== Dynamics of the Reaction H2 + H -> H + H2 ==
=== the Transition State and its Identification ===

The transition state on a potential energy surface diagram is defined as the point where the gradient of the potential ∂V(ri)/∂ri is zero in all directions. It is the point with the highest energy on the minimum energy path that connects the reactants and the products. 
For a symmetric potential energy surface, the transition state lies on the diagonal line where r1=r2. It is the point on that line with the lowest potential energy. The exact location of the transition state for a reaction with a symmetric potential energy surface can be found by running a minimum energy path simulation starting from a point where r1 = r2. The point where that minimum energy path ends is the transition state. [[File:MRD0158103301.png|frame|thumb|left|Figure 1: potential energy surface plot of Ha-Hb + Hc -> Ha + Hb-Hc, where the diagonal r1=r2 line is dotted in black and the transition state is circled in red]] 
A transition state is different from a local minimum due to the fact that it is a saddle point, meaning the surface around it has negative Gaussian curvature. This means that the maximal and minimal values of curvature (the second derivatives) around the transition state are of opposite signs, meaning they curve in opposite directions (up and down). A local minimum, on the other hand, has positive Gaussian curvature, meaning that the curvature is in the same direction all around (up in this case).
=== The Position of the Tranition State for the Reaction H2 + H -> H + H2 ===
For the reaction H2 + H -> H + H2, the transition state is at a point where the distances between neighboring hydrogen atoms are around 90.774 pm. This point was found using the method mentioned before - a minimum energy path simulation was run, starting from a point on the diagonal r1 = r2 line. The point where the minimum energy path simulation stops is the transition state, because it is the minimum energy point along the r1 = r2 line. It can not move any lower (towards the reactants or the products) because at that point the gradient is zero in all directions, including the minimum energy path. 
[[File:MRD0158103302.png|frame|thumb|left|Figure 2: internuclear distances of hydrogens for the aforementioned reaction plotted against time. This plot was constructed by simulating a minimum energy path from a starting point where r1 = r2 = 60 pm]] 
=== The Minimum Energy Path (MEP) ===
The minimum energy path, or reaction path is a theoretical reaction trajectory that corresponds to infinitely slow motion of reactants and products. On an MEP, the momenta and velocities are always zero, because the movement of reactants and products is on an infinitely long timescale. The MEP is different from the actual reaction trajectory because on the MEP, the system has no kinetic energy and the potential energy always goes down, unless the MEP is specifically connecting two points on the surface. This means that the system is always losing total energy to head directly to the point of lowest potential energy. In reality, total energy must always be conserved, so when the trajectory heads downward on the potential energy surface (losing potential energy), it must gain kinetic energy. This can be in the form of translational, rotational, vibrational, or electronic energy. For the given reaction, when a MEP is simulated from a point where r1 
[[File:MRD0158103303.png|frame|thumb|left|Figure 3: a comparison of the MEP (right) and the actual reaction path (left), where both paths start from a point that is minimally deviated from the transition state (r1=r2+1=rts+1)]] 
Testing different reaction trajectories for the reaction Ha-Hb + Hc -> Ha + Hb-Hc 
Running a simulation for this reaction over a long enough timespan gives the values of momenta required for the reaction to occur. For rab = 74 pm and rbc = 200 pm, the required value for the momentum p1 of Ha-Hb is between -3.1 and -1.6 g.mol-1.pm.fs-1. The required value for the momentum p2 of Hb-Hc is -5.1 g.mol-1.pm.fs-1. It might seem a fair assumption that all trajectories with the same starting positions, but with higher values of momenta than the minimum required values, would result in the reaction going through to the products. To test this assumption, a table has been constructed where various values for momenta have been tested with the same starting positions rab = 74 pm and rbc = 200 pm.
{| class="wikitable" border="1"
|+
! 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 || -414.28 || Yes || Ha-Hb is in the vibrational ground state, Hc approaches the molecule directly and forms Hb-Hc, which is vibrationally excited. Ha drifts away from Hb-Hc. Path crosses the activation energy barrier once || [[File:MRD0158103304.png|750px]]
|-
| -3.1 || -4.1 || -420.08 || No || Ha-Hb is in a vibrationally excited state, Hc approaches but is unable to displace Ha, and drifts away from Ha-Hb. Path does not cross the activation energy barrier. || [[File:MRD0158103305.png|750px]]
|-
| -3.1 || -5.1 || -413.97 || Yes || Ha-Hb is in a vibrationally excited state, Hc approaches and displaces Ha to produce a vibrationally excited Hb-Hc. Ha drifts away from Hb-Hc. Path crosses the activation energy barrier once. || [[File:MRD0158103306.png|750px]]
|-
| -5.1 || -10.1 || -357.2 || No || Ha-Hb is in the vibrational ground state. Hc approaches and displaces Ha to produce a vibrationally excited Hb-Hc, but Ha then displaces Hc to produce Ha-Hb once again. Hc drifts away from the vibrationally excited Ha-Hb. Path crosses the activation energy barrier twice. || [[File:Mrd0158103307.png|750px]]
|-
| -5.1 || -10.6 || -349.4 || Yes || Ha-Hb is in the vibrational ground state. Hc approaches and displaces Ha to produce a vibrationally excited Hb-Hc, but the repulsion between Hb-Hc overcomes the attraction and forces Hb and Hc to separate. Ha then approaches Hb to produce Ha-Hb once again. Hc then approaches and displaces Ha a second time to produce a vibrationally excited Hb-Hc. Ha drifts away from the vibrationally excited Hb-Hc. Path crosses the activation energy barrier three times. || [[File:MRD0158103308.png|750px]]
|-
|}
 The results from these simulations show that not all trajectories starting from the same positions are reactive, even those that have larger momenta than required. The right combination of momenta is needed for a reaction to occur. Transition State Theory Transition State Theory tries to predict the rate constant for a reaction based on the properties of the reactants and the structure of the transition state. It is based on an assumption that all trajectories along the reaction coordinate with a kinetic energy higher than the activation energy will be reactive. However, as we determined from the previous tests, it is possible for the reaction trajectory to pass the activation energy barrier to form products briefly, but then cross the barrier once more to form the reactants again. In those cases, the trajectory is not reactive, even though it has enough kinetic energy to overcome the activation energy barrier. For this reaction, there is also a possibility for the reaction to occur without having enough kinetic energy to pass the barrier. This is possible due to quantum tunneling, and quantum tunneling can only have an impact because a hydrogen atom is the lightest atom that can exist. This tunnelling, however, does not occur frequently enough to overcome the reduction in rate caused by the possibility of recrossing the activation energy barrier. This means that the actual rate of reaction will be smaller than the one determined by Transition State Theory.

== Dynamics of the Forward and Backward Reactions of F + H2 -> HF + H ==
The forward reaction is exothermic, whereas the backward reaction is endothermic. This is visible on the surface and contour plots below, as the forward reaction products are lower in potential energy than the reactants. This is because the H-F bond is stronger (565 kJ/Mol) than the H-H bond (432 kJ/mol), making the products HF and H more stable than the reactants H2 and F. [[File:MRD0158103309.png|frame|thumb|left|Figure 4: potential energy surface (left) and contour plot (right) for the reaction, where A is the fluorine atom, and B and C are the hydrogen atoms]] The location of the transition state is approximately AB = 181.1 pm and BC = 74.5 pm (distance between hydrogens is 74.5 pm and distance between fluorine and bond-forming hydrogen is 181.1 pm).
 [[File:MRD0158103310.png|frame|thumb|left|Figure 5: transition state of the reaction displayed on a contour plot]] The activation energy for the forwards reaction is ~1.05 kJ/mol and for the backwards reaction it is ~126.67 kJ/mol. [[File:MRD0158103311.png|frame|thumb|left|Figure 6: energies plotted against time for going from the transition state to the reactants (left) and to the products (right). This is from a MEP simulation, not a dynamics simulation. The activation energy is the difference between the minimum and maximum energy plateaus]] For the forward reaction, which is exothermic, the products have less potential energy than the reactants. Because total energy must always be conserved, this means that the products must have a higher kinetic energy than the reactants. Kinetic energy can be seen in the form of translational, rotational, or vibrational motion. In this reaction, potential energy is converted to kinetic energy in the form of vibrational motion. This is shown on the figure below. [[File:MRD0158103312.png|frame|thumb|left|Figure 7: contour plot of the forward reaction, where the starting positions are AB = 230 pm, BC = 74 pm, p1 = -2 g.mol-1.pm.fs-1, and p2 = 2 g.mol-1.pm.fs-1]] To confirm whether the reaction energy is being released vibrationally, one could probe the reaction via IR spectroscopy. If the reaction energy is not converted to vibrational energy, all the product molecules in the reaction would be in the ground state. An IR spectrum, in this situation, would only display the peaks going from the ground state to higher vibrational energy states. However, if the reaction energy does get converted to vibrational energy, some of the molecules would populate higher vibrational energy levels, and an IR spectrum would also show peaks going from the higher energy states to even higher energy states. Those peaks are called hot bands. Since the differences between neighboring energy levels for an anharmonic oscillator become smaller at higher energy levels, an IR spectrum would show smaller hot bands that are slightly shifted to smaller wavenumbers from the main peaks. If many IR spectra were taken throughout the reaction, one would at first see hot bands along with the main peaks, then the hot bands would shrink and the main bands grow as the product molecules relax to the ground state. Another way to determine whether vibrational excitation is occurring is to probe the reaction using infrared chemiluminescence. This is a technique for detecting changes in infrared emission during a reaction, and it works because radiative vibrational relaxation emits infrared radiation.

The required distribution of kinetic energy between translational and vibrational motion is different in an exothermic reaction and an endothermic reaction. For an exothermic reaction, where the products have a lower potential energy than the reactants, and the activation energy is small, it is better to have less vibrational energy and more translational energy. This is because the activation energy barrier is small and the transition state is directly ahead of the reactants on the potential energy surface plot. This means that the barrier can be easily overcome with enough translational motion, and then the trajectory will just fall into the lower energy products area. Having excess vibrational motion, in this case, could prevent the reaction from occurring because the movement of the vibrational motion on the potential energy surface plot is in a different direction than the minimum energy path. Because of this, excess vibrational energy will cause the trajectory of the reaction to move back and forth up the valley. At the transition state, this can increase the energy barrier that needs to be overcome, and might cause the trajectory to simply fall back towards the products.

For an endothermic reaction, the activation energy is large and a very large amount of translational energy is needed to overcome the barrier. On a potential energy surface, the transition state is also not directly ahead of the reactants on the MEP. It is behind the valley's corner, so approaching it directly requires a very large energy barrier to be overcome. Vibrational motion is more useful in this case because it does not move along the minimum energy path, and if the vibrational motion is timed right, it can more easily overcome the large activation energy barrier. This is because if the timing of the vibrational motion is right, the direction of the trajectory will curve behind the valley's corner in the exact right way for the trajectory to be heading directly towards the transition state. The effect of these conditions is illustrated on the figures below.
[[File:MRD0158103313.png|left|frame|Figure 8: an unreactive trajectory (left) with high translational energy but low vibrational energy, and a reactive trajectory (right) with low translational energy but high vibrational energy. These plots are for the endothermic reaction H + HF -> F + H2]]
[[File:MRD0158103314.png|left|frame|Figure 9: an unreactive trajectory (left) with high vibrational energy but low translational energy, and a reactive trajectory (right) with low vibrational energy and high translational energy. These plots are for the exothermic reaction F + H2 -> HF + H]]

MRD:ML9418

2020-05-21T16:51:42Z

Ml9418: /* Transition State */

== Transition State ==

The transition state on a potential energy surface diagram is defined as the point where the gradient of the potential ∂V(ri)/∂ri is zero in all directions. It is the point on the minimum energy path with the highest energy. 
A transition state can be identified by following the minimum energy path on the potential energy surface and finding the point with the highest potential energy. For a symmetric potential energy surface, the transition state lies on the diagonal line where r1=r2. It is the point on that line with the lowest potential energy. [[File:MRD0158103301.png]] 
A transition state is different from a local minimum due to the fact that it is a saddle point, meaning the surface around it has negative Gaussian curvature. This means that the maximal and minimal values of curvature around the transition state are of opposite signs, meaning they curve in opposite directions (up and down).

For the reaction H2 + H -> H + H2, the transition state is at a point where the distances between neighboring hydrogen atoms are around 90.8 pm.
[[File:MRD0158103302.png|thumb|Figure 2: internuclear distances of hydrogens A-B, B-C and A-C plotted against time]]

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