MRD:EP01189257

EXERCISE 1: H + H2 system

Dynamics from the transition state region

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

$\frac{\partial^2V}{\partial r_1\partial r_2} = 0$ where V is the potential energy.

The transition state corresponds to a saddle point on the potential energy surface.

We first note that at a stationary point (x₀, y₀) we have $\frac{\partial V}{\partial r_1} = 0$ and $\frac{\partial V}{\partial r_2} = 0$

and if $\frac{\partial^2V}{\partial r_1\partial r_1}\frac{\partial^2V}{\partial r_2\partial r_2} - (\frac{\partial^2V}{\partial r_1\partial r_2})^2 < 0$, we have a saddle point at (x₀, y₀) on a potential energy surface

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 the maximum on the minimum energy path linking reactants and the products.

Local minimumː If one starts a trajectory exactly at a local minimum with no initial momentum, it will remain there forever. If one changes the geometry by a small amount, the system will revert back to its original state (local minimum) if the change is small enough.

Transition stateː Conversely, if one starts a trajectory exactly at the transition state, with no initial momentum, it will remain there forever. However, if one changes the geometry by an infinitesimal amount in the direction of the products it will roll towards the products (and similarly for the reactants).

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 trajectoryː

r_ts = 0.907742 Å

As observed from the above screenshot, forces along AB and BC = 0, hence the system remains at the transition state without oscillating.

The internuclear distances (A-B, B-C and A-C) are constant over time, meaning that there is no change in geometry with respect to time, indicating a transition state position.

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

Both mep and the trajectory eventually resulted in H₁+ H₂-H₃, where there is a steady increase in distance between H₁ and H₂.

MEP: There was no oscillation of the H₂-H₃ bond in mep due to the fact that momenta/velocities are always reset to zero in each time step (as shown in below).

The trajectory: On the other hand, the trajectory involved a oscillation of the H₂-H₃ bond (as shown in below).

MEP for r₁ = r_ts + 0.01, r₂ = r_ts

Trajectory for r₁ = r_ts + 0.01, r₂ = r_ts

What would change if we used the initial conditions r₁ = r_ts and  r₂ = r_ts + 0.01 instead?:

At t = ∞	r1 =  rts + 0.01 and  r2 = rts	r1 = rts and  r2 = rts + 0.01
r1/ Å	∞	oscillates at 0.73
r2/ Å	oscillates at 0.73	∞
p1/ kgÅs-1	2.5	oscillates at 1.2
p2/ kgÅs-1	oscillates at 1.2	2.5

The system will end up with the formation of H₁-H₂ instead of H₂-H₃ bond (as shown below). For both mep and trajectory, values of r₁ and r₂ were swapped and so did p₁ and p₂. In summary, the reaction goes the opposite direction.

Trajectory for r₁ = r_ts, r₂ = r_ts + 0.01

Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed. What do you observe?:

By principle of microscopic reversibility,¹ the opposite pathway occurs where the system returns back to the transition state. However, note that the system did not end up exactly at the transition state because the initial conditions (final conditions from previous calculations) used were not exact.

Reactive and unreactive trajectories

p1	p2	Etot	Reactive?	Description of the dynamics
-1.25	-2.5	-99.018	Yes	The two molecules collide to form products (H1 and H2-H3), resulting in vibration of H2-H3 bond and the two products drifting away from each other
-1.5	-2.0	-100.456	No	The two molecules did not have enough kinetic energy, hence the system reverts back to the reactants (H1-H2 and H3), which drift away from each other
-1.5	-2.5	-98.956	Yes	The two molecules collide to form products (H1 and H2-H3), resulting in vibration of H2-H3 bond and the two products drifting away from each other
-2.5	-5.0	-84.956	No	In this interesting trajectory, the two reactants 'react' to form transient products (H1 and vibrating H2-H3). However, the system then reverts back to reactants (vibrating H1-H2 and H3).
-2.5	-5.2	-83.416	Yes	In this interesting trajectory, the two reactants 'react' to form products (H1 and vibrating H2-H3). However, the system then reverts back to reactants (vibrating H1-H2 and H3) for a limited amount of time. Consequently, the system resulted in products (H1 and vibrating H2-H3).

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?:

Reactants require enough kinetic energy to overcome the activation barrier to reaction. However, one must note that energy is not the only factor that can predict whether the trajectory is reactive. In the 2nd last example, despite its total energy being considerably higher than some of the other trajectories, it did not result in a reaction. Therefore, one must note that the success of a reaction depends not only on kinetic energy, but also on the momentum of the individual reactants.

State what are the main assumptions of Transition State Theory:

Main assumptions:¹

1. Electronic and nuclear motions are independent of each other (analogous to the Born-Oppenheimer approximation).
2. The distribution of reactant molecules among the states correspond to the Boltzmann distribution.
3. Once the molecular system crosses the transition state in the direction of products, it cannot turn around and reform reactants (fundamental assumption of TST).
4. In the transition state, motion along the reaction coordinate may be decoupled from the other motions and regarded classically as a translation.
5. Quasi-equilibrium between transition state and reactants

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

The third assumption which forbids the 'recrossing' of the activation barrier and hence assumes that all reactants that reach the transition state will become product. However, based on the results obtained, it can be observed that not all trajectories that reach the transition state will result in a reaction because some may revert back to reactants, which Transition State Theory fails to take into account. A consequence of this is that TST overestimates the reaction values hence one would expect the experimental values to be lower.

PES inspection

By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic):

F + H₂ → HF + H: Exothermic

H + HF → H₂ + F: Endothermic

How does this relate to the bond strength of the chemical species involved?:

The H-F bond (565 kJmol^-1) is stronger than the H-H bond (432 kJmol^-1) because of the polarization of charges in H and F resulting in stronger electrostatic attraction between them.²

Locate the approximate position of the transition stateː

Report the activation energy for both reactionsː

Reaction	ratom/ Å (distance between reactants)	rdiatomic/ Å (distance between atoms in diatomic molecule)	Activation energy/ kcal mol-1
F + H2 → HF + H	1.8107	0.74488	0.258
H + HF → H2 + F	0.74488	1.8107	30.255

Reaction dynamics

Identify a set of initial conditions that results in a reactive trajectory for the F + H₂, and look at the “Animation” and “Momenta vs Time”:

rFH/ Å	rHH/ Å	pFH	pHH
2.0	0.74	0	-3

As seen from the figure below, the trajectory is reactive

Reaction trajectory of F + H₂

In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy:

Momenta vs. Time of F + H₂

Energy vs. Time of F + H₂

As seen from the two graphs above:

1. At the start of the reaction (t = 0 s):

• non-zero momentum for H-H (both translational and vibrational)
• zero momentum for F
• kinetic energy and potential fluctuate due to oscillation in H-H

2. At the transition state (t ~ 2.0 s): Two reactants meet at the transition state at before the momentum is redistributed to the two products (H and HF.

3. After the reaction:

• non-zero momentum for H (only translational)
• non-zero momentum for HF (both translational and vibrational)
• kinetic energy and potential fluctuate kinetic energy and potential fluctuate at larger amplitudes due to oscillation in H-F instead of H-H

In other words, translational and vibrational energy of H₂ (reactant) has be released to vibrationally excite HF (product), resulting in:

1. a vibrationally and translationally excited HF
2. a translationally excited H

Explain how this could be confirmed experimentally:

Infrared spectroscopy can be used to obtain the IR spectrum of HF (IR-active 4460 cm^-1).³ On the other hand H₂ is not IR-active (no change in dipole moment) but is Raman active (change in polarizability) hence a Raman spectrum of H₂ can be obtained. These spectra would shed light on the vibrationally excited states of HF and H₂, and hence indicate transfer of vibrational energy from H₂ to HF.

In addition, calorimetry could measure the heat change of reaction.

Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state:

References

1. Steinfeld, J. I., Francisco, J. S. & Hase, W. L. Chemical kinetics and dynamics. (Prentice Hall, 1999).

2. Libretexts. Bond Energies. Chemistry LibreTexts (2019). Available at: https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Chemical_Bonding/Fundamentals_of_Chemical_Bonding/Bond_Energies. (Accessed: 6th May 2019)

3. Hydrogen Fluoride, HF. Vibrational Modes of Hydrogen Fluoride Available at: http://www.chem.purdue.edu/gchelp/vibs/hf.html. (Accessed: 6th May 2019)