File:MO phh3.PNG

2018-05-11T14:33:21Z

Mmo116:

MRD:mmo116

2018-05-11T14:33:05Z

Mmo116:

==Exercise 1: H + H2 system==

[[File:MOS initial traj.PNG|left|thumb|Figure 1: Internuclear Distances vs Time with '''r1'''=0.74, '''r2'''=2.3 and '''p1'''= 0, '''p2'''=-2.7]]
[[File:MOS first TS.PNG|thumb|Figure 2: Internuclear Distances vs Time with '''rts'''=0.915 and '''p1'''='''p2'''=0]]
[[File:MOS_1000_step_TS.PNG|left|thumb|Figure 3: Internuclear Distances vs Time with '''rts'''=0.910075 and '''p1'''='''p2'''=0]]
[[File:MOS reaction path MEP.PNG|thumb|Figure 4: Minimum Energy Path]]
[[File:MOS reaction path dynamics.PNG|left|thumb|Figure 5: Trajectory]]
[[File:MOS mep surface.PNG|thumb|Figure 6: Surface Plot mep]]
[[File:MOS dynamics surface.PNG|left|thumb|Figure 7: Surface Plot Trajectory]]
[[File:MOS dynamics momenta.PNG|thumb|Figure 8: Internuclear Momenta vs Time Trajectory]]
[[File:MOS reaction path dynamics switch.PNG|left|thumb|Figure 9: Trajectory switched]]
[[File:MOS dynamics momenta switch.PNG|thumb|Figure 10: Momenta switched]]
[[File:MOS contour reverse.PNG|left|thumb|Figure 11: Trajectory with swapped initial conditions]]
[[File:Reactive or unreactive 1.PNG|thumb|Figure 12: '''p1''' = -1.25, '''p2''' = -2.5]]
[[File:Reactive or unreactive 2.PNG|left|thumb|Figure 13: '''p1''' = -1.5, '''p2''' = -2.0]]
[[File:Reactive or unreactive 3.PNG|thumb|Figure 14: '''p1''' = -1.5, '''p2''' = -2.5]]
[[File:Reactive or unreactive 4.PNG|left|thumb|Figure 15: '''p1''' = -2.5, '''p2''' = -5.0]]
[[File:Reactive or unreactive 5.PNG|thumb|Figure 16: '''p1''' = -2.5, '''p2''' = -5.2]]

===What value do the different components of the 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.===

A potential energy surface shows how the potential energy of a system changes with changing interatomic distances. It can be used to follow a chemical reaction and to identify minimum energy states and transition states. The gradient of the potential energy surface (i.e. its first derivative) is 0 at both minima and transition states. The second derivatives are able to identify whether the stationary point is a saddle point, and hence a transition state, or a minima. If it has a positive value in both the x and y directions then the point is a minimum potential energy state and if it is positive in one direction and negative in the other, it is a transition structure. These can be sometimes be seen by eye where the transition state is the top of a peak in the trough and a minimum would be the very bottom of a trough.

===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.===

We're looking at H + H2 so all the atoms in the system are the same and it's symmetric. This means at the transition state '''r1'''='''r2'''='''rts'''. Both momenta will be 0 because it's a stationary point. Using a trial and error method, with an initial guess of '''rts '''where A-B and B-C cross in Figure 1 (0.915), the value of '''rts''' was estimated to be 0.910075. The condition given was that if the internuclear distances differed then it was not the transition state.

In order to get the best estimate, a graph with A-B and B-C with the exact same line is ideal. This is when you can only see one line at the bottom of the graph. The longer the lines stay the same, the better the estimate for the interatomic distance at the transition state. A result for this was when '''rts''' = 0.915 but when this was extended to 1000 steps from 500, it was showed that the they did not stay the same after about 2.8 seconds, so the estimate could be improved (Figure 2). The final value, which showed A-B and B-C being equal at 1000 steps was 0.910075 (Figure 3).

===Comment on how the minimum energy path (mep) and the trajectory differ.===
The starting positions were set as:

A-B distance: 0.920075

B-C distance: 0.910075

So slightly away from the transition state. The mep then shows the lowest energy pathway of the molecule B-C forming - so the distance between the two atoms gets smaller - and the distance between B and C gets larger (Figure 4). It is a straight line with no oscillations (Figure 6).

The trajectory is the actual path the atoms take. It's an oscillating line showing the same trends, just with vibrations in the bond (Figure 5) so in actuality they do not follow the minimum exactly
It looks to the atoms rolling up and down the surface plot (Figure 7).

Other differences between the mep and the trajectory is that the trajectory has much less smooth transitions. The distance between A and B increases much more quickly - the gradient is steeper in Figure 5 - and thus, it ends with much larger distances between the two atoms. It also is a straight line rather than a curved line like in the mep. This makes sense because after a certain distance the atom and the molecule stop interacting and so move off at a constant speed. The mep shows the rate of change of distance between the atom and molecule tailing off which is unrealistic as they will have gained momentum going in opposite directions and the mass does not change so they'll keep moving at a constant speed with out other forces acting on them.

For both the mep and the trajectory, the distance between B and C remains pretty much constant after the atom and the molecule split off from eachother - disregarding the slight oscillations in the trajectory. This is to be expected as there is a bond between the two so the distance will not change much and also explains why the gradient of the trajectory is greater than the mep - an actual bond has been formed.

The final values of distance at 2.5 seconds are:

'''r1'''= 9.0278

'''r2'''= 0.7680 (Figure 5)

The average momenta at large t is:

'''p1'''=2.4868

'''p2'''=1.2465 (Figure 8)

===Changed initial conditions so '''r1''' = '''rts''' and '''r2''' = '''rts'''+0.01 instead?===

The same pattern can be seen - just with A-B and B-C switched on the graph (Figures 9 and 10). This makes sense because of the symmetry in the system. The trajectory follows a path in the y-direction instead of the x but it is an equivalent path just with A and B as the molecule and C as the atom instead of B and C being bonded like before. The starting position was switched to the other side but by the same quantity so the direction is all that has changed.

===When initial positions are set to the final positions of the trajectory and the initial momenta values are the same as the final momenta values, with reversed signs.===
'''r1'''= 9.0278

'''r2'''= 0.7680

'''p1'''=-2.4868

'''p2'''=-1.2465

The contour plot (Figure 11) shows the trajectory with these switched conditions.

We've placed the atoms at their final positions from before and given them enough momentum in the right direction to get back to the transition state. What you actually see is when they get to the close to the transition state, they don't have enough energy to overcome the energy hill and move over to the other side and they look like they're starting to roll back down the hill again. As they've been given the same momenta (in the opposite direction) and position, you'd expect the system to have the exact right conditions to get the the transition state and stay at the stationary point. This would need the EXACT values of all parameters to be the case and we have only a good estimate for the actual transition state structure and the final positions from before.

===Reactive and unreactive trajectories===

{| class="wikitable" border="1"
|+
!'''p1''' !! '''p2''' !! Total Energy !! Reactive/Unreactive Trajectory
|-
| -1.25 || -2.5 || -99.018 || Reactive
|-
| -1.5 || -2.0 || -100.456 || Unreactive
|-
| -1.5 || -2.5 || -98.956 || Reactive
|-
| -2.5 || -5.0 || -84.965 || Unreactive
|-
| -2.5 || -5.2 || -83.416 || Reactive
|}

All paths start with A and B bonded to each other and C as a separate atom.

'''Figure 12:''' 1) There is no oscillation between A and B. They've been set at a certain distance apart and as they follow the path they have been given the same momentum. As the atoms get to the transition state, they begin to interact with each other and oscillation begins. They all have enough momentum to overcome the energy barrier at the transition state and so B and C are bonded, leaving A as a single atom. The reaction has occurred. This oscillates - the two atoms in the molecule have different momenta.

'''Figure 13:''' 2) Here molecule A-B has a slightly higher momentum but atom C has less. There is not enough energy to get to the transition structure, never mind over it. This means that although the molecule and the atom get close, they just repel each other when they do and so no reaction occurs.

'''Figure 14:''' (description)

'''Figure 15:''' (description)

'''Figure 16:''' (description)

===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?===

==Exercise 2: F - H - H system==

[[File:MO endo or exo.PNG|left|thumb|Figure 17: Surface Plot of F-H-H]]
[[File:MO ts initial.PNG|thumb|Figure 18: First attempt for transition state with '''r1'''=1.5, '''r2'''=0.5 and '''p1'''='''p2'''=0]]
[[File:MO ts estimate.PNG|left|thumb|Figure 19: Final estimate for transition state with '''r1'''=1.81412, '''r2'''=0.73915 and '''p1'''='''p2'''=0 ]]
[[File:MO activation energy 1.PNG|thumb|Figure 20: MEP Energy vs Time with '''r1'''=1.7, '''r2'''=0.73915 and '''p1'''='''p2'''=0]]
[[File:MO activation energy 2.PNG|left|thumb|Figure 21: MEP Energy vs Time with '''r1'''=1.81412, '''r2'''=0.79 and '''p1'''='''p2'''=0]]
[[File:MO reactive traj.PNG|thumb|Figure 22: Reactive Trajectory with '''r1'''=2.3, '''r2'''=0.74 and '''p1'''=-2.5 '''p2'''=-2]]
[[File:MO reactive traj energy.PNG|left|thumb|Figure 23: Kinetic and Potential Energy Plot of Reactive Trajectory]]
[[File:MO phh-3.PNG|thumb|Figure 24:]]
[[File:MO phh-2.8.PNG|left|thumb|Figure 25:]]
[[File:MO phh-2.5.PNG|thumb|Figure 26:]]
[[File:MO phh2.6.PNG|left|thumb|Figure 27:]]
[[File:MO phh3.PNG|thumb|Figure 28:]]

===PES inspection===

===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?===

Figure 17: A-B corresponds to HF and B-C corresponds to H2.

When A-B is small - the energy is lower than when B-C is small. This means that the F + H2 reaction is exothermic and H + HF reaction is endothermic, as it has to be due to it being the reverse reaction.

Bond breaking needs energy and bond energy releases energy. Because the H+HF reaction overall takes in energy and the F + H2 reaction releases energy overall, the HF bond is stronger than the H2 bond.

===Locate the approximate position of the transition state===
Hammond's postulate says the transition state structure will be closer to the products' structure in an endothermic reaction. Because AB is the bond that's broken and corresponds to the H-F bond length and it's the bond being broken, you'd expect the transition state to resemble H2 more closely, hence the BC distance will be smaller. The initial guess for the transition state was '''r1'''(AB)= 1.5 and '''r2'''(BC)= 0.5. The outcome of this can be seen in Figure 18. The best estimate would have no path line and just a cross so a much better guess was needed.

After a series of trial and error the best estimate was '''r1'''(AB)= 1.81412 and '''r2'''(BC)= 0.73915 (Figure 19)

===Report the activation energy for both reactions.===
The energy of the transition state = -103.739

An MEP with 9000 steps of the following conditions was calculated and an energy vs time plot was made (Figure 20)

AB = 1.7

BC = 0.73915

The final energy was -133.443 meaning the activation energy of the HF + H was -103.739-(-133.443)=29.704.

The second activation energy was calculated using the following conditions using the same method (Figure 21):

AB=1.81412

BC=0.79

The final energy was -103.009 meaning the activation energy of the H2 + F was -103.009-(-103.739)=0.73. This was the bets estimate however, the further the BC distance was from the transition state, the larger the activation energy became so it is hard to know which is right.

===Reaction Dynamics===

=== In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?===

The reactive trajectory I've plotted (Figure 22) is as follows:

AB=2.3 Å

BC=0.74 Å

AB Momentum = -2.3 kgm/s

BC Momentum = -2 kgm/s.

There are smaller oscillations between the H2 molecule than the HF molecule when it has been formed. This could clearly be seen in the animation also. Figure 23 shows both the kinetic and potential energies of the trajectory. You can see that the potential energy and the kinetic energy are mirror images of one another so when the kinetic energy increases the potential energy decreases - the energy is conserved. As the potential energy overall decreases and the kinetic energy increases, the mechanism for the release of reaction energy is from potential to kinetic energy. The kinetic energy increasing is to do with the release of heat by the exothermic reaction (we have calculated the H<sub2 + F reaction).

=== 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.===

File:MO phh-3.PNG

2018-05-11T14:32:56Z

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File:MO phh2.6.PNG

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File:MO phh-2.5.PNG

2018-05-11T14:32:14Z

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2018-05-11T13:54:47Z

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MRD:mmo116

2018-05-11T13:52:06Z

Mmo116: /* In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally? */

==Exercise 1: H + H2 system==

[[File:MOS initial traj.PNG|left|thumb|Figure 1: Internuclear Distances vs Time with '''r1'''=0.74, '''r2'''=2.3 and '''p1'''= 0, '''p2'''=-2.7]]
[[File:MOS first TS.PNG|thumb|Figure 2: Internuclear Distances vs Time with '''rts'''=0.915 and '''p1'''='''p2'''=0]]
[[File:MOS_1000_step_TS.PNG|left|thumb|Figure 3: Internuclear Distances vs Time with '''rts'''=0.910075 and '''p1'''='''p2'''=0]]
[[File:MOS reaction path MEP.PNG|thumb|Figure 4: Minimum Energy Path]]
[[File:MOS reaction path dynamics.PNG|left|thumb|Figure 5: Trajectory]]
[[File:MOS mep surface.PNG|thumb|Figure 6: Surface Plot mep]]
[[File:MOS dynamics surface.PNG|left|thumb|Figure 7: Surface Plot Trajectory]]
[[File:MOS dynamics momenta.PNG|thumb|Figure 8: Internuclear Momenta vs Time Trajectory]]
[[File:MOS reaction path dynamics switch.PNG|left|thumb|Figure 9: Trajectory switched]]
[[File:MOS dynamics momenta switch.PNG|thumb|Figure 10: Momenta switched]]
[[File:MOS contour reverse.PNG|left|thumb|Figure 11: Trajectory with swapped initial conditions]]
[[File:Reactive or unreactive 1.PNG|thumb|Figure 12: '''p1''' = -1.25, '''p2''' = -2.5]]
[[File:Reactive or unreactive 2.PNG|left|thumb|Figure 13: '''p1''' = -1.5, '''p2''' = -2.0]]
[[File:Reactive or unreactive 3.PNG|thumb|Figure 14: '''p1''' = -1.5, '''p2''' = -2.5]]
[[File:Reactive or unreactive 4.PNG|left|thumb|Figure 15: '''p1''' = -2.5, '''p2''' = -5.0]]
[[File:Reactive or unreactive 5.PNG|thumb|Figure 16: '''p1''' = -2.5, '''p2''' = -5.2]]

===What value do the different components of the 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.===

A potential energy surface shows how the potential energy of a system changes with changing interatomic distances. It can be used to follow a chemical reaction and to identify minimum energy states and transition states. The gradient of the potential energy surface (i.e. its first derivative) is 0 at both minima and transition states. The second derivatives are able to identify whether the stationary point is a saddle point, and hence a transition state, or a minima. If it has a positive value in both the x and y directions then the point is a minimum potential energy state and if it is positive in one direction and negative in the other, it is a transition structure. These can be sometimes be seen by eye where the transition state is the top of a peak in the trough and a minimum would be the very bottom of a trough.

===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.===

We're looking at H + H2 so all the atoms in the system are the same and it's symmetric. This means at the transition state '''r1'''='''r2'''='''rts'''. Both momenta will be 0 because it's a stationary point. Using a trial and error method, with an initial guess of '''rts '''where A-B and B-C cross in Figure 1 (0.915), the value of '''rts''' was estimated to be 0.910075. The condition given was that if the internuclear distances differed then it was not the transition state.

In order to get the best estimate, a graph with A-B and B-C with the exact same line is ideal. This is when you can only see one line at the bottom of the graph. The longer the lines stay the same, the better the estimate for the interatomic distance at the transition state. A result for this was when '''rts''' = 0.915 but when this was extended to 1000 steps from 500, it was showed that the they did not stay the same after about 2.8 seconds, so the estimate could be improved (Figure 2). The final value, which showed A-B and B-C being equal at 1000 steps was 0.910075 (Figure 3).

===Comment on how the minimum energy path (mep) and the trajectory differ.===
The starting positions were set as:

A-B distance: 0.920075

B-C distance: 0.910075

So slightly away from the transition state. The mep then shows the lowest energy pathway of the molecule B-C forming - so the distance between the two atoms gets smaller - and the distance between B and C gets larger (Figure 4). It is a straight line with no oscillations (Figure 6).

The trajectory is the actual path the atoms take. It's an oscillating line showing the same trends, just with vibrations in the bond (Figure 5) so in actuality they do not follow the minimum exactly
It looks to the atoms rolling up and down the surface plot (Figure 7).

Other differences between the mep and the trajectory is that the trajectory has much less smooth transitions. The distance between A and B increases much more quickly - the gradient is steeper in Figure 5 - and thus, it ends with much larger distances between the two atoms. It also is a straight line rather than a curved line like in the mep. This makes sense because after a certain distance the atom and the molecule stop interacting and so move off at a constant speed. The mep shows the rate of change of distance between the atom and molecule tailing off which is unrealistic as they will have gained momentum going in opposite directions and the mass does not change so they'll keep moving at a constant speed with out other forces acting on them.

For both the mep and the trajectory, the distance between B and C remains pretty much constant after the atom and the molecule split off from eachother - disregarding the slight oscillations in the trajectory. This is to be expected as there is a bond between the two so the distance will not change much and also explains why the gradient of the trajectory is greater than the mep - an actual bond has been formed.

The final values of distance at 2.5 seconds are:

'''r1'''= 9.0278

'''r2'''= 0.7680 (Figure 5)

The average momenta at large t is:

'''p1'''=2.4868

'''p2'''=1.2465 (Figure 8)

===Changed initial conditions so '''r1''' = '''rts''' and '''r2''' = '''rts'''+0.01 instead?===

The same pattern can be seen - just with A-B and B-C switched on the graph (Figures 9 and 10). This makes sense because of the symmetry in the system. The trajectory follows a path in the y-direction instead of the x but it is an equivalent path just with A and B as the molecule and C as the atom instead of B and C being bonded like before. The starting position was switched to the other side but by the same quantity so the direction is all that has changed.

===When initial positions are set to the final positions of the trajectory and the initial momenta values are the same as the final momenta values, with reversed signs.===
'''r1'''= 9.0278

'''r2'''= 0.7680

'''p1'''=-2.4868

'''p2'''=-1.2465

The contour plot (Figure 11) shows the trajectory with these switched conditions.

We've placed the atoms at their final positions from before and given them enough momentum in the right direction to get back to the transition state. What you actually see is when they get to the close to the transition state, they don't have enough energy to overcome the energy hill and move over to the other side and they look like they're starting to roll back down the hill again. As they've been given the same momenta (in the opposite direction) and position, you'd expect the system to have the exact right conditions to get the the transition state and stay at the stationary point. This would need the EXACT values of all parameters to be the case and we have only a good estimate for the actual transition state structure and the final positions from before.

===Reactive and unreactive trajectories===

{| class="wikitable" border="1"
|+
!'''p1''' !! '''p2''' !! Total Energy !! Reactive/Unreactive Trajectory
|-
| -1.25 || -2.5 || -99.018 || Reactive
|-
| -1.5 || -2.0 || -100.456 || Unreactive
|-
| -1.5 || -2.5 || -98.956 || Reactive
|-
| -2.5 || -5.0 || -84.965 || Unreactive
|-
| -2.5 || -5.2 || -83.416 || Reactive
|}

All paths start with A and B bonded to each other and C as a separate atom.

'''Figure 12:''' 1) There is no oscillation between A and B. They've been set at a certain distance apart and as they follow the path they have been given the same momentum. As the atoms get to the transition state, they begin to interact with each other and oscillation begins. They all have enough momentum to overcome the energy barrier at the transition state and so B and C are bonded, leaving A as a single atom. The reaction has occurred. This oscillates - the two atoms in the molecule have different momenta.

'''Figure 13:''' 2) Here molecule A-B has a slightly higher momentum but atom C has less. There is not enough energy to get to the transition structure, never mind over it. This means that although the molecule and the atom get close, they just repel each other when they do and so no reaction occurs.

'''Figure 14:''' (description)

'''Figure 15:''' (description)

'''Figure 16:''' (description)

===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?===

==Exercise 2: F - H - H system==

[[File:MO endo or exo.PNG|left|thumb|Figure 17: Surface Plot of F-H-H]]
[[File:MO ts initial.PNG|thumb|Figure 18: First attempt for transition state with '''r1'''=1.5, '''r2'''=0.5 and '''p1'''='''p2'''=0]]
[[File:MO ts estimate.PNG|left|thumb|Figure 19: Final estimate for transition state with '''r1'''=1.81412, '''r2'''=0.73915 and '''p1'''='''p2'''=0 ]]
[[File:MO activation energy 1.PNG|thumb|Figure 20: MEP Energy vs Time with '''r1'''=1.7, '''r2'''=0.73915 and '''p1'''='''p2'''=0]]
[[File:MO activation energy 2.PNG|left|thumb|Figure 21: MEP Energy vs Time with '''r1'''=1.81412, '''r2'''=0.79 and '''p1'''='''p2'''=0]]
[[File:MO reactive traj.PNG|thumb|Figure 22: Reactive Trajectory]]

===PES inspection===

===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?===

Figure 17: A-B corresponds to HF and B-C corresponds to H2.

When A-B is small - the energy is lower than when B-C is small. This means that the F + H2 reaction is exothermic and H + HF reaction is endothermic, as it has to be due to it being the reverse reaction.

Bond breaking needs energy and bond energy releases energy. Because the H+HF reaction overall takes in energy and the F + H2 reaction releases energy overall, the HF bond is stronger than the H2 bond.

===Locate the approximate position of the transition state===
Hammond's postulate says the transition state structure will be closer to the products' structure in an endothermic reaction. Because AB is the bond that's broken and corresponds to the H-F bond length and it's the bond being broken, you'd expect the transition state to resemble H2 more closely, hence the BC distance will be smaller. The initial guess for the transition state was '''r1'''(AB)= 1.5 and '''r2'''(BC)= 0.5. The outcome of this can be seen in Figure 18. The best estimate would have no path line and just a cross so a much better guess was needed.

After a series of trial and error the best estimate was '''r1'''(AB)= 1.81412 and '''r2'''(BC)= 0.73915 (Figure 19)

===Report the activation energy for both reactions.===
The energy of the transition state = -103.739

An MEP with 9000 steps of the following conditions was calculated and an energy vs time plot was made (Figure 20)

AB = 1.7

BC = 0.73915

The final energy was -133.443 meaning the activation energy of the HF + H was -103.739-(-133.443)=29.704.

The second activation energy was calculated using the following conditions using the same method (Figure 21):

AB=1.81412

BC=0.79

The final energy was -103.009 meaning the activation energy of the H2 + F was -103.009-(-103.739)=0.73. This was the bets estimate however, the further the BC distance was from the transition state, the larger the activation energy became so it is hard to know which is right.

===Reaction Dynamics===

=== In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?===

The reactive trajectory I've plotted (Figure 22) is as follows:

AB=2.3 Å

BC=0.74 Å

AB Momentum = -2.3 kgm/s

BC Momentum = -2 kgm/s.

There are smaller oscillations between the H2 molecule than the HF molecule after it has been formed.

File:MO reactive traj.PNG

2018-05-11T13:48:55Z

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