MRD:01524831.molecularreactiondynamics

Exercise 1: H + H₂ system

Figure 1. A surface plot showing the transition state of a H + H₂ system

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?

On a potential energy surface diagram , the transition state is mathematically defined as the saddle point. The saddle point is a stationary point on the plot, where along the reaction path it is a maxima and perpendicular to the reaction path the saddle point is perceived to be a minima. The transition point can be identified in this simple example when the distances between the 3 hydrogen atoms are of equal distance. As shown in figure 1, the reaction path is oscillating at the saddle point. This depicts the vibrational freedom of the system at the transition state. A saddle point is found by...

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 estimated r_ts is found to be around 91 pm. The transition state is defined as the position where internuclear distances are equal, and at a energy maxima in relation to the reaction path. From figure 1 it is evident that there are vibrations within the system, depicted by the oscillations on the ridge. The transition state position was found by determining which position the system stops oscillating, where the potential energy was a maximum, and thus the stable point of the transition state, the saddle point was found as shown in figure 2 and 3.

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

When calculating under map the reaction path is depicted different to when the calculation type is set to dynamics. What is absent in the map calculation is the motion of the atoms. The vibrational motion of the three hydrogen atoms are not depicted as shown in figure .... The initial condition is set so that the distance between AB is r_ts and the distance between BC deviates from r_ts by 1 pm, resulting in the reaction pathway following through, and advance towards the H_A + H_B-H_C product. The flat BC line indicates the fact that the mep calculation does not consider vibrational motion of atoms.

Look at the “Internuclear Distances vs Time” and “Momenta vs Time”. What would change if we used the initial conditions r₁ = r_ts and  r₂ = r_ts+1 pm instead?

The change of initial conditions to those above will change final product from H_A + H_B-H_C to H_A-H_B + H_C as it is the H_B-H_C bond distance that deviates away from the  r_ts. As a result the internuclear distances vs time plot will now show AB to be constant over time. Momentum remains unchanged as initial momentum is 0 for both scenarios.

Complete the table above 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. What can you conclude from the table?

p1/ g.mol-1.pm.fs-1	p2/ g.mol-1.pm.fs-1	Etot/kJ.mol-1	Reactive?	Description of the dynamics	Illustration of the trajectory
-2.56	-5.1	-414.280	Yes	Hydrogen atoms AB and C approach eachother. Notably, the vibrational motion present after the transition state is not present before.
-3.1	-4.1	-420.077	No	Hydrogen atoms AB approaches C, however before approaching the saddle point to AB swiftly moves away from C
-3.1	-5.1	-413.977	Yes	Hydrogen atom C approaches AB with a large amount of momentum, resulting in a new BC Hydrogen molecule forming, releasing the A hydrogen atom in the reaction.
-5.1	-10.1	-357.277	No	Hydrogen atoms AB and C approach each other very quickly. Upon impact, large internuclear distance and velocities are depicted, showing the B atom being tugged by both A and C. In the end, B forms a molecule AB and moves away from C.
-5.1	-10.6	-349.477	Yes	Upon impact, the B atom is tugged between the two atoms A and C, and is finally bonded to to atom C. A then moves away from BC.

The table above shows that a reaction isn't only reliant on the amount of kinetic energy in the system. As shown in the table above, there are instances where kinetic energy is higher than the activation energy, however no reaction takes place. One therefore needs to consider that successful collision are reliant on the system having enough energy to break and reform bonds, and also the orientation at which atoms collide.

Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

Transition State Theory has a fundamental assumption which will result in its predictions being different to experimental values. The assumption that all trajectories possessing enough kinetic energy to circumvent the activation barrier will be reactive will result in a larger R_cl and thus a larger rate prediction than the experimental outcome. As seen in the table above, not all collisions are reactive even if there is sufficient kinetic energy - the case of barrier recrossing perfectly illustrates this.

Exercise 2

By inspecting the potential energy surfaces, classify the F + H₂ and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

Figure 4. A surface plot showing the trajectory of F + H₂ to H + HF reaction]

The reaction is exothermic as shown in figure 4, there is a decrease in potential energy when passing the transition state and forming the new HF bond. This relates to bond-strength, H-H has a bond strength of 432 kJ/mol whilst H-F has a bond strength of 565 kJ/mol.(reference). As HF is a polar bond as a result of the very electronegative F, there is an ionic element to the bond, resulting in a much greater attraction between the two atoms, thus releasing a lot of energy on formation. As the energy released is greater than that absorbed, the reaction is exothermic as depicted in the trajectory shown in figure 4.

[[File:Surface_Plot_hf-h_hk3918.png|thumb|Figure 5. A surface plot showing the trajectory of the FH + H to F + H₂ reaction|300px]

Figure 5 on the other hand shows an endothermic reaction, illustrated by the increase in potential energy along the trajectory, indicating that the energy absorbed to break the stronger H-F bond is greater than the energy released when forming the weaker H-H bond.