MRD:lhl17

Exercise 1: H + H₂ 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 program calculates potential energy using It is a London-Eyring-Polanyi-Sato (LEPS) potential and calculates individual trajectories. By analysis of these reaction paths and using TST we can analyse reactions and the molecular reaction dynamics.

On a potential energy surface (PES) diagram, the transition state is mathematically defined as the point at which the following is true:

${ \partial V(r)\over \partial r} = 0$

While a one-dimensional potential energy curve (PEC) could simply be approximated by quadratic curves, the PES is multi-dimensional function (with the energy on the z axis and bond distances on the x/y axes). This means the TS, the maximum point on the minimum reactive trajectory, can be visually identified as the "saddle point" - a maximum point in one direction and minimum in the other.

However, to distinguish the TS from other local minima, both the second derivative and the determinant of the Hessian Matrix at the TS should be smaller than 0. In addition, the eigenvalues of the Hessian Matrix should have opposite positive/negative signs.

Question 2

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 trajectory.

The TS position for the H + H₂ system was found at r_AB = r_BC = 90.77 pm with p_AB = p_BC = 0 g.mol^-1.pm.fm^-1.

Internuclear Distances vs Time	Contour Plot	Force analysis
Starting the TS with no momentum with no forces on the atoms, they will remain in place and not move at all. Intermolecular distances remain constant which results in horizontal lines in the graph above.	The red cross indicates the initial conditions input, and it is clear that no bond movement is shown here. The transition state is totally symmetric (neither endo nor exothermic, and neither early nor late according to Hammond's postulate).	The exact point was confirmed by force analysis (where forces on the molecules are minimised).

Question 3

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

The minimum energy path (mep) is a very special trajectory that corresponds to infinitely slow motion (i.e. the momenta/velocities are always reset to zero in each time step). By slightly displacing the initial conditions from the TS in either direction, it possible to encourage product or reactant formation and observe how the trajectory evolves just away from the saddle. Like a ball rolling along a hilly surface, the system will gain momentum as soon as the simulation starts. as the system is not at a stable point on the PES it will fall to an energy minimum, then potential energy  is converted into kinetic. By comparing this same system in a MEP and Dynamic simulation, the produced trajectories should illustrate the differences between both approaches.

When calculating the minimum energy pathway using calculation type MEP and Dynamics, we obtain the following results (with r_AB = 90.77 pm, r_BC = 91.77 pm and p_AB = p_BC = 0 g.mol^-1.pm.fm^-1): 

Dynamic Calculation	MEP Calculation (5000 steps)	Comments
Contour plot of reaction path	Contour plot of reaction path	There is no vibrational energy for the MEP plot, unlike in the dynamic plot. This is an indication of how different MEP is algorithmically set - for MEP velocity is reset to 0 at each time step, so that kinetic energy always remains at 0. Thus, there is no vibrational energy in the system so no oscillation.
Momentum vs Time plot of reaction path	Momentum vs Time plot of reaction path	In Dynamics, we can see that the oscillation of the atoms has been included in the calculation, and increases over time.
Distance vs Time plot of reaction path	Distance vs Time plot of reaction path	In both cases, the atoms are moving away from each other. Note that in the dynamics velocities at which they are moving remain constant.

Question 4

Complete the table below 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?

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	Reaction path passes through the TS (at around 21 fs) and forms products. The vibrational oscillations in the product channel confirm formation of the the product.
-3.1	-4.1	-420.077	No	Reaction path unable to get over maximum at TS and form products. There is no successful collision and instead, the system rolls back down the potential to reactants.
-3.1	-5.1	-413.977	Yes	Momentum of incoming H atom increased compared to first example, and as before, the reaction path passes over the TS (at around 21 fs) to the product channel. More oscillations observed in the product than reactant channel.
-5.1	-10.1	-357.277	No	Initial momenta large enough for reaction path to reach the TS (at around 8 fs), but excess energy causes the product to cross the activation barrier twice. Even though the product bond does form, the system eventually returns to reactants (with significantly higher vibronic energy).
-5.1	-10.6	-349.477	Yes	Similar to the previous example, but with the momentum of the incoming H atom increased. The system fluctuates twice between reactant and product bond formation (with TS at 9, 15 and 26 fs), which suggests the TS in this scenario is far from the lowest energy saddle point.

The higher kinetic energy, hence momenta does not necessarily guarantee the reaction would proceed. Higher kinetic energy can mean that the reactants can go over the activation energy barrier, but with energy being too high, products newly formed might overcome the transition state barrier again in the reversed direction and reform reactants, in which case it is not reactive overall. 

Question 5

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

The Transition State Theory assumes that only the reactants are in constant equilibrium with the transition state structure (quasi-equilibrium), and that the energy of the particles follow a Boltzmann distribution.

In addition, it is assumed that once the reactants have trajectories with a kinetic energy along the reaction coordinate greater than the activation energy, that the products will definitely form, ignoring the possibility of barrier recrossing caused by excess vibrational energy. As seen in row 4, the starting materials may be reformed instead leading to an overall unreactive path.

Furthermore, the theory fails for some reactions at low temperatures due to quantum tunneling, where the product can still be formed even if the TS is not reached (the effect of this is negligible compared to barrier recrossing). The nuclei are also treated as classical point masses and don't have wavefunctions, which is another flaw of TST.

In the table above, only 5 possible individual trajectories of the H + H₂ reaction were investigated. The true reaction rate can be thought of as a "weighted average" of all such individual paths and so, with TST predicting unreactive trajectories to in face be reactive, it can be concluded that the theory overestimates the rate of reaction when compared with experimental values.

Exercise 2: F - H - H system

Question 6

By inspecting the potential energy surfaces, 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?

PES of F + H₂, showing early transition state

The enthalpy change of a reaction can be calculated using the following equation:

ΔH_rxn=ΣΔH_{bond breaking}-ΣΔH_{bond forming}

Using literature values for bond dissociation energies of H-H (436 kJ mol^-1) and H-F (569 kJ mol^-1) shows that the enthalpy released from breaking the H-F bond is greater than the energy required to break H-H bonds. Therefore, the reaction F + H₂ → HF + H is energetically favorable and exothermic overall. This is consistent with the very small atomic radius of the F atom, and therefore stronger attraction and a stronger bond overall. The PES shows that the reactants are higher in energy than the products (where AB distance, representing the H-F bond), indicating an exothermic reaction.

The reverse reaction, H + HF is therefore endothermic, as extra external energy is required in order to break the H-F bond.

Question 7

Locate the approximate position of the transition state.

According to Hammond's postulate the structure of the transition state will resemble the structure of the nearest stable species (either products or reactants). Taking the first reaction F + H₂ → HF + H, the transition state will resemble the reactants since it is an exothermic reaction. For HF + H → F + H₂, the opposite is true: the transition state will resemble the products for the endothermic reaction.

After testing several initial distances by trial and error, initial momentum of both reactants = 0, it was found that the transition state has the following internuclear distances: r_FH = 180.5 pm and r_HH = 74.5 pm.: -433.981 kJ mol^-1  

Internuclear distances	Contour	Forces
To find the TS, both momenta were set to zero and the distances were optimised until minimal oscillation was reached in the internuclear distances vs time graph	Unlike exercise 1, the system is clearly asymmetric.	the forces at the transition state are very close to 0 which means that the system will not move if no momentum or change in position is given to it.

Note that for these triatomics there are two bond lengths we can vary and thus at a saddle point there is one direction that decreases most in energy, and another that increases the most. For the transition structure of a polyatomic molecule with 3N-6 vibrational modes, there would exist one direction down and the remaining 3N-7 vibrational modes acting as sinks for the excess energy.

Question 8

Report the activation energy for both reactions.

The E_a was found by altering the TS distance for each of AB, then BC by 1 pm. The energy difference between the TS system with maximum displacement between the reactants (F-H=750 pm, energy=-435.100 kJ.mol^-1).

F + H2	HF + H
Ea: 1.158 kJ.mol-1.	Ea: 127 kJ.mol-1.

Question 9

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

To investigate a reactive trajectory for the reaction H₂ + F -> HF + H, the following initial conditions were chosen: 

rFH / pm	rHH / pm	pFH/gmol-1pmfs-1	pHH/gmol-1pmfs-1
200	74.5	-1.0	0

The conservation of energy suggests that total energy is constant through the course of reaction. From the Energy Vs Time graph of the exothermic H₂ and F reaction below, the loss in potential energy from the breaking of H₂ bonds is mirrored by the increase in kinetic energy. The kinetic energy can be of three forms: translational, rotational or vibrational.

Calorimetry can be used to measuring the transfer of both translational and kinetic energy from the products to the walls of the calorimeter in the form of heat. The heat transferred can then be measured in the form of change in temperature with a thermal sensor. We would expect an increase in temperature for exothermic reactions and a decrease for endothermic ones, and the energy can be calculated using ΔQ=cΔΤ.

Meanwhile, IR spectroscopy can be used to quantify the vibrational energy. Vibrational exictation and peak wavenumbers in the infrared spectrum can be used to identify the changes in E_vib of the IR active HF molecule. 

Contour plot	Momenta vs time	Surface Plot	Energy plot
The contour plot shows strong oscillations after the system passes through the transition state, which represents the system passes its excess energy into the vibration of HF.

Question 10

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.

Polanyi's Empirical rules state that vibrational energy is more efficient at activating a late transition state than translational energy (endothermic reaction - Hammonds Postulate) and that translational energy is more efficient in activating an early transition state in an exothermic reaction than vibrational energy. Therefore for a successful exothermic reaction, the molecules need an excess of translational (kinetic) energy compared to vibrational (oscillatory) energy.

To investigate thiis, a calculation starting on the side of the reactants of F + H₂, at the bottom of the well r_HH = 74 pm, with a momentum p_FH = -1.0 g.mol^-1.pm.fs^-1, and explore several values of p_HH in the range -6.1 to 6.1 g.mol^-1.pm.fs^-1 (explore values also close to these limits). What do you observe? Note that we are putting a significant amount of energy (much more than the activation energy) into the system on the H - H vibration.

The table below shows the reactivity and energy of the reaction H₂ + F with differing initial momenta. The initial positions are set to AB = 74 pm and BC= 200 pm where A and B are H and C is F. 

pHH/gmol-1pmfs-1	Etot/ kJ.mol-1	Reactive trajectory	Contour plot
-6.1	-402.5	No
-5		Yes
0	-433.6	Yes
5		No
6		No

For the same initial position, the momentum p_FH  ( represents the translational energy) was increased slightly to -1.6 g.mol^-1.pm.fs^-1,and p_HH  (representing vibrational energy decreased to 0.2 g.mol^-1.pm.fs^-1. Although the overall energy was similar to a successful trajectory (-432.4 here, ) the reaction did not go to completion. This shows that the type of energy in the system is important to determine whether the reaction is successful. In the reactions above with total energy being large but still having an unsuccessful reaction is mainly due to the reversibility where the products are able to cross back and become reactants again, as shown by the contour plots.

If you start a trajectory from close to the transition structure it will gain momentum as it moves in the reactant or product channel. At some molecular geometry in that channel, you could reverse the momentum (change the sign) and you should get a trajectory that goes back to the transition structure

Contour	Momentum	Energy