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?

On a potential energy surface (PES) diagram, the transition state is mathematically defined as the point at which ∂V(r)/∂r =0 is true, and can be visualised as the "saddle point". The saddle point only exists for 2D functions and it is important to considr both directions (the energy of the PES on the Z axies and distances on the x/y axes). This can be described as the maximum point on the minimum reactive trajectory (minimum energy path linking reactants and products) and can be distubguished from other local minima if the the second differential at the TS is less than 0.Furthermore, the determinant of the Hessian Matrix is smaller than 0, with the eigenvalues of the Hessian Matrix being one positive and one negative

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

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 transition state appears as a saddle point in the potential energy surface (PES) - i.e. in one direction, the point is a maximum point in the graph and in the other direction, it is a minimum.

Startin ght esystem at the TS with no momentum with no forces on the atoms, they will remain in place and not move at all. The point was found at r_AB = r_BC = 90.77 pm and p_AB = p_BC = 0 g.mol^-1.pm.fm^-1. The transiiton state was confirmed by force analysis (forces = 0). In this H + H₂ system, the transition state is totally symmetric and is netiher endo nor exothermic, and niehter early nor late accroding to Hammond's postulate.

Internuclear Distances vs Time	Contour Plot	Force analysis

Question 3

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

this should allow you to see how the trajectory evolves just away from the saddle. By nudging in either direction you can encourage product or reactant. 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): 

The program calculates potential energy using It is a London-Eyring-Polanyi-Sato (LEPS) potential and calculateds individual trahectories. By analysis of these reactionion paths and using TST we can analyse reactions and the molecular reaciont dynamcis. 

 The abstraction you are being invited to make is - think about a ball rolling along a hilly surface. Even if you start with no momentum, you might have started on a hill, so will gain some momentum as soon as the simulation starts.  if 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 

The trajectory is a sequence of A-B and B-C distances. So if you follow that line, you see how the 2 distance s are changing with time 

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.

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, 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 that the transition state structure will collapse back to into the reactant channel, as in row 4.

The experimental scenarios above show that excess kinetic energy can result in multiple crossing of the activation energy barrier through multiple forward and backward directions. Theoretically, the theory suggests only a single crossing of the activation energy barrier in the forward direction is possible.

Finally, the theory fails for some reactions at elevated temperatures due to more complex molecule motions or low temperatures due to quantum tunneling, where the product can still be formed even if the TS is not reached. However, the effect of this is negligible compared to barrier crossing. 

Here we're treating the nuclei as classical point masses... they don't have wavefunctions, so there are some things this model's missing 

Thus, it is clear that Transition state theory overestimates the rate of reaction.

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 the 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 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 = H + HF, the transition state will resemble the reactants since it is an exothermic reaction. This is the opposite for the second reaction: the transition state will resemble the products for the endothermic reaction. The saddle point is significantly smaller than for system 1.

For these triatomics where we have two bond lengths we can vary, then yes: at a saddle point there's one direction that goes most down in energy, and another that goes most up. For a polyatomic molecule with 3N-6 vibrational modes, at a saddle point / transition structure there'd be one direction down; 3N-7 going up

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  

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.

Question 8

Report the activation energy for both reactions.

Ea was found by setting the calculation type in MEP and alter the ts distance for one out of two distances by 1 pm. By taking the difference between the energy differnce between the TS and the energy of a system with maximum displacement between the reactants (F-H=750 pm, energy=-435.100 kJ.mol^-1).Ea = the difference between Ets and Ereactants.

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.

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This conversion from translational to vibrational energy can be tested experimentally by calorimetry, which would record an increase in temperature for exothermic reactions and a decrease for endothermic ones, or by IR spectroscopy using FTIR (Fourier transform infrared spectrometer). This would allow us to monitor vibrational excitations in a molecules and see them on the IR spectrum as bands of a certain frequency and overtones. If these are present, we know that the molecule must have some vibrational energy. 

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, and the energy calculated using ΔQ=cΔΤ.

Meanwhile, IR spectroscopy can be used to quantify the vibrational energy. Peaks and their wavenumbers in infrared spectrum can be used to identify the changes in vibrational energy of the IR active H-F 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

References

1. w
2. https://labs.chem.ucsb.edu/zakarian/armen/11---bonddissociationenergy.pdf