MRD:01512675

Exercise 1: H + H$_2$ system

Question 1: The Transition State Region

The transition state is defined as the maximum point of the minimum energy path between the reactants and products. Therefore, on the potential energy surface diagram, it exists as a minimax; a saddle point. This can therefore be described as the point where the partial differential of the potential energy with respect to all variables is zero.

In this case, due to a 3 dimensional surface plot, we have $\frac{\partial V}{\partial r_1} = \frac{\partial V}{\partial r_2} = 0$, where $r_1$ and $r_2$ are the respective distances between AB and BC.

The local minimum point would see that a deviation in any direction would be an increase in potential energy. However, the transition state is such that this would only happen in the axis where $r_1$ = $r_2$, and would decrease in energy in any deviation of the AB or BC position.

This is shown in the images shown below.

The photo on the left shows the energy potential curve in a 2D viewpoint, as a function of the AB distance $r_1$. We can see that there is a steep increase of the V($r_1$) function on either side of $r_1$ = 74 pm, indicating the minimum in a potential well at this point. This continues as a valley between the reactants and products path.

The photo on the right shows a clear maximum point being mapped out by the black line. Despite being at a minimum point on the potential valley floor, this is a maximum point, indicating a minimax state. This is therefore a saddle point, and the transition state. Due to the symmetry of the system only consisting of H atoms, the transition state is observed at a point where $r_1$ = $r_2$.

This can further be identified using the second partial derivative test; computing a Hessian matrix. If H <0, it must be a saddle.

Question 2: Locating the transition state

Due to the symmetry between the H and the \chem{H_2} molecule, and the fact that the interaction is happening at 180º, we know that $r_1$ must equal $r_2$ for any point on the potential for the transition state. As a result, if both particles carry zero initial velocity and are on the transition state only, they will drop down the potential well to the transition state distance $r_{12}$. Thus, in a scenario where $r_1$ = $r_2$, and $v_2$ = $v_1$ = 0, the transition state will act only in a 2D potential plot; the dynamics will reach the minimum in the potential well.

This distance $r_12$ is shown to be 90.775 pm, as there is no force along AB or BC at this distance when v = 0. Furthermore, the contour plot shows the final position to be the same as the initial position; it is a point of equilibrium. The internuclear distances vs time plot are also shown to constant here; this must therefore be the transition state.

Question 3: The Minimum Energy Path (MEP)

The MEP shows a clear difference in its plot and animation compared to the Dynamic calculation.

1. There is no vibration/ oscillation in the MEP plot. This makes sense; it would be going against minimum energy route, as the oscillations would mean that the particle is not always at the minimum of the potential well.
2. The MEP plot finishes once the minimum point of the energy potential surface is reached; after 193.1pm as the AB distance and 74 pm as the BC distance, the contour is consistent. However, we know that the particles must keep moving from this, as they carry opposite velocities. This is shown in the Dynamics plot, where the reaction path continues long after reaching the end of the x-label in the plot. This confirms that the MEP is relying only on the minimum of the potential energy surface; not the distances AB and BC.

This is shown in the comparison below.

It is also interesting to note the action of the MEP when we start from the reactants, as before, with enough energy to go to the products in the Dynamics. This is shown in the comparison below.

In this scenario, despite having enough initial velocity in one direction, the MEP doesn't follow that path. It reverses the direction of the initial velocity and instead ends up with no reaction taking place. This implies that it has followed the minimum energy path at every point in the plot, ignoring the vectors that the particle is following. This further exemplifies the difference between Dynamics and MEP, showing that MEP is only taking the mathematically logical route, whereas the Dynamics considers the physical process that is occuring.

Question 4: Reactive and Unreactive Trajectories

The following table assesses the plot for the initial positions r₁ = 74 pm and r₂ = 200 pm, with varying initial momenta.

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	Proceeds as normal. No initial collision, but repulsion between the two systems is overcome, etc.
-3.1	-4.1	-420.077	No	The initial velocity of the atom is far too slow to collide with the molecule. The two systems approach each other initially, but the repulsion between them forces them to move in opposite directions, and with more velocity than
-3.1	-5.1	-413.977	Yes	The two systems approach the transition state, and then the diatom splits, with the central atom pairing with the other terminal atom. They then proceed to move in opposite velocities.
-5.1	-10.1	-357.277	No	Initial velocity of particle C is so great that there is some repulsion between particle B and C, despite moving towards each other. As a result, particle B moves away from C after the initial attraction. It then oscillates around the transition state, moving from AB + C to A+ BC. It eventually settles with AB + C, the initial reactants. It can be seen by increasing the number of steps to 2000 that the final position of the two systems is greater separation. The particles AB and C are moving in opposite directions, so end up further apart than the initial conditions.
-5.1	-10.6	-349.477	Yes	Oscillations around the transition state after an initial collision. This is very similar initially to the -10.1 p2 path. However, the rebound collision for A and B is even faster, resulting in another rebound for B. B then collides with particle C as they are both moving in the same directions, and the final state is that of the products, A + BC

Question 5: Transition State Theory Predictions

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Exercise 2: F-H-H system

Question 1: Energetics and bond strength

Let's set the initial position as $r_1$ = $r_2$, where $r_1$ is F-H, and $r_2$ is H-H.

If we run a MEP plot from this position, the particle will end up at the lower energy state, regardless of the initial reactants or products.

We can see in the image below, the minimum energy path is such that $r_1$ is minimised, and $r_2$ is maximised. With $r_1$ as F-H, this implies that the lower energy state is H + H-F.

As a result, we can classify the F + H₂ --> H + HF reaction as exothermic, as it is progressing from a higher energy state to a lower energy state.

H + HF ---> F + H₂ would therefore be an endothermic process; it would require an input of energy to get this product.

This order of potential energies suggests that the bond strength between F-H is greater than for H₂.

Question 1: Transition state distance

As before, the transition state will be the saddle point on the surface plot. However, as the system is no longer symmetric as before, the transition state is no longer at the point where $r_1$ = $r_2$.

Hammond's postulate suggests that the transition state will be closer to the higher energy system in the reaction. We know that this is the case for F + H₂, so the transition state must be such that the bond distance for F-H is greater than for H₂.

Knowing this, we can surmise that if AB is F-H and BC is H₂, the distances at the transition state are: AB =184 pm and BC = 74.4 pm. This complies with the logic that the reaction where FH is the product is exothermic, as this is supported by Hammond's postulate.