MRD：01508610
Contents
EXERCISE 1: H + H2 system
Coordinates
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 rAB and rBC 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?
•The molecular geometry and chemical reaction dynamics are analysed with a potential energy surface (PES) where the necessary points can be evaluated and classified according to first and second derivatives of the energy with respect to position, V(r), which respectively are the gradient and the curvature. Stationary points are those with zero gradient (∂V(ri)/∂ri=0) and have physical meaning: energy minima (or local minima of the PES) correspond to physically stable chemical species (such as reactant and product) and saddle points representing transition states, the maximum on the lowest energy pathway (or minimum energy path) that linking reactant and product. Mathematically, a saddle point is a point on the surface of a function where the slopes in orthogonal directions are all zero, but which is not a local extremum of the function. 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 rAB and rBC are equal.
•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 ri where the force acting on) will be zero at transition state. Consequently, if one starts a trajectory exactly at the transition state, with no initial momentum, it will remain there forever. Therefore, the method to identify the transition state position from a PES is to change the values of rAB and rBC 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 rAB = rBC: 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.
By testing different initial conditions with rAB = rBC, and p1 = p2 = 0.0 g.mol-1.pm.fs-1, the transition state position (rts) is found to be the point where distances rAB and rBC 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 symmetrical, the transition state must have rAB = rBC. If we start a trajectory on the ridge rAB = rBC 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 rAB = rBC trajectories are shown below. In these graphical display the values of rAB(t), rBC(t) and rAC(t) are given (Y axis) against time t (X axis). At transition state, distance rAB and rBC (and thus rAC) are constant since all three atoms are stationary at the saddle point. The potential energy is maximized while the kinetic energy is zero. This geometry is transition structure.
In contrast, when initial conditions increased to rAB = rBC =100 pm, it shows an oscillatory behaviour for the whole trajectory corresponding to the vibration of atom A and atom C, since there are both potential and kinetic energy. The distance rAB and rBC are overlapped, since they have the same amplitude of vibration and the vibrations are symmetrical around atom B. The distance rAC) has doubled the amplitude of vibration (i.e. the sum or superposition of the amplitude). The oscillatory behaviour is changed after the transition state, at t≈270 fs, to a dissociation pattern where rAB (and thus rAC) keeps increasing with a slight oscillatory behaviour while rBC remained at the same level with a higher vibrational frequency. This means that the atom A leaves away from the molecule B-C with the translational energy while molecule B-C is left with vibrational energy, which resembles the reactant.
