Gd1118yr2
Contents
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?
On the surface diagram, the transition state is mathematically defined as the saddle point of the diagram where f (rAB) = f(rBC) = f(rABrBC) and f' (rAB) = f' (rBC) = f'(rABrBC)= 0.
Transition state is "the maximum on the minimum energy path linking reactants and the products". The minimum energy path is the black line, and the transition state is shown as the red dot in Fig 1, the maximum of the line.
A local minimum of the potential energy surface is the minimum energy of the surface, the lowest possible energy a system can have, while transition state is the maximum of the minimum energy path.
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 transition state, it is expected that rAB = rBC. From Fig 2, The intersection occurs at roughly 25.5 seconds after motion started. The bond distance at the time of transition state is about 90 pm.
To get more precise result, p1 and p2 are set to 0.0 g.mol-1.pm.fs-1. Internuclear distances is adjusted around 90 pm (Fig 3.), and the result indicate that rts = 90.8 pm where there is no molecular vibration.
| r | 75 pm | 90.8 pm | 100 pm |
|---|---|---|---|
| Internuclear Distances vs Time |  |
 |
|
Comment on how the mep and the trajectory you just calculated differ
| Plot type | Dynamics | MEP |
|---|---|---|
| Internuclear Distances vs Time |  |
|
| Momenta vs Time |  |
|
From Table 3. above, MEP ignores the vibrations of the product. In this case, the stretching of H2 molecule is ignored, indicated by the smooth line in "Internuclear Distances vs Time" plot. The zero momenta of B-C is also the result of non-stretching. The other thing that is different is the time for complete reaction. It takes about 8 fs in MEP, while 18fs is required in Dynamics.
The thing that the two models agree is that the reaction goes to completion towards the product with the formation of A and B-C.
| Plot type | Dynamics | MEP |
|---|---|---|
| Internuclear Distances vs Time |  |
|
| Momenta vs Time |  |
|
As can be seen from Table 3., after switching the initial features of r1 and r2, there is not any change in the shape of the diagrams. The only thing that differs from the previous experiment is that instead of forming A and B-C, the product now in this case is A-B and C.
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?
| p1/ g.mol-1.pm.fs-1 | p2/ g.mol-1.pm.fs-1 | Etot | Reactive? | Description of the dynamics | Illustration of the trajectory |
|---|---|---|---|---|---|
| -2.56 | -5.1 | -414.280 | Yes | Atom C approaches vibrating molecule AB, the vibration is distorted when r1 is close to transition. After the transition state, atom A and molecule
BC is formed. r2 increases with comparatively gentle vibration. |
|
| -3.1 | -4.1 | -420.077 | No | Atom C approaches vibrating molecule AB, the vibration is slightly distorted when r1 is small. Then AB and C separate without collision. No reaction occur. r1 increases with AB vibrating. | 
|
| -3.1 | -5.1 | -413.977 | Yes | Atom C approaches vibrating molecule AB, the vibration is distorted when r1 is close to transition. After the transition state, atom A and molecule
BC is formed. r2 increases with comparatively gentle vibration. |
|
| -5.1 | -10.1 | -357.277 | No | Atom C approaches vigorously vibrating molecule AB, the vibration is slightly distorted when r1 is small. Then AB and C separate after collision. No reaction occur. r1 increases with AB vibrating. | 
|
| -5.1 | -10.6 | -349.477 | Yes | Atom C approaches non-vibrating molecule AB, vibration is distorted close to transition state, atom A and molecule BC are formed. r2 then increases with vigorous vibration. | 
|
The results indicate that the hypothesis is wrong. Higher momenta and total kinetic energy not necessarily result in reaction. This can be concluded by comparing trial 1 & 2 and 4 &5. Reaction succeeded under condition 1,3 and 5. By comparing 1 & 2, 2 & 3 and 4 & 5, the difference between p1 and p2 seems to be an important factor for successful reaction. This can be further explored by keeping kinetic energy constant.
Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?
Overall, Transition State Theory overestimates the rate constant because it simplifies the model by only consider the translation in the system. Besides, it does not allow re-crossing. However, as can be seen from previous simulations, there were vibrations in molecules and re-crossing can possibly happen. Under these circumstances, the reaction path will be higher in energy than the minimum values on the potential surface, while the theory assumes that transition state as a saddle point on the potential surface which is " the maximum in minimum energies". Therefore, the experimental activation energy is higher than theoretical prediction. Thus, experimental rate constant is lower the value calculated based on Transition State Theory. Furthurmore, Transition State Theory implies classical mechanics and ignores tunneling effect which is important for light particles such hydrogen. (lit.[1])
Exercise 2
By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?
| H2+F | H+HF |
|---|---|
| A:H B:H C:F | A:F B:H C:H |
 |
|
As can be seen from Table 4., the reaction between H2 and F is exothermic , while it is endothermic for H and HF. This can be identified by looking at the reaction pathway. Reactants have lies higher in energy than product in H2 and F reaction, while it is the reverse in H and HF.
The result implies that H-H is weaker and longer than H-F. The energy change in the reaction is associated with bond breakage and formation. In this case, only H-H and H-F are either broken or formed. H-F stabilising effect override destabilising effect by H-H, hence stabilising the system in H2 + F reaction. The reaction is henceforth exothermic. It is the other way round for H+HF reaction.
Locate the approximate position of the transition state.
Hammond's postulate is used in the process. Since H+HF reaction is endothermic, the structure of transition state should resemble product H2. Similar to what was done to find the transition state of H-H-H, momenta were set to 0, length BC was set to 74 pm. By adjusting length AB and make minor changes to BC, transition state bond lengths F-H is found to be 181 pm, and that of H-H is 74.49 pm. By looking at "forces" section in Fig 4., both values give " 0 ", indicating that the transition state is reached.
Report the activation energy for both reactions
| H + HF | H-H-F | H2+F |
|---|---|---|
 |
 |
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The energies of reactants and transition states were shown in Table 5., The distance between reactants in each reaction were set to 1000 pm so that they are separated apart. Activation energy is the difference in energy between transition state and reactants. The calculated activation energy for H2 + F is 1.12 kJ/mol, and for H + HF is 126.72 kJ/mol.
| H2+F to HF+H | HF+H to H2+F |
|---|---|
| 1.12 | 126.72 |
- ↑ Chapter 10 of J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics 2nd ed., Prentice-Hall, 1998
