MRD:hys116
H + H2 System
Transition State
At both the minima and at the transition state, the gradient of the potential energy surface is 0. This is a saddle point with both the maxima of the transition state and the minima in the same position but they can be identified using their second derivative and the shape of the curvature of the plot. The second derivative of the transition state is less than 0 as it is a maximum and this can be identified from the potential energy curve as the curve between the base of the plot and the minimum energy path. The second derivative of the minima is greater than 0 and is the minimum point in the minimum energy path.
The transition state position has been estimated to be 0.9076. From the contour plot, this is where the atoms are in the same position, not falling off the ridge and there is no orthognoal gradient acting on the system. From the plot of internuclear distance vs time, this position shows no oscillation as there is no momentum in the system so the transition state is still. The internuclear distance of A-B is equal to B-C and this is seen in the plot of internuclear distance vs time in which the corresponding two lines are overlapping.
Calculating the Reaction Path
The MEP diagram shows a straight line for the trajectory whereas the dynamic trajectory is a wavy line. This is because it does not take into account oscillations of the bonds, as in each time step the velocity is reset to 0, so this provides a slightly less accurate path.
Ng611 (talk) 15:46, 1 June 2018 (BST) Accurate here is subjective based on what you're hoping to get from your calculation. An MEP allows you to accurately track the reaction channel, while a dynamics simulation provides a more realistic picture of what's going on -- they're both useful.
Reactive and Unreactive Trajectories
| p1 | p2 | Total Energy | Reactivity | Contour Plot | Description |
|---|---|---|---|---|---|
| -1.25 | -2.5 | -99.018 | Reactive |  |
C approaches A-B and collides. The collision has sufficient momentum to cross the transition state and for a reaction to occur. The A-B bond breaks and a B-C bond is formed and the molecules move in opposite directions. |
| -1.5 | -2.0 | -100.456 | Unreactive |  |
As before, C approaches A-B and collides. However, the collision does not have enough momentum for a reaction to occur so the trajectory does not cross the transition state and C moves away from the A-B bond, which remains intact. |
| -1.5 | -2.5 | -98.956 | Reactive |  |
This is very similar to the first situation as there is sufficient momentum for the collision to result in a reaction and the A-B bond to break. A-B has a slightly lower momentum than in the first situation so it takes slightly longer for the reaction to occur. |
| -2.5 | -5.0 | -84.956 | Uneactive |  |
This collision has sufficient momentum to cross the transition state and break the A-B bond. However there excess momentum so the trajectory recrosses the barrier, resulting in the reformation of the A-B bond and an overall unreactive situation. |
| -2.5 | -5.2 | -83.416 | Reactive |  |
This begins in the same way as the previous situation, with the trajectory crossing then recrossing the transition state barrier, however there is still excess momentum so the trajectory crosses the barrier again, resulting in an overall reactive situation. |
Transition State Theory
Transition State Theory is used to predict reaction rates and is based on a number of assumptions. It assumes that nuclei behave classically, and ignores the effect of quantum tunnelling. According to classical mechanics, a collision between reacting atoms/molecules must have energy greater than or equal to the activation energy in order for the reaction to occur. However, quantum tunnelling allows particles to tunnel through energy barriers that they cannot get over classically, meaning molecules can react, even if their collisions do not have energy ≥ Ea. Tunnelling is more likely in reactions with low activation energy.
Another assumption is that intermediates formed are long-lived, however this does not affect the results in this experiment as this is an elementary reaction with no intermediates. It is also assumed that the reaction trajectory is acrsoss the lowest energy saddle point, however at high temperatures molecules can be in higher vibrational energy levels so different, higher energy transition states may exist.
The experimental values obtained are likely to be similar to the Transition State Theory predictions however there will be small systematic differences due to quantum tunnelling and molecules in higher vibrational modes, which can be reduced by carrying out calculations at low temperatures. From the previous examples, the reactions are successful if they have energy equal to or greater than the activation energy so this theory gives accurate predictions for the previous examples.
Ng611 (talk) 15:47, 1 June 2018 (BST) What about recrossing of the transition state?
F-H-H System
PES inspection
F + H2 ---> HF + H
Using the inital conditions:
F H distance: 2.30
H-H distance: 0.74
F H momentum: -2.7
H-H momentum: 0
the following surface plot was produced:
The reaction is exothermic as the products are lower in energy than the reactants. This suggests that the H-F bond formed is stronger than the H-H bond broken.
H + HF ---> + H2 + F
Using the inital conditions:
H H distance: 2.30
H-F distance: 0.74
H H momentum: -5
H-F momentum: 0
The surface plot shown below was produced:
The reaction is endothermic as the products are higher in energy than the reactants. Again, this suggests that the H-F bond broken is stronger than the H-H bond formed.
Transition State
Transition state distances were estimated to be:
H-H distance = 0.7465 Å
H-F distance = 1.8095 Å
At these distances, the contour plot shows no reaction trajectory and the plot of internuclear distances vs time shows minimum oscillation.
Activation Energies
The activation energies were calculated by finding the difference between the energy of the transition state and the energy of the reactants.
Energy of the transition state: -103.751 kcal/mol
The energy of the reactants can be found by displacing the H-F bond length by 0.1 Å in each direction.
F + H2
When 0.1 Å is added, the reaction falls towards the F + H2 reaction. The last point of the trajectory; the beginning of this reaction, can be found and a MEP was plotted:
Energy of F + H2: -103.989 kcal/mol
Activation energy = -103.751 - (-103.989) = 0.238 kcal/mol
HF + H
Similarly, when 0.1 Å is subtracted from the H-F bond distance, the reaction falls towards the side of HF + F. Again, the last point of the trajectory was found and a MEP plotted:
Energy of H + HF: -133.762 kcal/mol
Activation energy = -103.751 - (-133.762) = 30.011 kcal/mol
Reaction Dynamics
Reaction Energy
The following conditions were found to result in a reactive trajectory shown in the contour plot below:
H F distance: 2.3
H-H distance: 0.74
H F momentum: -2
H-H momentum: 2.7
The H2 molecule approaches F and collides. H-H bond breaks and one H atom collides with F, then recrosses the barrier and collides with other H then crosses the barrier again to form H-F bond.
Graphs of momentum vs time and energy vs time are shown below:
The momentum graph shows that before the collision, that H2 (BC) has kinetic energy as shown by the oscillations in the graph. After the collision, the line for H2 shows no oscillation; this energy has been converted into potential energy. F-H (AB) shows the opposite pattern. Energy is conserved, so on collision, the kinetic energy of H2 is transferred to HF. This is supported by the energy v time graph which shows that potential and kinetic energy are mirror images, showing that all kinetic energy is converted into potential and vice versa; energy is conserved.
This could be determined experimentally by analysing the IR spectra of the reactants and the products as the energy of the oscillations of the bonds is conserved during the reaction.
Ng611 (talk) 15:48, 1 June 2018 (BST) Good suggestion! What would you expect to see in your IR spectrum that would allow you to determine how much vibrational energy is present?
Polanyi's empirical rules
Polanyi's empirical rules state that for reactions with late transition states, such as the endothermic reaction of HF + H, vibrational energy has a larger effect on the efficiency of the reaction than translational energy. For reactions with early transition states, such as the exothermic reaction of F + H2, translational energy has a larger effect than vibrational energy.
F + H2:
This reaction is exothermic so has an early transition state. Therefore increasing the translational energy of F and decreasing the vibrational energy of H2 increases the efficiency of the reaction.
When the vibrational energy is varied between -3 and 3, this effect can be seen. When the vibrational energy (HH momentum) is close to 0, the reaction is successful.
F H distance: 2
H2 distance: 0.74
F H momentum: -0.5
When the vibrational energy (HH momentum) is close to 0, the reaction is successful:
As vibrational energy increases (further away from 0), the reaction becomes unsuccessful
For pHH = -3:
For pHH = +3:
This is because the excess vibrational energy causes the trajectory to recross the transition state and results in a non-reactive collision.
For same positions but:
F H momentum: -0.8
H-H momentum: 0.1
This is reactive. In this example, the translational motion of F has been increased and the vibrational energy has been decreased. From Polanyi's rules, this increases the efficiency of the reaction, as supported by the contour plot above.
HF + H
This reaction is endothermic so has a late transition state. Therefore increasing the vibrational energy of HF and decreasing the translational energy of the incoming H increases the efficiency of the reaction as shown below
Initial conditions:
H-F distance: 0.91
H H distance: 2.3
The momentum of the incoming H atom was started at -2.5 and the HF momentum at 0.1.
When pHF (vibrational energy) is increased to 1 and pHH (translational energy) is decreased to -10, a reactive trajectory is found:
Ng611 (talk) 15:53, 1 June 2018 (BST) I'm not sure your final trajectory can really be considered 'reactive', as the kinetic energy is so high that it's basically ignoring the PES (and thus any products formed will probably not be closely associated with one another).
Ng611 (talk) 15:53, 1 June 2018 (BST) Overall, a reasonable report. There are a few questions (especially this final question, the question on the mechanism of energy release, and the question on TS theory) that certainly needed more thought. Overall though, a good report.