MRD:jw7815test123
Exercise 1: H + H2 system
Hi
What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.
In a Potential Energy Surface (PES), the first derivative refers to the gradient of a point on the surface; while the second derivative refers to the rate of change of gradient of a point on the surface, also known as the curvature. A positive curvature implies that the gradient changes from negative to zero to positive (minimum point), a negative curvature implies that the gradient changes from positive to zero to negative (maximum point), and zero curvature implies that the gradient changes from negative to zero to negative or positive to zero to positive (saddle point).
The gradient of the potential energy surface at a minimum and at a transitions structure are both zero.
The minima and transition structures can be differentiated by their curvatures because the minima has a positive curvature (minimum point), while the transition structure has zero curvature (saddle point).
Report best estimate
From the "Internuclear distance vs Time" plot, the best estimate was obtained when rts = 0.9075 Å. With reference to the above Figure, the value was obtained by varying the values of r1 = r2 until the internuclear distance vs time behaviour switches from "oscillating" to just a steady horizontal line. A steady horizontal line means that the internuclear distances are constant with time, and is expected because a transition state will remain in its position indefinitely when the momentum p1 and p2 is zero.
Comment on dynamics vs mep
Using r1 = rts + 0.1 and r2 = rts;
In mep, the trajectory follows the valley floor as a non-oscillating straight line because the velocity is reset to zero in each time step, and the trajectory will only move in the direction of the steepest slope (towards the product).
In dynamic, the trajectory follows the valley as an oscilllating line because it takes into accout the initial velocity, which will influence its subsequent motion in the next time step.
Reactive/Unreactive
In the following analyses, it is important to point out that the momentum we are adjusting is directly proportional to the kinetic energy required to overcome the activation barrier ( KE = p2/2m).
| Reaction | p1 | p2 | Reactive/Unreactive | Trajectory | Description |
|---|---|---|---|---|---|
| 1 | -1.25 | -2.5 | Reactive | re | As A approaches B-C, the AB momentum is high enough such that the kinetic energy is sufficient to overcome the activation barrier, thus the trajectory will pass through the transition state to form A-B, with a bond distance of around 0.70. There is greater oscillation once the transition state is passed, implying that the product A-B has more vibrational energy than the reactants. |
| 2 | -1.5 | -2.0 | Unreactive | rere | As A approaches B-C, the AB momentum is not high enough and there is insufficient kinetic energy to overcome the activation barrier for the trajectory to pass through the transition state to form A-B. The trajectory thus shows an initial decrease in A-B bond length until around 1.1 Å, but increases back to its initial value (to reactants) and is unreactive. |
| 3 | -1.5 | -2.5 | Reactive | rerere | As A approaches B-C, the AB momentum is high enough such that kinetic energy is sufficient to overcome the activation barrier, thus the trajectory will pass through the transition state to form A-B, with a bond distance of around 0.70. Unlike Reaction 1, the product A-B has only slightly more vibrational energy than the reactants as indicated by the greater degree of oscillation once it passes through the transition barrier. |
| 4 | -2.5 | -5.0 | UR | rererere | As A approaches B-C, the AB momentum is high enough such that kinetic energy is sufficient to overcome the activation barrier, however, while the trajectory initially crosses the transition barrier towards the product, it eventually recrosses back towards to reactants, and the reactant is reformed with a much higher vibration energy. |
| 5 | -2.5 | -5.2 | R | rerere | As A approaches B-C, the AB momentum is higher enough such that the kinetic energy is sufficient to overcome the activation barrier. In this case, the trajectory passes through the transition barrier to the products, but recrosses to the reactants and then back to the product A-B, which has a bond length of around 0.70 Å and possess a much greater vibrational energy than the reactants. |
Assumptions of Transition State Theory
The transition state theory is used to qualitatively understand how chemical reactions take place. The following are key assumptions made in this theory:
- The theory assumes that the atomic nuclei follow classical mechanics, where sufficient kinetic energy must be present to overcome the activation barrier. This is not true because in quantum mechanics, atoms with insufficient energy can still tunnel through the activation barrier without actually "overcoming" it.
- The theory assumes that barrier recrossing does not occur, as seen in reaction 4 and 5. This means that if a trajectory passes through the activation barrier, it must lead to products. However, from reaction 4, we see that this is not necessarily true - recrossing occured to reform the reactants. This highlights that even if a molecule possesses sufficient kinetic energy to overcome the activation barrier, there are other factors that must be considered for a successful reaction.
- The theory assumes that the intermediates have a sufficiently long lifetime to reach a Boltzmann distribution of energies before forming the products. In reality, intermediates are not long-lived and exist on a timescale of around 10-13 seconds, which is insufficient for the Boltzmann distribution to be reached.
- The theory assumes that the reaction trajectory passes through the transition state (saddle point on the PES) before reaching the product. This is not necessarily the case, and we see in reaction 5 that s possible for trajectories to result in product without having passes through the saddle point.
- The theory only assumes 3 possible states in the reaction - the reactants, transition complex, and the products, anything else is not considered.
Apart from these assumptions, it is important to highlight that the transition state theory may not apply at elevated temperatures because molecules begin to populate higher vibrational modes and result in complex motion which may lead to transition states far away from the lowest energy saddle point.
Exercise 2: F - H - H system
PES inspection, classify
| Bond | Bond Strength/kJ mol-1 |
|---|---|
| H-F | 565 |
| H-H | 432 |
For F + H2, the reaction is exothermic as the product energy is lower than the reactant energy as seen in the potential energy surface. This is in agreement with the prediction based on bond strengths, as the H-F bond being formed is stronger than the H-H bond being broken, resulting in a net lowering of energy.
For H + HF, the reaction is endothermic as the product energy is higher than the reactant energy as seen in the potential energy surface. Once again, this is to be expected as the H-H bond being formed is weaker than the H-F bond being broken, resulting in a net raising of energy.
Location of T.S
(INSERT PLOT)
From the "Internuclear distance vs Time" plot, a constant internuclear distance is obtained (as shown above) when rHH = 0.744 Å and rHF = 1.811 Å.
Energy conserved
initial conditions PIC
In the forward exothermic reaction F + H2 --> H + HF, the reactants posses more potential energy than the products; because of conservation of energy, through the course of the reaction, the potential energy decreases and is converted to kinetic energy. This reaction energy released in the form of kinetic energy and be converted to vibrational energy and translational energy.
Infrared Chemiluminescence
From the above trajectory, it can be seen that the product H-F has a greater vibrational energy than the reactant H-H represented by the greater degree of oscillations. The infrared emission of the vibrationally excited product can be measured using methods such as infrared chemiluminescence (REF)
Temperature Measurements
If reaction energy is released as thermal energy (heat), this can also be measured experimentally my monitoring the increase in temperature as the reaction progresses.
Polyani stuff
For an efficient reaction, the reactants must have the right distribution of kinetic energy between the vibrational and translational components.
Polanyi's empircal rules states that in an endothermic reaction, vibrational energy is more effective than translational energy in promoting a late transition state. According to Hammond's postulate, endothermic reactions have a later transition state. This means that for endothermic reactions (e.g. HF + H -> F + H2), it is more important to possess vibrational energy than translational energy. This is illustrated in the figures on the left: using conditions of low vibrational energy but high translational energy does not result in a reactive trajectory, however, by increasing the vibrational energy