MRD:sds2618

2020-05-14T12:28:28Z

Sds2618:

== Exercise 1: H + H2 system ==

=== Question 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?'''

On a potential energy surface, the transition state can be defined as a saddle point of

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

At the transition state, r1 = r2. From the "Internuclear Distances vs Time" plot of the initial H + H2 system, we can already estimate a transition state position around an internuclear distance of r1 = r2 = 90 pm at approximately 20 fs (corresponding to the point at which internuclear distances A-B and A-C intersect). After testing several initial conditions, it was found that for rAB = rBC = 90.77 pm and pAB = pBC = 0 g.mol-1.pm.fm-1 the system undergoes an extremely small systematic periodic vibration, symmetric on both atoms. Furthermore, the forces at rAB = rBC = 90.77 pm are very close to 0 (0.001 kJ.mol-1.pm-1</sup).

{| class="wikitable" border="1"
! Internuclear Distances vs Time for rAB = 230 pm, pAB = -5.2 g.mol-1.pm.fm-1 and rBC = 74 pm, pBC = 0 g.mol-1.pm.fm-1 || Plot of internuclear distances vs time for rAB = rBC = 90.77 pm and pAB = pBC = 0 g.mol-1.pm.fm-1|| Forces at rAB = rBC = 90.77 pm
|-
| [[File: D vs time-1.png|300px]] || [[File: D vs t TS HHHsds2618.png|300px]] || [[File:Screenshot forcessds2618.png|300px ]]
|}

We can therefore estimate rts = 90.77 pm.

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

When calculating the minimum energy pathway using calculation type MEP and Dynamics, we obtain the following results (with rAB = 90.77 pm, rBC = 91.77 pm and pAB = pBC = 0 g.mol-1.pm.fm-1):

{| class="wikitable" border="1"
! Contour plot of the reaction path using MEP !! Contour plot of the reaction path using Dynamics
|-
| [[File: Q3-MEPsds2618.png|300px]] || [[File: Q3-dynamicssds2618.png|300px]]
|-
| '''Plot of momentum vs time of the reaction path using MEP''' || '''Plot of momentum vs time of the reaction path using Dynamics'''
|-
| [[File: TS MEPsds2618.png|300px]] || [[File: TS Dynamicssds2618.png|300px]]
|}

When comparing these two results, we can see that the "Dynamics" calculation includes oscillation of the atoms involved in the reaction. This is visible in the contour plots: in MEP, a straight curve shows the reaction pathway but in Dynamics we can see oscillations along this path. This is clearly visible in the plots of momentum vs time. Since for MEP, the momentum is reset to 0 after each step, we see no change in momentum over the whole reaction path but in Dynamics, we can see the increase in momentum and oscillation along this reaction.

{| class="wikitable" border="1"
! Internuclear distances vs time where rAB = 90.77 pm, rBC = 91.77 pm !! Internuclear distances vs time where rAB = 91.77 pm, rBC = 90.77 pm
|-
| [[File: Q3-dynamics d vs tsds2618.png|300px]] || [[File: Extra-d vs t switchsds2618.png|300px]]
|-
| '''Momenta vs time where rAB = 90.77 pm, rBC = 91.77 pm''' || '''Momenta vs time where rAB = 91.77 pm, rBC = 90.77 pm'''
|-
| [[File: Extra-momenta vs t normalsds2618.png|300px]] || [[File: Extra-momenta vs t switchsds2618.png|300px]]
|}

Changing from '''rAB = rts''' and '''rBC = rts + 1''' to '''rAB = rts + 1''' and '''rBC = rts''' effectively reverses the direction of the reaction pathway observed. The former gives plots showing how the internuclear distances and momenta vary with time for the H + H2 system when atom A approaches molecule BC. The latter depicts how internuclear distances and momenta vary with time for the same system when atom C approaches molecule AB. This is observed in the graphs essentially as a "switch" between A-B and B-C curves.

From [[Media:D vs time-1.png| this graph]] we can obtain the final geometry of the system which was found to be rAB = 73 pm with pAB = 5.3 g.mol-1.pm.fm-1 and rBC = 278 pm with pBC = 5.1 g.mol-1.pm.fm-1 at t = 50 fs.
Let us now setup a system with initial positions rAB (0) = 73 pm and rBC (0) = 278 pm and with pAB (0) = -5.3 g.mol-1.pm.fm-1 and pBC (0) = -5.1 g.mol-1.pm.fm-1. We obtain the following results:

{| class="wikitable" border="1"
! Internuclear distances vs time with rAB(t=50 fm) = 73 pm, rBC(t=50 fm) = 278 pm, pAB(t=50 fm) = 5.3 g.mol-1.pm.fm-1 and pBC(t=50 fm) = 5.1 g.mol-1.pm.fm-1!! Internuclear distances vs time for rAB (0) = 73 pm and rBC (0) = 278 pm and with pAB (0) = -5.3 g.mol-1.pm.fm-1 and pBC (0) = -5.1 g.mol-1.pm.fm-1
|-
| [[File: D vs time-1.png|300px]] || [[File: D vs t inversion final and intialsds2618.png|300px]]
|-
|}

Similar to before, swapping initial positions for final positions of the same system as well as initial and final momenta (with inversion of the sign) amounts to flipping the trend of the graph of internuclear distances vs time. The graphs show the same reactions as mentioned above: atom A approaches molecule BC to form molecule AB and atom C (left hand side) and atom C approaches molecule AB to form molecule BC and atom A (right hand side). The transition states of the two reactions are located at t ~ 20 fs and ~ 30 fs but with the same rts = 90 pm.

=== Question 4 ===
'''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?'''

{| class="wikitable" border="1"
|+ Testing trajectories with r1 = 74 pm and r2 = 200 pm
! p1 /g.mol-1.pm.fm-1 !! p2 /g.mol-1.pm.fm-1 !! Etot /kj.mol-1 !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280 || Reactive || Atom A approaches molecule BC and passes through the transition state at ~ 21 fs. The reaction gives the following products: molecule AB and atom C|| [[File: Table 1sds2618.png|300px]] [[File: Table 1 d vs tsds2618.png|300px]]
|-
| -3.1 || -4.1 || -420.077 || Unreactive || Atom A approaches molecule BC but the momentum isn't enough for the two to collide. The reaction does not pass through the transition state, instead it regenerates the reactants since no collision has taken place. || [[File: Table 2sds2618.png|300px]] [[File: Table 2 d vs tsds2618.png|300px]]
|-
| -3.1 || -5.1 || -413.977 || Reactive || Atom A approaches molecule BC and passes through the transition state at ~ 21 fs. The reaction gives the following products: molecule AB and atom C || [[File: Table 3sds2618.png|300px]] [[File: Table 3 d vs tsds2618.png|300px]]
|-
| -5.1 || -10.1 || -357.277 || Unreactive (barrier recrossing) || Atom A approaches molecule BC and a bond between A and B forms (at t~ 8 fs). However, this bond still oscillates which breaks the AB bond and reforms the BC one (at t~ 35 fs), regenerating the reactants|| [[File: Table 4sds2618.png|300px]] [[File: Table 4 d vs tsds2618.png|300px]]
|-
| -5.1 || -10.6 || -349.477 || Reactive || Atom A approaches molecule BC and the AB bond forms (at t~ 8 fs). However, the oscillations of the newly formed bond are very strong (since the molecule was given such a high kinetic energy) which causes the AB bond to break to reform molecule BC temporarily (at t~ 15 fs). The oscillations of the BC bond continue and the BC bond breaks (at t~ 25 fs) which generates the products AB and single atom C|| [[File: Table 5sds2618.png|300px]] [[File: Table 5 d vs tsds2618.png|300px]]
|}

As concluded before, trajectories with -3.1 < p1 /g.mol-1.pm.fm-1 < -1.6 and p2 = -5.1 g.mol-1.pm.fm-1 are reactive: this is the case for the first and third reaction from the table.
However, if we increase the momenta outside the previously defined range, the initial positions need to be adjusted accordingly or the reaction will be unreactive. The system will either not have enough kinetic energy (too small momentum) to overcome the activation barrier thus never crossing the transition state (second reaction from the table) or the momentum will be too high and the system will cross the transition state several times, breaking and remaking the bonds between A, B and C (for the two final reactions of the table), this can result in reactive or unreactive trajectories depending on the momenta.

=== Question 5 ===
'''Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''

In order to apply the Transition State Theory, several assumptions are made one of which is that the system can only pass through the transition state once and can't return to the reactants. <ref name="Q5-1" />
As we have seen from the different trajectories of the table above, this is not always the case. If the initial momenta of the different components of the system are too large, the system can cross the transition state several times or not at all if the initial momenta are too small. This is the first limitation in predicting reaction rates using Transition State Theory which leads to an overestimation of the rate constant.

Furthermore, the assumption is made that motion along the reaction coordinate can be treated independently and classically as translational motion.<ref name="Q5-2" /> This will cause another discrepancy between theory and experimental results since in reality, quantum effects will contribute to the motion of the system and in some cases may even cause quantum tunneling, which enables the reactants to form the products without having to pass the energy barrier to go via the transition state (even though this effect is not very likely for this system, it is possible for certain systems). This leads to a higher rate constant since the system can form the products faster by "tunnelling past" the activation energy to form the products. However, this effect varies with the mass of the atoms and is the probability of tunnelling is usually quite small which is why in general, transition state theory overestimates the rate constant.

There are also assumptions made when calculating the partition functions (which include translational, rotational, vibrational and electronic contributions). <ref name="Q5-2" /> The equations used to determine translational and rotational contribution assume high temperatures (ie when the spacing between energy levels are much smaller than kT) and the vibrational contribution is determined using the basis of a simple harmonic oscillator, which is not always for high temperatures. <ref name="Q5-3" />

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

=== Question 1 ===
'''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? Locate the approximate position of the transition state. Report the activation energy for both reactions.'''

By evaluation of the potential energy surfaces of F + H2 and H + HF <ref name="Q1-1" /> it was found that the first reaction is exothermic and the second endothermic. This means that the F-H bond is stronger than the H-H bond since the products of the first reaction (H + HF) are more stable than the reactants (F + H2).

According to Hammond's postulate the structure of the transition state will resemble the structure of the nearest stable species (either products or reactants). Taking the first reaction F + H2 = H + HF, the transition state will resemble the reactants since it is an exothermic reaction. This is the opposite for the second reaction: the transition state will resemble the products for the endothermic reaction. In both cases, the structure of the transition state will be closer to that of "F + H2".

We can therefore assume that at the transition state the F-H distance will be larger than the H-H distance. After testing several initial distances, it was found that the transition state has the following internuclear distances: rFH = 182 pm and rHH = 74 pm. When setting the initial conditions to these parameters (with pFH = pHH = 0 g.mol-1.pm.fm-1), the Animation showed a small periodic oscillation of the atoms and the plot of Internuclear Distances vs Time was the following:
[[File: TS HF+Hsds2618.png|300px|thumb|center|Internuclear Distances vs Time of the transition state of HF + H.]]

We can see that the oscillations of all internuclear distances are "parallel" to each other which shows that no collision will take place with 0 momentum at these specific distances, which is the definition of the transition state.

To find the activation energies of both reactions, we will run the program (using MEP) starting from the a structure close to the transition state and drive the reaction towards HF + H to find Ea for H + HF = F + H2 and to find Ea for F + H2 = H + HF, we will drive the reaction towards F + H2. Considering the previously determined internuclear distances of the transition state determined previously, it was found that the following parameters drive the transition state towards FH + H: rFH = 170 pm, rHH = 74 pm and pFH = pHH = 0 g.mol-1.pm.fm-1. In order to drive the transition state towards F + H2 we will need to prolong the F-H distance and reduce the H-H distance slightly. After several tries, the following parameters were used to find the activation energy: rFH = 190 pm, rHH = 70 pm and pFH = pHH = 0 g.mol-1.pm.fm-1. Using these two initial conditions, we will plot a graph of Energy vs Time and determine the activation energies of both reactions by taking the difference of the maximum of the curve (energy of the transition state) and the minimum (energy of the products).

The following results were obtained:

{| class="wikitable" border="1"
! Energy vs Time (rFH = 150 pm, rHH = 74 pm, pFH = pHH = 0 g.mol-1.pm.fm-1) !! energy vs Time ( rFH = 190 pm, rHH = 70 pm and pFH = pHH = 0 g.mol-1.pm.fm-1)
|-
| [[File: Activation energy 1sds2618.png|300px]] || [[File: Activation energy 2sds2618.png|300px]]
|-
|}

According to these graphs, the activation for the reaction H + HF = H2 + F is approximately 127 kJ.mol-1 and that of H2 = HF + H is around 0.9 kJ.mol-1. The very small activation energy for the second reaction is due to the strong driving force to form the H-F bond since this is a very stable compound. The same can be used to explain the relatively large activation energy for the inverse reaction: the HF compound is much more stable and has a very strong bon which means a lot of energy is required to break that bond in order to form a weaker, less stable one between two H atoms.

=== Question 2 ===

'''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 the following initial conditions for the reaction F + H2 rHF = 200 pm, rHH = 74 pm, pHF = -1.0 g.mol-1.pm.fs-1 and pHH = 0 g.mol-1.pm.fs-1 (considering the activation energy for this reaction is very small, not much initial momentum is required since the reaction is "very spontaneous"), the following results were obtained:

{| class="wikitable" border="1"
! Contour plot !! Momenta vs Time
|-
| [[File: Energy transfer contour plotsds2618.png|200px]] [[File: Energy transfer 3d contoursds2618.png|200px]] || [[File: Energy transfer m vs tsds2618.png|300px]]
|}

According to these graphs, we can see that once the system has passed the transition state, the newly formed H-F bond has a very large momentum (around 17 g.mol-1.pm.fs-1 ). This is because the bond between H and F is of such different nature compared to that of HH. The newly formed HF molecule is so much more stable than the reactants which is why the system gains a lot of energy once it has overcome the very small activation energy. This means that, in order to conserve energy, the system passes its excess energy into the vibration of HF. This can be seen when we observe the contour plot of this reaction: after the system passes through the transition state (in this case several times), it oscillates strongly.

=== Question 3 ===

'''Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, with a momentum pFH = -1.0 g.mol-1.pm.fs-1, and explore several values of pHH in the range -6.1 to 6.1 g.mol-1.pm.fs-1 (explore values also close to these limits). What do you observe?'''

When testing different momenta of pHH close to the limits of -6.1 and 6 g.mol-1.pm.fs-1, we observe reactive and unreactive trajectories, however, most are unreactive. Some examples of tested trajectories are shown below:

{| class="wikitable" border="1"
! Contour plot pHH = -6.0 g.mol-1.pm.fm-1 !! Contour plot pHH = -5.3 g.mol-1.pm.fm-1 !! Contour plot pHH = 5.6 g.mol-1.pm.fm-1 !! Contour plot pHH = 6.0 g.mol-1.pm.fm-1
|-
| [[File: Ex2Q3sds2618.png|200px]] || [[File: Ex2Q3-5.3sds2618.png|200px]] || [[File: Ex2Q3 5.6sds2618.png|200px]] || [[File: Ex2Q3 6sds2618.png|200px]]
|}

The energy put in to the HH vibration is much larger than the required activation energy of this reaction. Therefore, the H-H bond vibrates strongly whilst approaching the F molecule which makes them collide. However, the H atom colliding with the F atom has such a strong momentum that it effectively "bounces" back to the H atom and thereby back to the reactants. This results in unreactive trajectories. Even when the trajectory results in a reactive pathway (for example when pHH = 5.6 g.mol-1.pm.fm-1), the trajectory of the system has many twists and turns which represent the system making and breaking the H-H as well as the H-F bond very often. This proves that if too much initial energy is given to the system, it has enough energy to overcome the activation energy for the forward reaction as well as for the reverse reaction. Furthermore, as seen before, when the system forms the HF bond it gains a large amount of energy in the form of vibration. If this vibrational energy is large enough to overcome the activation energy of the reverse reaction, the system could potentially revert to the reactants.

'''For the same initial position, increase slightly the momentum pFH = -1.6 g.mol-1.pm.fs-1, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.2 g.mol-1.pm.fs-1. What do you observe now?'''

The following trajectory resulted from the decrease of the overall energy:
{| class="wikitable" border="1"
|+Parameters rHF= 200 pm, rHH= 74 pm pFH = -1.6 g.mol-1.pm.fs-1 and pHH = 0.2 g.mol-1.pm.fs-1
! Contour plot !! Momentum vs Time
|-
| [[File: Ex2_reduced_overall_E_contoursds2618.png|300px]] || [[File:Overall reduced E m vs tsds2618.png|300px]]
|}

The Contour plot shows that even reducing the overall energy significantly will not result in a reactive trajectory. In this case, the HH molecule approaches the F atom and arrives at the transition state, the H-F bond forms but the strong vibrations of H-F bond (16 g.mol-1.pm.fs-1) make the system oscillate between the formation of the H-F bond and that of the H-H bond. In this case, the oscillations result in the system going back to the reactants.

=== Question 4 ===

'''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.'''

== References ==

<references>
<ref name="Q5-1">Laidler, Keith J, ''Chemical kinetics'', 3rd edition, '''1987''', 287-321.</ref>
<ref name="Q5-2">Laidler, Keith J, ''Chemical kinetics'', 3rd edition, '''1987''', 88-98.</ref>
<ref name="Q5-3">Laidler, Keith J, ''Chemical kinetics'', 3rd edition, '''1987''', 115-123.</ref>
<ref name="Q1-1">Laidler, Keith J, ''Chemical kinetics'', 3rd edition, '''1987''', 460-471.</ref>
</references>

MRD:sds2618

2020-05-14T11:48:52Z

Sds2618:

== Exercise 1: H + H2 system ==

=== Question 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?'''

On a potential energy surface, the transition state can be defined as a saddle point of

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

At the transition state, r1 = r2. From the "Internuclear Distances vs Time" plot of the initial H + H2 system, we can already estimate a transition state position around an internuclear distance of r1 = r2 = 90 pm at approximately 20 fs (corresponding to the point at which internuclear distances A-B and A-C intersect). After testing several initial conditions, it was found that for rAB = rBC = 90.77 pm and pAB = pBC = 0 g.mol-1.pm.fm-1 the system undergoes an extremely small systematic periodic vibration, symmetric on both atoms. Furthermore, the forces at rAB = rBC = 90.77 pm are very close to 0 (0.001 kJ.mol-1.pm-1</sup).

{| class="wikitable" border="1"
! Internuclear Distances vs Time for rAB = 230 pm, pAB = -5.2 g.mol-1.pm.fm-1 and rBC = 74 pm, pBC = 0 g.mol-1.pm.fm-1 || Plot of internuclear distances vs time for rAB = rBC = 90.77 pm and pAB = pBC = 0 g.mol-1.pm.fm-1|| Forces at rAB = rBC = 90.77 pm
|-
| [[File: D vs time-1.png|300px]] || [[File: D vs t TS HHHsds2618.png|300px]] || [[File:Screenshot forcessds2618.png|300px ]]
|}

We can therefore estimate rts = 90.77 pm.

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

When calculating the minimum energy pathway using calculation type MEP and Dynamics, we obtain the following results (with rAB = 90.77 pm, rBC = 91.77 pm and pAB = pBC = 0 g.mol-1.pm.fm-1):

{| class="wikitable" border="1"
! Contour plot of the reaction path using MEP !! Contour plot of the reaction path using Dynamics
|-
| [[File: Q3-MEPsds2618.png|300px]] || [[File: Q3-dynamicssds2618.png|300px]]
|-
| '''Plot of momentum vs time of the reaction path using MEP''' || '''Plot of momentum vs time of the reaction path using Dynamics'''
|-
| [[File: TS MEPsds2618.png|300px]] || [[File: TS Dynamicssds2618.png|300px]]
|}

When comparing these two results, we can see that the "Dynamics" calculation includes oscillation of the atoms involved in the reaction. This is visible in the contour plots: in MEP, a straight curve shows the reaction pathway but in Dynamics we can see oscillations along this path. This is clearly visible in the plots of momentum vs time. Since for MEP, the momentum is reset to 0 after each step, we see no change in momentum over the whole reaction path but in Dynamics, we can see the increase in momentum and oscillation along this reaction.

{| class="wikitable" border="1"
! Internuclear distances vs time where rAB = 90.77 pm, rBC = 91.77 pm !! Internuclear distances vs time where rAB = 91.77 pm, rBC = 90.77 pm
|-
| [[File: Q3-dynamics d vs tsds2618.png|300px]] || [[File: Extra-d vs t switchsds2618.png|300px]]
|-
| '''Momenta vs time where rAB = 90.77 pm, rBC = 91.77 pm''' || '''Momenta vs time where rAB = 91.77 pm, rBC = 90.77 pm'''
|-
| [[File: Extra-momenta vs t normalsds2618.png|300px]] || [[File: Extra-momenta vs t switchsds2618.png|300px]]
|}

Changing from '''rAB = rts''' and '''rBC = rts + 1''' to '''rAB = rts + 1''' and '''rBC = rts''' effectively reverses the direction of the reaction pathway observed. The former gives plots showing how the internuclear distances and momenta vary with time for the H + H2 system when atom A approaches molecule BC. The latter depicts how internuclear distances and momenta vary with time for the same system when atom C approaches molecule AB. This is observed in the graphs essentially as a "switch" between A-B and B-C curves.

From [[Media:D vs time-1.png| this graph]] we can obtain the final geometry of the system which was found to be rAB = 73 pm with pAB = 5.3 g.mol-1.pm.fm-1 and rBC = 278 pm with pBC = 5.1 g.mol-1.pm.fm-1 at t = 50 fs.
Let us now setup a system with initial positions rAB (0) = 73 pm and rBC (0) = 278 pm and with pAB (0) = -5.3 g.mol-1.pm.fm-1 and pBC (0) = -5.1 g.mol-1.pm.fm-1. We obtain the following results:

{| class="wikitable" border="1"
! Internuclear distances vs time with rAB(t=50 fm) = 73 pm, rBC(t=50 fm) = 278 pm, pAB(t=50 fm) = 5.3 g.mol-1.pm.fm-1 and pBC(t=50 fm) = 5.1 g.mol-1.pm.fm-1!! Internuclear distances vs time for rAB (0) = 73 pm and rBC (0) = 278 pm and with pAB (0) = -5.3 g.mol-1.pm.fm-1 and pBC (0) = -5.1 g.mol-1.pm.fm-1
|-
| [[File: D vs time-1.png|300px]] || [[File: D vs t inversion final and intialsds2618.png|300px]]
|-
|}

Similar to before, swapping initial positions for final positions of the same system as well as initial and final momenta (with inversion of the sign) amounts to flipping the trend of the graph of internuclear distances vs time. The graphs show the same reactions as mentioned above: atom A approaches molecule BC to form molecule AB and atom C (left hand side) and atom C approaches molecule AB to form molecule BC and atom A (right hand side). The transition states of the two reactions are located at t ~ 20 fs and ~ 30 fs but with the same rts = 90 pm.

=== Question 4 ===
'''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?'''

{| class="wikitable" border="1"
|+ Testing trajectories with r1 = 74 pm and r2 = 200 pm
! p1 /g.mol-1.pm.fm-1 !! p2 /g.mol-1.pm.fm-1 !! Etot /kj.mol-1 !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280 || Reactive || Atom A approaches molecule BC and passes through the transition state at ~ 21 fs. The reaction gives the following products: molecule AB and atom C|| [[File: Table 1sds2618.png|300px]] [[File: Table 1 d vs tsds2618.png|300px]]
|-
| -3.1 || -4.1 || -420.077 || Unreactive || Atom A approaches molecule BC but the momentum isn't enough for the two to collide. The reaction does not pass through the transition state, instead it regenerates the reactants since no collision has taken place. || [[File: Table 2sds2618.png|300px]] [[File: Table 2 d vs tsds2618.png|300px]]
|-
| -3.1 || -5.1 || -413.977 || Reactive || Atom A approaches molecule BC and passes through the transition state at ~ 21 fs. The reaction gives the following products: molecule AB and atom C || [[File: Table 3sds2618.png|300px]] [[File: Table 3 d vs tsds2618.png|300px]]
|-
| -5.1 || -10.1 || -357.277 || Unreactive (barrier recrossing) || Atom A approaches molecule BC and a bond between A and B forms (at t~ 8 fs). However, this bond still oscillates which breaks the AB bond and reforms the BC one (at t~ 35 fs), regenerating the reactants|| [[File: Table 4sds2618.png|300px]] [[File: Table 4 d vs tsds2618.png|300px]]
|-
| -5.1 || -10.6 || -349.477 || Reactive || Atom A approaches molecule BC and the AB bond forms (at t~ 8 fs). However, the oscillations of the newly formed bond are very strong (since the molecule was given such a high kinetic energy) which causes the AB bond to break to reform molecule BC temporarily (at t~ 15 fs). The oscillations of the BC bond continue and the BC bond breaks (at t~ 25 fs) which generates the products AB and single atom C|| [[File: Table 5sds2618.png|300px]] [[File: Table 5 d vs tsds2618.png|300px]]
|}

As concluded before, trajectories with -3.1 < p1 /g.mol-1.pm.fm-1 < -1.6 and p2 = -5.1 g.mol-1.pm.fm-1 are reactive: this is the case for the first and third reaction from the table.
However, if we increase the momenta outside the previously defined range, the initial positions need to be adjusted accordingly or the reaction will be unreactive. The system will either not have enough kinetic energy (too small momentum) to overcome the activation barrier thus never crossing the transition state (second reaction from the table) or the momentum will be too high and the system will cross the transition state several times, breaking and remaking the bonds between A, B and C (for the two final reactions of the table), this can result in reactive or unreactive trajectories depending on the momenta.

=== Question 5 ===
'''Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''

In order to apply the Transition State Theory, several assumptions are made one of which is that the system can only pass through the transition state once and can't return to the reactants. <ref name="Q5-1" />
As we have seen from the different trajectories of the table above, this is not always the case. If the initial momenta of the different components of the system are too large, the system can cross the transition state several times or not at all if the initial momenta are too small. This is the first limitation in predicting reaction rates using Transition State Theory.

Furthermore, the assumption is made that motion along the reaction coordinate can be treated independently and classically as translational motion.<ref name="Q5-2" /> This will cause another discrepancy between theory and experimental results since in reality, quantum effects will contribute to the motion of the system and in some cases may even cause quantum tunneling, which enables the reactants to form the products without having to pass the energy barrier to go via the transition state (even though this effect is not very likely for this system, it is possible for certain systems). However, the program used treats motion classically so there won't be a deviation between theory and experimental due to this assumption.

There are also assumptions made when calculating the partition functions (which include translational, rotational, vibrational and electronic contributions). <ref name="Q5-2" /> The equations used to determine translational and rotational contribution assume high temperatures (ie when the spacing between energy levels are much smaller than kT) and the vibrational contribution is determined using the basis of a simple harmonic oscillator, which is not always for high temperatures. <ref name="Q5-3" />

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

=== Question 1 ===
'''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? Locate the approximate position of the transition state. Report the activation energy for both reactions.'''

By evaluation of the potential energy surfaces of F + H2 and H + HF <ref name="Q1-1" /> it was found that the first reaction is exothermic and the second endothermic. This means that the F-H bond is stronger than the H-H bond since the products of the first reaction (H + HF) are more stable than the reactants (F + H2).

According to Hammond's postulate the structure of the transition state will resemble the structure of the nearest stable species (either products or reactants). Taking the first reaction F + H2 = H + HF, the transition state will resemble the reactants since it is an exothermic reaction. This is the opposite for the second reaction: the transition state will resemble the products for the endothermic reaction. In both cases, the structure of the transition state will be closer to that of "F + H2".

We can therefore assume that at the transition state the F-H distance will be larger than the H-H distance. After testing several initial distances, it was found that the transition state has the following internuclear distances: rFH = 182 pm and rHH = 74 pm. When setting the initial conditions to these parameters (with pFH = pHH = 0 g.mol-1.pm.fm-1), the Animation showed a small periodic oscillation of the atoms and the plot of Internuclear Distances vs Time was the following:
[[File: TS HF+Hsds2618.png|300px|thumb|center|Internuclear Distances vs Time of the transition state of HF + H.]]

We can see that the oscillations of all internuclear distances are "parallel" to each other which shows that no collision will take place with 0 momentum at these specific distances, which is the definition of the transition state.

To find the activation energies of both reactions, we will run the program (using MEP) starting from the a structure close to the transition state and drive the reaction towards HF + H to find Ea for H + HF = F + H2 and to find Ea for F + H2 = H + HF, we will drive the reaction towards F + H2. Considering the previously determined internuclear distances of the transition state determined previously, it was found that the following parameters drive the transition state towards FH + H: rFH = 170 pm, rHH = 74 pm and pFH = pHH = 0 g.mol-1.pm.fm-1. In order to drive the transition state towards F + H2 we will need to prolong the F-H distance and reduce the H-H distance slightly. After several tries, the following parameters were used to find the activation energy: rFH = 190 pm, rHH = 70 pm and pFH = pHH = 0 g.mol-1.pm.fm-1. Using these two initial conditions, we will plot a graph of Energy vs Time and determine the activation energies of both reactions by taking the difference of the maximum of the curve (energy of the transition state) and the minimum (energy of the products).

The following results were obtained:

{| class="wikitable" border="1"
! Energy vs Time (rFH = 150 pm, rHH = 74 pm, pFH = pHH = 0 g.mol-1.pm.fm-1) !! energy vs Time ( rFH = 190 pm, rHH = 70 pm and pFH = pHH = 0 g.mol-1.pm.fm-1)
|-
| [[File: Activation energy 1sds2618.png|300px]] || [[File: Activation energy 2sds2618.png|300px]]
|-
|}

According to these graphs, the activation for the reaction H + HF = H2 + F is approximately 127 kJ.mol-1 and that of H2 = HF + H is around 0.9 kJ.mol-1. The very small activation energy for the second reaction is due to the strong driving force to form the H-F bond since this is a very stable compound. The same can be used to explain the relatively large activation energy for the inverse reaction: the HF compound is much more stable and has a very strong bon which means a lot of energy is required to break that bond in order to form a weaker, less stable one between two H atoms.

=== Question 2 ===

'''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 the following initial conditions for the reaction F + H2 rHF = 200 pm, rHH = 74 pm, pHF = -1.0 g.mol-1.pm.fs-1 and pHH = 0 g.mol-1.pm.fs-1 (considering the activation energy for this reaction is very small, not much initial momentum is required since the reaction is "very spontaneous"), the following results were obtained:

{| class="wikitable" border="1"
! Contour plot !! Momenta vs Time
|-
| [[File: Energy transfer contour plotsds2618.png|200px]] [[File: Energy transfer 3d contoursds2618.png|200px]] || [[File: Energy transfer m vs tsds2618.png|300px]]
|}

According to these graphs, we can see that once the system has passed the transition state, the newly formed H-F bond has a very large momentum (around 17 g.mol-1.pm.fs-1 ). This is because the bond between H and F is of such different nature compared to that of HH. The newly formed HF molecule is so much more stable than the reactants which is why the system gains a lot of energy once it has overcome the very small activation energy. This means that, in order to conserve energy, the system passes its excess energy into the vibration of HF. This can be seen when we observe the contour plot of this reaction: after the system passes through the transition state (in this case several times), it oscillates strongly.

=== Question 3 ===

'''Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, with a momentum pFH = -1.0 g.mol-1.pm.fs-1, and explore several values of pHH in the range -6.1 to 6.1 g.mol-1.pm.fs-1 (explore values also close to these limits). What do you observe?'''

When testing different momenta of pHH close to the limits of -6.1 and 6 g.mol-1.pm.fs-1, we observe reactive and unreactive trajectories, however, most are unreactive. Some examples of tested trajectories are shown below:

{| class="wikitable" border="1"
! Contour plot pHH = -6.0 g.mol-1.pm.fm-1 !! Contour plot pHH = -5.3 g.mol-1.pm.fm-1 !! Contour plot pHH = 5.6 g.mol-1.pm.fm-1 !! Contour plot pHH = 6.0 g.mol-1.pm.fm-1
|-
| [[File: Ex2Q3sds2618.png|200px]] || [[File: Ex2Q3-5.3sds2618.png|200px]] || [[File: Ex2Q3 5.6sds2618.png|200px]] || [[File: Ex2Q3 6sds2618.png|200px]]
|}

The energy put in to the HH vibration is much larger than the required activation energy of this reaction. Therefore, the H-H bond vibrates strongly whilst approaching the F molecule which makes them collide. However, the H atom colliding with the F atom has such a strong momentum that it effectively "bounces" back to the H atom and thereby back to the reactants. This results in unreactive trajectories. Even when the trajectory results in a reactive pathway (for example when pHH = 5.6 g.mol-1.pm.fm-1), the trajectory of the system has many twists and turns which represent the system making and breaking the H-H as well as the H-F bond very often. This proves that if too much initial energy is given to the system, it has enough energy to overcome the activation energy for the forward reaction as well as for the reverse reaction. Furthermore, as seen before, when the system forms the HF bond it gains a large amount of energy in the form of vibration. If this vibrational energy is large enough to overcome the activation energy of the reverse reaction, the system could potentially revert to the reactants.

'''For the same initial position, increase slightly the momentum pFH = -1.6 g.mol-1.pm.fs-1, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.2 g.mol-1.pm.fs-1. What do you observe now?'''

The following trajectory resulted from the decrease of the overall energy:
{| class="wikitable" border="1"
|+Parameters rHF= 200 pm, rHH= 74 pm pFH = -1.6 g.mol-1.pm.fs-1 and pHH = 0.2 g.mol-1.pm.fs-1
! Contour plot !! Momentum vs Time
|-
| [[File: Ex2_reduced_overall_E_contoursds2618.png|300px]] || [[File:Overall reduced E m vs tsds2618.png|300px]]
|}

The Contour plot shows that even reducing the overall energy significantly will not result in a reactive trajectory. In this case, the HH molecule approaches the F atom and arrives at the transition state, the H-F bond forms but the strong vibrations of H-F bond (16 g.mol-1.pm.fs-1) make the system oscillate between the formation of the H-F bond and that of the H-H bond. In this case, the oscillations result in the system going back to the reactants.

=== Question 4 ===

'''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.'''

== References ==

<references>
<ref name="Q5-1">Laidler, Keith J, ''Chemical kinetics'', 3rd edition, '''1987''', 287-321.</ref>
<ref name="Q5-2">Laidler, Keith J, ''Chemical kinetics'', 3rd edition, '''1987''', 88-98.</ref>
<ref name="Q5-3">Laidler, Keith J, ''Chemical kinetics'', 3rd edition, '''1987''', 115-123.</ref>
<ref name="Q1-1">Laidler, Keith J, ''Chemical kinetics'', 3rd edition, '''1987''', 460-471.</ref>
</references>

MRD:sds2618

2020-05-14T11:22:00Z

Sds2618:

== Exercise 1: H + H2 system ==

=== Question 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?'''

On a potential energy surface, the transition state can be defined as a saddle point of

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

At the transition state, r1 = r2. From the "Internuclear Distances vs Time" plot of the initial H + H2 system, we can already estimate a transition state position around an internuclear distance of r1 = r2 = 90 pm at approximately 20 fs (corresponding to the point at which internuclear distances A-B and A-C intersect). After testing several initial conditions, it was found that for rAB = rBC = 90.77 pm and pAB = pBC = 0 g.mol-1.pm.fm-1 the system undergoes an extremely small systematic periodic vibration, symmetric on both atoms. Furthermore, the forces at rAB = rBC = 90.77 pm are very close to 0 (0.001 kJ.mol-1.pm-1</sup).

{| class="wikitable" border="1"
! Internuclear Distances vs Time for rAB = 230 pm, pAB = -5.2 g.mol-1.pm.fm-1 and rBC = 74 pm, pBC = 0 g.mol-1.pm.fm-1 || Plot of internuclear distances vs time for rAB = rBC = 90.77 pm and pAB = pBC = 0 g.mol-1.pm.fm-1|| Forces at rAB = rBC = 90.77 pm
|-
| [[File: D vs time-1.png|300px]] || [[File: D vs t TS HHHsds2618.png|300px]] || [[File:Screenshot forcessds2618.png|300px ]]
|}

We can therefore estimate rts = 90.77 pm.

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

When calculating the minimum energy pathway using calculation type MEP and Dynamics, we obtain the following results (with rAB = 90.77 pm, rBC = 91.77 pm and pAB = pBC = 0 g.mol-1.pm.fm-1):

{| class="wikitable" border="1"
! Contour plot of the reaction path using MEP !! Contour plot of the reaction path using Dynamics
|-
| [[File: Q3-MEPsds2618.png|300px]] || [[File: Q3-dynamicssds2618.png|300px]]
|-
| '''Plot of momentum vs time of the reaction path using MEP''' || '''Plot of momentum vs time of the reaction path using Dynamics'''
|-
| [[File: TS MEPsds2618.png|300px]] || [[File: TS Dynamicssds2618.png|300px]]
|}

When comparing these two results, we can see that the "Dynamics" calculation includes oscillation of the atoms involved in the reaction. This is visible in the contour plots: in MEP, a straight curve shows the reaction pathway but in Dynamics we can see oscillations along this path. This is clearly visible in the plots of momentum vs time. Since for MEP, the momentum is reset to 0 after each step, we see no change in momentum over the whole reaction path but in Dynamics, we can see the increase in momentum and oscillation along this reaction.

{| class="wikitable" border="1"
! Internuclear distances vs time where rAB = 90.77 pm, rBC = 91.77 pm !! Internuclear distances vs time where rAB = 91.77 pm, rBC = 90.77 pm
|-
| [[File: Q3-dynamics d vs tsds2618.png|300px]] || [[File: Extra-d vs t switchsds2618.png|300px]]
|-
| '''Momenta vs time where rAB = 90.77 pm, rBC = 91.77 pm''' || '''Momenta vs time where rAB = 91.77 pm, rBC = 90.77 pm'''
|-
| [[File: Extra-momenta vs t normalsds2618.png|300px]] || [[File: Extra-momenta vs t switchsds2618.png|300px]]
|}

Changing from '''rAB = rts''' and '''rBC = rts + 1''' to '''rAB = rts + 1''' and '''rBC = rts''' effectively reverses the direction of the reaction pathway observed. The former gives plots showing how the internuclear distances and momenta vary with time for the H + H2 system when atom A approaches molecule BC. The latter depicts how internuclear distances and momenta vary with time for the same system when atom C approaches molecule AB. This is observed in the graphs essentially as a "switch" between A-B and B-C curves.

From [[Media:D vs time-1.png| this graph]] we can obtain the final geometry of the system which was found to be rAB = 73 pm with pAB = 5.3 g.mol-1.pm.fm-1 and rBC = 278 pm with pBC = 5.1 g.mol-1.pm.fm-1 at t = 50 fs.
Let us now setup a system with initial positions rAB (0) = 73 pm and rBC (0) = 278 pm and with pAB (0) = -5.3 g.mol-1.pm.fm-1 and pBC (0) = -5.1 g.mol-1.pm.fm-1. We obtain the following results:

{| class="wikitable" border="1"
! Internuclear distances vs time with rAB(t=50 fm) = 73 pm, rBC(t=50 fm) = 278 pm, pAB(t=50 fm) = 5.3 g.mol-1.pm.fm-1 and pBC(t=50 fm) = 5.1 g.mol-1.pm.fm-1!! Internuclear distances vs time for rAB (0) = 73 pm and rBC (0) = 278 pm and with pAB (0) = -5.3 g.mol-1.pm.fm-1 and pBC (0) = -5.1 g.mol-1.pm.fm-1
|-
| [[File: D vs time-1.png|300px]] || [[File: D vs t inversion final and intialsds2618.png|300px]]
|-
|}

Similar to before, swapping initial positions for final positions of the same system as well as initial and final momenta (with inversion of the sign) amounts to flipping the trend of the graph of internuclear distances vs time. The graphs show the same reactions as mentioned above: atom A approaches molecule BC to form molecule AB and atom C (left hand side) and atom C approaches molecule AB to form molecule BC and atom A (right hand side). The transition states of the two reactions are located at t ~ 20 fs and ~ 30 fs but with the same rts = 90 pm.

=== Question 4 ===
'''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?'''

{| class="wikitable" border="1"
|+ Testing trajectories with r1 = 74 pm and r2 = 200 pm
! p1 /g.mol-1.pm.fm-1 !! p2 /g.mol-1.pm.fm-1 !! Etot /kj.mol-1 !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280 || Reactive || Atom A approaches molecule BC and passes through the transition state at ~ 21 fs. The reaction gives the following products: molecule AB and atom C|| [[File: Table 1sds2618.png|300px]] [[File: Table 1 d vs tsds2618.png|300px]]
|-
| -3.1 || -4.1 || -420.077 || Unreactive || Atom A approaches molecule BC but the momentum isn't enough for the two to collide. The reaction does not pass through the transition state, instead it regenerates the reactants since no collision has taken place. || [[File: Table 2sds2618.png|300px]] [[File: Table 2 d vs tsds2618.png|300px]]
|-
| -3.1 || -5.1 || -413.977 || Reactive || Atom A approaches molecule BC and passes through the transition state at ~ 21 fs. The reaction gives the following products: molecule AB and atom C || [[File: Table 3sds2618.png|300px]] [[File: Table 3 d vs tsds2618.png|300px]]
|-
| -5.1 || -10.1 || -357.277 || Unreactive (barrier recrossing) || Atom A approaches molecule BC and a bond between A and B forms (at t~ 8 fs). However, this bond still oscillates which breaks the AB bond and reforms the BC one (at t~ 35 fs), regenerating the reactants|| [[File: Table 4sds2618.png|300px]] [[File: Table 4 d vs tsds2618.png|300px]]
|-
| -5.1 || -10.6 || -349.477 || Reactive || Atom A approaches molecule BC and the AB bond forms (at t~ 8 fs). However, the oscillations of the newly formed bond are very strong (since the molecule was given such a high kinetic energy) which causes the AB bond to break to reform molecule BC temporarily (at t~ 15 fs). The oscillations of the BC bond continue and the BC bond breaks (at t~ 25 fs) which generates the products AB and single atom C|| [[File: Table 5sds2618.png|300px]] [[File: Table 5 d vs tsds2618.png|300px]]
|}

As concluded before, trajectories with -3.1 < p1 /g.mol-1.pm.fm-1 < -1.6 and p2 = -5.1 g.mol-1.pm.fm-1 are reactive: this is the case for the first and third reaction from the table.
However, if we increase the momenta outside the previously defined range, the initial positions need to be adjusted accordingly or the reaction will be unreactive. The system will either not have enough kinetic energy (too small momentum) to overcome the activation barrier thus never crossing the transition state (second reaction from the table) or the momentum will be too high and the system will oscillate several times, breaking and remaking the bonds between A, B and C (for the two final reactions of the table), this can result in reactive or unreactive trajectories depending on the momenta.

=== Question 5 ===
'''Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''

In order to apply the Transition State Theory, several assumptions are made one of which is that the system can only pass through the transition state once and can't return to the reactants. <ref name="Q5-1" />
As we have seen from the different trajectories of the table above, this is not always the case. If the initial momenta of the different components of the system are too large, the system can cross the transition state several times or not at all if the initial momenta are too small. This is the first limitation in predicting reaction rates using Transition State Theory.

Furthermore, the assumption is made that motion along the reaction coordinate can be treated independently and classically as translational motion.<ref name="Q5-2" /> This will cause another discrepancy between theory and experimental results since in reality, quantum effects will contribute to the motion of the system and in some cases may even cause quantum tunneling, which enables the reactants to form the products without having to pass the energy barrier to go via the transition state (even though this effect is not very likely for this system, it is possible for certain systems). However, the program used treats motion classically so there won't be a deviation between theory and experimental due to this assumption.

There are also assumptions made when calculating the partition functions (which include translational, rotational, vibrational and electronic contributions). <ref name="Q5-2" /> The equations used to determine translational and rotational contribution assume high temperatures (ie when the spacing between energy levels are much smaller than kT) and the vibrational contribution is determined using the basis of a simple harmonic oscillator, which is not always for high temperatures. <ref name="Q5-3" />

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

=== Question 1 ===
'''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? Locate the approximate position of the transition state. Report the activation energy for both reactions.'''

By evaluation of the potential energy surfaces of F + H2 and H + HF <ref name="Q1-1" /> it was found that the first reaction is exothermic and the second endothermic. This means that the F-H bond is stronger than the H-H bond since the products of the first reaction (H + HF) are more stable than the reactants (F + H2).

According to Hammond's postulate the structure of the transition state will resemble the structure of the nearest stable species (either products or reactants). Taking the first reaction F + H2 = H + HF, the transition state will resemble the reactants since it is an exothermic reaction. This is the opposite for the second reaction: the transition state will resemble the products for the endothermic reaction. In both cases, the structure of the transition state will be closer to that of "F + H2".

We can therefore assume that at the transition state the F-H distance will be larger than the H-H distance. After testing several initial distances, it was found that the transition state has the following internuclear distances: rFH = 182 pm and rHH = 74 pm. When setting the initial conditions to these parameters (with pFH = pHH = 0 g.mol-1.pm.fm-1), the Animation showed a small periodic oscillation of the atoms and the plot of Internuclear Distances vs Time was the following:
[[File: TS HF+Hsds2618.png|300px|thumb|center|Internuclear Distances vs Time of the transition state of HF + H.]]

We can see that the oscillations of all internuclear distances are "parallel" to each other which shows that no collision will take place with 0 momentum at these specific distances, which is the definition of the transition state.

To find the activation energies of both reactions, we will run the program (using MEP) starting from the a structure close to the transition state and drive the reaction towards HF + H to find Ea for H + HF = F + H2 and to find Ea for F + H2 = H + HF, we will drive the reaction towards F + H2. Considering the previously determined internuclear distances of the transition state determined previously, it was found that the following parameters drive the transition state towards FH + H: rFH = 170 pm, rHH = 74 pm and pFH = pHH = 0 g.mol-1.pm.fm-1. In order to drive the transition state towards F + H2 we will need to prolong the F-H distance and reduce the H-H distance slightly. After several tries, the following parameters were used to find the activation energy: rFH = 190 pm, rHH = 70 pm and pFH = pHH = 0 g.mol-1.pm.fm-1. Using these two initial conditions, we will plot a graph of Energy vs Time and determine the activation energies of both reactions by taking the difference of the maximum of the curve (energy of the transition state) and the minimum (energy of the products).

The following results were obtained:

{| class="wikitable" border="1"
! Energy vs Time (rFH = 150 pm, rHH = 74 pm, pFH = pHH = 0 g.mol-1.pm.fm-1) !! energy vs Time ( rFH = 190 pm, rHH = 70 pm and pFH = pHH = 0 g.mol-1.pm.fm-1)
|-
| [[File: Activation energy 1sds2618.png|300px]] || [[File: Activation energy 2sds2618.png|300px]]
|-
|}

According to these graphs, the activation for the reaction H + HF = H2 + F is approximately 127 kJ.mol-1 and that of H2 = HF + H is around 0.9 kJ.mol-1. The very small activation energy for the second reaction is due to the strong driving force to form the H-F bond since this is a very stable compound. The same can be used to explain the relatively large activation energy for the inverse reaction: the HF compound is much more stable and has a very strong bon which means a lot of energy is required to break that bond in order to form a weaker, less stable one between two H atoms.

=== Question 2 ===

'''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 the following initial conditions for the reaction F + H2 rHF = 200 pm, rHH = 74 pm, pHF = -1.0 g.mol-1.pm.fs-1 and pHH = 0 g.mol-1.pm.fs-1 (considering the activation energy for this reaction is very small, not much initial momentum is required since the reaction is "very spontaneous"), the following results were obtained:

{| class="wikitable" border="1"
! Contour plot !! Momenta vs Time
|-
| [[File: Energy transfer contour plotsds2618.png|200px]] [[File: Energy transfer 3d contoursds2618.png|200px]] || [[File: Energy transfer m vs tsds2618.png|300px]]
|}

According to these graphs, we can see that once the system has passed the transition state, the newly formed H-F bond has a very large momentum (around 17 g.mol-1.pm.fs-1 ). This is because the bond between H and F is of such different nature compared to that of HH. The newly formed HF molecule is so much more stable than the reactants which is why the system gains a lot of energy once it has overcome the very small activation energy. This means that, in order to conserve energy, the system passes its excess energy into the vibration of HF. This can be seen when we observe the contour plot of this reaction: after the system passes through the transition state (in this case several times), it oscillates strongly.

=== Question 3 ===

'''Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, with a momentum pFH = -1.0 g.mol-1.pm.fs-1, and explore several values of pHH in the range -6.1 to 6.1 g.mol-1.pm.fs-1 (explore values also close to these limits). What do you observe?'''

When testing different momenta of pHH close to the limits of -6.1 and 6 g.mol-1.pm.fs-1, we observe reactive and unreactive trajectories, however, most are unreactive. Some examples of tested trajectories are shown below:

{| class="wikitable" border="1"
! Contour plot pHH = -6.0 g.mol-1.pm.fm-1 !! Contour plot pHH = -5.3 g.mol-1.pm.fm-1 !! Contour plot pHH = 5.6 g.mol-1.pm.fm-1 !! Contour plot pHH = 6.0 g.mol-1.pm.fm-1
|-
| [[File: Ex2Q3sds2618.png|200px]] || [[File: Ex2Q3-5.3sds2618.png|200px]] || [[File: Ex2Q3 5.6sds2618.png|200px]] || [[File: Ex2Q3 6sds2618.png|200px]]
|}

The energy put in to the HH vibration is much larger than the required activation energy of this reaction. Therefore, the H-H bond vibrates strongly whilst approaching the F molecule which makes them collide. However, the H atom colliding with the F atom has such a strong momentum that it effectively "bounces" back to the H atom and thereby back to the reactants. This results in unreactive trajectories. Even when the trajectory results in a reactive pathway (for example when pHH = 5.6 g.mol-1.pm.fm-1), the trajectory of the system has many twists and turns which represent the system making and breaking the H-H as well as the H-F bond very often. This proves that if too much initial energy is given to the system, it has enough energy to overcome the activation energy for the forward reaction as well as for the reverse reaction. Furthermore, as seen before, when the system forms the HF bond it gains a large amount of energy in the form of vibration. If this vibrational energy is large enough to overcome the activation energy of the reverse reaction, the system could potentially revert to the reactants.

'''For the same initial position, increase slightly the momentum pFH = -1.6 g.mol-1.pm.fs-1, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.2 g.mol-1.pm.fs-1. What do you observe now?'''

The following trajectory resulted from the decrease of the overall energy:
{| class="wikitable" border="1"
|+Parameters rHF= 200 pm, rHH= 74 pm pFH = -1.6 g.mol-1.pm.fs-1 and pHH = 0.2 g.mol-1.pm.fs-1
! Contour plot !! Momentum vs Time
|-
| [[File: Ex2_reduced_overall_E_contoursds2618.png|300px]] || [[File:Overall reduced E m vs tsds2618.png|300px]]
|}

The Contour plot shows that even reducing the overall energy significantly will not result in a reactive trajectory. In this case, the HH molecule approaches the F atom and arrives at the transition state, the H-F bond forms but the strong vibrations of H-F bond (16 g.mol-1.pm.fs-1) make the system oscillate between the formation of the H-F bond and that of the H-H bond. In this case, the oscillations result in the system going back to the reactants.

=== Question 4 ===

'''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.'''

== References ==

<references>
<ref name="Q5-1">Laidler, Keith J, ''Chemical kinetics'', 3rd edition, '''1987''', 287-321.</ref>
<ref name="Q5-2">Laidler, Keith J, ''Chemical kinetics'', 3rd edition, '''1987''', 88-98.</ref>
<ref name="Q5-3">Laidler, Keith J, ''Chemical kinetics'', 3rd edition, '''1987''', 115-123.</ref>
<ref name="Q1-1">Laidler, Keith J, ''Chemical kinetics'', 3rd edition, '''1987''', 460-471.</ref>
</references>

MRD:sds2618

2020-05-14T11:16:32Z

Sds2618:

== Exercise 1: H + H2 system ==

=== Question 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?'''

On a potential energy surface, the transition state can be defined as a saddle point of

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

At the transition state, r1 = r2. From the "Internuclear Distances vs Time" plot of the initial H + H2 system, we can already estimate a transition state position around an internuclear distance of r1 = r2 = 90 pm at approximately 20 fs (corresponding to the point at which internuclear distances A-B and A-C intersect). After testing several initial conditions, it was found that for rAB = rBC = 90.77 pm and pAB = pBC = 0 g.mol-1.pm.fm-1 the system undergoes an extremely small systematic periodic vibration, symmetric on both atoms. Furthermore, the forces at rAB = rBC = 90.77 pm are very close to 0 (0.001 kJ.mol-1.pm-1</sup).

{| class="wikitable" border="1"
! Internuclear Distances vs Time for rAB = 230 pm, pAB = -5.2 g.mol-1.pm.fm-1 and rBC = 74 pm, pBC = 0 g.mol-1.pm.fm-1 || Plot of internuclear distances vs time for rAB = rBC = 90.77 pm and pAB = pBC = 0 g.mol-1.pm.fm-1|| Forces at rAB = rBC = 90.77 pm
|-
| [[File: D vs time-1.png|300px]] || [[File: D vs t TS HHHsds2618.png|300px]] || [[File:Screenshot forcessds2618.png|300px ]]
|}

We can therefore estimate rts = 90.77 pm.

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

When calculating the minimum energy pathway using calculation type MEP and Dynamics, we obtain the following results (with rAB = 90.77 pm, rBC = 91.77 pm and pAB = pBC = 0 g.mol-1.pm.fm-1):

{| class="wikitable" border="1"
! Contour plot of the reaction path using MEP !! Contour plot of the reaction path using Dynamics
|-
| [[File: Q3-MEPsds2618.png|300px]] || [[File: Q3-dynamicssds2618.png|300px]]
|-
| '''Plot of momentum vs time of the reaction path using MEP''' || '''Plot of momentum vs time of the reaction path using Dynamics'''
|-
| [[File: TS MEPsds2618.png|300px]] || [[File: TS Dynamicssds2618.png|300px]]
|}

When comparing these two results, we can see that the "Dynamics" calculation includes oscillation of the atoms involved in the reaction. This is very clearly visible in the plots of internuclear distances vs time: the distance between atoms varies as a smooth curve when using the MEP calculation as opposed to oscillating distances calculated with Dynamics (best visible for A-B but also present in A-C and B-C).

{| class="wikitable" border="1"
! Internuclear distances vs time where rAB = 90.77 pm, rBC = 91.77 pm !! Internuclear distances vs time where rAB = 91.77 pm, rBC = 90.77 pm
|-
| [[File: Q3-dynamics d vs tsds2618.png|300px]] || [[File: Extra-d vs t switchsds2618.png|300px]]
|-
| '''Momenta vs time where rAB = 90.77 pm, rBC = 91.77 pm''' || '''Momenta vs time where rAB = 91.77 pm, rBC = 90.77 pm'''
|-
| [[File: Extra-momenta vs t normalsds2618.png|300px]] || [[File: Extra-momenta vs t switchsds2618.png|300px]]
|}

Changing from '''rAB = rts''' and '''rBC = rts + 1''' to '''rAB = rts + 1''' and '''rBC = rts''' effectively reverses the direction of the reaction pathway observed. The former gives plots showing how the internuclear distances and momenta vary with time for the H + H2 system when atom A approaches molecule BC. The latter depicts how internuclear distances and momenta vary with time for the same system when atom C approaches molecule AB. This is observed in the graphs essentially as a "switch" between A-B and B-C curves.

From [[Media:D vs time-1.png| this graph]] we can obtain the final geometry of the system which was found to be rAB = 73 pm with pAB = 5.3 g.mol-1.pm.fm-1 and rBC = 278 pm with pBC = 5.1 g.mol-1.pm.fm-1 at t = 50 fs.
Let us now setup a system with initial positions rAB (0) = 73 pm and rBC (0) = 278 pm and with pAB (0) = -5.3 g.mol-1.pm.fm-1 and pBC (0) = -5.1 g.mol-1.pm.fm-1. We obtain the following results:

{| class="wikitable" border="1"
! Internuclear distances vs time with rAB(t=50 fm) = 73 pm, rBC(t=50 fm) = 278 pm, pAB(t=50 fm) = 5.3 g.mol-1.pm.fm-1 and pBC(t=50 fm) = 5.1 g.mol-1.pm.fm-1!! Internuclear distances vs time for rAB (0) = 73 pm and rBC (0) = 278 pm and with pAB (0) = -5.3 g.mol-1.pm.fm-1 and pBC (0) = -5.1 g.mol-1.pm.fm-1
|-
| [[File: D vs time-1.png|300px]] || [[File: D vs t inversion final and intialsds2618.png|300px]]
|-
|}

Similar to before, swapping initial positions for final positions of the same system as well as initial and final momenta (with inversion of the sign) amounts to flipping the trend of the graph of internuclear distances vs time. The graphs show the same reactions as mentioned above: atom A approaches molecule BC to form molecule AB and atom C (left hand side) and atom C approaches molecule AB to form molecule BC and atom A (right hand side). The transition states of the two reactions are located at t ~ 20 fs and ~ 30 fs but with the same rts = 90 pm.

=== Question 4 ===
'''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?'''

{| class="wikitable" border="1"
|+ Testing trajectories with r1 = 74 pm and r2 = 200 pm
! p1 /g.mol-1.pm.fm-1 !! p2 /g.mol-1.pm.fm-1 !! Etot /kj.mol-1 !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280 || Reactive || Atom A approaches molecule BC and passes through the transition state at ~ 21 fs. The reaction gives the following products: molecule AB and atom C|| [[File: Table 1sds2618.png|300px]] [[File: Table 1 d vs tsds2618.png|300px]]
|-
| -3.1 || -4.1 || -420.077 || Unreactive || Atom A approaches molecule BC but the momentum isn't enough for the two to collide. The reaction does not pass through the transition state, instead it regenerates the reactants since no collision has taken place. || [[File: Table 2sds2618.png|300px]] [[File: Table 2 d vs tsds2618.png|300px]]
|-
| -3.1 || -5.1 || -413.977 || Reactive || Atom A approaches molecule BC and passes through the transition state at ~ 21 fs. The reaction gives the following products: molecule AB and atom C || [[File: Table 3sds2618.png|300px]] [[File: Table 3 d vs tsds2618.png|300px]]
|-
| -5.1 || -10.1 || -357.277 || Unreactive (barrier recrossing) || Atom A approaches molecule BC and a bond between A and B forms (at t~ 8 fs). However, this bond still oscillates which breaks the AB bond and reforms the BC one (at t~ 35 fs), regenerating the reactants|| [[File: Table 4sds2618.png|300px]] [[File: Table 4 d vs tsds2618.png|300px]]
|-
| -5.1 || -10.6 || -349.477 || Reactive || Atom A approaches molecule BC and the AB bond forms (at t~ 8 fs). However, the oscillations of the newly formed bond are very strong (since the molecule was given such a high kinetic energy) which causes the AB bond to break to reform molecule BC temporarily (at t~ 15 fs). The oscillations of the BC bond continue and the BC bond breaks (at t~ 25 fs) which generates the products AB and single atom C|| [[File: Table 5sds2618.png|300px]] [[File: Table 5 d vs tsds2618.png|300px]]
|}

As concluded before, trajectories with -3.1 < p1 /g.mol-1.pm.fm-1 < -1.6 and p2 = -5.1 g.mol-1.pm.fm-1 are reactive: this is the case for the first and third reaction from the table.
However, if we increase the momenta outside the previously defined range, the initial positions need to be adjusted accordingly or the reaction will be unreactive. The system will either not have enough kinetic energy (too small momentum) to overcome the activation barrier thus never crossing the transition state (second reaction from the table) or the momentum will be too high and the system will oscillate several times, breaking and remaking the bonds between A, B and C (for the two final reactions of the table), this can result in reactive or unreactive trajectories depending on the momenta.

=== Question 5 ===
'''Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''

In order to apply the Transition State Theory, several assumptions are made one of which is that the system can only pass through the transition state once and can't return to the reactants. <ref name="Q5-1" />
As we have seen from the different trajectories of the table above, this is not always the case. If the initial momenta of the different components of the system are too large, the system can cross the transition state several times or not at all if the initial momenta are too small. This is the first limitation in predicting reaction rates using Transition State Theory.

Furthermore, the assumption is made that motion along the reaction coordinate can be treated independently and classically as translational motion.<ref name="Q5-2" /> This will cause another discrepancy between theory and experimental results since in reality, quantum effects will contribute to the motion of the system and in some cases may even cause quantum tunneling, which enables the reactants to form the products without having to pass the energy barrier to go via the transition state (even though this effect is not very likely for this system, it is possible for certain systems). However, the program used treats motion classically so there won't be a deviation between theory and experimental due to this assumption.

There are also assumptions made when calculating the partition functions (which include translational, rotational, vibrational and electronic contributions). <ref name="Q5-2" /> The equations used to determine translational and rotational contribution assume high temperatures (ie when the spacing between energy levels are much smaller than kT) and the vibrational contribution is determined using the basis of a simple harmonic oscillator, which is not always for high temperatures. <ref name="Q5-3" />

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

=== Question 1 ===
'''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? Locate the approximate position of the transition state. Report the activation energy for both reactions.'''

By evaluation of the potential energy surfaces of F + H2 and H + HF <ref name="Q1-1" /> it was found that the first reaction is exothermic and the second endothermic. This means that the F-H bond is stronger than the H-H bond since the products of the first reaction (H + HF) are more stable than the reactants (F + H2).

According to Hammond's postulate the structure of the transition state will resemble the structure of the nearest stable species (either products or reactants). Taking the first reaction F + H2 = H + HF, the transition state will resemble the reactants since it is an exothermic reaction. This is the opposite for the second reaction: the transition state will resemble the products for the endothermic reaction. In both cases, the structure of the transition state will be closer to that of "F + H2".

We can therefore assume that at the transition state the F-H distance will be larger than the H-H distance. After testing several initial distances, it was found that the transition state has the following internuclear distances: rFH = 182 pm and rHH = 74 pm. When setting the initial conditions to these parameters (with pFH = pHH = 0 g.mol-1.pm.fm-1), the Animation showed a small periodic oscillation of the atoms and the plot of Internuclear Distances vs Time was the following:
[[File: TS HF+Hsds2618.png|300px|thumb|center|Internuclear Distances vs Time of the transition state of HF + H.]]

We can see that the oscillations of all internuclear distances are "parallel" to each other which shows that no collision will take place with 0 momentum at these specific distances, which is the definition of the transition state.

To find the activation energies of both reactions, we will run the program (using MEP) starting from the a structure close to the transition state and drive the reaction towards HF + H to find Ea for H + HF = F + H2 and to find Ea for F + H2 = H + HF, we will drive the reaction towards F + H2. Considering the previously determined internuclear distances of the transition state determined previously, it was found that the following parameters drive the transition state towards FH + H: rFH = 170 pm, rHH = 74 pm and pFH = pHH = 0 g.mol-1.pm.fm-1. In order to drive the transition state towards F + H2 we will need to prolong the F-H distance and reduce the H-H distance slightly. After several tries, the following parameters were used to find the activation energy: rFH = 190 pm, rHH = 70 pm and pFH = pHH = 0 g.mol-1.pm.fm-1. Using these two initial conditions, we will plot a graph of Energy vs Time and determine the activation energies of both reactions by taking the difference of the maximum of the curve (energy of the transition state) and the minimum (energy of the products).

The following results were obtained:

{| class="wikitable" border="1"
! Energy vs Time (rFH = 150 pm, rHH = 74 pm, pFH = pHH = 0 g.mol-1.pm.fm-1) !! energy vs Time ( rFH = 190 pm, rHH = 70 pm and pFH = pHH = 0 g.mol-1.pm.fm-1)
|-
| [[File: Activation energy 1sds2618.png|300px]] || [[File: Activation energy 2sds2618.png|300px]]
|-
|}

According to these graphs, the activation for the reaction H + HF = H2 + F is approximately 127 kJ.mol-1 and that of H2 = HF + H is around 0.9 kJ.mol-1. The very small activation energy for the second reaction is due to the strong driving force to form the H-F bond since this is a very stable compound. The same can be used to explain the relatively large activation energy for the inverse reaction: the HF compound is much more stable and has a very strong bon which means a lot of energy is required to break that bond in order to form a weaker, less stable one between two H atoms.

=== Question 2 ===

'''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 the following initial conditions for the reaction F + H2 rHF = 200 pm, rHH = 74 pm, pHF = -1.0 g.mol-1.pm.fs-1 and pHH = 0 g.mol-1.pm.fs-1 (considering the activation energy for this reaction is very small, not much initial momentum is required since the reaction is "very spontaneous"), the following results were obtained:

{| class="wikitable" border="1"
! Contour plot !! Momenta vs Time
|-
| [[File: Energy transfer contour plotsds2618.png|200px]] [[File: Energy transfer 3d contoursds2618.png|200px]] || [[File: Energy transfer m vs tsds2618.png|300px]]
|}

According to these graphs, we can see that once the system has passed the transition state, the newly formed H-F bond has a very large momentum (around 17 g.mol-1.pm.fs-1 ). This is because the bond between H and F is of such different nature compared to that of HH. The newly formed HF molecule is so much more stable than the reactants which is why the system gains a lot of energy once it has overcome the very small activation energy. This means that, in order to conserve energy, the system passes its excess energy into the vibration of HF. This can be seen when we observe the contour plot of this reaction: after the system passes through the transition state (in this case several times), it oscillates strongly.

=== Question 3 ===

'''Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, with a momentum pFH = -1.0 g.mol-1.pm.fs-1, and explore several values of pHH in the range -6.1 to 6.1 g.mol-1.pm.fs-1 (explore values also close to these limits). What do you observe?'''

When testing different momenta of pHH close to the limits of -6.1 and 6 g.mol-1.pm.fs-1, we observe reactive and unreactive trajectories, however, most are unreactive. Some examples of tested trajectories are shown below:

{| class="wikitable" border="1"
! Contour plot pHH = -6.0 g.mol-1.pm.fm-1 !! Contour plot pHH = -5.3 g.mol-1.pm.fm-1 !! Contour plot pHH = 5.6 g.mol-1.pm.fm-1 !! Contour plot pHH = 6.0 g.mol-1.pm.fm-1
|-
| [[File: Ex2Q3sds2618.png|200px]] || [[File: Ex2Q3-5.3sds2618.png|200px]] || [[File: Ex2Q3 5.6sds2618.png|200px]] || [[File: Ex2Q3 6sds2618.png|200px]]
|}

The energy put in to the HH vibration is much larger than the required activation energy of this reaction. Therefore, the H-H bond vibrates strongly whilst approaching the F molecule which makes them collide. However, the H atom colliding with the F atom has such a strong momentum that it effectively "bounces" back to the H atom and thereby back to the reactants. This results in unreactive trajectories. Even when the trajectory results in a reactive pathway (for example when pHH = 5.6 g.mol-1.pm.fm-1), the trajectory of the system has many twists and turns which represent the system making and breaking the H-H as well as the H-F bond very often. This proves that if too much initial energy is given to the system, it has enough energy to overcome the activation energy for the forward reaction as well as for the reverse reaction. Furthermore, as seen before, when the system forms the HF bond it gains a large amount of energy in the form of vibration. If this vibrational energy is large enough to overcome the activation energy of the reverse reaction, the system could potentially revert to the reactants.

'''For the same initial position, increase slightly the momentum pFH = -1.6 g.mol-1.pm.fs-1, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.2 g.mol-1.pm.fs-1. What do you observe now?'''

The following trajectory resulted from the decrease of the overall energy:
{| class="wikitable" border="1"
|+Parameters rHF= 200 pm, rHH= 74 pm pFH = -1.6 g.mol-1.pm.fs-1 and pHH = 0.2 g.mol-1.pm.fs-1
! Contour plot !! Momentum vs Time
|-
| [[File: Ex2_reduced_overall_E_contoursds2618.png|300px]] || [[File:Overall reduced E m vs tsds2618.png|300px]]
|}

The Contour plot shows that even reducing the overall energy significantly will not result in a reactive trajectory. In this case, the HH molecule approaches the F atom and arrives at the transition state, the H-F bond forms but the strong vibrations of H-F bond (16 g.mol-1.pm.fs-1) make the system oscillate between the formation of the H-F bond and that of the H-H bond. In this case, the oscillations result in the system going back to the reactants.

=== Question 4 ===

'''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.'''

== References ==

<references>
<ref name="Q5-1">Laidler, Keith J, ''Chemical kinetics'', 3rd edition, '''1987''', 287-321.</ref>
<ref name="Q5-2">Laidler, Keith J, ''Chemical kinetics'', 3rd edition, '''1987''', 88-98.</ref>
<ref name="Q5-3">Laidler, Keith J, ''Chemical kinetics'', 3rd edition, '''1987''', 115-123.</ref>
<ref name="Q1-1">Laidler, Keith J, ''Chemical kinetics'', 3rd edition, '''1987''', 460-471.</ref>
</references>

File:TS Dynamicssds2618.png

2020-05-14T11:13:36Z

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File:Energy transfer contour plotsds2618.png

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2020-05-11T11:26:37Z

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File:Initial conditions m vs tsds2618.png

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