MRD:01336581

2019-06-03T13:32:33Z

Sw2711: /* Excercise 1 */

==Excercise 1==
The reaction transition state happens at a saddle point on the energy surface.

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

<math> {\partial^2 V \over r_1^2} </math>
and
<math> {\partial^2 V \over r_1^2} </math>
are of opposite sign.

 Good, but you need to explain what do those two maths functions mean. Why the first derivatives are zero and the second derivatives are opposite signs ? Do your r1, r2 represent the two distances AB and BC? if so, I'm afraid it is not really correct. [[User:Sw2711|Sw2711]] ([[User talk:Sw2711|talk]]) 14:32, 3 June 2019 (BST) 

The transition state is located between '''rts''' = 0.907742 Å and 0.907743 Å.
 How did you find the values? [[User:Sw2711|Sw2711]] ([[User talk:Sw2711|talk]]) 14:32, 3 June 2019 (BST) 

[[File:01336581-TS-internuclear-distance.png|400px]]

The dynamics simulation differs from the MEP simulation by showing the vibration of the product molecule.
 Okay, why is it? Is it the only difference you've observed? [[User:Sw2711|Sw2711]] ([[User talk:Sw2711|talk]]) 14:32, 3 June 2019 (BST)

[[File:01336581-Dynamics-MEP.png|400px]]

[[File:01336581-MEP.png|400px]]

For the initial conditions '''r1''' = 0.74 and '''r2''' = 2.0

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 || -99.018 || Yes || Approaches transtition state smoothly. The product molecule has a small vibrational energy. || [[File:01336581-1.25-2.5.png|130px]]
|-
| -1.5 || -2.0 || -100.456 || No || Reactant molecule has a small vibrational energy. It does not pass over the transition state and does not react. || [[File:01336581-1.5-2.png|130px]]
|-
| -1.5 || -2.5 || -98.956 || Yes || Reactant molecule has a small vibrational energy. It passes over the transiton state, and the product molecule has a slightly larger vibrational energy.|| [[File:01336581-1.5-2.5.png|130px]]
|-
| -2.5 || -5.0 || -84.956 || No || The reactants approach the transition state quickly, and pass over. They gain a large vibrationl energy, and the products vibrate back into the reactants. the reactant molecule is left with a large vibrational energy.|| [[File:01336581-2.5-5.png|130px]]
|-
| -2.5 || -5.2 || -83.416 || Yes || The reqactants approach each other quickly, and pass through the transition state, gaining a large vibrational energy. they vibrate back over the transition state,towards the reactants, and then back ovetr the tranition state again, leaving the product molecule with a large vibrational energy. || [[File:01336581-2.5-5.2.png|130px]]
|}

Transition state theory assumes that the reaction passes over the lowest energy point on the energy surface, i.e. the saddle point. This does not always happen if the reactants have higher energies. This can lead to recrossing the transition state. If the tranisiton state is recrossed, any rate calulations are unlikely to be accurate, since the time for the reaction to complete is longer than if the transition state was not recrossed. The transition state theory rate predictions form an upper bound for the rate constant.

 This part is good. But you need to be aware that there are a few more assumptions in TST. [[User:Sw2711|Sw2711]] ([[User talk:Sw2711|talk]]) 14:32, 3 June 2019 (BST)

==Excercise 2==
The potential energy surface shows that the reaction of F with H2 is exothermic, and the reaction of HF with H is endothermic.

[[File:01336581-FHHSurface Plot.png|300px]]

For this simulation the F atom is in positoin A, so the left hand side is the F + H2, and the right hand side is the HF + H side. the left hand side is higher in energy, so going from F + H2 to HF + H is exothermic. This shows that the H-F bond is stronger than the H-H bond, since overall energy is released on the formation of the H-F bond over the H-H bond.

The transition state is approximately at F-H = 1.81 Å and H-H = 0.745 Å.

A minimum energy pathway from these values gave a value for the activation energy of the HF + H reaction of 30 kcalmol-1.

[[File:01336581-HF-H_AE.png|300px]]

A minimum energy pathway calculated from F-H = 1.82 Å and H-H = 0.745 Å resulted in a value for the activation energy of ~0.25 kcalmol-1.

[[File:01336581-F-HH_AE.png|300px]]

A reactive pathway has initial conditions '''rFH''' = 2 Å, '''rHH''' = 0.74 Å, pFH = -0.8, pHH = 0.

[[File:01336581-FHH.png|300px]]

The reaction energy is released by the formation of the H-F bond. Quantum mechanically, the constructive overlap of electron wavefunctions is stabilising, which releases energy to be converted into vibrational or translational forms. This can be shown by removing the fluorine atom, leaving just the atomisation of a hydrogen molecule, which is endothermic.

Polanyi's rules state that vibrational energy is more useful for promoting reactions with a late transition state, and translational energy is more useful in promoting reactions with an early transition state. For the reaction between HF and H, the transition state is very late, so a small amount of energy in the vibration of the HF molecule results in a reaction, whereas a much larger amount of translational energy on the H atom results in an unreactive pathway. The inverse is true for the reaction between F and H2.

MRD:FMJ17

2019-05-30T11:58:20Z

Sw2711: /* Polanyi's Empirical Rules */

=Exercise 1=
===On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?===
The transition state is defined as the point at which the potential gradient is zero, δV(ri)/δ(ri)=0. Contrary to local minima in the enrgy surface, the transition state will be at the point along the reaction path where the energy is maximum.

 Not sure you fully understand this by reading your statement. Strictly, first of all, transition state is the maximum point along the MINIMUM energy reaction path way. Not any reaction pathway. Second, you also need to demonstrate what is going on with the second derivatives. Overall, a saddle point, in this case, the TS point, is a point where in a 3-dimensional space, it is both a maximum and a minimum at the same time, depending on which plane you are looking at. That differentiates from a local minimum or a maximum, because whichever plane you are looking at, it's always a minimum or a maximum. Please check the figure that I attached. [[File: SaddleWSK.png]] [[User:Sw2711|Sw2711]] ([[User talk:Sw2711|talk]]) 12:45, 30 May 2019 (BST) 

===Trajectories from r1 = r2: locating the transition state===
transition state position, rts=0.90775 Å
====Internuclear Dictances vs Time Plot====
[[File:Int_nuc_1_fmj.PNG]]
The straight lines indicate no vibrations. This indicates that the momenta are zero, confirming that the point corresponds to the transition state.

 Yes, you have confirmed it. But you need to tell your approach to 0.90775 A too.[[User:Sw2711|Sw2711]] ([[User talk:Sw2711|talk]]) 12:46, 30 May 2019 (BST)

===Calculating the Reaction Path===
====Minimum Energy Pathway Calculation====
The figure below shows a surface plot obtained using the MEP calculation type. The black line illustrates the reaction path (from the transition state to the products).
[[File:Surface_Plot_mep_fmj.png]]

Below is the contour plot, showing the same reaction path. (black line)

[[File:Contour_plot_mep_fmj.png]]

====Comparing Dynamics and MEP Calculation Types====
Below is a surface plot obtained by using the dynamics calculation type.
[[File:Surface_Plot_dyn_fmj.png]]

Again, the black line shows the the reaction path (from the transition state to the products). This calculation was made using the same initial conditions as the MEP calculation above.

The two trajectories calculated using the two calculation types are different. The dynamic calculation indicates vibrations (curves on the line), while the MEP calculation does not (no curves).

 Are there any more differences? What caused those differences? What's the difference in each algorithm? [[User:Sw2711|Sw2711]] ([[User talk:Sw2711|talk]]) 12:48, 30 May 2019 (BST)

===Reactive and unreactive trajectories===

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 ||-99.018 ||Yes||The hydrogen atom collides with the hydrogen molecule, the transition state forms. Eventually, the original H-H bond breaks and a new one is formed.||[[File:Contour_1_fmj.png|150px]]
|-
| -1.5 || -2.0 ||-100.456 ||No ||The hydrogen atom collides with the hydrogen molecule, the transition state does not form. The atom and the molecule bounce off each other and move in opposite directions. ||[[File:Contour_2_fmj.png|150px]]
|-
| -1.5 || -2.5 ||-98.956 ||Yes ||The hydrogen atom collides with the hydrogen molecule, the transition state forms. Eventually, the original H-H bond breaks and a new one is formed. ||[[File:Contour_3_fmj.png|150px]]
|-
| -2.5 || -5.0 ||-84.956 ||No ||The hydrogen atom collides with the hydrogen molecule, the transition state forms, but the original bond does not break. Instead, the molecule and the atom move away from each other. ||[[File:Contour_4_fmj.png|150px]]
|-
| -2.5 || -5.2 ||-83.416 ||Yes ||The hydrogen atom collides with the hydrogen molecule, the transition state forms, then break downs back into lower energy structures. It is then formed again, followed by the products. ||[[File:Contour_5_fmj.png|150px]]
|}

Good enough, but where is your answer for the question about the transition state theory? [[User:Sw2711|Sw2711]] ([[User talk:Sw2711|talk]]) 12:50, 30 May 2019 (BST)

=Exercise 2=
===Potential Energy Surface inspection of F-H-H system===
Below is a surface plot for the F-H-H system.
[[File:H2_f_surface_potential_fmj.png]]
It shows H-H and F as being more energetic than F-H and H. This indicates that the reaction between H-H and F is exothermic and that the H-F bond is stronger than the H-H bond. It also indicates that the reaction between H-F and H is endothermic, again because the H-F bond is stronger than the H-H bond.

 This part is good. [[User:Sw2711|Sw2711]] ([[User talk:Sw2711|talk]]) 12:51, 30 May 2019 (BST)

====Transition State====
Below is a surface plot showing the position of the transition state (black dot) of the H-H + F reaction.
[[File:F_h_TS_fmj.png]]

 Again, where is your approach? [[User:Sw2711|Sw2711]] ([[User talk:Sw2711|talk]]) 12:51, 30 May 2019 (BST) 

====Activation Energy====
The activation energy for the H-H + F reaction is 30 kcal/mol.

[[File:Activation_energy.png]]

The activation energy for the H-F + H reaction is 0.15 kcal/mol.

[[File:Act_energy_2_fmj.png]]

 This part is alright. But you still need to improve your methodology descriptions. Not everyone knows what those two pictures mean or how to interpret the pictures. [[User:Sw2711|Sw2711]] ([[User talk:Sw2711|talk]]) 12:52, 30 May 2019 (BST)

===Reaction dynamics===
A set of initial conditions which result in a reaction is shown below.

[[File:In_con_fmj.PNG]]

This is the momentum vs Time graph for the reaction trajectory resulting from the initial conditions above.

[[File:Momenta_time_fh_fmj.png]]

==== In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.====
The formation of HF is an exothermic process. The formation of the H-F bond releases more energy than is used to break the H-H bond. This can be confirmed by observing an increase in temperature corresponding to the release of energy.

 So why is that conserved? You need to explain better. [[User:Sw2711|Sw2711]] ([[User talk:Sw2711|talk]]) 12:56, 30 May 2019 (BST)

{| class="wikitable" border=1
! pFH !! pHH !! Reactive? !! Comments !! Plots
|-
| -0.5 || -3.0 ||Yes||The graph shows that the H-H distance increases with time and the H-F distance initially decreases, but then maintains a constant equilibrium value. This indicates the formation of the F-H bond ||[[File:Mom_1_fmj.png|150px]]
|-
| -0.5 || -2.8 ||Yes ||The graph now shows a reactive trajectory. ||[[File:Plot_2_fmj.png|150px]]
|-
| -0.5 || -2.0 ||No ||The graph shows that the H-H distance maintains a constant equilibrium value with time and the H-F distance initially decreases, but increases with time. This indicates that the F-H bond did not form. ||[[File:Plot2_fmj.png|150px]]
|-
| -0.5 || -1.0 ||No ||The graph shows that the H-H distance maintains a constant equilibrium value with time and the H-F distance initially decreases, but increases with time. This indicates that the F-H bond did not form. ||[[File:Plot3_fmj.png|150px]]
|-
| -0.5 || 0 ||No||The graph shows that the H-H distance maintains a constant equilibrium value with time and the H-F distance initially decreases, but increases with time. This indicates that the F-H bond did not form. ||[[File:Plot4_fmj.png|150px]]
|-
| -0.5 || 1.0 ||No ||The graph shows that the H-H distance maintains a constant equilibrium value with time and the H-F distance initially decreases, but increases with time. This indicates that the F-H bond did not form. ||[[File:Plot5_fmj.png|150px]]
|-
| -0.5 || 2.0 ||Yes ||The graph shows that the H-H distance increases with time and the H-F distance initially decreases, but then maintains a constant equilibrium value. This indicates the formation of the F-H bond ||[[File:Plot6_fmj.png|150px]]
|-
| -0.5 || 2.8 ||No ||The graph shows that the H-H distance maintains a constant equilibrium value with time and the H-F distance initially decreases, but increases with time. This indicates that the F-H bond did not form. ||[[File:Plot7_fmj.png |150px]]
|-
| -0.5 || 3.0 ||No ||The graph shows that the H-H distance maintains a constant equilibrium value with time and the H-F distance initially decreases, but increases with time. This indicates that the F-H bond did not form. ||[[File:Plot8_fmj.png|150px]]
|-
| -0.8 || 0.1 ||Yes||The graph shows that the H-H distance increases with time and the H-F distance initially decreases, but then maintains a constant equilibrium value. This indicates the formation of the F-H bond ||[[File:Plot9_fmj.png|150px]]
|-
|}

===Polanyi's Empirical Rules===
For a reaction with a late transition state, that is the transition state is closer in structure to the products than it is to the reactants, it's success will depend more on the vibrational energy of the reactant molecule than it will on the translational energy of said molecules. For a reaction with an early transition state, the opposite is true; translational energy will be at the main promoter of the reaction.
 So based on your experiments above, do you agree or disagree with the Polanyi's rules?[[User:Sw2711|Sw2711]] ([[User talk:Sw2711|talk]]) 12:58, 30 May 2019 (BST)

2019-05-30T11:35:31Z

Sw2711: /* Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state. */

== Exercise 1: H+H2 system ==

=== On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===

The transition state is mathematically defined as:<gallery>
File:ri3717_Eq.1
</gallery>where q1 and q2 are the reactant molecules, V is the potential energy.

The transition state can be identified by starting trajectories near the transition state and if they move towards reactants or products (even with no initial momentum) then it is the transition state. It can be distinguished from a local minimum by calculating the second derivative, which gives a negative value for a local minimum but a positive value for a local maximum (i.e. the transition state).

 Are you sure about your statement? TS is the maximum point on the minimum energy pathway. A saddle point, in this case, the TS point, is a point where in a 3-dimensional space, it is both a maximum and a minimum at the same time, depending on which plane you are looking at. That differentiates from a local minimum or a maximum, because whichever plane you are looking at, it's always a minimum or a maximum. Please check the figure that I attached. [[File: SaddleWSK.png]][[User:Sw2711|Sw2711]] ([[User talk:Sw2711|talk]]) 12:04, 30 May 2019 (BST)

=== Report your best estimate of the transition state position ('''rts''') for trajectories from r1 = r2 and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ===
The transition state position is the position at which the distance between A-B (r1) and B-C (r2) is equal, as it is symmetrical for this particular reaction. However, for trajectories from r1 = r2, the H atoms perform oscillation, so the transition state position needs to be estimated by testing different r values. The transition state position (rts) found with r1 = r2 is 0.908 Å. At this distance, there is almost no oscillation as shown in the Internuclear Distance vs Time plot below: <gallery>
File:Ri3717_plot1.png

</gallery>

This shows that at this value of r, the H atoms do not move with no initial momentum, which is characteristic of a transition state.
This part is good. [[User:Sw2711|Sw2711]] ([[User talk:Sw2711|talk]]) 12:05, 30 May 2019 (BST)

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

For trajectories with r values below:

r1 = rts + 0.01 = 0.909

r2 = rts = 0.908,

the MEP and Dynamics trajectories are compared.<gallery>
ri3717_ABC.png|Initial conditions of the three H atoms. Momenta= 0

</gallery>

Dynamics

Animation- B moves towards the C and A moves away from the rest. H atoms B and C continue to move to the right as they oscillate.

Internuclear distances- stays constant for a short time and A-B and A-C distances increase as they move away from each other, while B-C distance decreases and fluctuates as the molecule performs an oscillation. At large t, the final value of B-C distance is approx. 0.75.

Momenta- the momentum of A-B (p1(t)) increases with time; rapid increase at first then continues to slowly increase to a limit. The momentum of B-C (p2(t)) decreases first, then increases with fluctuation. At large t, p1(t) reaches a limit at 2.5 and p2(t) oscillates between 1.0-1.5, giving an average of 1.25.

The angle θ stays constant at 0.

MEP

Animation- a smooth movement of the H atom in the centre towards the H atom on the right as the H atom on the left accelerates towards left.

Internuclear distances- stays constant for a very short time and A-B and A-C distances increase as they move away from each other, while B-C distance decreases without fluctuation. At large t, the final value of B-C distance is approx. 0.75.

Momenta- no change, constant at 0.

When the r1 and r2 values are swapped, H atom A and C swap roles- A and B forms a molecule, C moves away to the right.

This part, it is very descriptive. Very good observation. But you haven't really shown much understanding, like what do you understand why these differences occur by using two different algorithms?

=== Complete the table below by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table. ===
This part, in general is good. But you need to improve on your descriptions. Just wondering, did you write all that by looking at the animations? It will be better if you could talk more, in terms of the energy, transition state, e.g. system #2 doesn't acquire enough activation energy to reach the transition point, therefore, the reaction doesn't occur. --- something like that, I'm sure you can do a better job than me. [[User:Sw2711|Sw2711]] ([[User talk:Sw2711|talk]]) 12:16, 30 May 2019 (BST)
{| class="wikitable" border="1"
!#
! p1 (kg.m/s) !! p2 (kg.m/s) !! Etot (kcal/mol) !! Reactive? !! Description of the dynamics
|-
|1
| -1.25 || -2.5 ||-99.0||yes||As molecule A-B approaches C, B is pulled away from A, forming molecule B-C. A moves away to the left, B-C moves to the right with oscillation.
|-
|2
| -1.5 || -2.0 ||-100.5||no||Molecule A-B and atom C slowly approaches each other, and when they get to a certain distance, they repel and moves away from each other.
|-
|3
| -1.5 || -2.5 ||-99.0||yes||As molecule A-B slowly approaches C, B is pulled away from A, forming molecule B-C. A moves away to the left, B-C moves to the right with oscillation.
|-
|4
| -2.5 || -5.0 ||-85.0||no||Molecule A-B approaches C and B is pulled away from A. B-C oscillates but gets too close and repel each other, pushing B back to A, forming molecule A-B.
|-
|5
| -2.5 || -5.2 ||-83.4||yes||Same as the one above happens, but molecule A-B oscillates again, gets too close and repel each other, pushing B back to C, finally forming molecule B-C.
|}
Although there is no simple trend in the reactivity and the initial conditions, It is evident that small changes to the initial conditions (momentum and hence the total energy) can affect the reactivity. For example based on the last two sets of initial conditions (conditions 4 and 5), we can see that increasing the magnitude of p2 by 0.2 kg.m/s increases the total energy by 1.6 kcal/mol, and enables the reaction between molecule A-B and atom C, forming atom A and molecule B-C. <gallery>
ri3717tableplot1.png|Initial conditions #1
</gallery><gallery>ri3717tableplot2.png|Initial conditions #2
</gallery><gallery>ri3717tableplot3.png|Initial conditions #3
</gallery><gallery>
ri3717tableplot4.png|Initial conditions #4
</gallery><gallery>
ri3717tableplot5.png|Initial conditions #5
</gallery>

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

The main assumptions of Transition State Theory are<ref>Bligaard, T., and J.k. Nørskov. “Heterogeneous Catalysis.” ''Chemical Bonding at Surfaces and Interfaces'', 2008, pp. 255–321., doi:10.1016/b978-044452837-7.50005-8.</ref>:

- the reactants are in thermal equilibrium and therefore the energy of the particles can be calculated by the Boltzmann equation

- the reactants and the transition state are in equilibrium

- reactants that has reached the transition state with some velocity towards the product will not form back the reactants again.

The transition state theory predictions for reaction rate values ignores the fact that the reaction may not proceed even with energy higher than the activation energy. For example, with initial conditions #4 discussed previously, the reactants approach each other with enough energy and reacts, but the product forms back the reactant species, so there is no reaction overall. This means that the transition state theory predicts reaction rate to be faster than the experimental rate.

This part is good. Please be aware that terms like 'Boltzmann'; 'equilibrium' are not very applicable in our case, because we only have three atoms in the system. [[User:Sw2711|Sw2711]] ([[User talk:Sw2711|talk]]) 12:12, 30 May 2019 (BST)

== Exercise 2-1: F-H-H system ==

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

H-F bond is stronger (bond strength~ 565 kJ/mol<ref name=":0">Yoder, Claude. “Wired Chemist.” ''Common Bond Energies (D) and Bond Length (r)'', www.wiredchemist.com/chemistry/data/bond_energies_lengths.html.</ref>) than H-H bond (bond strength~ 432 kJ/mol<ref name=":0" />) because although both are covalent bonds and have good orbital overlap due to the same or very similar sized orbitals. H-F is highly polar while H-H bond is purely covalent. This means that the formation of H-F bond is favourable in the reaction F + H2 even with the need to break the H-H bond and thus exothermic, while for the reverse reaction of H + HF, the breaking of H-F to make H-H bond is overall not stabilising and thus the reaction is endothermic.

 Did you get the conclusion by checking the values, your prior knowledge or the energy surfaces plot? Did you notice in the energy surface plot, the channel representing H-F + H system is lower in energy compared to H-H + F system. That's why from H-H + F -> H-F + H is from a higher energy surface to a lower energy surface, releasing energy. That's why it is exothermic. [[User:Sw2711|Sw2711]] ([[User talk:Sw2711|talk]]) 12:20, 30 May 2019 (BST)

=== Locate the approximate position of the transition state. ===
The transition state for the F-H-H system is not symmetrical, so the distance between F-H (r1) and H-H (r2) is not equal. With the initial conditions r1= 1.814 Å, and r2 = 0.745 Å, the three atoms F, H, H do not move as shown in the internuclear Distances vs Time plot below, which is indicative of a transition state. <gallery>
ri3717_plot34.png
</gallery>

The surface plot also shows the position of the transition state at r1= 1.814 Å, and r2 = 0.745 Å as the black spot on the graph is stationary. <gallery>
ri3717_plot35.png

</gallery>Focusing on the reaction of F + H2, which is exothermic, the transition state is a "early" transition state. based on the Hammond's postulate, the structure of the transition state is energetically closer and similar in structure to the reactants. This agrees with the internuclear distances (r1 and r2) obtained for the transition state since the distance between the two H atoms is much shorter than the distance between F and H.

 This part is good[[User:Sw2711|Sw2711]] ([[User talk:Sw2711|talk]]) 12:21, 30 May 2019 (BST)

=== Report the activation energy for both reactions. ===
Now choosing a point close to the transition state, the set of initial conditions is r1= 1.804 Å, and r2 = 0.745 Å. This yielded the following Energy vs Time graph:<gallery>
ri3717plot36.png
</gallery>The decrease in potential energy (hence the total energy) was determined from the graph as 29.9 kcal/mol. This is indicative of the energy difference between the transition state and the (F-H + H) state. Therefore this is the activation energy for the endothermic reaction between HF + H.

For the activation energy of the reaction F + H2, another set of initial conditions is tested: r1= 1.824 Å, and r2 = 0.745 Å. This yielded the following Energy vs Time graph:<gallery>
ri3717plot7.png
</gallery>The energy change is very small even at a very large value of t (t= 3000). The activation energy of the reaction F + H2 is estimated to be 0.2 kcal/mol.
This part is also very good. [[User:Sw2711|Sw2711]] ([[User talk:Sw2711|talk]]) 12:22, 30 May 2019 (BST)

=== Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Momenta vs Time”. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally. ===
Initial conditions: r1= 2.0, r2 = 0.7, p1= -0.7, p2 = 0

Observations from Animation: The H2 molecule moves towards F and one of the H atoms gets pulled towards F. Then it is bounced back to H, vibrates, and forms back H-F and H in the end.

As shown in the Momenta vs Time graph below, the overall momentum has increased after reaction. The reaction energy has been transferred to the products in the form of kinetic energy (vibration). <gallery>
ri3717plotmomentum.png|Momenta vs Time
</gallery>This could be confirmed experimentally by monitoring the change in temperature as the energy could be released to or absorbed from the environment as heat. As F+H2 is an exothermic reaction, the temperature is expected to increase on reaction.

 Good, but how does the change in moment from reactants to products prove the conservation of energy? What's the method called by measuring the change in temperature? [[User:Sw2711|Sw2711]] ([[User talk:Sw2711|talk]]) 12:31, 30 May 2019 (BST)

=== Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 0.74, with a momentum pFH = -0.5, and explore several values of pHH in the range -3 to 3 (explore values also close to these limits). What do you observe? Note that we are putting a significant amount of energy (much more than the activation energy) into the system on the H - H vibration. ===
For initial conditions rFH = 2.10, rHH = 0.74 and pFH = -0.5, different values of pHH between -3 and 3 were tested and results are shown below.
{| class="wikitable"
!#
!pHH (kg.m/s)
!Reaction?
!Observation
|-
|1
|<nowiki>-3.0</nowiki>
|no
|H-H molecule vibrates for a while, then one of the H atoms gets closer to F, forming H-F for a very short while. Then the H atom gets pushed back and forms back H-H.
|-
|2
|<nowiki>-2.9</nowiki>
|yes
|H2 molecule moves closer to the F atom as it vibrates, and eventually one of the H atoms forms a bond with F. The remaining H gets repelled and moves away, resulting in H-F and H.
|-
|3
|<nowiki>-2.0</nowiki>
|yes
|same as above but reaction happens more slowly.
|-
|4
|<nowiki>-1.0</nowiki>
|yes
|same as above but reaction happens more slowly.
|-
|5
|0
|no
|H2 molecule vibrates and moves towards the F atom until it gets repelled by F and moves away.
|-
|6
|1.0
|yes
|H2 molecule moves closer to the F atom as it vibrates, and eventually one of the H atoms forms a bond with F. The remaining H gets repelled and moves away, resulting in H-F and H.
|-
|7
|2.0
|yes
|Same as above but reaction happens more quickly.
|-
|8
|2.9
|yes
|Same as above but reaction happens more quickly.
|-
|9
|3.0
|no
|H-H molecule vibrates for a while, then one of the H atoms gets closer to F, forming H-F for a very short while. Then the H atom gets pushed back and forms back H-H.
|}

=== For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now? ===
Initial conditions: rFH = 2.10, rHH = 0.74 and pFH = -0.8, pHH = 0.1<gallery>
ri3717Hahb.png|Starting positions of F and H-H
</gallery>Observation: A vibrating hydrogen molecule slowly approaches an F atom. One of the H atoms (HA) gets pulled towards the F atom, producing F-HA and HB, but HA gets repelled and moves back towards HB, producing back the reactant, F and H-H. Then the HA atom gets repelled by HB and bonds to F again, ending with F-H and H.<gallery>
ri3717plot0.png|Contour plot of the reaction
</gallery>

 Very good effort for recording all these different trials. As I said above, you need to improve the way you describe your observations [[User:Sw2711|Sw2711]] ([[User talk:Sw2711|talk]]) 12:33, 30 May 2019 (BST)

== Exercise 2-2: H + HF System ==
=== 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. ===
Initial conditions #1: rFH = 0.92, rHH = 1.5 and pFH = -0.1, pHH = -2.5

I have set the initial conditions at the bottom of the entry channel (H + HF reactant conditions) with very low pFH, and obtained a reactive trajectory by decreasing pHH and increasing pFH, which is:

Initial conditions #2: rFH = 0.92, rHH = 1.5 and pFH = -10.0, pHH = -1.0 <gallery>
Screen_Shot_2019-05-23_at_14.41.03.png|Initial conditions (F-H, H)
</gallery><gallery>
Screen_Shot_2019-05-23_at_14.40.57.png|After reaction (F, H-H)
</gallery><gallery>
Screen_Shot_2019-05-23_at_14.39.12.png|Contour plot for the reaction between F-H and H, producing F and H-H
</gallery><gallery>
Screen_Shot_2019-05-23_at_14.38.15.png|Change in internuclear distances during and after reaction
</gallery>For a given amount of total energy, the distribution of energy between different modes can affect the efficiency of the reaction. For example, even with enough total energy to overcome the energy barrier for the reaction to occur, if there is not enough translational energy, the reactants will not come close enough to react.

The Polanyi's empirical rules state that in simple terms, translational energy is more effective in activating the reactants for exothermic reactions, while vibrational energy is more effective for endothermic reactions<ref>Bowman, Joel M. “Dynamics of the Reaction of Methane with Chlorine Atom on an Accurate Potential Energy Surface.” ''Science'', American Association for the Advancement of Science, 21 Oct. 2011, science.sciencemag.org/content/334/6054/343.</ref>. The relationship can also be shown as below:

Ea = mΔH + constant

where Ea= activation energy, m= indicative of the position of transition state on the reaction coordinate<ref>Wubbels, Gene G. “The Bell–Evans–Polanyi Principle and the Regioselectivity of Electrophilic Aromatic Substitution Reactions.” ''Tetrahedron Letters'', vol. 56, no. 13, 2015, pp. 1716–1719., doi:10.1016/j.tetlet.2015.02.070.</ref>. The position of the transition state is important because it determines which of the two modes, the translational energy or the vibrational energy has a larger influence on the efficiency of the reaction.

Looking at initial conditions #1 ( rFH = 0.92, rHH = 1.5 and pFH = -0.1, pHH = -2.5 ) in which the reaction did not take place, we can say based on the Polanyi's rules that since this is an endothermic reaction (H + H-F --> H-H + F), vibrational energy has a more significant impact on the reaction efficiency. the momentum in F-H bond is set very low at -0.1, and this is why the reaction does not take place in these conditions.

Now looking at initial conditions #2 ( rFH = 0.92, rHH = 1.5 and pFH = -10.0, pHH = -1.0 ), the reaction was successful and we can see why based on the Polanyi's rules. In initial conditions #2, the momentum in F-H bond is now set very high at -10.0, which provided the reactants with enough vibrational energy to react.

This part is very good. [[User:Sw2711|Sw2711]] ([[User talk:Sw2711|talk]]) 12:35, 30 May 2019 (BST)

== References ==

2019-05-30T11:05:36Z

Sw2711: /* Report your best estimate of the transition state position (rts) for trajectories from r1 = r2 and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. */

== Exercise 1: H+H2 system ==

=== On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===

The transition state is mathematically defined as:<gallery>
File:ri3717_Eq.1
</gallery>where q1 and q2 are the reactant molecules, V is the potential energy.

The transition state can be identified by starting trajectories near the transition state and if they move towards reactants or products (even with no initial momentum) then it is the transition state. It can be distinguished from a local minimum by calculating the second derivative, which gives a negative value for a local minimum but a positive value for a local maximum (i.e. the transition state).

 Are you sure about your statement? TS is the maximum point on the minimum energy pathway. A saddle point, in this case, the TS point, is a point where in a 3-dimensional space, it is both a maximum and a minimum at the same time, depending on which plane you are looking at. That differentiates from a local minimum or a maximum, because whichever plane you are looking at, it's always a minimum or a maximum. Please check the figure that I attached. [[File: SaddleWSK.png]][[User:Sw2711|Sw2711]] ([[User talk:Sw2711|talk]]) 12:04, 30 May 2019 (BST)

=== Report your best estimate of the transition state position ('''rts''') for trajectories from r1 = r2 and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ===
The transition state position is the position at which the distance between A-B (r1) and B-C (r2) is equal, as it is symmetrical for this particular reaction. However, for trajectories from r1 = r2, the H atoms perform oscillation, so the transition state position needs to be estimated by testing different r values. The transition state position (rts) found with r1 = r2 is 0.908 Å. At this distance, there is almost no oscillation as shown in the Internuclear Distance vs Time plot below: <gallery>
File:Ri3717_plot1.png

</gallery>

This shows that at this value of r, the H atoms do not move with no initial momentum, which is characteristic of a transition state.
This part is good. [[User:Sw2711|Sw2711]] ([[User talk:Sw2711|talk]]) 12:05, 30 May 2019 (BST)

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

For trajectories with r values below:

r1 = rts + 0.01 = 0.909

r2 = rts = 0.908,

the MEP and Dynamics trajectories are compared.<gallery>
ri3717_ABC.png|Initial conditions of the three H atoms. Momenta= 0

</gallery>

Dynamics

Animation- B moves towards the C and A moves away from the rest. H atoms B and C continue to move to the right as they oscillate.

Internuclear distances- stays constant for a short time and A-B and A-C distances increase as they move away from each other, while B-C distance decreases and fluctuates as the molecule performs an oscillation. At large t, the final value of B-C distance is approx. 0.75.

Momenta- the momentum of A-B (p1(t)) increases with time; rapid increase at first then continues to slowly increase to a limit. The momentum of B-C (p2(t)) decreases first, then increases with fluctuation. At large t, p1(t) reaches a limit at 2.5 and p2(t) oscillates between 1.0-1.5, giving an average of 1.25.

The angle θ stays constant at 0.

MEP

Animation- a smooth movement of the H atom in the centre towards the H atom on the right as the H atom on the left accelerates towards left.

Internuclear distances- stays constant for a very short time and A-B and A-C distances increase as they move away from each other, while B-C distance decreases without fluctuation. At large t, the final value of B-C distance is approx. 0.75.

Momenta- no change, constant at 0.

When the r1 and r2 values are swapped, H atom A and C swap roles- A and B forms a molecule, C moves away to the right.

=== Complete the table below by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table. ===
{| class="wikitable" border="1"
!#
! p1 (kg.m/s) !! p2 (kg.m/s) !! Etot (kcal/mol) !! Reactive? !! Description of the dynamics
|-
|1
| -1.25 || -2.5 ||-99.0||yes||As molecule A-B approaches C, B is pulled away from A, forming molecule B-C. A moves away to the left, B-C moves to the right with oscillation.
|-
|2
| -1.5 || -2.0 ||-100.5||no||Molecule A-B and atom C slowly approaches each other, and when they get to a certain distance, they repel and moves away from each other.
|-
|3
| -1.5 || -2.5 ||-99.0||yes||As molecule A-B slowly approaches C, B is pulled away from A, forming molecule B-C. A moves away to the left, B-C moves to the right with oscillation.
|-
|4
| -2.5 || -5.0 ||-85.0||no||Molecule A-B approaches C and B is pulled away from A. B-C oscillates but gets too close and repel each other, pushing B back to A, forming molecule A-B.
|-
|5
| -2.5 || -5.2 ||-83.4||yes||Same as the one above happens, but molecule A-B oscillates again, gets too close and repel each other, pushing B back to C, finally forming molecule B-C.
|}
Although there is no simple trend in the reactivity and the initial conditions, It is evident that small changes to the initial conditions (momentum and hence the total energy) can affect the reactivity. For example based on the last two sets of initial conditions (conditions 4 and 5), we can see that increasing the magnitude of p2 by 0.2 kg.m/s increases the total energy by 1.6 kcal/mol, and enables the reaction between molecule A-B and atom C, forming atom A and molecule B-C. <gallery>
ri3717tableplot1.png|Initial conditions #1
</gallery><gallery>ri3717tableplot2.png|Initial conditions #2
</gallery><gallery>ri3717tableplot3.png|Initial conditions #3
</gallery><gallery>
ri3717tableplot4.png|Initial conditions #4
</gallery><gallery>
ri3717tableplot5.png|Initial conditions #5
</gallery>

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

The main assumptions of Transition State Theory are<ref>Bligaard, T., and J.k. Nørskov. “Heterogeneous Catalysis.” ''Chemical Bonding at Surfaces and Interfaces'', 2008, pp. 255–321., doi:10.1016/b978-044452837-7.50005-8.</ref>:

- the reactants are in thermal equilibrium and therefore the energy of the particles can be calculated by the Boltzmann equation

- the reactants and the transition state are in equilibrium

- reactants that has reached the transition state with some velocity towards the product will not form back the reactants again.

The transition state theory predictions for reaction rate values ignores the fact that the reaction may not proceed even with energy higher than the activation energy. For example, with initial conditions #4 discussed previously, the reactants approach each other with enough energy and reacts, but the product forms back the reactant species, so there is no reaction overall. This means that the transition state theory predicts reaction rate to be faster than the experimental rate.

== Exercise 2-1: F-H-H system ==

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

H-F bond is stronger (bond strength~ 565 kJ/mol<ref name=":0">Yoder, Claude. “Wired Chemist.” ''Common Bond Energies (D) and Bond Length (r)'', www.wiredchemist.com/chemistry/data/bond_energies_lengths.html.</ref>) than H-H bond (bond strength~ 432 kJ/mol<ref name=":0" />) because although both are covalent bonds and have good orbital overlap due to the same or very similar sized orbitals. H-F is highly polar while H-H bond is purely covalent. This means that the formation of H-F bond is favourable in the reaction F + H2 even with the need to break the H-H bond and thus exothermic, while for the reverse reaction of H + HF, the breaking of H-F to make H-H bond is overall not stabilising and thus the reaction is endothermic.

=== Locate the approximate position of the transition state. ===
The transition state for the F-H-H system is not symmetrical, so the distance between F-H (r1) and H-H (r2) is not equal. With the initial conditions r1= 1.814 Å, and r2 = 0.745 Å, the three atoms F, H, H do not move as shown in the internuclear Distances vs Time plot below, which is indicative of a transition state. <gallery>
ri3717_plot34.png
</gallery>

The surface plot also shows the position of the transition state at r1= 1.814 Å, and r2 = 0.745 Å as the black spot on the graph is stationary. <gallery>
ri3717_plot35.png

</gallery>Focusing on the reaction of F + H2, which is exothermic, the transition state is a "early" transition state. based on the Hammond's postulate, the structure of the transition state is energetically closer and similar in structure to the reactants. This agrees with the internuclear distances (r1 and r2) obtained for the transition state since the distance between the two H atoms is much shorter than the distance between F and H.

=== Report the activation energy for both reactions. ===
Now choosing a point close to the transition state, the set of initial conditions is r1= 1.804 Å, and r2 = 0.745 Å. This yielded the following Energy vs Time graph:<gallery>
ri3717plot36.png
</gallery>The decrease in potential energy (hence the total energy) was determined from the graph as 29.9 kcal/mol. This is indicative of the energy difference between the transition state and the (F-H + H) state. Therefore this is the activation energy for the endothermic reaction between HF + H.

For the activation energy of the reaction F + H2, another set of initial conditions is tested: r1= 1.824 Å, and r2 = 0.745 Å. This yielded the following Energy vs Time graph:<gallery>
ri3717plot7.png
</gallery>The energy change is very small even at a very large value of t (t= 3000). The activation energy of the reaction F + H2 is estimated to be 0.2 kcal/mol.

=== Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Momenta vs Time”. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally. ===
Initial conditions: r1= 2.0, r2 = 0.7, p1= -0.7, p2 = 0

Observations from Animation: The H2 molecule moves towards F and one of the H atoms gets pulled towards F. Then it is bounced back to H, vibrates, and forms back H-F and H in the end.

As shown in the Momenta vs Time graph below, the overall momentum has increased after reaction. The reaction energy has been transferred to the products in the form of kinetic energy (vibration). <gallery>
ri3717plotmomentum.png|Momenta vs Time
</gallery>This could be confirmed experimentally by monitoring the change in temperature as the energy could be released to or absorbed from the environment as heat. As F+H2 is an exothermic reaction, the temperature is expected to increase on reaction.

=== Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 0.74, with a momentum pFH = -0.5, and explore several values of pHH in the range -3 to 3 (explore values also close to these limits). What do you observe? Note that we are putting a significant amount of energy (much more than the activation energy) into the system on the H - H vibration. ===
For initial conditions rFH = 2.10, rHH = 0.74 and pFH = -0.5, different values of pHH between -3 and 3 were tested and results are shown below.
{| class="wikitable"
!#
!pHH (kg.m/s)
!Reaction?
!Observation
|-
|1
|<nowiki>-3.0</nowiki>
|no
|H-H molecule vibrates for a while, then one of the H atoms gets closer to F, forming H-F for a very short while. Then the H atom gets pushed back and forms back H-H.
|-
|2
|<nowiki>-2.9</nowiki>
|yes
|H2 molecule moves closer to the F atom as it vibrates, and eventually one of the H atoms forms a bond with F. The remaining H gets repelled and moves away, resulting in H-F and H.
|-
|3
|<nowiki>-2.0</nowiki>
|yes
|same as above but reaction happens more slowly.
|-
|4
|<nowiki>-1.0</nowiki>
|yes
|same as above but reaction happens more slowly.
|-
|5
|0
|no
|H2 molecule vibrates and moves towards the F atom until it gets repelled by F and moves away.
|-
|6
|1.0
|yes
|H2 molecule moves closer to the F atom as it vibrates, and eventually one of the H atoms forms a bond with F. The remaining H gets repelled and moves away, resulting in H-F and H.
|-
|7
|2.0
|yes
|Same as above but reaction happens more quickly.
|-
|8
|2.9
|yes
|Same as above but reaction happens more quickly.
|-
|9
|3.0
|no
|H-H molecule vibrates for a while, then one of the H atoms gets closer to F, forming H-F for a very short while. Then the H atom gets pushed back and forms back H-H.
|}

=== For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now? ===
Initial conditions: rFH = 2.10, rHH = 0.74 and pFH = -0.8, pHH = 0.1<gallery>
ri3717Hahb.png|Starting positions of F and H-H
</gallery>Observation: A vibrating hydrogen molecule slowly approaches an F atom. One of the H atoms (HA) gets pulled towards the F atom, producing F-HA and HB, but HA gets repelled and moves back towards HB, producing back the reactant, F and H-H. Then the HA atom gets repelled by HB and bonds to F again, ending with F-H and H.<gallery>
ri3717plot0.png|Contour plot of the reaction
</gallery>

== Exercise 2-2: H + HF System ==
=== 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. ===
Initial conditions #1: rFH = 0.92, rHH = 1.5 and pFH = -0.1, pHH = -2.5

I have set the initial conditions at the bottom of the entry channel (H + HF reactant conditions) with very low pFH, and obtained a reactive trajectory by decreasing pHH and increasing pFH, which is:

Initial conditions #2: rFH = 0.92, rHH = 1.5 and pFH = -10.0, pHH = -1.0 <gallery>
Screen_Shot_2019-05-23_at_14.41.03.png|Initial conditions (F-H, H)
</gallery><gallery>
Screen_Shot_2019-05-23_at_14.40.57.png|After reaction (F, H-H)
</gallery><gallery>
Screen_Shot_2019-05-23_at_14.39.12.png|Contour plot for the reaction between F-H and H, producing F and H-H
</gallery><gallery>
Screen_Shot_2019-05-23_at_14.38.15.png|Change in internuclear distances during and after reaction
</gallery>For a given amount of total energy, the distribution of energy between different modes can affect the efficiency of the reaction. For example, even with enough total energy to overcome the energy barrier for the reaction to occur, if there is not enough translational energy, the reactants will not come close enough to react.

The Polanyi's empirical rules state that in simple terms, translational energy is more effective in activating the reactants for exothermic reactions, while vibrational energy is more effective for endothermic reactions<ref>Bowman, Joel M. “Dynamics of the Reaction of Methane with Chlorine Atom on an Accurate Potential Energy Surface.” ''Science'', American Association for the Advancement of Science, 21 Oct. 2011, science.sciencemag.org/content/334/6054/343.</ref>. The relationship can also be shown as below:

Ea = mΔH + constant

where Ea= activation energy, m= indicative of the position of transition state on the reaction coordinate<ref>Wubbels, Gene G. “The Bell–Evans–Polanyi Principle and the Regioselectivity of Electrophilic Aromatic Substitution Reactions.” ''Tetrahedron Letters'', vol. 56, no. 13, 2015, pp. 1716–1719., doi:10.1016/j.tetlet.2015.02.070.</ref>. The position of the transition state is important because it determines which of the two modes, the translational energy or the vibrational energy has a larger influence on the efficiency of the reaction.

Looking at initial conditions #1 ( rFH = 0.92, rHH = 1.5 and pFH = -0.1, pHH = -2.5 ) in which the reaction did not take place, we can say based on the Polanyi's rules that since this is an endothermic reaction (H + H-F --> H-H + F), vibrational energy has a more significant impact on the reaction efficiency. the momentum in F-H bond is set very low at -0.1, and this is why the reaction does not take place in these conditions.

Now looking at initial conditions #2 ( rFH = 0.92, rHH = 1.5 and pFH = -10.0, pHH = -1.0 ), the reaction was successful and we can see why based on the Polanyi's rules. In initial conditions #2, the momentum in F-H bond is now set very high at -10.0, which provided the reactants with enough vibrational energy to react.

== References ==

MRD:ri3717 summer 2019

2019-05-30T11:04:27Z

== Exercise 1: H+H2 system ==

=== On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===

The transition state is mathematically defined as:<gallery>
File:ri3717_Eq.1
</gallery>where q1 and q2 are the reactant molecules, V is the potential energy.

The transition state can be identified by starting trajectories near the transition state and if they move towards reactants or products (even with no initial momentum) then it is the transition state. It can be distinguished from a local minimum by calculating the second derivative, which gives a negative value for a local minimum but a positive value for a local maximum (i.e. the transition state).

 Are you sure about your statement? TS is the maximum point on the minimum energy pathway. A saddle point, in this case, the TS point, is a point where in a 3-dimensional space, it is both a maximum and a minimum at the same time, depending on which plane you are looking at. That differentiates from a local minimum or a maximum, because whichever plane you are looking at, it's always a minimum or a maximum. Please check the figure that I attached. [[File: SaddleWSK.png]][[User:Sw2711|Sw2711]] ([[User talk:Sw2711|talk]]) 12:04, 30 May 2019 (BST)

=== Report your best estimate of the transition state position ('''rts''') for trajectories from r1 = r2 and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ===
The transition state position is the position at which the distance between A-B (r1) and B-C (r2) is equal, as it is symmetrical for this particular reaction. However, for trajectories from r1 = r2, the H atoms perform oscillation, so the transition state position needs to be estimated by testing different r values. The transition state position (rts) found with r1 = r2 is 0.908 Å. At this distance, there is almost no oscillation as shown in the Internuclear Distance vs Time plot below: <gallery>
File:Ri3717_plot1.png

</gallery>

This shows that at this value of r, the H atoms do not move with no initial momentum, which is characteristic of a transition state.

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

For trajectories with r values below:

r1 = rts + 0.01 = 0.909

r2 = rts = 0.908,

the MEP and Dynamics trajectories are compared.<gallery>
ri3717_ABC.png|Initial conditions of the three H atoms. Momenta= 0

</gallery>

Dynamics

Animation- B moves towards the C and A moves away from the rest. H atoms B and C continue to move to the right as they oscillate.

Internuclear distances- stays constant for a short time and A-B and A-C distances increase as they move away from each other, while B-C distance decreases and fluctuates as the molecule performs an oscillation. At large t, the final value of B-C distance is approx. 0.75.

Momenta- the momentum of A-B (p1(t)) increases with time; rapid increase at first then continues to slowly increase to a limit. The momentum of B-C (p2(t)) decreases first, then increases with fluctuation. At large t, p1(t) reaches a limit at 2.5 and p2(t) oscillates between 1.0-1.5, giving an average of 1.25.

The angle θ stays constant at 0.

MEP

Animation- a smooth movement of the H atom in the centre towards the H atom on the right as the H atom on the left accelerates towards left.

Internuclear distances- stays constant for a very short time and A-B and A-C distances increase as they move away from each other, while B-C distance decreases without fluctuation. At large t, the final value of B-C distance is approx. 0.75.

Momenta- no change, constant at 0.

When the r1 and r2 values are swapped, H atom A and C swap roles- A and B forms a molecule, C moves away to the right.

=== Complete the table below by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table. ===
{| class="wikitable" border="1"
!#
! p1 (kg.m/s) !! p2 (kg.m/s) !! Etot (kcal/mol) !! Reactive? !! Description of the dynamics
|-
|1
| -1.25 || -2.5 ||-99.0||yes||As molecule A-B approaches C, B is pulled away from A, forming molecule B-C. A moves away to the left, B-C moves to the right with oscillation.
|-
|2
| -1.5 || -2.0 ||-100.5||no||Molecule A-B and atom C slowly approaches each other, and when they get to a certain distance, they repel and moves away from each other.
|-
|3
| -1.5 || -2.5 ||-99.0||yes||As molecule A-B slowly approaches C, B is pulled away from A, forming molecule B-C. A moves away to the left, B-C moves to the right with oscillation.
|-
|4
| -2.5 || -5.0 ||-85.0||no||Molecule A-B approaches C and B is pulled away from A. B-C oscillates but gets too close and repel each other, pushing B back to A, forming molecule A-B.
|-
|5
| -2.5 || -5.2 ||-83.4||yes||Same as the one above happens, but molecule A-B oscillates again, gets too close and repel each other, pushing B back to C, finally forming molecule B-C.
|}
Although there is no simple trend in the reactivity and the initial conditions, It is evident that small changes to the initial conditions (momentum and hence the total energy) can affect the reactivity. For example based on the last two sets of initial conditions (conditions 4 and 5), we can see that increasing the magnitude of p2 by 0.2 kg.m/s increases the total energy by 1.6 kcal/mol, and enables the reaction between molecule A-B and atom C, forming atom A and molecule B-C. <gallery>
ri3717tableplot1.png|Initial conditions #1
</gallery><gallery>ri3717tableplot2.png|Initial conditions #2
</gallery><gallery>ri3717tableplot3.png|Initial conditions #3
</gallery><gallery>
ri3717tableplot4.png|Initial conditions #4
</gallery><gallery>
ri3717tableplot5.png|Initial conditions #5
</gallery>

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

The main assumptions of Transition State Theory are<ref>Bligaard, T., and J.k. Nørskov. “Heterogeneous Catalysis.” ''Chemical Bonding at Surfaces and Interfaces'', 2008, pp. 255–321., doi:10.1016/b978-044452837-7.50005-8.</ref>:

- the reactants are in thermal equilibrium and therefore the energy of the particles can be calculated by the Boltzmann equation

- the reactants and the transition state are in equilibrium

- reactants that has reached the transition state with some velocity towards the product will not form back the reactants again.

The transition state theory predictions for reaction rate values ignores the fact that the reaction may not proceed even with energy higher than the activation energy. For example, with initial conditions #4 discussed previously, the reactants approach each other with enough energy and reacts, but the product forms back the reactant species, so there is no reaction overall. This means that the transition state theory predicts reaction rate to be faster than the experimental rate.

== Exercise 2-1: F-H-H system ==

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

H-F bond is stronger (bond strength~ 565 kJ/mol<ref name=":0">Yoder, Claude. “Wired Chemist.” ''Common Bond Energies (D) and Bond Length (r)'', www.wiredchemist.com/chemistry/data/bond_energies_lengths.html.</ref>) than H-H bond (bond strength~ 432 kJ/mol<ref name=":0" />) because although both are covalent bonds and have good orbital overlap due to the same or very similar sized orbitals. H-F is highly polar while H-H bond is purely covalent. This means that the formation of H-F bond is favourable in the reaction F + H2 even with the need to break the H-H bond and thus exothermic, while for the reverse reaction of H + HF, the breaking of H-F to make H-H bond is overall not stabilising and thus the reaction is endothermic.

=== Locate the approximate position of the transition state. ===
The transition state for the F-H-H system is not symmetrical, so the distance between F-H (r1) and H-H (r2) is not equal. With the initial conditions r1= 1.814 Å, and r2 = 0.745 Å, the three atoms F, H, H do not move as shown in the internuclear Distances vs Time plot below, which is indicative of a transition state. <gallery>
ri3717_plot34.png
</gallery>

The surface plot also shows the position of the transition state at r1= 1.814 Å, and r2 = 0.745 Å as the black spot on the graph is stationary. <gallery>
ri3717_plot35.png

</gallery>Focusing on the reaction of F + H2, which is exothermic, the transition state is a "early" transition state. based on the Hammond's postulate, the structure of the transition state is energetically closer and similar in structure to the reactants. This agrees with the internuclear distances (r1 and r2) obtained for the transition state since the distance between the two H atoms is much shorter than the distance between F and H.

=== Report the activation energy for both reactions. ===
Now choosing a point close to the transition state, the set of initial conditions is r1= 1.804 Å, and r2 = 0.745 Å. This yielded the following Energy vs Time graph:<gallery>
ri3717plot36.png
</gallery>The decrease in potential energy (hence the total energy) was determined from the graph as 29.9 kcal/mol. This is indicative of the energy difference between the transition state and the (F-H + H) state. Therefore this is the activation energy for the endothermic reaction between HF + H.

For the activation energy of the reaction F + H2, another set of initial conditions is tested: r1= 1.824 Å, and r2 = 0.745 Å. This yielded the following Energy vs Time graph:<gallery>
ri3717plot7.png
</gallery>The energy change is very small even at a very large value of t (t= 3000). The activation energy of the reaction F + H2 is estimated to be 0.2 kcal/mol.

=== Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Momenta vs Time”. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally. ===
Initial conditions: r1= 2.0, r2 = 0.7, p1= -0.7, p2 = 0

Observations from Animation: The H2 molecule moves towards F and one of the H atoms gets pulled towards F. Then it is bounced back to H, vibrates, and forms back H-F and H in the end.

As shown in the Momenta vs Time graph below, the overall momentum has increased after reaction. The reaction energy has been transferred to the products in the form of kinetic energy (vibration). <gallery>
ri3717plotmomentum.png|Momenta vs Time
</gallery>This could be confirmed experimentally by monitoring the change in temperature as the energy could be released to or absorbed from the environment as heat. As F+H2 is an exothermic reaction, the temperature is expected to increase on reaction.

=== Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 0.74, with a momentum pFH = -0.5, and explore several values of pHH in the range -3 to 3 (explore values also close to these limits). What do you observe? Note that we are putting a significant amount of energy (much more than the activation energy) into the system on the H - H vibration. ===
For initial conditions rFH = 2.10, rHH = 0.74 and pFH = -0.5, different values of pHH between -3 and 3 were tested and results are shown below.
{| class="wikitable"
!#
!pHH (kg.m/s)
!Reaction?
!Observation
|-
|1
|<nowiki>-3.0</nowiki>
|no
|H-H molecule vibrates for a while, then one of the H atoms gets closer to F, forming H-F for a very short while. Then the H atom gets pushed back and forms back H-H.
|-
|2
|<nowiki>-2.9</nowiki>
|yes
|H2 molecule moves closer to the F atom as it vibrates, and eventually one of the H atoms forms a bond with F. The remaining H gets repelled and moves away, resulting in H-F and H.
|-
|3
|<nowiki>-2.0</nowiki>
|yes
|same as above but reaction happens more slowly.
|-
|4
|<nowiki>-1.0</nowiki>
|yes
|same as above but reaction happens more slowly.
|-
|5
|0
|no
|H2 molecule vibrates and moves towards the F atom until it gets repelled by F and moves away.
|-
|6
|1.0
|yes
|H2 molecule moves closer to the F atom as it vibrates, and eventually one of the H atoms forms a bond with F. The remaining H gets repelled and moves away, resulting in H-F and H.
|-
|7
|2.0
|yes
|Same as above but reaction happens more quickly.
|-
|8
|2.9
|yes
|Same as above but reaction happens more quickly.
|-
|9
|3.0
|no
|H-H molecule vibrates for a while, then one of the H atoms gets closer to F, forming H-F for a very short while. Then the H atom gets pushed back and forms back H-H.
|}

=== For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now? ===
Initial conditions: rFH = 2.10, rHH = 0.74 and pFH = -0.8, pHH = 0.1<gallery>
ri3717Hahb.png|Starting positions of F and H-H
</gallery>Observation: A vibrating hydrogen molecule slowly approaches an F atom. One of the H atoms (HA) gets pulled towards the F atom, producing F-HA and HB, but HA gets repelled and moves back towards HB, producing back the reactant, F and H-H. Then the HA atom gets repelled by HB and bonds to F again, ending with F-H and H.<gallery>
ri3717plot0.png|Contour plot of the reaction
</gallery>

== Exercise 2-2: H + HF System ==
=== 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. ===
Initial conditions #1: rFH = 0.92, rHH = 1.5 and pFH = -0.1, pHH = -2.5

I have set the initial conditions at the bottom of the entry channel (H + HF reactant conditions) with very low pFH, and obtained a reactive trajectory by decreasing pHH and increasing pFH, which is:

Initial conditions #2: rFH = 0.92, rHH = 1.5 and pFH = -10.0, pHH = -1.0 <gallery>
Screen_Shot_2019-05-23_at_14.41.03.png|Initial conditions (F-H, H)
</gallery><gallery>
Screen_Shot_2019-05-23_at_14.40.57.png|After reaction (F, H-H)
</gallery><gallery>
Screen_Shot_2019-05-23_at_14.39.12.png|Contour plot for the reaction between F-H and H, producing F and H-H
</gallery><gallery>
Screen_Shot_2019-05-23_at_14.38.15.png|Change in internuclear distances during and after reaction
</gallery>For a given amount of total energy, the distribution of energy between different modes can affect the efficiency of the reaction. For example, even with enough total energy to overcome the energy barrier for the reaction to occur, if there is not enough translational energy, the reactants will not come close enough to react.

The Polanyi's empirical rules state that in simple terms, translational energy is more effective in activating the reactants for exothermic reactions, while vibrational energy is more effective for endothermic reactions<ref>Bowman, Joel M. “Dynamics of the Reaction of Methane with Chlorine Atom on an Accurate Potential Energy Surface.” ''Science'', American Association for the Advancement of Science, 21 Oct. 2011, science.sciencemag.org/content/334/6054/343.</ref>. The relationship can also be shown as below:

Ea = mΔH + constant

where Ea= activation energy, m= indicative of the position of transition state on the reaction coordinate<ref>Wubbels, Gene G. “The Bell–Evans–Polanyi Principle and the Regioselectivity of Electrophilic Aromatic Substitution Reactions.” ''Tetrahedron Letters'', vol. 56, no. 13, 2015, pp. 1716–1719., doi:10.1016/j.tetlet.2015.02.070.</ref>. The position of the transition state is important because it determines which of the two modes, the translational energy or the vibrational energy has a larger influence on the efficiency of the reaction.

Looking at initial conditions #1 ( rFH = 0.92, rHH = 1.5 and pFH = -0.1, pHH = -2.5 ) in which the reaction did not take place, we can say based on the Polanyi's rules that since this is an endothermic reaction (H + H-F --> H-H + F), vibrational energy has a more significant impact on the reaction efficiency. the momentum in F-H bond is set very low at -0.1, and this is why the reaction does not take place in these conditions.

Now looking at initial conditions #2 ( rFH = 0.92, rHH = 1.5 and pFH = -10.0, pHH = -1.0 ), the reaction was successful and we can see why based on the Polanyi's rules. In initial conditions #2, the momentum in F-H bond is now set very high at -10.0, which provided the reactants with enough vibrational energy to react.

== References ==

MRD:01364124

2019-05-24T16:20:29Z

Sw2711: /* Reaction Dynamics */

=Molecular Reaction Dynamics=

==Exercise 1: H + H2 System==

===Dynamics From the Transition State Region===

The transition state is defined as a saddle point where there is both a maximum and minimum depending upon the direction it is viewed from. At the transition state, the maximum and minimum will have differentials equal to zero. By taking the second partial derivative, we may then identify whether this point is a maximum or minimum. The potential energy surface diagram below illustrates how potential energy varies with interatomic distance between A, B and C. It may be seen that there is both a maximum and minimum as indicated by the curvature of the surface plot. Based on the surface plot, it may be observed that there is a greater curvature for points at the maximum in comparison to those at the minimum. This is due to the nature of the intrinsic chemical properties of the molecules in question.

 This part is good. [[User:Sw2711|Sw2711]] ([[User talk:Sw2711|talk]]) 17:08, 24 May 2019 (BST)
[[File:Exercise_1_PE_Surface_Diagram.PNG|300px]]

[[File:Traansition_State_Minimum.PNG|300px]]

===Trajectories from r1 = r2===

The best determined estimate of the transition state position (rts) was found to be 0.909. An initial estimate of 0.91 was found via the contour plot which showed that all atoms were at equal distance from one another and not falling off the ridge. By then establishing a plot of intermolecular distance against time, it was found that movement of atoms with time was little. Measurements up to a time of 1 second showed that the atoms did not fall off the ridge. However, extending this time to 5 seconds proved the opposite and thus accuracy was increased to a maximum of 0.909.

 It shows that you've understood the concept. Just wondering whether you have tried 0.908, 0.907, 0.906 etc? [[User:Sw2711|Sw2711]] ([[User talk:Sw2711|talk]]) 17:11, 24 May 2019 (BST)
[[File:Locating_Transition_State_Contour.PNG|300px]]

[[File:Locating_Transition_State_Time_Distance.PNG|300px]]

===Calculating the Reaction Path===

This part is generally good. [[User:Sw2711|Sw2711]] ([[User talk:Sw2711|talk]]) 17:12, 24 May 2019 (BST)
Calculation of the reaction path via a dynamic calculation setup results in a wavy line as shown in figure x. On the other hand, calculation via MEP results in a straight line. The MEP calculation produces a straight line as the velocity always resets to zero in each time step.

[[File:Trajectories_from_r1_Dynamics.PNG|300px]]

[[File:Trajectories_from_r1_MEP.PNG|300px]]

If an initial condition of r1 = rts and r2 = rts + 0.01 was instead used ie values reversed then the reverse of the graphs above would be shown. These graphs are shown below:

[[File:Trajectories_from_r1_MEP_reverse.PNG|300px]]

[[File:Trajectories_from_r1_Reverse.PNG|300px]]

As shown by the distance-time graph, the A-B and A-C distance increase with time whilst the B-C distance remains relatively constant where the atoms do not move away from one another. Additionally, the graph indicates that the A-B distance is approximately 10 whilst the A-C distance is 0.75.

Setting the initial condition instead to r1 = rts and r2 = rts + 0.01, the B-C distance would instead increase with time and the A-B distance remain relatively constant. which means? [[User:Sw2711|Sw2711]] ([[User talk:Sw2711|talk]]) 17:12, 24 May 2019 (BST)

===Reactive and Unreactive Trajectories===

Initial Positions r1 = 0.74 and r2 = 2.0 are kept constant throughout

* For the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, run trajectories with the following momenta combination and complete the table.
{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Graph Number
|-
| -1.25 || -2.5 || -99.02 || Reactive || Atom C closes towards A-B which then breaks and results in formation of a B-C bond. || 1
|-
| -1.5 || -2.0 || -100.46 || Un-Reactive || Atom C closes towards A-B but does not have sufficient momentum to break the A-B bond. Thus it draws away from the A-B which remains intact. || 2
|-
| -1.5 || -2.5 || -98.96 ||Reactive || Atom C closes towards A-B just in the same pathway as Graph 1, however in this instance atom c has less energy. As a result, the time taken to reach the transition state and consequently react with it is greater || 3
|-
| -2.5 || -5.0 || -84.96 ||Un-Reactive || where is your description for this one? || 4
|-
| -2.5 || -5.2 || -81.80 || Reactive || Atom C approaches the A-B bond and results in oscillation of the A-B but the A-B bond reforms. Momentum of Atom C however is much greater and therefore results in breakage of the A-B bond and consequent formation of the B-C bond || 5
|}

Graph 1
[[File:Contour_Graph_1_Table.PNG|300px]]

Graph 2
[[File:Contour_Graph_2_Table.PNG|300px]]

Graph 3
[[File:Contour_Graph_3_Table.PNG|300px]]

Graph 4
[[File:Contour_Graph_4_Table.PNG|300px]]

Graph 5
[[File:Contour_Graph_5_Table.PNG|300px]]

===Transition State Theory===

Transition State Theory (TST) can be used to understand the reaction rates of elementary chemical reactions. The theory makes an assumption of a special chemical equilibrium (quasi-equilibrium). This is an equilibrium between reactants and activated transition state complexes. The main aim of TST is to provide a qualitative explanation as to how chemical reactions take place. The main assumption of the transition state theory is that if a molecule has sufficient energy to reach the transition state, it will react. If it does not have enough energy to react then it will not react. Furthermore, classical mechanics rather than quantum mechanics is used to explain the transition state theory. This theoretically means that a reaction can take place without the atoms needing to reach the transition state.

 Good, but how does that link to the previous experiment. Do you agree with the TST? In addition, we only have 3 atoms in our system. Do you think terms like 'equilibrium', 'quantum' is applicable in our case? [[User:Sw2711|Sw2711]] ([[User talk:Sw2711|talk]]) 17:16, 24 May 2019 (BST)

==Exercise 2 F-H-F System==

===PES Inspection===

====F + H2 -> HF + H====

Conditions set for this reaction

F - H2 Distance (AB Distance): 2.00 A

H-H Bond Distance (BC Distance): 0.74 A

F-H Momentum: 2.90 kgms-1

H-H Momentum: -0.5 kgms-1

By observing the potential energy surface plot, it can be concluded that this reaction is exothermic as the products are higher in energy compared to the reactants.

 Product is higher in energy than reactant, so it's exothermic??? [[User:Sw2711|Sw2711]] ([[User talk:Sw2711|talk]]) 17:17, 24 May 2019 (BST)

[[File:F_H2_Transition_State.PNG|300px]]

====H + HF====

Conditions set for this reaction

F - H2 Distance (AB Distance): 2.0 A

H-H Bond Distance (BC Distance): 0.74 A

F - H2 Momentum: 2.55 kgms-1

H-H Momentum: -0.5 kgms-1

By observing the potential energy plot for this reaction, it may be concluded that this reaction is endothermic as the products are lower in energy in comparison to the reactants

 Do you really observed product lower in energy in the plot? [[User:Sw2711|Sw2711]] ([[User talk:Sw2711|talk]]) 17:18, 24 May 2019 (BST)

[[File:H_HF_Transition_State.PNG|300px]]

From both these contour diagrams, it may be determined that the H-F bond is stronger than the H-H. This is because for H + HF, the energy required to break the H-F bond is much greater than the energy released from the formation of the H-H bond, consequently the H-F bond is stronger than the H-H bond.
 Good. [[User:Sw2711|Sw2711]] ([[User talk:Sw2711|talk]]) 17:18, 24 May 2019 (BST)

====Position of the Transition State====

HF: 1.80 A

HH: 0.75 A

 You need some explanations. [[User:Sw2711|Sw2711]] ([[User talk:Sw2711|talk]]) 17:19, 24 May 2019 (BST)

====Activation Energy====

Activation Energy is the energy required to reach the transition state. In terms of a potential energy surface diagram, it is the difference in energy between the reactants and the transition state. In order to determine the energy of the reactants, a distance may be added or subtracted from the bond length of HF. By doing this, the contour plots the trajectory of the atoms. In this case, 0.1 A may be added to the H-F bond length such that the trajectory of the atom falls into the reactants.

 So what's the energy? [[User:Sw2711|Sw2711]] ([[User talk:Sw2711|talk]]) 17:19, 24 May 2019 (BST)

====Energy of the Transition State of F + H2====

===Reaction Dynamics===

The initial conditions that result in a reactive trajectory are as follows:

H-H Bond Distance: 0.74 A

H-F Bond Distance: 1.8 A

H -F Momentum: -0.5 kgms-1

H-H Momentum: 2.90 kgms-1

 Huh? so what? Where are your answers for the last two questions? [[User:Sw2711|Sw2711]] ([[User talk:Sw2711|talk]]) 17:20, 24 May 2019 (BST)

2019-05-24T16:13:54Z

Sw2711: /* Reactive and Unreactive Trajectories */

=Molecular Reaction Dynamics=

==Exercise 1: H + H2 System==

===Dynamics From the Transition State Region===

The transition state is defined as a saddle point where there is both a maximum and minimum depending upon the direction it is viewed from. At the transition state, the maximum and minimum will have differentials equal to zero. By taking the second partial derivative, we may then identify whether this point is a maximum or minimum. The potential energy surface diagram below illustrates how potential energy varies with interatomic distance between A, B and C. It may be seen that there is both a maximum and minimum as indicated by the curvature of the surface plot. Based on the surface plot, it may be observed that there is a greater curvature for points at the maximum in comparison to those at the minimum. This is due to the nature of the intrinsic chemical properties of the molecules in question.

 This part is good. [[User:Sw2711|Sw2711]] ([[User talk:Sw2711|talk]]) 17:08, 24 May 2019 (BST)
[[File:Exercise_1_PE_Surface_Diagram.PNG|300px]]

[[File:Traansition_State_Minimum.PNG|300px]]

===Trajectories from r1 = r2===

The best determined estimate of the transition state position (rts) was found to be 0.909. An initial estimate of 0.91 was found via the contour plot which showed that all atoms were at equal distance from one another and not falling off the ridge. By then establishing a plot of intermolecular distance against time, it was found that movement of atoms with time was little. Measurements up to a time of 1 second showed that the atoms did not fall off the ridge. However, extending this time to 5 seconds proved the opposite and thus accuracy was increased to a maximum of 0.909.

 It shows that you've understood the concept. Just wondering whether you have tried 0.908, 0.907, 0.906 etc? [[User:Sw2711|Sw2711]] ([[User talk:Sw2711|talk]]) 17:11, 24 May 2019 (BST)
[[File:Locating_Transition_State_Contour.PNG|300px]]

[[File:Locating_Transition_State_Time_Distance.PNG|300px]]

===Calculating the Reaction Path===

This part is generally good. [[User:Sw2711|Sw2711]] ([[User talk:Sw2711|talk]]) 17:12, 24 May 2019 (BST)
Calculation of the reaction path via a dynamic calculation setup results in a wavy line as shown in figure x. On the other hand, calculation via MEP results in a straight line. The MEP calculation produces a straight line as the velocity always resets to zero in each time step.

[[File:Trajectories_from_r1_Dynamics.PNG|300px]]

[[File:Trajectories_from_r1_MEP.PNG|300px]]

If an initial condition of r1 = rts and r2 = rts + 0.01 was instead used ie values reversed then the reverse of the graphs above would be shown. These graphs are shown below:

[[File:Trajectories_from_r1_MEP_reverse.PNG|300px]]

[[File:Trajectories_from_r1_Reverse.PNG|300px]]

As shown by the distance-time graph, the A-B and A-C distance increase with time whilst the B-C distance remains relatively constant where the atoms do not move away from one another. Additionally, the graph indicates that the A-B distance is approximately 10 whilst the A-C distance is 0.75.

Setting the initial condition instead to r1 = rts and r2 = rts + 0.01, the B-C distance would instead increase with time and the A-B distance remain relatively constant. which means? [[User:Sw2711|Sw2711]] ([[User talk:Sw2711|talk]]) 17:12, 24 May 2019 (BST)

===Reactive and Unreactive Trajectories===

Initial Positions r1 = 0.74 and r2 = 2.0 are kept constant throughout

* For the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, run trajectories with the following momenta combination and complete the table.
{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Graph Number
|-
| -1.25 || -2.5 || -99.02 || Reactive || Atom C closes towards A-B which then breaks and results in formation of a B-C bond. || 1
|-
| -1.5 || -2.0 || -100.46 || Un-Reactive || Atom C closes towards A-B but does not have sufficient momentum to break the A-B bond. Thus it draws away from the A-B which remains intact. || 2
|-
| -1.5 || -2.5 || -98.96 ||Reactive || Atom C closes towards A-B just in the same pathway as Graph 1, however in this instance atom c has less energy. As a result, the time taken to reach the transition state and consequently react with it is greater || 3
|-
| -2.5 || -5.0 || -84.96 ||Un-Reactive || where is your description for this one? || 4
|-
| -2.5 || -5.2 || -81.80 || Reactive || Atom C approaches the A-B bond and results in oscillation of the A-B but the A-B bond reforms. Momentum of Atom C however is much greater and therefore results in breakage of the A-B bond and consequent formation of the B-C bond || 5
|}

Graph 1
[[File:Contour_Graph_1_Table.PNG|300px]]

Graph 2
[[File:Contour_Graph_2_Table.PNG|300px]]

Graph 3
[[File:Contour_Graph_3_Table.PNG|300px]]

Graph 4
[[File:Contour_Graph_4_Table.PNG|300px]]

Graph 5
[[File:Contour_Graph_5_Table.PNG|300px]]

===Transition State Theory===

Transition State Theory (TST) can be used to understand the reaction rates of elementary chemical reactions. The theory makes an assumption of a special chemical equilibrium (quasi-equilibrium). This is an equilibrium between reactants and activated transition state complexes. The main aim of TST is to provide a qualitative explanation as to how chemical reactions take place. The main assumption of the transition state theory is that if a molecule has sufficient energy to reach the transition state, it will react. If it does not have enough energy to react then it will not react. Furthermore, classical mechanics rather than quantum mechanics is used to explain the transition state theory. This theoretically means that a reaction can take place without the atoms needing to reach the transition state.

==Exercise 2 F-H-F System==

===PES Inspection===

====F + H2 -> HF + H====

Conditions set for this reaction

F - H2 Distance (AB Distance): 2.00 A

H-H Bond Distance (BC Distance): 0.74 A

F-H Momentum: 2.90 kgms-1

H-H Momentum: -0.5 kgms-1

By observing the potential energy surface plot, it can be concluded that this reaction is exothermic as the products are higher in energy compared to the reactants.

[[File:F_H2_Transition_State.PNG|300px]]

====H + HF====

Conditions set for this reaction

F - H2 Distance (AB Distance): 2.0 A

H-H Bond Distance (BC Distance): 0.74 A

F - H2 Momentum: 2.55 kgms-1

H-H Momentum: -0.5 kgms-1

By observing the potential energy plot for this reaction, it may be concluded that this reaction is endothermic as the products are lower in energy in comparison to the reactants

[[File:H_HF_Transition_State.PNG|300px]]

From both these contour diagrams, it may be determined that the H-F bond is stronger than the H-H. This is because for H + HF, the energy required to break the H-F bond is much greater than the energy released from the formation of the H-H bond, consequently the H-F bond is stronger than the H-H bond.

====Position of the Transition State====

HF: 1.80 A

HH: 0.75 A

====Activation Energy====

Activation Energy is the energy required to reach the transition state. In terms of a potential energy surface diagram, it is the difference in energy between the reactants and the transition state. In order to determine the energy of the reactants, a distance may be added or subtracted from the bond length of HF. By doing this, the contour plots the trajectory of the atoms. In this case, 0.1 A may be added to the H-F bond length such that the trajectory of the atom falls into the reactants.

====Energy of the Transition State of F + H2====

===Reaction Dynamics===

The initial conditions that result in a reactive trajectory are as follows:

H-H Bond Distance: 0.74 A

H-F Bond Distance: 1.8 A

H -F Momentum: -0.5 kgms-1

H-H Momentum: 2.90 kgms-1