View source for MRD:Ja1915 Physical new

 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 value of the total gradient of the potential energy surface at a minimum is the minimum value for the total gradient. This point is known as the minimum point on the potential energy plot. The value of the total gradient of the potential energy surface at a transition structure is at a point known as a saddle point. This is because it is at the maximum of the potential surface energy which occurs at the minimum energy path. To distinguish between them, the second derivative of the potential energy surface at this point can be used. For a minumum, the second derivative will be bigger than 0. For a saddle point, the second derivative will be equal to 0.Figure 1 Displaying the plot of internuclear distance against time at the transition state



 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. 
It was found that the location of the transition state is: rts = 0.9077 Å. A saddle point represents the transition state in the potential energy surface. It is at a maximum of potential energy and therefore there is no the molecule does not have any kinetic energy. As a result the inter nuclear bond distance does not change at this point. The location of the transition state was therefore confirmed observing if the inter nuclear bond distance changed with time. As can be seen in figure (), the inter nuclear bond distance does not change therefore this point can be regarded as the transition state.

 Comment on how the mep and the trajectory you just calculated differ. 
The mep is the minimum energy path where atom A is moving away from the di-atom B-C. It can be seen in Figure 2 that once the optimum inter nuclear distance is reached, the energy does not change. However in Figure 3 which displays the dynamic representation of this process, the potential energy is oscillating after optimum inter nuclear distance has been reached.

Internuclear Distances vs Time: 

Final values of the positions r1(t) r2(t): 4.666 and 0.7451

Internuclear Momenta vs Time: 

Average momenta  p1(t) p2(t): p(1) = 2.485 and p(2) = 1.512

The plot gets reflected in the y=x axis because C moves away instead of A but the energy and momentum are the same as when A was moving.  

A calculation was set up for AB distance 4.666 and BC distance 0.7451 with 2.485 and 1.512 respectively. It was observed that atom A moves towards BC until it reaches the transition state + 0.01 distance for AB. After this point has been reached, atom A travels back to its original position taking the same path it took last time.     Figure 2 Displaying the MEP plotFigure 3 Displaying the dynamics plot

 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. 

p1
p2
Reactivity

-1.25
-2.5
Yes, it was reactive

-1.5
-2.0
No it was not reactive

-1.5
-2.5
Yes it was reactive

-2.5
-5.0
Yes it was reactive 

-2.5
-5.2
Yes it was reactive

 In the first case, the atom approached the diatom and displaced one of the atoms forming a diaatom with one of the atoms. The diatom's internuclear distance then increased and decreased as the bond vibrated. The distance between the new diatom and leaving atom increases with time. 
 The diatom is vibrating while the other atom approaches it. It does not have enough momentum to travel close enough to the atom to displace the diatom and therefore returns on the same path it took to this point. 
 The atom approached the diatom and displaced one of the atoms forming a diaatom with one of the atoms. However in this case, the atom approach was slower and the dissociation of the diatom/formation of new diatom was a slower process. The diatom's internuclear distance then increased and decreased as the bond vibrated. The distance between the new diatom and leaving atom increases with time. 
 ?
 In the first case, the atom approached the diatom and displaced one of the atoms forming a diaatom with one of the atoms. The diatom's internuclear distance then increased and decreased as the bond vibrated. The distance between the new diatom and leaving atom increases with time.  
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Figure 4 Figure 6Figure 5Figure 7
Figure 8Figure 9

 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? 
 For each elementary step in a multi-step reaction, the transition state theory assumes that each intermediate produced has a long enough lifetime that a Boltzmann distribution of energies is reached before the reaction continues to the next step. 
 Atoms and molecules will only react or reach the transition state when the atoms or molecules have enough energy. Each atomic nuclei is thus behaving classically. 
 The reaction will go through the lowest energy saddle possible on the potential energy surface.   Any reaction which goes through the transition state never crosses it again.  
https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4555010/ 
This is not the case as seen from figures () where the reaction goes past the transition state and then goes back towards the reactants. As a result, the predicted reaction rates using the transition state theory would be higher than the experimental values obtained in the conditions used for the earlier question. 
Small Note: If you take into quantum mechanics, it is possible for the particles to tunnel across the barrier. This means that molecules that don't have enough energy to react can still react, which disagrees with the second assumption shown. Therefore you would expect the rate of the experimental values obtained to be faster than those predicted from the Transition State theory. However this can't be quantified in this case as quantum mechanics has not been considered in the mechanism.  
 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: The reaction between F + H2 is exothermic as a strong HF bond. The HF bond is stronger than H2, therefore more energy is released from the HF bond than taken in to break the H2 bond. H + HF: The reaction between H + HF is endothermic as HF is a stronger bond than H2  and thus the reaction is endothermic. Figure 8

 Locate the approximate position of the transition state 
Hammond's postulate states that the transition state will resemble the state it is closer to. An exothermic reaction means the products are lower in energy than the reactants and the transition state is closest to the highest energy phase. Therefore in the first reaction, the transition state will resemble the reactant and so bond distances for the transition state will resemble the reactants bond distance. 

The transition state position was found to be: AB (1.81) and BC (0.745). 0.745

 Report the activation energy for both reactions. 
Energy of the reactants at the beginning of the reaction: -103.9

Energy of the transition state of the reaction: -103.7

Energy of the products at the end of the reaction: -133.9

Activation Energy of F + H2 reaction: 0.02 kcal/mol

Activation Energy of H + HF reaction: 30.0 kcal/mol 

 In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally? 
AB distance: 1.81 Momentum: -10

BC distance: 0.745 Momentum: 0
Figure 10
Since there is excess momentum in the reaction, the energy has to be released in order to obey the law of conservation of energy. In this reaction, this excess energy causes the new diatom HF to vibrate. This can be seen in figure () where the bond distance of HF (AB) is fluctuating with time. In order to observe this energy release experimentally, IR spectroscopy could be used where overtones in the IR spectrum of the HF molecule produced would be seen. This is because the excess energy is exciting the vibrational ground state in the molecule. Decay of these states... Also you could you calorimetry. You can also use the emission of a photon in a UV spectrometer?

AB distance: 2.0 Momentum: -0.5

BC distance: 0.74 Momentum: -3

No overall reaction but there is attraction between F and H forming a bond which then dissociates. 

 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 rules state that the vibrational energy is better at promoting a reaction where the transition state resembles the products than translational energy. The translational energy is better at promoting a reaction where the transition state resembles the reactants than the vibrational energy. Therefore if the reaction is exothermic, then it will have an early transition state and this means that translation will contribute more towards the reaction. This results in the total available energy being deposited mostly into vibrations in the product. This can be seen in figure () where in the reaction between F + H2 (exothermic) shows the product (HF) vibrating. http://www.pnas.org/content/105/35/12667.full