MRD:01535442

Molecular Reaction Dynamics: Applications to Triatomic systems

In this report, we will be investigating the reaction dynamics of two triatomic systems, H-H-H and F-H-H. This includes investigation of their transition states, reaction coordinates and potential energy surfaces, and how these affect the outcome of chemical reactions.

Exercise 1: H + H₂ System

Figure 1: Internuclear Distance vs Time Graph for H + H₂ system, showing estimate for r_ts

On a potential energy surface diagram, how is the transition state mathematically defined?

∂V(r_i)/∂r_i=0 defines the point in the potential enrgy surface diagram where to gradient is zero, being defined as the maximum on the minimum energy curve.

How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

The transitions state can be identified by mapping trajectories near the supposed transition state, and observe whether the line goes towards the products or reactants. It can be distinguished from a local minimum by looking at the second derivative, which will be positive when the function is a local minimum, and negative if it is a maximum point.

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

The best estimate for r_ts (the transition state position) is 90.8 pm. If you take a look at the internuclear distance vs time graph (Figure 1), it shows that across time, there is no oscillatory behaviour in the triatomic system that alters the distance from the r_ts, therefore 90.8 pm is the closest estimate for this position. Note that distance A-B and B-C are the same, so line A-B in the graph is behind line B-C.

Figure 2: Contour plot of system displaced from transtition state. Calculation type: Dynamics

Comment on how the mep and the trajectory you just calculated differ

The minimum energy path or mep takes a path through the potential surface where it does not move up or down the contour, rather it takes the the path of lowest energy througout (see Figure 2). The trajectory calculated by dynamics (Figure 3) follows the same overall path as the mep, though it's much more wavy. This is due to the vibration of the diatomic molecule. Conservation of energy is present in both as the dynamic trajectory oscillates between the same energy contours.

Figure 3: Contour plot of system displaced from transtion state. Calculation type: MEP

Complete the table below 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.

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	-414.280	Yes	The H2 (Hb-Hc) molecule has relatively lower momentum than Ha, and so there is little oscillatory behaviour. When it gets to the transition state, there is reintroduction of oscillatory behaviour in Ha-Hb, due to excess vibrational energy.
-3.1	-4.1	-420.077	No	The H2 (Hb-Hc) molecule has relatively similiar levels of momentum to Ha. Hb-Hc has more visible oscillations in it's trajectory. The path does not reach the transition state, as there is not enough kinetic energy for Ha to form a bond with Hb, and repulsion forces between them push them apart, oscillating back to the reactant snapshot.
-3.1	-5.1	-413.977	Yes	The H2 (Hb-Hc) molecule has relatively lower momentum than Ha, and there is some oscillatory behaviour with Hb-Hc. There is enough momentum for Ha to reach the transition state, as it overcomes the repulsion to form a new bond.There is then increased oscillatory behaviour in Ha-Hb, due to excess vibrational energy.
-5.1	-10.1	-357.277	No	The H2 (Hb-Hc) molecule has relatively lower momentum than Ha, and there is little oscillatory behaviour with Hb-Hc. There is enough momentum for Ha to reach the transition state, as it overcomes the repulsion to form a new bond. Yet, as momentum is high, there is large oscillations in the path of the new Ha-Hb, so much so that new bond then breaks and reforms the reactants, oscillating backwards.
-5.1	-10.6	-349.477	Yes	The H2 (Hb-Hc) molecule has relatively lower momentum than Ha, and there is little oscillatory behaviour with Hb-Hc. There is enough momentum for Ha to reach the transition state, and there is few larger oscillations where the new bond between Ha-Hb is formed, and then this breaks and the reactant Hb-Hc reforms. The reactant bond than breaks and reforms the new Ha-Hb molecule, and the path goes towards the products.

What can you conclude from the table?

In summary, the main influence on whether a trajectory will lead to a reaction is that the momentum p₂ must be larger than p₁, and around double the magnitude of p₁. The higher that values of momentum, the higher the kinetic energy in the system and the higher the total energy becomes.