MRD:ksg115
EXERCISE 1: H + H2 system
Gradient at the Transition State
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
The transition state is defined as the maximum on the minimum energy energy path. The transition structure is at a saddle point[1], also called a minimax point. Saddle points have a gradient equal to zero: ∂V(ri)/∂ri=0. It is possible to locate a transition state by setting conditions near an expected/predicted point and running calculation to see if there was any movement "rolling" on the potential energy surface. If the position does not invoke movement, then the gradient at that point is 0 and it is a minima. If the position invokes rolling, then the initial conditions would need to be altered to find point with a gradient equal to 0.
Nf710 (talk) 16:12, 2 June 2017 (BST) It is a maximum on the reaction coordinate.
Transition State Estimations
Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” screenshot for a relevant trajectory.
Using the initial conditions of r1=0.74, r2=2.3, AB momentum=0 and BC momentum=-2.7, we can look at the inter-nuclear distances vs time plot to get an approximate value for the transition state geometry. Figure 1 shows that at the transition state of 0.38 time units, r1=r2. The approximate value given for this distance is 0.9148 Å.
| Fig.1 - Surface plot for the initial conditions. |
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It is implied that at the transition state, each atom is stationary and thus momenta are equal to 0.
Changing the conditions to make the AB and BC momenta equal to 0, it can be possible to find a best estimate for the transition state geometry. Inserting r1=r2=0.91, the inter-nuclear distances vs time graph show how the lines are almost linear.
| Fig.2 - Internuclear Distance v. Time. |
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These values start and stop the trajectory at the transition state where it stays there indefinitely. The distance was optimized by trial and error to find the value where the lines were as linear as possible. My best estimate for the transition state geometry is when r1=r2=0.9075 Å and the energy being -98.8 kcal/mol.
| Fig.3 - Optimised, Internuclear Distance v. Time. |
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| Fig.4 - Transition State Energy. |
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Running a MEP Calculation
Comment on how the mep and the trajectory you just calculated differ.
The minimum energy path (mep) is a trajectory that corresponds to infinitely slow motion, where the velocity=0 at every next step. The following calculations have begun with the internuclear distance displaced slight, with r1=rts+0.01 and r2=rts. The momenta in both cases are zero. In the dynamic calculation, you can see a long reaction pathway in the form of an oscillating line. The oscillation shows how an atom can move up and down a potential well.
| Fig.5 - Dynamic Calculation with displaced r1. |
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However, this is not seen in mep as the calculation runs to always find the minimum energy point and hence a shorter line is seen, which also does not oscillate.
| Fig.6 - mep Calculation with displaced r1. |
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Take note of the final values of the positions r1(t) r2(t) and the average momenta p1(t) p2(t) at large t.
| r1(t) | r2(t) | p1(t) | p2(t) |
|---|---|---|---|
| 0.75 Å | 5.28 Å | 1.24 | 2.48 |
What would change if we used the initial conditions r1 = rts and r2 = rts+0.01 instead?
There would still be oscillation (in the dynamic calculation) and a similar energy pathway however the reaction direction would be different due to the change in displacement conditions, with the B-C distance decreasing as the transition state is reached.
| Fig.7 - mep Calculation with reversed displacement. |
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| Fig.8 - Dynamic Calculation with reversed displacement. |
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Reactive and Unreactive Trajectories
Complete the table by adding a column reporting if the trajectory is reactive or unreactive.
For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory.
Initial positions of r1=0.74, r2=2.0 with variable momenta p1, p2.
| p1 | p2 | Trajectory | Figure | Description |
|---|---|---|---|---|
| -1.25 | -2.5 | Reactive | 
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The trajectory is reactive. The reaction proceeds through the transition state to form products. Molecule BC is at a constant distance as A approaches the diatomic until collision where an AB molecule has formed and is now moving along the minimum energy pathway. The BC distance thus increases as it moves away from the newly-formed diatomic which vibrates, moving up and down the potential well. |
| -1.5 | -2.0 | Unreactive | 
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The trajectory is unreactive, Molecule A approaches BC, but only gets as close as approximately 1.16 Å to B before its velocity is reversed and it moves away from the diatomic. Molecule A does not have enough kinetic energy to overcome the activation energy barrier. The BC distance stays constant with minimum oscillation yet it does become perturbed slightly as A gets as close as it can to the molecule. |
| -1.5 | -2.5 | Reactive | 
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Compared to the scenario above, this scenario is reactive. This is because A now has enough kinetic energy to overcome the energy barrier of reaction and form molecule AB. Both BC and AB vibrate before and after the reaction has occurred, respectively. |
| -2.5 | -5.0 | Unreactive | 
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Molecule A approaches BC with enough momentum to cross the transition state, but because A has given so much vibrational energy to the new AB molecule, the oscillation of the bond allows the transition state to be crossed in the backwards direction, reforming the BC bond and A moving away. |
| -2.5 | -5.2 | Reactive | 
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Compared to Scenario D (above), this trajectory is reactive. Although there is a period where BC reforms after the initial A collision, the oscillation of BC allows A to cross the transition state again, forming AB. |
Transition State Theory
State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?
Transition State Theory (TST) eliminates the need for trajectory calculations. It requires that the only known regions of a potential energy surface are the reactants and the transition state. TST uses a boundary surface (critical dividing surface), between the reactants and products, which is placed through the saddle point. One assumption of TST is that all supermolecules that cross the boundary surface become products (which is plausible as once a supermolecule has passed the boundary line, it is a downhill pathway to the products) Other assumptions of TST: - During the reaction, the Boltzmann distribution of energy is maintained for the reactant molecules. - Supermolecules crossing the critical surface from reactants have a Boltzmann distribution of energy coresponding to the temperature of the reacting system. - Motion along the reaction coordinate can be treated clasically.
The TST predictions will be different from experimental values. The assumption of motion being treated in a clasical fashion fails for reactions involving light species (such as in a H + H(r2) system), where there is a possibility for quantum-mechanical tunelling. A correction must be applied to take this assumption into account[2].
EXERCISE 2: F - H - H system
Exothermic and Endothermic Reactions
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?
F + H2 --> HF + H is an exothermic reaction. According to Hammond's Postulate, in an exothermic reaction, the transition state will resemble the reactants more than the products.
| Fig.9 - A reactive trajectory for F + H2 --> HF + H. |
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The above figure shows a reactive trajectory between F and H-H. The reaction is successful as a H-F diatomic is made, which oscillates. Simultaneously, as the new HF bond forms, a H atom moves away from the newly-formed diatomic. The reaction trajectory shows how formation of HF is exothermic, as the products are lower in energy than the products. Following on from this, the backwards reaction, H + H-F --> H-H + F, is an endothermic reaction as the products now are at a higher energy than the reactants. This relates to the bond strengths of the molecules formed and broken. It shows that the H-F bond is stronger than the H-H bond as the formation of H-F is an exothermic process, meaning it requires more energy to break.
Transition State of the F - H - H system
Locate the approximate position of the transition state.
According to Hammond's Postulate, in an exothermic reaction, the transition state will resemble the reactants more than the products. The transition state was found by trial and error, with values of H-H being 0.75 Å and H-F 1.81 Å. By looking at the internuclear distance v. time graph, it was possible to alter the AB and BC distance to obtain linear lines. These represent the transition point, which should be fixed and not oscillating.
| Fig.10 - Surface plot for the transition state. |
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| Fig.11 - Internuclear Distance v. Time, F + H2 --> HF + H |
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Activation Energy
Report the activation energy for both reactions.
The activation energy for the exothermic reaction to form H-F is the difference between the energy of the reactants and the transition state. Figure 12 shows an mep calculation, which resulted in the value of the transition state energy being -103.3 kcal/mol.
| Fig.12 - mep Calculation, F + H2 --> HF + H |
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Figure 13 (below) shows a potential energy v. time graph, which is from a mep calculation used to find the lowest energy point of the reactants, H-H and F. The value of AB was displaced slightly (+0.01) so when the mep calculation ran, the reactants, which were at virtually the transition state, would essentially roll down the potential energy surface to the lowest point. The value calculated was found to be -104.02 kcal/mol.
| Fig.13 - Potential Energy v. Time, reactants energy for F + H2 --> HF + H |
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Using this information, the activation energy for the exothermic reaction is: -103.3 - -104.02 = 0.72 kcal/mol
Figure 14 shows the mep calculation for the products. The value calculated was found to be -133.9 kcal/mol.
| Fig.14 - mep Calculation for the products. |
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The activation energy for the endothermic reaction is:
-103.3 - -133.9 = 30.6 kcal/mol.
Mechanism of Reaction Energy
In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?
According to the law of conservation of energy, energy can only converted from one form to another, and cannot be created or destroyed. For a simple triatomic reaction, such as F + H-H, F approaches the diatomic with kinetic energy and as it gets closer, potential energy rises as a result of electronic repulsion and kinetic energy decreases. Kinetic energy is converted to potential energy. The maxima of this potential energy is reached at the transition state, where for an instant in time, all atoms are stationary and all energy is in the form of potential and no kinetic. Once the reaction has proceeded past the transition state, potential energy is converted to a) translational, as a H atom moves away from the newly formed H-F diatomic, and b) vibrational energy in the form of an oscillating diatomic H-F molecule.
Calorimetry can be used to find the overall energy change of a reaction. If energy is released in the form of heat (vibrational energy), the reaction is exothermic. This method could confirm experimentally that the reaction between F and H-H is exothermic. The vibrational energy can be probed further via infrared spectroscopy.
Distribution of Energy
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.
In an exothermic reaction, the transition state is nearer to the reactants than the products, via Hammond's Postulate. The potential energy surface (PES) is thus said to be attractive. The transition state is reached as the reactants approach each other, exhibiting translational energy which as they approach each other, is converted into potential energy. Following the TS, as a new bond forms and the bond distance decreases to an equilibrium position. However, potential energy is converted into vibrational energy in the form of an oscillating diatomic.
For endothermic reactions, the PES type determines the type of energy that would be most effective in resulting in a successful reaction. At an attractive surface, translational energy is most effective in promoting a reaction whereas for an unattractive surface (such as that in an endothermic reaction, with a late transition state), vibrational energy is more effective for inducing reactions[1].
Nf710 (talk) 16:36, 2 June 2017 (BST) I dont know what you mean about attractive surfaces. You should be saying early and late TSs. Also you havent made an attempt to understand or study Polyaanis rules. This is the whole point of the experiment.
References
[1] - Atkins, Peter, and Julio De Paula. Atkins' Physical Chemistry. 10th ed. Oxford: Oxford University Press, 2006, pg. 909-911.
[2] - Levine, I.N. Physical Chemistry. 1st ed. Auckland: McGraw-Hill, 1978, pg. 893.