MRD:thl3318

EXERCISE 1: 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 can be mathematically defined as a saddle point (which is also an unstable equilibrium). It can be identified by determining the point where the partial derivatives of the function potential with respect to the internuclear distances are equal to 0.

$\frac{\partial V}{\partial r_1} =0$ and $\frac{\partial V}{\partial r_2} =0$

However, simply looking at the results for the first derivative does not give any information with regards to whether the point is a local minimum, a local maximum, or a saddle point. Analyzing the 2nd derivative then allows for the determination of whether the point is a saddle point. This can be done using the following equation:

$D=\frac{\partial^2 V}{\partial r_1^2} \frac{\partial^2 V}{\partial r_2^2} - (\frac{\partial^2 V}{\partial r_1\partial r_2})^2$

If D<0, the point is therefore a saddle point and also 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.

My best estimate for the transition state position is at time=20fs when the A-B distance is equal to the B-C distance. The A-B and B-C distance at this point is determined to be 90.8pm. This point in the "Internuclear Distance vs Time" plot can be rationed as the transition state position

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

The mep and dynamics trajectory differ in many different ways. The first difference is the time it takes for the reaction to take place (in this case, for the distance to follow to the valley floor of the potential graph). In the mep trajectory, it takes around 30 steps (15 fs) for the reaction to proceed whereas in the dynamics trajectory this number is around 140 fs. The second difference is that the internuclear distance between the H₂ molecule and the lone H atom increases logarithmically in the mep calculation but increases linearly in the dynamics calculation.

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. What can you conclude from the table?

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 KJ.mol-1	Yes	Plot shows trajectory going from reactant, going over the minimum energy path via the transition state and finally to form the product.
-3.1	-4.1	-420.077 KJ.mol-1	No	Plot shows the trajectory going up the potential slope towards the transition state. However it does not have enough momentum to go over the curve and thus the trajectory returns to its original position. The reaction does not proceed and the reactants remain.
-3.1	-5.1	-413.997 KJ.mol-1	Yes	Plot shows that increasing the momentum of the H-H bond in the reactants results in more vibrational energy hence a more distinct wave pattern appears in the trajectory. Similar to the first plot, the trajectory shows the reactant forming the product by going over the minimum energy path via the transition state.
-5.1	-10.1	-357.277 KJ.mol-1	Yes
-5.1	-10.6	-349.477 KJ.mol-1	Yes

From the table above, I can conclude that whether or not a trajectory results in a reaction depends on the momentum p2. Changing the momentum p1 does not affect the result of the trajectory but if momentum p2 is not great enough the result will be unreactive.

Trajectory that re-crosses the TS region

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

Experimental value will be greater than the reaction rate the transition state theory predicts because it does not take into account quantum tunneling which is more significant at low temperatures. Transition State Theory states that as long as the energy is greater than the activation energy the reaction will proceed to the products but this does not take into account the possibility of the "rebounding" and thus the reaction not proceeding.

EXERCISE 2: F + H₂ system

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?

The F + H₂ reaction is exothermic whereas the H + HF reaction is endothermic. This means that the H-F bond strength is greater than the H-H bond strength and is lower in energy.

Locate the approximate position of the transition state.

H-H distance = 77pm H-F distance = 179.5pm 1step size = 0.51

Report the activation energy for both reactions.

transition state energy = -433.020

H-H energy = -435.100

H-F energy = -560.700

Activation energy for F + H₂ --> HF + H = 2.080

Activation energy for H + HF --> H₂ + F = 127.68

You will be able to report a reasonable estimate by performing a mep (with a sufficient number of steps) from a structure neighbouring the transition state, and choosing to plot the appropriate quantity as a function of "time".

In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.

Reaction energy is released in the form of translational motion also known as heat. This can be confirmed experimentally by measuring the temperatures before and after the reaction. In the case of the reaction of F + H₂, this can be confirmed by an increase in temperature due to the reaction being exothermic. In the case of the reaction H + HF, this can be confirmed by a decrease in the temperature due to the reaction being endothermic.