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 transition state is symmetrical, which means that the transition state is 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. (In this system, forces are both -0.001 KJ.mol^-1.pm^-1, nearly zero, when distances r_AB and r_BC are equal to 90.78 pm, and momentum p_AB and p_BC are equal to 0 g.mol^-1.pm.fs^-1. This is the point where the transition state lies.)

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