the previous case however the hydrogen rebounds twice into flourine but on the second rebounds remains returns back to oscillating with the original H atom
|-
| 2 || -0.5|| -98.753|| No || [[File: Mlw115_Surface_Plot_HHF_2_0.5.png|300px]] || The HF molecule oscillates within the energy well at high vibrational energy after collision
|-
| 1.5|| -0.5|| 103.754|| No || [[File: Mlw115_Surface_Plot_HHF_1.5_0.5.png|300px]] || Similarly to the previous situation
|-
|0.1||-0.8|| 103.364|| Yes|| [[File: Mlw115_Surface_Plot_HHF_0.1_0.8.png|300px]] || Despite having lower total energy than the previous cases the reaction goes to completion as the momentum between the reacting atoms is sufficient
|}

Despite the H2 energy being higher than the activation energy if the BC atoms do not collide with a high enough energy the reaction will return back over the peak and the original reactants will be produced.

===Polanyi's Empirical Rules ===

Polanyi's rules state that in reactions with an early transition state (most closely resembling the reactants) translational energy most effectively promotes a successful reaction. Conversely where the transition state is late (resembling products closely) vibrational energy, specifically of the reactants most effectively promotes a successful reaction. <ref>The Journal of Chemical Physics 138, 234104 (2013); https://doi.org/10.1063/1.4810007</ref>

====H-F + H====
rAB= 2 rBC = 0.74
{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -5 || 0.2 || -103.364 || No || [[File:Mlw115_Surface_Plot_HFH_5_0.2.png|300px]] || Approaching H rebounds and reaction doesn't complete
|-
| 2.21||0.37||
|}

Reversal of conditions:
AB D 5.07 M -2.22
BC D 1.19 M 0.37

Random Guessing:
AB D 5.0 M -1.5
BC D 1.2 M 1.0

[[File: Mlw115_Surface_Plot_HFH_-1.5_1.png| 400px]]

=References=
<references/>

MRD:mlw115RXNDy

2018-05-22T13:31:14Z

Mlw115: /* Calculating Activation Energy */

= Reaction Dynamics =

==Introduction==

The reactivity of atom-diatom systems are studied in order to gain insight on the internal degrees of freedom of the system. Particularly vibrational and translational energy are considered with respect to overcoming the transition state energy barrier.

The atoms are considered to behave classically and the interactions are expressed as potential energy surfaces.

= Exercise 1: H2 + H system=

==Dynamics from the transition state region ==

Where a reaction is modeled on a potential energy surface the lowest energy path between the reactants and products is seen as a "valley floor" of the surface. The transition structure is a saddle point along the minimum energy ridge.

The gradient of the potential energy surface is zero at the minima and so the first derivative with respect to the is zero. This is also the case for the peak of the saddle point.

δV(ri)/δri = 0

However using second derivatives
δV2(ri)/δri2 > 0 Minima
δV2(ri)/δri2 < 0 Maxima

The second derivative of a maximum point is negative, whereas the second derivative of a minimum is positive. As the saddle point is a maxima along the minimum path the second derivative is negative

==Estimating Transition State Position ==

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.908||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot 1 !! Surface Plot 2!! Internuclear distance vs Time
|-
|[[File:Mlw115_H2H_saddle.png|300px]]||[[File:mlw115_H2H_TS.png|300px]]||[[File:Mlw115_H2H_TS_INDvT.png|300px]]|
|}

The best guess of inter nuclear distances at the transition state was found using trial and error, to find the positions that would result in the minimum trajectory. As the system is symmetric the distances would be equal. The two surface plots are included to demonstrate the "reaction path" which involves only slight oscillations of the molecules. The internuclear distance vs time plot demonstrates no change in any bond lengths at the transition state so the overall momenta is 0 and potential energy is at the maximum.

== Reaction Path ==
===MEP vs Dynamics===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.909||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot (MEP) !! Surface Plot (Dynamics
|-
|[[File:Mlw115_H2H_MEP_surfaceplot.png|300px]]||[[File:mlw115_H2H_MEP_surfaceplot_dynamics.png|300px]]
|}

It can be seen in this plot that the trajectory follows the minimum energy valley floor without any obvious deviations or oscillation.

The trajectory oscillates in this plot, also in much less steps the AB bond distance increases considerably more. This plot is more realistic as it accounts for the motion of atoms during the reaction.

{| class="wikitable" border="1"
|+ Final Geometry
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| 2.486||1.167
|}

=== Reversed starting positions ===

What would change if we used the initial conditions r1 = rts and r2 = rts+0.01 instead? ****** DO THIS ******

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| -2.486||-1.167
|}

[[File:mlw115_H2H_dynamics_reversed_momenta.png|300px]]

From taking the previous runs final conditions and setting these as the initial conditions and reversing the momenta values the reaction seems to run in reverse, reaching an end exactly at the transition state.

===Determining Reactivity===

 rAB = 0.74 

 rBC = 2.0 

{| class="wikitable" border=1
! pAB !! pBC!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -1.25 || -2.5 || -99.018|| Yes || [[File:Mlw115_Surface_Plot_1.25_2.5.png|300px]] || Initially the HAHB distance changes very little as the vibrational energy is quite low, as the HC approaches it can be seen that the BC distance decreases linearly. The molecules pass over the transition state, after this point the HAHB distance decreases and the HBHC distance oscillates slightly, with a higher vibrational energy than the HAHB molecule. This overall process represents the HAHB bond breaking and forming a HBHC molecule.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mlw115_Surface_Plot_1.5_2.0.png|300px]] || The molecule approaches the maxima with some vibrational energy, shown by oscillations in the initial path, the energy of the system is not high enough to pass over the transition state energy maxima so the HC "rebounds" and the HBHC distance begins to increase. The HAHB molecule oscillations increase.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mlw115_Surface_Plot_1.5_2.5.png|300px]] || The reaction proceeds similarly to the first case with HC approaching and forming a bond with HB as there is sufficient energy to proceed over the transition state maxima. The vibrational energy throughout is slightly higher than the first case, demonstrated by higher oscillations in atom distances.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mlw115_Surface_Plot_2.5_5.png|300px]] || In this case the energy is high enough and the atoms are able to pass over the transition state energy maxima, however the vibrational energy of the HBHC system formed seems to be too high to form the molecule and the HC atom rebounds and the distance begins to decrease and so the HAHB is reformed but with considerably higher oscillations.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mlw115_Surface_Plot_2.5_5.2.png|300px]] || The HC approaches the oscillating HAHB system, the HB atom rebounds between the two atoms, as it can be seen on the surface plot the system passes over the transition state maxima, back over to the other side and back again as this occurs. Finally the HBHC bond is formed with a high vibrationally energy
|}

===Transition State Theory===

TS state theory assumes that if the energy is high enough to cross the Transition state energy maxima then the reaction will be able to go to completion. However as demonstrated in the penultimate example the system can cross back over the maxima and the initial product will be reformed resulting in an incomplete reaction. This therefore will incorrectly predict the rate of reaction to be higher than experimentally measured as not all collisions above the necessary energy will result in a successful reaction.

=Exercise 2: H-H-F System=

==PES Inspection==

For the forward reaction of H2 + F --> HF + H is exothermic, as shown in the surface plot below the energy of the reactants H2 is higher than the energy of the product HF. As the energy absorbed in breaking the H-H bond in the first case is less than the energy released forming the H-F bond which means that in total energy is released which results in an exothermic reaction.

{| class="wikitable" border=1
! Surface Plot H2+F->HF+H !! Surface Plot HF+H -> H2+F
|-
|[[File:mlw115_Surface_Plot_HHF_labelled.png|300px]]||[[File:mlw115_Surface_Plot_HHF_H2Prod.png|300px]]|
|}

==Determining Transition State ==

A = HA B = HB and C = F

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! Surface Plot !! Internuclear distance vs Time
|-
|[[File:Mlw115_Surface_Plot_TS_HHF.png|400px]]||[[File:Mlw115 INDvT HHF TS.png|350px]]|
|}

== Calculating Activation Energy ==

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! BC= 1.8135!! BC= 1.8135 Zoomed!! BC= 1.813
|
|[[File:Mlw115_HHF_EvT_1.8135.png|300px]] ‎||[[File:Mlw115_Plot_1.813_EvT_zoomT0_Resized.png|300px]]||[[File:Mlw115_HHF_EvT_1.813.png|300px]]
|}

pAB = pBC =0

distAB = 0.74

{| class="wikitable" border=1
|+1.813
|-
! !! Energy (Kcal/mol)
|-
|Top || -103.752
|-
|Bottom || -133.89
|-
|EA|| 30.138
|}

{| class="wikitable" border=1
|+1.8135
|-
! !! Energy (Kcal/mol)
|-
|Top of hump|| -103.772
|-
|Bottom of hump|| -103.742
|-
|EA|| 0.01
|}

==Reaction Dynamics==

A=H B=H C=F

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.74||0
|-
| BC|| 2.3||-2.7
|}

[[File:Mlw_Surface_Plot_HHF_Reactive.png]]

{| class="wikitable" border=1
!Energy!!
|-
|Kinetic|| 3.837
|-
|Potential|| -103.886

|-
|Total|| -100.049
|}

Internuclear momentum vs Time

[[File:Mlw_INMvT_HHF_Reactive.png]]

It can be seen from the animation that as the H-H (AB) bond is broken and the H-F (BC) bond is formed the vibrational energy (and so the kinetic energy) considerably increases for H-F. The internuclear momentum vs time graph demonstrates this, as the momentum is much higher for the H-F atom. This vibrational energy will be released as heat so experimentally an increase in temperature could confirm the reaction had occurred.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -3 || -0.5 || -100.339|| No || [[File:Mlw115_Surface_Plot_HHF_-3_0.5.png|300px]] || H2 approaches the fluorine atom slowly but when a collision is made the H rebounds straight back into the other H and they continue their path oscillating together in the opposite direction
|-

| 3 || -0.5 || -93.254 || No || [[File:Mlw115_Surface_Plot_HHF_3_0.5.png|300px]] || This is similar to the previous case however the hydrogen rebounds twice into flourine but on the second rebounds remains returns back to oscillating with the original H atom
|-
| 2 || -0.5|| -98.753|| No || [[File: Mlw115_Surface_Plot_HHF_2_0.5.png|300px]] || The HF molecule oscillates within the energy well at high vibrational energy after collision
|-
| 1.5|| -0.5|| 103.754|| No || [[File: Mlw115_Surface_Plot_HHF_1.5_0.5.png|300px]] || Similarly to the previous situation
|-
|0.1||-0.8|| 103.364|| Yes|| [[File: Mlw115_Surface_Plot_HHF_0.1_0.8.png|300px]] || Despite having lower total energy than the previous cases the reaction goes to completion as the momentum between the reacting atoms is sufficient
|}

Despite the H2 energy being higher than the activation energy if the BC atoms do not collide with a high enough energy the reaction will return back over the peak and the original reactants will be produced.

===Polanyi's Empirical Rules ===

Polanyi's rules state that in reactions with an early transition state (most closely resembling the reactants) translational energy most effectively promotes a successful reaction. Conversely where the transition state is late (resembling products closely) vibrational energy, specifically of the reactants most effectively promotes a successful reaction. <ref>The Journal of Chemical Physics 138, 234104 (2013); https://doi.org/10.1063/1.4810007</ref>

====H-F + H====
rAB= 2 rBC = 0.74
{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -5 || 0.2 || -103.364 || No || [[File:Mlw115_Surface_Plot_HFH_5_0.2.png|300px]] || Approaching H rebounds and reaction doesn't complete
|-
| 2.21||0.37||
|}

Reversal of conditions:
AB D 5.07 M -2.22
BC D 1.19 M 0.37

Random Guessing:
AB D 5.0 M -1.5
BC D 1.2 M 1.0

[[File: Mlw115_Surface_Plot_HFH_-1.5_1.png| 400px]]

=References=
<references/>

MRD:mlw115RXNDy

2018-05-22T13:30:55Z

Mlw115: /* Calculating Activation Energy */

= Reaction Dynamics =

==Introduction==

The reactivity of atom-diatom systems are studied in order to gain insight on the internal degrees of freedom of the system. Particularly vibrational and translational energy are considered with respect to overcoming the transition state energy barrier.

The atoms are considered to behave classically and the interactions are expressed as potential energy surfaces.

= Exercise 1: H2 + H system=

==Dynamics from the transition state region ==

Where a reaction is modeled on a potential energy surface the lowest energy path between the reactants and products is seen as a "valley floor" of the surface. The transition structure is a saddle point along the minimum energy ridge.

The gradient of the potential energy surface is zero at the minima and so the first derivative with respect to the is zero. This is also the case for the peak of the saddle point.

δV(ri)/δri = 0

However using second derivatives
δV2(ri)/δri2 > 0 Minima
δV2(ri)/δri2 < 0 Maxima

The second derivative of a maximum point is negative, whereas the second derivative of a minimum is positive. As the saddle point is a maxima along the minimum path the second derivative is negative

==Estimating Transition State Position ==

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.908||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot 1 !! Surface Plot 2!! Internuclear distance vs Time
|-
|[[File:Mlw115_H2H_saddle.png|300px]]||[[File:mlw115_H2H_TS.png|300px]]||[[File:Mlw115_H2H_TS_INDvT.png|300px]]|
|}

The best guess of inter nuclear distances at the transition state was found using trial and error, to find the positions that would result in the minimum trajectory. As the system is symmetric the distances would be equal. The two surface plots are included to demonstrate the "reaction path" which involves only slight oscillations of the molecules. The internuclear distance vs time plot demonstrates no change in any bond lengths at the transition state so the overall momenta is 0 and potential energy is at the maximum.

== Reaction Path ==
===MEP vs Dynamics===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.909||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot (MEP) !! Surface Plot (Dynamics
|-
|[[File:Mlw115_H2H_MEP_surfaceplot.png|300px]]||[[File:mlw115_H2H_MEP_surfaceplot_dynamics.png|300px]]
|}

It can be seen in this plot that the trajectory follows the minimum energy valley floor without any obvious deviations or oscillation.

The trajectory oscillates in this plot, also in much less steps the AB bond distance increases considerably more. This plot is more realistic as it accounts for the motion of atoms during the reaction.

{| class="wikitable" border="1"
|+ Final Geometry
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| 2.486||1.167
|}

=== Reversed starting positions ===

What would change if we used the initial conditions r1 = rts and r2 = rts+0.01 instead? ****** DO THIS ******

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| -2.486||-1.167
|}

[[File:mlw115_H2H_dynamics_reversed_momenta.png|300px]]

From taking the previous runs final conditions and setting these as the initial conditions and reversing the momenta values the reaction seems to run in reverse, reaching an end exactly at the transition state.

===Determining Reactivity===

 rAB = 0.74 

 rBC = 2.0 

{| class="wikitable" border=1
! pAB !! pBC!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -1.25 || -2.5 || -99.018|| Yes || [[File:Mlw115_Surface_Plot_1.25_2.5.png|300px]] || Initially the HAHB distance changes very little as the vibrational energy is quite low, as the HC approaches it can be seen that the BC distance decreases linearly. The molecules pass over the transition state, after this point the HAHB distance decreases and the HBHC distance oscillates slightly, with a higher vibrational energy than the HAHB molecule. This overall process represents the HAHB bond breaking and forming a HBHC molecule.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mlw115_Surface_Plot_1.5_2.0.png|300px]] || The molecule approaches the maxima with some vibrational energy, shown by oscillations in the initial path, the energy of the system is not high enough to pass over the transition state energy maxima so the HC "rebounds" and the HBHC distance begins to increase. The HAHB molecule oscillations increase.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mlw115_Surface_Plot_1.5_2.5.png|300px]] || The reaction proceeds similarly to the first case with HC approaching and forming a bond with HB as there is sufficient energy to proceed over the transition state maxima. The vibrational energy throughout is slightly higher than the first case, demonstrated by higher oscillations in atom distances.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mlw115_Surface_Plot_2.5_5.png|300px]] || In this case the energy is high enough and the atoms are able to pass over the transition state energy maxima, however the vibrational energy of the HBHC system formed seems to be too high to form the molecule and the HC atom rebounds and the distance begins to decrease and so the HAHB is reformed but with considerably higher oscillations.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mlw115_Surface_Plot_2.5_5.2.png|300px]] || The HC approaches the oscillating HAHB system, the HB atom rebounds between the two atoms, as it can be seen on the surface plot the system passes over the transition state maxima, back over to the other side and back again as this occurs. Finally the HBHC bond is formed with a high vibrationally energy
|}

===Transition State Theory===

TS state theory assumes that if the energy is high enough to cross the Transition state energy maxima then the reaction will be able to go to completion. However as demonstrated in the penultimate example the system can cross back over the maxima and the initial product will be reformed resulting in an incomplete reaction. This therefore will incorrectly predict the rate of reaction to be higher than experimentally measured as not all collisions above the necessary energy will result in a successful reaction.

=Exercise 2: H-H-F System=

==PES Inspection==

For the forward reaction of H2 + F --> HF + H is exothermic, as shown in the surface plot below the energy of the reactants H2 is higher than the energy of the product HF. As the energy absorbed in breaking the H-H bond in the first case is less than the energy released forming the H-F bond which means that in total energy is released which results in an exothermic reaction.

{| class="wikitable" border=1
! Surface Plot H2+F->HF+H !! Surface Plot HF+H -> H2+F
|-
|[[File:mlw115_Surface_Plot_HHF_labelled.png|300px]]||[[File:mlw115_Surface_Plot_HHF_H2Prod.png|300px]]|
|}

==Determining Transition State ==

A = HA B = HB and C = F

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! Surface Plot !! Internuclear distance vs Time
|-
|[[File:Mlw115_Surface_Plot_TS_HHF.png|400px]]||[[File:Mlw115 INDvT HHF TS.png|350px]]|
|}

== Calculating Activation Energy ==

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! BC= 1.8135!! BC= 1.8135 Zoomed!! BC= 1.813
|-
|[[File:Mlw115_HHF_EvT_1.8135.png|300px]] ‎||[[File:Mlw115_Plot_1.813_EvT_zoomT0_Resized.png|300px]]||[[File:Mlw115_HHF_EvT_1.813.png|350px]]
|}

pAB = pBC =0

distAB = 0.74

{| class="wikitable" border=1
|+1.813
|-
! !! Energy (Kcal/mol)
|-
|Top || -103.752
|-
|Bottom || -133.89
|-
|EA|| 30.138
|}

{| class="wikitable" border=1
|+1.8135
|-
! !! Energy (Kcal/mol)
|-
|Top of hump|| -103.772
|-
|Bottom of hump|| -103.742
|-
|EA|| 0.01
|}

==Reaction Dynamics==

A=H B=H C=F

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.74||0
|-
| BC|| 2.3||-2.7
|}

[[File:Mlw_Surface_Plot_HHF_Reactive.png]]

{| class="wikitable" border=1
!Energy!!
|-
|Kinetic|| 3.837
|-
|Potential|| -103.886

|-
|Total|| -100.049
|}

Internuclear momentum vs Time

[[File:Mlw_INMvT_HHF_Reactive.png]]

It can be seen from the animation that as the H-H (AB) bond is broken and the H-F (BC) bond is formed the vibrational energy (and so the kinetic energy) considerably increases for H-F. The internuclear momentum vs time graph demonstrates this, as the momentum is much higher for the H-F atom. This vibrational energy will be released as heat so experimentally an increase in temperature could confirm the reaction had occurred.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -3 || -0.5 || -100.339|| No || [[File:Mlw115_Surface_Plot_HHF_-3_0.5.png|300px]] || H2 approaches the fluorine atom slowly but when a collision is made the H rebounds straight back into the other H and they continue their path oscillating together in the opposite direction
|-

| 3 || -0.5 || -93.254 || No || [[File:Mlw115_Surface_Plot_HHF_3_0.5.png|300px]] || This is similar to the previous case however the hydrogen rebounds twice into flourine but on the second rebounds remains returns back to oscillating with the original H atom
|-
| 2 || -0.5|| -98.753|| No || [[File: Mlw115_Surface_Plot_HHF_2_0.5.png|300px]] || The HF molecule oscillates within the energy well at high vibrational energy after collision
|-
| 1.5|| -0.5|| 103.754|| No || [[File: Mlw115_Surface_Plot_HHF_1.5_0.5.png|300px]] || Similarly to the previous situation
|-
|0.1||-0.8|| 103.364|| Yes|| [[File: Mlw115_Surface_Plot_HHF_0.1_0.8.png|300px]] || Despite having lower total energy than the previous cases the reaction goes to completion as the momentum between the reacting atoms is sufficient
|}

Despite the H2 energy being higher than the activation energy if the BC atoms do not collide with a high enough energy the reaction will return back over the peak and the original reactants will be produced.

===Polanyi's Empirical Rules ===

Polanyi's rules state that in reactions with an early transition state (most closely resembling the reactants) translational energy most effectively promotes a successful reaction. Conversely where the transition state is late (resembling products closely) vibrational energy, specifically of the reactants most effectively promotes a successful reaction. <ref>The Journal of Chemical Physics 138, 234104 (2013); https://doi.org/10.1063/1.4810007</ref>

====H-F + H====
rAB= 2 rBC = 0.74
{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -5 || 0.2 || -103.364 || No || [[File:Mlw115_Surface_Plot_HFH_5_0.2.png|300px]] || Approaching H rebounds and reaction doesn't complete
|-
| 2.21||0.37||
|}

Reversal of conditions:
AB D 5.07 M -2.22
BC D 1.19 M 0.37

Random Guessing:
AB D 5.0 M -1.5
BC D 1.2 M 1.0

[[File: Mlw115_Surface_Plot_HFH_-1.5_1.png| 400px]]

=References=
<references/>

MRD:mlw115RXNDy

2018-05-22T13:30:39Z

Mlw115: /* Calculating Activation Energy */

= Reaction Dynamics =

==Introduction==

The reactivity of atom-diatom systems are studied in order to gain insight on the internal degrees of freedom of the system. Particularly vibrational and translational energy are considered with respect to overcoming the transition state energy barrier.

The atoms are considered to behave classically and the interactions are expressed as potential energy surfaces.

= Exercise 1: H2 + H system=

==Dynamics from the transition state region ==

Where a reaction is modeled on a potential energy surface the lowest energy path between the reactants and products is seen as a "valley floor" of the surface. The transition structure is a saddle point along the minimum energy ridge.

The gradient of the potential energy surface is zero at the minima and so the first derivative with respect to the is zero. This is also the case for the peak of the saddle point.

δV(ri)/δri = 0

However using second derivatives
δV2(ri)/δri2 > 0 Minima
δV2(ri)/δri2 < 0 Maxima

The second derivative of a maximum point is negative, whereas the second derivative of a minimum is positive. As the saddle point is a maxima along the minimum path the second derivative is negative

==Estimating Transition State Position ==

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.908||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot 1 !! Surface Plot 2!! Internuclear distance vs Time
|-
|[[File:Mlw115_H2H_saddle.png|300px]]||[[File:mlw115_H2H_TS.png|300px]]||[[File:Mlw115_H2H_TS_INDvT.png|300px]]|
|}

The best guess of inter nuclear distances at the transition state was found using trial and error, to find the positions that would result in the minimum trajectory. As the system is symmetric the distances would be equal. The two surface plots are included to demonstrate the "reaction path" which involves only slight oscillations of the molecules. The internuclear distance vs time plot demonstrates no change in any bond lengths at the transition state so the overall momenta is 0 and potential energy is at the maximum.

== Reaction Path ==
===MEP vs Dynamics===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.909||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot (MEP) !! Surface Plot (Dynamics
|-
|[[File:Mlw115_H2H_MEP_surfaceplot.png|300px]]||[[File:mlw115_H2H_MEP_surfaceplot_dynamics.png|300px]]
|}

It can be seen in this plot that the trajectory follows the minimum energy valley floor without any obvious deviations or oscillation.

The trajectory oscillates in this plot, also in much less steps the AB bond distance increases considerably more. This plot is more realistic as it accounts for the motion of atoms during the reaction.

{| class="wikitable" border="1"
|+ Final Geometry
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| 2.486||1.167
|}

=== Reversed starting positions ===

What would change if we used the initial conditions r1 = rts and r2 = rts+0.01 instead? ****** DO THIS ******

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| -2.486||-1.167
|}

[[File:mlw115_H2H_dynamics_reversed_momenta.png|300px]]

From taking the previous runs final conditions and setting these as the initial conditions and reversing the momenta values the reaction seems to run in reverse, reaching an end exactly at the transition state.

===Determining Reactivity===

 rAB = 0.74 

 rBC = 2.0 

{| class="wikitable" border=1
! pAB !! pBC!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -1.25 || -2.5 || -99.018|| Yes || [[File:Mlw115_Surface_Plot_1.25_2.5.png|300px]] || Initially the HAHB distance changes very little as the vibrational energy is quite low, as the HC approaches it can be seen that the BC distance decreases linearly. The molecules pass over the transition state, after this point the HAHB distance decreases and the HBHC distance oscillates slightly, with a higher vibrational energy than the HAHB molecule. This overall process represents the HAHB bond breaking and forming a HBHC molecule.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mlw115_Surface_Plot_1.5_2.0.png|300px]] || The molecule approaches the maxima with some vibrational energy, shown by oscillations in the initial path, the energy of the system is not high enough to pass over the transition state energy maxima so the HC "rebounds" and the HBHC distance begins to increase. The HAHB molecule oscillations increase.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mlw115_Surface_Plot_1.5_2.5.png|300px]] || The reaction proceeds similarly to the first case with HC approaching and forming a bond with HB as there is sufficient energy to proceed over the transition state maxima. The vibrational energy throughout is slightly higher than the first case, demonstrated by higher oscillations in atom distances.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mlw115_Surface_Plot_2.5_5.png|300px]] || In this case the energy is high enough and the atoms are able to pass over the transition state energy maxima, however the vibrational energy of the HBHC system formed seems to be too high to form the molecule and the HC atom rebounds and the distance begins to decrease and so the HAHB is reformed but with considerably higher oscillations.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mlw115_Surface_Plot_2.5_5.2.png|300px]] || The HC approaches the oscillating HAHB system, the HB atom rebounds between the two atoms, as it can be seen on the surface plot the system passes over the transition state maxima, back over to the other side and back again as this occurs. Finally the HBHC bond is formed with a high vibrationally energy
|}

===Transition State Theory===

TS state theory assumes that if the energy is high enough to cross the Transition state energy maxima then the reaction will be able to go to completion. However as demonstrated in the penultimate example the system can cross back over the maxima and the initial product will be reformed resulting in an incomplete reaction. This therefore will incorrectly predict the rate of reaction to be higher than experimentally measured as not all collisions above the necessary energy will result in a successful reaction.

=Exercise 2: H-H-F System=

==PES Inspection==

For the forward reaction of H2 + F --> HF + H is exothermic, as shown in the surface plot below the energy of the reactants H2 is higher than the energy of the product HF. As the energy absorbed in breaking the H-H bond in the first case is less than the energy released forming the H-F bond which means that in total energy is released which results in an exothermic reaction.

{| class="wikitable" border=1
! Surface Plot H2+F->HF+H !! Surface Plot HF+H -> H2+F
|-
|[[File:mlw115_Surface_Plot_HHF_labelled.png|300px]]||[[File:mlw115_Surface_Plot_HHF_H2Prod.png|300px]]|
|}

==Determining Transition State ==

A = HA B = HB and C = F

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! Surface Plot !! Internuclear distance vs Time
|-
|[[File:Mlw115_Surface_Plot_TS_HHF.png|400px]]||[[File:Mlw115 INDvT HHF TS.png|350px]]|
|}

== Calculating Activation Energy ==

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! BC= 1.8135!! BC= 1.8135 Zoomed!! BC= 1.813
|-
|[[File:Mlw115_HHF_EvT_1.8135.png|300px]] ‎||[[File:Mlw115_Plot_1.813_EvT_zoomT0_Resized.png|400px]]||[[File:Mlw115_HHF_EvT_1.813.png|350px]]
|}

pAB = pBC =0

distAB = 0.74

{| class="wikitable" border=1
|+1.813
|-
! !! Energy (Kcal/mol)
|-
|Top || -103.752
|-
|Bottom || -133.89
|-
|EA|| 30.138
|}

{| class="wikitable" border=1
|+1.8135
|-
! !! Energy (Kcal/mol)
|-
|Top of hump|| -103.772
|-
|Bottom of hump|| -103.742
|-
|EA|| 0.01
|}

==Reaction Dynamics==

A=H B=H C=F

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.74||0
|-
| BC|| 2.3||-2.7
|}

[[File:Mlw_Surface_Plot_HHF_Reactive.png]]

{| class="wikitable" border=1
!Energy!!
|-
|Kinetic|| 3.837
|-
|Potential|| -103.886

|-
|Total|| -100.049
|}

Internuclear momentum vs Time

[[File:Mlw_INMvT_HHF_Reactive.png]]

It can be seen from the animation that as the H-H (AB) bond is broken and the H-F (BC) bond is formed the vibrational energy (and so the kinetic energy) considerably increases for H-F. The internuclear momentum vs time graph demonstrates this, as the momentum is much higher for the H-F atom. This vibrational energy will be released as heat so experimentally an increase in temperature could confirm the reaction had occurred.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -3 || -0.5 || -100.339|| No || [[File:Mlw115_Surface_Plot_HHF_-3_0.5.png|300px]] || H2 approaches the fluorine atom slowly but when a collision is made the H rebounds straight back into the other H and they continue their path oscillating together in the opposite direction
|-

| 3 || -0.5 || -93.254 || No || [[File:Mlw115_Surface_Plot_HHF_3_0.5.png|300px]] || This is similar to the previous case however the hydrogen rebounds twice into flourine but on the second rebounds remains returns back to oscillating with the original H atom
|-
| 2 || -0.5|| -98.753|| No || [[File: Mlw115_Surface_Plot_HHF_2_0.5.png|300px]] || The HF molecule oscillates within the energy well at high vibrational energy after collision
|-
| 1.5|| -0.5|| 103.754|| No || [[File: Mlw115_Surface_Plot_HHF_1.5_0.5.png|300px]] || Similarly to the previous situation
|-
|0.1||-0.8|| 103.364|| Yes|| [[File: Mlw115_Surface_Plot_HHF_0.1_0.8.png|300px]] || Despite having lower total energy than the previous cases the reaction goes to completion as the momentum between the reacting atoms is sufficient
|}

Despite the H2 energy being higher than the activation energy if the BC atoms do not collide with a high enough energy the reaction will return back over the peak and the original reactants will be produced.

===Polanyi's Empirical Rules ===

Polanyi's rules state that in reactions with an early transition state (most closely resembling the reactants) translational energy most effectively promotes a successful reaction. Conversely where the transition state is late (resembling products closely) vibrational energy, specifically of the reactants most effectively promotes a successful reaction. <ref>The Journal of Chemical Physics 138, 234104 (2013); https://doi.org/10.1063/1.4810007</ref>

====H-F + H====
rAB= 2 rBC = 0.74
{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -5 || 0.2 || -103.364 || No || [[File:Mlw115_Surface_Plot_HFH_5_0.2.png|300px]] || Approaching H rebounds and reaction doesn't complete
|-
| 2.21||0.37||
|}

Reversal of conditions:
AB D 5.07 M -2.22
BC D 1.19 M 0.37

Random Guessing:
AB D 5.0 M -1.5
BC D 1.2 M 1.0

[[File: Mlw115_Surface_Plot_HFH_-1.5_1.png| 400px]]

=References=
<references/>

MRD:mlw115RXNDy

2018-05-22T13:30:15Z

Mlw115: /* Dynamics from the transition state region */

= Reaction Dynamics =

==Introduction==

The reactivity of atom-diatom systems are studied in order to gain insight on the internal degrees of freedom of the system. Particularly vibrational and translational energy are considered with respect to overcoming the transition state energy barrier.

The atoms are considered to behave classically and the interactions are expressed as potential energy surfaces.

= Exercise 1: H2 + H system=

==Dynamics from the transition state region ==

Where a reaction is modeled on a potential energy surface the lowest energy path between the reactants and products is seen as a "valley floor" of the surface. The transition structure is a saddle point along the minimum energy ridge.

The gradient of the potential energy surface is zero at the minima and so the first derivative with respect to the is zero. This is also the case for the peak of the saddle point.

δV(ri)/δri = 0

However using second derivatives
δV2(ri)/δri2 > 0 Minima
δV2(ri)/δri2 < 0 Maxima

The second derivative of a maximum point is negative, whereas the second derivative of a minimum is positive. As the saddle point is a maxima along the minimum path the second derivative is negative

==Estimating Transition State Position ==

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.908||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot 1 !! Surface Plot 2!! Internuclear distance vs Time
|-
|[[File:Mlw115_H2H_saddle.png|300px]]||[[File:mlw115_H2H_TS.png|300px]]||[[File:Mlw115_H2H_TS_INDvT.png|300px]]|
|}

The best guess of inter nuclear distances at the transition state was found using trial and error, to find the positions that would result in the minimum trajectory. As the system is symmetric the distances would be equal. The two surface plots are included to demonstrate the "reaction path" which involves only slight oscillations of the molecules. The internuclear distance vs time plot demonstrates no change in any bond lengths at the transition state so the overall momenta is 0 and potential energy is at the maximum.

== Reaction Path ==
===MEP vs Dynamics===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.909||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot (MEP) !! Surface Plot (Dynamics
|-
|[[File:Mlw115_H2H_MEP_surfaceplot.png|300px]]||[[File:mlw115_H2H_MEP_surfaceplot_dynamics.png|300px]]
|}

It can be seen in this plot that the trajectory follows the minimum energy valley floor without any obvious deviations or oscillation.

The trajectory oscillates in this plot, also in much less steps the AB bond distance increases considerably more. This plot is more realistic as it accounts for the motion of atoms during the reaction.

{| class="wikitable" border="1"
|+ Final Geometry
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| 2.486||1.167
|}

=== Reversed starting positions ===

What would change if we used the initial conditions r1 = rts and r2 = rts+0.01 instead? ****** DO THIS ******

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| -2.486||-1.167
|}

[[File:mlw115_H2H_dynamics_reversed_momenta.png|300px]]

From taking the previous runs final conditions and setting these as the initial conditions and reversing the momenta values the reaction seems to run in reverse, reaching an end exactly at the transition state.

===Determining Reactivity===

 rAB = 0.74 

 rBC = 2.0 

{| class="wikitable" border=1
! pAB !! pBC!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -1.25 || -2.5 || -99.018|| Yes || [[File:Mlw115_Surface_Plot_1.25_2.5.png|300px]] || Initially the HAHB distance changes very little as the vibrational energy is quite low, as the HC approaches it can be seen that the BC distance decreases linearly. The molecules pass over the transition state, after this point the HAHB distance decreases and the HBHC distance oscillates slightly, with a higher vibrational energy than the HAHB molecule. This overall process represents the HAHB bond breaking and forming a HBHC molecule.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mlw115_Surface_Plot_1.5_2.0.png|300px]] || The molecule approaches the maxima with some vibrational energy, shown by oscillations in the initial path, the energy of the system is not high enough to pass over the transition state energy maxima so the HC "rebounds" and the HBHC distance begins to increase. The HAHB molecule oscillations increase.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mlw115_Surface_Plot_1.5_2.5.png|300px]] || The reaction proceeds similarly to the first case with HC approaching and forming a bond with HB as there is sufficient energy to proceed over the transition state maxima. The vibrational energy throughout is slightly higher than the first case, demonstrated by higher oscillations in atom distances.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mlw115_Surface_Plot_2.5_5.png|300px]] || In this case the energy is high enough and the atoms are able to pass over the transition state energy maxima, however the vibrational energy of the HBHC system formed seems to be too high to form the molecule and the HC atom rebounds and the distance begins to decrease and so the HAHB is reformed but with considerably higher oscillations.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mlw115_Surface_Plot_2.5_5.2.png|300px]] || The HC approaches the oscillating HAHB system, the HB atom rebounds between the two atoms, as it can be seen on the surface plot the system passes over the transition state maxima, back over to the other side and back again as this occurs. Finally the HBHC bond is formed with a high vibrationally energy
|}

===Transition State Theory===

TS state theory assumes that if the energy is high enough to cross the Transition state energy maxima then the reaction will be able to go to completion. However as demonstrated in the penultimate example the system can cross back over the maxima and the initial product will be reformed resulting in an incomplete reaction. This therefore will incorrectly predict the rate of reaction to be higher than experimentally measured as not all collisions above the necessary energy will result in a successful reaction.

=Exercise 2: H-H-F System=

==PES Inspection==

For the forward reaction of H2 + F --> HF + H is exothermic, as shown in the surface plot below the energy of the reactants H2 is higher than the energy of the product HF. As the energy absorbed in breaking the H-H bond in the first case is less than the energy released forming the H-F bond which means that in total energy is released which results in an exothermic reaction.

{| class="wikitable" border=1
! Surface Plot H2+F->HF+H !! Surface Plot HF+H -> H2+F
|-
|[[File:mlw115_Surface_Plot_HHF_labelled.png|300px]]||[[File:mlw115_Surface_Plot_HHF_H2Prod.png|300px]]|
|}

==Determining Transition State ==

A = HA B = HB and C = F

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! Surface Plot !! Internuclear distance vs Time
|-
|[[File:Mlw115_Surface_Plot_TS_HHF.png|400px]]||[[File:Mlw115 INDvT HHF TS.png|350px]]|
|}

== Calculating Activation Energy ==

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! BC= 1.8135!! BC= 1.8135 Zoomed!! BC= 1.813
|-
|[[File:Mlw115_HHF_EvT_1.8135.png|300px]] ‎||[[File:Mlw115 HHF EvT 1.8135 zoom.png|400px]]||[[File:Mlw115_HHF_EvT_1.813.png|350px]]
|}

pAB = pBC =0

distAB = 0.74

{| class="wikitable" border=1
|+1.813
|-
! !! Energy (Kcal/mol)
|-
|Top || -103.752
|-
|Bottom || -133.89
|-
|EA|| 30.138
|}

{| class="wikitable" border=1
|+1.8135
|-
! !! Energy (Kcal/mol)
|-
|Top of hump|| -103.772
|-
|Bottom of hump|| -103.742
|-
|EA|| 0.01
|}

==Reaction Dynamics==

A=H B=H C=F

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.74||0
|-
| BC|| 2.3||-2.7
|}

[[File:Mlw_Surface_Plot_HHF_Reactive.png]]

{| class="wikitable" border=1
!Energy!!
|-
|Kinetic|| 3.837
|-
|Potential|| -103.886

|-
|Total|| -100.049
|}

Internuclear momentum vs Time

[[File:Mlw_INMvT_HHF_Reactive.png]]

It can be seen from the animation that as the H-H (AB) bond is broken and the H-F (BC) bond is formed the vibrational energy (and so the kinetic energy) considerably increases for H-F. The internuclear momentum vs time graph demonstrates this, as the momentum is much higher for the H-F atom. This vibrational energy will be released as heat so experimentally an increase in temperature could confirm the reaction had occurred.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -3 || -0.5 || -100.339|| No || [[File:Mlw115_Surface_Plot_HHF_-3_0.5.png|300px]] || H2 approaches the fluorine atom slowly but when a collision is made the H rebounds straight back into the other H and they continue their path oscillating together in the opposite direction
|-

| 3 || -0.5 || -93.254 || No || [[File:Mlw115_Surface_Plot_HHF_3_0.5.png|300px]] || This is similar to the previous case however the hydrogen rebounds twice into flourine but on the second rebounds remains returns back to oscillating with the original H atom
|-
| 2 || -0.5|| -98.753|| No || [[File: Mlw115_Surface_Plot_HHF_2_0.5.png|300px]] || The HF molecule oscillates within the energy well at high vibrational energy after collision
|-
| 1.5|| -0.5|| 103.754|| No || [[File: Mlw115_Surface_Plot_HHF_1.5_0.5.png|300px]] || Similarly to the previous situation
|-
|0.1||-0.8|| 103.364|| Yes|| [[File: Mlw115_Surface_Plot_HHF_0.1_0.8.png|300px]] || Despite having lower total energy than the previous cases the reaction goes to completion as the momentum between the reacting atoms is sufficient
|}

Despite the H2 energy being higher than the activation energy if the BC atoms do not collide with a high enough energy the reaction will return back over the peak and the original reactants will be produced.

===Polanyi's Empirical Rules ===

Polanyi's rules state that in reactions with an early transition state (most closely resembling the reactants) translational energy most effectively promotes a successful reaction. Conversely where the transition state is late (resembling products closely) vibrational energy, specifically of the reactants most effectively promotes a successful reaction. <ref>The Journal of Chemical Physics 138, 234104 (2013); https://doi.org/10.1063/1.4810007</ref>

====H-F + H====
rAB= 2 rBC = 0.74
{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -5 || 0.2 || -103.364 || No || [[File:Mlw115_Surface_Plot_HFH_5_0.2.png|300px]] || Approaching H rebounds and reaction doesn't complete
|-
| 2.21||0.37||
|}

Reversal of conditions:
AB D 5.07 M -2.22
BC D 1.19 M 0.37

Random Guessing:
AB D 5.0 M -1.5
BC D 1.2 M 1.0

[[File: Mlw115_Surface_Plot_HFH_-1.5_1.png| 400px]]

=References=
<references/>

File:Mlw115 Plot 1.813 EvT zoomT0 Resized.png

2018-05-22T13:30:13Z

Mlw115:

MRD:mlw115RXNDy

2018-05-22T13:26:29Z

Mlw115: /* Calculating Activation Energy */

= Reaction Dynamics =

==Introduction==

The reactivity of atom-diatom systems are studied in order to gain insight on the internal degrees of freedom of the system. Particularly vibrational and translational energy are considered with respect to overcoming the transition state energy barrier.

The atoms are considered to behave classically and the interactions are expressed as potential energy surfaces.

= Exercise 1: H2 + H system=

==Dynamics from the transition state region ==

Where a reaction is modeled on a potential energy surface the lowest energy path between the reactants and products is seen as a "valley floor" of the surface. The transition structure is a saddle point along the minimum energy ridge.

The gradient of the potential energy surface is zero at the minima and so the first derivative is zero. This is also the case for the peak of the saddle point.

δV(ri)/δri = 0

However using second derivatives
δV2(ri)/δri2 > 0 Minima
δV2(ri)/δri2 < 0 Maxima

The second derivative of a maximum point is negative, whereas the second derivative of a minimum is positive. As the saddle point is a maxima along the minimum path the second derivative is negative

==Estimating Transition State Position ==

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.908||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot 1 !! Surface Plot 2!! Internuclear distance vs Time
|-
|[[File:Mlw115_H2H_saddle.png|300px]]||[[File:mlw115_H2H_TS.png|300px]]||[[File:Mlw115_H2H_TS_INDvT.png|300px]]|
|}

The best guess of inter nuclear distances at the transition state was found using trial and error, to find the positions that would result in the minimum trajectory. As the system is symmetric the distances would be equal. The two surface plots are included to demonstrate the "reaction path" which involves only slight oscillations of the molecules. The internuclear distance vs time plot demonstrates no change in any bond lengths at the transition state so the overall momenta is 0 and potential energy is at the maximum.

== Reaction Path ==
===MEP vs Dynamics===

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.909||0
|-
| BC|| 0.908||0
|}

{| class="wikitable" border=1
! Surface Plot (MEP) !! Surface Plot (Dynamics
|-
|[[File:Mlw115_H2H_MEP_surfaceplot.png|300px]]||[[File:mlw115_H2H_MEP_surfaceplot_dynamics.png|300px]]
|}

It can be seen in this plot that the trajectory follows the minimum energy valley floor without any obvious deviations or oscillation.

The trajectory oscillates in this plot, also in much less steps the AB bond distance increases considerably more. This plot is more realistic as it accounts for the motion of atoms during the reaction.

{| class="wikitable" border="1"
|+ Final Geometry
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| 2.486||1.167
|}

=== Reversed starting positions ===

What would change if we used the initial conditions r1 = rts and r2 = rts+0.01 instead? ****** DO THIS ******

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 9.993||0.754
|-
| BC|| -2.486||-1.167
|}

[[File:mlw115_H2H_dynamics_reversed_momenta.png|300px]]

From taking the previous runs final conditions and setting these as the initial conditions and reversing the momenta values the reaction seems to run in reverse, reaching an end exactly at the transition state.

===Determining Reactivity===

 rAB = 0.74 

 rBC = 2.0 

{| class="wikitable" border=1
! pAB !! pBC!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -1.25 || -2.5 || -99.018|| Yes || [[File:Mlw115_Surface_Plot_1.25_2.5.png|300px]] || Initially the HAHB distance changes very little as the vibrational energy is quite low, as the HC approaches it can be seen that the BC distance decreases linearly. The molecules pass over the transition state, after this point the HAHB distance decreases and the HBHC distance oscillates slightly, with a higher vibrational energy than the HAHB molecule. This overall process represents the HAHB bond breaking and forming a HBHC molecule.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Mlw115_Surface_Plot_1.5_2.0.png|300px]] || The molecule approaches the maxima with some vibrational energy, shown by oscillations in the initial path, the energy of the system is not high enough to pass over the transition state energy maxima so the HC "rebounds" and the HBHC distance begins to increase. The HAHB molecule oscillations increase.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Mlw115_Surface_Plot_1.5_2.5.png|300px]] || The reaction proceeds similarly to the first case with HC approaching and forming a bond with HB as there is sufficient energy to proceed over the transition state maxima. The vibrational energy throughout is slightly higher than the first case, demonstrated by higher oscillations in atom distances.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:Mlw115_Surface_Plot_2.5_5.png|300px]] || In this case the energy is high enough and the atoms are able to pass over the transition state energy maxima, however the vibrational energy of the HBHC system formed seems to be too high to form the molecule and the HC atom rebounds and the distance begins to decrease and so the HAHB is reformed but with considerably higher oscillations.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Mlw115_Surface_Plot_2.5_5.2.png|300px]] || The HC approaches the oscillating HAHB system, the HB atom rebounds between the two atoms, as it can be seen on the surface plot the system passes over the transition state maxima, back over to the other side and back again as this occurs. Finally the HBHC bond is formed with a high vibrationally energy
|}

===Transition State Theory===

TS state theory assumes that if the energy is high enough to cross the Transition state energy maxima then the reaction will be able to go to completion. However as demonstrated in the penultimate example the system can cross back over the maxima and the initial product will be reformed resulting in an incomplete reaction. This therefore will incorrectly predict the rate of reaction to be higher than experimentally measured as not all collisions above the necessary energy will result in a successful reaction.

=Exercise 2: H-H-F System=

==PES Inspection==

For the forward reaction of H2 + F --> HF + H is exothermic, as shown in the surface plot below the energy of the reactants H2 is higher than the energy of the product HF. As the energy absorbed in breaking the H-H bond in the first case is less than the energy released forming the H-F bond which means that in total energy is released which results in an exothermic reaction.

{| class="wikitable" border=1
! Surface Plot H2+F->HF+H !! Surface Plot HF+H -> H2+F
|-
|[[File:mlw115_Surface_Plot_HHF_labelled.png|300px]]||[[File:mlw115_Surface_Plot_HHF_H2Prod.png|300px]]|
|}

==Determining Transition State ==

A = HA B = HB and C = F

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! Surface Plot !! Internuclear distance vs Time
|-
|[[File:Mlw115_Surface_Plot_TS_HHF.png|400px]]||[[File:Mlw115 INDvT HHF TS.png|350px]]|
|}

== Calculating Activation Energy ==

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 1.813
|}

{| class="wikitable" border=1
! BC= 1.8135!! BC= 1.8135 Zoomed!! BC= 1.813
|-
|[[File:Mlw115_HHF_EvT_1.8135.png|300px]] ‎||[[File:Mlw115 HHF EvT 1.8135 zoom.png|400px]]||[[File:Mlw115_HHF_EvT_1.813.png|350px]]
|}

pAB = pBC =0

distAB = 0.74

{| class="wikitable" border=1
|+1.813
|-
! !! Energy (Kcal/mol)
|-
|Top || -103.752
|-
|Bottom || -133.89
|-
|EA|| 30.138
|}

{| class="wikitable" border=1
|+1.8135
|-
! !! Energy (Kcal/mol)
|-
|Top of hump|| -103.772
|-
|Bottom of hump|| -103.742
|-
|EA|| 0.01
|}

==Reaction Dynamics==

A=H B=H C=F

{| class="wikitable" border="1"
|+ Initial Conditions
! !! Distance!! Momentum
|-
| AB|| 0.74||0
|-
| BC|| 2.3||-2.7
|}

[[File:Mlw_Surface_Plot_HHF_Reactive.png]]

{| class="wikitable" border=1
!Energy!!
|-
|Kinetic|| 3.837
|-
|Potential|| -103.886

|-
|Total|| -100.049
|}

Internuclear momentum vs Time

[[File:Mlw_INMvT_HHF_Reactive.png]]

It can be seen from the animation that as the H-H (AB) bond is broken and the H-F (BC) bond is formed the vibrational energy (and so the kinetic energy) considerably increases for H-F. The internuclear momentum vs time graph demonstrates this, as the momentum is much higher for the H-F atom. This vibrational energy will be released as heat so experimentally an increase in temperature could confirm the reaction had occurred.

{| class="wikitable" border=1
! Atoms !! Internuclear distance
|-
|AB|| 0.74
|-
|BC || 2
|}

{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -3 || -0.5 || -100.339|| No || [[File:Mlw115_Surface_Plot_HHF_-3_0.5.png|300px]] || H2 approaches the fluorine atom slowly but when a collision is made the H rebounds straight back into the other H and they continue their path oscillating together in the opposite direction
|-

| 3 || -0.5 || -93.254 || No || [[File:Mlw115_Surface_Plot_HHF_3_0.5.png|300px]] || This is similar to the previous case however the hydrogen rebounds twice into flourine but on the second rebounds remains returns back to oscillating with the original H atom
|-
| 2 || -0.5|| -98.753|| No || [[File: Mlw115_Surface_Plot_HHF_2_0.5.png|300px]] || The HF molecule oscillates within the energy well at high vibrational energy after collision
|-
| 1.5|| -0.5|| 103.754|| No || [[File: Mlw115_Surface_Plot_HHF_1.5_0.5.png|300px]] || Similarly to the previous situation
|-
|0.1||-0.8|| 103.364|| Yes|| [[File: Mlw115_Surface_Plot_HHF_0.1_0.8.png|300px]] || Despite having lower total energy than the previous cases the reaction goes to completion as the momentum between the reacting atoms is sufficient
|}

Despite the H2 energy being higher than the activation energy if the BC atoms do not collide with a high enough energy the reaction will return back over the peak and the original reactants will be produced.

===Polanyi's Empirical Rules ===

Polanyi's rules state that in reactions with an early transition state (most closely resembling the reactants) translational energy most effectively promotes a successful reaction. Conversely where the transition state is late (resembling products closely) vibrational energy, specifically of the reactants most effectively promotes a successful reaction. <ref>The Journal of Chemical Physics 138, 234104 (2013); https://doi.org/10.1063/1.4810007</ref>

====H-F + H====
rAB= 2 rBC = 0.74
{| class="wikitable" border=1
! pHH !! pHF!! Total Energy!! Reactive? !! Surface Plot !! Explanation
|-
| -5 || 0.2 || -103.364 || No || [[File:Mlw115_Surface_Plot_HFH_5_0.2.png|300px]] || Approaching H rebounds and reaction doesn't complete
|-
| 2.21||0.37||
|}

Reversal of conditions:
AB D 5.07 M -2.22
BC D 1.19 M 0.37

Random Guessing:
AB D 5.0 M -1.5
BC D 1.2 M 1.0

[[File: Mlw115_Surface_Plot_HFH_-1.5_1.png| 400px]]

=References=
<references/>