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MRD:celiab

2020-05-08T21:40:04Z

Cb2018: /* Activation Energy for F + H2 and H + HF reactions */

== EXERCISE 1: H + H2 ==

=== The Transition State ===
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?
* On PES diagrams, transition states are mathematically defined as points with 0 gradient with respect to position: stationary points.
* The transition state corresponds to the saddle point; the highest energy point on the reaction coordinate, which is the lowest energy pathway.
* We can distinguish it from a local minimum because on the reaction coordinate axis, the saddle point representing the transition state, will be a local maximum, but on the orthogonal axis it will be a local minimum. The minima are minimums on either direction because they represent stable reactants and products.

=== H + H2 reaction ===
[[File:Internuclear_distances_vs_time.png|250px|thumb|middle|Internuclear Distances vs Time plot ]]

===== Locating the Transition State =====
Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.
* The reaction is symmetric, hence, the transition state must have '''r1''' = '''r2''' by Hammond's Postulate
* By testing different initial conditions where '''rAB'''='''rBC '''and '''p1''' = '''p2''' = 0 g.mol-1.pm.fs-1, we can find the position of the Transition State
** The position of the transition state was found to be '''rts'''=90.775 pm
** This is where the system underwent small periodic symmetric vibrations keeping the distance between the atoms almost constant
* In the plot to the right, we see two straight lines, representing almost a stationary behaviour; thus we know we are at the symmetric transition state.

===== Calculating the Reaction Path =====
Comment on how the ''mep'' and the trajectory you just calculated differ.
* The minimum energy path is calculated from a system slightly displaced from the Transition state ('''r1''' = '''rts'''+1 pm, '''r2''' = '''rts''') and 0 initial momenta and follows the lowest energy path on a PES.
* As the reaction is symmetric it doesn't matter which atoms we label as A B C, so if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead, we get the same results but switched: AB will become BC and viceversa.
* The dynamic trajectory includes the momentum of the particles which causes oscillations about the trajectory path.
* The ''mep'' trajectory is roughly the same but it's smooth and doesn't oscillate, as the velocities are set to 0 in every step. This means we are artificially removing any gain in vibrational energy which results in oscillations. As the velocity is set 0 at every step, the kinetic energy is also 0.
[[File:picture 2 mep.png|350px|thumb|left|MEP contour plot ]]
[[File:picture 3 dynamic.png|350px|thumb|right|Dynamic contour plot ]]
[[File:Pic_4_dt.png|350px|thumb|left|MEP Internuclear Distances vs Time plot ]]
[[File:Pic_5_dt.png|350px|thumb|right| Dynamic Internuclear Distances vs Time plot ]]

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

For the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, the following trajectories with the following momenta combination were run:

(in the descriptions A B and C are used to describe the atoms colliding: AB is the initial H2 approaching H, represented by C)
{| class="wikitable"
!'''#'''
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|'''1'''
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.3</nowiki>
|yes
|The reaction dynamics proceed in a straight forward manner: AB and C approach each other, a collision occurs with the correct momentum for the reaction to proceed and a new bond BC is formed.
| [[File:Pic_a.png|260px]]
|-
|'''2'''
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420</nowiki>
|no
|AB approaches C and collides with it, but it doesn't have enough momentum to form a new bond so they are repelled. Momentum transfer does occur and C goes back with a greater momentum.
| [[File:Pic_b.png|260px]]
|-
|'''3'''
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414</nowiki>
|yes
|Oscillating AB collides with C and forms a new bond. The new H2 molecule, BC, has a higher vibrational frequency. A's direction is reversed.
| [[File:Pic_c.png|260px]]
|-
|'''4'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-375</nowiki>
|no
|AB approaches C and a BC bond forms, however, the system has too much momentum and C is repelled by the nucleus in B and bounces back. This leaves AB with greater vibrational energy and the distance BC increases as C is repelled.

| [[File:Pic_d.png|260px]]
|-
|'''5'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349</nowiki>
|yes
|AB collides with C and a new bond forms. However, unlike above, C still has enough energy to overcome the activation barrier. So, the BC bond forms, it breaks and reforms AB which then collides again with C to reform BC and leave as 2 entities: A + BC
| [[File:Pic_e.png|260px]]
|}

==== 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 allows us to investigate the progress of a reaction by splitting it into 2 parts: reactants and products.
* For a reaction to occur the reactants must cross into the product "space" by going over the saddle point – the transition state!
* With the Born-Oppenheimer approximation<ref>[https://www.iue.tuwien.ac.at/phd/schanovsky/thesisch2.html]</ref>, it is assumed there is a single saddle point (thus transition state) per reaction
* Transition State theory should only be used to estimate reaction rates at low temperature as it fails to correctly predict reaction rates for scenarios with high momentum (ie high temperature), where the transition state can be recrossed and thus predict reaction rates which are a poor match with experimental values
* We must also take into account that not every reaction with the correct energy will react, as for example, we must also take into account orientation, so, predicted rates with TS theory will be higher than experimental values.

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

=== PES inspection ===

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

This can be deduced by inspecting the PES surfaces bellow. (A=F, B=H, C=H)
# '''F + H2 → HF + H''' : Exothermic
#* Looking at the reactants: A + BC in the PES plot (small BC distance, large AB distance), we can see they have a higher potential energy than the products: AB + C. Which is confirmed by looking at bond strengths<ref>[http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html]</ref>: breaking the H–H bond (432 kJ/mol) vs making the H–F bond (565 kJ/mol), so energy is released during the reaction.
# '''HF + H → H2 + F''' : Endothermic
#* If we know look at: AB + C as reactants (small AB distance, large BC distance) we can see they have a lower potential energy than the products: BC + A. It is again rationalised by the fact that the H-F bond being broken in the reaction is stronger than the H-H bond being formed, so it requires an amount of energy to initiate the reaction.

[[File:H+hf_pes1L.png|275px|thumb|centre| F-H-H PES plot: AB is FH distance and BC is HH distance ]]

==== Transition state for F-H-H system ====
Locate the approximate position of the transition state.

From Hammond's postulate we know that in an endothermic reaction the transition state resembles the products and in an exothermic reaction it resembles the reactants. So, we can infer that the transition state will resemble '''F + H2'''.

The transition State for the F-H-H reaction was found at: 181.11 pm AB and 74.487 pm BC, with 0 initial momenta for both. This was done by varying the distances until they remained constant, confirmed by the forces along AB and BC being 0, marking a stationary point as the first derivative is 0.

[[File: HFTS_cp.png|275px|thumb|centre| F-H-H Contour plot when placed at Transition State ]]

==== Activation Energy for F + H2 and H + HF reactions ====
Report the activation energy for both reactions

The MEP was found for both reactions for a structure neighbouring the transition state: '''rts'''+~1 pm.

The activation energy is the difference between the saddle point's maximum energy and the minimum energy of the products:

* '''Reaction''' '''1''': F + H2 → HF + H
** ''TS Energy: '' - 433.8 kJ / mol
** ''Final Energy: '' - 434.5 kJ / mol
** Activation Energy: + 0.7 kJ / mol
[[File:Eactbiggraph.png|350px|thumb|centre|MEP Energy vs Time diagram of transition state for Reaction 1 (FH (AB) 180.11 pm and HH (BC) 75.487 pm) ]]
* '''Reaction''' '''2''': HF + H → H2 + F 
** ''TS Energy: - 434 kJ / mol''
** ''Final Energy: - 560 kJ / mol''
** Activation Energy: + 126 kJ / mol
[[File:H2+f_emep.png|350px|thumb|centre|MEP Energy vs Time diagram of transition state for Reaction 2 (FH (AB) 74.487 pm and HH (BC) 181.11 pm) ]]

=== Reaction Dynamics ===

==== Reactive trajectory for F + H2 → HF + H ====
In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.

Using initial conditions; rAB = 225 pm, rBC = 75 pm, pAB = -5 g.mol-1.pm.fs-1, pBC = -1 g.mol-1.pm.fs-1, F + H2 → HF + H is a successful reaction, and, as seen above, this is an exothermic reaction and thus energy is released. From the animation we can see that a vibrating H2 molecule (BC) approaches F (A) and an HF bond forms, however it breaks and reforms BC, which then collides again with the Fluorine atom to reform a highly energetic HF and leave as the final products: HF + H.

[[File:Goodreachf.png|350px|thumb|centre|Contour plot of successful reaction of formation of HF]]

The energy in the reaction is converted from potential energy to kinetic energy, which is shown by the highly vibrating HF molecule, compared to the initial H2. This is confirmed with a Momentum vs Time plot (shown bellow): the AB distance, representing the HF bond, oscillates at a much larger amplitude than the initial BC oscillations which represent the H2 molecule.
[[File:Memtumsuccreac.png|350px|thumb|centre|Momentum vs Time plot of formation of HF]]

Experimentally, this would result in an increase in temperature of our reaction mixture, which could be measured using calorimetry. Additionally, the HF vibrational modes would be highly excited and overtones would show in the IR spectra as multiples of the fundamental absorption frequency<ref>[https://chem.libretexts.org/Courses/Pacific_Union_College/Quantum_Chemistry/13%3A_Molecular_Spectroscopy/13.05%3A_Vibrational_Overtones ]</ref> (3962 cm-1)<ref>[http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html]</ref>. Because overtone bands are difficult to see, this could be done using use Low Temperature FTIR<ref>[https://pubs.acs.org/doi/abs/10.1021/j150647a021]</ref>.

==== Illustrating Polanyi's empirical rules [ Reaction 2: HF + H → H2 + F ] ====
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

Polyani's Rules<ref>[https://pubs.acs.org/doi/10.1021/jz301649w]</ref>state that:
# For an early transition state, translational energy is more efficient in promoting a reaction
# For a late transition state, vibrational energy is more important in promoting a reaction
So, for the endothermic reaction HF + H → H2 + F, compared to the reverse exothermic reaction, we need a higher amount of vibrational energy, as well as translational to overcome the transition state and for a reaction to occur.

This can be illustrated by comparing the following initial reaction conditions (''A=F, B=H, C=H, rAB = 75 pm, rBC = 225 pm''):
# '''pAB = -1, pBC = -5''' – Unsuccessful Reaction 2 (NB. under these conditions Reaction 1 is successful, as shown above)
# '''pAB = -2, pBC = -4''' – Successful Reaction 2 when we redistribute momentum to increase vibrational energy
(units of p: g.mol-1.pm.fs-1)

[[File:Unsuccessreac2pol.png|350px|thumb|left|Contour plot of unsuccessful formation of H2 + F with conditions parallel to a succesfull reverse reaction]][[File:Successreac2pol.png|350px|thumb|right|Contour plot of successful formation of H2 + F ]]

== References ==

MRD:celiab

2020-05-08T18:45:06Z

Cb2018:

== EXERCISE 1: H + H2 ==

=== The Transition State ===
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?
* On PES diagrams, transition states are mathematically defined as points with 0 gradient with respect to position: stationary points.
* The transition state corresponds to the saddle point; the highest energy point on the reaction coordinate, which is the lowest energy pathway.
* We can distinguish it from a local minimum because on the reaction coordinate axis, the saddle point representing the transition state, will be a local maximum, but on the orthogonal axis it will be a local minimum. The minima are minimums on either direction because they represent stable reactants and products.

=== H + H2 reaction ===
[[File:Internuclear_distances_vs_time.png|250px|thumb|middle|Internuclear Distances vs Time plot ]]

===== Locating the Transition State =====
Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.
* The reaction is symmetric, hence, the transition state must have '''r1''' = '''r2''' by Hammond's Postulate
* By testing different initial conditions where '''rAB'''='''rBC '''and '''p1''' = '''p2''' = 0 g.mol-1.pm.fs-1, we can find the position of the Transition State
** The position of the transition state was found to be '''rts'''=90.775 pm
** This is where the system underwent small periodic symmetric vibrations keeping the distance between the atoms almost constant
* In the plot to the right, we see two straight lines, representing almost a stationary behaviour; thus we know we are at the symmetric transition state.

===== Calculating the Reaction Path =====
Comment on how the ''mep'' and the trajectory you just calculated differ.
* The minimum energy path is calculated from a system slightly displaced from the Transition state ('''r1''' = '''rts'''+1 pm, '''r2''' = '''rts''') and 0 initial momenta and follows the lowest energy path on a PES.
* As the reaction is symmetric it doesn't matter which atoms we label as A B C, so if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead, we get the same results but switched: AB will become BC and viceversa.
* The dynamic trajectory includes the momentum of the particles which causes oscillations about the trajectory path.
* The ''mep'' trajectory is roughly the same but it's smooth and doesn't oscillate, as the velocities are set to 0 in every step. This means we are artificially removing any gain in vibrational energy which results in oscillations. As the velocity is set 0 at every step, the kinetic energy is also 0.
[[File:picture 2 mep.png|350px|thumb|left|MEP contour plot ]]
[[File:picture 3 dynamic.png|350px|thumb|right|Dynamic contour plot ]]
[[File:Pic_4_dt.png|350px|thumb|left|MEP Internuclear Distances vs Time plot ]]
[[File:Pic_5_dt.png|350px|thumb|right| Dynamic Internuclear Distances vs Time plot ]]

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

For the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, the following trajectories with the following momenta combination were run:

(in the descriptions A B and C are used to describe the atoms colliding: AB is the initial H2 approaching H, represented by C)
{| class="wikitable"
!'''#'''
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|'''1'''
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.3</nowiki>
|yes
|The reaction dynamics proceed in a straight forward manner: AB and C approach each other, a collision occurs with the correct momentum for the reaction to proceed and a new bond BC is formed.
| [[File:Pic_a.png|260px]]
|-
|'''2'''
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420</nowiki>
|no
|AB approaches C and collides with it, but it doesn't have enough momentum to form a new bond so they are repelled. Momentum transfer does occur and C goes back with a greater momentum.
| [[File:Pic_b.png|260px]]
|-
|'''3'''
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414</nowiki>
|yes
|Oscillating AB collides with C and forms a new bond. The new H2 molecule, BC, has a higher vibrational frequency. A's direction is reversed.
| [[File:Pic_c.png|260px]]
|-
|'''4'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-375</nowiki>
|no
|AB approaches C and a BC bond forms, however, the system has too much momentum and C is repelled by the nucleus in B and bounces back. This leaves AB with greater vibrational energy and the distance BC increases as C is repelled.

| [[File:Pic_d.png|260px]]
|-
|'''5'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349</nowiki>
|yes
|AB collides with C and a new bond forms. However, unlike above, C still has enough energy to overcome the activation barrier. So, the BC bond forms, it breaks and reforms AB which then collides again with C to reform BC and leave as 2 entities: A + BC
| [[File:Pic_e.png|260px]]
|}

==== 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 allows us to investigate the progress of a reaction by splitting it into 2 parts: reactants and products.
* For a reaction to occur the reactants must cross into the product "space" by going over the saddle point – the transition state!
* With the Born-Oppenheimer approximation<ref>[https://www.iue.tuwien.ac.at/phd/schanovsky/thesisch2.html]</ref>, it is assumed there is a single saddle point (thus transition state) per reaction
* Transition State theory should only be used to estimate reaction rates at low temperature as it fails to correctly predict reaction rates for scenarios with high momentum (ie high temperature), where the transition state can be recrossed and thus predict reaction rates which are a poor match with experimental values
* We must also take into account that not every reaction with the correct energy will react, as for example, we must also take into account orientation, so, predicted rates with TS theory will be higher than experimental values.

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

=== PES inspection ===

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

This can be deduced by inspecting the PES surfaces bellow. (A=F, B=H, C=H)
# '''F + H2 → HF + H''' : Exothermic
#* Looking at the reactants: A + BC in the PES plot (small BC distance, large AB distance), we can see they have a higher potential energy than the products: AB + C. Which is confirmed by looking at bond strengths<ref>[http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html]</ref>: breaking the H–H bond (432 kJ/mol) vs making the H–F bond (565 kJ/mol), so energy is released during the reaction.
# '''HF + H → H2 + F''' : Endothermic
#* If we know look at: AB + C as reactants (small AB distance, large BC distance) we can see they have a lower potential energy than the products: BC + A. It is again rationalised by the fact that the H-F bond being broken in the reaction is stronger than the H-H bond being formed, so it requires an amount of energy to initiate the reaction.

[[File:H+hf_pes1L.png|275px|thumb|centre| F-H-H PES plot: AB is FH distance and BC is HH distance ]]

==== Transition state for F-H-H system ====
Locate the approximate position of the transition state.

From Hammond's postulate we know that in an endothermic reaction the transition state resembles the products and in an exothermic reaction it resembles the reactants. So, we can infer that the transition state will resemble '''F + H2'''.

The transition State for the F-H-H reaction was found at: 181.11 pm AB and 74.487 pm BC, with 0 initial momenta for both. This was done by varying the distances until they remained constant, confirmed by the forces along AB and BC being 0, marking a stationary point as the first derivative is 0.

[[File: HFTS_cp.png|275px|thumb|centre| F-H-H Contour plot when placed at Transition State ]]

==== Activation Energy for F + H2 and H + HF reactions ====
Report the activation energy for both reactions

The MEP was found for both reactions for a structure neighbouring the transition state: '''rts'''+~1 pm.

The activation energy is the difference between the saddle point's maximum energy and the minimum energy of the products:

* '''Reaction''' '''1''': F + H2 → HF + H
** ''TS Energy: ''- 434 kJ / mol
** ''Final Energy: ''- 560 kJ / mol
** Activation Energy: + 126 kJ / mol
[[File:Eactbiggraph.png|350px|thumb|centre|MEP Energy vs Time diagram of transition state for Reaction 1 (FH (AB) 180.11 pm and HH (BC) 75.487 pm) ]]
* '''Reaction''' '''2''': HF + H → H2 + F 
** ''TS Energy: ''-433.8 kJ / mol
** ''Final Energy: ''-434.5 kJ / mol
** Activation Energy: 0.7 kJ / mol
[[File:H2+f_emep.png|350px|thumb|centre|MEP Energy vs Time diagram of transition state for Reaction 2 (FH (AB) 74.487 pm and HH (BC) 181.11 pm) ]]

=== Reaction Dynamics ===

==== Reactive trajectory for F + H2 → HF + H ====
In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.

Using initial conditions; rAB = 225 pm, rBC = 75 pm, pAB = -5 g.mol-1.pm.fs-1, pBC = -1 g.mol-1.pm.fs-1, F + H2 → HF + H is a successful reaction, and, as seen above, this is an exothermic reaction and thus energy is released. From the animation we can see that a vibrating H2 molecule (BC) approaches F (A) and an HF bond forms, however it breaks and reforms BC, which then collides again with the Fluorine atom to reform a highly energetic HF and leave as the final products: HF + H.

[[File:Goodreachf.png|350px|thumb|centre|Contour plot of successful reaction of formation of HF]]

The energy in the reaction is converted from potential energy to kinetic energy, which is shown by the highly vibrating HF molecule, compared to the initial H2. This is confirmed with a Momentum vs Time plot (shown bellow): the AB distance, representing the HF bond, oscillates at a much larger amplitude than the initial BC oscillations which represent the H2 molecule.
[[File:Memtumsuccreac.png|350px|thumb|centre|Momentum vs Time plot of formation of HF]]

Experimentally, this would result in an increase in temperature of our reaction mixture, which could be measured using calorimetry. Additionally, the HF vibrational modes would be highly excited and overtones would show in the IR spectra as multiples of the fundamental absorption frequency<ref>[https://chem.libretexts.org/Courses/Pacific_Union_College/Quantum_Chemistry/13%3A_Molecular_Spectroscopy/13.05%3A_Vibrational_Overtones ]</ref> (3962 cm-1)<ref>[http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html]</ref>. Because overtone bands are difficult to see, this could be done using use Low Temperature FTIR<ref>[https://pubs.acs.org/doi/abs/10.1021/j150647a021]</ref>.

==== Illustrating Polanyi's empirical rules [ Reaction 2: HF + H → H2 + F ] ====
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

Polyani's Rules<ref>[https://pubs.acs.org/doi/10.1021/jz301649w]</ref>state that:
# For an early transition state, translational energy is more efficient in promoting a reaction
# For a late transition state, vibrational energy is more important in promoting a reaction
So, for the endothermic reaction HF + H → H2 + F, compared to the reverse exothermic reaction, we need a higher amount of vibrational energy, as well as translational to overcome the transition state and for a reaction to occur.

This can be illustrated by comparing the following initial reaction conditions (''A=F, B=H, C=H, rAB = 75 pm, rBC = 225 pm''):
# '''pAB = -1, pBC = -5''' – Unsuccessful Reaction 2 (NB. under these conditions Reaction 1 is successful, as shown above)
# '''pAB = -2, pBC = -4''' – Successful Reaction 2 when we redistribute momentum to increase vibrational energy
(units of p: g.mol-1.pm.fs-1)

[[File:Unsuccessreac2pol.png|350px|thumb|left|Contour plot of unsuccessful formation of H2 + F with conditions parallel to a succesfull reverse reaction]][[File:Successreac2pol.png|350px|thumb|right|Contour plot of successful formation of H2 + F ]]

== References ==

MRD:celiab

2020-05-08T13:44:49Z

Cb2018:

== EXERCISE 1: H + H2 ==

=== The Transition State ===
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?
* On PES diagrams, transition states are mathematically defined as points with 0 gradient with respect to position: stationary points.
* The transition state corresponds to the saddle point; the highest energy point on the reaction coordinate, which is the lowest energy pathway.
* We can distinguish it from a local minimum because on the reaction coordinate axis, the saddle point representing the transition state, will be a local maximum, but on the orthogonal axis it will be a local minimum. The minima are minimums on either direction because they represent stable reactants and products.

=== H + H2 reaction ===
[[File:Internuclear_distances_vs_time.png|250px|thumb|middle|Internuclear Distances vs Time plot ]]

===== Locating the Transition State =====
Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.
* The reaction is symmetric, hence, the transition state must have '''r1''' = '''r2''' by Hammond's Postulate
* By testing different initial conditions where '''rAB'''='''rBC '''and '''p1''' = '''p2''' = 0 g.mol-1.pm.fs-1, we can find the position of the Transition State
** The position of the transition state was found to be '''rts'''=90.775 pm
** This is where the system underwent small periodic symmetric vibrations keeping the distance between the atoms almost constant
* In the plot to the right, we see two straight lines, representing almost a stationary behaviour; thus we know we are at the symmetric transition state.

===== Calculating the Reaction Path =====
Comment on how the ''mep'' and the trajectory you just calculated differ.
* The minimum energy path is calculated from a system slightly displaced from the Transition state ('''r1''' = '''rts'''+1 pm, '''r2''' = '''rts''') and 0 initial momenta and follows the lowest energy path on a PES.
* As the reaction is symmetric it doesn't matter which atoms we label as A B C, so if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead, we get the same results but switched: AB will become BC and viceversa.
* The dynamic trajectory includes the momentum of the particles which causes oscillations about the trajectory path.
* The ''mep'' trajectory is roughly the same but it's smooth and doesn't oscillate, as the velocities are set to 0 in every step. This means we are artificially removing any gain in vibrational energy which results in oscillations. As the velocity is set 0 at every step, the kinetic energy is also 0.
[[File:picture 2 mep.png|350px|thumb|left|MEP contour plot ]]
[[File:picture 3 dynamic.png|350px|thumb|right|Dynamic contour plot ]]
[[File:Pic_4_dt.png|350px|thumb|left|MEP Internuclear Distances vs Time plot ]]
[[File:Pic_5_dt.png|350px|thumb|right| Dynamic Internuclear Distances vs Time plot ]]

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

For the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, the following trajectories with the following momenta combination were run:

(in the descriptions A B and C are used to describe the atoms colliding: AB is the initial H2 approaching H, represented by C)
{| class="wikitable"
!'''#'''
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|'''1'''
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.3</nowiki>
|yes
|The reaction dynamics proceed in a straight forward manner: AB and C approach each other, a collision occurs with the correct momentum for the reaction to proceed and a new bond BC is formed.
| [[File:Pic_a.png|260px]]
|-
|'''2'''
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420</nowiki>
|no
|AB approaches C and collides with it, but it doesn't have enough momentum to form a new bond so they are repelled. Momentum transfer does occur and C goes back with a greater momentum.
| [[File:Pic_b.png|260px]]
|-
|'''3'''
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414</nowiki>
|yes
|Oscillating AB collides with C and forms a new bond. The new H2 molecule, BC, has a higher vibrational frequency. A's direction is reversed.
| [[File:Pic_c.png|260px]]
|-
|'''4'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-375</nowiki>
|no
|AB approaches C and a BC bond forms, however, the system has too much momentum and C is repelled by the nucleus in B and bounces back. This leaves AB with greater vibrational energy and the distance BC increases as C is repelled.

| [[File:Pic_d.png|260px]]
|-
|'''5'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349</nowiki>
|yes
|AB collides with C and a new bond forms. However, unlike above, C still has enough energy to overcome the activation barrier. So, the BC bond forms, it breaks and reforms AB which then collides again with C to reform BC and leave as 2 entities: A + BC
| [[File:Pic_e.png|260px]]
|}

==== 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 allows us to investigate the progress of a reaction by splitting it into 2 parts: reactants and products.
* For a reaction to occur the reactants must cross into the product "space" by going over the saddle point – the transition state!
* With the Born-Oppenheimer approximation<ref>[https://www.iue.tuwien.ac.at/phd/schanovsky/thesisch2.html]</ref>, it is assumed there is a single saddle point (thus transition state) per reaction
* Transition State theory should only be used to estimate reaction rates at low temperature as it fails to correctly predict reaction rates for scenarios with high momentum (ie high temperature), where the transition state can be recrossed and thus predict reaction rates which are a poor match with experimental values
* We must also take into account that not every reaction with the correct energy will react, as for example, we must also take into account orientation, so, predicted rates with TS theory will be higher than experimental values.

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

=== PES inspection ===

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

This can be deduced by inspecting the PES surfaces bellow. (A=F, B=H, C=H)
# '''F + H2 → HF + H''' : Exothermic
#* Looking at the reactants: A + BC in the PES plot (small BC distance, large AB distance), we can see they have a higher potential energy than the products: AB + C. Which is confirmed by looking at bond strengths<ref>[http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html]</ref>: breaking the H–H bond (432 kJ/mol) vs making the H–F bond (565 kJ/mol), so energy is released during the reaction.
# '''HF + H → H2 + F''' : Endothermic
#* If we know look at: AB + C as reactants (small AB distance, large BC distance) we can see they have a lower potential energy than the products: BC + A. It is again rationalised by the fact that the H-F bond being broken in the reaction is stronger than the H-H bond being formed, so it requires an amount of energy to initiate the reaction.

[[File:H+hf_pes1L.png|275px|thumb|centre| F-H-H PES plot: AB is FH distance and BC is HH distance ]]

==== Transition state for F-H-H system ====
Locate the approximate position of the transition state.

From Hammond's postulate we know that in an endothermic reaction the transition state resembles the products and in an exothermic reaction it resembles the reactants. So, we can infer that the transition state will resemble H + HF.

The transition State for the F-H-H reaction was found at: 181.11 pm AB and 74.487 pm BC, with 0 initial momenta for both. This was done by varying the distances until they remained constant, confirmed by the forces along AB and BC being 0, marking a stationary point as the first derivative is 0.

[[File: HFTS_cp.png|275px|thumb|centre| F-H-H Contour plot when placed at Transition State ]]

==== Activation Energy for F + H2 and H + HF reactions ====
Report the activation energy for both reactions

The MEP was found for both reactions for a structure neighbouring the transition state: '''rts'''+~1 pm.

The activation energy is the difference between the saddle point's maximum energy and the minimum energy of the products:

* '''Reaction''' '''1''': F + H2 → HF + H
** ''TS Energy: ''- 434 kJ / mol
** ''Final Energy: ''- 560 kJ / mol
** Activation Energy: + 126 kJ / mol
[[File:Eactbiggraph.png|350px|thumb|centre|MEP Energy vs Time diagram of transition state for Reaction 1 (FH (AB) 180.11 pm and HH (BC) 75.487 pm) ]]
* '''Reaction''' '''2''': HF + H → H2 + F 
** ''TS Energy: ''-433.8 kJ / mol
** ''Final Energy: ''-434.5 kJ / mol
** Activation Energy: 0.7 kJ / mol
[[File:H2+f_emep.png|350px|thumb|centre|MEP Energy vs Time diagram of transition state for Reaction 2 (FH (AB) 74.487 pm and HH (BC) 181.11 pm) ]]

=== Reaction Dynamics ===

==== Reactive trajectory for F + H2 → HF + H ====
In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.

Using initial conditions; rAB = 225 pm, rBC = 75 pm, pAB = -5 g.mol-1.pm.fs-1, pBC = -1 g.mol-1.pm.fs-1, F + H2 → HF + H is a successful reaction, and, as seen above, this is an exothermic reaction and thus energy is released. From the animation we can see that a vibrating H2 molecule (BC) approaches F (A) and an HF bond forms, however it breaks and reforms BC, which then collides again with the Fluorine atom to reform a highly energetic HF and leave as the final products: HF + H.

[[File:Goodreachf.png|350px|thumb|centre|Contour plot of successful reaction of formation of HF]]

The energy in the reaction is converted from potential energy to kinetic energy, which is shown by the highly vibrating HF molecule, compared to the initial H2. This is confirmed with a Momentum vs Time plot (shown bellow): the AB distance, representing the HF bond, oscillates at a much larger amplitude than the initial BC oscillations which represent the H2 molecule.
[[File:Memtumsuccreac.png|350px|thumb|centre|Momentum vs Time plot of formation of HF]]

Experimentally, this would result in an increase in temperature of our reaction mixture, which could be measured using calorimetry. Additionally, the HF vibrational modes would be highly excited and overtones would show in the IR spectra as multiples of the fundamental absorption frequency<ref>[https://chem.libretexts.org/Courses/Pacific_Union_College/Quantum_Chemistry/13%3A_Molecular_Spectroscopy/13.05%3A_Vibrational_Overtones ]</ref> (3962 cm-1)<ref>[http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html]</ref>. Because overtone bands are difficult to see, this could be done using use Low Temperature FTIR<ref>[https://pubs.acs.org/doi/abs/10.1021/j150647a021]</ref>.

==== Illustrating Polanyi's empirical rules [ Reaction 2: HF + H → H2 + F ] ====
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

Polyani's Rules<ref>[https://pubs.acs.org/doi/10.1021/jz301649w]</ref>state that:
# For an early transition state, translational energy is more efficient in promoting a reaction
# For a late transition state, vibrational energy is more important in promoting a reaction
So, for the endothermic reaction HF + H → H2 + F, compared to the reverse exothermic reaction, we need a higher amount of vibrational energy, as well as translational to overcome the transition state and for a reaction to occur.

This can be illustrated by comparing the following initial reaction conditions (''A=F, B=H, C=H, rAB = 75 pm, rBC = 225 pm''):
# '''pAB = -1, pBC = -5''' – Unsuccessful Reaction 2 (NB. under these conditions Reaction 1 is successful, as shown above)
# '''pAB = -2, pBC = -4''' – Successful Reaction 2 when we redistribute momentum to increase vibrational energy
(units of p: g.mol-1.pm.fs-1)

[[File:Unsuccessreac2pol.png|350px|thumb|left|Contour plot of unsuccessful formation of H2 + F with conditions parallel to a succesfull reverse reaction]][[File:Successreac2pol.png|350px|thumb|right|Contour plot of successful formation of H2 + F ]]

== References ==

MRD:celiab

2020-05-08T13:44:23Z

Cb2018:

== EXERCISE 1: H + H2 ==

=== The Transition State ===
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?
* On PES diagrams, transition states are mathematically defined as points with 0 gradient with respect to position: stationary points.
* The transition state corresponds to the saddle point; the highest energy point on the reaction coordinate, which is the lowest energy pathway.
* We can distinguish it from a local minimum because on the reaction coordinate axis, the saddle point representing the transition state, will be a local maximum, but on the orthogonal axis it will be a local minimum. The minima are minimums on either direction because they represent stable reactants and products.

=== H + H2 reaction ===
[[File:Internuclear_distances_vs_time.png|250px|thumb|middle|Internuclear Distances vs Time plot ]]

===== Locating the Transition State =====
Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.
* The reaction is symmetric, hence, the transition state must have '''r1''' = '''r2''' by Hammond's Postulate
* By testing different initial conditions where '''rAB'''='''rBC '''and '''p1''' = '''p2''' = 0 g.mol-1.pm.fs-1, we can find the position of the Transition State
** The position of the transition state was found to be '''rts'''=90.775 pm
** This is where the system underwent small periodic symmetric vibrations keeping the distance between the atoms almost constant
* In the plot to the right, we see two straight lines, representing almost a stationary behaviour; thus we know we are at the symmetric transition state.

===== Calculating the Reaction Path =====
Comment on how the ''mep'' and the trajectory you just calculated differ.
* The minimum energy path is calculated from a system slightly displaced from the Transition state ('''r1''' = '''rts'''+1 pm, '''r2''' = '''rts''') and 0 initial momenta and follows the lowest energy path on a PES.
* As the reaction is symmetric it doesn't matter which atoms we label as A B C, so if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead, we get the same results but switched: AB will become BC and viceversa.
* The dynamic trajectory includes the momentum of the particles which causes oscillations about the trajectory path.
* The ''mep'' trajectory is roughly the same but it's smooth and doesn't oscillate, as the velocities are set to 0 in every step. This means we are artificially removing any gain in vibrational energy which results in oscillations. As the velocity is set 0 at every step, the kinetic energy is also 0.
[[File:picture 2 mep.png|350px|thumb|left|MEP contour plot ]]
[[File:picture 3 dynamic.png|350px|thumb|right|Dynamic contour plot ]]
[[File:Pic_4_dt.png|350px|thumb|left|MEP Internuclear Distances vs Time plot ]]
[[File:Pic_5_dt.png|350px|thumb|right| Dynamic Internuclear Distances vs Time plot ]]

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

For the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, the following trajectories with the following momenta combination were run:

(in the descriptions A B and C are used to describe the atoms colliding: AB is the initial H2 approaching H, represented by C)
{| class="wikitable"
!'''#'''
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|'''1'''
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.3</nowiki>
|yes
|The reaction dynamics proceed in a straight forward manner: AB and C approach each other, a collision occurs with the correct momentum for the reaction to proceed and a new bond BC is formed.
| [[File:Pic_a.png|260px]]
|-
|'''2'''
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420</nowiki>
|no
|AB approaches C and collides with it, but it doesn't have enough momentum to form a new bond so they are repelled. Momentum transfer does occur and C goes back with a greater momentum.
| [[File:Pic_b.png|260px]]
|-
|'''3'''
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414</nowiki>
|yes
|Oscillating AB collides with C and forms a new bond. The new H2 molecule, BC, has a higher vibrational frequency. A's direction is reversed.
| [[File:Pic_c.png|260px]]
|-
|'''4'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-375</nowiki>
|no
|AB approaches C and a BC bond forms, however, the system has too much momentum and C is repelled by the nucleus in B and bounces back. This leaves AB with greater vibrational energy and the distance BC increases as C is repelled.

| [[File:Pic_d.png|260px]]
|-
|'''5'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349</nowiki>
|yes
|AB collides with C and a new bond forms. However, unlike above, C still has enough energy to overcome the activation barrier. So, the BC bond forms, it breaks and reforms AB which then collides again with C to reform BC and leave as 2 entities: A + BC
| [[File:Pic_e.png|260px]]
|}

==== 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 allows us to investigate the progress of a reaction by splitting it into 2 parts: reactants and products.
* For a reaction to occur the reactants must cross into the product "space" by going over the saddle point – the transition state!
* With the Born-Oppenheimer approximation<ref>[https://www.iue.tuwien.ac.at/phd/schanovsky/thesisch2.html]</ref>, it is assumed there is a single saddle point (thus transition state) per reaction
* Transition State theory should only be used to estimate reaction rates at low temperature as it fails to correctly predict reaction rates for scenarios with high momentum (ie high temperature), where the transition state can be recrossed and thus predict reaction rates which are a poor match with experimental values
* We must also take into account that not every reaction with the correct energy will react, as for example, we must also take into account orientation, so, predicted rates with TS theory will be higher than experimental values.

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

=== PES inspection ===

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

This can be deduced by inspecting the PES surfaces bellow. (A=F, B=H, C=H)
# '''F + H2 → HF + H''' : Exothermic
#* Looking at the reactants: A + BC in the PES plot (small BC distance, large AB distance), we can see they have a higher potential energy than the products: AB + C. Which is confirmed by looking at bond strengths<ref>[http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html]</ref>: breaking the H–H bond (432 kJ/mol) vs making the H–F bond (565 kJ/mol), so energy is released during the reaction.
# '''HF + H → H2 + F''' : Endothermic
#* If we know look at: AB + C as reactants (small AB distance, large BC distance) we can see they have a lower potential energy than the products: BC + A. It is again rationalised by the fact that the H-F bond being broken in the reaction is stronger than the H-H bond being formed, so it requires an amount of energy to initiate the reaction.

[[File:H+hf_pes1L.png|275px|thumb|centre| F-H-H PES plot: AB is FH distance and BC is HH distance ]]

==== Transition state for F-H-H system ====
Locate the approximate position of the transition state.

From Hammond's postulate we know that in an endothermic reaction the transition state resembles the products and in an exothermic reaction it resembles the reactants. So, we can infer that the transition state will resemble H + HF.

The transition State for the F-H-H reaction was found at: 181.11 pm AB and 74.487 pm BC, with 0 initial momenta for both. This was done by varying the distances until they remained constant, confirmed by the forces along AB and BC being 0, marking a stationary point as the first derivative is 0.

[[File: HFTS_cp.png|275px|thumb|centre| F-H-H Contour plot when placed at Transition State ]]

==== Activation Energy for F + H2 and H + HF reactions ====
Report the activation energy for both reactions

The MEP was found for both reactions for a structure neighbouring the transition state: '''rts'''+~1 pm.

The activation energy is the difference between the saddle point's maximum energy and the minimum energy of the products:

* '''Reaction''' '''1''': F + H2 → HF + H
** ''TS Energy: ''- 434 kJ / mol
** ''Final Energy: ''- 560 kJ / mol
** Activation Energy: + 126 kJ / mol
[[File:Eactbiggraph.png|350px|thumb|centre|MEP Energy vs Time diagram of transition state for Reaction 1 (FH (AB) 180.11 pm and HH (BC) 75.487 pm) ]]
* '''Reaction''' '''2''': HF + H → H2 + F 
** ''TS Energy: ''-433.8 kJ / mol
** ''Final Energy: ''-434.5 kJ / mol
** Activation Energy: 0.7 kJ / mol
[[File:H2+f_emep.png|350px|thumb|centre|MEP Energy vs Time diagram of transition state for Reaction 2 (FH (AB) 74.487 pm and HH (BC) 181.11 pm) ]]

=== Reaction Dynamics ===

==== Reactive trajectory for F + H2 → HF + H ====
In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.

Using initial conditions; rAB = 225 pm, rBC = 75 pm, pAB = -5 g.mol-1.pm.fs-1, pBC = -1 g.mol-1.pm.fs-1, F + H2 → HF + H is a successful reaction, and, as seen above, this is an exothermic reaction and thus energy is released. From the animation we can see that a vibrating H2 molecule (BC) approaches F (A) and an HF bond forms, however it breaks and reforms BC, which then collides again with the Fluorine atom to reform a highly energetic HF and leave as the final products: HF + H.

[[File:Goodreachf.png|350px|thumb|centre|Contour plot of successful reaction of formation of HF]]

The energy in the reaction is converted from potential energy to kinetic energy, which is shown by the highly vibrating HF molecule, compared to the initial H2. This is confirmed with a Momentum vs Time plot (shown bellow): the AB distance, representing the HF bond, oscillates at a much larger amplitude than the initial BC oscillations which represent the H2 molecule.
[[File:Memtumsuccreac.png|350px|thumb|centre|Momentum vs Time plot of formation of HF]]

Experimentally, this would result in an increase in temperature of our reaction mixture, which could be measured using calorimetry. Additionally, the HF vibrational modes would be highly excited and overtones would show in the IR spectra as multiples of the fundamental absorption frequency<ref>[https://chem.libretexts.org/Courses/Pacific_Union_College/Quantum_Chemistry/13%3A_Molecular_Spectroscopy/13.05%3A_Vibrational_Overtones ]</ref> (3962 cm-1)<ref>[http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html]</ref>. Because overtone bands are difficult to see, this could be done using use Low Temperature FTIR<ref>[https://pubs.acs.org/doi/abs/10.1021/j150647a021]</ref>.

==== Illustrating Polanyi's empirical rules [ Reaction 2: HF + H → H2 + F ] ====
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

Polyani's Rules<ref>[https://pubs.acs.org/doi/10.1021/jz301649w]</ref>state that:
# For an early transition state, translational energy is more efficient in promoting a reaction
# For a late transition state, vibrational energy is more important in promoting a reaction
So, for the endothermic reaction HF + H → H2 + F, compared to the reverse exothermic reaction, we need a higher amount of vibrational energy, as well as translational to overcome the transition state and for a reaction to occur.

This can be illustrated by comparing the following initial reaction conditions (''A=F, B=H, C=H, rAB = 75 pm, rBC = 225 pm''):
# '''pAB = -1, pBC = -5''' – Unsuccessful Reaction 2 (NB. under these conditions Reaction 1 is successful, as shown above)
# '''pAB = -2, pBC = -4''' – Successful Reaction 2 when we redistribute momentum to increase vibrational energy
(units of p: g.mol-1.pm.fs-1)

[[File:Unsuccessreac2pol.png|350px|thumb|left|Contour plot of unsuccessful formation of H2 + F with conditions parallel to a succesfull reverse reaction]][[File:Successreac2pol.png|350px|thumb|right|Contour plot of successful formation of H2 + F ]]

== References ==

MRD:celiab

2020-05-08T13:38:06Z

Cb2018: /* Reaction Dynamics */

== EXERCISE 1: H + H2 ==

=== The Transition State ===
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?
* On PES diagrams, transition states are mathematically defined as points with 0 gradient with respect to position: stationary points.
* The transition state corresponds to the saddle point; the highest energy point on the reaction coordinate, which is the lowest energy pathway.
* We can distinguish it from a local minimum because on the reaction coordinate axis, the saddle point representing the transition state, will be a local maximum, but on the orthogonal axis it will be a local minimum. The minima are minimums on either direction because they represent stable reactants and products.

=== H + H2 reaction ===
[[File:Internuclear_distances_vs_time.png|250px|thumb|middle|Internuclear Distances vs Time plot ]]

===== Locating the Transition State =====
Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.
* The reaction is symmetric, hence, the transition state must have '''r1''' = '''r2''' by Hammond's Postulate
* By testing different initial conditions where '''rAB'''='''rBC '''and '''p1''' = '''p2''' = 0 g.mol-1.pm.fs-1, we can find the position of the Transition State
** The position of the transition state was found to be '''rts'''=90.775 pm
** This is where the system underwent small periodic symmetric vibrations keeping the distance between the atoms almost constant
* In the plot to the right, we see two straight lines, representing almost a stationary behaviour; thus we know we are at the symmetric transition state.

===== Calculating the Reaction Path =====
Comment on how the ''mep'' and the trajectory you just calculated differ.
* The minimum energy path is calculated from a system slightly displaced from the Transition state ('''r1''' = '''rts'''+1 pm, '''r2''' = '''rts''') and 0 initial momenta and follows the lowest energy path on a PES.
* As the reaction is symmetric it doesn't matter which atoms we label as A B C, so if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead, we get the same results but switched: AB will become BC and viceversa.
* The dynamic trajectory includes the momentum of the particles which causes oscillations about the trajectory path.
* The ''mep'' trajectory is roughly the same but it's smooth and doesn't oscillate, as the velocities are set to 0 in every step. This means we are artificially removing any gain in vibrational energy which results in oscillations. As the velocity is set 0 at every step, the kinetic energy is also 0.
[[File:picture 2 mep.png|350px|thumb|left|MEP contour plot ]]
[[File:picture 3 dynamic.png|350px|thumb|right|Dynamic contour plot ]]
[[File:Pic_4_dt.png|350px|thumb|left|MEP Internuclear Distances vs Time plot ]]
[[File:Pic_5_dt.png|350px|thumb|right| Dynamic Internuclear Distances vs Time plot ]]

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

For the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, the following trajectories with the following momenta combination were run:

(in the descriptions A B and C are used to describe the atoms colliding: AB is the initial H2 approaching H, represented by C)
{| class="wikitable"
!'''#'''
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|'''1'''
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.3</nowiki>
|yes
|The reaction dynamics proceed in a straight forward manner: AB and C approach each other, a collision occurs with the correct momentum for the reaction to proceed and a new bond BC is formed.
| [[File:Pic_a.png|260px]]
|-
|'''2'''
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420</nowiki>
|no
|AB approaches C and collides with it, but it doesn't have enough momentum to form a new bond so they are repelled. Momentum transfer does occur and C goes back with a greater momentum.
| [[File:Pic_b.png|260px]]
|-
|'''3'''
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414</nowiki>
|yes
|Oscillating AB collides with C and forms a new bond. The new H2 molecule, BC, has a higher vibrational frequency. A's direction is reversed.
| [[File:Pic_c.png|260px]]
|-
|'''4'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-375</nowiki>
|no
|AB approaches C and a BC bond forms, however, the system has too much momentum and C is repelled by the nucleus in B and bounces back. This leaves AB with greater vibrational energy and the distance BC increases as C is repelled.

| [[File:Pic_d.png|260px]]
|-
|'''5'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349</nowiki>
|yes
|AB collides with C and a new bond forms. However, unlike above, C still has enough energy to overcome the activation barrier. So, the BC bond forms, it breaks and reforms AB which then collides again with C to reform BC and leave as 2 entities: A + BC
| [[File:Pic_e.png|260px]]
|}

==== 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 allows us to investigate the progress of a reaction by splitting it into 2 parts: reactants and products.
* For a reaction to occur the reactants must cross into the product "space" by going over the saddle point – the transition state!
* With the Born-Oppenheimer approximation<ref>[https://www.iue.tuwien.ac.at/phd/schanovsky/thesisch2.html]</ref>, it is assumed there is a single saddle point (thus transition state) per reaction
* Transition State theory should only be used to estimate reaction rates at low temperature as it fails to correctly predict reaction rates for scenarios with high momentum (ie high temperature), where the transition state can be recrossed and thus predict reaction rates which are a poor match with experimental values
* We must also take into account that not every reaction with the correct energy will react, as for example, we must also take into account orientation, so, predicted rates with TS theory will be higher than experimental values.

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

=== PES inspection ===

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

This can be deduced by inspecting the PES surfaces bellow. (A=F, B=H, C=H)
# '''F + H2 → HF + H''' : Exothermic
#* Looking at the reactants: A + BC in the PES plot (small BC distance, large AB distance), we can see they have a higher potential energy than the products: AB + C. Which is confirmed by looking at bond strengths<ref>[http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html]</ref>: breaking the H–H bond (432 kJ/mol) vs making the H–F bond (565 kJ/mol), so energy is released during the reaction.
# '''HF + H → H2 + F''' : Endothermic
#* If we know look at: AB + C as reactants (small AB distance, large BC distance) we can see they have a lower potential energy than the products: BC + A. It is again rationalised by the fact that the H-F bond being broken in the reaction is stronger than the H-H bond being formed, so it requires an amount of energy to initiate the reaction.

[[File:H+hf_pes1L.png|275px|thumb|centre| F-H-H PES plot: AB is FH distance and BC is HH distance ]]

==== Transition state for F-H-H system ====
Locate the approximate position of the transition state.

From Hammond's postulate we know that in an endothermic reaction the transition state resembles the products and in an exothermic reaction it resembles the reactants. So, we can infer that the transition state will resemble H + HF.

The transition State for the F-H-H reaction was found at: 181.11 pm AB and 74.487 pm BC, with 0 initial momenta for both. This was done by varying the distances until they remained constant, confirmed by the forces along AB and BC being 0, marking a stationary point as the first derivative is 0.

[[File: HFTS_cp.png|275px|thumb|centre| F-H-H Contour plot when placed at Transition State ]]

==== Activation Energy for F + H2 and H + HF reactions ====
Report the activation energy for both reactions

The MEP was found for both reactions for a structure neighbouring the transition state: '''rts'''+~1 pm.

The activation energy is the difference between the saddle point's maximum energy and the minimum energy of the products:

* '''Reaction''' '''1''': F + H2 → HF + H
** ''TS Energy: ''- 434 kJ / mol
** ''Final Energy: ''- 560 kJ / mol
** Activation Energy: + 126 kJ / mol
[[File:Eactbiggraph.png|350px|thumb|centre|MEP Energy vs Time diagram of transition state for Reaction 1 (FH (AB) 180.11 pm and HH (BC) 75.487 pm) ]]
* '''Reaction''' '''2''': HF + H → H2 + F 
** ''TS Energy: ''-433.8 kJ / mol
** ''Final Energy: ''-434.5 kJ / mol
** Activation Energy: 0.7 kJ / mol
[[File:H2+f_emep.png|350px|thumb|centre|MEP Energy vs Time diagram of transition state for Reaction 2 (FH (AB) 74.487 pm and HH (BC) 181.11 pm) ]]

=== Reaction Dynamics ===

==== Reactive trajectory for F + H2 → HF + H ====
In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.

Using initial conditions; rAB = 225 pm, rBC = 75 pm, pAB = -5 g.mol-1.pm.fs-1, pBC = -1 g.mol-1.pm.fs-1, F + H2 → HF + H is a successful reaction, and, as seen above, this is an exothermic reaction and thus energy is released. From the animation we can see that a vibrating H2 molecule (BC) approaches F (A) and an HF bond forms, however it breaks and reforms BC, which then collides again with the Fluorine atom to reform a highly energetic HF and leave as the final products: HF + H.

[[File:Goodreachf.png|350px|thumb|centre|Contour plot of successful reaction of formation of HF]]

The energy in the reaction is converted from potential energy to kinetic energy, which is shown by the highly vibrating HF molecule, compared to the initial H2. This is confirmed with a Momentum vs Time plot (shown bellow): the AB distance, representing the HF bond, oscillates at a much larger amplitude than the initial BC oscillations which represent the H2 molecule.
[[File:Memtumsuccreac.png|350px|thumb|centre|Momentum vs Time plot of formation of HF]]

Experimentally, this would result in an increase in temperature of our reaction mixture, which could be measured using calorimetry. Additionally, the HF vibrational modes would be highly excited and overtones would show in the IR spectra as multiples of the fundamental absorption frequency<ref>[https://chem.libretexts.org/Courses/Pacific_Union_College/Quantum_Chemistry/13%3A_Molecular_Spectroscopy/13.05%3A_Vibrational_Overtones ]</ref> (3962 cm-1)<ref>[http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html]</ref>. Because overtone bands are difficult to see, this could be done using use Low Temperature FTIR<ref>[https://pubs.acs.org/doi/abs/10.1021/j150647a021]</ref>.

==== Illustrating Polanyi's empirical rules [ Reaction 2: HF + H → H2 + F ] ====
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

Polyani's Rules<ref>[https://pubs.acs.org/doi/10.1021/jz301649w]</ref>state that:
# For an early transition state, translational energy is more efficient in promoting a reaction
# For a late transition state, vibrational energy is more important in promoting a reaction
So, for the endothermic reaction HF + H → H2 + F, compared to the reverse exothermic reaction, we need a higher amount of vibrational energy, as well as translational to overcome the transition state and for a reaction to occur.

This can be illustrated by comparing the following initial reaction conditions (''A=F, B=H, C=H, rAB = 75 pm, rBC = 225 pm''):
# '''pAB = -1, pBC = -5''' – Unsuccessful Reaction 2 (NB. under these conditions Reaction 1 is successful, as shown above)
# '''pAB = -2, pBC = -4''' – Successful Reaction 2 when we redistribute momentum to increase vibrational energy
(units of p: g.mol-1.pm.fs-1)

[[File:Unsuccessreac2pol.png|350px|thumb|left|Contour plot of unsuccessful formation of H2 + F with conditions parallel to a succesfull reverse reaction]][[File:Successreac2pol.png|350px|thumb|right|Contour plot of successful formation of H2 + F ]]

== References ==

MRD:celiab

2020-05-07T16:26:35Z

Cb2018: /* Reactive trajectory for H2 + F → HF + H */

== EXERCISE 1: H + H2 ==

=== The Transition State ===
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?
* On PES diagrams, transition states are mathematically defined as points with 0 gradient with respect to position: stationary points.
* The transition state corresponds to the saddle point; the highest energy point on the reaction coordinate, which is the lowest energy pathway.
* We can distinguish it from a local minimum because on the reaction coordinate axis, the saddle point representing the transition state, will be a local maximum, but on the orthogonal axis it will be a local minimum. The minima are minimums on either direction because they represent stable reactants and products.

=== H + H2 reaction ===
[[File:Internuclear_distances_vs_time.png|250px|thumb|middle|Internuclear Distances vs Time plot ]]

===== Locating the Transition State =====
Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.
* The reaction is symmetric, hence, the transition state must have '''r1''' = '''r2''' by Hammond's Postulate
* By testing different initial conditions where '''rAB'''='''rBC '''and '''p1''' = '''p2''' = 0 g.mol-1.pm.fs-1, we can find the position of the Transition State
** The position of the transition state was found to be '''rts'''=90.775 pm
** This is where the system underwent small periodic symmetric vibrations keeping the distance between the atoms almost constant
* In the plot to the right, we see two straight lines, representing almost a stationary behaviour; thus we know we are at the symmetric transition state.

===== Calculating the Reaction Path =====
Comment on how the ''mep'' and the trajectory you just calculated differ.
* The minimum energy path is calculated from a system slightly displaced from the Transition state ('''r1''' = '''rts'''+1 pm, '''r2''' = '''rts''') and 0 initial momenta and follows the lowest energy path on a PES.
* As the reaction is symmetric it doesn't matter which atoms we label as A B C, so if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead, we get the same results but switched: AB will become BC and viceversa.
* The dynamic trajectory includes the momentum of the particles which causes oscillations about the trajectory path.
* The ''mep'' trajectory is roughly the same but it's smooth and doesn't oscillate, as the velocities are set to 0 in every step. This means we are artificially removing any gain in vibrational energy which results in oscillations. As the velocity is set 0 at every step, the kinetic energy is also 0.
[[File:picture 2 mep.png|350px|thumb|left|MEP contour plot ]]
[[File:picture 3 dynamic.png|350px|thumb|right|Dynamic contour plot ]]
[[File:Pic_4_dt.png|350px|thumb|left|MEP Internuclear Distances vs Time plot ]]
[[File:Pic_5_dt.png|350px|thumb|right| Dynamic Internuclear Distances vs Time plot ]]

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

For the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, the following trajectories with the following momenta combination were run:

(in the descriptions A B and C are used to describe the atoms colliding: AB is the initial H2 approaching H, represented by C)
{| class="wikitable"
!'''#'''
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|'''1'''
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.3</nowiki>
|yes
|The reaction dynamics proceed in a straight forward manner: AB and C approach each other, a collision occurs with the correct momentum for the reaction to proceed and a new bond BC is formed.
| [[File:Pic_a.png|260px]]
|-
|'''2'''
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420</nowiki>
|no
|AB approaches C and collides with it, but it doesn't have enough momentum to form a new bond so they are repelled. Momentum transfer does occur and C goes back with a greater momentum.
| [[File:Pic_b.png|260px]]
|-
|'''3'''
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414</nowiki>
|yes
|Oscillating AB collides with C and forms a new bond. The new H2 molecule, BC, has a higher vibrational frequency. A's direction is reversed.
| [[File:Pic_c.png|260px]]
|-
|'''4'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-375</nowiki>
|no
|AB approaches C and a BC bond forms, however, the system has too much momentum and C is repelled by the nucleus in B and bounces back. This leaves AB with greater vibrational energy and the distance BC increases as C is repelled.

| [[File:Pic_d.png|260px]]
|-
|'''5'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349</nowiki>
|yes
|AB collides with C and a new bond forms. However, unlike above, C still has enough energy to overcome the activation barrier. So, the BC bond forms, it breaks and reforms AB which then collides again with C to reform BC and leave as 2 entities: A + BC
| [[File:Pic_e.png|260px]]
|}

==== 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 allows us to investigate the progress of a reaction by splitting it into 2 parts: reactants and products.
* For a reaction to occur the reactants must cross into the product "space" by going over the saddle point – the transition state!
* With the Born-Oppenheimer approximation<ref>[https://www.iue.tuwien.ac.at/phd/schanovsky/thesisch2.html]</ref>, it is assumed there is a single saddle point (thus transition state) per reaction
* Transition State theory should only be used to estimate reaction rates at low temperature as it fails to correctly predict reaction rates for scenarios with high momentum (ie high temperature), where the transition state can be recrossed and thus predict reaction rates which are a poor match with experimental values
* We must also take into account that not every reaction with the correct energy will react, as for example, we must also take into account orientation, so, predicted rates with TS theory will be higher than experimental values.

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

=== PES inspection ===

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

This can be deduced by inspecting the PES surfaces bellow. (A=F, B=H, C=H)
# '''F + H2 → HF + H''' : Exothermic
#* Looking at the reactants: A + BC in the PES plot (small BC distance, large AB distance), we can see they have a higher potential energy than the products: AB + C. Which is confirmed by looking at bond strengths<ref>[http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html]</ref>: breaking the H–H bond (432 kJ/mol) vs making the H–F bond (565 kJ/mol), so energy is released during the reaction.
# '''HF + H → H2 + F''' : Endothermic
#* If we know look at: AB + C as reactants (small AB distance, large BC distance) we can see they have a lower potential energy than the products: BC + A. It is again rationalised by the fact that the H-F bond being broken in the reaction is stronger than the H-H bond being formed, so it requires an amount of energy to initiate the reaction.

[[File:H+hf_pes1L.png|275px|thumb|centre| F-H-H PES plot: AB is FH distance and BC is HH distance ]]

==== Transition state for F-H-H system ====
Locate the approximate position of the transition state.

From Hammond's postulate we know that in an endothermic reaction the transition state resembles the products and in an exothermic reaction it resembles the reactants. So, we can infer that the transition state will resemble H + HF.

The transition State for the F-H-H reaction was found at: 181.11 pm AB and 74.487 pm BC, with 0 initial momenta for both. This was done by varying the distances until they remained constant, confirmed by the forces along AB and BC being 0, marking a stationary point as the first derivative is 0.

[[File: HFTS_cp.png|275px|thumb|centre| F-H-H Contour plot when placed at Transition State ]]

==== Activation Energy for F + H2 and H + HF reactions ====
Report the activation energy for both reactions

The MEP was found for both reactions for a structure neighbouring the transition state: '''rts'''+~1 pm.

The activation energy is the difference between the saddle point's maximum energy and the minimum energy of the products:

* '''Reaction''' '''1''': F + H2 → HF + H
** ''TS Energy: ''- 434 kJ / mol
** ''Final Energy: ''- 560 kJ / mol
** Activation Energy: + 126 kJ / mol
[[File:Eactbiggraph.png|350px|thumb|centre|MEP Energy vs Time diagram of transition state for Reaction 1 (FH (AB) 180.11 pm and HH (BC) 75.487 pm) ]]
* '''Reaction''' '''2''': HF + H → H2 + F 
** ''TS Energy: ''-433.8 kJ / mol
** ''Final Energy: ''-434.5 kJ / mol
** Activation Energy: 0.7 kJ / mol
[[File:H2+f_emep.png|350px|thumb|centre|MEP Energy vs Time diagram of transition state for Reaction 2 (FH (AB) 74.487 pm and HH (BC) 181.11 pm) ]]

=== Reaction Dynamics ===

==== Reactive trajectory for F + H2 → HF + H ====
In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.

Using initial conditions; rAB = 225 pm, rBC = 75 pm, pAB = -5 g.mol-1.pm.fs-1, pBC = -1 g.mol-1.pm.fs-1, F + H2 → HF + H is a successful reaction, and, as seen above, this is an exothermic reaction and thus energy is released. From the animation we can see that a vibrating H2 molecule (BC) approaches F (A) and an HF bond forms, however it breaks and reforms BC, which then collides again with the Fluorine atom to reform a highly energetic HF and leave as the final products: HF + H.

[[File:Goodreachf.png|350px|thumb|centre|Contour plot of successful reaction of formation of HF]]

The energy in the reaction is converted from potential energy to kinetic energy, which is shown by the highly vibrating HF molecule, compared to the initial H2. This is confirmed with a Momentum vs Time plot (shown bellow): the AB distance, representing the HF bond, oscillates at a much larger amplitude than the initial BC oscillations which represent the H2 molecule.
[[File:Memtumsuccreac.png|350px|thumb|centre|Momentum vs Time plot of formation of HF]]

Experimentally, this would result in an increase in temperature of our reaction mixture, which could be measured using calorimetry. Additionally, the HF vibrational modes would be highly excited and overtones would show in the IR spectra as multiples of the fundamental absorption frequency<ref>[https://chem.libretexts.org/Courses/Pacific_Union_College/Quantum_Chemistry/13%3A_Molecular_Spectroscopy/13.05%3A_Vibrational_Overtones ]</ref> (3962 cm-1)<ref>[http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html]</ref>. Because overtone bands are difficult to see, this could be done using use Low Temperature FTIR<ref>[https://pubs.acs.org/doi/abs/10.1021/j150647a021]</ref>.

==== Illustrating Polanyi's empirical rules [ Reaction 2: HF + H → H2 + F ] ====
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

Polyani's Rules<ref>[https://pubs.acs.org/doi/10.1021/jz301649w]</ref>state that:
# For an early transition state, translational energy is more efficient in promoting a reaction
# For a late transition state, vibrational energy is more important in promoting a reaction
So, for the endothermic reaction HF + H → H2 + F, compared to the reverse exothermic reaction, we need a higher amount of vibrational energy, as well as translational to overcome the transition state and for a reaction to occur.

This can be illustrated by comparing the following initial reaction conditions (''A=F, B=H, C=H, rAB = 75 pm, rBC = 225 pm''):
# '''pAB = -1, pBC = -5''' – Unsuccessful Reaction 2 (NB. under these conditions Reaction 1 is successful, as shown above)
# '''pAB = -2, pBC = -4''' – Successful Reaction 2 when we redistribute momentum to increase vibrational energy
(units of p: g.mol-1.pm.fs-1)

[[File:Unsuccessreac2pol.png|350px|thumb|left|Contour plot of unsuccessful formation of HF with conditions parallel to a succesfull reverse reaction]][[File:Successreac2pol.png|350px|thumb|right|Contour plot of successful formation of HF ]]

== References ==

MRD:celiab

2020-05-07T16:16:56Z

Cb2018: /* Activation Energy for F + H2 and H + HF reactions */

== EXERCISE 1: H + H2 ==

=== The Transition State ===
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?
* On PES diagrams, transition states are mathematically defined as points with 0 gradient with respect to position: stationary points.
* The transition state corresponds to the saddle point; the highest energy point on the reaction coordinate, which is the lowest energy pathway.
* We can distinguish it from a local minimum because on the reaction coordinate axis, the saddle point representing the transition state, will be a local maximum, but on the orthogonal axis it will be a local minimum. The minima are minimums on either direction because they represent stable reactants and products.

=== H + H2 reaction ===
[[File:Internuclear_distances_vs_time.png|250px|thumb|middle|Internuclear Distances vs Time plot ]]

===== Locating the Transition State =====
Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.
* The reaction is symmetric, hence, the transition state must have '''r1''' = '''r2''' by Hammond's Postulate
* By testing different initial conditions where '''rAB'''='''rBC '''and '''p1''' = '''p2''' = 0 g.mol-1.pm.fs-1, we can find the position of the Transition State
** The position of the transition state was found to be '''rts'''=90.775 pm
** This is where the system underwent small periodic symmetric vibrations keeping the distance between the atoms almost constant
* In the plot to the right, we see two straight lines, representing almost a stationary behaviour; thus we know we are at the symmetric transition state.

===== Calculating the Reaction Path =====
Comment on how the ''mep'' and the trajectory you just calculated differ.
* The minimum energy path is calculated from a system slightly displaced from the Transition state ('''r1''' = '''rts'''+1 pm, '''r2''' = '''rts''') and 0 initial momenta and follows the lowest energy path on a PES.
* As the reaction is symmetric it doesn't matter which atoms we label as A B C, so if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead, we get the same results but switched: AB will become BC and viceversa.
* The dynamic trajectory includes the momentum of the particles which causes oscillations about the trajectory path.
* The ''mep'' trajectory is roughly the same but it's smooth and doesn't oscillate, as the velocities are set to 0 in every step. This means we are artificially removing any gain in vibrational energy which results in oscillations. As the velocity is set 0 at every step, the kinetic energy is also 0.
[[File:picture 2 mep.png|350px|thumb|left|MEP contour plot ]]
[[File:picture 3 dynamic.png|350px|thumb|right|Dynamic contour plot ]]
[[File:Pic_4_dt.png|350px|thumb|left|MEP Internuclear Distances vs Time plot ]]
[[File:Pic_5_dt.png|350px|thumb|right| Dynamic Internuclear Distances vs Time plot ]]

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

For the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, the following trajectories with the following momenta combination were run:

(in the descriptions A B and C are used to describe the atoms colliding: AB is the initial H2 approaching H, represented by C)
{| class="wikitable"
!'''#'''
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|'''1'''
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.3</nowiki>
|yes
|The reaction dynamics proceed in a straight forward manner: AB and C approach each other, a collision occurs with the correct momentum for the reaction to proceed and a new bond BC is formed.
| [[File:Pic_a.png|260px]]
|-
|'''2'''
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420</nowiki>
|no
|AB approaches C and collides with it, but it doesn't have enough momentum to form a new bond so they are repelled. Momentum transfer does occur and C goes back with a greater momentum.
| [[File:Pic_b.png|260px]]
|-
|'''3'''
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414</nowiki>
|yes
|Oscillating AB collides with C and forms a new bond. The new H2 molecule, BC, has a higher vibrational frequency. A's direction is reversed.
| [[File:Pic_c.png|260px]]
|-
|'''4'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-375</nowiki>
|no
|AB approaches C and a BC bond forms, however, the system has too much momentum and C is repelled by the nucleus in B and bounces back. This leaves AB with greater vibrational energy and the distance BC increases as C is repelled.

| [[File:Pic_d.png|260px]]
|-
|'''5'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349</nowiki>
|yes
|AB collides with C and a new bond forms. However, unlike above, C still has enough energy to overcome the activation barrier. So, the BC bond forms, it breaks and reforms AB which then collides again with C to reform BC and leave as 2 entities: A + BC
| [[File:Pic_e.png|260px]]
|}

==== 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 allows us to investigate the progress of a reaction by splitting it into 2 parts: reactants and products.
* For a reaction to occur the reactants must cross into the product "space" by going over the saddle point – the transition state!
* With the Born-Oppenheimer approximation<ref>[https://www.iue.tuwien.ac.at/phd/schanovsky/thesisch2.html]</ref>, it is assumed there is a single saddle point (thus transition state) per reaction
* Transition State theory should only be used to estimate reaction rates at low temperature as it fails to correctly predict reaction rates for scenarios with high momentum (ie high temperature), where the transition state can be recrossed and thus predict reaction rates which are a poor match with experimental values
* We must also take into account that not every reaction with the correct energy will react, as for example, we must also take into account orientation, so, predicted rates with TS theory will be higher than experimental values.

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

=== PES inspection ===

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

This can be deduced by inspecting the PES surfaces bellow. (A=F, B=H, C=H)
# '''F + H2 → HF + H''' : Exothermic
#* Looking at the reactants: A + BC in the PES plot (small BC distance, large AB distance), we can see they have a higher potential energy than the products: AB + C. Which is confirmed by looking at bond strengths<ref>[http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html]</ref>: breaking the H–H bond (432 kJ/mol) vs making the H–F bond (565 kJ/mol), so energy is released during the reaction.
# '''HF + H → H2 + F''' : Endothermic
#* If we know look at: AB + C as reactants (small AB distance, large BC distance) we can see they have a lower potential energy than the products: BC + A. It is again rationalised by the fact that the H-F bond being broken in the reaction is stronger than the H-H bond being formed, so it requires an amount of energy to initiate the reaction.

[[File:H+hf_pes1L.png|275px|thumb|centre| F-H-H PES plot: AB is FH distance and BC is HH distance ]]

==== Transition state for F-H-H system ====
Locate the approximate position of the transition state.

From Hammond's postulate we know that in an endothermic reaction the transition state resembles the products and in an exothermic reaction it resembles the reactants. So, we can infer that the transition state will resemble H + HF.

The transition State for the F-H-H reaction was found at: 181.11 pm AB and 74.487 pm BC, with 0 initial momenta for both. This was done by varying the distances until they remained constant, confirmed by the forces along AB and BC being 0, marking a stationary point as the first derivative is 0.

[[File: HFTS_cp.png|275px|thumb|centre| F-H-H Contour plot when placed at Transition State ]]

==== Activation Energy for F + H2 and H + HF reactions ====
Report the activation energy for both reactions

The MEP was found for both reactions for a structure neighbouring the transition state: '''rts'''+~1 pm.

The activation energy is the difference between the saddle point's maximum energy and the minimum energy of the products:

* '''Reaction''' '''1''': F + H2 → HF + H
** ''TS Energy: ''- 434 kJ / mol
** ''Final Energy: ''- 560 kJ / mol
** Activation Energy: + 126 kJ / mol
[[File:Eactbiggraph.png|350px|thumb|centre|MEP Energy vs Time diagram of transition state for Reaction 1 (FH (AB) 180.11 pm and HH (BC) 75.487 pm) ]]
* '''Reaction''' '''2''': HF + H → H2 + F 
** ''TS Energy: ''-433.8 kJ / mol
** ''Final Energy: ''-434.5 kJ / mol
** Activation Energy: 0.7 kJ / mol
[[File:H2+f_emep.png|350px|thumb|centre|MEP Energy vs Time diagram of transition state for Reaction 2 (FH (AB) 74.487 pm and HH (BC) 181.11 pm) ]]

=== Reaction Dynamics ===

==== Reactive trajectory for H2 + F → HF + H ====
In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.

Using initial conditions; rAB = 225 pm, rBC = 75 pm, pAB = -5 g.mol-1.pm.fs-1, pBC = -1 g.mol-1.pm.fs-1, (NB. A=F, B=H, C=H): F + H2 → HF + H is a successful reaction. As seen above, this is an exothermic reaction and thus energy is released. From the animation we can see that a vibrating H2 molecule (BC) approaches F (A) and an HF bond forms, however it breaks and reforms BC, which then collides again with the Fluorine atom to reform a highly energetic HF and leave as the final products: HF + H.

[[File:Goodreachf.png|350px|thumb|centre|Contour plot of successful reaction of formation of HF]]

The energy in the reaction is converted from potential energy to kinetic energy, which is shown by the highly vibrating HF molecule, compared to the initial H2. This is confirmed with a Momentum vs Time plot (shown bellow): the AB distance, representing the HF bond, oscillates at a much larger amplitude than the initial BC oscillations which represent the H2 molecule.
[[File:Memtumsuccreac.png|350px|thumb|centre|Momentum vs Time plot of formation of HF]]

Experimentally, this would result in an increase in temperature of our reaction mixture, which could be measured using calorimetry. Additionally, the HF vibrational modes would be highly excited and overtones would show in the IR spectra as multiples of the fundamental absorption frequency<ref>[https://chem.libretexts.org/Courses/Pacific_Union_College/Quantum_Chemistry/13%3A_Molecular_Spectroscopy/13.05%3A_Vibrational_Overtones ]</ref> (3962 cm-1)<ref>[http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html]</ref>. Because overtone bands are difficult to see, this could be done using use Low Temperature FTIR<ref>[https://pubs.acs.org/doi/abs/10.1021/j150647a021]</ref>.

==== Illustrating Polanyi's empirical rules: Reaction 2 HF + H → H2 + F, ====
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

Polyani's Rules<ref>[https://pubs.acs.org/doi/10.1021/jz301649w]</ref>state that:
# For an early transition state, translational energy is more efficient in promoting a reaction
# For a late transition state, vibrational energy is more important in promoting a reaction
<nowiki> </nowiki>So, for the endothermic reaction HF + H → H2 + F, compared to the reverse exothermic reaction, we need a higher amount of vibrational energy, as well as translational to overcome the transition state and for a reaction to occur.

This can be illustrated by comparing the following initial reaction conditions (''A=F, B=H, C=H, rAB = 75 pm, rBC = 225 pm''):
# '''pAB = -1, pBC = -5''' – Unsuccessful Reaction 2 (NB. under these conditions Reaction 1 is successful, as shown above)
# '''pAB = -2, pBC = -4''' – Successful Reaction 2 when we redistribute momentum to increase vibrational energy
(units of p: g.mol-1.pm.fs-1)

[[File:Unsuccessreac2pol.png|350px|thumb|left|Contour plot of unsuccessful formation of HF with conditions parallel to a succesfull reverse reaction]][[File:Successreac2pol.png|350px|thumb|right|Contour plot of successful formation of HF ]]

== References ==

MRD:celiab

2020-05-07T16:13:50Z

Cb2018: /* Activation Energy for F + H2 and H + HF reactions */

== EXERCISE 1: H + H2 ==

=== The Transition State ===
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?
* On PES diagrams, transition states are mathematically defined as points with 0 gradient with respect to position: stationary points.
* The transition state corresponds to the saddle point; the highest energy point on the reaction coordinate, which is the lowest energy pathway.
* We can distinguish it from a local minimum because on the reaction coordinate axis, the saddle point representing the transition state, will be a local maximum, but on the orthogonal axis it will be a local minimum. The minima are minimums on either direction because they represent stable reactants and products.

=== H + H2 reaction ===
[[File:Internuclear_distances_vs_time.png|250px|thumb|middle|Internuclear Distances vs Time plot ]]

===== Locating the Transition State =====
Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.
* The reaction is symmetric, hence, the transition state must have '''r1''' = '''r2''' by Hammond's Postulate
* By testing different initial conditions where '''rAB'''='''rBC '''and '''p1''' = '''p2''' = 0 g.mol-1.pm.fs-1, we can find the position of the Transition State
** The position of the transition state was found to be '''rts'''=90.775 pm
** This is where the system underwent small periodic symmetric vibrations keeping the distance between the atoms almost constant
* In the plot to the right, we see two straight lines, representing almost a stationary behaviour; thus we know we are at the symmetric transition state.

===== Calculating the Reaction Path =====
Comment on how the ''mep'' and the trajectory you just calculated differ.
* The minimum energy path is calculated from a system slightly displaced from the Transition state ('''r1''' = '''rts'''+1 pm, '''r2''' = '''rts''') and 0 initial momenta and follows the lowest energy path on a PES.
* As the reaction is symmetric it doesn't matter which atoms we label as A B C, so if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead, we get the same results but switched: AB will become BC and viceversa.
* The dynamic trajectory includes the momentum of the particles which causes oscillations about the trajectory path.
* The ''mep'' trajectory is roughly the same but it's smooth and doesn't oscillate, as the velocities are set to 0 in every step. This means we are artificially removing any gain in vibrational energy which results in oscillations. As the velocity is set 0 at every step, the kinetic energy is also 0.
[[File:picture 2 mep.png|350px|thumb|left|MEP contour plot ]]
[[File:picture 3 dynamic.png|350px|thumb|right|Dynamic contour plot ]]
[[File:Pic_4_dt.png|350px|thumb|left|MEP Internuclear Distances vs Time plot ]]
[[File:Pic_5_dt.png|350px|thumb|right| Dynamic Internuclear Distances vs Time plot ]]

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

For the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, the following trajectories with the following momenta combination were run:

(in the descriptions A B and C are used to describe the atoms colliding: AB is the initial H2 approaching H, represented by C)
{| class="wikitable"
!'''#'''
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|'''1'''
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.3</nowiki>
|yes
|The reaction dynamics proceed in a straight forward manner: AB and C approach each other, a collision occurs with the correct momentum for the reaction to proceed and a new bond BC is formed.
| [[File:Pic_a.png|260px]]
|-
|'''2'''
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420</nowiki>
|no
|AB approaches C and collides with it, but it doesn't have enough momentum to form a new bond so they are repelled. Momentum transfer does occur and C goes back with a greater momentum.
| [[File:Pic_b.png|260px]]
|-
|'''3'''
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414</nowiki>
|yes
|Oscillating AB collides with C and forms a new bond. The new H2 molecule, BC, has a higher vibrational frequency. A's direction is reversed.
| [[File:Pic_c.png|260px]]
|-
|'''4'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-375</nowiki>
|no
|AB approaches C and a BC bond forms, however, the system has too much momentum and C is repelled by the nucleus in B and bounces back. This leaves AB with greater vibrational energy and the distance BC increases as C is repelled.

| [[File:Pic_d.png|260px]]
|-
|'''5'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349</nowiki>
|yes
|AB collides with C and a new bond forms. However, unlike above, C still has enough energy to overcome the activation barrier. So, the BC bond forms, it breaks and reforms AB which then collides again with C to reform BC and leave as 2 entities: A + BC
| [[File:Pic_e.png|260px]]
|}

==== 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 allows us to investigate the progress of a reaction by splitting it into 2 parts: reactants and products.
* For a reaction to occur the reactants must cross into the product "space" by going over the saddle point – the transition state!
* With the Born-Oppenheimer approximation<ref>[https://www.iue.tuwien.ac.at/phd/schanovsky/thesisch2.html]</ref>, it is assumed there is a single saddle point (thus transition state) per reaction
* Transition State theory should only be used to estimate reaction rates at low temperature as it fails to correctly predict reaction rates for scenarios with high momentum (ie high temperature), where the transition state can be recrossed and thus predict reaction rates which are a poor match with experimental values
* We must also take into account that not every reaction with the correct energy will react, as for example, we must also take into account orientation, so, predicted rates with TS theory will be higher than experimental values.

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

=== PES inspection ===

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

This can be deduced by inspecting the PES surfaces bellow. (A=F, B=H, C=H)
# '''F + H2 → HF + H''' : Exothermic
#* Looking at the reactants: A + BC in the PES plot (small BC distance, large AB distance), we can see they have a higher potential energy than the products: AB + C. Which is confirmed by looking at bond strengths<ref>[http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html]</ref>: breaking the H–H bond (432 kJ/mol) vs making the H–F bond (565 kJ/mol), so energy is released during the reaction.
# '''HF + H → H2 + F''' : Endothermic
#* If we know look at: AB + C as reactants (small AB distance, large BC distance) we can see they have a lower potential energy than the products: BC + A. It is again rationalised by the fact that the H-F bond being broken in the reaction is stronger than the H-H bond being formed, so it requires an amount of energy to initiate the reaction.

[[File:H+hf_pes1L.png|275px|thumb|centre| F-H-H PES plot: AB is FH distance and BC is HH distance ]]

==== Transition state for F-H-H system ====
Locate the approximate position of the transition state.

From Hammond's postulate we know that in an endothermic reaction the transition state resembles the products and in an exothermic reaction it resembles the reactants. So, we can infer that the transition state will resemble H + HF.

The transition State for the F-H-H reaction was found at: 181.11 pm AB and 74.487 pm BC, with 0 initial momenta for both. This was done by varying the distances until they remained constant, confirmed by the forces along AB and BC being 0, marking a stationary point as the first derivative is 0.

[[File: HFTS_cp.png|275px|thumb|centre| F-H-H Contour plot when placed at Transition State ]]

==== Activation Energy for F + H2 and H + HF reactions ====
Report the activation energy for both reactions

The MEP was found for both reactions for a structure neighbouring the transition state: '''rts'''+~1 pm.

The activation energy is the difference between the saddle point's maximum energy and the minimum energy of the products:

* '''Reaction''' '''1''': F + H2 → HF + H
** ''TS Energy: ''- 434 kJ / mol
** ''Final Energy: ''- 560 kJ / mol
** Activation Energy: + 126 kJ / mol
[[File:Eactbiggraph.png|350px|thumb|centre|MEP Energy vs Time diagram of transition state to HH + F state (FH (AB) 180.11 pm and HH (BC) 75.487 pm) ]]
* '''Reaction''' '''2''': HF + H → H2 + F 
** ''TS Energy: ''-433.8 kJ / mol
** ''Final Energy: ''-434.5 kJ / mol
** Activation Energy: 0.7 kJ / mol
[[File:H2+f_emep.png|350px|thumb|centre|MEP Energy vs Time diagram of transition state to form HF + H state (FH (AB) 181.11 pm and HH (BC) 74.487 pm) ]]

=== Reaction Dynamics ===

==== Reactive trajectory for H2 + F → HF + H ====
In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.

Using initial conditions; rAB = 225 pm, rBC = 75 pm, pAB = -5 g.mol-1.pm.fs-1, pBC = -1 g.mol-1.pm.fs-1, (NB. A=F, B=H, C=H): F + H2 → HF + H is a successful reaction. As seen above, this is an exothermic reaction and thus energy is released. From the animation we can see that a vibrating H2 molecule (BC) approaches F (A) and an HF bond forms, however it breaks and reforms BC, which then collides again with the Fluorine atom to reform a highly energetic HF and leave as the final products: HF + H.

[[File:Goodreachf.png|350px|thumb|centre|Contour plot of successful reaction of formation of HF]]

The energy in the reaction is converted from potential energy to kinetic energy, which is shown by the highly vibrating HF molecule, compared to the initial H2. This is confirmed with a Momentum vs Time plot (shown bellow): the AB distance, representing the HF bond, oscillates at a much larger amplitude than the initial BC oscillations which represent the H2 molecule.
[[File:Memtumsuccreac.png|350px|thumb|centre|Momentum vs Time plot of formation of HF]]

Experimentally, this would result in an increase in temperature of our reaction mixture, which could be measured using calorimetry. Additionally, the HF vibrational modes would be highly excited and overtones would show in the IR spectra as multiples of the fundamental absorption frequency<ref>[https://chem.libretexts.org/Courses/Pacific_Union_College/Quantum_Chemistry/13%3A_Molecular_Spectroscopy/13.05%3A_Vibrational_Overtones ]</ref> (3962 cm-1)<ref>[http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html]</ref>. Because overtone bands are difficult to see, this could be done using use Low Temperature FTIR<ref>[https://pubs.acs.org/doi/abs/10.1021/j150647a021]</ref>.

==== Illustrating Polanyi's empirical rules: Reaction 2 HF + H → H2 + F, ====
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

Polyani's Rules<ref>[https://pubs.acs.org/doi/10.1021/jz301649w]</ref>state that:
# For an early transition state, translational energy is more efficient in promoting a reaction
# For a late transition state, vibrational energy is more important in promoting a reaction
<nowiki> </nowiki>So, for the endothermic reaction HF + H → H2 + F, compared to the reverse exothermic reaction, we need a higher amount of vibrational energy, as well as translational to overcome the transition state and for a reaction to occur.

This can be illustrated by comparing the following initial reaction conditions (''A=F, B=H, C=H, rAB = 75 pm, rBC = 225 pm''):
# '''pAB = -1, pBC = -5''' – Unsuccessful Reaction 2 (NB. under these conditions Reaction 1 is successful, as shown above)
# '''pAB = -2, pBC = -4''' – Successful Reaction 2 when we redistribute momentum to increase vibrational energy
(units of p: g.mol-1.pm.fs-1)

[[File:Unsuccessreac2pol.png|350px|thumb|left|Contour plot of unsuccessful formation of HF with conditions parallel to a succesfull reverse reaction]][[File:Successreac2pol.png|350px|thumb|right|Contour plot of successful formation of HF ]]

== References ==

MRD:celiab

2020-05-07T16:13:28Z

Cb2018: /* EXERCISE 1: H + H2 */

== EXERCISE 1: H + H2 ==

=== The Transition State ===
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?
* On PES diagrams, transition states are mathematically defined as points with 0 gradient with respect to position: stationary points.
* The transition state corresponds to the saddle point; the highest energy point on the reaction coordinate, which is the lowest energy pathway.
* We can distinguish it from a local minimum because on the reaction coordinate axis, the saddle point representing the transition state, will be a local maximum, but on the orthogonal axis it will be a local minimum. The minima are minimums on either direction because they represent stable reactants and products.

=== H + H2 reaction ===
[[File:Internuclear_distances_vs_time.png|250px|thumb|middle|Internuclear Distances vs Time plot ]]

===== Locating the Transition State =====
Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.
* The reaction is symmetric, hence, the transition state must have '''r1''' = '''r2''' by Hammond's Postulate
* By testing different initial conditions where '''rAB'''='''rBC '''and '''p1''' = '''p2''' = 0 g.mol-1.pm.fs-1, we can find the position of the Transition State
** The position of the transition state was found to be '''rts'''=90.775 pm
** This is where the system underwent small periodic symmetric vibrations keeping the distance between the atoms almost constant
* In the plot to the right, we see two straight lines, representing almost a stationary behaviour; thus we know we are at the symmetric transition state.

===== Calculating the Reaction Path =====
Comment on how the ''mep'' and the trajectory you just calculated differ.
* The minimum energy path is calculated from a system slightly displaced from the Transition state ('''r1''' = '''rts'''+1 pm, '''r2''' = '''rts''') and 0 initial momenta and follows the lowest energy path on a PES.
* As the reaction is symmetric it doesn't matter which atoms we label as A B C, so if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead, we get the same results but switched: AB will become BC and viceversa.
* The dynamic trajectory includes the momentum of the particles which causes oscillations about the trajectory path.
* The ''mep'' trajectory is roughly the same but it's smooth and doesn't oscillate, as the velocities are set to 0 in every step. This means we are artificially removing any gain in vibrational energy which results in oscillations. As the velocity is set 0 at every step, the kinetic energy is also 0.
[[File:picture 2 mep.png|350px|thumb|left|MEP contour plot ]]
[[File:picture 3 dynamic.png|350px|thumb|right|Dynamic contour plot ]]
[[File:Pic_4_dt.png|350px|thumb|left|MEP Internuclear Distances vs Time plot ]]
[[File:Pic_5_dt.png|350px|thumb|right| Dynamic Internuclear Distances vs Time plot ]]

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

For the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, the following trajectories with the following momenta combination were run:

(in the descriptions A B and C are used to describe the atoms colliding: AB is the initial H2 approaching H, represented by C)
{| class="wikitable"
!'''#'''
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|'''1'''
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.3</nowiki>
|yes
|The reaction dynamics proceed in a straight forward manner: AB and C approach each other, a collision occurs with the correct momentum for the reaction to proceed and a new bond BC is formed.
| [[File:Pic_a.png|260px]]
|-
|'''2'''
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420</nowiki>
|no
|AB approaches C and collides with it, but it doesn't have enough momentum to form a new bond so they are repelled. Momentum transfer does occur and C goes back with a greater momentum.
| [[File:Pic_b.png|260px]]
|-
|'''3'''
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414</nowiki>
|yes
|Oscillating AB collides with C and forms a new bond. The new H2 molecule, BC, has a higher vibrational frequency. A's direction is reversed.
| [[File:Pic_c.png|260px]]
|-
|'''4'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-375</nowiki>
|no
|AB approaches C and a BC bond forms, however, the system has too much momentum and C is repelled by the nucleus in B and bounces back. This leaves AB with greater vibrational energy and the distance BC increases as C is repelled.

| [[File:Pic_d.png|260px]]
|-
|'''5'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349</nowiki>
|yes
|AB collides with C and a new bond forms. However, unlike above, C still has enough energy to overcome the activation barrier. So, the BC bond forms, it breaks and reforms AB which then collides again with C to reform BC and leave as 2 entities: A + BC
| [[File:Pic_e.png|260px]]
|}

==== 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 allows us to investigate the progress of a reaction by splitting it into 2 parts: reactants and products.
* For a reaction to occur the reactants must cross into the product "space" by going over the saddle point – the transition state!
* With the Born-Oppenheimer approximation<ref>[https://www.iue.tuwien.ac.at/phd/schanovsky/thesisch2.html]</ref>, it is assumed there is a single saddle point (thus transition state) per reaction
* Transition State theory should only be used to estimate reaction rates at low temperature as it fails to correctly predict reaction rates for scenarios with high momentum (ie high temperature), where the transition state can be recrossed and thus predict reaction rates which are a poor match with experimental values
* We must also take into account that not every reaction with the correct energy will react, as for example, we must also take into account orientation, so, predicted rates with TS theory will be higher than experimental values.

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

=== PES inspection ===

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

This can be deduced by inspecting the PES surfaces bellow. (A=F, B=H, C=H)
# '''F + H2 → HF + H''' : Exothermic
#* Looking at the reactants: A + BC in the PES plot (small BC distance, large AB distance), we can see they have a higher potential energy than the products: AB + C. Which is confirmed by looking at bond strengths<ref>[http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html]</ref>: breaking the H–H bond (432 kJ/mol) vs making the H–F bond (565 kJ/mol), so energy is released during the reaction.
# '''HF + H → H2 + F''' : Endothermic
#* If we know look at: AB + C as reactants (small AB distance, large BC distance) we can see they have a lower potential energy than the products: BC + A. It is again rationalised by the fact that the H-F bond being broken in the reaction is stronger than the H-H bond being formed, so it requires an amount of energy to initiate the reaction.

[[File:H+hf_pes1L.png|275px|thumb|centre| F-H-H PES plot: AB is FH distance and BC is HH distance ]]

==== Transition state for F-H-H system ====
Locate the approximate position of the transition state.

From Hammond's postulate we know that in an endothermic reaction the transition state resembles the products and in an exothermic reaction it resembles the reactants. So, we can infer that the transition state will resemble H + HF.

The transition State for the F-H-H reaction was found at: 181.11 pm AB and 74.487 pm BC, with 0 initial momenta for both. This was done by varying the distances until they remained constant, confirmed by the forces along AB and BC being 0, marking a stationary point as the first derivative is 0.

[[File: HFTS_cp.png|275px|thumb|centre| F-H-H Contour plot when placed at Transition State ]]

==== Activation Energy for F + H2 and H + HF reactions ====
Report the activation energy for both reactions

The MEP was found for both reactions for a structure neighbouring the transition state: '''rts'''+~1 pm.

The activation energy is the difference between the saddle point's maximum energy and the minimum energy of the products:

* '''Reaction''' '''1''': F + H2 → HF + H
** ''TS Energy: ''- 434 kJ / mol
** ''Final Energy: ''- 560 kJ / mol
** Activation Energy: + 126 kJ / mol
[[File:Eactbiggraph.png|350px|thumb|centre|Energy vs Time diagram of transition state to HH + F state (FH (AB) 180.11 pm and HH (BC) 75.487 pm) ]]
* '''Reaction''' '''2''': HF + H → H2 + F 
** ''TS Energy: ''-433.8 kJ / mol
** ''Final Energy: ''-434.5 kJ / mol
** Activation Energy: 0.7 kJ / mol
[[File:H2+f_emep.png|350px|thumb|centre|Energy vs Time diagram of transition state to form HF + H state (FH (AB) 181.11 pm and HH (BC) 74.487 pm) ]]

=== Reaction Dynamics ===

==== Reactive trajectory for H2 + F → HF + H ====
In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.

Using initial conditions; rAB = 225 pm, rBC = 75 pm, pAB = -5 g.mol-1.pm.fs-1, pBC = -1 g.mol-1.pm.fs-1, (NB. A=F, B=H, C=H): F + H2 → HF + H is a successful reaction. As seen above, this is an exothermic reaction and thus energy is released. From the animation we can see that a vibrating H2 molecule (BC) approaches F (A) and an HF bond forms, however it breaks and reforms BC, which then collides again with the Fluorine atom to reform a highly energetic HF and leave as the final products: HF + H.

[[File:Goodreachf.png|350px|thumb|centre|Contour plot of successful reaction of formation of HF]]

The energy in the reaction is converted from potential energy to kinetic energy, which is shown by the highly vibrating HF molecule, compared to the initial H2. This is confirmed with a Momentum vs Time plot (shown bellow): the AB distance, representing the HF bond, oscillates at a much larger amplitude than the initial BC oscillations which represent the H2 molecule.
[[File:Memtumsuccreac.png|350px|thumb|centre|Momentum vs Time plot of formation of HF]]

Experimentally, this would result in an increase in temperature of our reaction mixture, which could be measured using calorimetry. Additionally, the HF vibrational modes would be highly excited and overtones would show in the IR spectra as multiples of the fundamental absorption frequency<ref>[https://chem.libretexts.org/Courses/Pacific_Union_College/Quantum_Chemistry/13%3A_Molecular_Spectroscopy/13.05%3A_Vibrational_Overtones ]</ref> (3962 cm-1)<ref>[http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html]</ref>. Because overtone bands are difficult to see, this could be done using use Low Temperature FTIR<ref>[https://pubs.acs.org/doi/abs/10.1021/j150647a021]</ref>.

==== Illustrating Polanyi's empirical rules: Reaction 2 HF + H → H2 + F, ====
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

Polyani's Rules<ref>[https://pubs.acs.org/doi/10.1021/jz301649w]</ref>state that:
# For an early transition state, translational energy is more efficient in promoting a reaction
# For a late transition state, vibrational energy is more important in promoting a reaction
<nowiki> </nowiki>So, for the endothermic reaction HF + H → H2 + F, compared to the reverse exothermic reaction, we need a higher amount of vibrational energy, as well as translational to overcome the transition state and for a reaction to occur.

This can be illustrated by comparing the following initial reaction conditions (''A=F, B=H, C=H, rAB = 75 pm, rBC = 225 pm''):
# '''pAB = -1, pBC = -5''' – Unsuccessful Reaction 2 (NB. under these conditions Reaction 1 is successful, as shown above)
# '''pAB = -2, pBC = -4''' – Successful Reaction 2 when we redistribute momentum to increase vibrational energy
(units of p: g.mol-1.pm.fs-1)

[[File:Unsuccessreac2pol.png|350px|thumb|left|Contour plot of unsuccessful formation of HF with conditions parallel to a succesfull reverse reaction]][[File:Successreac2pol.png|350px|thumb|right|Contour plot of successful formation of HF ]]

== References ==

MRD:celiab

2020-05-07T16:07:18Z

Cb2018: /* Classifying F + H2 and H + HF reactions */

== EXERCISE 1: H + H2 ==

=== The Transition State ===
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?
* On PES diagrams, transition states are mathematically defined as points with 0 gradient with respect to position: stationary points.
* The transition state corresponds to the saddle point; the highest energy point on the reaction coordinate, which is the lowest energy pathway.
* We can distinguish it from a local minimum because on the reaction coordinate axis, the saddle point representing the transition state, will be a local maximum, but on the orthogonal axis it will be a local minimum. The minima are minimums on either direction because they represent stable reactants and products.

=== H + H2 reaction ===
[[File:Internuclear_distances_vs_time.png|250px|thumb|middle|Internuclear Distances vs Time plot ]]

===== Locating the Transition State =====
Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.
* The reaction is symmetric, hence, the transition state must have '''r1''' = '''r2''' by Hammond's Postulate
* By testing different initial conditions where '''rAB'''='''rBC '''and '''p1''' = '''p2''' = 0 g.mol-1.pm.fs-1, we can find the position of the Transition State
** The position of the transition state was found to be '''rts'''=90.775 pm
** This is where the system underwent small periodic symmetric vibrations keeping the distance between the atoms almost constant
* In the plot to the right, we see two straight lines, representing almost a stationary behaviour; thus we know we are at the symmetric transition state.

===== Calculating the Reaction Path =====
Comment on how the ''mep'' and the trajectory you just calculated differ.
* The minimum energy path is calculated from a system slightly displaced from the Transition state ('''r1''' = '''rts'''+1 pm, '''r2''' = '''rts''') and 0 initial momenta and follows the lowest energy path on a PES.
* As the reaction is symmetric it doesn't matter which atoms we label as A B C, so if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead, we get the same results but switched: AB will become BC and viceversa.
* The dynamic trajectory includes the momentum of the particles which causes oscillations about the trajectory path.
* The ''mep'' trajectory is roughly the same but it's smooth and doesn't oscillate, as the velocities are set to 0 in every step. This means we are artificially removing any gain in vibrational energy which results in oscillations. As the velocity is set 0 at every step, the kinetic energy is also 0.
[[File:picture 2 mep.png|350px|thumb|left|MEP contour plot ]]
[[File:picture 3 dynamic.png|350px|thumb|right|Dynamic contour plot ]]
[[File:Pic_4_dt.png|350px|thumb|left|MEP Internuclear Distances vs Time plot ]]
[[File:Pic_5_dt.png|350px|thumb|right| Dynamic Internuclear Distances vs Time plot ]]

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

For the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, the following trajectories with the following momenta combination were run:

(in the descriptions A B and C are used to describe the atoms colliding: AB is the initial H2 approaching H, represented by C)
{| class="wikitable"
!'''#'''
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|'''1'''
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.3</nowiki>
|yes
|The reaction dynamics proceed in a straight forward manner: AB and C approach each other, a collision occurs with the correct momentum for the reaction to proceed and a new bond BC is formed.
| [[File:Pic_a.png|260px]]
|-
|'''2'''
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420</nowiki>
|no
|AB approaches C and collides with it, but it doesn't have enough momentum to form a new bond so they are repelled. Momentum transfer does occur and C goes back with a greater momentum.
| [[File:Pic_b.png|260px]]
|-
|'''3'''
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414</nowiki>
|yes
|Oscillating AB collides with C and forms a new bond. The new H2 molecule, BC, has a higher vibrational frequency. A's direction is reversed.
| [[File:Pic_c.png|260px]]
|-
|'''4'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-375</nowiki>
|no
|AB approaches C and a BC bond forms, however, the system has too much momentum and C is repelled by the nucleus in B and bounces back. This leaves AB with greater vibrational energy and the distance BC increases as C is repelled.

| [[File:Pic_d.png|260px]]
|-
|'''5'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349</nowiki>
|yes
|AB collides with C and a new bond forms. However, unlike above, C still has enough energy to overcome the activation barrier. So, the BC bond forms, it breaks and reforms AB which then collides again with C to reform BC and leave as 2 entities: A + BC
| [[File:Pic_e.png|260px]]
|}

==== 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 allows us to investigate the progress of a reaction by splitting it into 2 parts: reactants and products.
* For a reaction to occur the reactants must cross into the product "space" by going over the saddle point – the transition state!
* With the Born-Oppenheimer approximation<ref>[https://www.iue.tuwien.ac.at/phd/schanovsky/thesisch2.html]</ref>, it is assumed there is a single saddle point (thus transition state) per reaction
* Transition State theory should only be used to estimate reaction rates at low temperature as it fails to correctly predict reaction rates for scenarios with high momentum (ie high temperature), where the transition state can be recrossed and thus predict reaction rates which are a poor match with experimental values
* We must also take into account that not every reaction with the correct energy will react, as for example, we must also take into account orientation, so, predicted rates with TS theory will be higher than experimental values.

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

=== PES inspection ===

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

This can be deduced by inspecting the PES surfaces bellow. (A=F, B=H, C=H)
# F + H2 → HF + H : Exothermic
#* Looking at the reactants: A + BC in the PES plot (small BC distance, large AB distance), we can see they have a higher potential energy than the products: AB + C. Which is confirmed by looking at bond strengths<ref>[http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html]</ref>: breaking the H–H bond (432 kJ/mol) vs making the H–F bond (565 kJ/mol), so energy is released during the reaction.
# HF + H → H2 + F : Endothermic
#* If we know look at: AB + C as reactants (small AB distance, large BC distance) we can see they have a lower potential energy than the products: BC + A. It is again rationalised by the fact that the H-F bond being broken in the reaction is stronger than the H-H bond being formed, so it requires an amount of energy to initiate the reaction.

[[File:H+hf_pes1L.png|275px|thumb|centre| F-H-H PES plot: AB is FH distance and BC is HH distance ]]

==== Transition state for F-H-H system ====
Locate the approximate position of the transition state.

From Hammond's postulate we know that in an endothermic reaction the transition state resembles the products and in an exothermic reaction it resembles the reactants. So, we can infer that the transition state will resemble H + HF.

The transition State for the F-H-H reaction was found at: 181.11 pm AB and 74.487 pm BC, with 0 initial momenta for both. This was done by varying the distances until they remained constant, confirmed by the forces along AB and BC being 0, marking a stationary point as the first derivative is 0.

[[File: HFTS_cp.png|275px|thumb|centre| F-H-H Contour plot when placed at Transition State ]]

==== Activation Energy for F + H2 and H + HF reactions ====
Report the activation energy for both reactions

The MEP was found for both reactions for a structure neighbouring the transition state: '''rts'''+~1 pm.

The activation energy is the difference between the saddle point's maximum energy and the minimum energy of the products:

* '''Reaction''' '''1''': F + H2 → HF + H
** ''TS Energy: ''- 434 kJ / mol
** ''Final Energy: ''- 560 kJ / mol
** Activation Energy: + 126 kJ / mol
[[File:Eactbiggraph.png|350px|thumb|centre|Energy vs Time diagram of transition state to HH + F state (FH (AB) 180.11 pm and HH (BC) 75.487 pm) ]]
* '''Reaction''' '''2''': HF + H → H2 + F 
** ''TS Energy: ''-433.8 kJ / mol
** ''Final Energy: ''-434.5 kJ / mol
** Activation Energy: 0.7 kJ / mol
[[File:H2+f_emep.png|350px|thumb|centre|Energy vs Time diagram of transition state to form HF + H state (FH (AB) 181.11 pm and HH (BC) 74.487 pm) ]]

=== Reaction Dynamics ===

==== Reactive trajectory for H2 + F → HF + H ====
In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.

Using initial conditions; rAB = 225 pm, rBC = 75 pm, pAB = -5 g.mol-1.pm.fs-1, pBC = -1 g.mol-1.pm.fs-1, (NB. A=F, B=H, C=H): F + H2 → HF + H is a successful reaction. As seen above, this is an exothermic reaction and thus energy is released. From the animation we can see that a vibrating H2 molecule (BC) approaches F (A) and an HF bond forms, however it breaks and reforms BC, which then collides again with the Fluorine atom to reform a highly energetic HF and leave as the final products: HF + H.

[[File:Goodreachf.png|350px|thumb|centre|Contour plot of successful reaction of formation of HF]]

The energy in the reaction is converted from potential energy to kinetic energy, which is shown by the highly vibrating HF molecule, compared to the initial H2. This is confirmed with a Momentum vs Time plot (shown bellow): the AB distance, representing the HF bond, oscillates at a much larger amplitude than the initial BC oscillations which represent the H2 molecule.
[[File:Memtumsuccreac.png|350px|thumb|centre|Momentum vs Time plot of formation of HF]]

Experimentally, this would result in an increase in temperature of our reaction mixture, which could be measured using calorimetry. Additionally, the HF vibrational modes would be highly excited and overtones would show in the IR spectra as multiples of the fundamental absorption frequency<ref>[https://chem.libretexts.org/Courses/Pacific_Union_College/Quantum_Chemistry/13%3A_Molecular_Spectroscopy/13.05%3A_Vibrational_Overtones ]</ref> (3962 cm-1)<ref>[http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html]</ref>. Because overtone bands are difficult to see, this could be done using use Low Temperature FTIR<ref>[https://pubs.acs.org/doi/abs/10.1021/j150647a021]</ref>.

==== Illustrating Polanyi's empirical rules: Reaction 2 HF + H → H2 + F, ====
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

Polyani's Rules<ref>[https://pubs.acs.org/doi/10.1021/jz301649w]</ref>state that:
# For an early transition state, translational energy is more efficient in promoting a reaction
# For a late transition state, vibrational energy is more important in promoting a reaction
<nowiki> </nowiki>So, for the endothermic reaction HF + H → H2 + F, compared to the reverse exothermic reaction, we need a higher amount of vibrational energy, as well as translational to overcome the transition state and for a reaction to occur.

This can be illustrated by comparing the following initial reaction conditions (''A=F, B=H, C=H, rAB = 75 pm, rBC = 225 pm''):
# '''pAB = -1, pBC = -5''' – Unsuccessful Reaction 2 (NB. under these conditions Reaction 1 is successful, as shown above)
# '''pAB = -2, pBC = -4''' – Successful Reaction 2 when we redistribute momentum to increase vibrational energy
(units of p: g.mol-1.pm.fs-1)

[[File:Unsuccessreac2pol.png|350px|thumb|left|Contour plot of unsuccessful formation of HF with conditions parallel to a succesfull reverse reaction]][[File:Successreac2pol.png|350px|thumb|right|Contour plot of successful formation of HF ]]

== References ==

MRD:celiab

2020-05-07T16:06:08Z

Cb2018: /* Transition State Theory */

== EXERCISE 1: H + H2 ==

=== The Transition State ===
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?
* On PES diagrams, transition states are mathematically defined as points with 0 gradient with respect to position: stationary points.
* The transition state corresponds to the saddle point; the highest energy point on the reaction coordinate, which is the lowest energy pathway.
* We can distinguish it from a local minimum because on the reaction coordinate axis, the saddle point representing the transition state, will be a local maximum, but on the orthogonal axis it will be a local minimum. The minima are minimums on either direction because they represent stable reactants and products.

=== H + H2 reaction ===
[[File:Internuclear_distances_vs_time.png|250px|thumb|middle|Internuclear Distances vs Time plot ]]

===== Locating the Transition State =====
Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.
* The reaction is symmetric, hence, the transition state must have '''r1''' = '''r2''' by Hammond's Postulate
* By testing different initial conditions where '''rAB'''='''rBC '''and '''p1''' = '''p2''' = 0 g.mol-1.pm.fs-1, we can find the position of the Transition State
** The position of the transition state was found to be '''rts'''=90.775 pm
** This is where the system underwent small periodic symmetric vibrations keeping the distance between the atoms almost constant
* In the plot to the right, we see two straight lines, representing almost a stationary behaviour; thus we know we are at the symmetric transition state.

===== Calculating the Reaction Path =====
Comment on how the ''mep'' and the trajectory you just calculated differ.
* The minimum energy path is calculated from a system slightly displaced from the Transition state ('''r1''' = '''rts'''+1 pm, '''r2''' = '''rts''') and 0 initial momenta and follows the lowest energy path on a PES.
* As the reaction is symmetric it doesn't matter which atoms we label as A B C, so if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead, we get the same results but switched: AB will become BC and viceversa.
* The dynamic trajectory includes the momentum of the particles which causes oscillations about the trajectory path.
* The ''mep'' trajectory is roughly the same but it's smooth and doesn't oscillate, as the velocities are set to 0 in every step. This means we are artificially removing any gain in vibrational energy which results in oscillations. As the velocity is set 0 at every step, the kinetic energy is also 0.
[[File:picture 2 mep.png|350px|thumb|left|MEP contour plot ]]
[[File:picture 3 dynamic.png|350px|thumb|right|Dynamic contour plot ]]
[[File:Pic_4_dt.png|350px|thumb|left|MEP Internuclear Distances vs Time plot ]]
[[File:Pic_5_dt.png|350px|thumb|right| Dynamic Internuclear Distances vs Time plot ]]

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

For the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, the following trajectories with the following momenta combination were run:

(in the descriptions A B and C are used to describe the atoms colliding: AB is the initial H2 approaching H, represented by C)
{| class="wikitable"
!'''#'''
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|'''1'''
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.3</nowiki>
|yes
|The reaction dynamics proceed in a straight forward manner: AB and C approach each other, a collision occurs with the correct momentum for the reaction to proceed and a new bond BC is formed.
| [[File:Pic_a.png|260px]]
|-
|'''2'''
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420</nowiki>
|no
|AB approaches C and collides with it, but it doesn't have enough momentum to form a new bond so they are repelled. Momentum transfer does occur and C goes back with a greater momentum.
| [[File:Pic_b.png|260px]]
|-
|'''3'''
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414</nowiki>
|yes
|Oscillating AB collides with C and forms a new bond. The new H2 molecule, BC, has a higher vibrational frequency. A's direction is reversed.
| [[File:Pic_c.png|260px]]
|-
|'''4'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-375</nowiki>
|no
|AB approaches C and a BC bond forms, however, the system has too much momentum and C is repelled by the nucleus in B and bounces back. This leaves AB with greater vibrational energy and the distance BC increases as C is repelled.

| [[File:Pic_d.png|260px]]
|-
|'''5'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349</nowiki>
|yes
|AB collides with C and a new bond forms. However, unlike above, C still has enough energy to overcome the activation barrier. So, the BC bond forms, it breaks and reforms AB which then collides again with C to reform BC and leave as 2 entities: A + BC
| [[File:Pic_e.png|260px]]
|}

==== 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 allows us to investigate the progress of a reaction by splitting it into 2 parts: reactants and products.
* For a reaction to occur the reactants must cross into the product "space" by going over the saddle point – the transition state!
* With the Born-Oppenheimer approximation<ref>[https://www.iue.tuwien.ac.at/phd/schanovsky/thesisch2.html]</ref>, it is assumed there is a single saddle point (thus transition state) per reaction
* Transition State theory should only be used to estimate reaction rates at low temperature as it fails to correctly predict reaction rates for scenarios with high momentum (ie high temperature), where the transition state can be recrossed and thus predict reaction rates which are a poor match with experimental values
* We must also take into account that not every reaction with the correct energy will react, as for example, we must also take into account orientation, so, predicted rates with TS theory will be higher than experimental values.

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

=== PES inspection ===

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

This can be deduced by inspecting the PES surfaces bellow, where F is labelled as A and B C labels H H.
# F + H2 → HF + H : Exothermic
#* Looking at the reactants: A + BC in the PES plot (small BC distance, large AB distance), we can see they have a higher potential energy than the products: AB + C. Which is confirmed by looking at bond strengths<ref>[http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html]</ref>: breaking the H–H bond (432 kJ/mol) vs making the H–F bond (565 kJ/mol), so energy is released during the reaction.
# HF + H → H2 + F : Endothermic
#* If we know look at: AB + C as reactants (small AB distance, large BC distance) we can see they have a lower potential energy than the products: BC + A. It is again rationalised by the fact that the H-F bond being broken in the reaction is stronger than the H-H bond being formed, so it requires an amount of energy to initiate the reaction.

[[File:H+hf_pes1L.png|275px|thumb|centre| F-H-H PES plot: AB is FH distance and BC is HH distance ]]

==== Transition state for F-H-H system ====
Locate the approximate position of the transition state.

From Hammond's postulate we know that in an endothermic reaction the transition state resembles the products and in an exothermic reaction it resembles the reactants. So, we can infer that the transition state will resemble H + HF.

The transition State for the F-H-H reaction was found at: 181.11 pm AB and 74.487 pm BC, with 0 initial momenta for both. This was done by varying the distances until they remained constant, confirmed by the forces along AB and BC being 0, marking a stationary point as the first derivative is 0.

[[File: HFTS_cp.png|275px|thumb|centre| F-H-H Contour plot when placed at Transition State ]]

==== Activation Energy for F + H2 and H + HF reactions ====
Report the activation energy for both reactions

The MEP was found for both reactions for a structure neighbouring the transition state: '''rts'''+~1 pm.

The activation energy is the difference between the saddle point's maximum energy and the minimum energy of the products:

* '''Reaction''' '''1''': F + H2 → HF + H
** ''TS Energy: ''- 434 kJ / mol
** ''Final Energy: ''- 560 kJ / mol
** Activation Energy: + 126 kJ / mol
[[File:Eactbiggraph.png|350px|thumb|centre|Energy vs Time diagram of transition state to HH + F state (FH (AB) 180.11 pm and HH (BC) 75.487 pm) ]]
* '''Reaction''' '''2''': HF + H → H2 + F 
** ''TS Energy: ''-433.8 kJ / mol
** ''Final Energy: ''-434.5 kJ / mol
** Activation Energy: 0.7 kJ / mol
[[File:H2+f_emep.png|350px|thumb|centre|Energy vs Time diagram of transition state to form HF + H state (FH (AB) 181.11 pm and HH (BC) 74.487 pm) ]]

=== Reaction Dynamics ===

==== Reactive trajectory for H2 + F → HF + H ====
In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.

Using initial conditions; rAB = 225 pm, rBC = 75 pm, pAB = -5 g.mol-1.pm.fs-1, pBC = -1 g.mol-1.pm.fs-1, (NB. A=F, B=H, C=H): F + H2 → HF + H is a successful reaction. As seen above, this is an exothermic reaction and thus energy is released. From the animation we can see that a vibrating H2 molecule (BC) approaches F (A) and an HF bond forms, however it breaks and reforms BC, which then collides again with the Fluorine atom to reform a highly energetic HF and leave as the final products: HF + H.

[[File:Goodreachf.png|350px|thumb|centre|Contour plot of successful reaction of formation of HF]]

The energy in the reaction is converted from potential energy to kinetic energy, which is shown by the highly vibrating HF molecule, compared to the initial H2. This is confirmed with a Momentum vs Time plot (shown bellow): the AB distance, representing the HF bond, oscillates at a much larger amplitude than the initial BC oscillations which represent the H2 molecule.
[[File:Memtumsuccreac.png|350px|thumb|centre|Momentum vs Time plot of formation of HF]]

Experimentally, this would result in an increase in temperature of our reaction mixture, which could be measured using calorimetry. Additionally, the HF vibrational modes would be highly excited and overtones would show in the IR spectra as multiples of the fundamental absorption frequency<ref>[https://chem.libretexts.org/Courses/Pacific_Union_College/Quantum_Chemistry/13%3A_Molecular_Spectroscopy/13.05%3A_Vibrational_Overtones ]</ref> (3962 cm-1)<ref>[http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html]</ref>. Because overtone bands are difficult to see, this could be done using use Low Temperature FTIR<ref>[https://pubs.acs.org/doi/abs/10.1021/j150647a021]</ref>.

==== Illustrating Polanyi's empirical rules: Reaction 2 HF + H → H2 + F, ====
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

Polyani's Rules<ref>[https://pubs.acs.org/doi/10.1021/jz301649w]</ref>state that:
# For an early transition state, translational energy is more efficient in promoting a reaction
# For a late transition state, vibrational energy is more important in promoting a reaction
<nowiki> </nowiki>So, for the endothermic reaction HF + H → H2 + F, compared to the reverse exothermic reaction, we need a higher amount of vibrational energy, as well as translational to overcome the transition state and for a reaction to occur.

This can be illustrated by comparing the following initial reaction conditions (''A=F, B=H, C=H, rAB = 75 pm, rBC = 225 pm''):
# '''pAB = -1, pBC = -5''' – Unsuccessful Reaction 2 (NB. under these conditions Reaction 1 is successful, as shown above)
# '''pAB = -2, pBC = -4''' – Successful Reaction 2 when we redistribute momentum to increase vibrational energy
(units of p: g.mol-1.pm.fs-1)

[[File:Unsuccessreac2pol.png|350px|thumb|left|Contour plot of unsuccessful formation of HF with conditions parallel to a succesfull reverse reaction]][[File:Successreac2pol.png|350px|thumb|right|Contour plot of successful formation of HF ]]

== References ==

MRD:celiab

2020-05-07T16:03:28Z

Cb2018: /* Reactive and unreactive trajectories */

== EXERCISE 1: H + H2 ==

=== The Transition State ===
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?
* On PES diagrams, transition states are mathematically defined as points with 0 gradient with respect to position: stationary points.
* The transition state corresponds to the saddle point; the highest energy point on the reaction coordinate, which is the lowest energy pathway.
* We can distinguish it from a local minimum because on the reaction coordinate axis, the saddle point representing the transition state, will be a local maximum, but on the orthogonal axis it will be a local minimum. The minima are minimums on either direction because they represent stable reactants and products.

=== H + H2 reaction ===
[[File:Internuclear_distances_vs_time.png|250px|thumb|middle|Internuclear Distances vs Time plot ]]

===== Locating the Transition State =====
Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.
* The reaction is symmetric, hence, the transition state must have '''r1''' = '''r2''' by Hammond's Postulate
* By testing different initial conditions where '''rAB'''='''rBC '''and '''p1''' = '''p2''' = 0 g.mol-1.pm.fs-1, we can find the position of the Transition State
** The position of the transition state was found to be '''rts'''=90.775 pm
** This is where the system underwent small periodic symmetric vibrations keeping the distance between the atoms almost constant
* In the plot to the right, we see two straight lines, representing almost a stationary behaviour; thus we know we are at the symmetric transition state.

===== Calculating the Reaction Path =====
Comment on how the ''mep'' and the trajectory you just calculated differ.
* The minimum energy path is calculated from a system slightly displaced from the Transition state ('''r1''' = '''rts'''+1 pm, '''r2''' = '''rts''') and 0 initial momenta and follows the lowest energy path on a PES.
* As the reaction is symmetric it doesn't matter which atoms we label as A B C, so if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead, we get the same results but switched: AB will become BC and viceversa.
* The dynamic trajectory includes the momentum of the particles which causes oscillations about the trajectory path.
* The ''mep'' trajectory is roughly the same but it's smooth and doesn't oscillate, as the velocities are set to 0 in every step. This means we are artificially removing any gain in vibrational energy which results in oscillations. As the velocity is set 0 at every step, the kinetic energy is also 0.
[[File:picture 2 mep.png|350px|thumb|left|MEP contour plot ]]
[[File:picture 3 dynamic.png|350px|thumb|right|Dynamic contour plot ]]
[[File:Pic_4_dt.png|350px|thumb|left|MEP Internuclear Distances vs Time plot ]]
[[File:Pic_5_dt.png|350px|thumb|right| Dynamic Internuclear Distances vs Time plot ]]

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

For the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, the following trajectories with the following momenta combination were run:

(in the descriptions A B and C are used to describe the atoms colliding: AB is the initial H2 approaching H, represented by C)
{| class="wikitable"
!'''#'''
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|'''1'''
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.3</nowiki>
|yes
|The reaction dynamics proceed in a straight forward manner: AB and C approach each other, a collision occurs with the correct momentum for the reaction to proceed and a new bond BC is formed.
| [[File:Pic_a.png|260px]]
|-
|'''2'''
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420</nowiki>
|no
|AB approaches C and collides with it, but it doesn't have enough momentum to form a new bond so they are repelled. Momentum transfer does occur and C goes back with a greater momentum.
| [[File:Pic_b.png|260px]]
|-
|'''3'''
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414</nowiki>
|yes
|Oscillating AB collides with C and forms a new bond. The new H2 molecule, BC, has a higher vibrational frequency. A's direction is reversed.
| [[File:Pic_c.png|260px]]
|-
|'''4'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-375</nowiki>
|no
|AB approaches C and a BC bond forms, however, the system has too much momentum and C is repelled by the nucleus in B and bounces back. This leaves AB with greater vibrational energy and the distance BC increases as C is repelled.

| [[File:Pic_d.png|260px]]
|-
|'''5'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349</nowiki>
|yes
|AB collides with C and a new bond forms. However, unlike above, C still has enough energy to overcome the activation barrier. So, the BC bond forms, it breaks and reforms AB which then collides again with C to reform BC and leave as 2 entities: A + BC
| [[File:Pic_e.png|260px]]
|}

==== 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 allows us to investigate the progress of a reaction by splitting it into 2 parts: reactants and products.
* For a reaction to occur the reactants must cross into the product "space" by going over the saddle point – the transition state!
* With the Born-Oppenheimer approximation, it is assumed there is a single saddle point (thus transition state) per reaction
* Transition State theory should only be used to estimate reaction rates at low temperature as it fails to correctly predict reaction rates for scenarios with high momentum (ie high temperature), where the transition state can be recrossed and thus predict reaction rates which are a poor match with experimental values
* We must also take into account that not every reaction with the correct energy will react, as orientation is important too so predicted rates with TS theory will be higher than experimental values.

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

=== PES inspection ===

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

This can be deduced by inspecting the PES surfaces bellow, where F is labelled as A and B C labels H H.
# F + H2 → HF + H : Exothermic
#* Looking at the reactants: A + BC in the PES plot (small BC distance, large AB distance), we can see they have a higher potential energy than the products: AB + C. Which is confirmed by looking at bond strengths<ref>[http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html]</ref>: breaking the H–H bond (432 kJ/mol) vs making the H–F bond (565 kJ/mol), so energy is released during the reaction.
# HF + H → H2 + F : Endothermic
#* If we know look at: AB + C as reactants (small AB distance, large BC distance) we can see they have a lower potential energy than the products: BC + A. It is again rationalised by the fact that the H-F bond being broken in the reaction is stronger than the H-H bond being formed, so it requires an amount of energy to initiate the reaction.

[[File:H+hf_pes1L.png|275px|thumb|centre| F-H-H PES plot: AB is FH distance and BC is HH distance ]]

==== Transition state for F-H-H system ====
Locate the approximate position of the transition state.

From Hammond's postulate we know that in an endothermic reaction the transition state resembles the products and in an exothermic reaction it resembles the reactants. So, we can infer that the transition state will resemble H + HF.

The transition State for the F-H-H reaction was found at: 181.11 pm AB and 74.487 pm BC, with 0 initial momenta for both. This was done by varying the distances until they remained constant, confirmed by the forces along AB and BC being 0, marking a stationary point as the first derivative is 0.

[[File: HFTS_cp.png|275px|thumb|centre| F-H-H Contour plot when placed at Transition State ]]

==== Activation Energy for F + H2 and H + HF reactions ====
Report the activation energy for both reactions

The MEP was found for both reactions for a structure neighbouring the transition state: '''rts'''+~1 pm.

The activation energy is the difference between the saddle point's maximum energy and the minimum energy of the products:

* '''Reaction''' '''1''': F + H2 → HF + H
** ''TS Energy: ''- 434 kJ / mol
** ''Final Energy: ''- 560 kJ / mol
** Activation Energy: + 126 kJ / mol
[[File:Eactbiggraph.png|350px|thumb|centre|Energy vs Time diagram of transition state to HH + F state (FH (AB) 180.11 pm and HH (BC) 75.487 pm) ]]
* '''Reaction''' '''2''': HF + H → H2 + F 
** ''TS Energy: ''-433.8 kJ / mol
** ''Final Energy: ''-434.5 kJ / mol
** Activation Energy: 0.7 kJ / mol
[[File:H2+f_emep.png|350px|thumb|centre|Energy vs Time diagram of transition state to form HF + H state (FH (AB) 181.11 pm and HH (BC) 74.487 pm) ]]

=== Reaction Dynamics ===

==== Reactive trajectory for H2 + F → HF + H ====
In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.

Using initial conditions; rAB = 225 pm, rBC = 75 pm, pAB = -5 g.mol-1.pm.fs-1, pBC = -1 g.mol-1.pm.fs-1, (NB. A=F, B=H, C=H): F + H2 → HF + H is a successful reaction. As seen above, this is an exothermic reaction and thus energy is released. From the animation we can see that a vibrating H2 molecule (BC) approaches F (A) and an HF bond forms, however it breaks and reforms BC, which then collides again with the Fluorine atom to reform a highly energetic HF and leave as the final products: HF + H.

[[File:Goodreachf.png|350px|thumb|centre|Contour plot of successful reaction of formation of HF]]

The energy in the reaction is converted from potential energy to kinetic energy, which is shown by the highly vibrating HF molecule, compared to the initial H2. This is confirmed with a Momentum vs Time plot (shown bellow): the AB distance, representing the HF bond, oscillates at a much larger amplitude than the initial BC oscillations which represent the H2 molecule.
[[File:Memtumsuccreac.png|350px|thumb|centre|Momentum vs Time plot of formation of HF]]

Experimentally, this would result in an increase in temperature of our reaction mixture, which could be measured using calorimetry. Additionally, the HF vibrational modes would be highly excited and overtones would show in the IR spectra as multiples of the fundamental absorption frequency<ref>[https://chem.libretexts.org/Courses/Pacific_Union_College/Quantum_Chemistry/13%3A_Molecular_Spectroscopy/13.05%3A_Vibrational_Overtones ]</ref> (3962 cm-1)<ref>[http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html]</ref>. Because overtone bands are difficult to see, this could be done using use Low Temperature FTIR<ref>[https://pubs.acs.org/doi/abs/10.1021/j150647a021]</ref>.

==== Illustrating Polanyi's empirical rules: Reaction 2 HF + H → H2 + F, ====
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

Polyani's Rules<ref>[https://pubs.acs.org/doi/10.1021/jz301649w]</ref>state that:
# For an early transition state, translational energy is more efficient in promoting a reaction
# For a late transition state, vibrational energy is more important in promoting a reaction
<nowiki> </nowiki>So, for the endothermic reaction HF + H → H2 + F, compared to the reverse exothermic reaction, we need a higher amount of vibrational energy, as well as translational to overcome the transition state and for a reaction to occur.

This can be illustrated by comparing the following initial reaction conditions (''A=F, B=H, C=H, rAB = 75 pm, rBC = 225 pm''):
# '''pAB = -1, pBC = -5''' – Unsuccessful Reaction 2 (NB. under these conditions Reaction 1 is successful, as shown above)
# '''pAB = -2, pBC = -4''' – Successful Reaction 2 when we redistribute momentum to increase vibrational energy
(units of p: g.mol-1.pm.fs-1)

[[File:Unsuccessreac2pol.png|350px|thumb|left|Contour plot of unsuccessful formation of HF with conditions parallel to a succesfull reverse reaction]][[File:Successreac2pol.png|350px|thumb|right|Contour plot of successful formation of HF ]]

== References ==

MRD:celiab

2020-05-07T16:02:19Z

Cb2018:

== EXERCISE 1: H + H2 ==

=== The Transition State ===
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?
* On PES diagrams, transition states are mathematically defined as points with 0 gradient with respect to position: stationary points.
* The transition state corresponds to the saddle point; the highest energy point on the reaction coordinate, which is the lowest energy pathway.
* We can distinguish it from a local minimum because on the reaction coordinate axis, the saddle point representing the transition state, will be a local maximum, but on the orthogonal axis it will be a local minimum. The minima are minimums on either direction because they represent stable reactants and products.

=== H + H2 reaction ===
[[File:Internuclear_distances_vs_time.png|250px|thumb|middle|Internuclear Distances vs Time plot ]]

===== Locating the Transition State =====
Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.
* The reaction is symmetric, hence, the transition state must have '''r1''' = '''r2''' by Hammond's Postulate
* By testing different initial conditions where '''rAB'''='''rBC '''and '''p1''' = '''p2''' = 0 g.mol-1.pm.fs-1, we can find the position of the Transition State
** The position of the transition state was found to be '''rts'''=90.775 pm
** This is where the system underwent small periodic symmetric vibrations keeping the distance between the atoms almost constant
* In the plot to the right, we see two straight lines, representing almost a stationary behaviour; thus we know we are at the symmetric transition state.

===== Calculating the Reaction Path =====
Comment on how the ''mep'' and the trajectory you just calculated differ.
* The minimum energy path is calculated from a system slightly displaced from the Transition state ('''r1''' = '''rts'''+1 pm, '''r2''' = '''rts''') and 0 initial momenta and follows the lowest energy path on a PES.
* As the reaction is symmetric it doesn't matter which atoms we label as A B C, so if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead, we get the same results but switched: AB will become BC and viceversa.
* The dynamic trajectory includes the momentum of the particles which causes oscillations about the trajectory path.
* The ''mep'' trajectory is roughly the same but it's smooth and doesn't oscillate, as the velocities are set to 0 in every step. This means we are artificially removing any gain in vibrational energy which results in oscillations. As the velocity is set 0 at every step, the kinetic energy is also 0.
[[File:picture 2 mep.png|350px|thumb|left|MEP contour plot ]]
[[File:picture 3 dynamic.png|350px|thumb|right|Dynamic contour plot ]]
[[File:Pic_4_dt.png|350px|thumb|left|MEP Internuclear Distances vs Time plot ]]
[[File:Pic_5_dt.png|350px|thumb|right| Dynamic Internuclear Distances vs Time plot ]]

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

For the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, the following trajectories with the following momenta combination were run:

(in the descriptions A B and C are used to describe the atoms colliding: AB is the initial H2 approaching H, represented by C)
{| class="wikitable"
!'''#'''
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|'''1'''
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.3</nowiki>
|yes
|The reaction dynamics proceed in a straight forward manner: AB and C approach each other, a collision occurs with the correct momentum for the reaction to proceed and a new bond BC is formed.
| [[File:Pic_a.png|260px]]
|-
|'''2'''
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420</nowiki>
|no
|AB approaches C and collides with it, but it doesn't have enough momentum to form a new bond so they are repelled. Momentum transfer does occur and C goes back with a greater momentum.
| [[File:Pic_b.png|260px]]
|-
|'''3'''
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414</nowiki>
|yes
|Oscillating AB collides with C and forms a new bond. The new H2 molecule, BC, has a higher vibrational frequency. A's direction is reversed.
| [[File:Pic_c.png|260px]]
|-
|'''4'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-375</nowiki>
|no
|AB approaches C and a BC bond forms, however, the system has too much momentum and C is repelled by the nucleus in B and bounces back. This leaves AB with greater vibrational energy and the distance BC increases as C is repelled.

| [[File:Pic_d.png|260px]]
|-
|'''5'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349</nowiki>
|yes
|AB collides with C and a new bond forms. However, unlike above, C still has enough energy to overcome the activation barrier. So, the BC bond forms, it breaks and reforms AB which then collides again with C to reform BC and leave as 2 entities: A + BC
| [[File:Pic_e.png|260px]]
|}

Previous calculations have shown that that trajectories with initial conditions in the range '''r1''' =74 pm, '''r2''' = 200 pm, with -3.1 < '''p1'''/ g.mol-1.pm.fs-1 < -1.6 and '''p2''' = -5.1 g.mol-1.pm.fs-1 are reactive
* This is confirmed in combinations #1, #2 and #3.
From this we could hypothesise that all trajectories starting with the same positions but with higher values of momenta (higher kinetic energy) would be reactive as there would be enough kinetic energy to overcome the activation barrier. From the table we can see that this is not necessarily true:
* Combination #4 vs Combination #5.

==== 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 allows us to investigate the progress of a reaction by splitting it into 2 parts: reactants and products.
* For a reaction to occur the reactants must cross into the product "space" by going over the saddle point – the transition state!
* With the Born-Oppenheimer approximation, it is assumed there is a single saddle point (thus transition state) per reaction
* Transition State theory should only be used to estimate reaction rates at low temperature as it fails to correctly predict reaction rates for scenarios with high momentum (ie high temperature), where the transition state can be recrossed and thus predict reaction rates which are a poor match with experimental values
* We must also take into account that not every reaction with the correct energy will react, as orientation is important too so predicted rates with TS theory will be higher than experimental values.

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

=== PES inspection ===

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

This can be deduced by inspecting the PES surfaces bellow, where F is labelled as A and B C labels H H.
# F + H2 → HF + H : Exothermic
#* Looking at the reactants: A + BC in the PES plot (small BC distance, large AB distance), we can see they have a higher potential energy than the products: AB + C. Which is confirmed by looking at bond strengths<ref>[http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html]</ref>: breaking the H–H bond (432 kJ/mol) vs making the H–F bond (565 kJ/mol), so energy is released during the reaction.
# HF + H → H2 + F : Endothermic
#* If we know look at: AB + C as reactants (small AB distance, large BC distance) we can see they have a lower potential energy than the products: BC + A. It is again rationalised by the fact that the H-F bond being broken in the reaction is stronger than the H-H bond being formed, so it requires an amount of energy to initiate the reaction.

[[File:H+hf_pes1L.png|275px|thumb|centre| F-H-H PES plot: AB is FH distance and BC is HH distance ]]

==== Transition state for F-H-H system ====
Locate the approximate position of the transition state.

From Hammond's postulate we know that in an endothermic reaction the transition state resembles the products and in an exothermic reaction it resembles the reactants. So, we can infer that the transition state will resemble H + HF.

The transition State for the F-H-H reaction was found at: 181.11 pm AB and 74.487 pm BC, with 0 initial momenta for both. This was done by varying the distances until they remained constant, confirmed by the forces along AB and BC being 0, marking a stationary point as the first derivative is 0.

[[File: HFTS_cp.png|275px|thumb|centre| F-H-H Contour plot when placed at Transition State ]]

==== Activation Energy for F + H2 and H + HF reactions ====
Report the activation energy for both reactions

The MEP was found for both reactions for a structure neighbouring the transition state: '''rts'''+~1 pm.

The activation energy is the difference between the saddle point's maximum energy and the minimum energy of the products:

* '''Reaction''' '''1''': F + H2 → HF + H
** ''TS Energy: ''- 434 kJ / mol
** ''Final Energy: ''- 560 kJ / mol
** Activation Energy: + 126 kJ / mol
[[File:Eactbiggraph.png|350px|thumb|centre|Energy vs Time diagram of transition state to HH + F state (FH (AB) 180.11 pm and HH (BC) 75.487 pm) ]]
* '''Reaction''' '''2''': HF + H → H2 + F 
** ''TS Energy: ''-433.8 kJ / mol
** ''Final Energy: ''-434.5 kJ / mol
** Activation Energy: 0.7 kJ / mol
[[File:H2+f_emep.png|350px|thumb|centre|Energy vs Time diagram of transition state to form HF + H state (FH (AB) 181.11 pm and HH (BC) 74.487 pm) ]]

=== Reaction Dynamics ===

==== Reactive trajectory for H2 + F → HF + H ====
In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.

Using initial conditions; rAB = 225 pm, rBC = 75 pm, pAB = -5 g.mol-1.pm.fs-1, pBC = -1 g.mol-1.pm.fs-1, (NB. A=F, B=H, C=H): F + H2 → HF + H is a successful reaction. As seen above, this is an exothermic reaction and thus energy is released. From the animation we can see that a vibrating H2 molecule (BC) approaches F (A) and an HF bond forms, however it breaks and reforms BC, which then collides again with the Fluorine atom to reform a highly energetic HF and leave as the final products: HF + H.

[[File:Goodreachf.png|350px|thumb|centre|Contour plot of successful reaction of formation of HF]]

The energy in the reaction is converted from potential energy to kinetic energy, which is shown by the highly vibrating HF molecule, compared to the initial H2. This is confirmed with a Momentum vs Time plot (shown bellow): the AB distance, representing the HF bond, oscillates at a much larger amplitude than the initial BC oscillations which represent the H2 molecule.
[[File:Memtumsuccreac.png|350px|thumb|centre|Momentum vs Time plot of formation of HF]]

Experimentally, this would result in an increase in temperature of our reaction mixture, which could be measured using calorimetry. Additionally, the HF vibrational modes would be highly excited and overtones would show in the IR spectra as multiples of the fundamental absorption frequency<ref>[https://chem.libretexts.org/Courses/Pacific_Union_College/Quantum_Chemistry/13%3A_Molecular_Spectroscopy/13.05%3A_Vibrational_Overtones ]</ref> (3962 cm-1)<ref>[http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html]</ref>. Because overtone bands are difficult to see, this could be done using use Low Temperature FTIR<ref>[https://pubs.acs.org/doi/abs/10.1021/j150647a021]</ref>.

==== Illustrating Polanyi's empirical rules: Reaction 2 HF + H → H2 + F, ====
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

Polyani's Rules<ref>[https://pubs.acs.org/doi/10.1021/jz301649w]</ref>state that:
# For an early transition state, translational energy is more efficient in promoting a reaction
# For a late transition state, vibrational energy is more important in promoting a reaction
<nowiki> </nowiki>So, for the endothermic reaction HF + H → H2 + F, compared to the reverse exothermic reaction, we need a higher amount of vibrational energy, as well as translational to overcome the transition state and for a reaction to occur.

This can be illustrated by comparing the following initial reaction conditions (''A=F, B=H, C=H, rAB = 75 pm, rBC = 225 pm''):
# '''pAB = -1, pBC = -5''' – Unsuccessful Reaction 2 (NB. under these conditions Reaction 1 is successful, as shown above)
# '''pAB = -2, pBC = -4''' – Successful Reaction 2 when we redistribute momentum to increase vibrational energy
(units of p: g.mol-1.pm.fs-1)

[[File:Unsuccessreac2pol.png|350px|thumb|left|Contour plot of unsuccessful formation of HF with conditions parallel to a succesfull reverse reaction]][[File:Successreac2pol.png|350px|thumb|right|Contour plot of successful formation of HF ]]

== References ==

MRD:celiab

2020-05-07T16:01:32Z

Cb2018: /* Illustrating Polanyi's empirical rules: Reaction 2 HF + H → H2 + F, */

== EXERCISE 1: H + H2 ==

=== The Transition State ===
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?
* On PES diagrams, transition states are mathematically defined as points with 0 gradient with respect to position: stationary points.
* The transition state corresponds to the saddle point; the highest energy point on the reaction coordinate, which is the lowest energy pathway.
* We can distinguish it from a local minimum because on the reaction coordinate axis, the saddle point representing the transition state, will be a local maximum, but on the orthogonal axis it will be a local minimum. The minima are minimums on either direction because they represent stable reactants and products.

=== H + H2 reaction ===
[[File:Internuclear_distances_vs_time.png|250px|thumb|middle|Internuclear Distances vs Time plot ]]

===== Locating the Transition State =====
Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.
* The reaction is symmetric, hence, the transition state must have '''r1''' = '''r2''' by Hammond's Postulate
* By testing different initial conditions where '''rAB'''='''rBC '''and '''p1''' = '''p2''' = 0 g.mol-1.pm.fs-1, we can find the position of the Transition State
** The position of the transition state was found to be '''rts'''=90.775 pm
** This is where the system underwent small periodic symmetric vibrations keeping the distance between the atoms almost constant
* In the plot to the right, we see two straight lines, representing almost a stationary behaviour; thus we know we are at the symmetric transition state.

===== Calculating the Reaction Path =====
Comment on how the ''mep'' and the trajectory you just calculated differ.
* The minimum energy path is calculated from a system slightly displaced from the Transition state ('''r1''' = '''rts'''+1 pm, '''r2''' = '''rts''') and 0 initial momenta and follows the lowest energy path on a PES.
* As the reaction is symmetric it doesn't matter which atoms we label as A B C, so if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead, we get the same results but switched: AB will become BC and viceversa.
* The dynamic trajectory includes the momentum of the particles which causes oscillations about the trajectory path.
* The ''mep'' trajectory is roughly the same but it's smooth and doesn't oscillate, as the velocities are set to 0 in every step. This means we are artificially removing any gain in vibrational energy which results in oscillations. As the velocity is set 0 at every step, the kinetic energy is also 0.
[[File:picture 2 mep.png|350px|thumb|left|MEP contour plot ]]
[[File:picture 3 dynamic.png|350px|thumb|right|Dynamic contour plot ]]
[[File:Pic_4_dt.png|350px|thumb|left|MEP Internuclear Distances vs Time plot ]]
[[File:Pic_5_dt.png|350px|thumb|right| Dynamic Internuclear Distances vs Time plot ]]

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

For the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, the following trajectories with the following momenta combination were run:

(in the descriptions A B and C are used to describe the atoms colliding: AB is the initial H2 approaching H, represented by C)
{| class="wikitable"
!'''#'''
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|'''1'''
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.3</nowiki>
|yes
|The reaction dynamics proceed in a straight forward manner: AB and C approach each other, a collision occurs with the correct momentum for the reaction to proceed and a new bond BC is formed.
| [[File:Pic_a.png|260px]]
|-
|'''2'''
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420</nowiki>
|no
|AB approaches C and collides with it, but it doesn't have enough momentum to form a new bond so they are repelled. Momentum transfer does occur and C goes back with a greater momentum.
| [[File:Pic_b.png|260px]]
|-
|'''3'''
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414</nowiki>
|yes
|Oscillating AB collides with C and forms a new bond. The new H2 molecule, BC, has a higher vibrational frequency. A's direction is reversed.
| [[File:Pic_c.png|260px]]
|-
|'''4'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-375</nowiki>
|no
|AB approaches C and a BC bond forms, however, the system has too much momentum and C is repelled by the nucleus in B and bounces back. This leaves AB with greater vibrational energy and the distance BC increases as C is repelled.

| [[File:Pic_d.png|260px]]
|-
|'''5'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349</nowiki>
|yes
|AB collides with C and a new bond forms. However, unlike above, C still has enough energy to overcome the activation barrier. So, the BC bond forms, it breaks and reforms AB which then collides again with C to reform BC and leave as 2 entities: A + BC
| [[File:Pic_e.png|260px]]
|}

Previous calculations have shown that that trajectories with initial conditions in the range '''r1''' =74 pm, '''r2''' = 200 pm, with -3.1 < '''p1'''/ g.mol-1.pm.fs-1 < -1.6 and '''p2''' = -5.1 g.mol-1.pm.fs-1 are reactive
* This is confirmed in combinations #1, #2 and #3.
From this we could hypothesise that all trajectories starting with the same positions but with higher values of momenta (higher kinetic energy) would be reactive as there would be enough kinetic energy to overcome the activation barrier. From the table we can see that this is not necessarily true:
* Combination #4 vs Combination #5.

==== 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 allows us to investigate the progress of a reaction by splitting it into 2 parts: reactants and products.
* For a reaction to occur the reactants must cross into the product "space" by going over the saddle point – the transition state!
* With the Born-Oppenheimer approximation, it is assumed there is a single saddle point (thus transition state) per reaction
* Transition State theory should only be used to estimate reaction rates at low temperature as it fails to correctly predict reaction rates for scenarios with high momentum (ie high temperature), where the transition state can be recrossed and thus predict reaction rates which are a poor match with experimental values
* We must also take into account that not every reaction with the correct energy will react, as orientation is important too so predicted rates with TS theory will be higher than experimental values.

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

=== PES inspection ===

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

This can be deduced by inspecting the PES surfaces bellow, where F is labelled as A and B C labels H H.
# F + H2 → HF + H : Exothermic
#* Looking at the reactants: A + BC in the PES plot (small BC distance, large AB distance), we can see they have a higher potential energy than the products: AB + C. Which is confirmed by looking at bond strengths<ref>[http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html]</ref>: breaking the H–H bond (432 kJ/mol) vs making the H–F bond (565 kJ/mol), so energy is released during the reaction.
# HF + H → H2 + F : Endothermic
#* If we know look at: AB + C as reactants (small AB distance, large BC distance) we can see they have a lower potential energy than the products: BC + A. It is again rationalised by the fact that the H-F bond being broken in the reaction is stronger than the H-H bond being formed, so it requires an amount of energy to initiate the reaction.

[[File:H+hf_pes1L.png|275px|thumb|centre| F-H-H PES plot: AB is FH distance and BC is HH distance ]]

==== Transition state for F-H-H system ====
Locate the approximate position of the transition state.

From Hammond's postulate we know that in an endothermic reaction the transition state resembles the products and in an exothermic reaction it resembles the reactants. So, we can infer that the transition state will resemble H + HF.

The transition State for the F-H-H reaction was found at: 181.11 pm AB and 74.487 pm BC, with 0 initial momenta for both. This was done by varying the distances until they remained constant, confirmed by the forces along AB and BC being 0, marking a stationary point as the first derivative is 0.

[[File: HFTS_cp.png|275px|thumb|centre| F-H-H Contour plot when placed at Transition State ]]

==== Activation Energy for F + H2 and H + HF reactions ====
Report the activation energy for both reactions

The MEP was found for both reactions for a structure neighbouring the transition state: '''rts'''+~1 pm.

The activation energy is the difference between the saddle point's maximum energy and the minimum energy of the products:

* '''Reaction''' '''1''': F + H2 → HF + H
** ''TS Energy: ''- 434 kJ / mol
** ''Final Energy: ''- 560 kJ / mol
** Activation Energy: + 126 kJ / mol
[[File:Eactbiggraph.png|350px|thumb|centre|Energy vs Time diagram of transition state to HH + F state (FH (AB) 180.11 pm and HH (BC) 75.487 pm) ]]
* '''Reaction''' '''2''': HF + H → H2 + F 
** ''TS Energy: ''-433.8 kJ / mol
** ''Final Energy: ''-434.5 kJ / mol
** Activation Energy: 0.7 kJ / mol
[[File:H2+f_emep.png|350px|thumb|centre|Energy vs Time diagram of transition state to form HF + H state (FH (AB) 181.11 pm and HH (BC) 74.487 pm) ]]

=== Reaction Dynamics ===

==== Reactive trajectory for H2 + F → HF + H ====
In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.

Using initial conditions; rAB = 225 pm, rBC = 75 pm, pAB = -5 g.mol-1.pm.fs-1, pBC = -1 g.mol-1.pm.fs-1, (NB. A=F, B=H, C=H): F + H2 → HF + H is a successful reaction. As seen above, this is an exothermic reaction and thus energy is released. From the animation we can see that a vibrating H2 molecule (BC) approaches F (A) and an HF bond forms, however it breaks and reforms BC, which then collides again with the Fluorine atom to reform a highly energetic HF and leave as the final products: HF + H.

[[File:Goodreachf.png|350px|thumb|centre|Contour plot of successful reaction of formation of HF]]

The energy in the reaction is converted from potential energy to kinetic energy, which is shown by the highly vibrating HF molecule, compared to the initial H2. This is confirmed with a Momentum vs Time plot (shown bellow): the AB distance, representing the HF bond, oscillates at a much larger amplitude than the initial BC oscillations which represent the H2 molecule.
[[File:Memtumsuccreac.png|350px|thumb|centre|Momentum vs Time plot of formation of HF]]

Experimentally, this would result in an increase in temperature of our reaction mixture, which could be measured using calorimetry. Additionally, the HF vibrational modes would be highly excited and overtones would show in the IR spectra as multiples of the fundamental absorption frequency<ref>[https://chem.libretexts.org/Courses/Pacific_Union_College/Quantum_Chemistry/13%3A_Molecular_Spectroscopy/13.05%3A_Vibrational_Overtones ]</ref> (3962 cm-1)<ref>[http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html]</ref>. Because overtone bands are difficult to see, this could be done using use Low Temperature FTIR<ref>[https://pubs.acs.org/doi/abs/10.1021/j150647a021]</ref>.

==== Illustrating Polanyi's empirical rules: Reaction 2 HF + H → H2 + F, ====
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

Polyani's Rules<ref>[https://pubs.acs.org/doi/10.1021/jz301649w]</ref>state that:
# For an early transition state, translational energy is more efficient in promoting a reaction
# For a late transition state, vibrational energy is more important in promoting a reaction
<nowiki> </nowiki>So, for the endothermic reaction HF + H → H2 + F, compared to the reverse exothermic reaction, we need a higher amount of vibrational energy, as well as translational to overcome the transition state and for a reaction to occur.

This can be illustrated by comparing the following initial reaction conditions (''A=F, B=H, C=H, rAB = 75 pm, rBC = 225 pm''):
# '''pAB = -1, pBC = -5''' – Unsuccessful Reaction 2 (NB. under these conditions Reaction 1 is successful, as shown above)
# '''pAB = -2, pBC = -4''' – Successful Reaction 2 when we redistribute momentum to increase vibrational energy
(units of p: g.mol-1.pm.fs-1)

[[File:Unsuccessreac2pol.png|350px|thumb|left|Contour plot of unsuccessful formation of HF with conditions parallel to a succesfull reverse reaction]][[File:Successreac2pol.png|350px|thumb|right|Contour plot of successful formation of HF ]]

== References ==

MRD:celiab

2020-05-07T15:56:39Z

Cb2018: /* Illustrating Polanyi's empirical rules: Reaction 2 HF + H → H2 + F, */

== EXERCISE 1: H + H2 ==

=== The Transition State ===
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?
* On PES diagrams, transition states are mathematically defined as points with 0 gradient with respect to position: stationary points.
* The transition state corresponds to the saddle point; the highest energy point on the reaction coordinate, which is the lowest energy pathway.
* We can distinguish it from a local minimum because on the reaction coordinate axis, the saddle point representing the transition state, will be a local maximum, but on the orthogonal axis it will be a local minimum. The minima are minimums on either direction because they represent stable reactants and products.

=== H + H2 reaction ===
[[File:Internuclear_distances_vs_time.png|250px|thumb|middle|Internuclear Distances vs Time plot ]]

===== Locating the Transition State =====
Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.
* The reaction is symmetric, hence, the transition state must have '''r1''' = '''r2''' by Hammond's Postulate
* By testing different initial conditions where '''rAB'''='''rBC '''and '''p1''' = '''p2''' = 0 g.mol-1.pm.fs-1, we can find the position of the Transition State
** The position of the transition state was found to be '''rts'''=90.775 pm
** This is where the system underwent small periodic symmetric vibrations keeping the distance between the atoms almost constant
* In the plot to the right, we see two straight lines, representing almost a stationary behaviour; thus we know we are at the symmetric transition state.

===== Calculating the Reaction Path =====
Comment on how the ''mep'' and the trajectory you just calculated differ.
* The minimum energy path is calculated from a system slightly displaced from the Transition state ('''r1''' = '''rts'''+1 pm, '''r2''' = '''rts''') and 0 initial momenta and follows the lowest energy path on a PES.
* As the reaction is symmetric it doesn't matter which atoms we label as A B C, so if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead, we get the same results but switched: AB will become BC and viceversa.
* The dynamic trajectory includes the momentum of the particles which causes oscillations about the trajectory path.
* The ''mep'' trajectory is roughly the same but it's smooth and doesn't oscillate, as the velocities are set to 0 in every step. This means we are artificially removing any gain in vibrational energy which results in oscillations. As the velocity is set 0 at every step, the kinetic energy is also 0.
[[File:picture 2 mep.png|350px|thumb|left|MEP contour plot ]]
[[File:picture 3 dynamic.png|350px|thumb|right|Dynamic contour plot ]]
[[File:Pic_4_dt.png|350px|thumb|left|MEP Internuclear Distances vs Time plot ]]
[[File:Pic_5_dt.png|350px|thumb|right| Dynamic Internuclear Distances vs Time plot ]]

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

For the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, the following trajectories with the following momenta combination were run:

(in the descriptions A B and C are used to describe the atoms colliding: AB is the initial H2 approaching H, represented by C)
{| class="wikitable"
!'''#'''
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|'''1'''
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.3</nowiki>
|yes
|The reaction dynamics proceed in a straight forward manner: AB and C approach each other, a collision occurs with the correct momentum for the reaction to proceed and a new bond BC is formed.
| [[File:Pic_a.png|260px]]
|-
|'''2'''
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420</nowiki>
|no
|AB approaches C and collides with it, but it doesn't have enough momentum to form a new bond so they are repelled. Momentum transfer does occur and C goes back with a greater momentum.
| [[File:Pic_b.png|260px]]
|-
|'''3'''
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414</nowiki>
|yes
|Oscillating AB collides with C and forms a new bond. The new H2 molecule, BC, has a higher vibrational frequency. A's direction is reversed.
| [[File:Pic_c.png|260px]]
|-
|'''4'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-375</nowiki>
|no
|AB approaches C and a BC bond forms, however, the system has too much momentum and C is repelled by the nucleus in B and bounces back. This leaves AB with greater vibrational energy and the distance BC increases as C is repelled.

| [[File:Pic_d.png|260px]]
|-
|'''5'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349</nowiki>
|yes
|AB collides with C and a new bond forms. However, unlike above, C still has enough energy to overcome the activation barrier. So, the BC bond forms, it breaks and reforms AB which then collides again with C to reform BC and leave as 2 entities: A + BC
| [[File:Pic_e.png|260px]]
|}

Previous calculations have shown that that trajectories with initial conditions in the range '''r1''' =74 pm, '''r2''' = 200 pm, with -3.1 < '''p1'''/ g.mol-1.pm.fs-1 < -1.6 and '''p2''' = -5.1 g.mol-1.pm.fs-1 are reactive
* This is confirmed in combinations #1, #2 and #3.
From this we could hypothesise that all trajectories starting with the same positions but with higher values of momenta (higher kinetic energy) would be reactive as there would be enough kinetic energy to overcome the activation barrier. From the table we can see that this is not necessarily true:
* Combination #4 vs Combination #5.

==== 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 allows us to investigate the progress of a reaction by splitting it into 2 parts: reactants and products.
* For a reaction to occur the reactants must cross into the product "space" by going over the saddle point – the transition state!
* With the Born-Oppenheimer approximation, it is assumed there is a single saddle point (thus transition state) per reaction
* Transition State theory should only be used to estimate reaction rates at low temperature as it fails to correctly predict reaction rates for scenarios with high momentum (ie high temperature), where the transition state can be recrossed and thus predict reaction rates which are a poor match with experimental values
* We must also take into account that not every reaction with the correct energy will react, as orientation is important too so predicted rates with TS theory will be higher than experimental values.

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

=== PES inspection ===

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

This can be deduced by inspecting the PES surfaces bellow, where F is labelled as A and B C labels H H.
# F + H2 → HF + H : Exothermic
#* Looking at the reactants: A + BC in the PES plot (small BC distance, large AB distance), we can see they have a higher potential energy than the products: AB + C. Which is confirmed by looking at bond strengths<ref>[http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html]</ref>: breaking the H–H bond (432 kJ/mol) vs making the H–F bond (565 kJ/mol), so energy is released during the reaction.
# HF + H → H2 + F : Endothermic
#* If we know look at: AB + C as reactants (small AB distance, large BC distance) we can see they have a lower potential energy than the products: BC + A. It is again rationalised by the fact that the H-F bond being broken in the reaction is stronger than the H-H bond being formed, so it requires an amount of energy to initiate the reaction.

[[File:H+hf_pes1L.png|275px|thumb|centre| F-H-H PES plot: AB is FH distance and BC is HH distance ]]

==== Transition state for F-H-H system ====
Locate the approximate position of the transition state.

From Hammond's postulate we know that in an endothermic reaction the transition state resembles the products and in an exothermic reaction it resembles the reactants. So, we can infer that the transition state will resemble H + HF.

The transition State for the F-H-H reaction was found at: 181.11 pm AB and 74.487 pm BC, with 0 initial momenta for both. This was done by varying the distances until they remained constant, confirmed by the forces along AB and BC being 0, marking a stationary point as the first derivative is 0.

[[File: HFTS_cp.png|275px|thumb|centre| F-H-H Contour plot when placed at Transition State ]]

==== Activation Energy for F + H2 and H + HF reactions ====
Report the activation energy for both reactions

The MEP was found for both reactions for a structure neighbouring the transition state: '''rts'''+~1 pm.

The activation energy is the difference between the saddle point's maximum energy and the minimum energy of the products:

* '''Reaction''' '''1''': F + H2 → HF + H
** ''TS Energy: ''- 434 kJ / mol
** ''Final Energy: ''- 560 kJ / mol
** Activation Energy: + 126 kJ / mol
[[File:Eactbiggraph.png|350px|thumb|centre|Energy vs Time diagram of transition state to HH + F state (FH (AB) 180.11 pm and HH (BC) 75.487 pm) ]]
* '''Reaction''' '''2''': HF + H → H2 + F 
** ''TS Energy: ''-433.8 kJ / mol
** ''Final Energy: ''-434.5 kJ / mol
** Activation Energy: 0.7 kJ / mol
[[File:H2+f_emep.png|350px|thumb|centre|Energy vs Time diagram of transition state to form HF + H state (FH (AB) 181.11 pm and HH (BC) 74.487 pm) ]]

=== Reaction Dynamics ===

==== Reactive trajectory for H2 + F → HF + H ====
In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.

Using initial conditions; rAB = 225 pm, rBC = 75 pm, pAB = -5 g.mol-1.pm.fs-1, pBC = -1 g.mol-1.pm.fs-1, (NB. A=F, B=H, C=H): F + H2 → HF + H is a successful reaction. As seen above, this is an exothermic reaction and thus energy is released. From the animation we can see that a vibrating H2 molecule (BC) approaches F (A) and an HF bond forms, however it breaks and reforms BC, which then collides again with the Fluorine atom to reform a highly energetic HF and leave as the final products: HF + H.

[[File:Goodreachf.png|350px|thumb|centre|Contour plot of successful reaction of formation of HF]]

The energy in the reaction is converted from potential energy to kinetic energy, which is shown by the highly vibrating HF molecule, compared to the initial H2. This is confirmed with a Momentum vs Time plot (shown bellow): the AB distance, representing the HF bond, oscillates at a much larger amplitude than the initial BC oscillations which represent the H2 molecule.
[[File:Memtumsuccreac.png|350px|thumb|centre|Momentum vs Time plot of formation of HF]]

Experimentally, this would result in an increase in temperature of our reaction mixture, which could be measured using calorimetry. Additionally, the HF vibrational modes would be highly excited and overtones would show in the IR spectra as multiples of the fundamental absorption frequency<ref><nowiki>https://chem.libretexts.org/Courses/Pacific_Union_College/Quantum_Chemistry/13%3A_Molecular_Spectroscopy/13.05%3A_Vibrational_Overtones</nowiki></ref> (3962<ref>[http://vpl.astro.washington.edu/spectra/hf.htm]</ref> cm-1). Because overtone bands are difficult to see, this could be done using use Low Temperature FTIR<ref>[https://pubs.acs.org/doi/abs/10.1021/j150647a021]</ref>.

==== Illustrating Polanyi's empirical rules: Reaction 2 HF + H → H2 + F, ====
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

Polyani's Rules<ref>[https://pubs.acs.org/doi/10.1021/jz301649w]</ref>state that:
# For an early transition state, translational energy is more efficient in promoting a reaction
# For a late transition state, vibrational energy is more important in promoting a reaction
<nowiki> </nowiki>So, for the endothermic reaction HF + H → H2 + F, compared to the reverse exothermic reaction, we need a higher amount of vibrational energy, as well as translational to overcome the transition state and for a reaction to occur.

This can be illustrated by comparing the following initial reaction conditions (''A=F, B=H, C=H, rAB = 75 pm, rBC = 225 pm''):
# '''pAB = -1, pBC = -5''' – Unsuccessful Reaction 2 (NB. under these conditions Reaction 1 is successful, as shown above)
# '''pAB = -2, pBC = -4''' – Successful Reaction 2 when we redistribute momentum to increase vibrational energy
(units of p: g.mol-1.pm.fs-1)

[[File:Unsuccessreac2pol.png|350px|thumb|left|Contour plot of unsuccessful formation of HF with conditions parallel to a succesfull reverse reaction]][[File:Successreac2pol.png|350px|thumb|right|Contour plot of successful formation of HF ]]

MRD:celiab

2020-05-07T15:49:07Z

Cb2018: /* Illustrating Polanyi's empirical rules: Reaction 2 HF + H → H2 + F, */

== EXERCISE 1: H + H2 ==

=== The Transition State ===
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?
* On PES diagrams, transition states are mathematically defined as points with 0 gradient with respect to position: stationary points.
* The transition state corresponds to the saddle point; the highest energy point on the reaction coordinate, which is the lowest energy pathway.
* We can distinguish it from a local minimum because on the reaction coordinate axis, the saddle point representing the transition state, will be a local maximum, but on the orthogonal axis it will be a local minimum. The minima are minimums on either direction because they represent stable reactants and products.

=== H + H2 reaction ===
[[File:Internuclear_distances_vs_time.png|250px|thumb|middle|Internuclear Distances vs Time plot ]]

===== Locating the Transition State =====
Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.
* The reaction is symmetric, hence, the transition state must have '''r1''' = '''r2''' by Hammond's Postulate
* By testing different initial conditions where '''rAB'''='''rBC '''and '''p1''' = '''p2''' = 0 g.mol-1.pm.fs-1, we can find the position of the Transition State
** The position of the transition state was found to be '''rts'''=90.775 pm
** This is where the system underwent small periodic symmetric vibrations keeping the distance between the atoms almost constant
* In the plot to the right, we see two straight lines, representing almost a stationary behaviour; thus we know we are at the symmetric transition state.

===== Calculating the Reaction Path =====
Comment on how the ''mep'' and the trajectory you just calculated differ.
* The minimum energy path is calculated from a system slightly displaced from the Transition state ('''r1''' = '''rts'''+1 pm, '''r2''' = '''rts''') and 0 initial momenta and follows the lowest energy path on a PES.
* As the reaction is symmetric it doesn't matter which atoms we label as A B C, so if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead, we get the same results but switched: AB will become BC and viceversa.
* The dynamic trajectory includes the momentum of the particles which causes oscillations about the trajectory path.
* The ''mep'' trajectory is roughly the same but it's smooth and doesn't oscillate, as the velocities are set to 0 in every step. This means we are artificially removing any gain in vibrational energy which results in oscillations. As the velocity is set 0 at every step, the kinetic energy is also 0.
[[File:picture 2 mep.png|350px|thumb|left|MEP contour plot ]]
[[File:picture 3 dynamic.png|350px|thumb|right|Dynamic contour plot ]]
[[File:Pic_4_dt.png|350px|thumb|left|MEP Internuclear Distances vs Time plot ]]
[[File:Pic_5_dt.png|350px|thumb|right| Dynamic Internuclear Distances vs Time plot ]]

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

For the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, the following trajectories with the following momenta combination were run:

(in the descriptions A B and C are used to describe the atoms colliding: AB is the initial H2 approaching H, represented by C)
{| class="wikitable"
!'''#'''
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|'''1'''
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.3</nowiki>
|yes
|The reaction dynamics proceed in a straight forward manner: AB and C approach each other, a collision occurs with the correct momentum for the reaction to proceed and a new bond BC is formed.
| [[File:Pic_a.png|260px]]
|-
|'''2'''
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420</nowiki>
|no
|AB approaches C and collides with it, but it doesn't have enough momentum to form a new bond so they are repelled. Momentum transfer does occur and C goes back with a greater momentum.
| [[File:Pic_b.png|260px]]
|-
|'''3'''
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414</nowiki>
|yes
|Oscillating AB collides with C and forms a new bond. The new H2 molecule, BC, has a higher vibrational frequency. A's direction is reversed.
| [[File:Pic_c.png|260px]]
|-
|'''4'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-375</nowiki>
|no
|AB approaches C and a BC bond forms, however, the system has too much momentum and C is repelled by the nucleus in B and bounces back. This leaves AB with greater vibrational energy and the distance BC increases as C is repelled.

| [[File:Pic_d.png|260px]]
|-
|'''5'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349</nowiki>
|yes
|AB collides with C and a new bond forms. However, unlike above, C still has enough energy to overcome the activation barrier. So, the BC bond forms, it breaks and reforms AB which then collides again with C to reform BC and leave as 2 entities: A + BC
| [[File:Pic_e.png|260px]]
|}

Previous calculations have shown that that trajectories with initial conditions in the range '''r1''' =74 pm, '''r2''' = 200 pm, with -3.1 < '''p1'''/ g.mol-1.pm.fs-1 < -1.6 and '''p2''' = -5.1 g.mol-1.pm.fs-1 are reactive
* This is confirmed in combinations #1, #2 and #3.
From this we could hypothesise that all trajectories starting with the same positions but with higher values of momenta (higher kinetic energy) would be reactive as there would be enough kinetic energy to overcome the activation barrier. From the table we can see that this is not necessarily true:
* Combination #4 vs Combination #5.

==== 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 allows us to investigate the progress of a reaction by splitting it into 2 parts: reactants and products.
* For a reaction to occur the reactants must cross into the product "space" by going over the saddle point – the transition state!
* With the Born-Oppenheimer approximation, it is assumed there is a single saddle point (thus transition state) per reaction
* Transition State theory should only be used to estimate reaction rates at low temperature as it fails to correctly predict reaction rates for scenarios with high momentum (ie high temperature), where the transition state can be recrossed and thus predict reaction rates which are a poor match with experimental values
* We must also take into account that not every reaction with the correct energy will react, as orientation is important too so predicted rates with TS theory will be higher than experimental values.

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

=== PES inspection ===

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

This can be deduced by inspecting the PES surfaces bellow, where F is labelled as A and B C labels H H.
# F + H2 → HF + H : Exothermic
#* Looking at the reactants: A + BC in the PES plot (small BC distance, large AB distance), we can see they have a higher potential energy than the products: AB + C. Which is confirmed by looking at bond strengths: breaking the H–H bond (432 kJ / mol) vs making the H–F bond (565 kJ / mol), so energy is released during the reaction.
# HF + H → H2 + F : Endothermic
#* If we know look at: AB + C as reactants (small AB distance, large BC distance) we can see they have a lower potential energy than the products: BC + A. It is again rationalised by the fact that the H-F bond being broken in the reaction is stronger than the H-H bond being formed, so it requires an amount of energy to initiate the reaction.

[[File:H+hf_pes1L.png|275px|thumb|centre| F-H-H PES plot: AB is FH distance and BC is HH distance ]]

==== Transition state for F-H-H system ====
Locate the approximate position of the transition state.

From Hammond's postulate we know that in an endothermic reaction the transition state resembles the products and in an exothermic reaction it resembles the reactants. So, we can infer that the transition state will resemble H + HF.

The transition State for the F-H-H reaction was found at: 181.11 pm AB and 74.487 pm BC, with 0 initial momenta for both. This was done by varying the distances until they remained constant, confirmed by the forces along AB and BC being 0, marking a stationary point as the first derivative is 0.

[[File: HFTS_cp.png|275px|thumb|centre| F-H-H Contour plot when placed at Transition State ]]

==== Activation Energy for F + H2 and H + HF reactions ====
Report the activation energy for both reactions

The MEP was found for both reactions for a structure neighbouring the transition state: '''rts'''+~1 pm.

The activation energy is the difference between the saddle point's maximum energy and the minimum energy of the products:

* '''Reaction''' '''1''': F + H2 → HF + H
** ''TS Energy: ''- 434 kJ / mol
** ''Final Energy: ''- 560 kJ / mol
** Activation Energy: + 126 kJ / mol
[[File:Eactbiggraph.png|350px|thumb|centre|Energy vs Time diagram of transition state to HH + F state (FH (AB) 180.11 pm and HH (BC) 75.487 pm) ]]
* '''Reaction''' '''2''': HF + H → H2 + F 
** ''TS Energy: ''-433.8 kJ / mol
** ''Final Energy: ''-434.5 kJ / mol
** Activation Energy: 0.7 kJ / mol
[[File:H2+f_emep.png|350px|thumb|centre|Energy vs Time diagram of transition state to form HF + H state (FH (AB) 181.11 pm and HH (BC) 74.487 pm) ]]

=== Reaction Dynamics ===

==== Reactive trajectory for H2 + F → HF + H ====
In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.

Using initial conditions; rAB = 225 pm, rBC = 75 pm, pAB = -5 g.mol-1.pm.fs-1, pBC = -1 g.mol-1.pm.fs-1, (NB. A=F, B=H, C=H): F + H2 → HF + H is a successful reaction. As seen above, this is an exothermic reaction and thus energy is released. From the animation we can see that a vibrating H2 molecule (BC) approaches F (A) and an HF bond forms, however it breaks and reforms BC, which then collides again with the Fluorine atom to reform a highly energetic HF and leave as the final products: HF + H.

[[File:Goodreachf.png|350px|thumb|centre|Contour plot of successful reaction of formation of HF]]

The energy in the reaction is converted from potential energy to kinetic energy, which is shown by the highly vibrating HF molecule, compared to the initial H2. This is confirmed with a Momentum vs Time plot (shown bellow): the AB distance, representing the HF bond, oscillates at a much larger amplitude than the initial BC oscillations which represent the H2 molecule.
[[File:Memtumsuccreac.png|350px|thumb|centre|Momentum vs Time plot of formation of HF]]

Experimentally, this would result in an increase in temperature of our reaction mixture, which could be measured using calorimetry. Additionally, the HF vibrational modes would be highly excited and overtones would show in the IR spectra as multiples of the fundamental absorption frequency (3962 cm-1). Because overtone bands are difficult to see, this could be done using use Low Temperature FTIR.

==== Illustrating Polanyi's empirical rules: Reaction 2 HF + H → H2 + F, ====
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

Polyani's Rules state that:
# For an early transition state, translational energy is more efficient in promoting a reaction
For a late transition state, vibrational energy is more important in promoting a reaction

So, for the endothermic reaction HF + H → H2 + F, compared to the reverse exothermic reaction, we need a higher amount of vibrational energy, as well as translational to overcome the transition state and for a reaction to occur.

This can be illustrated by comparing the following initial reaction conditions (''A=F, B=H, C=H, rAB = 75 pm, rBC = 225 pm''):
# '''pAB = -1, pBC = -5''' – Unsuccessful Reaction 2 (NB. under these conditions Reaction 1 is successful, as shown above)
# '''pAB = -2, pBC = -4''' – Successful Reaction 2 when we redistribute momentum to increase vibrational energy
(units of p: g.mol-1.pm.fs-1)

[[File:Unsuccessreac2pol.png|350px|thumb|left|Contour plot of unsuccessful formation of HF with conditions parallel to a succesfull reverse reaction]][[File:Successreac2pol.png|350px|thumb|right|Contour plot of successful formation of HF ]]

MRD:celiab

2020-05-07T15:48:13Z

Cb2018: /* Illustrating Polanyi's empirical rules: Reaction 2 HF + H → H2 + F, */

== EXERCISE 1: H + H2 ==

=== The Transition State ===
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?
* On PES diagrams, transition states are mathematically defined as points with 0 gradient with respect to position: stationary points.
* The transition state corresponds to the saddle point; the highest energy point on the reaction coordinate, which is the lowest energy pathway.
* We can distinguish it from a local minimum because on the reaction coordinate axis, the saddle point representing the transition state, will be a local maximum, but on the orthogonal axis it will be a local minimum. The minima are minimums on either direction because they represent stable reactants and products.

=== H + H2 reaction ===
[[File:Internuclear_distances_vs_time.png|250px|thumb|middle|Internuclear Distances vs Time plot ]]

===== Locating the Transition State =====
Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.
* The reaction is symmetric, hence, the transition state must have '''r1''' = '''r2''' by Hammond's Postulate
* By testing different initial conditions where '''rAB'''='''rBC '''and '''p1''' = '''p2''' = 0 g.mol-1.pm.fs-1, we can find the position of the Transition State
** The position of the transition state was found to be '''rts'''=90.775 pm
** This is where the system underwent small periodic symmetric vibrations keeping the distance between the atoms almost constant
* In the plot to the right, we see two straight lines, representing almost a stationary behaviour; thus we know we are at the symmetric transition state.

===== Calculating the Reaction Path =====
Comment on how the ''mep'' and the trajectory you just calculated differ.
* The minimum energy path is calculated from a system slightly displaced from the Transition state ('''r1''' = '''rts'''+1 pm, '''r2''' = '''rts''') and 0 initial momenta and follows the lowest energy path on a PES.
* As the reaction is symmetric it doesn't matter which atoms we label as A B C, so if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead, we get the same results but switched: AB will become BC and viceversa.
* The dynamic trajectory includes the momentum of the particles which causes oscillations about the trajectory path.
* The ''mep'' trajectory is roughly the same but it's smooth and doesn't oscillate, as the velocities are set to 0 in every step. This means we are artificially removing any gain in vibrational energy which results in oscillations. As the velocity is set 0 at every step, the kinetic energy is also 0.
[[File:picture 2 mep.png|350px|thumb|left|MEP contour plot ]]
[[File:picture 3 dynamic.png|350px|thumb|right|Dynamic contour plot ]]
[[File:Pic_4_dt.png|350px|thumb|left|MEP Internuclear Distances vs Time plot ]]
[[File:Pic_5_dt.png|350px|thumb|right| Dynamic Internuclear Distances vs Time plot ]]

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

For the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, the following trajectories with the following momenta combination were run:

(in the descriptions A B and C are used to describe the atoms colliding: AB is the initial H2 approaching H, represented by C)
{| class="wikitable"
!'''#'''
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|'''1'''
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.3</nowiki>
|yes
|The reaction dynamics proceed in a straight forward manner: AB and C approach each other, a collision occurs with the correct momentum for the reaction to proceed and a new bond BC is formed.
| [[File:Pic_a.png|260px]]
|-
|'''2'''
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420</nowiki>
|no
|AB approaches C and collides with it, but it doesn't have enough momentum to form a new bond so they are repelled. Momentum transfer does occur and C goes back with a greater momentum.
| [[File:Pic_b.png|260px]]
|-
|'''3'''
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414</nowiki>
|yes
|Oscillating AB collides with C and forms a new bond. The new H2 molecule, BC, has a higher vibrational frequency. A's direction is reversed.
| [[File:Pic_c.png|260px]]
|-
|'''4'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-375</nowiki>
|no
|AB approaches C and a BC bond forms, however, the system has too much momentum and C is repelled by the nucleus in B and bounces back. This leaves AB with greater vibrational energy and the distance BC increases as C is repelled.

| [[File:Pic_d.png|260px]]
|-
|'''5'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349</nowiki>
|yes
|AB collides with C and a new bond forms. However, unlike above, C still has enough energy to overcome the activation barrier. So, the BC bond forms, it breaks and reforms AB which then collides again with C to reform BC and leave as 2 entities: A + BC
| [[File:Pic_e.png|260px]]
|}

Previous calculations have shown that that trajectories with initial conditions in the range '''r1''' =74 pm, '''r2''' = 200 pm, with -3.1 < '''p1'''/ g.mol-1.pm.fs-1 < -1.6 and '''p2''' = -5.1 g.mol-1.pm.fs-1 are reactive
* This is confirmed in combinations #1, #2 and #3.
From this we could hypothesise that all trajectories starting with the same positions but with higher values of momenta (higher kinetic energy) would be reactive as there would be enough kinetic energy to overcome the activation barrier. From the table we can see that this is not necessarily true:
* Combination #4 vs Combination #5.

==== 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 allows us to investigate the progress of a reaction by splitting it into 2 parts: reactants and products.
* For a reaction to occur the reactants must cross into the product "space" by going over the saddle point – the transition state!
* With the Born-Oppenheimer approximation, it is assumed there is a single saddle point (thus transition state) per reaction
* Transition State theory should only be used to estimate reaction rates at low temperature as it fails to correctly predict reaction rates for scenarios with high momentum (ie high temperature), where the transition state can be recrossed and thus predict reaction rates which are a poor match with experimental values
* We must also take into account that not every reaction with the correct energy will react, as orientation is important too so predicted rates with TS theory will be higher than experimental values.

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

=== PES inspection ===

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

This can be deduced by inspecting the PES surfaces bellow, where F is labelled as A and B C labels H H.
# F + H2 → HF + H : Exothermic
#* Looking at the reactants: A + BC in the PES plot (small BC distance, large AB distance), we can see they have a higher potential energy than the products: AB + C. Which is confirmed by looking at bond strengths: breaking the H–H bond (432 kJ / mol) vs making the H–F bond (565 kJ / mol), so energy is released during the reaction.
# HF + H → H2 + F : Endothermic
#* If we know look at: AB + C as reactants (small AB distance, large BC distance) we can see they have a lower potential energy than the products: BC + A. It is again rationalised by the fact that the H-F bond being broken in the reaction is stronger than the H-H bond being formed, so it requires an amount of energy to initiate the reaction.

[[File:H+hf_pes1L.png|275px|thumb|centre| F-H-H PES plot: AB is FH distance and BC is HH distance ]]

==== Transition state for F-H-H system ====
Locate the approximate position of the transition state.

From Hammond's postulate we know that in an endothermic reaction the transition state resembles the products and in an exothermic reaction it resembles the reactants. So, we can infer that the transition state will resemble H + HF.

The transition State for the F-H-H reaction was found at: 181.11 pm AB and 74.487 pm BC, with 0 initial momenta for both. This was done by varying the distances until they remained constant, confirmed by the forces along AB and BC being 0, marking a stationary point as the first derivative is 0.

[[File: HFTS_cp.png|275px|thumb|centre| F-H-H Contour plot when placed at Transition State ]]

==== Activation Energy for F + H2 and H + HF reactions ====
Report the activation energy for both reactions

The MEP was found for both reactions for a structure neighbouring the transition state: '''rts'''+~1 pm.

The activation energy is the difference between the saddle point's maximum energy and the minimum energy of the products:

* '''Reaction''' '''1''': F + H2 → HF + H
** ''TS Energy: ''- 434 kJ / mol
** ''Final Energy: ''- 560 kJ / mol
** Activation Energy: + 126 kJ / mol
[[File:Eactbiggraph.png|350px|thumb|centre|Energy vs Time diagram of transition state to HH + F state (FH (AB) 180.11 pm and HH (BC) 75.487 pm) ]]
* '''Reaction''' '''2''': HF + H → H2 + F 
** ''TS Energy: ''-433.8 kJ / mol
** ''Final Energy: ''-434.5 kJ / mol
** Activation Energy: 0.7 kJ / mol
[[File:H2+f_emep.png|350px|thumb|centre|Energy vs Time diagram of transition state to form HF + H state (FH (AB) 181.11 pm and HH (BC) 74.487 pm) ]]

=== Reaction Dynamics ===

==== Reactive trajectory for H2 + F → HF + H ====
In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.

Using initial conditions; rAB = 225 pm, rBC = 75 pm, pAB = -5 g.mol-1.pm.fs-1, pBC = -1 g.mol-1.pm.fs-1, (NB. A=F, B=H, C=H): F + H2 → HF + H is a successful reaction. As seen above, this is an exothermic reaction and thus energy is released. From the animation we can see that a vibrating H2 molecule (BC) approaches F (A) and an HF bond forms, however it breaks and reforms BC, which then collides again with the Fluorine atom to reform a highly energetic HF and leave as the final products: HF + H.

[[File:Goodreachf.png|350px|thumb|centre|Contour plot of successful reaction of formation of HF]]

The energy in the reaction is converted from potential energy to kinetic energy, which is shown by the highly vibrating HF molecule, compared to the initial H2. This is confirmed with a Momentum vs Time plot (shown bellow): the AB distance, representing the HF bond, oscillates at a much larger amplitude than the initial BC oscillations which represent the H2 molecule.
[[File:Memtumsuccreac.png|350px|thumb|centre|Momentum vs Time plot of formation of HF]]

Experimentally, this would result in an increase in temperature of our reaction mixture, which could be measured using calorimetry. Additionally, the HF vibrational modes would be highly excited and overtones would show in the IR spectra as multiples of the fundamental absorption frequency (3962 cm-1). Because overtone bands are difficult to see, this could be done using use Low Temperature FTIR.

==== Illustrating Polanyi's empirical rules: Reaction 2 HF + H → H2 + F, ====
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

Polyani's Rules state that:
# For an early transition state, translational energy is more efficient in promoting a reaction
For a late transition state, vibrational energy is more important in promoting a reaction

So, for the endothermic reaction HF + H → H2 + F, compared to the reverse exothermic reaction, we need a higher amount of vibrational energy, as well as translational to overcome the transition state and for a reaction to occur.

This can be illustrated by comparing the following initial reaction conditions (''A=F, B=H, C=H, rAB = 75 pm, rBC = 225 pm''):
# '''pAB = -1, pBC = -5''' – Unsuccessful Reaction 2 (NB. under these conditions Reaction 1 is successful, as shown above)
# '''pAB = -2, pBC = -4''' – Successful Reaction 2 when we redistribute momentum to increase vibrational energy
[[File:Unsuccessreac2pol.png|350px|thumb|left|Contour plot of unsuccessful formation of HF with conditions parallel to a succesfull reverse reaction]](units of p: g.mol-1.pm.fs-1)[[File:Successreac2pol.png|350px|thumb|right|Contour plot of successful formation of HF ]]

MRD:celiab

2020-05-07T15:45:38Z

Cb2018: /* Reaction Dynamics */

== EXERCISE 1: H + H2 ==

=== The Transition State ===
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?
* On PES diagrams, transition states are mathematically defined as points with 0 gradient with respect to position: stationary points.
* The transition state corresponds to the saddle point; the highest energy point on the reaction coordinate, which is the lowest energy pathway.
* We can distinguish it from a local minimum because on the reaction coordinate axis, the saddle point representing the transition state, will be a local maximum, but on the orthogonal axis it will be a local minimum. The minima are minimums on either direction because they represent stable reactants and products.

=== H + H2 reaction ===
[[File:Internuclear_distances_vs_time.png|250px|thumb|middle|Internuclear Distances vs Time plot ]]

===== Locating the Transition State =====
Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.
* The reaction is symmetric, hence, the transition state must have '''r1''' = '''r2''' by Hammond's Postulate
* By testing different initial conditions where '''rAB'''='''rBC '''and '''p1''' = '''p2''' = 0 g.mol-1.pm.fs-1, we can find the position of the Transition State
** The position of the transition state was found to be '''rts'''=90.775 pm
** This is where the system underwent small periodic symmetric vibrations keeping the distance between the atoms almost constant
* In the plot to the right, we see two straight lines, representing almost a stationary behaviour; thus we know we are at the symmetric transition state.

===== Calculating the Reaction Path =====
Comment on how the ''mep'' and the trajectory you just calculated differ.
* The minimum energy path is calculated from a system slightly displaced from the Transition state ('''r1''' = '''rts'''+1 pm, '''r2''' = '''rts''') and 0 initial momenta and follows the lowest energy path on a PES.
* As the reaction is symmetric it doesn't matter which atoms we label as A B C, so if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead, we get the same results but switched: AB will become BC and viceversa.
* The dynamic trajectory includes the momentum of the particles which causes oscillations about the trajectory path.
* The ''mep'' trajectory is roughly the same but it's smooth and doesn't oscillate, as the velocities are set to 0 in every step. This means we are artificially removing any gain in vibrational energy which results in oscillations. As the velocity is set 0 at every step, the kinetic energy is also 0.
[[File:picture 2 mep.png|350px|thumb|left|MEP contour plot ]]
[[File:picture 3 dynamic.png|350px|thumb|right|Dynamic contour plot ]]
[[File:Pic_4_dt.png|350px|thumb|left|MEP Internuclear Distances vs Time plot ]]
[[File:Pic_5_dt.png|350px|thumb|right| Dynamic Internuclear Distances vs Time plot ]]

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

For the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, the following trajectories with the following momenta combination were run:

(in the descriptions A B and C are used to describe the atoms colliding: AB is the initial H2 approaching H, represented by C)
{| class="wikitable"
!'''#'''
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|'''1'''
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.3</nowiki>
|yes
|The reaction dynamics proceed in a straight forward manner: AB and C approach each other, a collision occurs with the correct momentum for the reaction to proceed and a new bond BC is formed.
| [[File:Pic_a.png|260px]]
|-
|'''2'''
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420</nowiki>
|no
|AB approaches C and collides with it, but it doesn't have enough momentum to form a new bond so they are repelled. Momentum transfer does occur and C goes back with a greater momentum.
| [[File:Pic_b.png|260px]]
|-
|'''3'''
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414</nowiki>
|yes
|Oscillating AB collides with C and forms a new bond. The new H2 molecule, BC, has a higher vibrational frequency. A's direction is reversed.
| [[File:Pic_c.png|260px]]
|-
|'''4'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-375</nowiki>
|no
|AB approaches C and a BC bond forms, however, the system has too much momentum and C is repelled by the nucleus in B and bounces back. This leaves AB with greater vibrational energy and the distance BC increases as C is repelled.

| [[File:Pic_d.png|260px]]
|-
|'''5'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349</nowiki>
|yes
|AB collides with C and a new bond forms. However, unlike above, C still has enough energy to overcome the activation barrier. So, the BC bond forms, it breaks and reforms AB which then collides again with C to reform BC and leave as 2 entities: A + BC
| [[File:Pic_e.png|260px]]
|}

Previous calculations have shown that that trajectories with initial conditions in the range '''r1''' =74 pm, '''r2''' = 200 pm, with -3.1 < '''p1'''/ g.mol-1.pm.fs-1 < -1.6 and '''p2''' = -5.1 g.mol-1.pm.fs-1 are reactive
* This is confirmed in combinations #1, #2 and #3.
From this we could hypothesise that all trajectories starting with the same positions but with higher values of momenta (higher kinetic energy) would be reactive as there would be enough kinetic energy to overcome the activation barrier. From the table we can see that this is not necessarily true:
* Combination #4 vs Combination #5.

==== 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 allows us to investigate the progress of a reaction by splitting it into 2 parts: reactants and products.
* For a reaction to occur the reactants must cross into the product "space" by going over the saddle point – the transition state!
* With the Born-Oppenheimer approximation, it is assumed there is a single saddle point (thus transition state) per reaction
* Transition State theory should only be used to estimate reaction rates at low temperature as it fails to correctly predict reaction rates for scenarios with high momentum (ie high temperature), where the transition state can be recrossed and thus predict reaction rates which are a poor match with experimental values
* We must also take into account that not every reaction with the correct energy will react, as orientation is important too so predicted rates with TS theory will be higher than experimental values.

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

=== PES inspection ===

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

This can be deduced by inspecting the PES surfaces bellow, where F is labelled as A and B C labels H H.
# F + H2 → HF + H : Exothermic
#* Looking at the reactants: A + BC in the PES plot (small BC distance, large AB distance), we can see they have a higher potential energy than the products: AB + C. Which is confirmed by looking at bond strengths: breaking the H–H bond (432 kJ / mol) vs making the H–F bond (565 kJ / mol), so energy is released during the reaction.
# HF + H → H2 + F : Endothermic
#* If we know look at: AB + C as reactants (small AB distance, large BC distance) we can see they have a lower potential energy than the products: BC + A. It is again rationalised by the fact that the H-F bond being broken in the reaction is stronger than the H-H bond being formed, so it requires an amount of energy to initiate the reaction.

[[File:H+hf_pes1L.png|275px|thumb|centre| F-H-H PES plot: AB is FH distance and BC is HH distance ]]

==== Transition state for F-H-H system ====
Locate the approximate position of the transition state.

From Hammond's postulate we know that in an endothermic reaction the transition state resembles the products and in an exothermic reaction it resembles the reactants. So, we can infer that the transition state will resemble H + HF.

The transition State for the F-H-H reaction was found at: 181.11 pm AB and 74.487 pm BC, with 0 initial momenta for both. This was done by varying the distances until they remained constant, confirmed by the forces along AB and BC being 0, marking a stationary point as the first derivative is 0.

[[File: HFTS_cp.png|275px|thumb|centre| F-H-H Contour plot when placed at Transition State ]]

==== Activation Energy for F + H2 and H + HF reactions ====
Report the activation energy for both reactions

The MEP was found for both reactions for a structure neighbouring the transition state: '''rts'''+~1 pm.

The activation energy is the difference between the saddle point's maximum energy and the minimum energy of the products:

* '''Reaction''' '''1''': F + H2 → HF + H
** ''TS Energy: ''- 434 kJ / mol
** ''Final Energy: ''- 560 kJ / mol
** Activation Energy: + 126 kJ / mol
[[File:Eactbiggraph.png|350px|thumb|centre|Energy vs Time diagram of transition state to HH + F state (FH (AB) 180.11 pm and HH (BC) 75.487 pm) ]]
* '''Reaction''' '''2''': HF + H → H2 + F 
** ''TS Energy: ''-433.8 kJ / mol
** ''Final Energy: ''-434.5 kJ / mol
** Activation Energy: 0.7 kJ / mol
[[File:H2+f_emep.png|350px|thumb|centre|Energy vs Time diagram of transition state to form HF + H state (FH (AB) 181.11 pm and HH (BC) 74.487 pm) ]]

=== Reaction Dynamics ===

==== Reactive trajectory for H2 + F → HF + H ====
In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.

Using initial conditions; rAB = 225, rBC = 75, pAB = -5, pBC = -1, (NB. A=F, B=H, C=H): F + H2 → HF + H is a successful reaction. As seen above, this is an exothermic reaction and thus energy is released. From the animation we can see that a vibrating H2 molecule (BC) approaches F (A) and an HF bond forms, however it breaks and reforms BC, which then collides again with the Fluorine atom to reform a highly energetic HF and leave as the final products: HF + H.

[[File:Goodreachf.png|350px|thumb|centre|Contour plot of successful reaction of formation of HF]]

The energy in the reaction is converted from potential energy to kinetic energy, which is shown by the highly vibrating HF molecule, compared to the initial H2. This is confirmed with a Momentum vs Time plot (shown bellow): the AB distance, representing the HF bond, oscillates at a much larger amplitude than the initial BC oscillations which represent the H2 molecule.
[[File:Memtumsuccreac.png|350px|thumb|centre|Momentum vs Time plot of formation of HF]]

Experimentally, this would result in an increase in temperature of our reaction mixture, which could be measured using calorimetry. Additionally, the HF vibrational modes would be highly excited and overtones would show in the IR spectra as multiples of the fundamental absorption frequency (3962 cm-1). Because overtone bands are difficult to see, this could be done using use Low Temperature FTIR.

==== Illustrating Polanyi's empirical rules: Reaction 2 HF + H → H2 + F, ====
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

Polyani's Rules state that:
# For an early transition state, translational energy is more efficient in promoting a reaction
# For a late transition state, vibrational energy is more important in promoting a reaction
So, for the endothermic reaction HF + H → H2 + F, compared to the reverse exothermic reaction, we need a higher amount of vibrational energy, as well as translational to overcome the transition state and for a reaction to occur.

This can be illustrated by comparing the following initial reaction conditions (''A=F, B=H, C=H, rAB = 75, rBC = 225''):
# pAB = -1, pBC = -5 – Unsuccessful Reaction 2 (NB. under these conditions Reaction 1 is successful, as shown above)
# pAB = -2, pBC = -4 – Successful Reaction 2 when we redistribute momentum to increase vibrational energy
[[File:Unsuccessreac2pol.png|350px|thumb|left|Contour plot of unsuccessful formation of HF with conditions parallel to a succesfull reverse reaction]] [[File:Successreac2pol.png|350px|thumb|right|Contour plot of successful formation of HF ]]

File:Successreac2pol.png

2020-05-07T15:44:34Z

Cb2018:

File:Unsuccessreac2pol.png

2020-05-07T15:44:11Z

Cb2018:

File:Memtumsuccreac.png

2020-05-07T15:41:53Z

Cb2018:

File:Goodreachf.png

2020-05-07T15:40:44Z

Cb2018:

MRD:celiab

2020-05-07T15:33:38Z

Cb2018: /* HF + H → H2 + F : Illustrating Polanyi's empirical rules */

== EXERCISE 1: H + H2 ==

=== The Transition State ===
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?
* On PES diagrams, transition states are mathematically defined as points with 0 gradient with respect to position: stationary points.
* The transition state corresponds to the saddle point; the highest energy point on the reaction coordinate, which is the lowest energy pathway.
* We can distinguish it from a local minimum because on the reaction coordinate axis, the saddle point representing the transition state, will be a local maximum, but on the orthogonal axis it will be a local minimum. The minima are minimums on either direction because they represent stable reactants and products.

=== H + H2 reaction ===
[[File:Internuclear_distances_vs_time.png|250px|thumb|middle|Internuclear Distances vs Time plot ]]

===== Locating the Transition State =====
Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.
* The reaction is symmetric, hence, the transition state must have '''r1''' = '''r2''' by Hammond's Postulate
* By testing different initial conditions where '''rAB'''='''rBC '''and '''p1''' = '''p2''' = 0 g.mol-1.pm.fs-1, we can find the position of the Transition State
** The position of the transition state was found to be '''rts'''=90.775 pm
** This is where the system underwent small periodic symmetric vibrations keeping the distance between the atoms almost constant
* In the plot to the right, we see two straight lines, representing almost a stationary behaviour; thus we know we are at the symmetric transition state.

===== Calculating the Reaction Path =====
Comment on how the ''mep'' and the trajectory you just calculated differ.
* The minimum energy path is calculated from a system slightly displaced from the Transition state ('''r1''' = '''rts'''+1 pm, '''r2''' = '''rts''') and 0 initial momenta and follows the lowest energy path on a PES.
* As the reaction is symmetric it doesn't matter which atoms we label as A B C, so if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead, we get the same results but switched: AB will become BC and viceversa.
* The dynamic trajectory includes the momentum of the particles which causes oscillations about the trajectory path.
* The ''mep'' trajectory is roughly the same but it's smooth and doesn't oscillate, as the velocities are set to 0 in every step. This means we are artificially removing any gain in vibrational energy which results in oscillations. As the velocity is set 0 at every step, the kinetic energy is also 0.
[[File:picture 2 mep.png|350px|thumb|left|MEP contour plot ]]
[[File:picture 3 dynamic.png|350px|thumb|right|Dynamic contour plot ]]
[[File:Pic_4_dt.png|350px|thumb|left|MEP Internuclear Distances vs Time plot ]]
[[File:Pic_5_dt.png|350px|thumb|right| Dynamic Internuclear Distances vs Time plot ]]

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

For the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, the following trajectories with the following momenta combination were run:

(in the descriptions A B and C are used to describe the atoms colliding: AB is the initial H2 approaching H, represented by C)
{| class="wikitable"
!'''#'''
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|'''1'''
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.3</nowiki>
|yes
|The reaction dynamics proceed in a straight forward manner: AB and C approach each other, a collision occurs with the correct momentum for the reaction to proceed and a new bond BC is formed.
| [[File:Pic_a.png|260px]]
|-
|'''2'''
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420</nowiki>
|no
|AB approaches C and collides with it, but it doesn't have enough momentum to form a new bond so they are repelled. Momentum transfer does occur and C goes back with a greater momentum.
| [[File:Pic_b.png|260px]]
|-
|'''3'''
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414</nowiki>
|yes
|Oscillating AB collides with C and forms a new bond. The new H2 molecule, BC, has a higher vibrational frequency. A's direction is reversed.
| [[File:Pic_c.png|260px]]
|-
|'''4'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-375</nowiki>
|no
|AB approaches C and a BC bond forms, however, the system has too much momentum and C is repelled by the nucleus in B and bounces back. This leaves AB with greater vibrational energy and the distance BC increases as C is repelled.

| [[File:Pic_d.png|260px]]
|-
|'''5'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349</nowiki>
|yes
|AB collides with C and a new bond forms. However, unlike above, C still has enough energy to overcome the activation barrier. So, the BC bond forms, it breaks and reforms AB which then collides again with C to reform BC and leave as 2 entities: A + BC
| [[File:Pic_e.png|260px]]
|}

Previous calculations have shown that that trajectories with initial conditions in the range '''r1''' =74 pm, '''r2''' = 200 pm, with -3.1 < '''p1'''/ g.mol-1.pm.fs-1 < -1.6 and '''p2''' = -5.1 g.mol-1.pm.fs-1 are reactive
* This is confirmed in combinations #1, #2 and #3.
From this we could hypothesise that all trajectories starting with the same positions but with higher values of momenta (higher kinetic energy) would be reactive as there would be enough kinetic energy to overcome the activation barrier. From the table we can see that this is not necessarily true:
* Combination #4 vs Combination #5.

==== 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 allows us to investigate the progress of a reaction by splitting it into 2 parts: reactants and products.
* For a reaction to occur the reactants must cross into the product "space" by going over the saddle point – the transition state!
* With the Born-Oppenheimer approximation, it is assumed there is a single saddle point (thus transition state) per reaction
* Transition State theory should only be used to estimate reaction rates at low temperature as it fails to correctly predict reaction rates for scenarios with high momentum (ie high temperature), where the transition state can be recrossed and thus predict reaction rates which are a poor match with experimental values
* We must also take into account that not every reaction with the correct energy will react, as orientation is important too so predicted rates with TS theory will be higher than experimental values.

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

=== PES inspection ===

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

This can be deduced by inspecting the PES surfaces bellow, where F is labelled as A and B C labels H H.
# F + H2 → HF + H : Exothermic
#* Looking at the reactants: A + BC in the PES plot (small BC distance, large AB distance), we can see they have a higher potential energy than the products: AB + C. Which is confirmed by looking at bond strengths: breaking the H–H bond (432 kJ / mol) vs making the H–F bond (565 kJ / mol), so energy is released during the reaction.
# HF + H → H2 + F : Endothermic
#* If we know look at: AB + C as reactants (small AB distance, large BC distance) we can see they have a lower potential energy than the products: BC + A. It is again rationalised by the fact that the H-F bond being broken in the reaction is stronger than the H-H bond being formed, so it requires an amount of energy to initiate the reaction.

[[File:H+hf_pes1L.png|275px|thumb|centre| F-H-H PES plot: AB is FH distance and BC is HH distance ]]

==== Transition state for F-H-H system ====
Locate the approximate position of the transition state.

From Hammond's postulate we know that in an endothermic reaction the transition state resembles the products and in an exothermic reaction it resembles the reactants. So, we can infer that the transition state will resemble H + HF.

The transition State for the F-H-H reaction was found at: 181.11 pm AB and 74.487 pm BC, with 0 initial momenta for both. This was done by varying the distances until they remained constant, confirmed by the forces along AB and BC being 0, marking a stationary point as the first derivative is 0.

[[File: HFTS_cp.png|275px|thumb|centre| F-H-H Contour plot when placed at Transition State ]]

==== Activation Energy for F + H2 and H + HF reactions ====
Report the activation energy for both reactions

The MEP was found for both reactions for a structure neighbouring the transition state: '''rts'''+~1 pm.

The activation energy is the difference between the saddle point's maximum energy and the minimum energy of the products:

* '''Reaction''' '''1''': F + H2 → HF + H
** ''TS Energy: ''- 434 kJ / mol
** ''Final Energy: ''- 560 kJ / mol
** Activation Energy: + 126 kJ / mol
[[File:Eactbiggraph.png|350px|thumb|centre|Energy vs Time diagram of transition state to HH + F state (FH (AB) 180.11 pm and HH (BC) 75.487 pm) ]]
* '''Reaction''' '''2''': HF + H → H2 + F 
** ''TS Energy: ''-433.8 kJ / mol
** ''Final Energy: ''-434.5 kJ / mol
** Activation Energy: 0.7 kJ / mol
[[File:H2+f_emep.png|350px|thumb|centre|Energy vs Time diagram of transition state to form HF + H state (FH (AB) 181.11 pm and HH (BC) 74.487 pm) ]]

=== Reaction Dynamics ===

==== Reactive trajectory for H2 + F → HF + H ====
In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.

Using initial conditions; rAB = 225, rBC = 75, pAB = -5, pBC = -1, (NB. A=F, B=H, C=H): F + H2 → HF + H is a successful reaction. As seen above, this is an exothermic reaction and thus energy is released. From the animation we can see that a vibrating H2 molecule (BC) approaches F (A) and an HF bond forms, however it breaks and reforms BC, which then collides again with the Fluorine atom to reform a highly energetic HF and leave as the final products: HF + H.

The energy in the reaction is converted from potential energy to kinetic energy, which is shown by the highly vibrating HF molecule, compared to the initial H2. This is confirmed with a Momentum vs Time plot (shown bellow): the AB distance, representing the HF bond, oscillates at a much larger amplitude than the initial BC oscillations which represent the H2 molecule.

Experimentally, this would result in an increase in temperature of our reaction mixture, which could be measured using calorimetry. Additionally, the HF vibrational modes would be highly excited and overtones would show in the IR spectra as multiples of the fundamental absorption frequency (3962 cm-1). Because overtone bands are difficult to see, this could be done using use Low Temperature FTIR.

==== Illustrating Polanyi's empirical rules: Reaction 2 HF + H → H2 + F, ====
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

Polyani's Rules state that:
# For an early transition state, translational energy is more efficient in promoting a reaction
# For a late transition state, vibrational energy is more important in promoting a reaction
So, for the endothermic reaction HF + H → H2 + F, compared to the reverse exothermic reaction, we need a higher amount of vibrational energy, as well as translational to overcome the transition state and for a reaction to occur.

This can be illustrated by comparing the following initial reaction conditions (''A=F, B=H, C=H, rAB = 75, rBC = 225''):
# pAB = -1, pBC = -5 – Unsuccessful Reaction 2 (NB. under these conditions Reaction 1 is successful, as shown above)
# pAB = -2, pBC = -4 – Successful Reaction 2 when we redistribute momentum to increase vibrational energy

MRD:celiab

2020-05-07T15:06:58Z

Cb2018: /* Activation Energy for F + H2 and H + HF reactions */

== EXERCISE 1: H + H2 ==

=== The Transition State ===
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?
* On PES diagrams, transition states are mathematically defined as points with 0 gradient with respect to position: stationary points.
* The transition state corresponds to the saddle point; the highest energy point on the reaction coordinate, which is the lowest energy pathway.
* We can distinguish it from a local minimum because on the reaction coordinate axis, the saddle point representing the transition state, will be a local maximum, but on the orthogonal axis it will be a local minimum. The minima are minimums on either direction because they represent stable reactants and products.

=== H + H2 reaction ===
[[File:Internuclear_distances_vs_time.png|250px|thumb|middle|Internuclear Distances vs Time plot ]]

===== Locating the Transition State =====
Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.
* The reaction is symmetric, hence, the transition state must have '''r1''' = '''r2''' by Hammond's Postulate
* By testing different initial conditions where '''rAB'''='''rBC '''and '''p1''' = '''p2''' = 0 g.mol-1.pm.fs-1, we can find the position of the Transition State
** The position of the transition state was found to be '''rts'''=90.775 pm
** This is where the system underwent small periodic symmetric vibrations keeping the distance between the atoms almost constant
* In the plot to the right, we see two straight lines, representing almost a stationary behaviour; thus we know we are at the symmetric transition state.

===== Calculating the Reaction Path =====
Comment on how the ''mep'' and the trajectory you just calculated differ.
* The minimum energy path is calculated from a system slightly displaced from the Transition state ('''r1''' = '''rts'''+1 pm, '''r2''' = '''rts''') and 0 initial momenta and follows the lowest energy path on a PES.
* As the reaction is symmetric it doesn't matter which atoms we label as A B C, so if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead, we get the same results but switched: AB will become BC and viceversa.
* The dynamic trajectory includes the momentum of the particles which causes oscillations about the trajectory path.
* The ''mep'' trajectory is roughly the same but it's smooth and doesn't oscillate, as the velocities are set to 0 in every step. This means we are artificially removing any gain in vibrational energy which results in oscillations. As the velocity is set 0 at every step, the kinetic energy is also 0.
[[File:picture 2 mep.png|350px|thumb|left|MEP contour plot ]]
[[File:picture 3 dynamic.png|350px|thumb|right|Dynamic contour plot ]]
[[File:Pic_4_dt.png|350px|thumb|left|MEP Internuclear Distances vs Time plot ]]
[[File:Pic_5_dt.png|350px|thumb|right| Dynamic Internuclear Distances vs Time plot ]]

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

For the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, the following trajectories with the following momenta combination were run:

(in the descriptions A B and C are used to describe the atoms colliding: AB is the initial H2 approaching H, represented by C)
{| class="wikitable"
!'''#'''
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|'''1'''
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.3</nowiki>
|yes
|The reaction dynamics proceed in a straight forward manner: AB and C approach each other, a collision occurs with the correct momentum for the reaction to proceed and a new bond BC is formed.
| [[File:Pic_a.png|260px]]
|-
|'''2'''
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420</nowiki>
|no
|AB approaches C and collides with it, but it doesn't have enough momentum to form a new bond so they are repelled. Momentum transfer does occur and C goes back with a greater momentum.
| [[File:Pic_b.png|260px]]
|-
|'''3'''
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414</nowiki>
|yes
|Oscillating AB collides with C and forms a new bond. The new H2 molecule, BC, has a higher vibrational frequency. A's direction is reversed.
| [[File:Pic_c.png|260px]]
|-
|'''4'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-375</nowiki>
|no
|AB approaches C and a BC bond forms, however, the system has too much momentum and C is repelled by the nucleus in B and bounces back. This leaves AB with greater vibrational energy and the distance BC increases as C is repelled.

| [[File:Pic_d.png|260px]]
|-
|'''5'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349</nowiki>
|yes
|AB collides with C and a new bond forms. However, unlike above, C still has enough energy to overcome the activation barrier. So, the BC bond forms, it breaks and reforms AB which then collides again with C to reform BC and leave as 2 entities: A + BC
| [[File:Pic_e.png|260px]]
|}

Previous calculations have shown that that trajectories with initial conditions in the range '''r1''' =74 pm, '''r2''' = 200 pm, with -3.1 < '''p1'''/ g.mol-1.pm.fs-1 < -1.6 and '''p2''' = -5.1 g.mol-1.pm.fs-1 are reactive
* This is confirmed in combinations #1, #2 and #3.
From this we could hypothesise that all trajectories starting with the same positions but with higher values of momenta (higher kinetic energy) would be reactive as there would be enough kinetic energy to overcome the activation barrier. From the table we can see that this is not necessarily true:
* Combination #4 vs Combination #5.

==== 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 allows us to investigate the progress of a reaction by splitting it into 2 parts: reactants and products.
* For a reaction to occur the reactants must cross into the product "space" by going over the saddle point – the transition state!
* With the Born-Oppenheimer approximation, it is assumed there is a single saddle point (thus transition state) per reaction
* Transition State theory should only be used to estimate reaction rates at low temperature as it fails to correctly predict reaction rates for scenarios with high momentum (ie high temperature), where the transition state can be recrossed and thus predict reaction rates which are a poor match with experimental values
* We must also take into account that not every reaction with the correct energy will react, as orientation is important too so predicted rates with TS theory will be higher than experimental values.

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

=== PES inspection ===

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

This can be deduced by inspecting the PES surfaces bellow, where F is labelled as A and B C labels H H.
# F + H2 → HF + H : Exothermic
#* Looking at the reactants: A + BC in the PES plot (small BC distance, large AB distance), we can see they have a higher potential energy than the products: AB + C. Which is confirmed by looking at bond strengths: breaking the H–H bond (432 kJ / mol) vs making the H–F bond (565 kJ / mol), so energy is released during the reaction.
# HF + H → H2 + F : Endothermic
#* If we know look at: AB + C as reactants (small AB distance, large BC distance) we can see they have a lower potential energy than the products: BC + A. It is again rationalised by the fact that the H-F bond being broken in the reaction is stronger than the H-H bond being formed, so it requires an amount of energy to initiate the reaction.

[[File:H+hf_pes1L.png|275px|thumb|centre| F-H-H PES plot: AB is FH distance and BC is HH distance ]]

==== Transition state for F-H-H system ====
Locate the approximate position of the transition state.

From Hammond's postulate we know that in an endothermic reaction the transition state resembles the products and in an exothermic reaction it resembles the reactants. So, we can infer that the transition state will resemble H + HF.

The transition State for the F-H-H reaction was found at: 181.11 pm AB and 74.487 pm BC, with 0 initial momenta for both. This was done by varying the distances until they remained constant, confirmed by the forces along AB and BC being 0, marking a stationary point as the first derivative is 0.

[[File: HFTS_cp.png|275px|thumb|centre| F-H-H Contour plot when placed at Transition State ]]

==== Activation Energy for F + H2 and H + HF reactions ====
Report the activation energy for both reactions

The MEP was found for both reactions for a structure neighbouring the transition state: '''rts'''+~1 pm.

The activation energy is the difference between the saddle point's maximum energy and the minimum energy of the products:

* '''Reaction''' '''1''': F + H2 → HF + H
** ''TS Energy: ''- 434 kJ / mol
** ''Final Energy: ''- 560 kJ / mol
** Activation Energy: + 126 kJ / mol
[[File:Eactbiggraph.png|350px|thumb|centre|Energy vs Time diagram of transition state to HH + F state (FH (AB) 180.11 pm and HH (BC) 75.487 pm) ]]
* '''Reaction''' '''2''': HF + H → H2 + F 
** ''TS Energy: ''-433.8 kJ / mol
** ''Final Energy: ''-434.5 kJ / mol
** Activation Energy: 0.7 kJ / mol
[[File:H2+f_emep.png|350px|thumb|centre|Energy vs Time diagram of transition state to form HF + H state (FH (AB) 181.11 pm and HH (BC) 74.487 pm) ]]

=== Reaction Dynamics ===

==== Reactive trajectory for H2 + F → HF + H ====
In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.

Using initial conditions; rAB = 225, rBC = 75, pAB = -5, pBC = -1, (NB. A=F, B=H, C=H): F + H2 → HF + H is a successful reaction. As seen above, this is an exothermic reaction and thus energy is released. From the animation we can see that a vibrating H2 molecule (BC) approaches F (A) and an HF bond forms, however it breaks and reforms BC, which then collides again with the Fluorine atom to reform a highly energetic HF and leave as the final products: HF + H.

The energy in the reaction is converted from potential energy to kinetic energy, which is shown by the highly vibrating HF molecule, compared to the initial H2. This is confirmed with a Momentum vs Time plot (shown bellow): the AB distance, representing the HF bond, oscillates at a much larger amplitude than the initial BC oscillations which represent the H2 molecule.

Experimentally, this would result in an increase in temperature of our reaction mixture, which could be measured using calorimetry. Additionally, the HF vibrational modes would be highly excited and overtones would show in the IR spectra as multiples of the fundamental absorption frequency (3962 cm-1). Because overtone bands are difficult to see, this could be done using use Low Temperature FTIR.

==== HF + H → H2 + F : Illustrating Polanyi's empirical rules ====
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

Polyani's Rules state that:
# For an early transition state, translational energy is more efficient in promoting a reaction
# For a late transition state, vibrational energy is more important in promoting a reaction
So, for the endothermic reaction HF + H → H2 + F, compared to the reverse exothermic reaction, we need a higher amount of vibrational energy, as well as translational to overcome the transition state and for a reaction to occur.

MRD:celiab

2020-05-07T15:05:43Z

Cb2018: /* Illustrating Polanyi's empirical rules */

== EXERCISE 1: H + H2 ==

=== The Transition State ===
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?
* On PES diagrams, transition states are mathematically defined as points with 0 gradient with respect to position: stationary points.
* The transition state corresponds to the saddle point; the highest energy point on the reaction coordinate, which is the lowest energy pathway.
* We can distinguish it from a local minimum because on the reaction coordinate axis, the saddle point representing the transition state, will be a local maximum, but on the orthogonal axis it will be a local minimum. The minima are minimums on either direction because they represent stable reactants and products.

=== H + H2 reaction ===
[[File:Internuclear_distances_vs_time.png|250px|thumb|middle|Internuclear Distances vs Time plot ]]

===== Locating the Transition State =====
Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.
* The reaction is symmetric, hence, the transition state must have '''r1''' = '''r2''' by Hammond's Postulate
* By testing different initial conditions where '''rAB'''='''rBC '''and '''p1''' = '''p2''' = 0 g.mol-1.pm.fs-1, we can find the position of the Transition State
** The position of the transition state was found to be '''rts'''=90.775 pm
** This is where the system underwent small periodic symmetric vibrations keeping the distance between the atoms almost constant
* In the plot to the right, we see two straight lines, representing almost a stationary behaviour; thus we know we are at the symmetric transition state.

===== Calculating the Reaction Path =====
Comment on how the ''mep'' and the trajectory you just calculated differ.
* The minimum energy path is calculated from a system slightly displaced from the Transition state ('''r1''' = '''rts'''+1 pm, '''r2''' = '''rts''') and 0 initial momenta and follows the lowest energy path on a PES.
* As the reaction is symmetric it doesn't matter which atoms we label as A B C, so if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead, we get the same results but switched: AB will become BC and viceversa.
* The dynamic trajectory includes the momentum of the particles which causes oscillations about the trajectory path.
* The ''mep'' trajectory is roughly the same but it's smooth and doesn't oscillate, as the velocities are set to 0 in every step. This means we are artificially removing any gain in vibrational energy which results in oscillations. As the velocity is set 0 at every step, the kinetic energy is also 0.
[[File:picture 2 mep.png|350px|thumb|left|MEP contour plot ]]
[[File:picture 3 dynamic.png|350px|thumb|right|Dynamic contour plot ]]
[[File:Pic_4_dt.png|350px|thumb|left|MEP Internuclear Distances vs Time plot ]]
[[File:Pic_5_dt.png|350px|thumb|right| Dynamic Internuclear Distances vs Time plot ]]

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

For the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, the following trajectories with the following momenta combination were run:

(in the descriptions A B and C are used to describe the atoms colliding: AB is the initial H2 approaching H, represented by C)
{| class="wikitable"
!'''#'''
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|'''1'''
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.3</nowiki>
|yes
|The reaction dynamics proceed in a straight forward manner: AB and C approach each other, a collision occurs with the correct momentum for the reaction to proceed and a new bond BC is formed.
| [[File:Pic_a.png|260px]]
|-
|'''2'''
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420</nowiki>
|no
|AB approaches C and collides with it, but it doesn't have enough momentum to form a new bond so they are repelled. Momentum transfer does occur and C goes back with a greater momentum.
| [[File:Pic_b.png|260px]]
|-
|'''3'''
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414</nowiki>
|yes
|Oscillating AB collides with C and forms a new bond. The new H2 molecule, BC, has a higher vibrational frequency. A's direction is reversed.
| [[File:Pic_c.png|260px]]
|-
|'''4'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-375</nowiki>
|no
|AB approaches C and a BC bond forms, however, the system has too much momentum and C is repelled by the nucleus in B and bounces back. This leaves AB with greater vibrational energy and the distance BC increases as C is repelled.

| [[File:Pic_d.png|260px]]
|-
|'''5'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349</nowiki>
|yes
|AB collides with C and a new bond forms. However, unlike above, C still has enough energy to overcome the activation barrier. So, the BC bond forms, it breaks and reforms AB which then collides again with C to reform BC and leave as 2 entities: A + BC
| [[File:Pic_e.png|260px]]
|}

Previous calculations have shown that that trajectories with initial conditions in the range '''r1''' =74 pm, '''r2''' = 200 pm, with -3.1 < '''p1'''/ g.mol-1.pm.fs-1 < -1.6 and '''p2''' = -5.1 g.mol-1.pm.fs-1 are reactive
* This is confirmed in combinations #1, #2 and #3.
From this we could hypothesise that all trajectories starting with the same positions but with higher values of momenta (higher kinetic energy) would be reactive as there would be enough kinetic energy to overcome the activation barrier. From the table we can see that this is not necessarily true:
* Combination #4 vs Combination #5.

==== 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 allows us to investigate the progress of a reaction by splitting it into 2 parts: reactants and products.
* For a reaction to occur the reactants must cross into the product "space" by going over the saddle point – the transition state!
* With the Born-Oppenheimer approximation, it is assumed there is a single saddle point (thus transition state) per reaction
* Transition State theory should only be used to estimate reaction rates at low temperature as it fails to correctly predict reaction rates for scenarios with high momentum (ie high temperature), where the transition state can be recrossed and thus predict reaction rates which are a poor match with experimental values
* We must also take into account that not every reaction with the correct energy will react, as orientation is important too so predicted rates with TS theory will be higher than experimental values.

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

=== PES inspection ===

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

This can be deduced by inspecting the PES surfaces bellow, where F is labelled as A and B C labels H H.
# F + H2 → HF + H : Exothermic
#* Looking at the reactants: A + BC in the PES plot (small BC distance, large AB distance), we can see they have a higher potential energy than the products: AB + C. Which is confirmed by looking at bond strengths: breaking the H–H bond (432 kJ / mol) vs making the H–F bond (565 kJ / mol), so energy is released during the reaction.
# HF + H → H2 + F : Endothermic
#* If we know look at: AB + C as reactants (small AB distance, large BC distance) we can see they have a lower potential energy than the products: BC + A. It is again rationalised by the fact that the H-F bond being broken in the reaction is stronger than the H-H bond being formed, so it requires an amount of energy to initiate the reaction.

[[File:H+hf_pes1L.png|275px|thumb|centre| F-H-H PES plot: AB is FH distance and BC is HH distance ]]

==== Transition state for F-H-H system ====
Locate the approximate position of the transition state.

From Hammond's postulate we know that in an endothermic reaction the transition state resembles the products and in an exothermic reaction it resembles the reactants. So, we can infer that the transition state will resemble H + HF.

The transition State for the F-H-H reaction was found at: 181.11 pm AB and 74.487 pm BC, with 0 initial momenta for both. This was done by varying the distances until they remained constant, confirmed by the forces along AB and BC being 0, marking a stationary point as the first derivative is 0.

[[File: HFTS_cp.png|275px|thumb|centre| F-H-H Contour plot when placed at Transition State ]]

==== Activation Energy for F + H2 and H + HF reactions ====
Report the activation energy for both reactions

The MEP was found for both reactions for a structure neighbouring the transition state: '''rts'''+~1 pm.

The activation energy is the difference between the saddle point's maximum energy and the minimum energy of the products:

* '''Reaction''' '''1''': F + H2 → HF + H
** ''TS Energy: ''- 434 kJ / mol
** ''Final Energy: ''- 560 kJ / mol
** Activation Energy: + 126 kJ / mol
[[File:Eactbiggraph.png|250px|thumb|centre|Energy vs Time diagram of transition state to HH + F state (FH (AB) 180.11 pm and HH (BC) 75.487 pm) ]]
* '''Reaction''' '''2''': HF + H → H2 + F 
** ''TS Energy: ''-433.8 kJ / mol
** ''Final Energy: ''-434.5 kJ / mol
** Activation Energy: 0.7 kJ / mol
[[File:H2+f_emep.png|250px|thumb|centre|Energy vs Time diagram of transition state to form HF + H state (FH (AB) 181.11 pm and HH (BC) 74.487 pm) ]]

=== Reaction Dynamics ===

==== Reactive trajectory for H2 + F → HF + H ====
In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.

Using initial conditions; rAB = 225, rBC = 75, pAB = -5, pBC = -1, (NB. A=F, B=H, C=H): F + H2 → HF + H is a successful reaction. As seen above, this is an exothermic reaction and thus energy is released. From the animation we can see that a vibrating H2 molecule (BC) approaches F (A) and an HF bond forms, however it breaks and reforms BC, which then collides again with the Fluorine atom to reform a highly energetic HF and leave as the final products: HF + H.

The energy in the reaction is converted from potential energy to kinetic energy, which is shown by the highly vibrating HF molecule, compared to the initial H2. This is confirmed with a Momentum vs Time plot (shown bellow): the AB distance, representing the HF bond, oscillates at a much larger amplitude than the initial BC oscillations which represent the H2 molecule.

Experimentally, this would result in an increase in temperature of our reaction mixture, which could be measured using calorimetry. Additionally, the HF vibrational modes would be highly excited and overtones would show in the IR spectra as multiples of the fundamental absorption frequency (3962 cm-1). Because overtone bands are difficult to see, this could be done using use Low Temperature FTIR.

==== HF + H → H2 + F : Illustrating Polanyi's empirical rules ====
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

Polyani's Rules state that:
# For an early transition state, translational energy is more efficient in promoting a reaction
# For a late transition state, vibrational energy is more important in promoting a reaction
So, for the endothermic reaction HF + H → H2 + F, compared to the reverse exothermic reaction, we need a higher amount of vibrational energy, as well as translational to overcome the transition state and for a reaction to occur.

MRD:celiab

2020-05-07T14:33:10Z

Cb2018:

== EXERCISE 1: H + H2 ==

=== The Transition State ===
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?
* On PES diagrams, transition states are mathematically defined as points with 0 gradient with respect to position: stationary points.
* The transition state corresponds to the saddle point; the highest energy point on the reaction coordinate, which is the lowest energy pathway.
* We can distinguish it from a local minimum because on the reaction coordinate axis, the saddle point representing the transition state, will be a local maximum, but on the orthogonal axis it will be a local minimum. The minima are minimums on either direction because they represent stable reactants and products.

=== H + H2 reaction ===
[[File:Internuclear_distances_vs_time.png|250px|thumb|middle|Internuclear Distances vs Time plot ]]

===== Locating the Transition State =====
Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.
* The reaction is symmetric, hence, the transition state must have '''r1''' = '''r2''' by Hammond's Postulate
* By testing different initial conditions where '''rAB'''='''rBC '''and '''p1''' = '''p2''' = 0 g.mol-1.pm.fs-1, we can find the position of the Transition State
** The position of the transition state was found to be '''rts'''=90.775 pm
** This is where the system underwent small periodic symmetric vibrations keeping the distance between the atoms almost constant
* In the plot to the right, we see two straight lines, representing almost a stationary behaviour; thus we know we are at the symmetric transition state.

===== Calculating the Reaction Path =====
Comment on how the ''mep'' and the trajectory you just calculated differ.
* The minimum energy path is calculated from a system slightly displaced from the Transition state ('''r1''' = '''rts'''+1 pm, '''r2''' = '''rts''') and 0 initial momenta and follows the lowest energy path on a PES.
* As the reaction is symmetric it doesn't matter which atoms we label as A B C, so if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead, we get the same results but switched: AB will become BC and viceversa.
* The dynamic trajectory includes the momentum of the particles which causes oscillations about the trajectory path.
* The ''mep'' trajectory is roughly the same but it's smooth and doesn't oscillate, as the velocities are set to 0 in every step. This means we are artificially removing any gain in vibrational energy which results in oscillations. As the velocity is set 0 at every step, the kinetic energy is also 0.
[[File:picture 2 mep.png|350px|thumb|left|MEP contour plot ]]
[[File:picture 3 dynamic.png|350px|thumb|right|Dynamic contour plot ]]
[[File:Pic_4_dt.png|350px|thumb|left|MEP Internuclear Distances vs Time plot ]]
[[File:Pic_5_dt.png|350px|thumb|right| Dynamic Internuclear Distances vs Time plot ]]

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

For the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, the following trajectories with the following momenta combination were run:

(in the descriptions A B and C are used to describe the atoms colliding: AB is the initial H2 approaching H, represented by C)
{| class="wikitable"
!'''#'''
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|'''1'''
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.3</nowiki>
|yes
|The reaction dynamics proceed in a straight forward manner: AB and C approach each other, a collision occurs with the correct momentum for the reaction to proceed and a new bond BC is formed.
| [[File:Pic_a.png|260px]]
|-
|'''2'''
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420</nowiki>
|no
|AB approaches C and collides with it, but it doesn't have enough momentum to form a new bond so they are repelled. Momentum transfer does occur and C goes back with a greater momentum.
| [[File:Pic_b.png|260px]]
|-
|'''3'''
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414</nowiki>
|yes
|Oscillating AB collides with C and forms a new bond. The new H2 molecule, BC, has a higher vibrational frequency. A's direction is reversed.
| [[File:Pic_c.png|260px]]
|-
|'''4'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-375</nowiki>
|no
|AB approaches C and a BC bond forms, however, the system has too much momentum and C is repelled by the nucleus in B and bounces back. This leaves AB with greater vibrational energy and the distance BC increases as C is repelled.

| [[File:Pic_d.png|260px]]
|-
|'''5'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349</nowiki>
|yes
|AB collides with C and a new bond forms. However, unlike above, C still has enough energy to overcome the activation barrier. So, the BC bond forms, it breaks and reforms AB which then collides again with C to reform BC and leave as 2 entities: A + BC
| [[File:Pic_e.png|260px]]
|}

Previous calculations have shown that that trajectories with initial conditions in the range '''r1''' =74 pm, '''r2''' = 200 pm, with -3.1 < '''p1'''/ g.mol-1.pm.fs-1 < -1.6 and '''p2''' = -5.1 g.mol-1.pm.fs-1 are reactive
* This is confirmed in combinations #1, #2 and #3.
From this we could hypothesise that all trajectories starting with the same positions but with higher values of momenta (higher kinetic energy) would be reactive as there would be enough kinetic energy to overcome the activation barrier. From the table we can see that this is not necessarily true:
* Combination #4 vs Combination #5.

==== 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 allows us to investigate the progress of a reaction by splitting it into 2 parts: reactants and products.
* For a reaction to occur the reactants must cross into the product "space" by going over the saddle point – the transition state!
* With the Born-Oppenheimer approximation, it is assumed there is a single saddle point (thus transition state) per reaction
* Transition State theory should only be used to estimate reaction rates at low temperature as it fails to correctly predict reaction rates for scenarios with high momentum (ie high temperature), where the transition state can be recrossed and thus predict reaction rates which are a poor match with experimental values
* We must also take into account that not every reaction with the correct energy will react, as orientation is important too so predicted rates with TS theory will be higher than experimental values.

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

=== PES inspection ===

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

This can be deduced by inspecting the PES surfaces bellow, where F is labelled as A and B C labels H H.
# F + H2 → HF + H : Exothermic
#* Looking at the reactants: A + BC in the PES plot (small BC distance, large AB distance), we can see they have a higher potential energy than the products: AB + C. Which is confirmed by looking at bond strengths: breaking the H–H bond (432 kJ / mol) vs making the H–F bond (565 kJ / mol), so energy is released during the reaction.
# HF + H → H2 + F : Endothermic
#* If we know look at: AB + C as reactants (small AB distance, large BC distance) we can see they have a lower potential energy than the products: BC + A. It is again rationalised by the fact that the H-F bond being broken in the reaction is stronger than the H-H bond being formed, so it requires an amount of energy to initiate the reaction.

[[File:H+hf_pes1L.png|275px|thumb|centre| F-H-H PES plot: AB is FH distance and BC is HH distance ]]

==== Transition state for F-H-H system ====
Locate the approximate position of the transition state.

From Hammond's postulate we know that in an endothermic reaction the transition state resembles the products and in an exothermic reaction it resembles the reactants. So, we can infer that the transition state will resemble H + HF.

The transition State for the F-H-H reaction was found at: 181.11 pm AB and 74.487 pm BC, with 0 initial momenta for both. This was done by varying the distances until they remained constant, confirmed by the forces along AB and BC being 0, marking a stationary point as the first derivative is 0.

[[File: HFTS_cp.png|275px|thumb|centre| F-H-H Contour plot when placed at Transition State ]]

==== Activation Energy for F + H2 and H + HF reactions ====
Report the activation energy for both reactions

The MEP was found for both reactions for a structure neighbouring the transition state: '''rts'''+~1 pm.

The activation energy is the difference between the saddle point's maximum energy and the minimum energy of the products:

* '''Reaction''' '''1''': F + H2 → HF + H
** ''TS Energy: ''- 434 kJ / mol
** ''Final Energy: ''- 560 kJ / mol
** Activation Energy: + 126 kJ / mol
[[File:Eactbiggraph.png|250px|thumb|centre|Energy vs Time diagram of transition state to HH + F state (FH (AB) 180.11 pm and HH (BC) 75.487 pm) ]]
* '''Reaction''' '''2''': HF + H → H2 + F 
** ''TS Energy: ''-433.8 kJ / mol
** ''Final Energy: ''-434.5 kJ / mol
** Activation Energy: 0.7 kJ / mol
[[File:H2+f_emep.png|250px|thumb|centre|Energy vs Time diagram of transition state to form HF + H state (FH (AB) 181.11 pm and HH (BC) 74.487 pm) ]]

=== Reaction Dynamics ===

==== Reactive trajectory for H2 + F → HF + H ====
In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.

Using initial conditions; rAB = 225, rBC = 75, pAB = -5, pBC = -1, (NB. A=F, B=H, C=H): F + H2 → HF + H is a successful reaction. As seen above, this is an exothermic reaction and thus energy is released. From the animation we can see that a vibrating H2 molecule (BC) approaches F (A) and an HF bond forms, however it breaks and reforms BC, which then collides again with the Fluorine atom to reform a highly energetic HF and leave as the final products: HF + H.

The energy in the reaction is converted from potential energy to kinetic energy, which is shown by the highly vibrating HF molecule, compared to the initial H2. This is confirmed with a Momentum vs Time plot (shown bellow): the AB distance, representing the HF bond, oscillates at a much larger amplitude than the initial BC oscillations which represent the H2 molecule.

Experimentally, this would result in an increase in temperature of our reaction mixture, which could be measured using calorimetry. Additionally, the HF vibrational modes would be highly excited and overtones would show in the IR spectra as multiples of the fundamental absorption frequency (3962 cm-1). Because overtone bands are difficult to see, this could be done using use Low Temperature FTIR.

==== Illustrating Polanyi's empirical rules ====
Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state

MRD:celiab

2020-05-07T13:53:57Z

Cb2018:

== EXERCISE 1: H + H2 ==

=== The Transition State ===
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?
* On PES diagrams, transition states are mathematically defined as points with 0 gradient with respect to position: stationary points.
* The transition state corresponds to the saddle point; the highest energy point on the reaction coordinate, which is the lowest energy pathway.
* We can distinguish it from a local minimum because on the reaction coordinate axis, the saddle point representing the transition state, will be a local maximum, but on the orthogonal axis it will be a local minimum. The minima are minimums on either direction because they represent stable reactants and products.

=== H + H2 reaction ===
[[File:Internuclear_distances_vs_time.png|250px|thumb|middle|Internuclear Distances vs Time plot ]]

===== Locating the Transition State =====
Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.
* The reaction is symmetric, hence, the transition state must have '''r1''' = '''r2''' by Hammond's Postulate
* By testing different initial conditions where '''rAB'''='''rBC '''and '''p1''' = '''p2''' = 0 g.mol-1.pm.fs-1, we can find the position of the Transition State
** The position of the transition state was found to be '''rts'''=90.775 pm
** This is where the system underwent small periodic symmetric vibrations keeping the distance between the atoms almost constant
* In the plot to the right, we see two straight lines, representing almost a stationary behaviour; thus we know we are at the symmetric transition state.

===== Calculating the Reaction Path =====
Comment on how the ''mep'' and the trajectory you just calculated differ.
* The minimum energy path is calculated from a system slightly displaced from the Transition state ('''r1''' = '''rts'''+1 pm, '''r2''' = '''rts''') and 0 initial momenta and follows the lowest energy path on a PES.
* As the reaction is symmetric it doesn't matter which atoms we label as A B C, so if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead, we get the same results but switched: AB will become BC and viceversa.
* The dynamic trajectory includes the momentum of the particles which causes oscillations about the trajectory path.
* The ''mep'' trajectory is roughly the same but it's smooth and doesn't oscillate, as the velocities are set to 0 in every step. This means we are artificially removing any gain in vibrational energy which results in oscillations. As the velocity is set 0 at every step, the kinetic energy is also 0.
[[File:picture 2 mep.png|350px|thumb|left|MEP contour plot ]]
[[File:picture 3 dynamic.png|350px|thumb|right|Dynamic contour plot ]]
[[File:Pic_4_dt.png|350px|thumb|left|MEP Internuclear Distances vs Time plot ]]
[[File:Pic_5_dt.png|350px|thumb|right| Dynamic Internuclear Distances vs Time plot ]]

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

For the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, the following trajectories with the following momenta combination were run:

(in the descriptions A B and C are used to describe the atoms colliding: AB is the initial H2 approaching H, represented by C)
{| class="wikitable"
!'''#'''
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|'''1'''
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.3</nowiki>
|yes
|The reaction dynamics proceed in a straight forward manner: AB and C approach each other, a collision occurs with the correct momentum for the reaction to proceed and a new bond BC is formed.
| [[File:Pic_a.png|260px]]
|-
|'''2'''
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420</nowiki>
|no
|AB approaches C and collides with it, but it doesn't have enough momentum to form a new bond so they are repelled. Momentum transfer does occur and C goes back with a greater momentum.
| [[File:Pic_b.png|260px]]
|-
|'''3'''
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414</nowiki>
|yes
|Oscillating AB collides with C and forms a new bond. The new H2 molecule, BC, has a higher vibrational frequency. A's direction is reversed.
| [[File:Pic_c.png|260px]]
|-
|'''4'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-375</nowiki>
|no
|AB approaches C and a BC bond forms, however, the system has too much momentum and C is repelled by the nucleus in B and bounces back. This leaves AB with greater vibrational energy and the distance BC increases as C is repelled.

| [[File:Pic_d.png|260px]]
|-
|'''5'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349</nowiki>
|yes
|AB collides with C and a new bond forms. However, unlike above, C still has enough energy to overcome the activation barrier. So, the BC bond forms, it breaks and reforms AB which then collides again with C to reform BC and leave as 2 entities: A + BC:
| [[File:Pic_e.png|260px]]
|}

Previous calculations have shown that that trajectories with initial conditions in the range '''r1''' =74 pm, '''r2''' = 200 pm, with -3.1 < '''p1'''/ g.mol-1.pm.fs-1 < -1.6 and '''p2''' = -5.1 g.mol-1.pm.fs-1 are reactive
* This is confirmed in combinations #1, #2 and #3.
From this we could hypothesise that all trajectories starting with the same positions but with higher values of momenta (higher kinetic energy) would be reactive as there would be enough kinetic energy to overcome the activation barrier. From the table we can see that this is not necessarily true:
* Combination #4 vs Combination #5.

==== 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 allows us to investigate the progress of a reaction by splitting it into 2 parts: reactants and products.
* For a reaction to occur the reactants must cross into the product "space" by going over the saddle point – the transition state!
* With the Born-Oppenheimer approximation, it is assumed there is a single saddle point (thus transition state) per reaction
* Transition State theory should only be used to estimate reaction rates at low temperature as it fails to correctly predict reaction rates for scenarios with high momentum (ie high temperature), where the transition state can be recrossed and thus predict reaction rates which are a poor match with experimental values
* We must also take into account that not every reaction with the correct energy will react, as orientation is important too so predicted rates with TS theory will be higher than experimental values.

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

=== PES inspection ===

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

This can be deduced by inspecting the PES surfaces bellow, where F is labelled as A and B C labels H H.
# F + H2 → HF + H : Exothermic
#* Looking at the reactants: A + BC in the PES plot (small BC distance, large AB distance), we can see they have a higher potential energy than the products: AB + C. Which is confirmed by looking at bond strengths: breaking the H–H bond (432 kJ / mol) vs making the H–F bond (565 kJ / mol), so energy is released during the reaction.
# HF + H → H2 + F : Endothermic
#* If we know look at: AB + C as reactants (small AB distance, large BC distance) we can see they have a lower potential energy than the products: BC + A. It is again rationalised by the fact that the H-F bond being broken in the reaction is stronger than the H-H bond being formed, so it requires an amount of energy to initiate the reaction.

[[File:H+hf_pes1L.png|275px|thumb|centre| F-H-H PES plot: AB is FH distance and BC is HH distance ]]

==== Transition state for F-H-H system ====
Locate the approximate position of the transition state.

From Hammond's postulate we know that in an endothermic reaction the transition state resembles the products and in an exothermic reaction it resembles the reactants. So, we can infer that the transition state will resemble H + HF.

The transition State for the F-H-H reaction was found at: 181.11 pm AB and 74.487 pm BC, with 0 initial momenta for both. This was done by varying the distances until they remained constant, confirmed by the forces along AB and BC being 0, marking a stationary point as the first derivative is 0.

[[File: HFTS_cp.png|275px|thumb|centre| F-H-H Contour plot when placed at Transition State ]]

==== Activation Energy for F + H2 and H + HF reactions ====
Report the activation energy for both reactions

The MEP was found for both reactions for a structure neighbouring the transition state: '''rts'''+1 pm.

The activation energy is the difference between the saddle point's maximum energy and the minimum energy of the products:

* Reaction 1: F + H2 → HF + H
** ''TS Energy: ''- 434 kJ / mol
** ''Final Energy: ''- 560 kJ / mol
** Activation Energy: + 126 kJ / mol
[[File:Eactbiggraph.png|250px|thumb|centre|Energy vs Time diagram of transition state to HH + F state (FH (AB) 180.11 pm and HH (BC) 75.487 pm) ]]
* Reaction 2: HF + H → H2 + F 
** ''TS Energy: ''-433.8 kJ / mol
** ''Final Energy: ''-434.5 kJ / mol
** Activation Energy: 0.7 kJ / mol
[[File:H2+f_emep.png|250px|thumb|centre|Energy vs Time diagram of transition state to form HF + H state (FH (AB) 181.11 pm and HH (BC) 74.487 pm) ]]

=== Reaction Dynamics ===

==== Reactive trajectory for H2 + F → HF + H ====
In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.

==== Illustrating Polanyi's empirical rules ====
Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state

File:H2+f emep.png

2020-05-07T13:45:25Z

Cb2018:

File:Eactbiggraph.png

2020-05-07T13:44:22Z

Cb2018:

MRD:celiab

2020-05-07T13:43:20Z

Cb2018: /* Activation Energy for F + H2 and H + HF reactions */

== EXERCISE 1: H + H2 ==

=== The Transition State ===
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?
* On PES diagrams, transition states are mathematically defined as points with 0 gradient with respect to position: stationary points.
* The transition state corresponds to the saddle point; the highest energy point on the reaction coordinate, which is the lowest energy pathway.
* We can distinguish it from a local minimum because on the reaction coordinate axis, the saddle point representing the transition state, will be a local maximum, but on the orthogonal axis it will be a local minimum. The minima are minimums on either direction because they represent stable reactants and products.

=== H + H2 reaction ===
[[File:Internuclear_distances_vs_time.png|250px|thumb|middle|Internuclear Distances vs Time plot ]]

===== Locating the Transition State =====
Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.
* The reaction is symmetric, hence, the transition state must have '''r1''' = '''r2''' by Hammond's Postulate
* By testing different initial conditions where '''rAB'''='''rBC '''and '''p1''' = '''p2''' = 0 g.mol-1.pm.fs-1, we can find the position of the Transition State
** The position of the transition state was found to be '''rts'''=90.775 pm
** This is where the system underwent small periodic symmetric vibrations keeping the distance between the atoms almost constant
* In the plot to the right, we see two straight lines, representing almost a stationary behaviour; thus we know we are at the symmetric transition state.

===== Calculating the Reaction Path =====
Comment on how the ''mep'' and the trajectory you just calculated differ.
* The minimum energy path is calculated from a system slightly displaced from the Transition state ('''r1''' = '''rts'''+1 pm, '''r2''' = '''rts''') and 0 initial momenta and follows the lowest energy path on a PES.
* As the reaction is symmetric it doesn't matter which atoms we label as A B C, so if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead, we get the same results but switched: AB will become BC and viceversa.
* The dynamic trajectory includes the momentum of the particles which causes oscillations about the trajectory path.
* The ''mep'' trajectory is roughly the same but it's smooth and doesn't oscillate, as the velocities are set to 0 in every step. This means we are artificially removing any gain in vibrational energy which results in oscillations. As the velocity is set 0 at every step, the kinetic energy is also 0.
[[File:picture 2 mep.png|350px|thumb|left|MEP contour plot ]]
[[File:picture 3 dynamic.png|350px|thumb|right|Dynamic contour plot ]]
[[File:Pic_4_dt.png|350px|thumb|left|MEP Internuclear Distances vs Time plot ]]
[[File:Pic_5_dt.png|350px|thumb|right| Dynamic Internuclear Distances vs Time plot ]]

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

For the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, the following trajectories with the following momenta combination were run:

(in the descriptions A B and C are used to describe the atoms colliding: AB is the initial H2 approaching H, represented by C)
{| class="wikitable"
!'''#'''
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|'''1'''
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.3</nowiki>
|yes
|The reaction dynamics proceed in a straight forward manner: AB and C approach each other, a collision occurs with the correct momentum for the reaction to proceed and a new bond BC is formed.
| [[File:Pic_a.png|260px]]
|-
|'''2'''
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420</nowiki>
|no
|AB approaches C and collides with it, but it doesn't have enough momentum to form a new bond so they are repelled. Momentum transfer does occur and C goes back with a greater momentum.
| [[File:Pic_b.png|260px]]
|-
|'''3'''
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414</nowiki>
|yes
|Oscillating AB collides with C and forms a new bond. The new H2 molecule, BC, has a higher vibrational frequency. A's direction is reversed.
| [[File:Pic_c.png|260px]]
|-
|'''4'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-375</nowiki>
|no
|AB approaches C and a BC bond forms, however, the system has too much momentum and C is repelled by the nucleus in B and bounces back. This leaves AB with greater vibrational energy and the distance BC increases as C is repelled.

| [[File:Pic_d.png|260px]]
|-
|'''5'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349</nowiki>
|yes
|AB collides with C and a new bond forms. However, unlike above, C still has enough energy to overcome the activation barrier. So, the BC bond forms, it breaks and reforms AB which then collides again with C to reform BC and leave as 2 entities: A + BC:
| [[File:Pic_e.png|260px]]
|}

Previous calculations have shown that that trajectories with initial conditions in the range '''r1''' =74 pm, '''r2''' = 200 pm, with -3.1 < '''p1'''/ g.mol-1.pm.fs-1 < -1.6 and '''p2''' = -5.1 g.mol-1.pm.fs-1 are reactive
* This is confirmed in combinations #1, #2 and #3.
From this we could hypothesise that all trajectories starting with the same positions but with higher values of momenta (higher kinetic energy) would be reactive as there would be enough kinetic energy to overcome the activation barrier. From the table we can see that this is not necessarily true:
* Combination #4 vs Combination #5.

==== 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 allows us to investigate the progress of a reaction by splitting it into 2 parts: reactants and products.
* For a reaction to occur the reactants must cross into the product "space" by going over the saddle point – the transition state!
* With the Born-Oppenheimer approximation, it is assumed there is a single saddle point (thus transition state) per reaction
* Transition State theory should only be used to estimate reaction rates at low temperature as it fails to correctly predict reaction rates for scenarios with high momentum (ie high temperature), where the transition state can be recrossed and thus predict reaction rates which are a poor match with experimental values
* We must also take into account that not every reaction with the correct energy will react, as orientation is important too so predicted rates with TS theory will be higher than experimental values.

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

=== PES inspection ===

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

This can be deduced by inspecting the PES surfaces bellow, where F is labelled as A and B C labels H H.
# F + H2 → HF + H : Exothermic
#* Looking at the reactants: A + BC in the PES plot (small BC distance, large AB distance), we can see they have a higher potential energy than the products: AB + C. Which is confirmed by looking at bond strengths: breaking the H–H bond (432 kJ / mol) vs making the H–F bond (565 kJ / mol), so energy is released during the reaction.
# HF + H → H2 + F : Endothermic
#* If we know look at: AB + C as reactants (small AB distance, large BC distance) we can see they have a lower potential energy than the products: BC + A. It is again rationalised by the fact that the H-F bond being broken in the reaction is stronger than the H-H bond being formed, so it requires an amount of energy to initiate the reaction.

[[File:H+hf_pes1L.png|275px|thumb|centre| F-H-H PES plot: AB is FH distance and BC is HH distance ]]

==== Transition state for F-H-H system ====
Locate the approximate position of the transition state.

From Hammond's postulate we know that in an endothermic reaction the transition state resembles the products and in an exothermic reaction it resembles the reactants. So, we can infer that the transition state will resemble H + HF.

The transition State for the F-H-H reaction was found at: 181.11 pm AB and 74.487 pm BC, with 0 initial momenta for both. This was done by varying the distances until they remained constant, confirmed by the forces along AB and BC being 0, marking a stationary point as the first derivative is 0.

[[File: HFTS_cp.png|275px|thumb|centre| F-H-H Contour plot when placed at Transition State ]]

==== Activation Energy for F + H2 and H + HF reactions ====
Report the activation energy for both reactions

The MEP was found for both reactions for a structure neighbouring the transition state: '''rts'''+1 pm.

The activation energy is the difference between the saddle point's maximum energy and the minimum energy of the products , so:
* Reaction 1: F + H2 → HF + H (FH (AB) 180.11 pm and HH (BC) 75.487 pm)
** ''TS Energy: ''- 434 kJ / mol
** ''Final Energy: ''- 560 kJ / mol
** Activation Energy: + 126 kJ / mol
* Reaction 2: HF + H → H2 + F (FH (AB) 181.11 pm and HH (BC) 74.487 pm)
** ''TS Energy: ''-433.8 kJ / mol
** ''Final Energy: ''-434.5 kJ / mol
** Activation Energy: 0.7 kJ / mol

=== Reaction Dynamics ===

==== Reactive trajectory for H2 + F → HF + H ====
In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.

==== Illustrating Polanyi's empirical rules ====
Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state

MRD:celiab

2020-05-07T12:26:53Z

Cb2018:

== EXERCISE 1: H + H2 ==

=== The Transition State ===
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?
* On PES diagrams, transition states are mathematically defined as points with 0 gradient with respect to position: stationary points.
* The transition state corresponds to the saddle point; the highest energy point on the reaction coordinate, which is the lowest energy pathway.
* We can distinguish it from a local minimum because on the reaction coordinate axis, the saddle point representing the transition state, will be a local maximum, but on the orthogonal axis it will be a local minimum. The minima are minimums on either direction because they represent stable reactants and products.

=== H + H2 reaction ===
[[File:Internuclear_distances_vs_time.png|250px|thumb|middle|Internuclear Distances vs Time plot ]]

===== Locating the Transition State =====
Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.
* The reaction is symmetric, hence, the transition state must have '''r1''' = '''r2''' by Hammond's Postulate
* By testing different initial conditions where '''rAB'''='''rBC '''and '''p1''' = '''p2''' = 0 g.mol-1.pm.fs-1, we can find the position of the Transition State
** The position of the transition state was found to be '''rts'''=90.775 pm
** This is where the system underwent small periodic symmetric vibrations keeping the distance between the atoms almost constant
* In the plot to the right, we see two straight lines, representing almost a stationary behaviour; thus we know we are at the symmetric transition state.

===== Calculating the Reaction Path =====
Comment on how the ''mep'' and the trajectory you just calculated differ.
* The minimum energy path is calculated from a system slightly displaced from the Transition state ('''r1''' = '''rts'''+1 pm, '''r2''' = '''rts''') and 0 initial momenta and follows the lowest energy path on a PES.
* As the reaction is symmetric it doesn't matter which atoms we label as A B C, so if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead, we get the same results but switched: AB will become BC and viceversa.
* The dynamic trajectory includes the momentum of the particles which causes oscillations about the trajectory path.
* The ''mep'' trajectory is roughly the same but it's smooth and doesn't oscillate, as the velocities are set to 0 in every step. This means we are artificially removing any gain in vibrational energy which results in oscillations. As the velocity is set 0 at every step, the kinetic energy is also 0.
[[File:picture 2 mep.png|350px|thumb|left|MEP contour plot ]]
[[File:picture 3 dynamic.png|350px|thumb|right|Dynamic contour plot ]]
[[File:Pic_4_dt.png|350px|thumb|left|MEP Internuclear Distances vs Time plot ]]
[[File:Pic_5_dt.png|350px|thumb|right| Dynamic Internuclear Distances vs Time plot ]]

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

For the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, the following trajectories with the following momenta combination were run:

(in the descriptions A B and C are used to describe the atoms colliding: AB is the initial H2 approaching H, represented by C)
{| class="wikitable"
!'''#'''
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|'''1'''
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.3</nowiki>
|yes
|The reaction dynamics proceed in a straight forward manner: AB and C approach each other, a collision occurs with the correct momentum for the reaction to proceed and a new bond BC is formed.
| [[File:Pic_a.png|260px]]
|-
|'''2'''
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420</nowiki>
|no
|AB approaches C and collides with it, but it doesn't have enough momentum to form a new bond so they are repelled. Momentum transfer does occur and C goes back with a greater momentum.
| [[File:Pic_b.png|260px]]
|-
|'''3'''
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414</nowiki>
|yes
|Oscillating AB collides with C and forms a new bond. The new H2 molecule, BC, has a higher vibrational frequency. A's direction is reversed.
| [[File:Pic_c.png|260px]]
|-
|'''4'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-375</nowiki>
|no
|AB approaches C and a BC bond forms, however, the system has too much momentum and C is repelled by the nucleus in B and bounces back. This leaves AB with greater vibrational energy and the distance BC increases as C is repelled.

| [[File:Pic_d.png|260px]]
|-
|'''5'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349</nowiki>
|yes
|AB collides with C and a new bond forms. However, unlike above, C still has enough energy to overcome the activation barrier. So, the BC bond forms, it breaks and reforms AB which then collides again with C to reform BC and leave as 2 entities: A + BC:
| [[File:Pic_e.png|260px]]
|}

Previous calculations have shown that that trajectories with initial conditions in the range '''r1''' =74 pm, '''r2''' = 200 pm, with -3.1 < '''p1'''/ g.mol-1.pm.fs-1 < -1.6 and '''p2''' = -5.1 g.mol-1.pm.fs-1 are reactive
* This is confirmed in combinations #1, #2 and #3.
From this we could hypothesise that all trajectories starting with the same positions but with higher values of momenta (higher kinetic energy) would be reactive as there would be enough kinetic energy to overcome the activation barrier. From the table we can see that this is not necessarily true:
* Combination #4 vs Combination #5.

==== 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 allows us to investigate the progress of a reaction by splitting it into 2 parts: reactants and products.
* For a reaction to occur the reactants must cross into the product "space" by going over the saddle point – the transition state!
* With the Born-Oppenheimer approximation, it is assumed there is a single saddle point (thus transition state) per reaction
* Transition State theory should only be used to estimate reaction rates at low temperature as it fails to correctly predict reaction rates for scenarios with high momentum (ie high temperature), where the transition state can be recrossed and thus predict reaction rates which are a poor match with experimental values
* We must also take into account that not every reaction with the correct energy will react, as orientation is important too so predicted rates with TS theory will be higher than experimental values.

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

=== PES inspection ===

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

This can be deduced by inspecting the PES surfaces bellow, where F is labelled as A and B C labels H H.
# F + H2 → HF + H : Exothermic
#* Looking at the reactants: A + BC in the PES plot (small BC distance, large AB distance), we can see they have a higher potential energy than the products: AB + C. Which is confirmed by looking at bond strengths: breaking the H–H bond (432 kJ / mol) vs making the H–F bond (565 kJ / mol), so energy is released during the reaction.
# HF + H → H2 + F : Endothermic
#* If we know look at: AB + C as reactants (small AB distance, large BC distance) we can see they have a lower potential energy than the products: BC + A. It is again rationalised by the fact that the H-F bond being broken in the reaction is stronger than the H-H bond being formed, so it requires an amount of energy to initiate the reaction.

[[File:H+hf_pes1L.png|275px|thumb|centre| F-H-H PES plot: AB is FH distance and BC is HH distance ]]

==== Transition state for F-H-H system ====
Locate the approximate position of the transition state.

From Hammond's postulate we know that in an endothermic reaction the transition state resembles the products and in an exothermic reaction it resembles the reactants. So, we can infer that the transition state will resemble H + HF.

The transition State for the F-H-H reaction was found at: 181.11 pm AB and 74.487 pm BC, with 0 initial momenta for both. This was done by varying the distances until they remained constant, confirmed by the forces along AB and BC being 0, marking a stationary point as the first derivative is 0.

[[File: HFTS_cp.png|275px|thumb|centre| F-H-H Contour plot when placed at Transition State ]]

==== Activation Energy for F + H2 and H + HF reactions ====
Report the activation energy for both reactions

The MEP was found for both reactions for a structure neighbouring the transition state: '''rts'''+1 pm

The activation energy is the difference between the saddle point's maximum energy and the minimum energy of the products , so:
* Reaction 1:
** ''Initial Energy: ''
** ''Product Energy:''
** Activation Energy:
* Reaction 2:
** ''Initial Energy: ''
** ''Product Energy:''
** Activation Energy:

=== Reaction Dynamics ===

==== Reactive trajectory for H2 + F → HF + H ====
In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.

==== Illustrating Polanyi's empirical rules ====
Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state

MRD:celiab

2020-05-07T12:25:47Z

Cb2018: /* Transition state for F-H-H system */

== EXERCISE 1: H + H2 ==

=== The Transition State ===
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?
* On PES diagrams, transition states are mathematically defined as points with 0 gradient with respect to position: stationary points.
* The transition state corresponds to the saddle point; the highest energy point on the reaction coordinate, which is the lowest energy pathway.
* We can distinguish it from a local minimum because on the reaction coordinate axis, the saddle point representing the transition state, will be a local maximum, but on the orthogonal axis it will be a local minimum. The minima are minimums on either direction because they represent stable reactants and products.

=== H + H2 reaction ===
[[File:Internuclear_distances_vs_time.png|250px|thumb|middle|Internuclear Distances vs Time plot ]]

===== Locating the Transition State =====
Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.
* The reaction is symmetric, hence, the transition state must have '''r1''' = '''r2''' by Hammond's Postulate
* By testing different initial conditions where '''rAB'''='''rBC '''and '''p1''' = '''p2''' = 0 g.mol-1.pm.fs-1, we can find the position of the Transition State
** The position of the transition state was found to be '''rts'''=90.775 pm
** This is where the system underwent small periodic symmetric vibrations keeping the distance between the atoms almost constant
* In the plot to the right, we see two straight lines, representing almost a stationary behaviour; thus we know we are at the symmetric transition state.

===== Calculating the Reaction Path =====
Comment on how the ''mep'' and the trajectory you just calculated differ.
* The minimum energy path is calculated from a system slightly displaced from the Transition state ('''r1''' = '''rts'''+1 pm, '''r2''' = '''rts''') and 0 initial momenta and follows the lowest energy path on a PES.
* As the reaction is symmetric it doesn't matter which atoms we label as A B C, so if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead, we get the same results but switched: AB will become BC and viceversa.
* The dynamic trajectory includes the momentum of the particles which causes oscillations about the trajectory path.
* The ''mep'' trajectory is roughly the same but it's smooth and doesn't oscillate, as the velocities are set to 0 in every step. This means we are artificially removing any gain in vibrational energy which results in oscillations. As the velocity is set 0 at every step, the kinetic energy is also 0.
[[File:picture 2 mep.png|350px|thumb|left|MEP contour plot ]]
[[File:picture 3 dynamic.png|350px|thumb|right|Dynamic contour plot ]]
[[File:Pic_4_dt.png|350px|thumb|left|MEP Internuclear Distances vs Time plot ]]
[[File:Pic_5_dt.png|350px|thumb|right| Dynamic Internuclear Distances vs Time plot ]]

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

For the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, the following trajectories with the following momenta combination were run:

(in the descriptions A B and C are used to describe the atoms colliding: AB is the initial H2 approaching H, represented by C)
{| class="wikitable"
!'''#'''
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|'''1'''
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.3</nowiki>
|yes
|The reaction dynamics proceed in a straight forward manner: AB and C approach each other, a collision occurs with the correct momentum for the reaction to proceed and a new bond BC is formed.
| [[File:Pic_a.png|260px]]
|-
|'''2'''
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420</nowiki>
|no
|AB approaches C and collides with it, but it doesn't have enough momentum to form a new bond so they are repelled. Momentum transfer does occur and C goes back with a greater momentum.
| [[File:Pic_b.png|260px]]
|-
|'''3'''
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414</nowiki>
|yes
|Oscillating AB collides with C and forms a new bond. The new H2 molecule, BC, has a higher vibrational frequency. A's direction is reversed.
| [[File:Pic_c.png|260px]]
|-
|'''4'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-375</nowiki>
|no
|AB approaches C and a BC bond forms, however, the system has too much momentum and C is repelled by the nucleus in B and bounces back. This leaves AB with greater vibrational energy and the distance BC increases as C is repelled.

| [[File:Pic_d.png|260px]]
|-
|'''5'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349</nowiki>
|yes
|AB collides with C and a new bond forms. However, unlike above, C still has enough energy to overcome the activation barrier. So, the BC bond forms, it breaks and reforms AB which then collides again with C to reform BC and leave as 2 entities: A + BC:
| [[File:Pic_e.png|260px]]
|}

Previous calculations have shown that that trajectories with initial conditions in the range '''r1''' =74 pm, '''r2''' = 200 pm, with -3.1 < '''p1'''/ g.mol-1.pm.fs-1 < -1.6 and '''p2''' = -5.1 g.mol-1.pm.fs-1 are reactive
* This is confirmed in combinations #1, #2 and #3.
From this we could hypothesise that all trajectories starting with the same positions but with higher values of momenta (higher kinetic energy) would be reactive as there would be enough kinetic energy to overcome the activation barrier. From the table we can see that this is not necessarily true:
* Combination #4 vs Combination #5.

==== 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 allows us to investigate the progress of a reaction by splitting it into 2 parts: reactants and products.
* For a reaction to occur the reactants must cross into the product "space" by going over the saddle point – the transition state!
* With the Born-Oppenheimer approximation, it is assumed there is a single saddle point (thus transition state) per reaction
* Transition State theory should only be used to estimate reaction rates at low temperature as it fails to correctly predict reaction rates for scenarios with high momentum (ie high temperature), where the transition state can be recrossed and thus predict reaction rates which are a poor match with experimental values
* We must also take into account that not every reaction with the correct energy will react, as orientation is important too so predicted rates with TS theory will be higher than experimental values.

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

=== PES inspection ===

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

This can be deduced by inspecting the PES surfaces bellow, where F is labelled as A and B C labels H H.
# F + H2 → HF + H : Exothermic
#* Looking at the reactants: A + BC in the PES plot (small BC distance, large AB distance), we can see they have a higher potential energy than the products: AB + C. Which is confirmed by looking at bond strengths: breaking the H–H bond (432 kJ / mol) vs making the H–F bond (565 kJ / mol), so energy is released during the reaction.
# HF + H → H2 + F : Endothermic
#* If we know look at: AB + C as reactants (small AB distance, large BC distance) we can see they have a lower potential energy than the products: BC + A. It is again rationalised by the fact that the H-F bond being broken in the reaction is stronger than the H-H bond being formed, so it requires an amount of energy to initiate the reaction.

[[File:H+hf_pes1L.png|300px|thumb|centre| F-H-H PES plot: AB is FH distance and BC is HH distance ]]

==== Transition state for F-H-H system ====
Locate the approximate position of the transition state.

From Hammond's postulate we know that in an endothermic reaction the transition state resembles the products and in an exothermic reaction it resembles the reactants. So, we can infer that the transition state will resemble H + HF.

The transition State for the F-H-H reaction was found at: 181.11 pm AB and 74.487 pm BC, with 0 initial momenta for both. This was done by varying the distances until they remained constant, confirmed by the forces along AB and BC being 0, marking a stationary point as the first derivative is 0.

[[File: HFTS_cp.png]]
PICTURE contour zoom

==== Activation Energy for F + H2 and H + HF reactions ====
Report the activation energy for both reactions

The MEP was found for both reactions for a structure neighbouring the transition state: '''rts'''+1 pm

The activation energy is the difference between the saddle point's maximum energy and the minimum energy of the products , so:
* Reaction 1:
** ''Initial Energy: ''
** ''Product Energy:''
** Activation Energy:
* Reaction 2:
** ''Initial Energy: ''
** ''Product Energy:''
** Activation Energy:

=== Reaction Dynamics ===

==== Reactive trajectory for H2 + F → HF + H ====
In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.

==== Illustrating Polanyi's empirical rules ====
Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state

File:HFTS cp.png

2020-05-07T12:25:05Z

Cb2018:

MRD:celiab

2020-05-07T12:23:38Z

Cb2018: /* PES inspection */

== EXERCISE 1: H + H2 ==

=== The Transition State ===
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?
* On PES diagrams, transition states are mathematically defined as points with 0 gradient with respect to position: stationary points.
* The transition state corresponds to the saddle point; the highest energy point on the reaction coordinate, which is the lowest energy pathway.
* We can distinguish it from a local minimum because on the reaction coordinate axis, the saddle point representing the transition state, will be a local maximum, but on the orthogonal axis it will be a local minimum. The minima are minimums on either direction because they represent stable reactants and products.

=== H + H2 reaction ===
[[File:Internuclear_distances_vs_time.png|250px|thumb|middle|Internuclear Distances vs Time plot ]]

===== Locating the Transition State =====
Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.
* The reaction is symmetric, hence, the transition state must have '''r1''' = '''r2''' by Hammond's Postulate
* By testing different initial conditions where '''rAB'''='''rBC '''and '''p1''' = '''p2''' = 0 g.mol-1.pm.fs-1, we can find the position of the Transition State
** The position of the transition state was found to be '''rts'''=90.775 pm
** This is where the system underwent small periodic symmetric vibrations keeping the distance between the atoms almost constant
* In the plot to the right, we see two straight lines, representing almost a stationary behaviour; thus we know we are at the symmetric transition state.

===== Calculating the Reaction Path =====
Comment on how the ''mep'' and the trajectory you just calculated differ.
* The minimum energy path is calculated from a system slightly displaced from the Transition state ('''r1''' = '''rts'''+1 pm, '''r2''' = '''rts''') and 0 initial momenta and follows the lowest energy path on a PES.
* As the reaction is symmetric it doesn't matter which atoms we label as A B C, so if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead, we get the same results but switched: AB will become BC and viceversa.
* The dynamic trajectory includes the momentum of the particles which causes oscillations about the trajectory path.
* The ''mep'' trajectory is roughly the same but it's smooth and doesn't oscillate, as the velocities are set to 0 in every step. This means we are artificially removing any gain in vibrational energy which results in oscillations. As the velocity is set 0 at every step, the kinetic energy is also 0.
[[File:picture 2 mep.png|350px|thumb|left|MEP contour plot ]]
[[File:picture 3 dynamic.png|350px|thumb|right|Dynamic contour plot ]]
[[File:Pic_4_dt.png|350px|thumb|left|MEP Internuclear Distances vs Time plot ]]
[[File:Pic_5_dt.png|350px|thumb|right| Dynamic Internuclear Distances vs Time plot ]]

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

For the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, the following trajectories with the following momenta combination were run:

(in the descriptions A B and C are used to describe the atoms colliding: AB is the initial H2 approaching H, represented by C)
{| class="wikitable"
!'''#'''
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|'''1'''
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.3</nowiki>
|yes
|The reaction dynamics proceed in a straight forward manner: AB and C approach each other, a collision occurs with the correct momentum for the reaction to proceed and a new bond BC is formed.
| [[File:Pic_a.png|260px]]
|-
|'''2'''
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420</nowiki>
|no
|AB approaches C and collides with it, but it doesn't have enough momentum to form a new bond so they are repelled. Momentum transfer does occur and C goes back with a greater momentum.
| [[File:Pic_b.png|260px]]
|-
|'''3'''
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414</nowiki>
|yes
|Oscillating AB collides with C and forms a new bond. The new H2 molecule, BC, has a higher vibrational frequency. A's direction is reversed.
| [[File:Pic_c.png|260px]]
|-
|'''4'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-375</nowiki>
|no
|AB approaches C and a BC bond forms, however, the system has too much momentum and C is repelled by the nucleus in B and bounces back. This leaves AB with greater vibrational energy and the distance BC increases as C is repelled.

| [[File:Pic_d.png|260px]]
|-
|'''5'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349</nowiki>
|yes
|AB collides with C and a new bond forms. However, unlike above, C still has enough energy to overcome the activation barrier. So, the BC bond forms, it breaks and reforms AB which then collides again with C to reform BC and leave as 2 entities: A + BC:
| [[File:Pic_e.png|260px]]
|}

Previous calculations have shown that that trajectories with initial conditions in the range '''r1''' =74 pm, '''r2''' = 200 pm, with -3.1 < '''p1'''/ g.mol-1.pm.fs-1 < -1.6 and '''p2''' = -5.1 g.mol-1.pm.fs-1 are reactive
* This is confirmed in combinations #1, #2 and #3.
From this we could hypothesise that all trajectories starting with the same positions but with higher values of momenta (higher kinetic energy) would be reactive as there would be enough kinetic energy to overcome the activation barrier. From the table we can see that this is not necessarily true:
* Combination #4 vs Combination #5.

==== 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 allows us to investigate the progress of a reaction by splitting it into 2 parts: reactants and products.
* For a reaction to occur the reactants must cross into the product "space" by going over the saddle point – the transition state!
* With the Born-Oppenheimer approximation, it is assumed there is a single saddle point (thus transition state) per reaction
* Transition State theory should only be used to estimate reaction rates at low temperature as it fails to correctly predict reaction rates for scenarios with high momentum (ie high temperature), where the transition state can be recrossed and thus predict reaction rates which are a poor match with experimental values
* We must also take into account that not every reaction with the correct energy will react, as orientation is important too so predicted rates with TS theory will be higher than experimental values.

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

=== PES inspection ===

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

This can be deduced by inspecting the PES surfaces bellow, where F is labelled as A and B C labels H H.
# F + H2 → HF + H : Exothermic
#* Looking at the reactants: A + BC in the PES plot (small BC distance, large AB distance), we can see they have a higher potential energy than the products: AB + C. Which is confirmed by looking at bond strengths: breaking the H–H bond (432 kJ / mol) vs making the H–F bond (565 kJ / mol), so energy is released during the reaction.
# HF + H → H2 + F : Endothermic
#* If we know look at: AB + C as reactants (small AB distance, large BC distance) we can see they have a lower potential energy than the products: BC + A. It is again rationalised by the fact that the H-F bond being broken in the reaction is stronger than the H-H bond being formed, so it requires an amount of energy to initiate the reaction.

[[File:H+hf_pes1L.png|300px|thumb|centre| F-H-H PES plot: AB is FH distance and BC is HH distance ]]

==== Transition state for F-H-H system ====
Locate the approximate position of the transition state.

From Hammond's postulate we know that in an endothermic reaction the transition state resembles the products and in an exothermic reaction it resembles the reactants. So, we can infer that the transition state will resemble H + HF.

The transition State for the F-H-H reaction was found at: 181.11 pm AB and 74.487 pm BC, with 0 initial momenta for both. This was done by varying the distances until they remained constant, confirmed by the forces along AB and BC being 0, marking a stationary point as the first derivative is 0.

PICTURE contour zoom

==== Activation Energy for F + H2 and H + HF reactions ====
Report the activation energy for both reactions

The MEP was found for both reactions for a structure neighbouring the transition state: '''rts'''+1 pm

The activation energy is the difference between the saddle point's maximum energy and the minimum energy of the products , so:
* Reaction 1:
** ''Initial Energy: ''
** ''Product Energy:''
** Activation Energy:
* Reaction 2:
** ''Initial Energy: ''
** ''Product Energy:''
** Activation Energy:

=== Reaction Dynamics ===

==== Reactive trajectory for H2 + F → HF + H ====
In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.

==== Illustrating Polanyi's empirical rules ====
Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state

MRD:celiab

2020-05-07T12:23:18Z

Cb2018: /* PES inspection */

== EXERCISE 1: H + H2 ==

=== The Transition State ===
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?
* On PES diagrams, transition states are mathematically defined as points with 0 gradient with respect to position: stationary points.
* The transition state corresponds to the saddle point; the highest energy point on the reaction coordinate, which is the lowest energy pathway.
* We can distinguish it from a local minimum because on the reaction coordinate axis, the saddle point representing the transition state, will be a local maximum, but on the orthogonal axis it will be a local minimum. The minima are minimums on either direction because they represent stable reactants and products.

=== H + H2 reaction ===
[[File:Internuclear_distances_vs_time.png|250px|thumb|middle|Internuclear Distances vs Time plot ]]

===== Locating the Transition State =====
Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.
* The reaction is symmetric, hence, the transition state must have '''r1''' = '''r2''' by Hammond's Postulate
* By testing different initial conditions where '''rAB'''='''rBC '''and '''p1''' = '''p2''' = 0 g.mol-1.pm.fs-1, we can find the position of the Transition State
** The position of the transition state was found to be '''rts'''=90.775 pm
** This is where the system underwent small periodic symmetric vibrations keeping the distance between the atoms almost constant
* In the plot to the right, we see two straight lines, representing almost a stationary behaviour; thus we know we are at the symmetric transition state.

===== Calculating the Reaction Path =====
Comment on how the ''mep'' and the trajectory you just calculated differ.
* The minimum energy path is calculated from a system slightly displaced from the Transition state ('''r1''' = '''rts'''+1 pm, '''r2''' = '''rts''') and 0 initial momenta and follows the lowest energy path on a PES.
* As the reaction is symmetric it doesn't matter which atoms we label as A B C, so if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead, we get the same results but switched: AB will become BC and viceversa.
* The dynamic trajectory includes the momentum of the particles which causes oscillations about the trajectory path.
* The ''mep'' trajectory is roughly the same but it's smooth and doesn't oscillate, as the velocities are set to 0 in every step. This means we are artificially removing any gain in vibrational energy which results in oscillations. As the velocity is set 0 at every step, the kinetic energy is also 0.
[[File:picture 2 mep.png|350px|thumb|left|MEP contour plot ]]
[[File:picture 3 dynamic.png|350px|thumb|right|Dynamic contour plot ]]
[[File:Pic_4_dt.png|350px|thumb|left|MEP Internuclear Distances vs Time plot ]]
[[File:Pic_5_dt.png|350px|thumb|right| Dynamic Internuclear Distances vs Time plot ]]

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

For the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, the following trajectories with the following momenta combination were run:

(in the descriptions A B and C are used to describe the atoms colliding: AB is the initial H2 approaching H, represented by C)
{| class="wikitable"
!'''#'''
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|'''1'''
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.3</nowiki>
|yes
|The reaction dynamics proceed in a straight forward manner: AB and C approach each other, a collision occurs with the correct momentum for the reaction to proceed and a new bond BC is formed.
| [[File:Pic_a.png|260px]]
|-
|'''2'''
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420</nowiki>
|no
|AB approaches C and collides with it, but it doesn't have enough momentum to form a new bond so they are repelled. Momentum transfer does occur and C goes back with a greater momentum.
| [[File:Pic_b.png|260px]]
|-
|'''3'''
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414</nowiki>
|yes
|Oscillating AB collides with C and forms a new bond. The new H2 molecule, BC, has a higher vibrational frequency. A's direction is reversed.
| [[File:Pic_c.png|260px]]
|-
|'''4'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-375</nowiki>
|no
|AB approaches C and a BC bond forms, however, the system has too much momentum and C is repelled by the nucleus in B and bounces back. This leaves AB with greater vibrational energy and the distance BC increases as C is repelled.

| [[File:Pic_d.png|260px]]
|-
|'''5'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349</nowiki>
|yes
|AB collides with C and a new bond forms. However, unlike above, C still has enough energy to overcome the activation barrier. So, the BC bond forms, it breaks and reforms AB which then collides again with C to reform BC and leave as 2 entities: A + BC:
| [[File:Pic_e.png|260px]]
|}

Previous calculations have shown that that trajectories with initial conditions in the range '''r1''' =74 pm, '''r2''' = 200 pm, with -3.1 < '''p1'''/ g.mol-1.pm.fs-1 < -1.6 and '''p2''' = -5.1 g.mol-1.pm.fs-1 are reactive
* This is confirmed in combinations #1, #2 and #3.
From this we could hypothesise that all trajectories starting with the same positions but with higher values of momenta (higher kinetic energy) would be reactive as there would be enough kinetic energy to overcome the activation barrier. From the table we can see that this is not necessarily true:
* Combination #4 vs Combination #5.

==== 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 allows us to investigate the progress of a reaction by splitting it into 2 parts: reactants and products.
* For a reaction to occur the reactants must cross into the product "space" by going over the saddle point – the transition state!
* With the Born-Oppenheimer approximation, it is assumed there is a single saddle point (thus transition state) per reaction
* Transition State theory should only be used to estimate reaction rates at low temperature as it fails to correctly predict reaction rates for scenarios with high momentum (ie high temperature), where the transition state can be recrossed and thus predict reaction rates which are a poor match with experimental values
* We must also take into account that not every reaction with the correct energy will react, as orientation is important too so predicted rates with TS theory will be higher than experimental values.

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

=== PES inspection ===

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

This can be deduced by inspecting the PES surfaces bellow, where F is labelled as A and B C labels H H.
# F + H2 → HF + H : Exothermic
#* Looking at the reactants: A + BC in the PES plot (small BC distance, large AB distance), we can see they have a higher potential energy than the products: AB + C. Which is confirmed by looking at bond strengths: breaking the H–H bond (432 kJ / mol) vs making the H–F bond (565 kJ / mol), so energy is released during the reaction.
# HF + H → H2 + F : Endothermic
#* If we know look at: AB + C as reactants (small AB distance, large BC distance) we can see they have a lower potential energy than the products: BC + A. It is again rationalised by the fact that the H-F bond being broken in the reaction is stronger than the H-H bond being formed, so it requires an amount of energy to initiate the reaction.

[[File:H+hf_pes1L.png|200px|thumb|centre| F-H-H PES plot: AB is FH distance and BC is HH distance ]]

F H H energy surface where AB is FH distance and BC is HH distance
==== Transition state for F-H-H system ====
Locate the approximate position of the transition state.

From Hammond's postulate we know that in an endothermic reaction the transition state resembles the products and in an exothermic reaction it resembles the reactants. So, we can infer that the transition state will resemble H + HF.

The transition State for the F-H-H reaction was found at: 181.11 pm AB and 74.487 pm BC, with 0 initial momenta for both. This was done by varying the distances until they remained constant, confirmed by the forces along AB and BC being 0, marking a stationary point as the first derivative is 0.

PICTURE contour zoom

==== Activation Energy for F + H2 and H + HF reactions ====
Report the activation energy for both reactions

The MEP was found for both reactions for a structure neighbouring the transition state: '''rts'''+1 pm

The activation energy is the difference between the saddle point's maximum energy and the minimum energy of the products , so:
* Reaction 1:
** ''Initial Energy: ''
** ''Product Energy:''
** Activation Energy:
* Reaction 2:
** ''Initial Energy: ''
** ''Product Energy:''
** Activation Energy:

=== Reaction Dynamics ===

==== Reactive trajectory for H2 + F → HF + H ====
In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.

==== Illustrating Polanyi's empirical rules ====
Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state

MRD:celiab

2020-05-07T12:22:13Z

Cb2018: /* PES inspection */

== EXERCISE 1: H + H2 ==

=== The Transition State ===
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?
* On PES diagrams, transition states are mathematically defined as points with 0 gradient with respect to position: stationary points.
* The transition state corresponds to the saddle point; the highest energy point on the reaction coordinate, which is the lowest energy pathway.
* We can distinguish it from a local minimum because on the reaction coordinate axis, the saddle point representing the transition state, will be a local maximum, but on the orthogonal axis it will be a local minimum. The minima are minimums on either direction because they represent stable reactants and products.

=== H + H2 reaction ===
[[File:Internuclear_distances_vs_time.png|250px|thumb|middle|Internuclear Distances vs Time plot ]]

===== Locating the Transition State =====
Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.
* The reaction is symmetric, hence, the transition state must have '''r1''' = '''r2''' by Hammond's Postulate
* By testing different initial conditions where '''rAB'''='''rBC '''and '''p1''' = '''p2''' = 0 g.mol-1.pm.fs-1, we can find the position of the Transition State
** The position of the transition state was found to be '''rts'''=90.775 pm
** This is where the system underwent small periodic symmetric vibrations keeping the distance between the atoms almost constant
* In the plot to the right, we see two straight lines, representing almost a stationary behaviour; thus we know we are at the symmetric transition state.

===== Calculating the Reaction Path =====
Comment on how the ''mep'' and the trajectory you just calculated differ.
* The minimum energy path is calculated from a system slightly displaced from the Transition state ('''r1''' = '''rts'''+1 pm, '''r2''' = '''rts''') and 0 initial momenta and follows the lowest energy path on a PES.
* As the reaction is symmetric it doesn't matter which atoms we label as A B C, so if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead, we get the same results but switched: AB will become BC and viceversa.
* The dynamic trajectory includes the momentum of the particles which causes oscillations about the trajectory path.
* The ''mep'' trajectory is roughly the same but it's smooth and doesn't oscillate, as the velocities are set to 0 in every step. This means we are artificially removing any gain in vibrational energy which results in oscillations. As the velocity is set 0 at every step, the kinetic energy is also 0.
[[File:picture 2 mep.png|350px|thumb|left|MEP contour plot ]]
[[File:picture 3 dynamic.png|350px|thumb|right|Dynamic contour plot ]]
[[File:Pic_4_dt.png|350px|thumb|left|MEP Internuclear Distances vs Time plot ]]
[[File:Pic_5_dt.png|350px|thumb|right| Dynamic Internuclear Distances vs Time plot ]]

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

For the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, the following trajectories with the following momenta combination were run:

(in the descriptions A B and C are used to describe the atoms colliding: AB is the initial H2 approaching H, represented by C)
{| class="wikitable"
!'''#'''
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|'''1'''
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.3</nowiki>
|yes
|The reaction dynamics proceed in a straight forward manner: AB and C approach each other, a collision occurs with the correct momentum for the reaction to proceed and a new bond BC is formed.
| [[File:Pic_a.png|260px]]
|-
|'''2'''
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420</nowiki>
|no
|AB approaches C and collides with it, but it doesn't have enough momentum to form a new bond so they are repelled. Momentum transfer does occur and C goes back with a greater momentum.
| [[File:Pic_b.png|260px]]
|-
|'''3'''
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414</nowiki>
|yes
|Oscillating AB collides with C and forms a new bond. The new H2 molecule, BC, has a higher vibrational frequency. A's direction is reversed.
| [[File:Pic_c.png|260px]]
|-
|'''4'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-375</nowiki>
|no
|AB approaches C and a BC bond forms, however, the system has too much momentum and C is repelled by the nucleus in B and bounces back. This leaves AB with greater vibrational energy and the distance BC increases as C is repelled.

| [[File:Pic_d.png|260px]]
|-
|'''5'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349</nowiki>
|yes
|AB collides with C and a new bond forms. However, unlike above, C still has enough energy to overcome the activation barrier. So, the BC bond forms, it breaks and reforms AB which then collides again with C to reform BC and leave as 2 entities: A + BC:
| [[File:Pic_e.png|260px]]
|}

Previous calculations have shown that that trajectories with initial conditions in the range '''r1''' =74 pm, '''r2''' = 200 pm, with -3.1 < '''p1'''/ g.mol-1.pm.fs-1 < -1.6 and '''p2''' = -5.1 g.mol-1.pm.fs-1 are reactive
* This is confirmed in combinations #1, #2 and #3.
From this we could hypothesise that all trajectories starting with the same positions but with higher values of momenta (higher kinetic energy) would be reactive as there would be enough kinetic energy to overcome the activation barrier. From the table we can see that this is not necessarily true:
* Combination #4 vs Combination #5.

==== 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 allows us to investigate the progress of a reaction by splitting it into 2 parts: reactants and products.
* For a reaction to occur the reactants must cross into the product "space" by going over the saddle point – the transition state!
* With the Born-Oppenheimer approximation, it is assumed there is a single saddle point (thus transition state) per reaction
* Transition State theory should only be used to estimate reaction rates at low temperature as it fails to correctly predict reaction rates for scenarios with high momentum (ie high temperature), where the transition state can be recrossed and thus predict reaction rates which are a poor match with experimental values
* We must also take into account that not every reaction with the correct energy will react, as orientation is important too so predicted rates with TS theory will be higher than experimental values.

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

=== PES inspection ===

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

This can be deduced by inspecting the PES surfaces bellow, where F is labelled as A and B C labels H H.
# F + H2 → HF + H : Exothermic
#* Looking at the reactants: A + BC in the PES plot (small BC distance, large AB distance), we can see they have a higher potential energy than the products: AB + C. Which is confirmed by looking at bond strengths: breaking the H–H bond (432 kJ / mol) vs making the H–F bond (565 kJ / mol), so energy is released during the reaction.
# HF + H → H2 + F : Endothermic
#* If we know look at: AB + C as reactants (small AB distance, large BC distance) we can see they have a lower potential energy than the products: BC + A. It is again rationalised by the fact that the H-F bond being broken in the reaction is stronger than the H-H bond being formed, so it requires an amount of energy to initiate the reaction.
PES PES PES
[[File:H+hf_pes1L.png|350px|thumb|right| F-H-H PES plot: AB is FH distance and BC is HH distance ]]

F H H energy surface where AB is FH distance and BC is HH distance
==== Transition state for F-H-H system ====
Locate the approximate position of the transition state.

From Hammond's postulate we know that in an endothermic reaction the transition state resembles the products and in an exothermic reaction it resembles the reactants. So, we can infer that the transition state will resemble H + HF.

The transition State for the F-H-H reaction was found at: 181.11 pm AB and 74.487 pm BC, with 0 initial momenta for both. This was done by varying the distances until they remained constant, confirmed by the forces along AB and BC being 0, marking a stationary point as the first derivative is 0.

PICTURE contour zoom

==== Activation Energy for F + H2 and H + HF reactions ====
Report the activation energy for both reactions

The MEP was found for both reactions for a structure neighbouring the transition state: '''rts'''+1 pm

The activation energy is the difference between the saddle point's maximum energy and the minimum energy of the products , so:
* Reaction 1:
** ''Initial Energy: ''
** ''Product Energy:''
** Activation Energy:
* Reaction 2:
** ''Initial Energy: ''
** ''Product Energy:''
** Activation Energy:

=== Reaction Dynamics ===

==== Reactive trajectory for H2 + F → HF + H ====
In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.

==== Illustrating Polanyi's empirical rules ====
Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state

File:H+hf pes1L.png

2020-05-07T12:21:24Z

Cb2018:

MRD:celiab

2020-05-07T11:45:09Z

Cb2018: /* Activation Energy for F + H2 and H + HF reactions */

== EXERCISE 1: H + H2 ==

=== The Transition State ===
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?
* On PES diagrams, transition states are mathematically defined as points with 0 gradient with respect to position: stationary points.
* The transition state corresponds to the saddle point; the highest energy point on the reaction coordinate, which is the lowest energy pathway.
* We can distinguish it from a local minimum because on the reaction coordinate axis, the saddle point representing the transition state, will be a local maximum, but on the orthogonal axis it will be a local minimum. The minima are minimums on either direction because they represent stable reactants and products.

=== H + H2 reaction ===
[[File:Internuclear_distances_vs_time.png|250px|thumb|middle|Internuclear Distances vs Time plot ]]

===== Locating the Transition State =====
Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.
* The reaction is symmetric, hence, the transition state must have '''r1''' = '''r2''' by Hammond's Postulate
* By testing different initial conditions where '''rAB'''='''rBC '''and '''p1''' = '''p2''' = 0 g.mol-1.pm.fs-1, we can find the position of the Transition State
** The position of the transition state was found to be '''rts'''=90.775 pm
** This is where the system underwent small periodic symmetric vibrations keeping the distance between the atoms almost constant
* In the plot to the right, we see two straight lines, representing almost a stationary behaviour; thus we know we are at the symmetric transition state.

===== Calculating the Reaction Path =====
Comment on how the ''mep'' and the trajectory you just calculated differ.
* The minimum energy path is calculated from a system slightly displaced from the Transition state ('''r1''' = '''rts'''+1 pm, '''r2''' = '''rts''') and 0 initial momenta and follows the lowest energy path on a PES.
* As the reaction is symmetric it doesn't matter which atoms we label as A B C, so if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead, we get the same results but switched: AB will become BC and viceversa.
* The dynamic trajectory includes the momentum of the particles which causes oscillations about the trajectory path.
* The ''mep'' trajectory is roughly the same but it's smooth and doesn't oscillate, as the velocities are set to 0 in every step. This means we are artificially removing any gain in vibrational energy which results in oscillations. As the velocity is set 0 at every step, the kinetic energy is also 0.
[[File:picture 2 mep.png|350px|thumb|left|MEP contour plot ]]
[[File:picture 3 dynamic.png|350px|thumb|right|Dynamic contour plot ]]
[[File:Pic_4_dt.png|350px|thumb|left|MEP Internuclear Distances vs Time plot ]]
[[File:Pic_5_dt.png|350px|thumb|right| Dynamic Internuclear Distances vs Time plot ]]

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

For the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, the following trajectories with the following momenta combination were run:

(in the descriptions A B and C are used to describe the atoms colliding: AB is the initial H2 approaching H, represented by C)
{| class="wikitable"
!'''#'''
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|'''1'''
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.3</nowiki>
|yes
|The reaction dynamics proceed in a straight forward manner: AB and C approach each other, a collision occurs with the correct momentum for the reaction to proceed and a new bond BC is formed.
| [[File:Pic_a.png|260px]]
|-
|'''2'''
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420</nowiki>
|no
|AB approaches C and collides with it, but it doesn't have enough momentum to form a new bond so they are repelled. Momentum transfer does occur and C goes back with a greater momentum.
| [[File:Pic_b.png|260px]]
|-
|'''3'''
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414</nowiki>
|yes
|Oscillating AB collides with C and forms a new bond. The new H2 molecule, BC, has a higher vibrational frequency. A's direction is reversed.
| [[File:Pic_c.png|260px]]
|-
|'''4'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-375</nowiki>
|no
|AB approaches C and a BC bond forms, however, the system has too much momentum and C is repelled by the nucleus in B and bounces back. This leaves AB with greater vibrational energy and the distance BC increases as C is repelled.

| [[File:Pic_d.png|260px]]
|-
|'''5'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349</nowiki>
|yes
|AB collides with C and a new bond forms. However, unlike above, C still has enough energy to overcome the activation barrier. So, the BC bond forms, it breaks and reforms AB which then collides again with C to reform BC and leave as 2 entities: A + BC:
| [[File:Pic_e.png|260px]]
|}

Previous calculations have shown that that trajectories with initial conditions in the range '''r1''' =74 pm, '''r2''' = 200 pm, with -3.1 < '''p1'''/ g.mol-1.pm.fs-1 < -1.6 and '''p2''' = -5.1 g.mol-1.pm.fs-1 are reactive
* This is confirmed in combinations #1, #2 and #3.
From this we could hypothesise that all trajectories starting with the same positions but with higher values of momenta (higher kinetic energy) would be reactive as there would be enough kinetic energy to overcome the activation barrier. From the table we can see that this is not necessarily true:
* Combination #4 vs Combination #5.

==== 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 allows us to investigate the progress of a reaction by splitting it into 2 parts: reactants and products.
* For a reaction to occur the reactants must cross into the product "space" by going over the saddle point – the transition state!
* With the Born-Oppenheimer approximation, it is assumed there is a single saddle point (thus transition state) per reaction
* Transition State theory should only be used to estimate reaction rates at low temperature as it fails to correctly predict reaction rates for scenarios with high momentum (ie high temperature), where the transition state can be recrossed and thus predict reaction rates which are a poor match with experimental values
* We must also take into account that not every reaction with the correct energy will react, as orientation is important too so predicted rates with TS theory will be higher than experimental values.

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

=== PES inspection ===

==== Classifying F + H2 and H + HF reactions ====
By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?
# H2 + F → HF + H : Exothermic
# HF + H → H2 + F : Endothermic
This can be deduced by inspecting the PES surface bellow, where F is labelled as A and BC labels HH.
* If we look at small AB distances and large BC we are considering the reactants: HF + H, and we can see that the potential is lower than for the products (large AB small BC), so reaction 2 is endothermic, as the products are at a higher energy than the reactants
* Hence, this makes the reverse reaction, Reaction 1, exothermic
* This can be explained by looking at the bond strengths of the H–F bond (565 kJ / mol) and the H–H bond (432 kJ / mol). We can see that we need to give the system extra energy as the energy released when making H2 isn't enough, hence making this an endothermic reaction.

PES PES PES

==== Transition state for F-H-H system ====
Locate the approximate position of the transition state.

The transition State for the F-H-H reaction was found at: 181.11 pm AB and 74.487 pm BC, with 0 initial momenta for both. This was done by varying the distances until they remained constant, confirmed by the forces along AB and BC being 0, marking a stationary point as the first derivative is 0.

PICTURE contour zoom

==== Activation Energy for F + H2 and H + HF reactions ====
Report the activation energy for both reactions

The MEP was found for both reactions for a structure neighbouring the transition state: '''rts'''+1 pm

The activation energy is the difference between the saddle point's maximum energy and the minimum energy of the products , so:
* Reaction 1:
** ''Initial Energy: ''
** ''Product Energy:''
** Activation Energy:
* Reaction 2:
** ''Initial Energy: ''
** ''Product Energy:''
** Activation Energy:

=== Reaction Dynamics ===

==== Reactive trajectory for H2 + F → HF + H ====
In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.

==== Illustrating Polanyi's empirical rules ====
Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state

MRD:celiab

2020-05-07T11:22:25Z

Cb2018: /* EXERCISE 2: F - H - H system */

MRD:celiab

2020-05-07T10:12:08Z

Cb2018:

MRD:celiab

2020-05-07T10:10:51Z

Cb2018: /* Reactive and unreactive trajectories */

MRD:celiab

2020-05-07T10:10:22Z

Cb2018: /* Reactive and unreactive trajectories */

MRD:celiab

2020-05-07T10:10:08Z

Cb2018: /* Reactive and unreactive trajectories */

MRD:celiab

2020-05-07T10:09:23Z

Cb2018: /* The Transition State */

MRD:celiab

2020-05-05T15:05:25Z

Cb2018: /* Transition State Theory */

== EXERCISE 1: H + H2 ==

=== The Transition State ===
* On PES diagrams, transition states are mathematically defined as points with 0 gradient with respect to position: stationary points.
* The transition state corresponds to the saddle point; the highest energy point on the reaction coordinate, which is the lowest energy pathway.
* We can distinguish it from a local minimum because on the reaction coordinate axis, the saddle point representing the transition state, will be a local maximum, but on the orthogonal axis it will be a local minimum. The minima are minimums on either direction because they represent stable reactants and products.

=== H + H2 reaction ===
[[File:Internuclear_distances_vs_time.png|250px|thumb|middle|Internuclear Distances vs Time plot ]]

===== Locating the Transition State =====

* The reaction is symmetric, hence, the transition state must have '''r1''' = '''r2''' by Hammond's Postulate
* By testing different initial conditions where '''rAB'''='''rBC '''and '''p1''' = '''p2''' = 0 g.mol-1.pm.fs-1, we can find the position of the Transition State
** The position of the transition state was found to be '''rts'''=90.8 pm
** This is where the system underwent small periodic symmetric vibrations keeping the distance between the atoms almost constant
* In the plot to the right, we see two straight lines, representing almost a stationary behaviour; thus we know we are at the symmetric transition state.

===== Calculating the Reaction Path =====
* The minimum energy path is calculated from a system slightly displaced from the Transition state ('''r1''' = '''rts'''+1 pm, '''r2''' = '''rts''') and 0 initial momenta and follows the lowest energy path on a PES.
* As the reaction is symmetric it doesn't matter which atoms we label as A B C, so if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead, we get the same results but switched: AB will become BC and viceversa.
* The dynamic trajectory includes the momentum of the particles which causes oscillations about the trajectory path.
* The ''mep'' the trajectory is roughly the same but it's smooth and doesn't oscillate, as the velocities are set to 0 in every step. This means we are artificially removing any gain in vibrational energy which results in oscillations. As the velocity is set 0 at every step, the kinetic energy is also 0.
[[File:picture 2 mep.png|350px|thumb|left|MEP contour plot ]]
[[File:picture 3 dynamic.png|350px|thumb|right|Dynamic contour plot ]]
[[File:Pic_4_dt.png|350px|thumb|left|MEP Internuclear Distances vs Time plot ]]
[[File:Pic_5_dt.png|350px|thumb|right| Dynamic Internuclear Distances vs Time plot ]]

<nowiki> </nowiki>The final values for the trajectory (at 1000 time steps of 0.1 fs) are:
{| class="wikitable"
!
!Dynamics
!MEP
|-
|'''r1(t) / pm'''
|801
|217
|-
|'''r2(t) / pm'''
|73.6
|74.0
|-
|'''p1(t) / g.mol-1.pm.fs-1'''
|5.07
|0
|-
|'''p2(t) / g.mol-1.pm.fs-1'''
|<nowiki>-2.86</nowiki>
|0
|}
When we set up a calculation where the initial positions correspond to the final positions of the trajectory above and we have the same final momenta values but with their signs reversed, we can see the whole process reverses and the system approaches the initial conditions of the transition state. If we increase the time step, we are able to see the whole reaction happening again as it goes through the transition state.

===== Reactive and unreactive trajectories =====
For the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, the following trajectories with the following momenta combination were run:

(in the descriptions A B and C are used to describe the atoms colliding: AB is the initial H2 approaching H, represented by C)
{| class="wikitable"
!'''#'''
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|'''1'''
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.3</nowiki>
|yes
|The reaction dynamics proceed in a straight forward manner: AB and C approach each other, a collision occurs with the correct momentum for the reaction to proceed and a new bond BC is formed.
| [[File:Pic_a.png|260px]]
|-
|'''2'''
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420</nowiki>
|no
|AB approaches C and collides with it, but it doesn't have enough momentum to form a new bond so they are repelled. Momentum transfer does occur and C goes back with a greater momentum.
| [[File:Pic_b.png|260px]]
|-
|'''3'''
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414</nowiki>
|yes
|Oscillating AB collides with C and forms a new bond. The new H2 molecule, BC, has a higher vibrational frequency. A's direction is reversed.
| [[File:Pic_c.png|260px]]
|-
|'''4'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-375</nowiki>
|no
|AB approaches C and a BC bond forms, however, the system has too much momentum and C is repelled by the nucleus in B and bounces back. This leaves AB with greater vibrational energy and the distance BC increases as C is repelled.

| [[File:Pic_d.png|260px]]
|-
|'''5'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349</nowiki>
|yes
|AB collides with C and a new bond forms. However, unlike above, C still has enough energy to overcome the activation barrier. So, the BC bond forms, it breaks and reforms AB which then collides again with C to reform BC and leave as 2 entities: A + BC:
| [[File:Pic_e.png|260px]]
|}

Previous calculations have shown that that trajectories with initial conditions in the range '''r1''' =74 pm, '''r2''' = 200 pm, with -3.1 < '''p1'''/ g.mol-1.pm.fs-1 < -1.6 and '''p2''' = -5.1 g.mol-1.pm.fs-1 are reactive
* This is confirmed in combinations #1, #2 and #3.
From this we could hypothesise that all trajectories starting with the same positions but with higher values of momenta (higher kinetic energy) would be reactive as there would be enough kinetic energy to overcome the activation barrier. From the table we can see that this is not necessarily true:
* Combination #4 vs Combination #5.

==== Transition State Theory ====
Transition State theory allows us to investigate the progress of a reaction by splitting it into 2 parts: reactants and products.
* For a reaction to occur the reactants must cross into the product "space" by going over the saddle point – the transition state!
* With the Born-Oppenheimer approximation, it is assumed there is a single saddle point (thus transition state) per reaction
* Transition State theory should only be used to estimate reaction rates at low temperatures
* Fails to correctly predict reaction rates for scenarios with high momentum (ie high temperature), where the transition state can be recrossed – poor match with experimental values
* We must also take into account that not every reaction with the correct energy will react, as orientation is important too – predicted rates higher tan experimental

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

MRD:celiab

2020-05-05T14:16:27Z

Cb2018: /* Calculating the Reaction Path */

MRD:celiab

2020-05-05T14:13:11Z

Cb2018: /* Calculating the Reaction Path */

== EXERCISE 1: H + H2 ==

=== The Transition State ===
* On PES diagrams, transition states are mathematically defined as points with 0 gradient with respect to position: stationary points.
* The transition state corresponds to the saddle point; the highest energy point on the reaction coordinate, which is the lowest energy pathway.
* We can distinguish it from a local minimum because on the reaction coordinate axis, the saddle point representing the transition state, will be a local maximum, but on the orthogonal axis it will be a local minimum. The minima are minimums on either direction because they represent stable reactants and products.

=== H + H2 reaction ===
[[File:Internuclear_distances_vs_time.png|250px|thumb|middle|Internuclear Distances vs Time plot ]]

===== Locating the Transition State =====

* The reaction is symmetric, hence, the transition state must have '''r1''' = '''r2''' by Hammond's Postulate
* By testing different initial conditions where '''rAB'''='''rBC '''and '''p1''' = '''p2''' = 0 g.mol-1.pm.fs-1, we can find the position of the Transition State
** The position of the transition state was found to be '''rts'''=90.8 pm
** This is where the system underwent small periodic symmetric vibrations keeping the distance between the atoms almost constant
* In the plot to the right, we see two straight lines, representing almost a stationary behaviour; thus we know we are at the symmetric transition state.

===== Calculating the Reaction Path =====
* The minimum energy path is calculated from a system slightly displaced from the Transition state ('''r1''' = '''rts'''+1 pm, '''r2''' = '''rts''') and 0 initial momenta and follows the lowest energy path on a PES.
* As the reaction is symmetric it doesn't matter which atoms we label as A B C, so if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead, we get the same results but switched
* The dynamic trajectory includes the momentum of the particles which causes oscillations about the trajectory path.
* The ''mep'' the trajectory is roughly the same but it's smooth and doesn't oscillate, as the velocities are set to 0 in every step. This means we are artificially removing any gain in vibrational energy which results in oscillations. As the velocity is set 0 at every step, the kinetic energy is also 0.
[[File:picture 2 mep.png|350px|thumb|left|MEP contour plot ]]
[[File:picture 3 dynamic.png|350px|thumb|right|Dynamic contour plot ]]
[[File:Pic_4_dt.png|350px|thumb|left|MEP Internuclear Distances vs Time plot ]]
[[File:Pic_5_dt.png|350px|thumb|right| Dynamic Internuclear Distances vs Time plot ]]

<nowiki> </nowiki>The final values for the trajectory (at 1000 time steps of 0.1 fs) are:
{| class="wikitable"
!
!Dynamics
!MEP
|-
|'''r1(t) / pm'''
|801
|217
|-
|'''r2(t) / pm'''
|73.6
|74.0
|-
|'''p1(t) / g.mol-1.pm.fs-1'''
|5.07
|0
|-
|'''p2(t) / g.mol-1.pm.fs-1'''
|<nowiki>-2.86</nowiki>
|0
|}
When we set up a calculation where the initial positions correspond to the final positions of the trajectory above and we have the same final momenta values but with their signs reversed, we can see the whole process reverses and the system approaches the initial conditions of the transition state. If we increase the time step, we are able to see the whole reaction happening again as it goes through the transition state.

===== Reactive and unreactive trajectories =====
For the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, the following trajectories with the following momenta combination were run:

(in the descriptions A B and C are used to describe the atoms colliding: AB is the initial H2 approaching H, represented by C)
{| class="wikitable"
!'''#'''
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|'''1'''
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.3</nowiki>
|yes
|The reaction dynamics proceed in a straight forward manner: AB and C approach each other, a collision occurs with the correct momentum for the reaction to proceed and a new bond BC is formed.
| [[File:Pic_a.png|260px]]
|-
|'''2'''
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420</nowiki>
|no
|AB approaches C and collides with it, but it doesn't have enough momentum to form a new bond so they are repelled. Momentum transfer does occur and C goes back with a greater momentum.
| [[File:Pic_b.png|260px]]
|-
|'''3'''
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414</nowiki>
|yes
|Oscillating AB collides with C and forms a new bond. The new H2 molecule, BC, has a higher vibrational frequency. A's direction is reversed.
| [[File:Pic_c.png|260px]]
|-
|'''4'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-375</nowiki>
|no
|AB approaches C and a BC bond forms, however, the system has too much momentum and C is repelled by the nucleus in B and bounces back. This leaves AB with greater vibrational energy and the distance BC increases as C is repelled.

| [[File:Pic_d.png|260px]]
|-
|'''5'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349</nowiki>
|yes
|AB collides with C and a new bond forms. However, unlike above, C still has enough energy to overcome the activation barrier. So, the BC bond forms, it breaks and reforms AB which then collides again with C to reform BC and leave as 2 entities: A + BC:
| [[File:Pic_e.png|260px]]
|}

Previous calculations have shown that that trajectories with initial conditions in the range '''r1''' =74 pm, '''r2''' = 200 pm, with -3.1 < '''p1'''/ g.mol-1.pm.fs-1 < -1.6 and '''p2''' = -5.1 g.mol-1.pm.fs-1 are reactive
* This is confirmed in combinations #1, #2 and #3.
From this we could hypothesise that all trajectories starting with the same positions but with higher values of momenta (higher kinetic energy) would be reactive as there would be enough kinetic energy to overcome the activation barrier. From the table we can see that this is not necessarily true:
* Combination #4 vs Combination #5.

==== Transition State Theory ====
Transition State theory allows us to investigate the progress of a reaction by splitting it into 2 parts: reactants and products.
* For a reaction to occur the reactants must cross into the product "space" by going over the saddle point – the transition state!
* With the Born-Oppenheimer approximation, it is assumed there is a single saddle point (thus transition state) per reaction
* Transition State theory should only be used to estimate reaction rates at low temperatures
* Fails to correctly predict reaction rates for scenarios with high momentum (ie high temperature), where the transition state can be recrossed – poor match with experimental values

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

MRD:celiab

2020-05-05T14:12:09Z

Cb2018: /* Transition State Theory */

== EXERCISE 1: H + H2 ==

=== The Transition State ===
* On PES diagrams, transition states are mathematically defined as points with 0 gradient with respect to position: stationary points.
* The transition state corresponds to the saddle point; the highest energy point on the reaction coordinate, which is the lowest energy pathway.
* We can distinguish it from a local minimum because on the reaction coordinate axis, the saddle point representing the transition state, will be a local maximum, but on the orthogonal axis it will be a local minimum. The minima are minimums on either direction because they represent stable reactants and products.

=== H + H2 reaction ===
[[File:Internuclear_distances_vs_time.png|250px|thumb|middle|Internuclear Distances vs Time plot ]]

===== Locating the Transition State =====

* The reaction is symmetric, hence, the transition state must have '''r1''' = '''r2''' by Hammond's Postulate
* By testing different initial conditions where '''rAB'''='''rBC '''and '''p1''' = '''p2''' = 0 g.mol-1.pm.fs-1, we can find the position of the Transition State
** The position of the transition state was found to be '''rts'''=90.8 pm
** This is where the system underwent small periodic symmetric vibrations keeping the distance between the atoms almost constant
* In the plot to the right, we see two straight lines, representing almost a stationary behaviour; thus we know we are at the symmetric transition state.

===== Calculating the Reaction Path =====
* The minimum energy path is calculated from a system slightly displaced from the Transition state ('''r1''' = '''rts'''+1 pm, '''r2''' = '''rts''') and 0 initial momenta and follows the lowest energy path on a PES.
* As the reaction is symmetric it doesn't matter which atoms we label as A B C, so if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead, we get the same results but switched
* The dynamic trajectory includes the momentum of the particles which causes oscillations about the trajectory path.
* The ''mep'' the trajectory is roughly the same but it's smooth and doesn't oscillate, as the velocities are set to 0 in every step. This means we are artificially removing any gain in vibrational energy which results in oscillations. As the velocity is set 0 at every step, the kinetic energy is also 0.
[[File:picture 2 mep.png|350px|thumb|left|MEP contour plot ]]
[[File:picture 3 dynamic.png|350px|thumb|right|Dynamic contour plot ]]
[[File:Pic_4_dt.png|350px|thumb|left|MEP Internuclear Distances vs Time plot ]]
[[File:Pic_5_dt.png|350px|thumb|right| Dynamic Internuclear Distances vs Time plot ]]

<nowiki> </nowiki>The final values for the trajectory (at 1000 time steps of 0.1 fs) are:
{| class="wikitable"
!
!Dynamics
!MEP
|-
|'''r1(t) / pm'''
|801
|217
|-
|'''r2(t) / pm'''
|73.6
|74.0
|-
|'''p1(t) / g.mol-1.pm.fs-1'''
|5.07
|0
|-
|'''p2(t) / g.mol-1.pm.fs-1'''
|<nowiki>-2.86</nowiki>
|0
|}
When we set up a calculation where the initial positions correspond to the final positions of the trajectory above and we have the same final momenta values but with their signs reversed, we can see the whole process reverses and the system approaches the initial conditions of the transition state. If we increase the time step, we are able to see the whole reaction happening again as it goes through the transition state.

===== Reactive and unreactive trajectories =====
For the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, the following trajectories with the following momenta combination were run:

(in the descriptions A B and C are used to describe the atoms colliding: AB is the initial H2 approaching H, represented by C)
{| class="wikitable"
!'''#'''
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|'''1'''
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.3</nowiki>
|yes
|The reaction dynamics proceed in a straight forward manner: AB and C approach each other, a collision occurs with the correct momentum for the reaction to proceed and a new bond BC is formed.
| [[File:Pic_a.png|260px]]
|-
|'''2'''
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420</nowiki>
|no
|AB approaches C and collides with it, but it doesn't have enough momentum to form a new bond so they are repelled. Momentum transfer does occur and C goes back with a greater momentum.
| [[File:Pic_b.png|260px]]
|-
|'''3'''
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414</nowiki>
|yes
|Oscillating AB collides with C and forms a new bond. The new H2 molecule, BC, has a higher vibrational frequency. A's direction is reversed.
| [[File:Pic_c.png|260px]]
|-
|'''4'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-375</nowiki>
|no
|AB approaches C and a BC bond forms, however, the system has too much momentum and C is repelled by the nucleus in B and bounces back. This leaves AB with greater vibrational energy and the distance BC increases as C is repelled.

| [[File:Pic_d.png|260px]]
|-
|'''5'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349</nowiki>
|yes
|AB collides with C and a new bond forms. However, unlike above, C still has enough energy to overcome the activation barrier. So, the BC bond forms, it breaks and reforms AB which then collides again with C to reform BC and leave as 2 entities: A + BC:
| [[File:Pic_e.png|260px]]
|}

Previous calculations have shown that that trajectories with initial conditions in the range '''r1''' =74 pm, '''r2''' = 200 pm, with -3.1 < '''p1'''/ g.mol-1.pm.fs-1 < -1.6 and '''p2''' = -5.1 g.mol-1.pm.fs-1 are reactive
* This is confirmed in combinations #1, #2 and #3.
From this we could hypothesise that all trajectories starting with the same positions but with higher values of momenta (higher kinetic energy) would be reactive as there would be enough kinetic energy to overcome the activation barrier. From the table we can see that this is not necessarily true:
* Combination #4 vs Combination #5.

==== Transition State Theory ====
Transition State theory allows us to investigate the progress of a reaction by splitting it into 2 parts: reactants and products.
* For a reaction to occur the reactants must cross into the product "space" by going over the saddle point – the transition state!
* With the Born-Oppenheimer approximation, it is assumed there is a single saddle point (thus transition state) per reaction
* Transition State theory should only be used to estimate reaction rates at low temperatures
* Fails to correctly predict reaction rates for scenarios with high momentum (ie high temperature), where the transition state can be recrossed – poor match with experimental values

MRD:celiab

2020-05-05T14:06:27Z

Cb2018:

== EXERCISE 1: H + H2 ==

=== The Transition State ===
* On PES diagrams, transition states are mathematically defined as points with 0 gradient with respect to position: stationary points.
* The transition state corresponds to the saddle point; the highest energy point on the reaction coordinate, which is the lowest energy pathway.
* We can distinguish it from a local minimum because on the reaction coordinate axis, the saddle point representing the transition state, will be a local maximum, but on the orthogonal axis it will be a local minimum. The minima are minimums on either direction because they represent stable reactants and products.

=== H + H2 reaction ===
[[File:Internuclear_distances_vs_time.png|250px|thumb|middle|Internuclear Distances vs Time plot ]]

===== Locating the Transition State =====

* The reaction is symmetric, hence, the transition state must have '''r1''' = '''r2''' by Hammond's Postulate
* By testing different initial conditions where '''rAB'''='''rBC '''and '''p1''' = '''p2''' = 0 g.mol-1.pm.fs-1, we can find the position of the Transition State
** The position of the transition state was found to be '''rts'''=90.8 pm
** This is where the system underwent small periodic symmetric vibrations keeping the distance between the atoms almost constant
* In the plot to the right, we see two straight lines, representing almost a stationary behaviour; thus we know we are at the symmetric transition state.

===== Calculating the Reaction Path =====
* The minimum energy path is calculated from a system slightly displaced from the Transition state ('''r1''' = '''rts'''+1 pm, '''r2''' = '''rts''') and 0 initial momenta and follows the lowest energy path on a PES.
* As the reaction is symmetric it doesn't matter which atoms we label as A B C, so if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead, we get the same results but switched
* The dynamic trajectory includes the momentum of the particles which causes oscillations about the trajectory path.
* The ''mep'' the trajectory is roughly the same but it's smooth and doesn't oscillate, as the velocities are set to 0 in every step. This means we are artificially removing any gain in vibrational energy which results in oscillations. As the velocity is set 0 at every step, the kinetic energy is also 0.
[[File:picture 2 mep.png|350px|thumb|left|MEP contour plot ]]
[[File:picture 3 dynamic.png|350px|thumb|right|Dynamic contour plot ]]
[[File:Pic_4_dt.png|350px|thumb|left|MEP Internuclear Distances vs Time plot ]]
[[File:Pic_5_dt.png|350px|thumb|right| Dynamic Internuclear Distances vs Time plot ]]

<nowiki> </nowiki>The final values for the trajectory (at 1000 time steps of 0.1 fs) are:
{| class="wikitable"
!
!Dynamics
!MEP
|-
|'''r1(t) / pm'''
|801
|217
|-
|'''r2(t) / pm'''
|73.6
|74.0
|-
|'''p1(t) / g.mol-1.pm.fs-1'''
|5.07
|0
|-
|'''p2(t) / g.mol-1.pm.fs-1'''
|<nowiki>-2.86</nowiki>
|0
|}
When we set up a calculation where the initial positions correspond to the final positions of the trajectory above and we have the same final momenta values but with their signs reversed, we can see the whole process reverses and the system approaches the initial conditions of the transition state. If we increase the time step, we are able to see the whole reaction happening again as it goes through the transition state.

===== Reactive and unreactive trajectories =====
For the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, the following trajectories with the following momenta combination were run:

(in the descriptions A B and C are used to describe the atoms colliding: AB is the initial H2 approaching H, represented by C)
{| class="wikitable"
!'''#'''
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|'''1'''
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.3</nowiki>
|yes
|The reaction dynamics proceed in a straight forward manner: AB and C approach each other, a collision occurs with the correct momentum for the reaction to proceed and a new bond BC is formed.
| [[File:Pic_a.png|260px]]
|-
|'''2'''
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420</nowiki>
|no
|AB approaches C and collides with it, but it doesn't have enough momentum to form a new bond so they are repelled. Momentum transfer does occur and C goes back with a greater momentum.
| [[File:Pic_b.png|260px]]
|-
|'''3'''
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414</nowiki>
|yes
|Oscillating AB collides with C and forms a new bond. The new H2 molecule, BC, has a higher vibrational frequency. A's direction is reversed.
| [[File:Pic_c.png|260px]]
|-
|'''4'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-375</nowiki>
|no
|AB approaches C and a BC bond forms, however, the system has too much momentum and C is repelled by the nucleus in B and bounces back. This leaves AB with greater vibrational energy and the distance BC increases as C is repelled.

| [[File:Pic_d.png|260px]]
|-
|'''5'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349</nowiki>
|yes
|AB collides with C and a new bond forms. However, unlike above, C still has enough energy to overcome the activation barrier. So, the BC bond forms, it breaks and reforms AB which then collides again with C to reform BC and leave as 2 entities: A + BC:
| [[File:Pic_e.png|260px]]
|}

Previous calculations have shown that that trajectories with initial conditions in the range '''r1''' =74 pm, '''r2''' = 200 pm, with -3.1 < '''p1'''/ g.mol-1.pm.fs-1 < -1.6 and '''p2''' = -5.1 g.mol-1.pm.fs-1 are reactive
* This is confirmed in combinations #1, #2 and #3.
From this we could hypothesise that all trajectories starting with the same positions but with higher values of momenta (higher kinetic energy) would be reactive as there would be enough kinetic energy to overcome the activation barrier. From the table we can see that this is not necessarily true:
* Combination #4 vs Combination #5.

==== Transition State Theory ====
Transition State theory allows us to investigate the progress of a reaction by splitting it into 2 parts: reactants and products.
* For a reaction to occur the reactants must cross into the product "space" by going over the saddle point – the transition state!

Transition state theory splits the reaction into 2 distinct areas: the reactant and product space. It states that the reactants must cross the energy threshold of the saddle in the transition area to reach the product space. It also assumes atoms obey the laws of classical mechanics using the Born-Oppenheimer approximation, and that each reaction only has 1 saddle point and 1 transition state.

We can see from the above scenarios that for molecules with high momentum, the transition state can be recrossed, so the transition state prediction for reaction rate values will not provide an close approximation to experimental vales run at high temperatures. Therefore, the transition state theory should only be used for reactions at low temperature conditions for predicting reaction rates.

MRD:celiab

2020-05-05T13:43:01Z

Cb2018: /* Calculating the Reaction Path */

== EXERCISE 1: H + H2 ==

=== The Transition State ===
* On PES diagrams, transition states are mathematically defined as points with 0 gradient with respect to position: stationary points.
* The transition state corresponds to the saddle point; the highest energy point on the reaction coordinate, which is the lowest energy pathway.
* We can distinguish it from a local minimum because on the reaction coordinate axis, the saddle point representing the transition state, will be a local maximum, but on the orthogonal axis it will be a local minimum. The minima are minimums on either direction because they represent stable reactants and products.

=== H + H2 reaction ===
[[File:Internuclear_distances_vs_time.png|250px|thumb|middle|Internuclear Distances vs Time plot ]]

===== Locating the Transition State =====

* The reaction is symmetric, hence, the transition state must have '''r1''' = '''r2''' by Hammond's Postulate
* By testing different initial conditions where '''rAB'''='''rBC '''and '''p1''' = '''p2''' = 0 g.mol-1.pm.fs-1, we can find the position of the Transition State
** The position of the transition state was found to be '''rts'''=90.8 pm
** This is where the system underwent small periodic symmetric vibrations keeping the distance between the atoms almost constant
* In the plot to the right, we see two straight lines, representing almost a stationary behaviour; thus we know we are at the symmetric transition state.

===== Calculating the Reaction Path =====
* The minimum energy path is calculated from a system slightly displaced from the Transition state ('''r1''' = '''rts'''+1 pm, '''r2''' = '''rts''') and 0 initial momenta and follows the lowest energy path on a PES.
* As the reaction is symmetric it doesn't matter which atoms we label as A B C, so if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead, we get the same results but switched
* The dynamic trajectory includes the momentum of the particles which causes oscillations about the trajectory path.
* The ''mep'' the trajectory is roughly the same but it's smooth and doesn't oscillate, as the velocities are set to 0 in every step. This means we are artificially removing any gain in vibrational energy which results in oscillations. As the velocity is set 0 at every step, the kinetic energy is also 0.
[[File:picture 2 mep.png|350px|thumb|left|MEP contour plot ]]
[[File:picture 3 dynamic.png|350px|thumb|right|Dynamic contour plot ]]
[[File:Pic_4_dt.png|350px|thumb|left|MEP Internuclear Distances vs Time plot ]]
[[File:Pic_5_dt.png|350px|thumb|right| Dynamic Internuclear Distances vs Time plot ]]

<nowiki> </nowiki>The final values for the trajectory (at 1000 time steps of 0.1 fs) are:
{| class="wikitable"
!
!Dynamics
!MEP
|-
|'''r1(t) / pm'''
|801
|217
|-
|'''r2(t) / pm'''
|73.6
|74.0
|-
|'''p1(t) / g.mol-1.pm.fs-1'''
|5.07
|0
|-
|'''p2(t) / g.mol-1.pm.fs-1'''
|<nowiki>-2.86</nowiki>
|0
|}
When we set up a calculation where the initial positions correspond to the final positions of the trajectory above and we have the same final momenta values but with their signs reversed, we can see the whole process reverses and the system approaches the initial conditions of the transition state. If we increase the time step, we are able to see the whole reaction happening again as it goes through the transition state.

===== Reactive and unreactive trajectories =====
For the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, the following trajectories with the following momenta combination were run:
{| class="wikitable"
!'''#'''
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|'''1'''
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.3</nowiki>
|yes
|The reaction dynamics proceed in a straight forward manner: AB and C approach each other, a collision occurs, momentum is transferred and A's and C's directions are reversed. BC moves in the positive direction in an oscillating fashion.
| [[File:Pic_a.png|260px]]
|-
|'''2'''
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420</nowiki>
|no
|H2 approaches H but doesn't get close enough for the reaction to proceed, hence the direction of all the atoms involved are reversed and they all go back towards their initial position. Momentum transfer does occur and C goes back with a greater momentum, thus AB, as it moves back, oscillates faster ( smaller distances between oscillating atoms).
| [[File:Pic_b.png|260px]]
|-
|'''3'''
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414</nowiki>
|yes
|As H2 approaches H it slows down, goes through a symmetric transition state and the reaction proceeds with BC moving off in the positive direction and with A's momentum reversed.
| [[File:Pic_c.png|260px]]
|-
|'''4'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-375</nowiki>
|no
|As H2 approaches H, atoms B and C bounce off each other 3 times, and even though the BC bond forms, the system reverts backs to the reactants and the directions of AB and C are reversed.
| [[File:Pic_d.png|260px]]
|-
|'''5'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349</nowiki>
|yes
|As H2 approaches H, B bounces off C and then off A, then the reaction between B and C proceeds and BC moves away from A, whose direction is reversed.
| [[File:Pic_e.png|260px]]
|}

Previous calculations have shown that that trajectories with initial conditions in the range '''r1''' =74 pm, '''r2''' = 200 pm, with -3.1 < '''p1'''/ g.mol-1.pm.fs-1 < -1.6 and '''p2''' = -5.1 g.mol-1.pm.fs-1 are reactive
* This is confirmed in combinations #1, #2 and #3.
From this we could hypothesise that all trajectories starting with the same positions but with higher values of momenta (higher kinetic energy) would be reactive as there would be enough kinetic energy to overcome the activation barrier. From the table we can see that this is not necessarily true:
* Combination #4 has a much larger momentum, but still doesn't react. This is because it has so much energy that it can bounce off atom C multiple times, lose the extra KE, reverse its direction and leave unreacted (but with a smaller overall KE).

*Combination #5 on the other hand, has a slightly greater momentum than #4 and reacts. This is because after the collisions the AB still has enough energy to overcome the activation barrier and proceed with the reaction.