MRD:01512675

Exercise 1: H + H$_2$ system

Question 1: 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 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. An example of this is the position of the reactants or products on the PES.

However, the transition state would see a decrease in potential energy along the reaction coordinate, and an increase in the coordinate orthogonal to this. This can be seen if we visualise the PES.

A 2D viewpoint of the PES, with the x-axis being the AB distance, and the y-axis being V($r_1$). We see an established potential well. This runs across the entire PES as a valley of potential well. In this viewpoint, the transition state is seen as a minimum point.

A 3D visualisation of the PES, showing how the transition state is a maximum. The black curve is the reaction path, and maximum of this curve is the saddle point transition state.

Due to the symmetry of this system, the transition state saddle point is observed at a point where $r_1$ = $r_2$, but this is not necessarily always the case. In some asymmetric systems, the transition state will depend on the relative energy of the products and reactants; this can be further described by Hammond's postulate.

The saddle point of any-dimensional curve can be identified by computing a Hessian matrix.

$H_{x,y} = \begin{pmatrix} \frac{\partial^{2} V}{\partial x^{2}} & \frac{\partial^{2} V}{\partial x\partial y } \\ \frac{\partial^{2} V}{\partial x \partial y } & \frac{\partial^{2} V}{\partial y^{2} } \\ \end{pmatrix}$

Upon finding the determinant (H). If H <0, it must be a saddle point, so to find the transition state, we just need to find the coordinates where this is the case.

Question 2: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.

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.

If we were to further set the momenta to zero for all 3 particles on the transition state, we know that the internuclear distances between the atoms will not change. This must hold true, as all the derivatives of the potential are zero; there are no forces acting on the particles. As an equilibrium point, a plot of internuclear distances vs time will be constant for the transition state.

This was found for $r_{12}$ = 90.775 pm. The internuclear distances vs time plot confirms this; it is shown below.

Question 3: Comment on how the MEP and the trajectory you just calculated differ.

The Minimum Energy Path (MEP) plots the reaction path assuming that the particles have negligible intertial motion. It therefore ignores the vibration potential of the system, and hence only runs along the "valley" of the PES, rather than oscillating along the potential well as it does in the dynamical calculation. This makes sense; it would be the lowest energy path, but ignores the momenta of the particles at each point, and how it changes.

The differences are shown in the following two images; both of them tend towards a complete reaction from the transition state as predicted, but the dynamics plot shows oscillation along the reaction path in the y-axis as it moves along.

This shows the reaction path in a 'Dynamics' plot at an initial AB distance =91.775, and BC = 90.775.

This shows the reaction path in a 'MEP' plot at an initial AB distance =91.775, and BC = 90.775.

Furthermore, the MEP reaction path finishes at 193.1pm as the AB distance and 74 pm as the BC distance, whereas for the dynamical plot, the reaction path will continue after this. This makes intuitive physical sense; both have opposite velocities so will tend away from each other. However, for MEP, with no inertial motion, this is discounted, so the particles cease motion at this point, as there is negligible difference in energy at this point.

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

Transition state theory can be used to determine the rate of a reaction, but it relies on 5 assumptions:

1. Once the kinetic energy has crossed the potential barrier, the reaction will proceed; there is no recrossing of the barrier.
2. There are no quantum effects {specifically, no quantum tunnelling will take place}
3. It is modelled on the Boltzmann distribution
4. ask
5. ask

Assumptions 1 and 2 are broken in these reaction dynamics, as we are considering classical collisions with some forces of attraction and repulsion between the atoms. As a result, despite the fact that crossing of the barrier can occur in a certain scenario, that activation barrier may be recrossed later. This is shown in the reaction pathway for the momenta p1 = -5.1 and p2 = -10.1, where despite the initial crossing of the barrier and consequent overcoming of the energy barrier, the system is not reactive. This therefore breaks from the mold of TST.

As TST suggests that some reactions that are not reactive are in fact reactive, it suggests that TST will overestimate how many reactions are successful. As a result, we can surmise that TST will overestimate the rate of reactions, compared to the GUI that we are using.

If the GUI agreed with assumption 1, TST would underestimate the reaction rate, as Tunnelling of the atoms would lead to a higher rate than the classical picture, as not all of the atoms have to reach the potential barrier to overcome it.

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 2: 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.

We can once again, further confirm that this is the saddle point by using the same point that the forces along AB and BC at this point are both zero.

Question 3: Activation energy

To find the activation energies of each product F-H and H₂ (2 mutually exclusive scenarios), we simply need to find the difference between the transition state energy and the energy of the products. This can be calculated for the reaction only (excluding vibration) by using MEP to identify the shortest route.

Due to the asymmetry of the atoms, we have 2 activation energies.

Starting from the $r_{ts}$where AB =184 pm and BC = 74.4 pm, with a perturbation of ± 2 pm in the AB axis, and a step count of 2500, we get these MEP results.

The plots above show the activation energy for the enothermic formation of the product F + H₂ . The plots below show the activation energy for the exothermic formation of the products F-H + H.

Product	Energy of Transition State ($kJmol^{-1}$)	Energy of product	ΔE
F-H + H	-433.991	-434.155	0.164
F + H2	-433.979	-556.895	122.916

This is because we know that the energy of the transition state is the maximum energy state on the reaction path, and doesn't consider oscillations along the potential well; so we can just use the MEP.

These results are consistent with our understanding of the system, and the contour plot; it is an extremely early TS for the exothermic formation of HF + H, so the activation energy for this reaction is extremely low compared to the reverse, endothermic process. This can be further shown by the contour plot; the difference in energy is shown to be almost negligible as it is along the same contour line.

Question 4: The release of Reaction Energy

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Question 5: Efficiency of reaction

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