MRD:lhl17
Contents
Molecular Reaction Dynamics: Applications to Triatomic Systems
Exercise 1: H + H2 system
Question 1
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?
On a potential energy surface diagram, the transition state is mathematically defined as the point at which ∂V(r)/∂r =0 is true. This can be described as the maximum point on the minimum reactive trajectory (minimum energy path linking reactants and products).
The transition state appears as a saddle point in the potential energy surface (PES) - i.e. in one direction, the point is a maximum point in the graph and in the other direction, it is a minimum. The point was found at rAB = rBC = 90.77 pm and pAB = pBC = 0 g.mol-1.pm.fm-1
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.
| Internuclear Distances vs Time | Contour Plot | ||
|---|---|---|---|
| 90.77 | |||
Comment on how the mep and the trajectory you just calculated differ.
When calculating the minimum energy pathway using calculation type MEP and Dynamics, we obtain the following results (with rAB = 90.77 pm, rBC = 91.77 pm and pAB = pBC = 0 g.mol-1.pm.fm-1):
| Dynamic Calculation | MEP Calculation (5000 steps) | ||
|---|---|---|---|
| p1/ g.mol-1.pm.fs-1 | p2/ g.mol-1.pm.fs-1 | Etot | Reactive? | Description of the dynamics | Illustration of the trajectory |
|---|---|---|---|---|---|
| -2.56 | -5.1 | -414.280 | Yes | Reaction path can pass through the TS and form products. The vibrational oscillations in the product channel confirm formation of the the product. | |
| -3.1 | -4.1 | -420.077 | No | Reaction path unable to get over maximum at TS and form products. Instead, the system roll back down the potential to reactants. | |
| -3.1 | -5.1 | -413.977 | Yes | Momentum of incoming H atom increased compared to first example, and the reaction path is able to pass over the TS to the product channel. More oscillations observed in the product than reactant channel. | |
| -5.1 | -10.1 | -357.277 | No | Initial momenta large enough for reaction path to reach the TS, but excess energy causes the product to cross the activation barrier twice and eventually returns to reactant (with significantly higher vibronic energy). | |
| -5.1 | -10.6 | -349.477 | Yes | Similar to the previous example, but with the momentum of the incoming H atom increased. The system fluctuates twice between reactant and product bond formation, which suggests the TS in this scenario is far from the lowest energy saddle point. |
Complete the table above 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?
Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?
The Transition State Theory assumes that only the reactants are in constant equilibrium with the transition state structure, and that the energy of the particles follow a Boltzmann distribution. In addition, it is assumed that once the reactants have trajectories with a kinetic energy along the reaction coordinate greater than the activation energy, it will be reactive, and that the products will definitely form, ignoring the possibility that the transition state structure will collapse back to into the reactant channel, as in row 4.
The experimental scenarios above show that excess kinetic energy can result in multiple crossing of the activation energy barrier through multiple forward and backward directions. Theoretically, the theory suggests only a single crossing of the activation energy barrier in the forward direction is possible.
Finally, the theory fails for some reactions at elevated temperatures due to more complex molecule motions or low temperatures due to quantum tunneling, whereby the transition state need not be reached to form products.
Thus, it is clear that Transition state theory overestimates the rate of reaction.
Exercise 2: F - H - H system
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?
| Reaction | F + H2 | H + HF |
|---|---|---|
| Type of reaction | Exothermic | Endothermic |
| TS | -433.981 | As above |
| Reactiant Energy | energy graph straight | energy graph with step |
| Activation energy |
Locate the approximate position of the transition state.
| TS | ||
|---|---|---|
Report the activation energy for both reactions.
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.
| Contour plot | Momenta vs time | |
|---|---|---|
- Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, with a momentum pFH = -1.0 g.mol-1.pm.fs-1, and explore several values of pHH in the range -6.1 to 6.1 g.mol-1.pm.fs-1 (explore values also close to these limits). What do you observe? Note that we are putting a significant amount of energy (much more than the activation energy) into the system on the H - H vibration.
| pHH/gmol-1pmfs-1 | Relative trajectory | Contour plot | |
|---|---|---|---|
| -6.1 | |||
| -5 | |||
| 5 | |||
| 6 |
- For the same initial position, increase slightly the momentum pFH = -1.6 g.mol-1.pm.fs-1, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.2 g.mol-1.pm.fs-1. What do you observe now?
| Contour | Momentum | Energy | |
|---|---|---|---|
Let us now focus on the reverse reaction, H + HF.
- Setup initial conditions starting at the bottom of the entry channel, with very low vibrational motion on on the H - F bond, and an arbitrarily high value of pHH above the activation energy (an H atom colliding with a high kinetic energy).
- Try to obtain a reactive trajectory by decreasing the momentum of the incoming H atom and increasing the energy of the H - F vibration. (It may be difficult to find the initial conditions that generate a reactive trajectory for this reaction. Using the inversion of momentum procedure for a trajectory starting on the transition state can be useful in this case. Working on the Skew Plot framework could also be helpful.)
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.
Polanyi's Empirical rules state that vibrational energy is more efficient at activating a late transition state than translational energy (endothermic reaction - Hammonds Postulate) and that translational energy is more efficient in activating an early transition state in an exothermic reaction than vibrational energy. Therefore for a successful exothermic reaction, the molecules need an excess of translational (kinetic) energy compared to vibrational (oscillatory) energy.
In the PES of a chemical reaction, there exists an energetic barrier (the transition state) to reach the product which is the transition state (saddle point - highest value of potential energy) in the potential energy surface. F + H2 reaction is exothermic so the transition state is early according to Hammonds Postulate, the reverse reaction (FH +H) is endothermic and the transition state is late and resembles the products more (Hammonds Postulate).
References
- w
- w
