MRD：01508610

EXERCISE 1: H + H2 system

We will be looking at collision and reaction between diatomic molecule of hydrogen and an atom of H in a linear configuration in the gas phase. In simple terms, the atom A collides with the molecule B-C and forms a new molecule with B, while C is detached as a separate atom. The interatomic distances between A and B, B and C are labelled as r_AB and r_BC respectively.

Dynamics from the transition state region

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?

•In this H + H2 system, the surface is symmetrical, which means that the transition state must be exactly half way between the reactants and products. In another words, the transition state lies somewhere on the diagonal line where the distances r_AB and r_BC are equal. Actually, it is not easy to quickly find the transition state on a potential energy surface (PES) diagram. However, if the 2D diagram is simplified to a 1D reaction profile diagram, the transition state will be a saddle point (i.e. a local maxima; a stationary point) as a key feature.

•The hydrogen atoms will have no movement at the saddle point since the force on atoms (i.e. the negative derivative of PES with respect to the coordinate r_i where the force acting on) will be zero at transition state. Therefore, the method to identify the coordinates of the transition state structure from a PES is to change the values of r_AB and r_BC until the forces along AB and BC are both zero. In fact, forces are negative on one side of the PES and positive one the other side. The transition state is located where forces across zero.

Trajectories from r_AB = r_BC: locating the transition state

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

By testing different initial conditions with r_AB = r_BC, and p₁ = p₂ = 0.0 g.mol^-1.pm.fs^-1, the transition state position (r_ts) is found to be the point where distances r_AB and r_BC are equal to 90.78 pm. At that position, forces are both -0.001 KJ.mol^-1.pm^-1 (close to zero).

Since the H + H2 surface is symmetric, the transition state must have r_AB = r_BC. If we start a trajectory on the ridge r_AB = r_BC there is no gradient in the direction at right angles to the ridge, thus the trajectory will oscillate on the ridge and never fall off. To verify the transition state position, “Internuclear Distances vs Time” plots for some relevant r_AB = r_BC trajectories are shown below.

At transition state, all three atoms should be stationary and distance rBC and rAC should be constant, which is consistent with Figure 1. While when initial conditions are rAB = rBC =100 pm, 

This is because that transition state is usually a single spot on the ridge r_AB = r_BC in contour plot.

File:Figure internuc.vs.time.gjs123

Figure 1: A “Internuclear Distances vs Time” plot at transition state.

File:Figure 2 100pm internuc-time

Figure 2: A “Internuclear Distances vs Time” plot at r_AB = r_BC =100 pm.

• • • In these graphical display the values of r_BC(t), r_AB(t) are given (Y axis) against time t (X axis). Notice that r_AB decreases from the original value of 230 pm during the first 30 fs, then it has an oscillatory behaviour for the remainder of the trajectory corresponding to H2-H3 vibration. In contrast r1 (rBC here) has a slight oscillatory behaviour initially, then after the transition state, it grows in value as H1 leaves. Notice that r1(t) = r2(t) at ≈ 25 fs, and in this case is quite close to the transition state structure (this is not the case for all trajectories).