MRD:sj2814

Molecular Reaction Dynamics: Application to Triatomic Systems

Exercise 1: H + H ₂ 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.

The potential energy surface (PES) has a total gradient of 0 at the minimum and the transition state structure. The second derivative of the minimum will be a value greater than 0 while the second derivative of the transition state will be lower than 0. This is because the transition state is the maximum on the minimum energy path.

Nf710 (talk) 10:48, 26 May 2017 (BST) This needs to be in the reaction coordiate. You do this by changing the basis of corinadte system by a linear combination of the current coordinates. (this is what Guassian does)

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.

The length of bonds AB and BC would be equal in the transition state. This is observed when the internuclear distance for both is 0.9176 Å at 0.38 time units (Figure 1). To calculate a more accurate r_ts, momentum of AB and BC were set to 0 and distance was set to 0.9176 Å. The equilibrium line of the oscillating internuclear distance of AB and BC was observed to be 9.078 Å. Further simulations were carried out in which the resulting equilibrium values were set as the distance until a constant internuclear distance was found at 0.90775 Å (Figure 2). This figure was also confirmed through the animated geometry.

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

The dynamics calculations indicate vibrational motion of the atoms through internuclear distance oscillations on the graph whereas the mep does not show this and follows a linear curve instead. Furthermore, hydrogen A approaching and hydrogen C moving away are slower in the mep; at 2.5 time units, bond distance of AC is 1.99 Å in the mep calculations and 9.745 Å in the dynamics calculations.

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	Reactive/Unreactive	Trajectory	Figure
-1.25	-2.5	Reactive	BC internuclear momentum is greater overall than AB internuclear momentum. Hydrogen A has sufficient kinetic energy to proceed over the transition state and displace hydrogen hydrogen C hence this is a reactive trajectory.	Figure 3
-1.5	-2.0	Unreactive	BC internuclear momentum is not enough to overcome the activation barrier as is demonstrated by a lower kinetic energy. This is the reason why hydrogen A approaches hydrogen B but is unable to displace hydrogen C hence this is an unreactive trajectory.	Figure 4
-1.5	-2.5	Reactive	Hydrogen A possesses enough momentum to allow the system to proceed over the transition state and vibrational kinetic energy is observed after circa 4.5 time units hence this is reactive.	Figure 5
-2.5	-5.0	Unreactive	Hydrogen A has enough kinetic energy to partially displace hydrogen C twice i.e. the transition state has been achieved twice however eventually, BC regains enough internuclear momentum to cause hydrogen A to be pushed away as is shown by the oscillations of BC as opposed to the steady AB. This is known as barrier recrossing and the trajectory is unreactive.	Figure 6
-2.5	-5.2	Reactive	Hydrogen A approaches and initially has insufficient kinetic energy to cause hydrogen C to leave completely. The two oscillate around hydrogen B a couple of times before hydrogen C leaves. This is a slower but reactive trajectory.	Figure 7

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

If initial positions correspond to the final positions from the calculations done so far and the same momenta values are used but with reverse signs, the backward reaction would occur. Hydrogen C would approach hydrogen B as hydrogen A moves away.

State 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?

Transition state theory assumptions:

Once the transition state has been achieved, the reactants will result in the products.
The transition state will be a higher energy species than the reactants or the products.
Each step is slow enough to reach a certain Boltzmann distribution before proceeding to the next intermediate.
All atomic nuclei behave according to classical mechanics, not quantum mechanics.
The reaction will proceed through the lowest energy saddle point on the PES.

Due to these, reaction rate predictions are likely to be lower than experimental observations.

Exercise 2: F - H - H system

PES Inspection:

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?

F + HH --> HF + H (Figure 8):

The PES graph where A is fluorine while B and C are hydrogen atoms; F + HH --> HF + H. Distance AB was set to 1.68 Å and distance BC to 0.76 Å with both momenta at 0. From the local minima, this demonstrates that the HF bond releases energy when formed which means this is an exothermic process.

H + HF --> HH + F (Figure 9):

The PES graph where C is fluorine while A and B are hydrogen atoms; H + HF --> HH + F. Distance AB was set to 2.00 Å and distance BC to 0.92 Å with both momenta at 0. From the local minima, this demonstrates that the HF bonds intake energy when broken which means this is an endothermic process.

HF has a lower potential energy implying that more energy is released and it is a stronger bond than HH. The values observed correspond to expected values. This is because HF is bonded ionically whereas HH is bonded covalently and ionic bonding is stronger than covalent bonding. This corresponds with literature values which state that the bond dissociation enthalpy of HF is 565 kJ mol^-1whereas HH is 436 kJ mol^-1 at 298 K.^[3]

Locate the approximate position of the transition state.

Hammond’s postulate states that in an exothermic reaction like this one, the transition state will be closer in energy to the reactants thus the will resemble the reactants. For F + HH --> H + HF, the transition state will resemble HH and for H + HF --> F + HH, the transition state will resemble HF. The approximate values lie at 0.75 Å for HH and 1.68 Å for the internuclear distance between HF (Figure 9b).

Report the activation energy for both reactions.

Figure 10a and 10b show that the saddle point is at -133.9 kcal/mol and when the corresponding values are substituted in to give a potential energy vs time graph, the transition state can be observed to be at -103.8 kcal/mol. This indicates that the activation energy for H + HF --> F + HH is +30.1 kcal/mol.

Reaction Dynamics:

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

Figure 11 shows that as time progresses, vibrational energy increases for F + HH --> H + HF. Following that, considerations regarding rotational energy and translational energy can be made. There is no change in rotational energy as the angle of attack is linear. From this, it can be inferred that translational energy is likely to increase also. The exothermic nature of the reaction means energy must be converted to either translational or vibrational energy. This can be confirmed using IR spectrometry and calorimetry.

Nf710 (talk) 10:58, 26 May 2017 (BST) This is not quite correct. You start with initial kinetic (translational & vibrational). you go over the TS (you may have extra kenetic) due to the exthermic nature you have lots of potential energy at this point, which gets converted into translational and any excess vibrational. (Almost almost always excess).

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.

Of the two reactions occurring here, F + HH --> H + HF will be more exothermic than H + HF --> F + HH hence will have an earlier transition state. According to Polyani's rules, this means that this reaction would be a result of greater translational energy. Similarly, for H + HF --> F + HH, there is a later transition state which indicates that this reaction is the result of greater vibrational energy. [4]

Nf710 (talk) 10:59, 26 May 2017 (BST) You should have proved this with some examples from the HF system

References

P. Atkins & J. de Paula, Atkins’ Physical Chemistry, Oxford University Press, 2006, Chapter 24, page 880
T. Fueno, The Transition State: A Theoretical Approach, Gordon and Breach Science Publishers, 1999, Chapter 1, page 1 – 3
P. Atkins & J. de Paula, Atkins’ Physical Chemistry, Oxford University Press, 2006, Data Section, page 1011
Z. Zhang, Y. Zhou, D. Zhang, G. Czako, J. Bowman, J. Phys. Chem. Lett., 2012, 3, page 3416-3419