MRD:01496292
- 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 by the point at which F=∂V(ri)/∂ri=0. Therefore, in a potential energy surface plot, we are looking for the point where ∂V(r1)/∂r1=0 and ∂V(r2)/∂r2=0. (transition state pic) Because the energy goes down along the minimum energy path for a reaction, then if the reactants starts a trajectory at the transition state with no initial momentum, it will never leave that point in space because no force is acting on the system. However, ∂V(ri)/∂ri=0 can also indicate the existence of a local minimum point. To identify whether the stable point is at the transition state or at a local minimum, we can follow the calculations below:
1. If (fr1r1 - f2r2r2) > 0 and fr1r1 < 0 then the point is a local maximum
2. If (fr1r1 - f2r2r2) > 0 and fr1r1 > 0 then the point is a local minimum.
3. If (fr1r1 - f2r2r2) < 0, then the function is at the transition state (saddle point).
Trajectories from r1 = r2: locating the transition state
Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.
The best estimate of the transition state position is: rAB = rBC = rts = 90.775 pm.
For this reaction, we have a totally symmetric transition state i.e. rAB = rBC = rts . In the condition specified, the trajectory will only oscillate on the ridge and never fall off.
Hence, to find the position at the transition state we need to get rid of the oscillation and concentrate on the minimum potential energy point in this oscillation. To achieve this we could look at the ''internuclear Distance vs Time'' plot - if there is no oscillation at all, this plot would have the distance between AB and BC unchanged (two flat lines). Therefore, by altering the AB BC distances and looking for the best flat line, an estimate of rts can be found. Furthermore, we can also identify the existence of saddle point by looking at the hessian matrix: if one of the eigenvalue is positive with the other being negative, then it is a saddle point; all positive - minimum point; all positive - maximum point.
- Calculating the reaction path
Comment on how the mep and the trajectory you just calculated differ.
For the energy contour plot, it can be observed that unlike the dynamics plot there is no vibrational energy for the MEP plot. This is an indication of how different MEP is algorithmically set - for MEP velocity is reset to 0 at each time step, so that kinetic energy always remains at 0. Thus, there is no vibrational energy in the system so no oscillation.
(pictures of energy v time) 
For MEP, since KE=0 and PE decreases going downhill as the reaction proceeds away from the transition state, so total energy decreases. For dynamics, total energy remains constant - the loss in potential energy overtime has completely transferred into kinetic energy.
If we look at the Internuclear Distance v Time plot for MEP and dynamics, we can see that in both cases atoms are moving away from each other and in the dynamics velocities at which they are moving remain constant.
Take note of the final values of the positions r1(t) r2(t) and p1(t) p2(t) for your trajectory for large enough t.
The values of r1(t) r2(t) tends to infinity for large enough t whilst p1(t) p2(t) tends to remain the same average values (same amplitude but with momentum oscillates at higher frequencies for A-B.
Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.
What do you observe?
We could get the other plot by taking reflection along x axis and y axis from either one of the plots.
Reactive and unreactive trajectories
For the initial positions r1 = 74 pm and r2 = 200 pm, run trajectories with the following momenta combination and complete 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 | ||||
| -3.1 | -4.1 | ||||
| -3.1 | -5.1 | ||||
| -5.1 | -10.1 | ||||
| -5.1 | -10.6 |
