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== MRD Lab - afs17 ==
r1=rbc
r2=rab
== Exercise 1: H + H2 system ==

===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?===

Figure 1 shows a potential energy surface diagram, the transition state is the maximum on the minimum energy path linking reactants and products, it is a saddle point.

[[File:Q1afs17.png|thumb|center|Figure 1 - Potential Energy Surface Plot]]

The transition state can be defined mathematically as the point at which where the gradient of the potential energy surface equals zero ∂V(ri)/∂ri=0. The local minima's also have ∂V(ri)/∂ri=0 therefore a second partial derivative of potential energy with respect to r1 and r2 must be taken to ensure the correct point is taken as the transition state<ref> ChemLibre Texts [Accessed May 2019] https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Quantum_Mechanics/11%3A_Molecules/Potential_Energy_Surface</ref>.

The transition state is a saddle point on the potential energy surface diagram, the transition state is the first partial derivative with respect to r1 and r2 and must be equal to zero. Any minima or maxima will also have the same value of zero, to check if the critical point identified is the transition state (saddle point) the values of the partial second derivative with respect to r1 multiplied by the partial second derivative with respect to r2 minus the second partial. derivative with respect r2 and r1. If the value is less than zero then the point selected is the saddle point and is the transition state. The value of a local minima will be greater than zero and the partial second derivative with respect to r1 will be greater than zero.

For saddlepoints: <math>\frac{\partial^2 V}{\partial r_1^2} \frac{\partial^2 V}{\partial r_2^2} - (\frac{\partial^2 V}{\partial r_1 \partial r_2})^2 < 0</math>

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

Using figure 2 enabled the position of the transition state to be estimated at approximately r1=r2=0.9 Å which is where the cross lies. The transition state position was obtained by using various values of r1 and r2 from 0.9 Å until the graph of internuclear distances against time shown in Figure 3 appeared as a horizontal line. This indicated that the internuclear distance was constant and molecules just oscillated periodically as they are at the transition state. Therefore the position of the transition state was estimated to be r1=r2=0.908 Å with p1=p2=0.

[[File:Q1aafs17.png|thumb|centre|Figure 2 - Contour plot showing the transition state position at approximately r1=r2=0.9 Å]]
[[File:F2afs17.png|thumb|centre|Figure 3 - Internuclear Distances Against Time Graph for r1=r2=0.908 Å, p1=p2=0]]

===Comment on how the ''mep (minimum energy pathway)'' and the trajectory differ.===

====Calculating Reaction Pathway====

The initial conditions were changed such that r1 = rts +0.01 = 0.907, r2=rts=0.908 with p1=p2=0. The calculation was changed from dynamic to MEP and the trajectory follows the valley floor in both figure 4 and 5.

[[File:Afs17mep.png|thumb|centre|Figure 4 - MEP pathway with 500 steps with r1 = rts +0.01 = 0.907, r2=rts=0.908 with p1=p2=0]]

[[File:Afs17t2.png |thumb|centre|Figure 5 - MEP pathway with 2000 steps to show complete trajectory with r1 = rts +0.01 = 0.907, r2=rts=0.908 with p1=p2=0]]

====Trajectories from r1= rts====

[[File:Afs17dyn.png|thumb|centre|Figure 6 - Dynamic calculation with r1 = rts +0.01 = 0.907, r2=rts=0.908 with p1=p2=0]]

[[File:Afs17w.png|thumb|centre|Figure 7 - Dynamic calculation with r1 = rts +0.01 = 0.907, r2=rts=0.908 with p1=p2=0]]

Figure 4 and 5 show the calculation using MEP whilst Figure 6 and 7 show it using the dynamic calculation, with the dynamic calculation the molecules can oscillate which is shown in the line not being smooth instead periodically oscillating. The trajectories in both calculations follow the same pathway of increasing up to the transition state at the maxima, the MEP pathway is a theoretical pathway based on the molecules being perfect and not oscillating - with infinitely slow motion. In the MEP calculations momenta are always reset to zero in each time step which means that it doesn't consider the momentum input at the beginning, it only considered the input momentum in the first step. The dynamic pathway is more realistic as it considers motion of the atoms during a reaction, as well as considering mass and being in the gas phase and it doesn't reset momentum to zero at each time step.

===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?===

For the initial positions r1 = 0.74 and r2 = 2.0

r1=rbc
r2=rab

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 || - 99.119|| Yes ||Productive reaction in which the H2 molecule comes towards the H atom and the transition state is reached. Vibrates after the reaction but not before suggesting the initial diatomic molecule was stable.|| [[File:3afs.png|thumb]]
|-
| -1.5 || -2.0 ||-100.456|| No || The H2 molecule is vibrating but doesn't react, doesn't reach transition state, falls back as not enough kinetic energy to overcome the activation energy barrier|| [[File:4afs17.png|thumb]]
|-
| -1.5 || -2.5 || -98.956 || Yes|| Reactive as BC is produced and successfully reaches transition state, both AB and BC have vibrational energy and so the whole trajectory oscillates. This system has a higher energy than the first two and hence overcomes activation energy barrier.|| [[File:5afs17.png|thumb]]
|-
| -2.5 || -5.0 || - 84.956 || No|| Unproductive reaction, no product formed at the end. AB dissociates and forms BC temporarily before returning to AB and C. Reactants must have contained enough energy to exceed transition state and activation barrier but there could be too much vibrational energy from the collision which could have caused bond dissociation. At one point there are 3 atoms separate with A moving towards B to reform the reactant. This reaction has the most energy out of all of the previous which could explain why the bond was able to break.|| [[File:6afs17.png|thumb]]
|-
| -2.5 || -5.2 || -83.416 || Yes ||This is productive, but the reactants AB have no vibrational energy as the trajectory doesn't oscillate but it still reaches transition state to form products. The product BC then dissociates to reform AB which has a lot of vibrational energy and which is converted to kinetic energy in the reactants. When the product BC dissociates, atom B heads back towards A so reactant AB reforms - AB is vibrationally excited hence the large oscillations on the trajectory. Then AB breaks apart again reforming product BC which is an example of barrier recrossing. || [[File:8afs17.png|thumb]]
|}

===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?===

In Transition State Theory (TST) the reactants are separated from products to find an expression for thermal rate constant given a potential energy surface for the reaction. <ref>J. I. Steinfield, J. S. Francisco, W. L. Hase, Chemical Kinetics and Dynamics, Pearson, 1998 </ref>

1) Born-Oppenheimer approximation is applied to separate nuclear and electronic motion - as nuclei are much heavier than electrons.

2) Transition states obey a Maxwell-Boltzmann distribution.

3) Molecular systems that have crossed the transition state to the products cannot reform reactants.

4) Even without an equilibrium between reactant and product, the transition states are distributed according to Maxwell-Boltzmann distribution.

5) Motion of the transition states along the reaction coordinate may be separated from other motions and treated classically as a translation.

TST ignores that vibrations are quantised or that tunnelling can occur through quantum mechanics. It also assumes that in order for a reaction to occur the species must collide with enough energy to be successful (cross transition state barrier) despite the possibility that tunnelling can occur.<ref> D. C. Elton, Transition State Theory for Physicists, 2013.</ref>
Another issue with TST is that at higher temperatures the trajectory doesn't always pass through the lowest energy saddle point (transition state) in the surface graph as higher energy vibrational modes are populated.

For the first three systems, the predicted rates with TST are likely to match experimental theory as reactants simply pass through transition state barrier to form products or don't have enough energy to react.

For the last two systems the rates are will differ from predicted values as these systems show barrier-recrossing where products form and inform by recrossing transition state barrier - forbidden in TST.

TST also doesn't hold up for systems far from equilibrium - Boltzmann distribution breaks down.

== Exercise 2: F-H-H system ==

===PES Inspection===

====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?====

''F + HA-HB -> F-HA + HB''

[[File:e2afs17.png|thumb|Figure 9 - R+Surface plot of reaction of F + HA-HB -> F-HA + HB]]

The reaction above is exothermic as the reactants are at a higher energy level than the products. This relates to the bond strength

====Locate the approximate position of the transition state.====

====Report the activation energy for both reactions.====

===Reaction Dynamics===

====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.====

====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.====
==References==
{{Reflist}}

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