Exericse 1: H + H₂ System

Triatomic Systems

In a triatomic system where an atom, A, reacts with a diatomic molecule, BC, the distance between the atoms A and B, r₁, decreases as A approaches BC, with the internuclear distance between B and C, r₂, remaining effectively constant, barring vibrational oscillations.^[1] Upon a successful collision, r₁ will be constant due to bond A-B being formed, and r₂ will increase as bond B-C is broken and C moves away, as shown in Figure 1.^[1] The system is collinear, with the bond angle between the two bonds A-B and B-C being fixed at 180°.^[2] The reactant and product molecules are at potential energy minima, with potential energy therefore being dependent on the interatomic distances between A and B, and, B and C.

**Figure 1 -** Triatomic system, where atom A reacts with the diatomic molecule BC. (Diagram adapted from Imperial College London, https://wiki.ch.ic.ac.uk/wiki/index.php?title=CP3MD, (accessed May 2018.)^[1]

Potential Energy Surfaces

A potential energy surface is a visual representation of the relationship between the relative positions of the reactants and products and the potential energy of the system.^[1] The reaction path, or trajectory, represents the potential energy and internuclear distances as a function of the reaction coordinate and so depicts the changes in these quantities as the reaction proceeds.

The potential energy surface in Figure 1 was calculated for the reaction system H + H₂ using the following parameters:

Calculation Type	Steps	r₁ / Å	r₂ / Å	p₁	p₂
Dynamics	500	2.30	0.74	-2.7	0.0

The reaction path passes through each potential energy minimum, including the transition state, the saddle point at which the potential energy is at its maximum (i.e. the highest energy minimum), as shown in Figure 2, and contains components of both internuclear distances.^[1]^[2]

At each local minimum for the reactant or product, the first derivative with respect to the appropriate axis, ∂V(r₁)/∂r₁, for the products or ∂V(r₂)/∂r₂, for the reactants will equal 0 and its second derivative ∂V²(r₁)/∂²r₁ or ∂V²(r₂)/∂²r₂ will be negative.^[1] However, at the saddle point representing the transition state, there is a maximum point along the direction of the reaction path (at a 45° angle to the axes), but a minimum point in the orthogonal axis, hence the second derivative for the maximum will be negative.^[1]

Ng611 (talk) 15:23, 1 June 2018 (BST) Excellent definition of a TS. A diagram would make this definition even clearer!

It should also be noted that the reaction pathway is straightest in the region centered around the transition state as this is the point at which bonds are in the process of being broken and formed, hence a lesser degree of association and therefore minimal vibrational oscillations.

Ng611 (talk) 15:23, 1 June 2018 (BST) Interesting observation, as this is a key assumption of TS theory (that motion within the TS can be treated as purely translational).

Locating the Transition State

The best estimate for of the transition state position is r_ts = 0.9077 Å. The transition state is symmetric, reflecting the symmetrical potential energy surface, and so the two internuclear distances, r₁ and r₂, must be equal to achieve the transitions state.^[1] When each species is given a momentum of 0, the potential energy and the internuclear distances are all fixed, as indicated by straight horizontal lines across time in Figure 3, with no vibrational oscillations and fluctuations in potential energy observed.

The reaction path is therefore a single point on the potential energy surface which sits on the saddle point at the transition state, with no movement towards reactants or products shown in Figure 4.

Calculating the Reaction Path

Determination of the transition state allows the minimum energy path, or MEP, to be determined.^[1]

The reaction path starting from a slight displacement from the transition state (r₁ = r_ts + 0.01 and r₂ = r_ts) was calculated using the following parameters:

Calculation Type	Steps	r₁ / Å	r₂ / Å	p₁	p₂
MEP	50 000	0.9087	0.9077	0	0

The potential surface plot, with the MEP, is shown in Figures 6 and 7:

The calculation type was changed to dynamics and the same initial conditions applied to include changes in momentum and fluctuations in energy and obtain a more realistic model of the behaviour of the species during the reaction.^[1]

Calculation Type	Steps	r₁ / Å	r₂ / Å	p₁	p₂
Dynamics	500	0.9087	0.9077	0	0

The resultant potential surface plot is shown in Figures 8 and 9:

The MEP does not show vibrational oscillations, simply following the bottom of the channel and passing straight through each potential energy minimum with no consideration of how the momentum (namely the velocity) of the species changes during the course of the reaction. The calculated trajectory, however, does take into account momentum changes and oscillates about the minima. Moreover, many more steps (a factor of roughly 100 times more) are required in the MEP calculation for the reaction to go to completion and produce stable products than in the dynamics calculation.

The final values of the positions, r₁(t) and r₁(t), and the mean momenta after a longer period of time, t, p₁(t) and p₁(t) are listed below:

r₁(2.46412) / Å	r₂(2.46412) / Å	p₁(2.46412)	p₂(2.46412)
7.65947	0.75424	2.48490	1.25338

The internuclear distances vs time and internuclear momentum vs time graphs from which the positions and momenta were obtained respectively are given in Figure 10 and 11:

These plots also help to describe the reaction, with distance r₁ (and A-C) seen to increase as A moves away from BC, after a period of time at which the positions are effectively constant, and B-C decreasing slightly before plateauing, with the presence of oscillations signalling that the bond B-C has formed and vibrations can now occur.

The internuclear distances vs time and internuclear momentum vs time graphs corresponding to the MEP calculation are shown in Figures 12 and 13.

As the MEP moves towards the structure of the reactants, the internuclear distance r₁ increases rapidly and r₂ decreases rapidly before tending towards a limit and plateauing, while the system also has 0 momentum at all times adue to the velocity being reset to 0 at each step.^[1]

If the initial conditions were r₁ = r_ts and r₂ = r_ts + 0.01 instead, the reaction path would move in the opposite direction, exiting via the other channel formed by the potential energy minima where distance AB is constant and distance BC increases, as shown in Figure 14. In other words, while in the first case the transition state moves towards the reactants A + BC, the opposite would be true if the displacement was instead such that A started off closer to B than C was to B, favouring the formation of products AB + C. (Note that in this section, the terms 'reactants' and 'products' have been used with respect to the forward reaction as described by Figure 1.)

As the involved atoms are identical, in the case of the inversely distorted transition state, r₁ = r_ts and r₂ = r_ts + 0.01, r₁(t) would take the exact value of r₂(t) in the previous calculation and vice versa, with the graph for A-B appearing as the original B-C and B-C appearing as A-B. The same effect would also be observed in the momenta plots, as shown in Figures 15 and 16:

A calculation was carried out where the initial positions were set to the final positions determined above and the same final momenta with reversed signs were given to the species as follows:

Calculation Type	Steps	r₁ / Å	r₂ / Å	p₁	p₂
Dynamics	500	7.65947	0.75424	-2.48490	-1.25338

The potential energy surface, internuclear distances vs time and internuclear momentum vs time plots produced are shown in Figures 17-19.

In this scenario, the species start off as reactants and have just the right starting positions and momenta to reach the slightly displaced transition state. This process is the reverse of the original situation in which the slightly displaced transition state formed the reactants. However, a perfect symmetrical relationship between the cases is not actually observed, with the reaction path reaching the transition states but still having enough energy for further interactions to occur, resulting in a slight turn back towards the reactants. This can be attributed to errors in determining the final positions and mean momenta from the internuclear distances vs time and internuclear momenta vs time graphs repectively, as well as rounding.

Reactive and Unreactive Trajectories

Calculations where r₁ = 0.74 and r₂ = 2.0 with various combination of momenta as shown below were run to determine the conditions, namely range of momenta, required for a successful reactive trajectory.

p₁	p₂	Total Energy / kcal	Reactive Trajectory?	Comments
-1.25	-2.5	-99.018	Yes	The trajectory is reactive as the reaction path can get over the maximum at the transition state and moves into the product channel at which point the products are well defined. The straight trajectory centred around the MEP shows that molecule AB does not vibrate until C collides into it and provides it with sufficient momenta to result in the subsequently vibrational oscillations of molecule BC past the transition state.
-1.5	-2.0	-100.456	No	The trajectory is not reactive as the reaction path does not move past the transition state, turning back before the transition state once the potential energy exceeds the kinetic energy of the species as determined by the given momenta.
-1.5	-2.5	-98.956	Yes	The trajectory is reactive in a similar way to the first case, although a higher initial momentum for AB results in larger vibrations in the diatomic molecule AB.
-2.5	-5.0	-84.956	No	The trajectory is not reactive, despite the initial momenta being large enough for the reaction path to attain the transition state, A-B-C, and cross the activation energy barrier twice to eventually end up as reactants. Once past the transition state, C moves away from A and B as its initial momentum is not large enough to form stable products and the reaction path re-crosses the transition state back into the reactant channel, forming AB + C. Vibrational oscillations are much larger due to the larger initial momentum of AB.
-2.5	-5.2	-83.416	Yes	The trajectory is reactive, with the scenario analogous to that previously, except that the momenta and kinetic energies of the reactive species are now large enough such that three crossings of the activation energy barrier are possible and B can be transferred onto C to form stables products A + BC, with the reaction path travelling down the products channel.

Transition State Theory

Transition state theory assumes that only the reactants, states of the system on one side of the activation barrier, are kept at equilibrium with the configuration at the transition state.^[3] Once the transition state is attained, the reaction path will proceed to form stable products irreversibly and with certainty, with only a single crossing of the activation energy barrier in the forward direction possible.^[3] The rate of the reaction, or the rate constant, is therefore directly proportional to the rate at which colliding molecules reach and pass through the transition state.^[3] As shown above, this is only an approximation since the barrier can be recrossed if the particles have sufficiently large momenta and thus kinetic energy and only when the reaction path is well past the transition state and the product species are well separated from each other can the reaction be deemed to have been successful.^[3]

Transition state theory is also does not take into account observed quantum effects such as tunneling through the activation energy barrier, which avoids having to travel through the saddle point.^[4]

Exercise 2: F-H-H System

PES Inspection

The energetics of the reactions F + H₂ and the reverse H + HF were investigated, with the following properties observed:

Reaction	Transition State Energy / kcal mol^-1	Reactants Energy / kcal mol^-1	Activation Energy / kcal mol^-1	Classification
F + H₂	- 103.753	- 133.945	+ 30.192	Exothermic
H + HF	- 103.744	- 103.987	+ 0.243	Endothermic

F + H₂ is exothermic as the H-F bond is stronger than the H-H bond by virtue of the higher electronegativity of fluorine relative to hydrogen (Pauling Electronegativities of: F = 3.98 and H = 2.20).^[5] The fluorine atom in H-F pulls electrons closer to itself, leading to a highly polarised H-F bond and a strong ionic contribution to the bonding strength as well as increased orbital overlap. The H-F bond is therefore stronger than the purely covalent H-H bond, (bond dissociation energies of: H-F = + 567 kJ mol¹ and H-H = + 436 kJ mol¹.^[6] The formation of the H-F bond in the forward reaction F + H₂ is therefore energetically favourable and overcompensates for the loss of the H-H bond, with an exothermic release of energy.^[5] The reverse reaction, H + HF, is endothermic and so requires the input of energy as the energy released by the formation of the H-H bond does not outweigh the energy used to break the H-F bond.^[5]

According to Hammond's postulate, the transition state between two interconverting structures will most closely resemble the structure it is closer in energy to.^[7] For an exothermic reaction, the transition state would be early and closely resemble the reactants, therefore having only a marginally higher energy than the reactants, i.e. a small activation energy.^[7]

Calculation Type	Steps	r_H-F / Å	r_H-H / Å	p_HF	p_HH
Dynamics	500	1.811	0.745	0	0

The approximate position of the transition state for both F + H₂ and H + HF, was found to be r_H-F = 1.811 Å and r_H-H = 0.745 Å.

The potential surface plot in Figure 20 and energy vs time plot in Figure 21 shows that the transition state has been attained, with no change in potential energy as expected at the saddle point and no movement towards either products or reactants. The potential surface plot also shows the minima in the reactants channel to be higher in energy than the products channel, indicating an exothermic reaction in the event of a reactive trajectory.

The position of the transition state approaching from the reverse reaction H + HF is shown in Figure 22. The relative distances between the atoms at the transition state are the same, but in this direction, the transition state closely resembles the products, which are now higher in energy than the reactants, hence an endothermic reaction.

The following MEP calculations were set up and run from a structure neighbouring the transition state to determine the activation energy of both reactions:

Calculation Type	Steps	r_H-F / Å	r_H-H / Å	p_HF	p_HH
MEP	200 000	1.801	0.745	0	0

Calculation Type	Steps	r_H-F / Å	r_H-H / Å	p_HF	p_HH
MEP	200 000	1.821	0.745	0	0

Figure 23 shows the MEP from the distorted transition state favouring the products, while Figure 24 shows the change in potential energy against time from near the transition state to the products.

As the transition state is much closer to the reactants, the decrease in energy is very large and occurs very quickly.

Figures 25 and 26 show the MEP from the transition state favouring the reactants and the change in potential energy against time from the transition state to the reactants.

There is much less of a potential energy gradient to a stable configuration this time, hence a smaller decrease in energy over a longer period of time.

Reaction Dynamics

A calculation with the following conditions resulted in a reactive trajectory for the F + H₂ reaction, as shown by Figure 27:

Calculation Type	Steps	r_H-F / Å	r_H-H / Å	p_HF	p_HH
Dynamics	1400	2.39	0.74	-1	-0.1

The plot of internuclear momentum vs time in Figure 28 reveals that the H-H bond vibrates until the fluorine atom collides with and transfers its momentum to the H₂ molecule. Once past the transition state and stable products are formed, it is the H-F bond that vibrates, but with a larger amplitude of oscillation than H-H. The gradient in internuclear distance against time also increases, hence increased velocity and translational energy. These observations indicate an increase in kinetic energy of the system which, in accordance with the law of conservation energy, must have been converted from potential energy. This is demonstrated by Figure 29, which shows a decrease in potential energy but a gain in kinetic energy of the system from the products to the reactants at the transition state.

Such energy changes can be observed and quantified by calorimetry as increased average motion of particles equates to an increase in temperature, with the loss of potential energy corresponding to enthalpy loss of the system as expected for an exothermic reaction.^[5]

Ng611 (talk) 15:25, 1 June 2018 (BST) True, but would calorimetry distinguish between vibrational KE and translational KE?

The Polanyi Rules

The Polanyi rules state for reactants given momentum along the direction of the reaction path, translational energy, as opposed to vibrational energy, is preferred to overcome the early transition state in an exothemic reaction.^[8] The opposite is true for endothermic reaction with late transition states, with sufficient vibrational energy over translational energy resulting in the successful formation of stable products.^[8]

Trajectories were calculated for the exothermic reaction F + H₂ using 1000 steps and different values of momentum for HH, with all other conditions fixed such that only the vibrational energy of the H₂ molecule was varied.

r_H-F / Å	r_H-H / Å	p_HF	p_HH	Reactive Trajectory?	Comments
2.3	0.74	-0.5	-3	No	The H₂ molecule has a large vibrational energy and crosses the transition state but does not form stable products as the path doubles back after entering the products channel, and forms stable reactants.
2.3	0.74	-0.5	-1	No	The H₂ molecule has a smaller vibrational energy than in the previous case but does not reach the transition state, turning back towards the reactants whil still in the reactants channel.
2.3	0.74	-0.5	0	No	The H₂ molecule has no momentum given to it and so has vibrational oscillations of a very small amplitude. The transition state is not attained and the reaction path stops its approach in the reactants channel at a similar point to the previous case.
2.3	0.74	-0.5	2	No	The H₂ molecule has a large enough vibration energy to cross the transition state and enters the products channel but then turns back towards the reactants. The collision has therefore been unsuccessful and the reaction has not occurred.
2.3	0.74	-0.5	3	No	The H₂ molecule vibrates with the same energy and amplitude as in the first case (the largest of the set), but also does not form stable products despite passing the transition state and crossing into the products channel twice, eventually forming stable reactants.

With the same initial position maintained, the momentum p_FH was increased to -0.8 and the momentum p_HH reduced to 0.1. The resulting contour plot and reaction path calculated using 2000 steps is shown in Figure 28.

The increased translational energy of the system (and reduced vibrational energy) allows the transition state to be overcome and forms stable products, hence a successful collision and a reactive trajectory.

The endothermic reverse reaction H + HF was investigated by running trajectories firstly p_HF increased and p_HH fixed, and then the other way around. 1000 steps were used for each calculation where possible.

r_H-F / Å	r_H-H / Å	p_HF	p_HH	Reactive Trajectory?	Comments
0.919784	2.3	1	-1	No	The particles have insufficient energy to reach the transition state, with the reaction path turning back in the reactants channel.
0.919784	2.3	5	-1	No	Despite an increase in vibrational energy, observed by oscillations of wider amplitude, the transition state is still not attained and the reaction path ends up in the product channel.
0.919784	2.3	10	-1	Yes	This is a reactive trajectory as the reaction path reaches the transition state, albeit chaotically with several interconversions between the reactants and products structures. However, stable products are formed as the reaction path terminates well into the products channel.
0.919784	2.3	1	-5	No	The reaction path does not reach the transition state and turns back towards the reactants, hence an unreactive trajectory.
0.919784	2.3	1	-10	Yes	The reaction path does reach the products channel, although the transition state is bypassed. This anomaly is likely to be a limitation of the program used to calculate the trajectories as a result of approximations.

Since the increasing the vibration energy of the H-F bond for this endothermic reaction while keeping the translation energy fixed at a low value tends to result in the formation of stable products, the Polanyi rule has been demonstrated to hold true in this reverse case as well as for the forward exothermic reaction, F+ H₂. It should be noted that the Polanyi rule is a guide and may not necessarily apply to all reactions, possibly due to the redistribution of kinetic energy between translational and vibrational forms by molecules. However, the most probable cause for the anomalous trend observed for H + HF is approximations made for simplification in the program used for trajectory calculations.

Ng611 (talk) 15:27, 1 June 2018 (BST) A good overall explanation of Polanyi's rules and good examples given. Well done!

Ng611 (talk) 15:27, 1 June 2018 (BST) Overall a very thorough and well thought out report. Very well done!

References

↑ ^1.00 ^1.01 ^1.02 ^1.03 ^1.04 ^1.05 ^1.06 ^1.07 ^1.08 ^1.09 ^1.10 Imperial College London, https://wiki.ch.ic.ac.uk/wiki/index.php?title=CP3MD, (accessed May 2018).
↑ ^2.0 ^2.1 J. I. Steinfeld, J. S. Francisco and W. L. Hase, Chemical Kinetics and Dynamics, Prentice Hall, Englewood Cliffs, 1989, pp. 220-221.
↑ ^3.0 ^3.1 ^3.2 ^3.3 R. D. Levine, Molecular Reaction Dynamics, Cambridge University Press, Cambridge, 2005, pp. 202-203.
↑ G. D. Billing and K. V. Mikkelsen, Introduction to Molecular Dynamics and Chemical Kinetics, John Wiley & Sons, New York, 1996, p. 31.
↑ ^5.0 ^5.1 ^5.2 ^5.3 P. Atkins and J. de Paula, Atkins' Physical Chemistry, Oxford University Press, Oxford, 10th edn., 2014, pp. 65, 71 and 986.
↑ D. Shriver et al., Inorganic Chemistry, W. H. Freeman and Company, New York, 6th edn., 2014, pp. 59 and 458.
↑ ^7.0 ^7.1 J. Clayden, N. Greeves and S. Warren, Organic Chemistry, Oxford University Press, Oxford, 2nd edn., 2012, pp. 989.
↑ ^8.0 ^8.1 C. Daniel et al., J. Phys. Chem., 1993, 97, 12485-12490.

[CP3MD-1] 1.00 ^1.01 ^1.02 ^1.03 ^1.04 ^1.05 ^1.06 ^1.07 ^1.08 ^1.09 ^1.10 Imperial College London, https://wiki.ch.ic.ac.uk/wiki/index.php?title=CP3MD, (accessed May 2018).

[CKAD-2] 2.0 ^2.1 J. I. Steinfeld, J. S. Francisco and W. L. Hase, Chemical Kinetics and Dynamics, Prentice Hall, Englewood Cliffs, 1989, pp. 220-221.

[MRD-3] 3.0 ^3.1 ^3.2 ^3.3 R. D. Levine, Molecular Reaction Dynamics, Cambridge University Press, Cambridge, 2005, pp. 202-203.

[MDACK-4] G. D. Billing and K. V. Mikkelsen, Introduction to Molecular Dynamics and Chemical Kinetics, John Wiley & Sons, New York, 1996, p. 31.

[Physical_Chemistry-5] 5.0 ^5.1 ^5.2 ^5.3 P. Atkins and J. de Paula, Atkins' Physical Chemistry, Oxford University Press, Oxford, 10th edn., 2014, pp. 65, 71 and 986.

[Inorganic_Chemistry-6] D. Shriver et al., Inorganic Chemistry, W. H. Freeman and Company, New York, 6th edn., 2014, pp. 59 and 458.

[OC-7] 7.0 ^7.1 J. Clayden, N. Greeves and S. Warren, Organic Chemistry, Oxford University Press, Oxford, 2nd edn., 2012, pp. 989.

[J._Phys._Chem.-8] 8.0 ^8.1 C. Daniel et al., J. Phys. Chem., 1993, 97, 12485-12490.

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Exericse 1: H + H2 System