== Introduction ==
This lab studies the trajectories and relative behaviour of triatomic systems by simulation via lepsgui.py and looks into the importance of right energy distribution. H-H-H and F-H-H systems have been investigated specifically.

== EXERCISE 1: H + H2 system ==
'''Dynamics from the transition state region'''

In a potential energy surface, the transition state is the first-order saddle point where it is a minimum along all coordinates except one. To distinguish it from a local minimum, the hessian matrix could be found where the eigenvalues should be one positive and one negative. In contrast, the eigenvalues of a local minimum would be both positive and the second derivative would be positive. (see fig.1 below)

[[File:H-Hcontour.png|thumb|center|upright=2|Fig.1 contour plot at the transition state position where it is represented by a cross]]
[[File:H-Hdistanceplot.png|thumb|center|upright=2|Fig.2 internuclear distance vs time graph at the transition state position]]

The estimate for the transition state position rts=90.8pm. From the "Internuclear Distance VS Time" plot, it can be seen that rAB is equal to rBC in this case and the distance is constant, which means the reaction is now at transition state.

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

In the minimum energy path, it corresponds to infinitely slow motion around the transition state and the momenta are set to zero, so the internuclear velocity of A-B, A-C and B-C remained zero for the mep trajectory, whereas in dynamics trajectory the velocities start to oscillate in a fixed frequency after reaching the transition state. Besides, the intermolecular distance in mep calculation remains unchanged for the newly formed H2 molecule which means it is not oscillating.

[[File:mep-VT.png|thumb|center|upright=2|Fig.3 internuclear velocity vs time graph at the transition state position in mep]]
[[File:mep-DT.png|thumb|center|upright=2|Fig.4 internuclear distance vs time graph at the transition state position in mep]]

With the initial positions being r1 = 74 pm and r2 = 200 pm, a set of different momenta were used to run the trajectories (see table below). The results in the table emphasise the importance of the right energy distribution into the vibrational and translational mode and indicate the fact that just having high initial energy is not enough for the reaction to take place because of the possibility that the reactants may be reformed by recrossing the barrier.

{| class="wikitable"
! p1/ g.mol-1.pm.fs-1
! p2/ g.mol-1.pm.fs-1
! Etot/kJ.mol-1
! Reactive?
! Description of the dynamics
! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280 || Yes || As H(A) approaches the bonded Hs, it forms a bond with H(B) and the original B-C bond breaks. Then the bonded A-B hydrogen molecule starts to vibrate and moves back together. || see fig.5
|-
| -3.1 || -4.1 || -420.077 || No || H(A) is trying to approach the other two bonded Hs until it is stopped at some point and not able to form the bond as it is not reactive enough. Then H(A) moves back again alone. || see fig.6
|-
| -3.1 || -5.1 || -413.977 || Yes || H(A) is approaching the two bonded Hs until the reaction happens and it forms a bond with H(B). Then the newly formed H2 molecule moves back together and vibrates. || see fig.7
|-
| -5.1 || -10.1 || -357.277 || No || As H(A) approaches the bonded Hs, it forms a bond with H(B) and the bond starts to vibrate. But then the bond breaks, and H(B) moves back to bond with H(C) again. || see fig.8
|-
| -5.1 || -10.6 || -349.477 || Yes || As H(A) is approaching the two bonded Hs, it first collides with H(B) but then H(B) bounces back before it collides with H(A) again to form a new A-B bond. After that, the A-B hydrogen molecule moves back together and vibrates. || see fig.9
|}

[[File:contour1.jpeg|thumb|center|upright=2|Fig.10 illustration of the trajectory at p1=-2.56, p2=-5.1]]
[[File:contour2.jpeg|thumb|center|upright=2|Fig.11 illustration of the trajectory at p1=-3.1, p2=-4.1]]
[[File:contour3.jpeg|thumb|center|upright=2|Fig.12 illustration of the trajectory at p1=-3.1, p2=-5.1]]
[[File:contour4.jpeg|thumb|center|upright=2|Fig.13 illustration of the trajectory at p1=-5.1, p2=-10.1]]
[[File:contour5.jpeg|thumb|center|upright=2|Fig.14 illustration of the trajectory at p1=-5.1, p2=-10.6]]

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?

[[File:exothermic-mdlab.jpg]]

It can be seen from the surface plot of the F+H2 reaction above that the energy of the products is smaller than that of the reactants, which means the F+H2 reaction is exothermic. On the other hand, Hammond's postulate states that the transition state of a reaction resembles the reactant or the product depending on to which it is closer in energy and in an exothermic reaction the transition state is closer to the reactant, which illustrates that the reaction is exothermic in this case. In terms of the relative bond strengths it indicates that the energy required to break the original bond is less than the energy released in forming the new bond, and in this case it means that the H-H bond is weaker than the H-F bond formed. When it comes to H+HF, the reaction is endothermic as the H2 bond formed is weaker than the HF bond.

Report the activation energy for both reactions.

[[File:activationenergy-plot.jpg]]

By changing the AB or BC distance by 1pm away from the transition state and setting the calculation type to MEP, an energy VS time plot of the H+HF reaction can be obtained as above. The activation energy is equal to the energy of transition state minus that of reactants, and from this we can calculate the energy difference from the plot which is 126.7kJ/mol. In the F+H2 reaction, the energy of the reactants is found to be -435.8kJ/mol, giving rise to the activation energy of 1.8kJ/mol.

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.

When the F+H2 reaction happens, the reaction is exothermic and the potential energy is converted into vibrational energy. This could be confirmed using IR spectroscopy. When energy is released, the electrons are excited from the ground state to higher levels and the subsequent relaxation would give rise to two peaks shown in the IR spectrum. The intensity of the overtone would decrease whereas that of the fundamental would increase as time goes on. An energy calorimeter could also be used.

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.

The Polanyi's rule states that the translational energy is better at promoting the reaction to overcome the barrier in reactions with early transition states, i.e. exothermic reactions. The translational energy helps promote exothermic reactions as it aligns with the direction that forms the products, and the trajectory would fall back into the lower energy state. However, the vibrational motion is in a different direction from the reaction path, which makes it harder for the trajectory to cross the barrier.
By setting the initial conditions of rHH = 74 pm, rFH = 230pm, pFH = -1.0 g.mol-1.pm.fs-1 and exploring values of pHH in the range -6.1 to 6.1 g.mol-1.pm.fs-1, it can be seen that a lot of energy has been put into vibration and a high kinetic energy is required for the trajectory to be reactive. In contrast, when the exothermic reaction F+H2 happens and the transition state in early, a relatively low kinetic energy is required for the reaction to happen, which is in line with the Polanyi's empirical rule.

[[File:unreactive-trajectory.png|thumb|center|upright=2|Fig. illustration of an unreactive trajectory at p1=-1, p2=-3]]
[[File:reactive-trajectory.png|thumb|center|upright=2|Fig. illustration of a reactive trajectory at early transition state where p1=-5, p2=-3]]

File:Mep-DT.png

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