MRD:cyy113

EXERCISE 1: H + H₂ system

What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

At both the minimum and the transition structure, it can be observed that the gradients are zero. Mathematically, $\frac{\partial V}{\partial r_{AB}}=0$. The transition structure has the highest potential energy and lies on a maximum point on this curve, while the minima has the lowest potential energy and lies on the minimum point. They can be mathematically distinguished by taking the second derivative $\frac{{\partial }^{2}V}{\partial r_{AB}^2}$. $\frac{{\partial }^{2}V}{\partial r_{AB}^2}>0$ and $\frac{{\partial }^{2}V}{\partial r_{AB}^2}<0$ for a minimum and maximum point respectively.

Report your best estimate of the transition state position (r_ts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” screenshot for a relevant trajectory.

To determine the point corresponding to the transition structure, the following properties of the transition structure were considered:

1. The potential of the system is at a maximum point.
2. The transition state lies along the reactive trajectory (black line)
3. The molecule vibrates negligibly.

Hence, using the data cursor, the point where trajectory intersects within the region of the maximum point on the potential energy surface was estimated to be 0.9066 Angstroms. (fulfilling conditions 1 and 2). The intersection occurred near a maximum point and this was was largely expected as the transition state is defined as a structure with the highest potential energy. (Figure 1)

<insert figure 1>

In order to obtain a more accurate value, the AB distance and BC distances were first set to be 0.9066 Angstroms, and the momentum was set to be 0, since at the transition structure all atoms A, B and C remain stationary. AB and BC distances were optimized until the internuclear distance was constant with time in order to fulfill condition 3. (Figure 2) The analysis shows that r_ts was best optimised to be 0.90775 Angstroms.

<insert figure 2>

Analysis of Figure 2

r_AB = r_BC = 0.90775 and r_AC = 1.185. It was expected that r_AC is twice of r_AB or r_BC, since atoms A, B and C were in contact with each other in the transition structure and all atoms A, B and C are hydrogen atoms. The blue r_AB line could not be observed as it overlapped with the red r_BC line.

In addition, the animation showed that the oscillation of AB, BC and AC bond distances with time were so slow that it was not possible to observe a change in the values. (Figure 3)

<insert figure 3>

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

<insert and expand on explanation> reference: https://books.google.co.uk/books?id=rS3yCAAAQBAJ&pg=PA120&lpg=PA120&dq=minimum+energy+path+no+vibration&source=bl&ots=mQWXOYWl1X&sig=WIoM2ZZj-UQzCi5xu2tpZ3nJIVw&hl=en&sa=X&ved=0ahUKEwjoibvXwfnTAhWrK8AKHfNnBUYQ6AEILTAA#v=onepage&q=minimum%20energy%20path%20no%20vibration&f=false

In the MEP calculation, vibrational energies are assumed to flow in and out of the system during the reaction such that it remains zero throughout. The dynamic calculation showed an oscillating trajectory while the MEP calculation showed a non-oscillating one. This was because as the central atom B vibrates, r_AB and r_BC changes in an inverse relationship. (i.e. if r_AB increases then r_BC decreases, and vice-versa).

In addition, the range of the trajectory was markedly shorter for the same step size of (value). This suggests that in the MEP calculation, each step size corresponds to a smaller fraction of the molecule's trajectory. The presence of vibrational energies in a dynamic calculation makes it hard to determine accurately the activation energy. Running an MEP calculation removes these vibrational energies and greatly simplifies the determination of activation energy.

<insert figure 4 and 4A>

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

The same shape is obtained, but AB becomes BC and vice versa.

• 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?

This can be envisioned as the backward reaction. In this case, the enthalpy of reaction is still zero and the activation energy remains the same since there the reactants were the same as products.

• For the initial positions r₁ = 0.74 and r₂ = 2.0, run trajectories with the following momenta combination:

Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory.

p1	p2	Trajectory	Internuclear distance with Time	Animation
-1.25	-2.5	Fig 6A	Fig 7A	Fig 8A
-1.5	-2.0	Fig 6B	Fig 7B	Fig 8B
-1.5	-2.5	Fig 6C	Fig 7C	Fig 8C
-2.5	-5.0	Fig 6D	Fig 7D	Fig 8D
-2.5	-5.2	Fig 6E	Fig 7E	Fig 8E

State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

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 while the H + HF reaction (backward reaction) is endothermic, as can be observed in the potential surfaces below. This is expected, as the H-F bond includes ionic contributions not found in H-H bonds as a result of the electronegativity of the F atom.

Locate the approximate position of the transition state.

Since the F atom is highly reactive, its energy is extremely close to the transition state energy, hence it was difficult to determine a rough estimate of rHH and rHF at which the transition state occurred visually. Using the data cursor, it was observed that the energy remained around -103.7 kcal/mol before decreasing further. It was likely that the energy levels were so close that the potential energy surface was not able to resolve the minor differences. Hence, to obtain a rough estimate of rHH and rHF in the transition state, the maximum value of rHH and rHF was chosen such that the energy still remained at -103.7 kcal/mol but anymore deviations would have caused the energy to decrease further. The set of values were determined to be rHF=1.82 and rHH=0.74.

Next, this set of values were optimised by plotting the internuclear distance vs time in an MEP calculation. At the transition state, the internuclear distance does not vary with time and a straight horizontal line will be observed. The values were optimised within this region and found to be rHF=1.8105 and rHH = 0.745. This agreed with the Hammond's Postulate, which states that for an exothermic reaction, the transition state resembles the products. it can be observed that rHH=0.745 was extremely close to the initial rHH set at 0.74.

Report the activation energy for both reactions.

Using this optimised set of parameters of rHH and rHF, the potential energy vs time curve was plotted in an MEP calculation. Removal of vibrational energy near the transition state allows the activation energy to be determined simply without the complication of energy fluctuations due to vibrations. The activation energies were determined to be 0.21 and 30.1 kcal/mol respectively for the forward and backward reaction. It is noted that the step size used in the forward reaction calculation was much higher and rHF was slightly increased to 1.8305.

Identify a set of initial conditions that results in a reactive trajectory for the F + H₂, and look at the “Animation” and “Internuclear Momenta vs Time”. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?

In comparison to the initial H + H2 reaction, F is more reactive than the H atom. As such, rHF and momentum of HF was reduced in comparison to rHH and momentum of H2 in order to prevent barrier recrossing past the transition state. Indeed, it can be observed that the reaction occurs with the following set of parameters: rHH = 0.745, rHF = 1.55, pHF = -0.4, pHH = -1.2.

Because energy must be conserved, the excess reaction energy for an exothermic reaction will be released into exciting the products into higher translational, vibrational, rotational and electronic energy levels. The populations of the products in these levels can be calculated via the partition functions as follows:

(INSERT PARTITION FUNCTIONS)

Translation is the only mode that gives rise to kinetic energy. As such, the average change in translational energy can be simply measured as the change in temperature of the system. Vibrational and rotational energy levels result in the internal energy of the species and any changes can be measured via rovibrational spectroscopy in the IR region. Electronic energy levels can be measured in an emission spectrum within the UV/vis region. While there are reactions which gave rise to an electronically excited product (i.e. chemiluminescent reactions), in this H+H2 or F+H2 reaction it is unlikely to occur, since the energy gaps between the low-lying orbitals (1s to 2s and 2p to 3s for H and F respectively) is too great.

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.

The effect of translational and vibrational energy on a reaction can be explored through Polanyi's empirical rules which apply when an atom attacks a diatomic species. In this set of rules, if a reaction has a late transition state, it will be promoted by a high vibrational energy and a low translational energy, and the reverse is true. Understood in terms of Hammond's Postulate, an exothermic reaction (hence a late transition state resembling the products) is more likely to proceed if it has a high vibrational and low translational energy. The reactions F + H2 and H + HF studied in this example can be used to confirm if Polanyi's empirical rules in the table below:

Reaction	rHF	rHH	pHF	pHH	Trajectory	Comments
F + H2					13
F + H2					A
H + HF					B
H + HF					C

References:

http://pubs.acs.org/doi/pdf/10.1021/jz301649w

https://www.dartmouth.edu/~chemlab/chem6/hspect/full_text/chemistry.html

https://en.wikipedia.org/wiki/Emission_spectrum

https://en.wikipedia.org/wiki/Energy_level

http://teacher.pas.rochester.edu/phy121/lecturenotes/Chapter18/Chapter18.html

http://pubs.acs.org/doi/pdf/10.1021/j100349a009