MRD:01340064
Comp Lab
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 maximum on the minimum energy path linking reactants and the products.
The point on the potential energy surface can be identified as where the gradient of the potential is zero, and the energy goes down most steeply along the minimum energy path linking reactants and products.
At the transition state, if you change the geometry by a small amount in the direction of the products it will roll towards the products. The same thing would happen for for the reactants. Therefore another method of locating the transition state would be to start trajectories near it and see whether they roll to the reactants or products.
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.
My best estimate of the transition state position is 0.9076 Å
The first figure shows the internuclear distances vs time plot when r1=r2=0.908. The second figure shows the system rolling towards the reactants after a minor geometry change of 0.001 A towards the reactants.
Comment on how the mep and the trajectory you just calculated differ.
The mep follows the valley floor to H1+ H2-H3 without any indication of vibration in the molecule, the dynamic trajectory shows a more realistic account of the motion of the atoms as it shows the vibration in the molecule.
Complete the table 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 | p2 | Etot | Reactive? | Description of the dynamics | |
|---|---|---|---|---|---|
| -1.25 | -2.5 | -98.956 | Yes | 
|
The trajectory runs through the transition state to the products |
| -1.5 | -2.0 | -100.456 | No | 
|
The trajectory stops before reaching the transition state |
| -1.5 | -2.5 | -98.956 | Yes | 
|
The trajectory runs through the transition state to the products |
| -2.5 | -5.0 | -84.956 | No | 
|
Unreactive trajectory -the system crosses the transition state region, the bond in the product forms but then reverts back to the reactants. |
| -2.5 | -5.2 | -83.416 | Yes | 
|
The trajectory bounces at the barrier first and then runs through the transition state to the products |
Not all trajectories starting with the same positions but with higher values of momenta are reactive.
State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?
Transition state theory assumptions:
- The reactants are in constant equilibrium with the transition state.
- The energy follows a Boltzmann distribution.
- Once reactants become the transition state, the reactants do not regenerate.
Experimental values will have lower reaction rate values.
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?
It can be seen that the reaction of H2 + F is exothermic as the products are lower in energy than the reactants in the surface plot. It can also therefore be seen that the reverse reaction of HF + H is endothermic. This is due to the fact that the HF bond is much stronger than the HH bond, because of the difference in electronegativity of H and F. The HF bond is has an ionic contribution which makes it stronger, causing more energy to be released when HF forms.
Locate the approximate position of the transition state.
Looking at the Internuclear Distance vs Time plot for the H2 + F trajectory, an approximate time frame for the transition state can be observed where the lines BC and AB overlap (around 0.7 seconds).
To find an accurate position of the transition state, the momenta were set to zero and the distances BC and AB adjusted until no oscillation was observed in either line, which would correspond to the activated complex at the top of the transition state.
This was found to occur with a HH bond distance of 0.745 Å and a HF bond distance of 1.811 Å
Report the activation energy for both reactions.
An MEP calculation with one of the distances slightly adjusted from the transition state was determined. The trajectory therefore rolled to the valley floor of either the reactants or products. The activation energy was then found to be the difference in potential energy on an Energy vs Time graph. H2 + F --> HF + H Activation energy = 0.200 kcal/mol HF + H --> H2 + F Activation energy = 29.969 kcal/mol
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.
A reactive trajectory for the F + H2 was identified with the initial conditions of 0.79 Å HH, momentum= 0.78 and 1.8 Å HF, momentum=-0.5.
The contour plot shows that the product HF continues to oscillate once it forms as it possesses vibrational energy. The Energy vs Time graph shows that energy is conserved during the reaction as the kinetic energy is at maximum when the potential energy is minimum and the opposite is also true. The vibrationally excited product HF will release its energy as heat, colliding with other molecules of gas. This could be confirmed experimentally by use of a gas-phase FTIR spectrum of the product which would show the vibrational exciited state relax to the ground state, releasing a photon corresponding to the different in electronic states.
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.
| p(HH) Vibrational Energy | p(HF) Translational Energy | Plot |
|---|---|---|
| -0.5 | -2.0 |
|
| -2.8 | -0.5 |
|
| p(HH) Vibrational Energy | p(HF) Translational Energy | Plot |
|---|---|---|
| -6.7 | -0.4 |
|
| -0.4 | -6.7 | 
|
The first table shows that an increased amount translational energy with respect to the vibrational energy acts to drive the exothermic reaction to HF, with an early transition state. An increased amount of vibrational energy however does not lead to a reactive trajectory. The second table shows that the opposite is true for the reverse reaction, with excess vibrational energy driving the endothermic reaction to H2 , with a late transition state, but no reactive trajectory for an excess of translational energy. These results show Polyanyi's empirical rules.
