ChemWiki - User contributions [en]

MRD:sc3916

2018-05-25T15:25:04Z

Sc3916: /* References */

=Molecular Reaction Dynamics Report=

==Potential Energy Surfaces==

'''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.'''

The transition structure is at a saddle point - the intersection of the minimum of a curve in one plane, and the maximum of a curve in the other. In this case, the minimum curve is on the plane corresponding to RAB=RBC, and the maximum is on the perpendicular plane. The gradient of both these curves is equal to zero.

This is different from when the potential energy surface is at a minimum - in one plane the gradient = 0, but not in the other. This allows us to distinguish between the transition structure and minima by looking at the curvature in '''both''' planes.

[[File:Saddle_Point_Visualisation_sc3916.png|thumb|none|500px|Saddle Point Visualisation]]

'''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.'''

'''rts= 0.9075 angstroms =rAB=rBC'''

To estimate this, a rough estimate was found by looking at the saddle point. At the saddle point, gradient = 0 in both planes - so if the system is at exactly this point, the atoms will remain stationary if it has no initial momentum. If it isn't at this point, it will start to oscillate - and the further away from the saddle point, the greater the oscillation. Therefore by plotting this internuclear distance vs time, and adjusting the distance by small amounts until there was very little oscillation seen, a more precise value was found.

[[File: Sc3916_1.PNG|thumb|300px|none|Internuclear Distances vs Time]]

==Reaction trajectories==

[[File:Sc3916_mep.PNG|thumb|400px|left|Minimum Energy Path from from r1 = rts+δ, r2 = rts]]

[[File:Sc3916_dynamics.PNG|thumb|400px|center|Dynamic Energy Path from r1 = rts+δ, r2 = rts]]

'''Comment on how the MEP and the trajectory you just calculated differ.'''

The minimum energy path represents an infinitely slow reaction trajectory. We start off a very small distance from the transition structure, and then the trajectory "rolls" down the potential energy surface, following where the gradient is most negative. Since velocity resets to 0 every time interval, so does momentum (p=mv). This means at every infinitesimally small step, this pathway will minimise its energy. This is not how it works in reality.

The dynamic trajectory is much closer to reality. As the trajectory starts to move down the potential energy surface it gains momentum. This means it will carry on rolling up the curve on the other side of the minimum (which increases its potential energy), leading to oscillations. This reflects the vibrational motion of atoms, which MEP does not include.

Dynamics Internuclear distance vs time plot: 
Final AB distance = 0.8 
Final BC distance = 9.0 

Dynamics Internuclear Momenta vs time plot: 
Final average momentum for AB = 1.25 
Final average momentum for BC = 2.5 

This shows the reaction path from the above calculation leads to A-B + C

If we switch around initial values, so now rAB=rts+δ and rBC=rts, the reaction path will output the same values, except the products are B-C + A.

'''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.'''

The reaction proceeds from left to right across the below plot, A-B + C -> A + B-C. This reaction is essentially the reverse of the previous trajectory (allowing the pathway to "roll" down from the transition state), however it crosses over, to form the products.

[[File:Sc3916_dynamics2.PNG|thumb|400px|none| initial conditions: 
rAB = 0.8 
rBC = 9.0 
pAB = -1.25 
pBC = -2.5]]

==Reactive and unreactive trajectories==
The following table is for the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, with varying momentum.

{| class="wikitable" border=1
! p1 !! p2 !! E(kcal/mol) !! Reactive?
|-
| -1.25 || -2.5 || -99.119 || Yes
|-
| -1.5 || -2.0 || -100.456 || No
|-
| -1.5 || -2.5 || -98.956 || Yes
|-
| -2.5 || -5.0 || -84.956 || No
|-
| -2.5 || -5.2 || -83.416 || Yes
|}

[[File:Sc3916_rxn1.PNG|thumb|300px|none|Initial p1=-1.25, r2=-2.5 
the trajectory goes smoothly up to the transition state, then starts oscillating as it rolls down to the products]]

[[File:Sc3916_rxn2.PNG|thumb|300px|none|Initial p1=-1.5, r2=-2.0 
the trajectory goes up towards the transition state smoothly (but not very far), and then rolls back down the way it came, with oscillating]]

[[File:Sc3916_rxn3.PNG|thumb|300px|none|Initial p1=-1.5, r2=-2.5 
the trajectory goes smoothly up to the transition state, then starts oscillating as it rolls down to the products]]

[[File:Sc3916_rxn4b.PNG|thumb|300px|none|Initial p1=-2.5, r2=-5.0 
the trajectory goes smoothly up to the transition state, crosses it, but recrosses it and reverts back to the reactants, with very large oscillations]]

[[File:Sc3916_rxn5.PNG|thumb|300px|none|Initial p1=-2.5, r2=-5.2 
the trajectory crosses the transition state, forming the products, then recrosses it to the reactant side, and then recrosses it again, forming the products finally]]

'''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?'''

4 major assumptions are used. Firstly, the Born-Oppenheimer approximation is used, and that the effects of quantum tunneling are negligible (there is a non-zero probability that reactants can convert to products without reaching the activation energy). The reactants are assumed to have a Boltzmann distribution - which is perfectly valid at low temperatures. The final assumption is that "once the system attains the transition state, with a velocity towards the product configuration, it will not reenter the initial state region again."[https://doi.org/10.1016/B978-044452837-7.50005-8 1]

This last assumption does not fit with our simulations. Both the 4th, and 5th plots contain the trajectory crossing over the transition state and back again. Since there will be more trajectories (like the 4th) which involve crossing over and are still unreactive - which are not predicted by Transition State Theory - experimental rate values will be slower than those predicted by the theory.

==EXERCISE 2: F - H - H system==

'''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 is exothermic - the products have a more negative energy than the reactants. Because the enthalpy change only involves breaking a H-H bond and making a H-F bond, ΔH = +H(H-H) - H(H-F).

Since ΔH < 0:

H(H-F)>H(H-H)

This means HF has a greater bond enthalpy (and therefore strength) than H2. It can also be seen that H + HF is endothermic, which leads to the same conclusion.

'''Locate the approximate position of the transition state.'''

Since F + H2 is exothermic, we can assume the transition state is close in structure to the reactants (according to Hammond's postulate). Starting from near the reactants, MEP calculations with a large number of steps were done, varying the starting position to try minimise how far the trajectory moves from that position. This is because at the transition point, gradient will equal zero in both planes, so the MEP will not increase from there - so the shorter the MEP (from a constant number of steps), the closer we are to the transition state. With this method, a good approximation was found.

rts: 
RHH = 0.7449 angstroms 
RHF = 1.8108 angstroms

(0.7448778, 1.810754) is the super accurate value, but since this is just a model, this won't exactly match the real transition state position, so there is not much point doing it to this many decimal places.

'''Report the activation energy for both reactions.'''

By performing an MEP from next to the transition structure, the path from the TS to the reactants, or products, was found. By subtracting the final energy from the TS, the Ea was obtained. Since the activation energy for H2 + F is very small, the MEP has a very small gradient, and as such the calculation required an order of magnitude more steps in the calculation (than HF + H) to reach the reactants. Conversely, in HF + H a step change is seen.

[[File:Mep_energyvstime_sc3916.PNG|thumb|left|initial conditions: 
RHH=0.745 
RHF=1.83]]

[[File:3916_energyvstime_hf2.PNG|thumb|center|initial conditions: 
RHH=0.745 
RHF=1.8]]

Ea(H2 + F Ea) = 0.256 kcal/mol 
Ea(HF + H Ea) = 30.202 kcal/mol 

=== Reaction dynamics ===
* Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Internuclear Momenta vs Time”. {{fontcolor1|blue|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

[[File:H2+f_surface__3916.PNG|thumb|none|F + H2, initial conditions: 
RHH (AB) = 0.74 angstroms 
RHF (BC) = 2 angstroms 
AB momentum = +3 
BC momentum = -5]]

[[File:H2+F_MOMENTAVTIME3916.PNG|thumb|none|F + H2 Momentum vs Time Plot]]

The animation of this reaction clearly shows the velocity of the lone H atom formed has increased, and vibrations occurring in HF. This can be seen in the momentum vs time plot - with the increased AB and AC momentum reflecting the increase in velocity, and BC momentum oscillation reflecting the vibrations of the H-F bond. Since translational and vibrational energy contribute to kinetic energy, we know that the energy released by this exothermic reaction (a decrease in enthalpy) is conserved in the form of increased kinetic energy. This can be seen in the below graph, and is easily shown experimentally by the resultant increase in temperature. To determine the extent vibrational energy contributes to this, we can use IR spectroscopy. If there is a lot of vibrational energy, there will be a relatively high population in the first vibrational level, and we will see overtone bands in the IR spectrum.

[[File:H2+F_ENERGYVSTIME3916.PNG|thumb|none|F + H2 Energy vs Time Plot]]

* Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 0.74, with a momentum pFH = -0.5, and explore several values of pHH in the range -3 to 3 (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.

[[File:SC3916-3CONTOUR.PNG|thumb|none|F + H2, initial conditions A: 
rHH=0.74 rHF=1.83 
pHH=-3.0 pHF=-0.5 ]]

[[File:SC3916-2CONTOUR.PNG|thumb|none|F + H2, initial conditions B: 
rHH=0.74 rHF=1.83 
pHH=-2.0 pHF=-0.5 ]]

[[File:SC3916-1.5CONTOUR.PNG|thumb|none|F + H2, initial conditions C: 
rHH=0.74 rHF=1.83 
pHH=-1 pHF=-0.5 ]]

[[File:SC39160CONTOUR.PNG|thumb|none|F + H2, initial conditions D: 
rHH=0.74 rHF=1.83 
pHH=0.0 pHF=-0.5 ]]

[[File:Sc3916_redone1.PNG|thumb|none|F + H2, initial conditions E: 
rHH=0.74 rHF=1.83 
pHH=+1.0 pHF=-0.5 ]]

[[File:Sc3916_redone2.PNG|thumb|none|F + H2, initial conditions F: 
rHH=0.74 rHF=1.83 
pHH=+3.0 pHF=-0.5 ]]

* For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now?

[[File:Sc3916_redonelast.PNG|thumb|none|F + H2, initial conditions G: 
rHH=0.74 rHF=1.83 
pHH=+0.1 pHF=-0.8 ]]

*Let us now focus on the reverse reaction, H + HF (an H atom with high kinetic energy colliding with HF).

[[File:Sc3916_hh-10hf-.5.PNG|thumb|none|HF + H, initial conditions H: 
rHH=0.9 rHF=1.83 
pHH=-10 pHF=-5 ]]

[[File:Sc3916_hh-1.1h4-13.PNG|thumb|none|HF + H, initial conditions I: 
rHH=0.9 rHF=1.83 
pHH=-1.1 pHF=-13 ]]

The cases studied are an illustration of Polanyi's empirical rules. {{fontcolor|blue|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, for a reaction with an early transition state, vibrational energy is more effective in overcoming the TS - so initial conditions with higher vibrational energy are more likely to result in reaction, so are more efficient. For a late transition state, translational energy is more effective in overcoming it - so atoms/molecules with a higher translational energy create a more efficient reaction.[http://doi.org/10.1021/jz301649w 2]

This can be seen in the above reaction trajectories. H2 + F has an early transition state (Hammond's postulate tells us that in an endothermic reaction, the transition state most resembles the reactants). Reaction trajectories with high initial vibrational energy (shown by pHH, conditions A, B, F) are unreactive trajectories. Conversely, those with low vibrational energy (C, D, E) are reactive. F also has increased kinetic energy on top of low vibrational energy, and this too is reactive.

The reverse reaction, HF + H, has a late transition state (from Hammond's postulate) - so initial conditions with greater vibrational energy will be more efficient. This can be seen in the reactive trajectory with conditions I. Conversely, conditions H has a much greater translational energy, and is unreactive.

==References==

1. T. Bligaard, J.K. Nørskov, Chemical Bonding at Surfaces and Interfaces, 2008, p.255–321, https://doi.org/10.1016/B978-044452837-7.50005-8 [accessed 24/5/18]

2. Z. Zhang, Y. Zhou, D.Zhang, G.Czakó, J. Bowman, J. Phys. Chem. Lett., 2012, 3 (23), pp 3416–3419
https://pubs.acs.org/doi/abs/10.1021/jz301649w [accessed 25/5/18]

MRD:sc3916

2018-05-25T15:23:25Z

Sc3916:

MRD:sc3916

2018-05-25T15:15:39Z

Sc3916: /* EXERCISE 2: F - H - H system */

MRD:sc3916

2018-05-25T15:11:50Z

Sc3916: /* EXERCISE 2: F - H - H system */

=Molecular Reaction Dynamics Report=

==Potential Energy Surfaces==

'''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.'''

The transition structure is at a saddle point - the intersection of the minimum of a curve in one plane, and the maximum of a curve in the other. In this case, the minimum curve is on the plane corresponding to RAB=RBC, and the maximum is on the perpendicular plane. The gradient of both these curves is equal to zero.

This is different from when the potential energy surface is at a minimum - in one plane the gradient = 0, but not in the other. This allows us to distinguish between the transition structure and minima by looking at the curvature in '''both''' planes.

[[File:Saddle_Point_Visualisation_sc3916.png|thumb|none|500px|Saddle Point Visualisation]]

'''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.'''

'''rts= 0.9075 angstroms =rAB=rBC'''

To estimate this, a rough estimate was found by looking at the saddle point. At the saddle point, gradient = 0 in both planes - so if the system is at exactly this point, the atoms will remain stationary if it has no initial momentum. If it isn't at this point, it will start to oscillate - and the further away from the saddle point, the greater the oscillation. Therefore by plotting this internuclear distance vs time, and adjusting the distance by small amounts until there was very little oscillation seen, a more precise value was found.

[[File: Sc3916_1.PNG|thumb|300px|none|Internuclear Distances vs Time]]

==Reaction trajectories==

[[File:Sc3916_mep.PNG|thumb|400px|left|Minimum Energy Path from from r1 = rts+δ, r2 = rts]]

[[File:Sc3916_dynamics.PNG|thumb|400px|center|Dynamic Energy Path from r1 = rts+δ, r2 = rts]]

'''Comment on how the MEP and the trajectory you just calculated differ.'''

The minimum energy path represents an infinitely slow reaction trajectory. We start off a very small distance from the transition structure, and then the trajectory "rolls" down the potential energy surface, following where the gradient is most negative. Since velocity resets to 0 every time interval, so does momentum (p=mv). This means at every infinitesimally small step, this pathway will minimise its energy. This is not how it works in reality.

The dynamic trajectory is much closer to reality. As the trajectory starts to move down the potential energy surface it gains momentum. This means it will carry on rolling up the curve on the other side of the minimum (which increases its potential energy), leading to oscillations. This reflects the vibrational motion of atoms, which MEP does not include.

Dynamics Internuclear distance vs time plot: 
Final AB distance = 0.8 
Final BC distance = 9.0 

Dynamics Internuclear Momenta vs time plot: 
Final average momentum for AB = 1.25 
Final average momentum for BC = 2.5 

This shows the reaction path from the above calculation leads to A-B + C

If we switch around initial values, so now rAB=rts+δ and rBC=rts, the reaction path will output the same values, except the products are B-C + A.

'''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.'''

The reaction proceeds from left to right across the below plot, A-B + C -> A + B-C. This reaction is essentially the reverse of the previous trajectory (allowing the pathway to "roll" down from the transition state), however it crosses over, to form the products.

[[File:Sc3916_dynamics2.PNG|thumb|400px|none| initial conditions: 
rAB = 0.8 
rBC = 9.0 
pAB = -1.25 
pBC = -2.5]]

==Reactive and unreactive trajectories==
The following table is for the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, with varying momentum.

{| class="wikitable" border=1
! p1 !! p2 !! E(kcal/mol) !! Reactive?
|-
| -1.25 || -2.5 || -99.119 || Yes
|-
| -1.5 || -2.0 || -100.456 || No
|-
| -1.5 || -2.5 || -98.956 || Yes
|-
| -2.5 || -5.0 || -84.956 || No
|-
| -2.5 || -5.2 || -83.416 || Yes
|}

[[File:Sc3916_rxn1.PNG|thumb|300px|none|Initial p1=-1.25, r2=-2.5 
the trajectory goes smoothly up to the transition state, then starts oscillating as it rolls down to the products]]

[[File:Sc3916_rxn2.PNG|thumb|300px|none|Initial p1=-1.5, r2=-2.0 
the trajectory goes up towards the transition state smoothly (but not very far), and then rolls back down the way it came, with oscillating]]

[[File:Sc3916_rxn3.PNG|thumb|300px|none|Initial p1=-1.5, r2=-2.5 
the trajectory goes smoothly up to the transition state, then starts oscillating as it rolls down to the products]]

[[File:Sc3916_rxn4b.PNG|thumb|300px|none|Initial p1=-2.5, r2=-5.0 
the trajectory goes smoothly up to the transition state, crosses it, but recrosses it and reverts back to the reactants, with very large oscillations]]

[[File:Sc3916_rxn5.PNG|thumb|300px|none|Initial p1=-2.5, r2=-5.2 
the trajectory crosses the transition state, forming the products, then recrosses it to the reactant side, and then recrosses it again, forming the products finally]]

'''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?'''

4 major assumptions are used. Firstly, the Born-Oppenheimer approximation is used, and that the effects of quantum tunneling are negligible (there is a non-zero probability that reactants can convert to products without reaching the activation energy). The reactants are assumed to have a Boltzmann distribution - which is perfectly valid at low temperatures. The final assumption is that "once the system attains the transition state, with a velocity towards the product configuration, it will not reenter the initial state region again."[https://doi.org/10.1016/B978-044452837-7.50005-8 1]

This last assumption does not fit with our simulations. Both the 4th, and 5th plots contain the trajectory crossing over the transition state and back again. Since there will be more trajectories (like the 4th) which involve crossing over and are still unreactive - which are not predicted by Transition State Theory - experimental rate values will be slower than those predicted by the theory.

==EXERCISE 2: F - H - H system==

'''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 is exothermic - the products have a more negative energy than the reactants. Because the enthalpy change only involves breaking a H-H bond and making a H-F bond, ΔH = +H(H-H) - H(H-F).

Since ΔH < 0:

H(H-F)>H(H-H)

This means HF has a greater bond enthalpy (and therefore strength) than H2. It can also be seen that H + HF is endothermic, which leads to the same conclusion.

'''Locate the approximate position of the transition state.'''

Since F + H2 is exothermic, we can assume the transition state is close in structure to the reactants (according to Hammond's postulate). Starting from near the reactants, MEP calculations with a large number of steps were done, varying the starting position to try minimise how far the trajectory moves from that position. This is because at the transition point, gradient will equal zero in both planes, so the MEP will not increase from there - so the shorter the MEP (from a constant number of steps), the closer we are to the transition state. With this method, a good approximation was found.

rts: 
RHH = 0.7449 angstroms 
RHF = 1.8108 angstroms

(0.7448778, 1.810754) is the super accurate one

'''Report the activation energy for both reactions.'''

By performing an MEP from next to the transition structure, the path from the TS to the reactants, or products, was found. By subtracting the final energy from the TS, the Ea was obtained. Since the activation energy for H2 + F is very small, the MEP has a very small gradient, and as such the calculation required an order of magnitude more steps in the calculation (than HF + H) to reach the reactants. Conversely, in HF + H a step change is seen.

[[File:Mep_energyvstime_sc3916.PNG|thumb|left|initial conditions: 
RHH=0.745 
RHF=1.83]]

[[File:3916_energyvstime_hf2.PNG|thumb|center|initial conditions: 
RHH=0.745 
RHF=1.8]]

Ea(H2 + F Ea) = 0.256 kcal/mol 
Ea(HF + H Ea) = 30.202 kcal/mol 

=== Reaction dynamics ===
* Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Internuclear Momenta vs Time”. {{fontcolor1|blue|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

[[File:H2+f_surface__3916.PNG|thumb|none|F + H2, initial conditions: 
RHH (AB) = 0.74 angstroms 
RHF (BC) = 2 angstroms 
AB momentum = +3 
BC momentum = -5]]

[[File:H2+F_MOMENTAVTIME3916.PNG|thumb|none|F + H2 Momentum vs Time Plot]]

The animation of this reaction clearly shows the velocity of the lone H atom formed has increased, and vibrations occurring in HF. This can be seen in the momentum vs time plot - with the increased AB and AC momentum reflecting the increase in velocity, and BC momentum oscillation reflecting the vibrations of the H-F bond. Since translational and vibrational energy contribute to kinetic energy, we know that the energy released by this exothermic reaction (a decrease in enthalpy) is conserved in the form of increased kinetic energy. This can be seen in the below graph, and is easily shown experimentally by the resultant increase in temperature. To determine the extent vibrational energy contributes to this, we can use IR spectroscopy. If there is a lot of vibrational energy, there will be a relatively high population in the first vibrational level, and we will see overtone bands in the IR spectrum.

[[File:H2+F_ENERGYVSTIME3916.PNG|thumb|none|F + H2 Energy vs Time Plot]]

* Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 0.74, with a momentum pFH = -0.5, and explore several values of pHH in the range -3 to 3 (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.

[[File:SC3916-3CONTOUR.PNG|thumb|none|F + H2, initial conditions A: 
rHH=0.74 rHF=1.83 
pHH=-3.0 pHF=-0.5 ]]

[[File:SC3916-2CONTOUR.PNG|thumb|none|F + H2, initial conditions B: 
rHH=0.74 rHF=1.83 
pHH=-2.0 pHF=-0.5 ]]

[[File:SC3916-1.5CONTOUR.PNG|thumb|none|F + H2, initial conditions C: 
rHH=0.74 rHF=1.83 
pHH=-1 pHF=-0.5 ]]

[[File:SC39160CONTOUR.PNG|thumb|none|F + H2, initial conditions D: 
rHH=0.74 rHF=1.83 
pHH=0.0 pHF=-0.5 ]]

[[File:Sc3916_redone1.PNG|thumb|none|F + H2, initial conditions E: 
rHH=0.74 rHF=1.83 
pHH=+1.0 pHF=-0.5 ]]

[[File:Sc3916_redone2.PNG|thumb|none|F + H2, initial conditions F: 
rHH=0.74 rHF=1.83 
pHH=+3.0 pHF=-0.5 ]]

* For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now?

[[File:Sc3916_redonelast.PNG|thumb|none|F + H2, initial conditions G: 
rHH=0.74 rHF=1.83 
pHH=+0.1 pHF=-0.8 ]]

*Let us now focus on the reverse reaction, H + HF (an H atom with high kinetic energy colliding with HF).

[[File:Sc3916_hh-10hf-.5.PNG|thumb|none|HF + H, initial conditions H: 
rHH=0.9 rHF=1.83 
pHH=-10 pHF=-5 ]]

[[File:Sc3916_hh-1.1h4-13.PNG|thumb|none|HF + H, initial conditions I: 
rHH=0.9 rHF=1.83 
pHH=-1.1 pHF=-13 ]]

The cases studied are an illustration of Polanyi's empirical rules. {{fontcolor|blue|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, for a reaction with an early transition state, vibrational energy is more effective in overcoming the TS - so initial conditions with higher vibrational energy are more likely to result in reaction, so are more efficient. For a late transition state, translational energy is more effective in overcoming it - so atoms/molecules with a higher translational energy create a more efficient reaction.[http://doi.org/10.1021/jz301649w 2]

This can be seen in the above reaction trajectories. H2 + F has an early transition state (Hammond's postulate tells us that in an endothermic reaction, the transition state most resembles the reactants). Reaction trajectories with high initial vibrational energy (shown by pHH, conditions A, B, F) are unreactive trajectories. Conversely, those with low vibrational energy (C, D, E) are reactive. F also has increased kinetic energy on top of low vibrational energy, and this too is reactive.

The reverse reaction, HF + H, has a late transition state (from Hammond's postulate) - so initial conditions with greater vibrational energy will be more efficient. This can be seen in the reactive trajectory with conditions I. Conversely, conditions H has a much greater translational energy, and is unreactive.

MRD:sc3916

2018-05-25T14:56:55Z

Sc3916: /* EXERCISE 2: F - H - H system */

=Molecular Reaction Dynamics Report=

==Potential Energy Surfaces==

'''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.'''

The transition structure is at a saddle point - the intersection of the minimum of a curve in one plane, and the maximum of a curve in the other. In this case, the minimum curve is on the plane corresponding to RAB=RBC, and the maximum is on the perpendicular plane. The gradient of both these curves is equal to zero.

This is different from when the potential energy surface is at a minimum - in one plane the gradient = 0, but not in the other. This allows us to distinguish between the transition structure and minima by looking at the curvature in '''both''' planes.

[[File:Saddle_Point_Visualisation_sc3916.png|thumb|none|500px|Saddle Point Visualisation]]

'''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.'''

'''rts= 0.9075 angstroms =rAB=rBC'''

To estimate this, a rough estimate was found by looking at the saddle point. At the saddle point, gradient = 0 in both planes - so if the system is at exactly this point, the atoms will remain stationary if it has no initial momentum. If it isn't at this point, it will start to oscillate - and the further away from the saddle point, the greater the oscillation. Therefore by plotting this internuclear distance vs time, and adjusting the distance by small amounts until there was very little oscillation seen, a more precise value was found.

[[File: Sc3916_1.PNG|thumb|300px|none|Internuclear Distances vs Time]]

==Reaction trajectories==

[[File:Sc3916_mep.PNG|thumb|400px|left|Minimum Energy Path from from r1 = rts+δ, r2 = rts]]

[[File:Sc3916_dynamics.PNG|thumb|400px|center|Dynamic Energy Path from r1 = rts+δ, r2 = rts]]

'''Comment on how the MEP and the trajectory you just calculated differ.'''

The minimum energy path represents an infinitely slow reaction trajectory. We start off a very small distance from the transition structure, and then the trajectory "rolls" down the potential energy surface, following where the gradient is most negative. Since velocity resets to 0 every time interval, so does momentum (p=mv). This means at every infinitesimally small step, this pathway will minimise its energy. This is not how it works in reality.

The dynamic trajectory is much closer to reality. As the trajectory starts to move down the potential energy surface it gains momentum. This means it will carry on rolling up the curve on the other side of the minimum (which increases its potential energy), leading to oscillations. This reflects the vibrational motion of atoms, which MEP does not include.

Dynamics Internuclear distance vs time plot: 
Final AB distance = 0.8 
Final BC distance = 9.0 

Dynamics Internuclear Momenta vs time plot: 
Final average momentum for AB = 1.25 
Final average momentum for BC = 2.5 

This shows the reaction path from the above calculation leads to A-B + C

If we switch around initial values, so now rAB=rts+δ and rBC=rts, the reaction path will output the same values, except the products are B-C + A.

'''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.'''

The reaction proceeds from left to right across the below plot, A-B + C -> A + B-C. This reaction is essentially the reverse of the previous trajectory (allowing the pathway to "roll" down from the transition state), however it crosses over, to form the products.

[[File:Sc3916_dynamics2.PNG|thumb|400px|none| initial conditions: 
rAB = 0.8 
rBC = 9.0 
pAB = -1.25 
pBC = -2.5]]

==Reactive and unreactive trajectories==
The following table is for the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, with varying momentum.

{| class="wikitable" border=1
! p1 !! p2 !! E(kcal/mol) !! Reactive?
|-
| -1.25 || -2.5 || -99.119 || Yes
|-
| -1.5 || -2.0 || -100.456 || No
|-
| -1.5 || -2.5 || -98.956 || Yes
|-
| -2.5 || -5.0 || -84.956 || No
|-
| -2.5 || -5.2 || -83.416 || Yes
|}

[[File:Sc3916_rxn1.PNG|thumb|300px|none|Initial p1=-1.25, r2=-2.5 
the trajectory goes smoothly up to the transition state, then starts oscillating as it rolls down to the products]]

[[File:Sc3916_rxn2.PNG|thumb|300px|none|Initial p1=-1.5, r2=-2.0 
the trajectory goes up towards the transition state smoothly (but not very far), and then rolls back down the way it came, with oscillating]]

[[File:Sc3916_rxn3.PNG|thumb|300px|none|Initial p1=-1.5, r2=-2.5 
the trajectory goes smoothly up to the transition state, then starts oscillating as it rolls down to the products]]

[[File:Sc3916_rxn4b.PNG|thumb|300px|none|Initial p1=-2.5, r2=-5.0 
the trajectory goes smoothly up to the transition state, crosses it, but recrosses it and reverts back to the reactants, with very large oscillations]]

[[File:Sc3916_rxn5.PNG|thumb|300px|none|Initial p1=-2.5, r2=-5.2 
the trajectory crosses the transition state, forming the products, then recrosses it to the reactant side, and then recrosses it again, forming the products finally]]

'''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?'''

4 major assumptions are used. Firstly, the Born-Oppenheimer approximation is used, and that the effects of quantum tunneling are negligible (there is a non-zero probability that reactants can convert to products without reaching the activation energy). The reactants are assumed to have a Boltzmann distribution - which is perfectly valid at low temperatures. The final assumption is that "once the system attains the transition state, with a velocity towards the product configuration, it will not reenter the initial state region again."[https://doi.org/10.1016/B978-044452837-7.50005-8 1]

This last assumption does not fit with our simulations. Both the 4th, and 5th plots contain the trajectory crossing over the transition state and back again. Since there will be more trajectories (like the 4th) which involve crossing over and are still unreactive - which are not predicted by Transition State Theory - experimental rate values will be slower than those predicted by the theory.

==EXERCISE 2: F - H - H system==

'''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 is exothermic - the products have a more negative energy than the reactants. Because the enthalpy change only involves breaking a H-H bond and making a H-F bond, ΔH = +H(H-H) - H(H-F).

Since ΔH < 0:

H(H-F)>H(H-H)

This means HF has a greater bond enthalpy (and therefore strength) than H2. It can also be seen that H + HF is endothermic, which leads to the same conclusion.

'''Locate the approximate position of the transition state.'''

rts: 
RHH = 0.745 angstroms 
RHF = 1.811 angstroms

(0.7448778, 1.810754) is the super accurate one

'''Report the activation energy for both reactions.'''

By performing an MEP from next to the transition structure, the path from the TS to the reactants, or products, was found. By subtracting the final energy from the TS, the Ea was obtained. Since the activation energy for H2 + F is very small, the MEP has a very small gradient, and as such the calculation required an order of magnitude more steps in the calculation (than HF + H) to reach the reactants. Conversely, in HF + H a step change is seen.

[[File:Mep_energyvstime_sc3916.PNG|thumb|left|initial conditions: 
RHH=0.745 
RHF=1.83]]

[[File:3916_energyvstime_hf2.PNG|thumb|center|initial conditions: 
RHH=0.745 
RHF=1.8]]

Ea(H2 + F Ea) = 0.256 kcal/mol 
Ea(HF + H Ea) = 30.202 kcal/mol 

=== Reaction dynamics ===
* Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Internuclear Momenta vs Time”. {{fontcolor1|blue|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

[[File:H2+f_surface__3916.PNG|thumb|none|F + H2, initial conditions: 
RHH (AB) = 0.74 angstroms 
RHF (BC) = 2 angstroms 
AB momentum = +3 
BC momentum = -5]]

[[File:H2+F_MOMENTAVTIME3916.PNG|thumb|none|F + H2 Momentum vs Time Plot]]

The animation of this reaction clearly shows the velocity of the lone H atom formed has increased, and vibrations occurring in HF. This can be seen in the momentum vs time plot - with the increased AB and AC momentum reflecting the increase in velocity, and BC momentum oscillation reflecting the vibrations of the H-F bond. Since translational and vibrational energy contribute to kinetic energy, we know that the energy released by this exothermic reaction (a decrease in enthalpy) is conserved in the form of increased kinetic energy. This can be seen in the below graph, and is easily shown experimentally by the resultant increase in temperature. To determine the extent vibrational energy contributes to this, we can use IR spectroscopy. If there is a lot of vibrational energy, there will be a relatively high population in the first vibrational level, and we will see overtone bands in the IR spectrum.

[[File:H2+F_ENERGYVSTIME3916.PNG|thumb|none|F + H2 Energy vs Time Plot]]

* Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 0.74, with a momentum pFH = -0.5, and explore several values of pHH in the range -3 to 3 (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.

[[File:SC3916-3CONTOUR.PNG|thumb|none|F + H2, initial conditions A: 
rHH=0.74 rHF=1.83 
pHH=-3.0 pHF=-0.5 ]]

[[File:SC3916-2CONTOUR.PNG|thumb|none|F + H2, initial conditions B: 
rHH=0.74 rHF=1.83 
pHH=-2.0 pHF=-0.5 ]]

[[File:SC3916-1.5CONTOUR.PNG|thumb|none|F + H2, initial conditions C: 
rHH=0.74 rHF=1.83 
pHH=-1 pHF=-0.5 ]]

[[File:SC39160CONTOUR.PNG|thumb|none|F + H2, initial conditions D: 
rHH=0.74 rHF=1.83 
pHH=0.0 pHF=-0.5 ]]

[[File:Sc3916_redone1.PNG|thumb|none|F + H2, initial conditions E: 
rHH=0.74 rHF=1.83 
pHH=+1.0 pHF=-0.5 ]]

[[File:Sc3916_redone2.PNG|thumb|none|F + H2, initial conditions F: 
rHH=0.74 rHF=1.83 
pHH=+3.0 pHF=-0.5 ]]

* For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now?

[[File:Sc3916_redonelast.PNG|thumb|none|F + H2, initial conditions G: 
rHH=0.74 rHF=1.83 
pHH=+0.1 pHF=-0.8 ]]

*Let us now focus on the reverse reaction, H + HF (an H atom with high kinetic energy colliding with HF).

[[File:Sc3916_hh-10hf-.5.PNG|thumb|none|HF + H, initial conditions H: 
rHH=0.9 rHF=1.83 
pHH=-10 pHF=-5 ]]

[[File:Sc3916_hh-1.1h4-13.PNG|thumb|none|HF + H, initial conditions I: 
rHH=0.9 rHF=1.83 
pHH=-1.1 pHF=-13 ]]

The cases studied are an illustration of Polanyi's empirical rules. {{fontcolor|blue|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, for a reaction with an early transition state, vibrational energy is more effective in overcoming the TS - so initial conditions with higher vibrational energy are more likely to result in reaction, so are more efficient. For a late transition state, translational energy is more effective in overcoming it - so atoms/molecules with a higher translational energy create a more efficient reaction.[http://doi.org/10.1021/jz301649w 2]

This can be seen in the above reaction trajectories. H2 + F has an early transition state (Hammond's postulate tells us that in an endothermic reaction, the transition state most resembles the reactants). Reaction trajectories with high initial vibrational energy (shown by pHH, conditions A, B, F) are unreactive trajectories. Conversely, those with low vibrational energy (C, D, E) are reactive. F also has increased kinetic energy on top of low vibrational energy, and this too is reactive.

The reverse reaction, HF + H, has a late transition state (from Hammond's postulate) - so initial conditions with greater vibrational energy will be more efficient. This can be seen in the reactive trajectory with conditions I. Conversely, conditions H has a much greater translational energy, and is unreactive.

MRD:sc3916

2018-05-25T14:56:26Z

Sc3916: /* Reaction dynamics */

=Molecular Reaction Dynamics Report=

==Potential Energy Surfaces==

'''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.'''

The transition structure is at a saddle point - the intersection of the minimum of a curve in one plane, and the maximum of a curve in the other. In this case, the minimum curve is on the plane corresponding to RAB=RBC, and the maximum is on the perpendicular plane. The gradient of both these curves is equal to zero.

This is different from when the potential energy surface is at a minimum - in one plane the gradient = 0, but not in the other. This allows us to distinguish between the transition structure and minima by looking at the curvature in '''both''' planes.

[[File:Saddle_Point_Visualisation_sc3916.png|thumb|none|500px|Saddle Point Visualisation]]

'''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.'''

'''rts= 0.9075 angstroms =rAB=rBC'''

To estimate this, a rough estimate was found by looking at the saddle point. At the saddle point, gradient = 0 in both planes - so if the system is at exactly this point, the atoms will remain stationary if it has no initial momentum. If it isn't at this point, it will start to oscillate - and the further away from the saddle point, the greater the oscillation. Therefore by plotting this internuclear distance vs time, and adjusting the distance by small amounts until there was very little oscillation seen, a more precise value was found.

[[File: Sc3916_1.PNG|thumb|300px|none|Internuclear Distances vs Time]]

==Reaction trajectories==

[[File:Sc3916_mep.PNG|thumb|400px|left|Minimum Energy Path from from r1 = rts+δ, r2 = rts]]

[[File:Sc3916_dynamics.PNG|thumb|400px|center|Dynamic Energy Path from r1 = rts+δ, r2 = rts]]

'''Comment on how the MEP and the trajectory you just calculated differ.'''

The minimum energy path represents an infinitely slow reaction trajectory. We start off a very small distance from the transition structure, and then the trajectory "rolls" down the potential energy surface, following where the gradient is most negative. Since velocity resets to 0 every time interval, so does momentum (p=mv). This means at every infinitesimally small step, this pathway will minimise its energy. This is not how it works in reality.

The dynamic trajectory is much closer to reality. As the trajectory starts to move down the potential energy surface it gains momentum. This means it will carry on rolling up the curve on the other side of the minimum (which increases its potential energy), leading to oscillations. This reflects the vibrational motion of atoms, which MEP does not include.

Dynamics Internuclear distance vs time plot: 
Final AB distance = 0.8 
Final BC distance = 9.0 

Dynamics Internuclear Momenta vs time plot: 
Final average momentum for AB = 1.25 
Final average momentum for BC = 2.5 

This shows the reaction path from the above calculation leads to A-B + C

If we switch around initial values, so now rAB=rts+δ and rBC=rts, the reaction path will output the same values, except the products are B-C + A.

'''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.'''

The reaction proceeds from left to right across the below plot, A-B + C -> A + B-C. This reaction is essentially the reverse of the previous trajectory (allowing the pathway to "roll" down from the transition state), however it crosses over, to form the products.

[[File:Sc3916_dynamics2.PNG|thumb|400px|none| initial conditions: 
rAB = 0.8 
rBC = 9.0 
pAB = -1.25 
pBC = -2.5]]

==Reactive and unreactive trajectories==
The following table is for the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, with varying momentum.

{| class="wikitable" border=1
! p1 !! p2 !! E(kcal/mol) !! Reactive?
|-
| -1.25 || -2.5 || -99.119 || Yes
|-
| -1.5 || -2.0 || -100.456 || No
|-
| -1.5 || -2.5 || -98.956 || Yes
|-
| -2.5 || -5.0 || -84.956 || No
|-
| -2.5 || -5.2 || -83.416 || Yes
|}

[[File:Sc3916_rxn1.PNG|thumb|300px|none|Initial p1=-1.25, r2=-2.5 
the trajectory goes smoothly up to the transition state, then starts oscillating as it rolls down to the products]]

[[File:Sc3916_rxn2.PNG|thumb|300px|none|Initial p1=-1.5, r2=-2.0 
the trajectory goes up towards the transition state smoothly (but not very far), and then rolls back down the way it came, with oscillating]]

[[File:Sc3916_rxn3.PNG|thumb|300px|none|Initial p1=-1.5, r2=-2.5 
the trajectory goes smoothly up to the transition state, then starts oscillating as it rolls down to the products]]

[[File:Sc3916_rxn4b.PNG|thumb|300px|none|Initial p1=-2.5, r2=-5.0 
the trajectory goes smoothly up to the transition state, crosses it, but recrosses it and reverts back to the reactants, with very large oscillations]]

[[File:Sc3916_rxn5.PNG|thumb|300px|none|Initial p1=-2.5, r2=-5.2 
the trajectory crosses the transition state, forming the products, then recrosses it to the reactant side, and then recrosses it again, forming the products finally]]

'''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?'''

4 major assumptions are used. Firstly, the Born-Oppenheimer approximation is used, and that the effects of quantum tunneling are negligible (there is a non-zero probability that reactants can convert to products without reaching the activation energy). The reactants are assumed to have a Boltzmann distribution - which is perfectly valid at low temperatures. The final assumption is that "once the system attains the transition state, with a velocity towards the product configuration, it will not reenter the initial state region again."[https://doi.org/10.1016/B978-044452837-7.50005-8 1]

This last assumption does not fit with our simulations. Both the 4th, and 5th plots contain the trajectory crossing over the transition state and back again. Since there will be more trajectories (like the 4th) which involve crossing over and are still unreactive - which are not predicted by Transition State Theory - experimental rate values will be slower than those predicted by the theory.

==EXERCISE 2: F - H - H system==

'''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 is exothermic - the products have a more negative energy than the reactants. Because the enthalpy change only involves breaking a H-H bond and making a H-F bond, ΔH = +H(H-H) - H(H-F).

Since ΔH < 0:

H(H-F)>H(H-H)

This means HF has a greater bond enthalpy (and therefore strength) than H2. It can also be seen that H + HF is endothermic, which leads to the same conclusion.

'''Locate the approximate position of the transition state.'''

rts: 
RHH = 0.745 angstroms 
RHF = 1.811 angstroms

(0.7448778, 1.810754) is the super accurate one

'''Report the activation energy for both reactions.'''

By performing an MEP from next to the transition structure, the path from the TS to the reactants, or products, was found. By subtracting the final energy from the TS, the Ea was obtained. Since the activation energy for H2 + F is very small, the MEP has a very small gradient, and as such the calculation required an order of magnitude more steps in the calculation (than HF + H) to reach the reactants. Conversely, in HF + H a step change is seen.

[[File:Mep_energyvstime_sc3916.PNG|thumb|left|initial conditions: 
RHH=0.745 
RHF=1.83]]

[[File:3916_energyvstime_hf2.PNG|thumb|center|initial conditions: 
RHH=0.745 
RHF=1.8]]

Ea(H2 + F Ea) = 0.256 kcal/mol 
Ea(HF + H Ea) = 30.202 kcal/mol 

=== Reaction dynamics ===
* Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Internuclear Momenta vs Time”. {{fontcolor1|blue|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

[[File:H2+f_surface__3916.PNG|thumb|none|F + H2, initial conditions: 
RHH (AB) = 0.74 angstroms 
RHF (BC) = 2 angstroms 
AB momentum = +3 
BC momentum = -5]]

[[File:H2+F_MOMENTAVTIME3916.PNG|thumb|none|F + H2 Momentum vs Time Plot]]

The animation of this reaction clearly shows the velocity of the lone H atom formed has increased, and vibrations occurring in HF. This can be seen in the momentum vs time plot - with the increased AB and AC momentum reflecting the increase in velocity, and BC momentum oscillation reflecting the vibrations of the H-F bond. Since translational and vibrational energy contribute to kinetic energy, we know that the energy released by this exothermic reaction (a decrease in enthalpy) is conserved in the form of increased kinetic energy. This can be seen in the below graph, and is easily shown experimentally by the resultant increase in temperature. To determine the extent vibrational energy contributes to this, we can use IR spectroscopy. If there is a lot of vibrational energy, there will be a relatively high population in the first vibrational level, and we will see overtone bands in the IR spectrum.

[[File:H2+F_ENERGYVSTIME3916.PNG|thumb|none|F + H2 Energy vs Time Plot]]

* Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 0.74, with a momentum pFH = -0.5, and explore several values of pHH in the range -3 to 3 (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.

[[File:SC3916-3CONTOUR.PNG|thumb|none|F + H2, initial conditions A: 
rHH=0.74 rHF=1.83 
pHH=-3.0 pHF=-0.5 ]]

[[File:SC3916-2CONTOUR.PNG|thumb|none|F + H2, initial conditions B: 
rHH=0.74 rHF=1.83 
pHH=-2.0 pHF=-0.5 ]]

[[File:SC3916-1.5CONTOUR.PNG|thumb|none|F + H2, initial conditions C: 
rHH=0.74 rHF=1.83 
pHH=-1 pHF=-0.5 ]]

[[File:SC39160CONTOUR.PNG|thumb|none|F + H2, initial conditions D: 
rHH=0.74 rHF=1.83 
pHH=0.0 pHF=-0.5 ]]

[[File:Sc3916_redone1.PNG|thumb|none|F + H2, initial conditions E: 
rHH=0.74 rHF=1.83 
pHH=+1.0 pHF=-0.5 ]]

[[File:Sc3916_redone2.PNG|thumb|none|F + H2, initial conditions F: 
rHH=0.74 rHF=1.83 
pHH=+3.0 pHF=-0.5 ]]

* For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now?

[[File:Sc3916_redonelast.PNG|thumb|none|F + H2, initial conditions G: 
rHH=0.74 rHF=1.83 
pHH=+0.1 pHF=-0.8 ]]

*Let us now focus on the reverse reaction, H + HF (an H atom with high kinetic energy colliding with HF).

[[File:Sc3916_hh-10hf-.5.PNG|thumb|none|HF + H, initial conditions H: 
rHH=0.9 rHF=1.83 
pHH=-10 pHF=-5 ]]

[[File:Sc3916_hh-1.1h4-13.PNG|thumb|none|HF + H, initial conditions I: 
rHH=0.9 rHF=1.83 
pHH=-1.1 pHF=-13 ]]

The cases studied are an illustration of Polanyi's empirical rules. {{fontcolor|blue|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, for a reaction with an early transition state, vibrational energy is more effective in overcoming the TS - so initial conditions with higher vibrational energy are more likely to result in reaction, so are more efficient. For a late transition state, translational energy is more effective in overcoming it - so atoms/molecules with a higher translational energy create a more efficient reaction.[http://doi.org/10.1021/jz301649w 2]

This can be seen in the above reaction trajectories. H2 + F has an early transition state (Hammond's postulate tells us that in an endothermic reaction, the transition state most resembles the reactants). Reaction trajectories with high initial vibrational energy (shown by pHH, conditions A, B, F) are unreactive trajectories. Conversely, those with low vibrational energy (C, D, E) are reactive. F also has increased kinetic energy on top of low vibrational energy, and this too is reactive.

The reverse reaction, HF + H, has a late transition state (from Hammond's postulate) - so initial conditions with greater vibrational energy will be more efficient. This can be seen in the reactive trajectory with conditions I. Conversely, conditions H has a much greater translational energy, and is unreactive.

A good reference about transition state theory is chapter 10 of J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics Prentice-Hall.

MRD:sc3916

2018-05-25T14:55:43Z

Sc3916: /* Reaction dynamics */

=Molecular Reaction Dynamics Report=

==Potential Energy Surfaces==

'''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.'''

The transition structure is at a saddle point - the intersection of the minimum of a curve in one plane, and the maximum of a curve in the other. In this case, the minimum curve is on the plane corresponding to RAB=RBC, and the maximum is on the perpendicular plane. The gradient of both these curves is equal to zero.

This is different from when the potential energy surface is at a minimum - in one plane the gradient = 0, but not in the other. This allows us to distinguish between the transition structure and minima by looking at the curvature in '''both''' planes.

[[File:Saddle_Point_Visualisation_sc3916.png|thumb|none|500px|Saddle Point Visualisation]]

'''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.'''

'''rts= 0.9075 angstroms =rAB=rBC'''

To estimate this, a rough estimate was found by looking at the saddle point. At the saddle point, gradient = 0 in both planes - so if the system is at exactly this point, the atoms will remain stationary if it has no initial momentum. If it isn't at this point, it will start to oscillate - and the further away from the saddle point, the greater the oscillation. Therefore by plotting this internuclear distance vs time, and adjusting the distance by small amounts until there was very little oscillation seen, a more precise value was found.

[[File: Sc3916_1.PNG|thumb|300px|none|Internuclear Distances vs Time]]

==Reaction trajectories==

[[File:Sc3916_mep.PNG|thumb|400px|left|Minimum Energy Path from from r1 = rts+δ, r2 = rts]]

[[File:Sc3916_dynamics.PNG|thumb|400px|center|Dynamic Energy Path from r1 = rts+δ, r2 = rts]]

'''Comment on how the MEP and the trajectory you just calculated differ.'''

The minimum energy path represents an infinitely slow reaction trajectory. We start off a very small distance from the transition structure, and then the trajectory "rolls" down the potential energy surface, following where the gradient is most negative. Since velocity resets to 0 every time interval, so does momentum (p=mv). This means at every infinitesimally small step, this pathway will minimise its energy. This is not how it works in reality.

The dynamic trajectory is much closer to reality. As the trajectory starts to move down the potential energy surface it gains momentum. This means it will carry on rolling up the curve on the other side of the minimum (which increases its potential energy), leading to oscillations. This reflects the vibrational motion of atoms, which MEP does not include.

Dynamics Internuclear distance vs time plot: 
Final AB distance = 0.8 
Final BC distance = 9.0 

Dynamics Internuclear Momenta vs time plot: 
Final average momentum for AB = 1.25 
Final average momentum for BC = 2.5 

This shows the reaction path from the above calculation leads to A-B + C

If we switch around initial values, so now rAB=rts+δ and rBC=rts, the reaction path will output the same values, except the products are B-C + A.

'''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.'''

The reaction proceeds from left to right across the below plot, A-B + C -> A + B-C. This reaction is essentially the reverse of the previous trajectory (allowing the pathway to "roll" down from the transition state), however it crosses over, to form the products.

[[File:Sc3916_dynamics2.PNG|thumb|400px|none| initial conditions: 
rAB = 0.8 
rBC = 9.0 
pAB = -1.25 
pBC = -2.5]]

==Reactive and unreactive trajectories==
The following table is for the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, with varying momentum.

{| class="wikitable" border=1
! p1 !! p2 !! E(kcal/mol) !! Reactive?
|-
| -1.25 || -2.5 || -99.119 || Yes
|-
| -1.5 || -2.0 || -100.456 || No
|-
| -1.5 || -2.5 || -98.956 || Yes
|-
| -2.5 || -5.0 || -84.956 || No
|-
| -2.5 || -5.2 || -83.416 || Yes
|}

[[File:Sc3916_rxn1.PNG|thumb|300px|none|Initial p1=-1.25, r2=-2.5 
the trajectory goes smoothly up to the transition state, then starts oscillating as it rolls down to the products]]

[[File:Sc3916_rxn2.PNG|thumb|300px|none|Initial p1=-1.5, r2=-2.0 
the trajectory goes up towards the transition state smoothly (but not very far), and then rolls back down the way it came, with oscillating]]

[[File:Sc3916_rxn3.PNG|thumb|300px|none|Initial p1=-1.5, r2=-2.5 
the trajectory goes smoothly up to the transition state, then starts oscillating as it rolls down to the products]]

[[File:Sc3916_rxn4b.PNG|thumb|300px|none|Initial p1=-2.5, r2=-5.0 
the trajectory goes smoothly up to the transition state, crosses it, but recrosses it and reverts back to the reactants, with very large oscillations]]

[[File:Sc3916_rxn5.PNG|thumb|300px|none|Initial p1=-2.5, r2=-5.2 
the trajectory crosses the transition state, forming the products, then recrosses it to the reactant side, and then recrosses it again, forming the products finally]]

'''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?'''

4 major assumptions are used. Firstly, the Born-Oppenheimer approximation is used, and that the effects of quantum tunneling are negligible (there is a non-zero probability that reactants can convert to products without reaching the activation energy). The reactants are assumed to have a Boltzmann distribution - which is perfectly valid at low temperatures. The final assumption is that "once the system attains the transition state, with a velocity towards the product configuration, it will not reenter the initial state region again."[https://doi.org/10.1016/B978-044452837-7.50005-8 1]

This last assumption does not fit with our simulations. Both the 4th, and 5th plots contain the trajectory crossing over the transition state and back again. Since there will be more trajectories (like the 4th) which involve crossing over and are still unreactive - which are not predicted by Transition State Theory - experimental rate values will be slower than those predicted by the theory.

==EXERCISE 2: F - H - H system==

'''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 is exothermic - the products have a more negative energy than the reactants. Because the enthalpy change only involves breaking a H-H bond and making a H-F bond, ΔH = +H(H-H) - H(H-F).

Since ΔH < 0:

H(H-F)>H(H-H)

This means HF has a greater bond enthalpy (and therefore strength) than H2. It can also be seen that H + HF is endothermic, which leads to the same conclusion.

'''Locate the approximate position of the transition state.'''

rts: 
RHH = 0.745 angstroms 
RHF = 1.811 angstroms

(0.7448778, 1.810754) is the super accurate one

'''Report the activation energy for both reactions.'''

By performing an MEP from next to the transition structure, the path from the TS to the reactants, or products, was found. By subtracting the final energy from the TS, the Ea was obtained. Since the activation energy for H2 + F is very small, the MEP has a very small gradient, and as such the calculation required an order of magnitude more steps in the calculation (than HF + H) to reach the reactants. Conversely, in HF + H a step change is seen.

[[File:Mep_energyvstime_sc3916.PNG|thumb|left|initial conditions: 
RHH=0.745 
RHF=1.83]]

[[File:3916_energyvstime_hf2.PNG|thumb|center|initial conditions: 
RHH=0.745 
RHF=1.8]]

Ea(H2 + F Ea) = 0.256 kcal/mol 
Ea(HF + H Ea) = 30.202 kcal/mol 

=== Reaction dynamics ===
* Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Internuclear Momenta vs Time”. {{fontcolor1|blue|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

[[File:H2+f_surface__3916.PNG|thumb|none|F + H2, initial conditions: 
RHH (AB) = 0.74 angstroms 
RHF (BC) = 2 angstroms 
AB momentum = +3 
BC momentum = -5]]

[[File:H2+F_MOMENTAVTIME3916.PNG|thumb|none|F + H2 Momentum vs Time Plot]]

The animation of this reaction clearly shows the velocity of the lone H atom formed has increased, and vibrations occurring in HF. This can be seen in the momentum vs time plot - with the increased AB and AC momentum reflecting the increase in velocity, and BC momentum oscillation reflecting the vibrations of the H-F bond. Since translational and vibrational energy contribute to kinetic energy, we know that the energy released by this exothermic reaction (a decrease in enthalpy) is conserved in the form of increased kinetic energy. This can be seen in the below graph, and is easily shown experimentally by the resultant increase in temperature. To determine the extent vibrational energy contributes to this, we can use IR spectroscopy. If there is a lot of vibrational energy, there will be a relatively high population in the first vibrational level, and we will see overtone bands in the IR spectrum.

[[File:H2+F_ENERGYVSTIME3916.PNG|thumb|none|F + H2 Energy vs Time Plot]]

* Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 0.74, with a momentum pFH = -0.5, and explore several values of pHH in the range -3 to 3 (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.

[[File:SC3916-3CONTOUR.PNG|thumb|none|F + H2, initial conditions A: 
rHH=0.74 rHF=1.83 
pHH=-3.0 pHF=-0.5 ]]

[[File:SC3916-2CONTOUR.PNG|thumb|none|F + H2, initial conditions B: 
rHH=0.74 rHF=1.83 
pHH=-2.0 pHF=-0.5 ]]

[[File:SC3916-1.5CONTOUR.PNG|thumb|none|F + H2, initial conditions C: 
rHH=0.74 rHF=1.83 
pHH=-1 pHF=-0.5 ]]

[[File:SC39160CONTOUR.PNG|thumb|none|F + H2, initial conditions D: 
rHH=0.74 rHF=1.83 
pHH=0.0 pHF=-0.5 ]]

[[File:Sc3916_redone1.PNG|thumb|none|F + H2, initial conditions E: 
rHH=0.74 rHF=1.83 
pHH=+1.0 pHF=-0.5 ]]

[[File:Sc3916_redone2.PNG|thumb|none|F + H2, initial conditions F: 
rHH=0.74 rHF=1.83 
pHH=+3.0 pHF=-0.5 ]]

* For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now?

[[File:Sc3916_redonelast.PNG|thumb|none|F + H2, initial conditions G: 
rHH=0.74 rHF=1.83 
pHH=+0.1 pHF=-0.8 ]]

*Let us now focus on the reverse reaction, H + HF (an H atom with high kinetic energy colliding with HF).

[[File:Sc3916_hh-10hf-.5.PNG|thumb|none|HF + H, initial conditions H: 
rHH=0.9 rHF=1.83 
pHH=-10 pHF=-5 ]]

[[File:Sc3916_hh-1.1h4-13.PNG|thumb|none|HF + H, initial conditions I: 
rHH=0.9 rHF=1.83 
pHH=-1.1 pHF=-13 ]]

The cases studied are an illustration of Polanyi's empirical rules. {{fontcolor|blue|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, for a reaction with an early transition state, vibrational energy is more effective in overcoming the TS - so initial conditions with higher vibrational energy are more likely to result in reaction, so are more efficient. For a late transition state, translational energy is more effective in overcoming it - so atoms/molecules with a higher translational energy create a more efficient reaction.[[http://doi.org/10.1021/jz301649w 2]]

This can be seen in the above reaction trajectories. H2 + F has an early transition state (Hammond's postulate tells us that in an endothermic reaction, the transition state most resembles the reactants). Reaction trajectories with high initial vibrational energy (shown by pHH, conditions A, B, F) are unreactive trajectories. Conversely, those with low vibrational energy (C, D, E) are reactive. F also has increased kinetic energy on top of low vibrational energy, and this too is reactive.

The reverse reaction, HF + H, has a late transition state (from Hammond's postulate) - so initial conditions with greater vibrational energy will be more efficient. This can be seen in the reactive trajectory with conditions I. Conversely, conditions H has a much greater translational energy, and is unreactive.

A good reference about transition state theory is chapter 10 of J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics Prentice-Hall.

MRD:sc3916

2018-05-25T14:52:57Z

Sc3916: /* Reaction dynamics */

=Molecular Reaction Dynamics Report=

==Potential Energy Surfaces==

'''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.'''

The transition structure is at a saddle point - the intersection of the minimum of a curve in one plane, and the maximum of a curve in the other. In this case, the minimum curve is on the plane corresponding to RAB=RBC, and the maximum is on the perpendicular plane. The gradient of both these curves is equal to zero.

This is different from when the potential energy surface is at a minimum - in one plane the gradient = 0, but not in the other. This allows us to distinguish between the transition structure and minima by looking at the curvature in '''both''' planes.

[[File:Saddle_Point_Visualisation_sc3916.png|thumb|none|500px|Saddle Point Visualisation]]

'''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.'''

'''rts= 0.9075 angstroms =rAB=rBC'''

To estimate this, a rough estimate was found by looking at the saddle point. At the saddle point, gradient = 0 in both planes - so if the system is at exactly this point, the atoms will remain stationary if it has no initial momentum. If it isn't at this point, it will start to oscillate - and the further away from the saddle point, the greater the oscillation. Therefore by plotting this internuclear distance vs time, and adjusting the distance by small amounts until there was very little oscillation seen, a more precise value was found.

[[File: Sc3916_1.PNG|thumb|300px|none|Internuclear Distances vs Time]]

==Reaction trajectories==

[[File:Sc3916_mep.PNG|thumb|400px|left|Minimum Energy Path from from r1 = rts+δ, r2 = rts]]

[[File:Sc3916_dynamics.PNG|thumb|400px|center|Dynamic Energy Path from r1 = rts+δ, r2 = rts]]

'''Comment on how the MEP and the trajectory you just calculated differ.'''

The minimum energy path represents an infinitely slow reaction trajectory. We start off a very small distance from the transition structure, and then the trajectory "rolls" down the potential energy surface, following where the gradient is most negative. Since velocity resets to 0 every time interval, so does momentum (p=mv). This means at every infinitesimally small step, this pathway will minimise its energy. This is not how it works in reality.

The dynamic trajectory is much closer to reality. As the trajectory starts to move down the potential energy surface it gains momentum. This means it will carry on rolling up the curve on the other side of the minimum (which increases its potential energy), leading to oscillations. This reflects the vibrational motion of atoms, which MEP does not include.

Dynamics Internuclear distance vs time plot: 
Final AB distance = 0.8 
Final BC distance = 9.0 

Dynamics Internuclear Momenta vs time plot: 
Final average momentum for AB = 1.25 
Final average momentum for BC = 2.5 

This shows the reaction path from the above calculation leads to A-B + C

If we switch around initial values, so now rAB=rts+δ and rBC=rts, the reaction path will output the same values, except the products are B-C + A.

'''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.'''

The reaction proceeds from left to right across the below plot, A-B + C -> A + B-C. This reaction is essentially the reverse of the previous trajectory (allowing the pathway to "roll" down from the transition state), however it crosses over, to form the products.

[[File:Sc3916_dynamics2.PNG|thumb|400px|none| initial conditions: 
rAB = 0.8 
rBC = 9.0 
pAB = -1.25 
pBC = -2.5]]

==Reactive and unreactive trajectories==
The following table is for the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, with varying momentum.

{| class="wikitable" border=1
! p1 !! p2 !! E(kcal/mol) !! Reactive?
|-
| -1.25 || -2.5 || -99.119 || Yes
|-
| -1.5 || -2.0 || -100.456 || No
|-
| -1.5 || -2.5 || -98.956 || Yes
|-
| -2.5 || -5.0 || -84.956 || No
|-
| -2.5 || -5.2 || -83.416 || Yes
|}

[[File:Sc3916_rxn1.PNG|thumb|300px|none|Initial p1=-1.25, r2=-2.5 
the trajectory goes smoothly up to the transition state, then starts oscillating as it rolls down to the products]]

[[File:Sc3916_rxn2.PNG|thumb|300px|none|Initial p1=-1.5, r2=-2.0 
the trajectory goes up towards the transition state smoothly (but not very far), and then rolls back down the way it came, with oscillating]]

[[File:Sc3916_rxn3.PNG|thumb|300px|none|Initial p1=-1.5, r2=-2.5 
the trajectory goes smoothly up to the transition state, then starts oscillating as it rolls down to the products]]

[[File:Sc3916_rxn4b.PNG|thumb|300px|none|Initial p1=-2.5, r2=-5.0 
the trajectory goes smoothly up to the transition state, crosses it, but recrosses it and reverts back to the reactants, with very large oscillations]]

[[File:Sc3916_rxn5.PNG|thumb|300px|none|Initial p1=-2.5, r2=-5.2 
the trajectory crosses the transition state, forming the products, then recrosses it to the reactant side, and then recrosses it again, forming the products finally]]

'''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?'''

4 major assumptions are used. Firstly, the Born-Oppenheimer approximation is used, and that the effects of quantum tunneling are negligible (there is a non-zero probability that reactants can convert to products without reaching the activation energy). The reactants are assumed to have a Boltzmann distribution - which is perfectly valid at low temperatures. The final assumption is that "once the system attains the transition state, with a velocity towards the product configuration, it will not reenter the initial state region again."[https://doi.org/10.1016/B978-044452837-7.50005-8 1]

This last assumption does not fit with our simulations. Both the 4th, and 5th plots contain the trajectory crossing over the transition state and back again. Since there will be more trajectories (like the 4th) which involve crossing over and are still unreactive - which are not predicted by Transition State Theory - experimental rate values will be slower than those predicted by the theory.

==EXERCISE 2: F - H - H system==

'''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 is exothermic - the products have a more negative energy than the reactants. Because the enthalpy change only involves breaking a H-H bond and making a H-F bond, ΔH = +H(H-H) - H(H-F).

Since ΔH < 0:

H(H-F)>H(H-H)

This means HF has a greater bond enthalpy (and therefore strength) than H2. It can also be seen that H + HF is endothermic, which leads to the same conclusion.

'''Locate the approximate position of the transition state.'''

rts: 
RHH = 0.745 angstroms 
RHF = 1.811 angstroms

(0.7448778, 1.810754) is the super accurate one

'''Report the activation energy for both reactions.'''

By performing an MEP from next to the transition structure, the path from the TS to the reactants, or products, was found. By subtracting the final energy from the TS, the Ea was obtained. Since the activation energy for H2 + F is very small, the MEP has a very small gradient, and as such the calculation required an order of magnitude more steps in the calculation (than HF + H) to reach the reactants. Conversely, in HF + H a step change is seen.

[[File:Mep_energyvstime_sc3916.PNG|thumb|left|initial conditions: 
RHH=0.745 
RHF=1.83]]

[[File:3916_energyvstime_hf2.PNG|thumb|center|initial conditions: 
RHH=0.745 
RHF=1.8]]

Ea(H2 + F Ea) = 0.256 kcal/mol 
Ea(HF + H Ea) = 30.202 kcal/mol 

=== Reaction dynamics ===
* Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Internuclear Momenta vs Time”. {{fontcolor1|blue|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

[[File:H2+f_surface__3916.PNG|thumb|none|F + H2, initial conditions: 
RHH (AB) = 0.74 angstroms 
RHF (BC) = 2 angstroms 
AB momentum = +3 
BC momentum = -5]]

[[File:H2+F_MOMENTAVTIME3916.PNG|thumb|none|F + H2 Momentum vs Time Plot]]

The animation of this reaction clearly shows the velocity of the lone H atom formed has increased, and vibrations occurring in HF. This can be seen in the momentum vs time plot - with the increased AB and AC momentum reflecting the increase in velocity, and BC momentum oscillation reflecting the vibrations of the H-F bond. Since translational and vibrational energy contribute to kinetic energy, we know that the energy released by this exothermic reaction (a decrease in enthalpy) is conserved in the form of increased kinetic energy. This can be seen in the below graph, and is easily shown experimentally by the resultant increase in temperature. To determine the extent vibrational energy contributes to this, we can use IR spectroscopy. If there is a lot of vibrational energy, there will be a relatively high population in the first vibrational level, and we will see overtone bands in the IR spectrum.

[[File:H2+F_ENERGYVSTIME3916.PNG|thumb|none|F + H2 Energy vs Time Plot]]

* Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 0.74, with a momentum pFH = -0.5, and explore several values of pHH in the range -3 to 3 (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.

[[File:SC3916-3CONTOUR.PNG|thumb|none|F + H2, initial conditions A: 
rHH=0.74 rHF=1.83 
pHH=-3.0 pHF=-0.5 ]]

[[File:SC3916-2CONTOUR.PNG|thumb|none|F + H2, initial conditions B: 
rHH=0.74 rHF=1.83 
pHH=-2.0 pHF=-0.5 ]]

[[File:SC3916-1.5CONTOUR.PNG|thumb|none|F + H2, initial conditions C: 
rHH=0.74 rHF=1.83 
pHH=-1 pHF=-0.5 ]]

[[File:SC39160CONTOUR.PNG|thumb|none|F + H2, initial conditions D: 
rHH=0.74 rHF=1.83 
pHH=0.0 pHF=-0.5 ]]

[[File:Sc3916_redone1.PNG|thumb|none|F + H2, initial conditions E: 
rHH=0.74 rHF=1.83 
pHH=+1.0 pHF=-0.5 ]]

[[File:Sc3916_redone2.PNG|thumb|none|F + H2, initial conditions F: 
rHH=0.74 rHF=1.83 
pHH=+3.0 pHF=-0.5 ]]

* For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now?

[[File:Sc3916_redonelast.PNG|thumb|none|F + H2, initial conditions G: 
rHH=0.74 rHF=1.83 
pHH=+0.1 pHF=-0.8 ]]

*Let us now focus on the reverse reaction, H + HF (an H atom with high kinetic energy colliding with HF).

[[File:Sc3916_hh-10hf-.5.PNG|thumb|none|HF + H, initial conditions H: 
rHH=0.9 rHF=1.83 
pHH=-10 pHF=-5 ]]

[[File:Sc3916_hh-1.1h4-13.PNG|thumb|none|HF + H, initial conditions I: 
rHH=0.9 rHF=1.83 
pHH=-1.1 pHF=-13 ]]

The cases studied are an illustration of Polanyi's empirical rules. {{fontcolor|blue|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, for a reaction with an early transition state, vibrational energy is more effective in overcoming the TS - so initial conditions with higher vibrational energy are more likely to result in reaction, so are more efficient. For a late transition state, translational energy is more effective in overcoming it - so atoms/molecules with a higher translational energy create a more efficient reaction.

This can be seen in the above reaction trajectories. H2 + F has an early transition state (Hammond's postulate tells us that in an endothermic reaction, the transition state most resembles the reactants). Reaction trajectories with high initial vibrational energy (shown by pHH, conditions A, B, F) are unreactive trajectories. Conversely, those with low vibrational energy (C, D, E) are reactive. F also has increased kinetic energy on top of low vibrational energy, and this too is reactive.

The reverse reaction, HF + H, has a late transition state (from Hammond's postulate) - so initial conditions with greater vibrational energy will be more efficient. This can be seen in the reactive trajectory with conditions I. Conversely, conditions H has a much greater translational energy, and is unreactive.

A good reference about transition state theory is chapter 10 of J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics Prentice-Hall.

MRD:sc3916

2018-05-25T14:43:06Z

Sc3916: /* Reaction dynamics */

=Molecular Reaction Dynamics Report=

==Potential Energy Surfaces==

'''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.'''

The transition structure is at a saddle point - the intersection of the minimum of a curve in one plane, and the maximum of a curve in the other. In this case, the minimum curve is on the plane corresponding to RAB=RBC, and the maximum is on the perpendicular plane. The gradient of both these curves is equal to zero.

This is different from when the potential energy surface is at a minimum - in one plane the gradient = 0, but not in the other. This allows us to distinguish between the transition structure and minima by looking at the curvature in '''both''' planes.

[[File:Saddle_Point_Visualisation_sc3916.png|thumb|none|500px|Saddle Point Visualisation]]

'''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.'''

'''rts= 0.9075 angstroms =rAB=rBC'''

To estimate this, a rough estimate was found by looking at the saddle point. At the saddle point, gradient = 0 in both planes - so if the system is at exactly this point, the atoms will remain stationary if it has no initial momentum. If it isn't at this point, it will start to oscillate - and the further away from the saddle point, the greater the oscillation. Therefore by plotting this internuclear distance vs time, and adjusting the distance by small amounts until there was very little oscillation seen, a more precise value was found.

[[File: Sc3916_1.PNG|thumb|300px|none|Internuclear Distances vs Time]]

==Reaction trajectories==

[[File:Sc3916_mep.PNG|thumb|400px|left|Minimum Energy Path from from r1 = rts+δ, r2 = rts]]

[[File:Sc3916_dynamics.PNG|thumb|400px|center|Dynamic Energy Path from r1 = rts+δ, r2 = rts]]

'''Comment on how the MEP and the trajectory you just calculated differ.'''

The minimum energy path represents an infinitely slow reaction trajectory. We start off a very small distance from the transition structure, and then the trajectory "rolls" down the potential energy surface, following where the gradient is most negative. Since velocity resets to 0 every time interval, so does momentum (p=mv). This means at every infinitesimally small step, this pathway will minimise its energy. This is not how it works in reality.

The dynamic trajectory is much closer to reality. As the trajectory starts to move down the potential energy surface it gains momentum. This means it will carry on rolling up the curve on the other side of the minimum (which increases its potential energy), leading to oscillations. This reflects the vibrational motion of atoms, which MEP does not include.

Dynamics Internuclear distance vs time plot: 
Final AB distance = 0.8 
Final BC distance = 9.0 

Dynamics Internuclear Momenta vs time plot: 
Final average momentum for AB = 1.25 
Final average momentum for BC = 2.5 

This shows the reaction path from the above calculation leads to A-B + C

If we switch around initial values, so now rAB=rts+δ and rBC=rts, the reaction path will output the same values, except the products are B-C + A.

'''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.'''

The reaction proceeds from left to right across the below plot, A-B + C -> A + B-C. This reaction is essentially the reverse of the previous trajectory (allowing the pathway to "roll" down from the transition state), however it crosses over, to form the products.

[[File:Sc3916_dynamics2.PNG|thumb|400px|none| initial conditions: 
rAB = 0.8 
rBC = 9.0 
pAB = -1.25 
pBC = -2.5]]

==Reactive and unreactive trajectories==
The following table is for the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, with varying momentum.

{| class="wikitable" border=1
! p1 !! p2 !! E(kcal/mol) !! Reactive?
|-
| -1.25 || -2.5 || -99.119 || Yes
|-
| -1.5 || -2.0 || -100.456 || No
|-
| -1.5 || -2.5 || -98.956 || Yes
|-
| -2.5 || -5.0 || -84.956 || No
|-
| -2.5 || -5.2 || -83.416 || Yes
|}

[[File:Sc3916_rxn1.PNG|thumb|300px|none|Initial p1=-1.25, r2=-2.5 
the trajectory goes smoothly up to the transition state, then starts oscillating as it rolls down to the products]]

[[File:Sc3916_rxn2.PNG|thumb|300px|none|Initial p1=-1.5, r2=-2.0 
the trajectory goes up towards the transition state smoothly (but not very far), and then rolls back down the way it came, with oscillating]]

[[File:Sc3916_rxn3.PNG|thumb|300px|none|Initial p1=-1.5, r2=-2.5 
the trajectory goes smoothly up to the transition state, then starts oscillating as it rolls down to the products]]

[[File:Sc3916_rxn4b.PNG|thumb|300px|none|Initial p1=-2.5, r2=-5.0 
the trajectory goes smoothly up to the transition state, crosses it, but recrosses it and reverts back to the reactants, with very large oscillations]]

[[File:Sc3916_rxn5.PNG|thumb|300px|none|Initial p1=-2.5, r2=-5.2 
the trajectory crosses the transition state, forming the products, then recrosses it to the reactant side, and then recrosses it again, forming the products finally]]

'''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?'''

4 major assumptions are used. Firstly, the Born-Oppenheimer approximation is used, and that the effects of quantum tunneling are negligible (there is a non-zero probability that reactants can convert to products without reaching the activation energy). The reactants are assumed to have a Boltzmann distribution - which is perfectly valid at low temperatures. The final assumption is that "once the system attains the transition state, with a velocity towards the product configuration, it will not reenter the initial state region again."[https://doi.org/10.1016/B978-044452837-7.50005-8 1]

This last assumption does not fit with our simulations. Both the 4th, and 5th plots contain the trajectory crossing over the transition state and back again. Since there will be more trajectories (like the 4th) which involve crossing over and are still unreactive - which are not predicted by Transition State Theory - experimental rate values will be slower than those predicted by the theory.

==EXERCISE 2: F - H - H system==

'''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 is exothermic - the products have a more negative energy than the reactants. Because the enthalpy change only involves breaking a H-H bond and making a H-F bond, ΔH = +H(H-H) - H(H-F).

Since ΔH < 0:

H(H-F)>H(H-H)

This means HF has a greater bond enthalpy (and therefore strength) than H2. It can also be seen that H + HF is endothermic, which leads to the same conclusion.

'''Locate the approximate position of the transition state.'''

rts: 
RHH = 0.745 angstroms 
RHF = 1.811 angstroms

(0.7448778, 1.810754) is the super accurate one

'''Report the activation energy for both reactions.'''

By performing an MEP from next to the transition structure, the path from the TS to the reactants, or products, was found. By subtracting the final energy from the TS, the Ea was obtained. Since the activation energy for H2 + F is very small, the MEP has a very small gradient, and as such the calculation required an order of magnitude more steps in the calculation (than HF + H) to reach the reactants. Conversely, in HF + H a step change is seen.

[[File:Mep_energyvstime_sc3916.PNG|thumb|left|initial conditions: 
RHH=0.745 
RHF=1.83]]

[[File:3916_energyvstime_hf2.PNG|thumb|center|initial conditions: 
RHH=0.745 
RHF=1.8]]

Ea(H2 + F Ea) = 0.256 kcal/mol 
Ea(HF + H Ea) = 30.202 kcal/mol 

=== Reaction dynamics ===
* Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Internuclear Momenta vs Time”. {{fontcolor1|blue|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

[[File:H2+f_surface__3916.PNG|thumb|none|F + H2, initial conditions: 
RHH (AB) = 0.74 angstroms 
RHF (BC) = 2 angstroms 
AB momentum = +3 
BC momentum = -5]]

[[File:H2+F_MOMENTAVTIME3916.PNG|thumb|none|F + H2 Momentum vs Time Plot]]

The animation of this reaction clearly shows the velocity of the lone H atom formed has increased, and vibrations occurring in HF. This can be seen in the momentum vs time plot - with the increased AB and AC momentum reflecting the increase in velocity, and BC momentum oscillation reflecting the vibrations of the H-F bond. Since translational and vibrational energy contribute to kinetic energy, we know that the energy released by this exothermic reaction (a decrease in enthalpy) is conserved in the form of increased kinetic energy. This can be seen in the below graph, and is easily shown experimentally by the resultant increase in temperature. To determine the extent vibrational energy contributes to this, we can use IR spectroscopy. If there is a lot of vibrational energy, there will be a relatively high population in the first vibrational level, and we will see overtone bands in the IR spectrum.

[[File:H2+F_ENERGYVSTIME3916.PNG|thumb|none|F + H2 Energy vs Time Plot]]

* Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 0.74, with a momentum pFH = -0.5, and explore several values of pHH in the range -3 to 3 (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.

[[File:SC3916-3CONTOUR.PNG|thumb|none|F + H2, initial conditions A: 
rHH=0.74 rHF=1.83 
pHH=-3.0 pHF=-0.5 ]]

[[File:SC3916-2CONTOUR.PNG|thumb|none|F + H2, initial conditions B: 
rHH=0.74 rHF=1.83 
pHH=-2.0 pHF=-0.5 ]]

[[File:SC3916-1.5CONTOUR.PNG|thumb|none|F + H2, initial conditions C: 
rHH=0.74 rHF=1.83 
pHH=-1 pHF=-0.5 ]]

[[File:SC39160CONTOUR.PNG|thumb|none|F + H2, initial conditions D: 
rHH=0.74 rHF=1.83 
pHH=0.0 pHF=-0.5 ]]

[[File:Sc3916_redone1.PNG|thumb|none|F + H2, initial conditions E: 
rHH=0.74 rHF=1.83 
pHH=+1.0 pHF=-0.5 ]]

[[File:Sc3916_redone2.PNG|thumb|none|F + H2, initial conditions F: 
rHH=0.74 rHF=1.83 
pHH=+3.0 pHF=-0.5 ]]

* For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now?

[[File:Sc3916_redonelast.PNG|thumb|none|F + H2, initial conditions G: 
rHH=0.74 rHF=1.83 
pHH=+0.1 pHF=-0.8 ]]

*Let us now focus on the reverse reaction, H + HF (an H atom with high kinetic energy colliding with HF).

[[File:Sc3916_hh-10hf-.5.PNG|thumb|none|HF + H, initial conditions H: 
rHH=0.9 rHF=1.83 
pHH=-10 pHF=-5 ]]

[[File:Sc3916_hh-1.1h4-13.PNG|thumb|none|HF + H, initial conditions I: 
rHH=0.9 rHF=1.83 
pHH=-1.1 pHF=-13 ]]

The cases studied are an illustration of Polanyi's empirical rules. {{fontcolor|blue|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, for a reaction with an early transition state, vibrational energy is more effective in overcoming the TS - so initial conditions with higher vibrational energy are more likely to result in reaction, so are more efficient. For a late transition state, translational energy is more effective in overcoming it - so atoms/molecules with a higher translational energy create a more efficient reaction.

This can be seen in the above reaction trajectories. H2 + F has an early transition state (Hammond's postulate tells us that in an endothermic reaction, the transition state most resembles the reactants). Reaction trajectories with high initial vibrational energy (shown by pHH, conditions A, B, F) are unreactive trajectories. Conversely, those with low vibrational energy (C, D, E) are reactive. F also has increased kinetic energy on top of low vibrational energy, and this too is reactive.

The reverse reaction , HF + H, has a late transition state (from Hammond's postulate) - so initial conditions with greater translational energy will be more efficient. This can be seen in the reactive trajectory with conditions I. Conversely conditions H,

A good reference about transition state theory is chapter 10 of J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics Prentice-Hall.

MRD:sc3916

2018-05-25T14:17:50Z

Sc3916: /* Reaction dynamics */

=Molecular Reaction Dynamics Report=

==Potential Energy Surfaces==

'''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.'''

The transition structure is at a saddle point - the intersection of the minimum of a curve in one plane, and the maximum of a curve in the other. In this case, the minimum curve is on the plane corresponding to RAB=RBC, and the maximum is on the perpendicular plane. The gradient of both these curves is equal to zero.

This is different from when the potential energy surface is at a minimum - in one plane the gradient = 0, but not in the other. This allows us to distinguish between the transition structure and minima by looking at the curvature in '''both''' planes.

[[File:Saddle_Point_Visualisation_sc3916.png|thumb|none|500px|Saddle Point Visualisation]]

'''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.'''

'''rts= 0.9075 angstroms =rAB=rBC'''

To estimate this, a rough estimate was found by looking at the saddle point. At the saddle point, gradient = 0 in both planes - so if the system is at exactly this point, the atoms will remain stationary if it has no initial momentum. If it isn't at this point, it will start to oscillate - and the further away from the saddle point, the greater the oscillation. Therefore by plotting this internuclear distance vs time, and adjusting the distance by small amounts until there was very little oscillation seen, a more precise value was found.

[[File: Sc3916_1.PNG|thumb|300px|none|Internuclear Distances vs Time]]

==Reaction trajectories==

[[File:Sc3916_mep.PNG|thumb|400px|left|Minimum Energy Path from from r1 = rts+δ, r2 = rts]]

[[File:Sc3916_dynamics.PNG|thumb|400px|center|Dynamic Energy Path from r1 = rts+δ, r2 = rts]]

'''Comment on how the MEP and the trajectory you just calculated differ.'''

The minimum energy path represents an infinitely slow reaction trajectory. We start off a very small distance from the transition structure, and then the trajectory "rolls" down the potential energy surface, following where the gradient is most negative. Since velocity resets to 0 every time interval, so does momentum (p=mv). This means at every infinitesimally small step, this pathway will minimise its energy. This is not how it works in reality.

The dynamic trajectory is much closer to reality. As the trajectory starts to move down the potential energy surface it gains momentum. This means it will carry on rolling up the curve on the other side of the minimum (which increases its potential energy), leading to oscillations. This reflects the vibrational motion of atoms, which MEP does not include.

Dynamics Internuclear distance vs time plot: 
Final AB distance = 0.8 
Final BC distance = 9.0 

Dynamics Internuclear Momenta vs time plot: 
Final average momentum for AB = 1.25 
Final average momentum for BC = 2.5 

This shows the reaction path from the above calculation leads to A-B + C

If we switch around initial values, so now rAB=rts+δ and rBC=rts, the reaction path will output the same values, except the products are B-C + A.

'''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.'''

The reaction proceeds from left to right across the below plot, A-B + C -> A + B-C. This reaction is essentially the reverse of the previous trajectory (allowing the pathway to "roll" down from the transition state), however it crosses over, to form the products.

[[File:Sc3916_dynamics2.PNG|thumb|400px|none| initial conditions: 
rAB = 0.8 
rBC = 9.0 
pAB = -1.25 
pBC = -2.5]]

==Reactive and unreactive trajectories==
The following table is for the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, with varying momentum.

{| class="wikitable" border=1
! p1 !! p2 !! E(kcal/mol) !! Reactive?
|-
| -1.25 || -2.5 || -99.119 || Yes
|-
| -1.5 || -2.0 || -100.456 || No
|-
| -1.5 || -2.5 || -98.956 || Yes
|-
| -2.5 || -5.0 || -84.956 || No
|-
| -2.5 || -5.2 || -83.416 || Yes
|}

[[File:Sc3916_rxn1.PNG|thumb|300px|none|Initial p1=-1.25, r2=-2.5 
the trajectory goes smoothly up to the transition state, then starts oscillating as it rolls down to the products]]

[[File:Sc3916_rxn2.PNG|thumb|300px|none|Initial p1=-1.5, r2=-2.0 
the trajectory goes up towards the transition state smoothly (but not very far), and then rolls back down the way it came, with oscillating]]

[[File:Sc3916_rxn3.PNG|thumb|300px|none|Initial p1=-1.5, r2=-2.5 
the trajectory goes smoothly up to the transition state, then starts oscillating as it rolls down to the products]]

[[File:Sc3916_rxn4b.PNG|thumb|300px|none|Initial p1=-2.5, r2=-5.0 
the trajectory goes smoothly up to the transition state, crosses it, but recrosses it and reverts back to the reactants, with very large oscillations]]

[[File:Sc3916_rxn5.PNG|thumb|300px|none|Initial p1=-2.5, r2=-5.2 
the trajectory crosses the transition state, forming the products, then recrosses it to the reactant side, and then recrosses it again, forming the products finally]]

'''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?'''

4 major assumptions are used. Firstly, the Born-Oppenheimer approximation is used, and that the effects of quantum tunneling are negligible (there is a non-zero probability that reactants can convert to products without reaching the activation energy). The reactants are assumed to have a Boltzmann distribution - which is perfectly valid at low temperatures. The final assumption is that "once the system attains the transition state, with a velocity towards the product configuration, it will not reenter the initial state region again."[https://doi.org/10.1016/B978-044452837-7.50005-8 1]

This last assumption does not fit with our simulations. Both the 4th, and 5th plots contain the trajectory crossing over the transition state and back again. Since there will be more trajectories (like the 4th) which involve crossing over and are still unreactive - which are not predicted by Transition State Theory - experimental rate values will be slower than those predicted by the theory.

==EXERCISE 2: F - H - H system==

'''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 is exothermic - the products have a more negative energy than the reactants. Because the enthalpy change only involves breaking a H-H bond and making a H-F bond, ΔH = +H(H-H) - H(H-F).

Since ΔH < 0:

H(H-F)>H(H-H)

This means HF has a greater bond enthalpy (and therefore strength) than H2. It can also be seen that H + HF is endothermic, which leads to the same conclusion.

'''Locate the approximate position of the transition state.'''

rts: 
RHH = 0.745 angstroms 
RHF = 1.811 angstroms

(0.7448778, 1.810754) is the super accurate one

'''Report the activation energy for both reactions.'''

By performing an MEP from next to the transition structure, the path from the TS to the reactants, or products, was found. By subtracting the final energy from the TS, the Ea was obtained. Since the activation energy for H2 + F is very small, the MEP has a very small gradient, and as such the calculation required an order of magnitude more steps in the calculation (than HF + H) to reach the reactants. Conversely, in HF + H a step change is seen.

[[File:Mep_energyvstime_sc3916.PNG|thumb|left|initial conditions: 
RHH=0.745 
RHF=1.83]]

[[File:3916_energyvstime_hf2.PNG|thumb|center|initial conditions: 
RHH=0.745 
RHF=1.8]]

Ea(H2 + F Ea) = 0.256 kcal/mol 
Ea(HF + H Ea) = 30.202 kcal/mol 

=== Reaction dynamics ===
* Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Internuclear Momenta vs Time”. {{fontcolor1|blue|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

[[File:H2+f_surface__3916.PNG|thumb|none|F + H2, initial conditions: 
RHH (AB) = 0.74 angstroms 
RHF (BC) = 2 angstroms 
AB momentum = +3 
BC momentum = -5]]

[[File:H2+F_MOMENTAVTIME3916.PNG|thumb|none|F + H2 Momentum vs Time Plot]]

The animation of this reaction clearly shows the velocity of the lone H atom formed has increased, and vibrations occurring in HF. This can be seen in the momentum vs time plot - with the increased AB and AC momentum reflecting the increase in velocity, and BC momentum oscillation reflecting the vibrations of the H-F bond. Since translational and vibrational energy contribute to kinetic energy, we know that the energy released by this exothermic reaction (a decrease in enthalpy) is conserved in the form of increased kinetic energy. This can be seen in the below graph, and is easily shown experimentally by the resultant increase in temperature. To determine the extent vibrational energy contributes to this, we can use IR spectroscopy. If there is a lot of vibrational energy, there will be a relatively high population in the first vibrational level, and we will see overtone bands in the IR spectrum.

[[File:H2+F_ENERGYVSTIME3916.PNG|thumb|none|F + H2 Energy vs Time Plot]]

* Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 0.74, with a momentum pFH = -0.5, and explore several values of pHH in the range -3 to 3 (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.

[[File:SC3916-3CONTOUR.PNG|thumb|none|F + H2, initial conditions A: 
rHH=0.74 rHF=1.83 
pHH=-3.0 pHF=-0.5 ]]

[[File:SC3916-2CONTOUR.PNG|thumb|none|F + H2, initial conditions B: 
rHH=0.74 rHF=1.83 
pHH=-2.0 pHF=-0.5 ]]

[[File:SC3916-1.5CONTOUR.PNG|thumb|none|F + H2, initial conditions C: 
rHH=0.74 rHF=1.83 
pHH=-1 pHF=-0.5 ]]

[[File:SC39160CONTOUR.PNG|thumb|none|F + H2, initial conditions D: 
rHH=0.74 rHF=1.83 
pHH=0.0 pHF=-0.5 ]]

[[File:Sc3916_redone1.PNG|thumb|none|F + H2, initial conditions E: 
rHH=0.74 rHF=1.83 
pHH=+1.0 pHF=-0.5 ]]

[[File:Sc3916_redone2.PNG|thumb|none|F + H2, initial conditions F: 
rHH=0.74 rHF=1.83 
pHH=+3.0 pHF=-0.5 ]]

* For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now?

[[File:Sc3916_redonelast.PNG|thumb|none|F + H2, initial conditions G: 
rHH=0.74 rHF=1.83 
pHH=+0.1 pHF=-0.8 ]]

*Let us now focus on the reverse reaction, H + HF (an H atom with high kinetic energy colliding with HF).

[[File:Sc3916_hh-10hf-.5.PNG|thumb|none|HF + H, initial conditions H: 
rHH=0.9 rHF=1.83 
pHH=-10 pHF=-5 ]]

[[File:Sc3916_hh-1.1h4-13.PNG|thumb|none|HF + H, initial conditions I: 
rHH=0.9 rHF=1.83 
pHH=-1.1 pHF=-13 ]]

The cases studied are an illustration of Polanyi's empirical rules. {{fontcolor|blue|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, for a reaction with an early transition state (such as F + H2), vibrational energy is more effective in overcoming the TS. For a late transition state (such as HF + H), translational energy is more effective in overcoming it.
This can be seen in the above reaction trajectories.

A good reference about transition state theory is chapter 10 of J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics Prentice-Hall.

File:Sc3916 hh-1.1h4-13.PNG

2018-05-25T14:05:49Z

Sc3916:

File:Sc3916 hf2-0.5-20.PNG

2018-05-25T13:57:07Z

Sc3916:

File:Sc3916 hh-10hf-.5.PNG

2018-05-25T13:55:50Z

Sc3916:

MRD:sc3916

2018-05-25T13:44:37Z

Sc3916: /* Reaction dynamics */

=Molecular Reaction Dynamics Report=

==Potential Energy Surfaces==

'''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.'''

The transition structure is at a saddle point - the intersection of the minimum of a curve in one plane, and the maximum of a curve in the other. In this case, the minimum curve is on the plane corresponding to RAB=RBC, and the maximum is on the perpendicular plane. The gradient of both these curves is equal to zero.

This is different from when the potential energy surface is at a minimum - in one plane the gradient = 0, but not in the other. This allows us to distinguish between the transition structure and minima by looking at the curvature in '''both''' planes.

[[File:Saddle_Point_Visualisation_sc3916.png|thumb|none|500px|Saddle Point Visualisation]]

'''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.'''

'''rts= 0.9075 angstroms =rAB=rBC'''

To estimate this, a rough estimate was found by looking at the saddle point. At the saddle point, gradient = 0 in both planes - so if the system is at exactly this point, the atoms will remain stationary if it has no initial momentum. If it isn't at this point, it will start to oscillate - and the further away from the saddle point, the greater the oscillation. Therefore by plotting this internuclear distance vs time, and adjusting the distance by small amounts until there was very little oscillation seen, a more precise value was found.

[[File: Sc3916_1.PNG|thumb|300px|none|Internuclear Distances vs Time]]

==Reaction trajectories==

[[File:Sc3916_mep.PNG|thumb|400px|left|Minimum Energy Path from from r1 = rts+δ, r2 = rts]]

[[File:Sc3916_dynamics.PNG|thumb|400px|center|Dynamic Energy Path from r1 = rts+δ, r2 = rts]]

'''Comment on how the MEP and the trajectory you just calculated differ.'''

The minimum energy path represents an infinitely slow reaction trajectory. We start off a very small distance from the transition structure, and then the trajectory "rolls" down the potential energy surface, following where the gradient is most negative. Since velocity resets to 0 every time interval, so does momentum (p=mv). This means at every infinitesimally small step, this pathway will minimise its energy. This is not how it works in reality.

The dynamic trajectory is much closer to reality. As the trajectory starts to move down the potential energy surface it gains momentum. This means it will carry on rolling up the curve on the other side of the minimum (which increases its potential energy), leading to oscillations. This reflects the vibrational motion of atoms, which MEP does not include.

Dynamics Internuclear distance vs time plot: 
Final AB distance = 0.8 
Final BC distance = 9.0 

Dynamics Internuclear Momenta vs time plot: 
Final average momentum for AB = 1.25 
Final average momentum for BC = 2.5 

This shows the reaction path from the above calculation leads to A-B + C

If we switch around initial values, so now rAB=rts+δ and rBC=rts, the reaction path will output the same values, except the products are B-C + A.

'''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.'''

The reaction proceeds from left to right across the below plot, A-B + C -> A + B-C. This reaction is essentially the reverse of the previous trajectory (allowing the pathway to "roll" down from the transition state), however it crosses over, to form the products.

[[File:Sc3916_dynamics2.PNG|thumb|400px|none| initial conditions: 
rAB = 0.8 
rBC = 9.0 
pAB = -1.25 
pBC = -2.5]]

==Reactive and unreactive trajectories==
The following table is for the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, with varying momentum.

{| class="wikitable" border=1
! p1 !! p2 !! E(kcal/mol) !! Reactive?
|-
| -1.25 || -2.5 || -99.119 || Yes
|-
| -1.5 || -2.0 || -100.456 || No
|-
| -1.5 || -2.5 || -98.956 || Yes
|-
| -2.5 || -5.0 || -84.956 || No
|-
| -2.5 || -5.2 || -83.416 || Yes
|}

[[File:Sc3916_rxn1.PNG|thumb|300px|none|Initial p1=-1.25, r2=-2.5 
the trajectory goes smoothly up to the transition state, then starts oscillating as it rolls down to the products]]

[[File:Sc3916_rxn2.PNG|thumb|300px|none|Initial p1=-1.5, r2=-2.0 
the trajectory goes up towards the transition state smoothly (but not very far), and then rolls back down the way it came, with oscillating]]

[[File:Sc3916_rxn3.PNG|thumb|300px|none|Initial p1=-1.5, r2=-2.5 
the trajectory goes smoothly up to the transition state, then starts oscillating as it rolls down to the products]]

[[File:Sc3916_rxn4b.PNG|thumb|300px|none|Initial p1=-2.5, r2=-5.0 
the trajectory goes smoothly up to the transition state, crosses it, but recrosses it and reverts back to the reactants, with very large oscillations]]

[[File:Sc3916_rxn5.PNG|thumb|300px|none|Initial p1=-2.5, r2=-5.2 
the trajectory crosses the transition state, forming the products, then recrosses it to the reactant side, and then recrosses it again, forming the products finally]]

'''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?'''

4 major assumptions are used. Firstly, the Born-Oppenheimer approximation is used, and that the effects of quantum tunneling are negligible (there is a non-zero probability that reactants can convert to products without reaching the activation energy). The reactants are assumed to have a Boltzmann distribution - which is perfectly valid at low temperatures. The final assumption is that "once the system attains the transition state, with a velocity towards the product configuration, it will not reenter the initial state region again."[https://doi.org/10.1016/B978-044452837-7.50005-8 1]

This last assumption does not fit with our simulations. Both the 4th, and 5th plots contain the trajectory crossing over the transition state and back again. Since there will be more trajectories (like the 4th) which involve crossing over and are still unreactive - which are not predicted by Transition State Theory - experimental rate values will be slower than those predicted by the theory.

==EXERCISE 2: F - H - H system==

'''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 is exothermic - the products have a more negative energy than the reactants. Because the enthalpy change only involves breaking a H-H bond and making a H-F bond, ΔH = +H(H-H) - H(H-F).

Since ΔH < 0:

H(H-F)>H(H-H)

This means HF has a greater bond enthalpy (and therefore strength) than H2. It can also be seen that H + HF is endothermic, which leads to the same conclusion.

'''Locate the approximate position of the transition state.'''

rts: 
RHH = 0.745 angstroms 
RHF = 1.811 angstroms

(0.7448778, 1.810754) is the super accurate one

'''Report the activation energy for both reactions.'''

By performing an MEP from next to the transition structure, the path from the TS to the reactants, or products, was found. By subtracting the final energy from the TS, the Ea was obtained. Since the activation energy for H2 + F is very small, the MEP has a very small gradient, and as such the calculation required an order of magnitude more steps in the calculation (than HF + H) to reach the reactants. Conversely, in HF + H a step change is seen.

[[File:Mep_energyvstime_sc3916.PNG|thumb|left|initial conditions: 
RHH=0.745 
RHF=1.83]]

[[File:3916_energyvstime_hf2.PNG|thumb|center|initial conditions: 
RHH=0.745 
RHF=1.8]]

Ea(H2 + F Ea) = 0.256 kcal/mol 
Ea(HF + H Ea) = 30.202 kcal/mol 

=== Reaction dynamics ===
* Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Internuclear Momenta vs Time”. {{fontcolor1|blue|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

[[File:H2+f_surface__3916.PNG|thumb|none|F + H2, initial conditions: 
RHH (AB) = 0.74 angstroms 
RHF (BC) = 2 angstroms 
AB momentum = +3 
BC momentum = -5]]

[[File:H2+F_MOMENTAVTIME3916.PNG|thumb|none|F + H2 Momentum vs Time Plot]]

The animation of this reaction clearly shows the velocity of the lone H atom formed has increased, and vibrations occurring in HF. This can be seen in the momentum vs time plot - with the increased AB and AC momentum reflecting the increase in velocity, and BC momentum oscillation reflecting the vibrations of the H-F bond. Since translational and vibrational energy contribute to kinetic energy, we know that the energy released by this exothermic reaction (a decrease in enthalpy) is conserved in the form of increased kinetic energy. This can be seen in the below graph, and is easily shown experimentally by the resultant increase in temperature. To determine the extent vibrational energy contributes to this, we can use IR spectroscopy. If there is a lot of vibrational energy, there will be a relatively high population in the first vibrational level, and we will see overtone bands in the IR spectrum.

[[File:H2+F_ENERGYVSTIME3916.PNG|thumb|none|F + H2 Energy vs Time Plot]]

* Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 0.74, with a momentum pFH = -0.5, and explore several values of pHH in the range -3 to 3 (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.

[[File:SC3916-3CONTOUR.PNG|thumb|none|F + H2, initial conditions: 
rHH=0.74 rHF=1.83 
pHH=-3.0 pHF=-0.5 ]]

[[File:SC3916-2CONTOUR.PNG|thumb|none|F + H2, initial conditions: 
rHH=0.74 rHF=1.83 
pHH=-2.0 pHF=-0.5 ]]

[[File:SC3916-1.5CONTOUR.PNG|thumb|none|F + H2, initial conditions: 
rHH=0.74 rHF=1.83 
pHH=-1 pHF=-0.5 ]]

[[File:SC39160CONTOUR.PNG|thumb|none|F + H2, initial conditions: 
rHH=0.74 rHF=1.83 
pHH=0.0 pHF=-0.5 ]]

[[File:Sc3916_redone1.PNG|thumb|none|F + H2, initial conditions: 
rHH=0.74 rHF=1.83 
pHH=+1.0 pHF=-0.5 ]]

[[File:Sc3916_redone2.PNG|thumb|none|F + H2, initial conditions: 
rHH=0.74 rHF=1.83 
pHH=+3.0 pHF=-0.5 ]]

* For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now?

[[File:Sc3916_redonelast.PNG|thumb|none|F + H2, initial conditions: 
rHH=0.74 rHF=1.83 
pHH=+0.1 pHF=-0.8 ]]

*Let us now focus on the reverse reaction, H + HF (an H atom with high kinetic energy colliding with HF).

[[File:Sc3916_hf_contour.PNG|thumb|none|HF + H, initial conditions: 
rHH=0.74 rHF=1.83 
pHH=-0.5 pHF=-10 ]]

The cases studied are an illustration of Polanyi's empirical rules. {{fontcolor|blue|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.}}

Early barrier:
Kinetic energy effective in overcoming barrier
Vibrational energy ineffective

Late barrier:
Vibrational energy effective in overcoming barrier
Kinetic energy ineffective

File:Sc3916 redonelast.PNG

2018-05-25T13:44:11Z

Sc3916:

File:Sc3916 redone2.PNG

2018-05-25T13:43:36Z

Sc3916:

File:Sc3916 redone1.PNG

2018-05-25T13:42:51Z

Sc3916:

MRD:sc3916

2018-05-25T13:32:53Z

Sc3916: /* Reaction dynamics */

=Molecular Reaction Dynamics Report=

==Potential Energy Surfaces==

'''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.'''

The transition structure is at a saddle point - the intersection of the minimum of a curve in one plane, and the maximum of a curve in the other. In this case, the minimum curve is on the plane corresponding to RAB=RBC, and the maximum is on the perpendicular plane. The gradient of both these curves is equal to zero.

This is different from when the potential energy surface is at a minimum - in one plane the gradient = 0, but not in the other. This allows us to distinguish between the transition structure and minima by looking at the curvature in '''both''' planes.

[[File:Saddle_Point_Visualisation_sc3916.png|thumb|none|500px|Saddle Point Visualisation]]

'''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.'''

'''rts= 0.9075 angstroms =rAB=rBC'''

To estimate this, a rough estimate was found by looking at the saddle point. At the saddle point, gradient = 0 in both planes - so if the system is at exactly this point, the atoms will remain stationary if it has no initial momentum. If it isn't at this point, it will start to oscillate - and the further away from the saddle point, the greater the oscillation. Therefore by plotting this internuclear distance vs time, and adjusting the distance by small amounts until there was very little oscillation seen, a more precise value was found.

[[File: Sc3916_1.PNG|thumb|300px|none|Internuclear Distances vs Time]]

==Reaction trajectories==

[[File:Sc3916_mep.PNG|thumb|400px|left|Minimum Energy Path from from r1 = rts+δ, r2 = rts]]

[[File:Sc3916_dynamics.PNG|thumb|400px|center|Dynamic Energy Path from r1 = rts+δ, r2 = rts]]

'''Comment on how the MEP and the trajectory you just calculated differ.'''

The minimum energy path represents an infinitely slow reaction trajectory. We start off a very small distance from the transition structure, and then the trajectory "rolls" down the potential energy surface, following where the gradient is most negative. Since velocity resets to 0 every time interval, so does momentum (p=mv). This means at every infinitesimally small step, this pathway will minimise its energy. This is not how it works in reality.

The dynamic trajectory is much closer to reality. As the trajectory starts to move down the potential energy surface it gains momentum. This means it will carry on rolling up the curve on the other side of the minimum (which increases its potential energy), leading to oscillations. This reflects the vibrational motion of atoms, which MEP does not include.

Dynamics Internuclear distance vs time plot: 
Final AB distance = 0.8 
Final BC distance = 9.0 

Dynamics Internuclear Momenta vs time plot: 
Final average momentum for AB = 1.25 
Final average momentum for BC = 2.5 

This shows the reaction path from the above calculation leads to A-B + C

If we switch around initial values, so now rAB=rts+δ and rBC=rts, the reaction path will output the same values, except the products are B-C + A.

'''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.'''

The reaction proceeds from left to right across the below plot, A-B + C -> A + B-C. This reaction is essentially the reverse of the previous trajectory (allowing the pathway to "roll" down from the transition state), however it crosses over, to form the products.

[[File:Sc3916_dynamics2.PNG|thumb|400px|none| initial conditions: 
rAB = 0.8 
rBC = 9.0 
pAB = -1.25 
pBC = -2.5]]

==Reactive and unreactive trajectories==
The following table is for the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, with varying momentum.

{| class="wikitable" border=1
! p1 !! p2 !! E(kcal/mol) !! Reactive?
|-
| -1.25 || -2.5 || -99.119 || Yes
|-
| -1.5 || -2.0 || -100.456 || No
|-
| -1.5 || -2.5 || -98.956 || Yes
|-
| -2.5 || -5.0 || -84.956 || No
|-
| -2.5 || -5.2 || -83.416 || Yes
|}

[[File:Sc3916_rxn1.PNG|thumb|300px|none|Initial p1=-1.25, r2=-2.5 
the trajectory goes smoothly up to the transition state, then starts oscillating as it rolls down to the products]]

[[File:Sc3916_rxn2.PNG|thumb|300px|none|Initial p1=-1.5, r2=-2.0 
the trajectory goes up towards the transition state smoothly (but not very far), and then rolls back down the way it came, with oscillating]]

[[File:Sc3916_rxn3.PNG|thumb|300px|none|Initial p1=-1.5, r2=-2.5 
the trajectory goes smoothly up to the transition state, then starts oscillating as it rolls down to the products]]

[[File:Sc3916_rxn4b.PNG|thumb|300px|none|Initial p1=-2.5, r2=-5.0 
the trajectory goes smoothly up to the transition state, crosses it, but recrosses it and reverts back to the reactants, with very large oscillations]]

[[File:Sc3916_rxn5.PNG|thumb|300px|none|Initial p1=-2.5, r2=-5.2 
the trajectory crosses the transition state, forming the products, then recrosses it to the reactant side, and then recrosses it again, forming the products finally]]

'''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?'''

4 major assumptions are used. Firstly, the Born-Oppenheimer approximation is used, and that the effects of quantum tunneling are negligible (there is a non-zero probability that reactants can convert to products without reaching the activation energy). The reactants are assumed to have a Boltzmann distribution - which is perfectly valid at low temperatures. The final assumption is that "once the system attains the transition state, with a velocity towards the product configuration, it will not reenter the initial state region again."[https://doi.org/10.1016/B978-044452837-7.50005-8 1]

This last assumption does not fit with our simulations. Both the 4th, and 5th plots contain the trajectory crossing over the transition state and back again. Since there will be more trajectories (like the 4th) which involve crossing over and are still unreactive - which are not predicted by Transition State Theory - experimental rate values will be slower than those predicted by the theory.

==EXERCISE 2: F - H - H system==

'''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 is exothermic - the products have a more negative energy than the reactants. Because the enthalpy change only involves breaking a H-H bond and making a H-F bond, ΔH = +H(H-H) - H(H-F).

Since ΔH < 0:

H(H-F)>H(H-H)

This means HF has a greater bond enthalpy (and therefore strength) than H2. It can also be seen that H + HF is endothermic, which leads to the same conclusion.

'''Locate the approximate position of the transition state.'''

rts: 
RHH = 0.745 angstroms 
RHF = 1.811 angstroms

(0.7448778, 1.810754) is the super accurate one

'''Report the activation energy for both reactions.'''

By performing an MEP from next to the transition structure, the path from the TS to the reactants, or products, was found. By subtracting the final energy from the TS, the Ea was obtained. Since the activation energy for H2 + F is very small, the MEP has a very small gradient, and as such the calculation required an order of magnitude more steps in the calculation (than HF + H) to reach the reactants. Conversely, in HF + H a step change is seen.

[[File:Mep_energyvstime_sc3916.PNG|thumb|left|initial conditions: 
RHH=0.745 
RHF=1.83]]

[[File:3916_energyvstime_hf2.PNG|thumb|center|initial conditions: 
RHH=0.745 
RHF=1.8]]

Ea(H2 + F Ea) = 0.256 kcal/mol 
Ea(HF + H Ea) = 30.202 kcal/mol 

=== Reaction dynamics ===
* Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Internuclear Momenta vs Time”. {{fontcolor1|blue|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

[[File:H2+f_surface__3916.PNG|thumb|none|F + H2, initial conditions: 
RHH (AB) = 0.745 angstroms 
RHF (BC) = 2 angstroms 
AB momentum = +3 
BC momentum = -5]]

[[File:H2+F_MOMENTAVTIME3916.PNG|thumb|none|F + H2 Momentum vs Time Plot]]

The animation of this reaction clearly shows the velocity of the lone H atom formed has increased, and vibrations occurring in HF. This can be seen in the momentum vs time plot - with the increased AB and AC momentum reflecting the increase in velocity, and BC momentum oscillation reflecting the vibrations of the H-F bond. Since translational and vibrational energy contribute to kinetic energy, we know that the energy released by this exothermic reaction (a decrease in enthalpy) is conserved in the form of increased kinetic energy. This can be seen in the below graph, and is easily shown experimentally by the resultant increase in temperature. To determine the extent vibrational energy contributes to this, we can use IR spectroscopy. If there is a lot of vibrational energy, there will be a relatively high population in the first vibrational level, and we will see overtone bands in the IR spectrum.

[[File:H2+F_ENERGYVSTIME3916.PNG|thumb|none|F + H2 Energy vs Time Plot]]

* Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 0.74, with a momentum pFH = -0.5, and explore several values of pHH in the range -3 to 3 (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.

[[File:SC3916-3CONTOUR.PNG|thumb|none|F + H2, initial conditions: 
rHH=0.745 rHF=1.83 
pHH=-3.0 pHF=-0.5 ]]

[[File:SC3916-2CONTOUR.PNG|thumb|none|F + H2, initial conditions: 
rHH=0.745 rHF=1.83 
pHH=-2.0 pHF=-0.5 ]]

[[File:SC3916-1.5CONTOUR.PNG|thumb|none|F + H2, initial conditions: 
rHH=0.745 rHF=1.83 
pHH=-1 pHF=-0.5 ]]

[[File:SC39160CONTOUR.PNG|thumb|none|F + H2, initial conditions: 
rHH=0.745 rHF=1.83 
pHH=0.0 pHF=-0.5 ]]

REDO BELOW (SWITCH BC AND AB AXES) TO FIT WITH ABOVE

[[File:SC3916+1.PNG|thumb|none|F + H2, initial conditions: 
rHH=0.745 rHF=1.83 
pHH=+1.0 pHF=-0.5 ]]

REDO BELOW (SWITCH BC AND AB AXES) TO FIT WITH ABOVE

[[File:SC3916+3.PNG|thumb|none|F + H2, initial conditions: 
rHH=0.745 rHF=1.83 
pHH=+3.0 pHF=-0.5 ]]

* For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now?
REDO BELOW (SWITCH BC AND AB AXES) TO FIT WITH ABOVE
[[File:SC3916_-0.8+0.1.PNG|thumb|none|F + H2, initial conditions: 
rHH=0.745 rHF=1.83 
pHH=+0.1 pHF=-0.8 ]]

*Let us now focus on the reverse reaction, H + HF (an H atom with high kinetic energy colliding with HF).

[[File:Sc3916_hf_contour.PNG|thumb|none|HF + H, initial conditions: 
rHH=0.745 rHF=1.83 
pHH=-0.5 pHF=-10 ]]

The cases studied are an illustration of Polanyi's empirical rules. {{fontcolor|blue|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.}}

Early barrier:
Kinetic energy effective in overcoming barrier
Vibrational energy ineffective

Late barrier:
Vibrational energy effective in overcoming barrier
Kinetic energy ineffective

MRD:sc3916

2018-05-25T13:28:44Z

Sc3916: /* Reaction dynamics */

=Molecular Reaction Dynamics Report=

==Potential Energy Surfaces==

'''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.'''

The transition structure is at a saddle point - the intersection of the minimum of a curve in one plane, and the maximum of a curve in the other. In this case, the minimum curve is on the plane corresponding to RAB=RBC, and the maximum is on the perpendicular plane. The gradient of both these curves is equal to zero.

This is different from when the potential energy surface is at a minimum - in one plane the gradient = 0, but not in the other. This allows us to distinguish between the transition structure and minima by looking at the curvature in '''both''' planes.

[[File:Saddle_Point_Visualisation_sc3916.png|thumb|none|500px|Saddle Point Visualisation]]

'''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.'''

'''rts= 0.9075 angstroms =rAB=rBC'''

To estimate this, a rough estimate was found by looking at the saddle point. At the saddle point, gradient = 0 in both planes - so if the system is at exactly this point, the atoms will remain stationary if it has no initial momentum. If it isn't at this point, it will start to oscillate - and the further away from the saddle point, the greater the oscillation. Therefore by plotting this internuclear distance vs time, and adjusting the distance by small amounts until there was very little oscillation seen, a more precise value was found.

[[File: Sc3916_1.PNG|thumb|300px|none|Internuclear Distances vs Time]]

==Reaction trajectories==

[[File:Sc3916_mep.PNG|thumb|400px|left|Minimum Energy Path from from r1 = rts+δ, r2 = rts]]

[[File:Sc3916_dynamics.PNG|thumb|400px|center|Dynamic Energy Path from r1 = rts+δ, r2 = rts]]

'''Comment on how the MEP and the trajectory you just calculated differ.'''

The minimum energy path represents an infinitely slow reaction trajectory. We start off a very small distance from the transition structure, and then the trajectory "rolls" down the potential energy surface, following where the gradient is most negative. Since velocity resets to 0 every time interval, so does momentum (p=mv). This means at every infinitesimally small step, this pathway will minimise its energy. This is not how it works in reality.

The dynamic trajectory is much closer to reality. As the trajectory starts to move down the potential energy surface it gains momentum. This means it will carry on rolling up the curve on the other side of the minimum (which increases its potential energy), leading to oscillations. This reflects the vibrational motion of atoms, which MEP does not include.

Dynamics Internuclear distance vs time plot: 
Final AB distance = 0.8 
Final BC distance = 9.0 

Dynamics Internuclear Momenta vs time plot: 
Final average momentum for AB = 1.25 
Final average momentum for BC = 2.5 

This shows the reaction path from the above calculation leads to A-B + C

If we switch around initial values, so now rAB=rts+δ and rBC=rts, the reaction path will output the same values, except the products are B-C + A.

'''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.'''

The reaction proceeds from left to right across the below plot, A-B + C -> A + B-C. This reaction is essentially the reverse of the previous trajectory (allowing the pathway to "roll" down from the transition state), however it crosses over, to form the products.

[[File:Sc3916_dynamics2.PNG|thumb|400px|none| initial conditions: 
rAB = 0.8 
rBC = 9.0 
pAB = -1.25 
pBC = -2.5]]

==Reactive and unreactive trajectories==
The following table is for the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, with varying momentum.

{| class="wikitable" border=1
! p1 !! p2 !! E(kcal/mol) !! Reactive?
|-
| -1.25 || -2.5 || -99.119 || Yes
|-
| -1.5 || -2.0 || -100.456 || No
|-
| -1.5 || -2.5 || -98.956 || Yes
|-
| -2.5 || -5.0 || -84.956 || No
|-
| -2.5 || -5.2 || -83.416 || Yes
|}

[[File:Sc3916_rxn1.PNG|thumb|300px|none|Initial p1=-1.25, r2=-2.5 
the trajectory goes smoothly up to the transition state, then starts oscillating as it rolls down to the products]]

[[File:Sc3916_rxn2.PNG|thumb|300px|none|Initial p1=-1.5, r2=-2.0 
the trajectory goes up towards the transition state smoothly (but not very far), and then rolls back down the way it came, with oscillating]]

[[File:Sc3916_rxn3.PNG|thumb|300px|none|Initial p1=-1.5, r2=-2.5 
the trajectory goes smoothly up to the transition state, then starts oscillating as it rolls down to the products]]

[[File:Sc3916_rxn4b.PNG|thumb|300px|none|Initial p1=-2.5, r2=-5.0 
the trajectory goes smoothly up to the transition state, crosses it, but recrosses it and reverts back to the reactants, with very large oscillations]]

[[File:Sc3916_rxn5.PNG|thumb|300px|none|Initial p1=-2.5, r2=-5.2 
the trajectory crosses the transition state, forming the products, then recrosses it to the reactant side, and then recrosses it again, forming the products finally]]

'''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?'''

4 major assumptions are used. Firstly, the Born-Oppenheimer approximation is used, and that the effects of quantum tunneling are negligible (there is a non-zero probability that reactants can convert to products without reaching the activation energy). The reactants are assumed to have a Boltzmann distribution - which is perfectly valid at low temperatures. The final assumption is that "once the system attains the transition state, with a velocity towards the product configuration, it will not reenter the initial state region again."[https://doi.org/10.1016/B978-044452837-7.50005-8 1]

This last assumption does not fit with our simulations. Both the 4th, and 5th plots contain the trajectory crossing over the transition state and back again. Since there will be more trajectories (like the 4th) which involve crossing over and are still unreactive - which are not predicted by Transition State Theory - experimental rate values will be slower than those predicted by the theory.

==EXERCISE 2: F - H - H system==

'''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 is exothermic - the products have a more negative energy than the reactants. Because the enthalpy change only involves breaking a H-H bond and making a H-F bond, ΔH = +H(H-H) - H(H-F).

Since ΔH < 0:

H(H-F)>H(H-H)

This means HF has a greater bond enthalpy (and therefore strength) than H2. It can also be seen that H + HF is endothermic, which leads to the same conclusion.

'''Locate the approximate position of the transition state.'''

rts: 
RHH = 0.745 angstroms 
RHF = 1.811 angstroms

(0.7448778, 1.810754) is the super accurate one

'''Report the activation energy for both reactions.'''

By performing an MEP from next to the transition structure, the path from the TS to the reactants, or products, was found. By subtracting the final energy from the TS, the Ea was obtained. Since the activation energy for H2 + F is very small, the MEP has a very small gradient, and as such the calculation required an order of magnitude more steps in the calculation (than HF + H) to reach the reactants. Conversely, in HF + H a step change is seen.

[[File:Mep_energyvstime_sc3916.PNG|thumb|left|initial conditions: 
RHH=0.745 
RHF=1.83]]

[[File:3916_energyvstime_hf2.PNG|thumb|center|initial conditions: 
RHH=0.745 
RHF=1.8]]

Ea(H2 + F Ea) = 0.256 kcal/mol 
Ea(HF + H Ea) = 30.202 kcal/mol 

=== Reaction dynamics ===
* Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Internuclear Momenta vs Time”. {{fontcolor1|blue|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

[[File:H2+f_surface__3916.PNG|thumb|none|F + H2, initial conditions: 
RHH (AB) = 0.745 angstroms 
RHF (BC) = 2 angstroms 
AB momentum = +3 
BC momentum = -5]]

[[File:H2+F_MOMENTAVTIME3916.PNG|thumb|none|F + H2 Momentum vs Time Plot]]

The animation of this reaction clearly shows the velocity of the lone H atom formed has increased, and vibrations occurring in HF. This can be seen in the momentum vs time plot - with the increased AB and AC momentum reflecting the increase in velocity, and BC momentum oscillation reflecting the vibrations of the H-F bond. Since translational and vibrational energy contribute to kinetic energy, we know that the energy released by this exothermic reaction (a decrease in enthalpy) is conserved in the form of increased kinetic energy. This can be seen in the below graph, and is easily shown experimentally by the resultant increase in temperature. To determine the extent vibrational energy contributes to this, we can use IR spectroscopy. If there is a lot of vibrational energy, there will be a relatively high population in the first vibrational level, and we will see overtone bands in the IR spectrum.

[[File:H2+F_ENERGYVSTIME3916.PNG|thumb|none|F + H2 Energy vs Time Plot]]

* Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 0.74, with a momentum pFH = -0.5, and explore several values of pHH in the range -3 to 3 (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.

[[File:SC3916-3CONTOUR.PNG|thumb|none|F + H2, initial conditions: 
rHH=0.745 rHF=1.83 
pHH=-3.0 pHF=-0.5 ]]

[[File:SC3916-2CONTOUR.PNG|thumb|none|F + H2, initial conditions: 
rHH=0.745 rHF=1.83 
pHH=-2.0 pHF=-0.5 ]]

[[File:SC3916-1.5CONTOUR.PNG|thumb|none|F + H2, initial conditions: 
rHH=0.745 rHF=1.83 
pHH=-1 pHF=-0.5 ]]

[[File:SC39160CONTOUR.PNG|thumb|none|F + H2, initial conditions: 
rHH=0.745 rHF=1.83 
pHH=0.0 pHF=-0.5 ]]

REDO BELOW (SWITCH BC AND AB AXES) TO FIT WITH ABOVE

[[File:SC3916+1.PNG|thumb|none|F + H2, initial conditions: 
rHH=0.745 rHF=1.83 
pHH=+1.0 pHF=-0.5 ]]

REDO BELOW (SWITCH BC AND AB AXES) TO FIT WITH ABOVE

[[File:SC3916+3.PNG|thumb|none|F + H2, initial conditions: 
rHH=0.745 rHF=1.83 
pHH=+3.0 pHF=-0.5 ]]

* For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now?
REDO BELOW (SWITCH BC AND AB AXES) TO FIT WITH ABOVE
[[File:SC3916_-0.8+0.1.PNG|thumb|none|F + H2, initial conditions: 
rHH=0.745 rHF=1.83 
pHH=+0.1 pHF=-0.8 ]]

*Let us now focus on the reverse reaction, H + HF (an H atom with high kinetic energy colliding with HF).

[[File:Sc3916_hf_contour.PNG|thumb|none|F + H2, initial conditions: 
rHH=0.745 rHF=1.83 
pHH=-0.5 pHF=-10 ]]

The cases studied are an illustration of Polanyi's empirical rules. {{fontcolor|blue|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.}}

Early barrier:
Kinetic energy effective in overcoming barrier
Vibrational energy ineffective

Late barrier:
Vibrational energy effective in overcoming barrier
Kinetic energy ineffective

MRD:sc3916

2018-05-25T13:17:30Z

Sc3916: /* Reaction dynamics */

=Molecular Reaction Dynamics Report=

==Potential Energy Surfaces==

'''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.'''

The transition structure is at a saddle point - the intersection of the minimum of a curve in one plane, and the maximum of a curve in the other. In this case, the minimum curve is on the plane corresponding to RAB=RBC, and the maximum is on the perpendicular plane. The gradient of both these curves is equal to zero.

This is different from when the potential energy surface is at a minimum - in one plane the gradient = 0, but not in the other. This allows us to distinguish between the transition structure and minima by looking at the curvature in '''both''' planes.

[[File:Saddle_Point_Visualisation_sc3916.png|thumb|none|500px|Saddle Point Visualisation]]

'''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.'''

'''rts= 0.9075 angstroms =rAB=rBC'''

To estimate this, a rough estimate was found by looking at the saddle point. At the saddle point, gradient = 0 in both planes - so if the system is at exactly this point, the atoms will remain stationary if it has no initial momentum. If it isn't at this point, it will start to oscillate - and the further away from the saddle point, the greater the oscillation. Therefore by plotting this internuclear distance vs time, and adjusting the distance by small amounts until there was very little oscillation seen, a more precise value was found.

[[File: Sc3916_1.PNG|thumb|300px|none|Internuclear Distances vs Time]]

==Reaction trajectories==

[[File:Sc3916_mep.PNG|thumb|400px|left|Minimum Energy Path from from r1 = rts+δ, r2 = rts]]

[[File:Sc3916_dynamics.PNG|thumb|400px|center|Dynamic Energy Path from r1 = rts+δ, r2 = rts]]

'''Comment on how the MEP and the trajectory you just calculated differ.'''

The minimum energy path represents an infinitely slow reaction trajectory. We start off a very small distance from the transition structure, and then the trajectory "rolls" down the potential energy surface, following where the gradient is most negative. Since velocity resets to 0 every time interval, so does momentum (p=mv). This means at every infinitesimally small step, this pathway will minimise its energy. This is not how it works in reality.

The dynamic trajectory is much closer to reality. As the trajectory starts to move down the potential energy surface it gains momentum. This means it will carry on rolling up the curve on the other side of the minimum (which increases its potential energy), leading to oscillations. This reflects the vibrational motion of atoms, which MEP does not include.

Dynamics Internuclear distance vs time plot: 
Final AB distance = 0.8 
Final BC distance = 9.0 

Dynamics Internuclear Momenta vs time plot: 
Final average momentum for AB = 1.25 
Final average momentum for BC = 2.5 

This shows the reaction path from the above calculation leads to A-B + C

If we switch around initial values, so now rAB=rts+δ and rBC=rts, the reaction path will output the same values, except the products are B-C + A.

'''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.'''

The reaction proceeds from left to right across the below plot, A-B + C -> A + B-C. This reaction is essentially the reverse of the previous trajectory (allowing the pathway to "roll" down from the transition state), however it crosses over, to form the products.

[[File:Sc3916_dynamics2.PNG|thumb|400px|none| initial conditions: 
rAB = 0.8 
rBC = 9.0 
pAB = -1.25 
pBC = -2.5]]

==Reactive and unreactive trajectories==
The following table is for the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, with varying momentum.

{| class="wikitable" border=1
! p1 !! p2 !! E(kcal/mol) !! Reactive?
|-
| -1.25 || -2.5 || -99.119 || Yes
|-
| -1.5 || -2.0 || -100.456 || No
|-
| -1.5 || -2.5 || -98.956 || Yes
|-
| -2.5 || -5.0 || -84.956 || No
|-
| -2.5 || -5.2 || -83.416 || Yes
|}

[[File:Sc3916_rxn1.PNG|thumb|300px|none|Initial p1=-1.25, r2=-2.5 
the trajectory goes smoothly up to the transition state, then starts oscillating as it rolls down to the products]]

[[File:Sc3916_rxn2.PNG|thumb|300px|none|Initial p1=-1.5, r2=-2.0 
the trajectory goes up towards the transition state smoothly (but not very far), and then rolls back down the way it came, with oscillating]]

[[File:Sc3916_rxn3.PNG|thumb|300px|none|Initial p1=-1.5, r2=-2.5 
the trajectory goes smoothly up to the transition state, then starts oscillating as it rolls down to the products]]

[[File:Sc3916_rxn4b.PNG|thumb|300px|none|Initial p1=-2.5, r2=-5.0 
the trajectory goes smoothly up to the transition state, crosses it, but recrosses it and reverts back to the reactants, with very large oscillations]]

[[File:Sc3916_rxn5.PNG|thumb|300px|none|Initial p1=-2.5, r2=-5.2 
the trajectory crosses the transition state, forming the products, then recrosses it to the reactant side, and then recrosses it again, forming the products finally]]

'''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?'''

4 major assumptions are used. Firstly, the Born-Oppenheimer approximation is used, and that the effects of quantum tunneling are negligible (there is a non-zero probability that reactants can convert to products without reaching the activation energy). The reactants are assumed to have a Boltzmann distribution - which is perfectly valid at low temperatures. The final assumption is that "once the system attains the transition state, with a velocity towards the product configuration, it will not reenter the initial state region again."[https://doi.org/10.1016/B978-044452837-7.50005-8 1]

This last assumption does not fit with our simulations. Both the 4th, and 5th plots contain the trajectory crossing over the transition state and back again. Since there will be more trajectories (like the 4th) which involve crossing over and are still unreactive - which are not predicted by Transition State Theory - experimental rate values will be slower than those predicted by the theory.

==EXERCISE 2: F - H - H system==

'''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 is exothermic - the products have a more negative energy than the reactants. Because the enthalpy change only involves breaking a H-H bond and making a H-F bond, ΔH = +H(H-H) - H(H-F).

Since ΔH < 0:

H(H-F)>H(H-H)

This means HF has a greater bond enthalpy (and therefore strength) than H2. It can also be seen that H + HF is endothermic, which leads to the same conclusion.

'''Locate the approximate position of the transition state.'''

rts: 
RHH = 0.745 angstroms 
RHF = 1.811 angstroms

(0.7448778, 1.810754) is the super accurate one

'''Report the activation energy for both reactions.'''

By performing an MEP from next to the transition structure, the path from the TS to the reactants, or products, was found. By subtracting the final energy from the TS, the Ea was obtained. Since the activation energy for H2 + F is very small, the MEP has a very small gradient, and as such the calculation required an order of magnitude more steps in the calculation (than HF + H) to reach the reactants. Conversely, in HF + H a step change is seen.

[[File:Mep_energyvstime_sc3916.PNG|thumb|left|initial conditions: 
RHH=0.745 
RHF=1.83]]

[[File:3916_energyvstime_hf2.PNG|thumb|center|initial conditions: 
RHH=0.745 
RHF=1.8]]

Ea(H2 + F Ea) = 0.256 kcal/mol 
Ea(HF + H Ea) = 30.202 kcal/mol 

=== Reaction dynamics ===
* Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Internuclear Momenta vs Time”. {{fontcolor1|blue|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

[[File:H2+f_surface__3916.PNG|thumb|none|F + H2, initial conditions: 
RHH (AB) = 0.745 angstroms 
RHF (BC) = 2 angstroms 
AB momentum = +3 
BC momentum = -5]]

[[File:H2+F_MOMENTAVTIME3916.PNG|thumb|none|F + H2 Momentum vs Time Plot]]

The animation of this reaction clearly shows the velocity of the lone H atom formed has increased, and vibrations occurring in HF. This can be seen in the momentum vs time plot - with the increased AB and AC momentum reflecting the increase in velocity, and BC momentum oscillation reflecting the vibrations of the H-F bond. Since translational and vibrational energy contribute to kinetic energy, it is safe to assume that the energy released by this exothermic reaction (a decrease in enthalpy) is conserved in the form of increased kinetic energy. This can be seen in the below graph. An increase in average kinetic energy results in increased temperature - which fits with ouris easy to measure experimentally.

[[File:H2+F_ENERGYVSTIME3916.PNG|thumb|none|F + H2 Energy vs Time Plot]]

* Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 0.74, with a momentum pFH = -0.5, and explore several values of pHH in the range -3 to 3 (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.

[[File:SC3916-3CONTOUR.PNG|thumb|none|F + H2, initial conditions: 
rHH=0.745 rHF=1.83 
pHH=-3.0 pHF=-0.5 ]]

[[File:SC3916-2CONTOUR.PNG|thumb|none|F + H2, initial conditions: 
rHH=0.745 rHF=1.83 
pHH=-2.0 pHF=-0.5 ]]

[[File:SC3916-1.5CONTOUR.PNG|thumb|none|F + H2, initial conditions: 
rHH=0.745 rHF=1.83 
pHH=-1 pHF=-0.5 ]]

[[File:SC39160CONTOUR.PNG|thumb|none|F + H2, initial conditions: 
rHH=0.745 rHF=1.83 
pHH=0.0 pHF=-0.5 ]]

REDO BELOW (SWITCH BC AND AB AXES) TO FIT WITH ABOVE

[[File:SC3916+1.PNG|thumb|none|F + H2, initial conditions: 
rHH=0.745 rHF=1.83 
pHH=+1.0 pHF=-0.5 ]]

REDO BELOW (SWITCH BC AND AB AXES) TO FIT WITH ABOVE

[[File:SC3916+3.PNG|thumb|none|F + H2, initial conditions: 
rHH=0.745 rHF=1.83 
pHH=+3.0 pHF=-0.5 ]]

* For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now?
REDO BELOW (SWITCH BC AND AB AXES) TO FIT WITH ABOVE
[[File:SC3916_-0.8+0.1.PNG|thumb|none|F + H2, initial conditions: 
rHH=0.745 rHF=1.83 
pHH=+0.1 pHF=-0.8 ]]

*Let us now focus on the reverse reaction, H + HF (an H atom with high kinetic energy colliding with HF).

[[File:Sc3916_hf_contour.PNG|thumb|none|F + H2, initial conditions: 
rHH=0.745 rHF=1.83 
pHH=-0.5 pHF=-10 ]]

The cases studied are an illustration of Polanyi's empirical rules. {{fontcolor|blue|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.}}

Early barrier:
Kinetic energy effective in overcoming barrier
Vibrational energy ineffective

Late barrier:
Vibrational energy effective in overcoming barrier
Kinetic energy ineffective

MRD:sc3916

2018-05-24T19:00:51Z

Sc3916: /* EXERCISE 2: F - H - H system */

=Molecular Reaction Dynamics Report=

==Potential Energy Surfaces==

'''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.'''

The transition structure is at a saddle point - the intersection of the minimum of a curve in one plane, and the maximum of a curve in the other. In this case, the minimum curve is on the plane corresponding to RAB=RBC, and the maximum is on the perpendicular plane. The gradient of both these curves is equal to zero.

This is different from when the potential energy surface is at a minimum - in one plane the gradient = 0, but not in the other. This allows us to distinguish between the transition structure and minima by looking at the curvature in '''both''' planes.

[[File:Saddle_Point_Visualisation_sc3916.png|thumb|none|500px|Saddle Point Visualisation]]

'''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.'''

'''rts= 0.9075 angstroms =rAB=rBC'''

To estimate this, a rough estimate was found by looking at the saddle point. At the saddle point, gradient = 0 in both planes - so if the system is at exactly this point, the atoms will remain stationary if it has no initial momentum. If it isn't at this point, it will start to oscillate - and the further away from the saddle point, the greater the oscillation. Therefore by plotting this internuclear distance vs time, and adjusting the distance by small amounts until there was very little oscillation seen, a more precise value was found.

[[File: Sc3916_1.PNG|thumb|300px|none|Internuclear Distances vs Time]]

==Reaction trajectories==

[[File:Sc3916_mep.PNG|thumb|400px|left|Minimum Energy Path from from r1 = rts+δ, r2 = rts]]

[[File:Sc3916_dynamics.PNG|thumb|400px|center|Dynamic Energy Path from r1 = rts+δ, r2 = rts]]

'''Comment on how the MEP and the trajectory you just calculated differ.'''

The minimum energy path represents an infinitely slow reaction trajectory. We start off a very small distance from the transition structure, and then the trajectory "rolls" down the potential energy surface, following where the gradient is most negative. Since velocity resets to 0 every time interval, so does momentum (p=mv). This means at every infinitesimally small step, this pathway will minimise its energy. This is not how it works in reality.

The dynamic trajectory is much closer to reality. As the trajectory starts to move down the potential energy surface it gains momentum. This means it will carry on rolling up the curve on the other side of the minimum (which increases its potential energy), leading to oscillations. This reflects the vibrational motion of atoms, which MEP does not include.

Dynamics Internuclear distance vs time plot: 
Final AB distance = 0.8 
Final BC distance = 9.0 

Dynamics Internuclear Momenta vs time plot: 
Final average momentum for AB = 1.25 
Final average momentum for BC = 2.5 

This shows the reaction path from the above calculation leads to A-B + C

If we switch around initial values, so now rAB=rts+δ and rBC=rts, the reaction path will output the same values, except the products are B-C + A.

'''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.'''

The reaction proceeds from left to right across the below plot, A-B + C -> A + B-C. This reaction is essentially the reverse of the previous trajectory (allowing the pathway to "roll" down from the transition state), however it crosses over, to form the products.

[[File:Sc3916_dynamics2.PNG|thumb|400px|none| initial conditions: 
rAB = 0.8 
rBC = 9.0 
pAB = -1.25 
pBC = -2.5]]

==Reactive and unreactive trajectories==
The following table is for the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, with varying momentum.

{| class="wikitable" border=1
! p1 !! p2 !! E(kcal/mol) !! Reactive?
|-
| -1.25 || -2.5 || -99.119 || Yes
|-
| -1.5 || -2.0 || -100.456 || No
|-
| -1.5 || -2.5 || -98.956 || Yes
|-
| -2.5 || -5.0 || -84.956 || No
|-
| -2.5 || -5.2 || -83.416 || Yes
|}

[[File:Sc3916_rxn1.PNG|thumb|300px|none|Initial p1=-1.25, r2=-2.5 
the trajectory goes smoothly up to the transition state, then starts oscillating as it rolls down to the products]]

[[File:Sc3916_rxn2.PNG|thumb|300px|none|Initial p1=-1.5, r2=-2.0 
the trajectory goes up towards the transition state smoothly (but not very far), and then rolls back down the way it came, with oscillating]]

[[File:Sc3916_rxn3.PNG|thumb|300px|none|Initial p1=-1.5, r2=-2.5 
the trajectory goes smoothly up to the transition state, then starts oscillating as it rolls down to the products]]

[[File:Sc3916_rxn4b.PNG|thumb|300px|none|Initial p1=-2.5, r2=-5.0 
the trajectory goes smoothly up to the transition state, crosses it, but recrosses it and reverts back to the reactants, with very large oscillations]]

[[File:Sc3916_rxn5.PNG|thumb|300px|none|Initial p1=-2.5, r2=-5.2 
the trajectory crosses the transition state, forming the products, then recrosses it to the reactant side, and then recrosses it again, forming the products finally]]

'''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?'''

4 major assumptions are used. Firstly, the Born-Oppenheimer approximation is used, and that the effects of quantum tunneling are negligible (there is a non-zero probability that reactants can convert to products without reaching the activation energy). The reactants are assumed to have a Boltzmann distribution - which is perfectly valid at low temperatures. The final assumption is that "once the system attains the transition state, with a velocity towards the product configuration, it will not reenter the initial state region again."[https://doi.org/10.1016/B978-044452837-7.50005-8 1]

This last assumption does not fit with our simulations. Both the 4th, and 5th plots contain the trajectory crossing over the transition state and back again. Since there will be more trajectories (like the 4th) which involve crossing over and are still unreactive - which are not predicted by Transition State Theory - experimental rate values will be slower than those predicted by the theory.

==EXERCISE 2: F - H - H system==

'''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 is exothermic - the products have a more negative energy than the reactants. Because the enthalpy change only involves breaking a H-H bond and making a H-F bond, ΔH = +H(H-H) - H(H-F).

Since ΔH < 0:

H(H-F)>H(H-H)

This means HF has a greater bond enthalpy (and therefore strength) than H2. It can also be seen that H + HF is endothermic, which leads to the same conclusion.

'''Locate the approximate position of the transition state.'''

rts: 
RHH = 0.745 angstroms 
RHF = 1.811 angstroms

(0.7448778, 1.810754) is the super accurate one

'''Report the activation energy for both reactions.'''

By performing an MEP from next to the transition structure, the path from the TS to the reactants, or products, was found. By subtracting the final energy from the TS, the Ea was obtained. Since the activation energy for H2 + F is very small, the MEP has a very small gradient, and as such the calculation required an order of magnitude more steps in the calculation (than HF + H) to reach the reactants. Conversely, in HF + H a step change is seen.

[[File:Mep_energyvstime_sc3916.PNG|thumb|left|initial conditions: 
RHH=0.745 
RHF=1.83]]

[[File:3916_energyvstime_hf2.PNG|thumb|center|initial conditions: 
RHH=0.745 
RHF=1.8]]

Ea(H2 + F Ea) = 0.256 kcal/mol 
Ea(HF + H Ea) = 30.202 kcal/mol 

=== Reaction dynamics ===
* Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Internuclear Momenta vs Time”. {{fontcolor1|blue|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

[[File:H2+f_surface__3916.PNG|thumb|none|F + H2, initial conditions: 
RHH (AB) = 0.745 angstroms 
RHF (BC) = 2 angstroms 
AB momentum = +3 
BC momentum = -5]]

[[File:H2+F_MOMENTAVTIME3916.PNG|thumb|none|F + H2 Momentum vs Time Plot]]

The animation of this reaction clearly shows the velocity of the lone H atom formed has increased, and vibrations occurring in HF. This can be seen in the momentum vs time plot - with the increased AB and AC momentum reflecting the increase in velocity, and BC momentum oscillation reflecting the vibrations of the H-F bond. Since translational and vibrational energy contribute to kinetic energy, it is safe to assume that the energy released by this exothermic reaction (a decrease in enthalpy) is conserved in the form of increased kinetic energy. This can be seen in the below graph. An increase in average kinetic energy results in increased temperature - which is easy to measure experimentally.

[[File:H2+F_ENERGYVSTIME3916.PNG|thumb|none|F + H2 Energy vs Time Plot]]

* Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 0.74, with a momentum pFH = -0.5, and explore several values of pHH in the range -3 to 3 (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.

[[File:SC3916-3CONTOUR.PNG|thumb|none|F + H2, initial conditions: 
rHH=0.745 rHF=1.83 
pHH=-3.0 pHF=-0.5 ]]

[[File:SC3916-2CONTOUR.PNG|thumb|none|F + H2, initial conditions: 
rHH=0.745 rHF=1.83 
pHH=-2.0 pHF=-0.5 ]]

[[File:SC3916-1.5CONTOUR.PNG|thumb|none|F + H2, initial conditions: 
rHH=0.745 rHF=1.83 
pHH=-1 pHF=-0.5 ]]

[[File:SC39160CONTOUR.PNG|thumb|none|F + H2, initial conditions: 
rHH=0.745 rHF=1.83 
pHH=0.0 pHF=-0.5 ]]

REDO BELOW (SWITCH BC AND AB AXES) TO FIT WITH ABOVE

[[File:SC3916+1.PNG|thumb|none|F + H2, initial conditions: 
rHH=0.745 rHF=1.83 
pHH=+1.0 pHF=-0.5 ]]

REDO BELOW (SWITCH BC AND AB AXES) TO FIT WITH ABOVE

[[File:SC3916+3.PNG|thumb|none|F + H2, initial conditions: 
rHH=0.745 rHF=1.83 
pHH=+3.0 pHF=-0.5 ]]

* For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now?
REDO BELOW (SWITCH BC AND AB AXES) TO FIT WITH ABOVE
[[File:SC3916_-0.8+0.1.PNG|thumb|none|F + H2, initial conditions: 
rHH=0.745 rHF=1.83 
pHH=+0.1 pHF=-0.8 ]]

*Let us now focus on the reverse reaction, H + HF (an H atom with high kinetic energy colliding with HF).

[[File:Sc3916_hf_contour.PNG|thumb|none|F + H2, initial conditions: 
rHH=0.745 rHF=1.83 
pHH=-0.5 pHF=-10 ]]

The cases studied are an illustration of Polanyi's empirical rules. {{fontcolor|blue|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.}}

Early barrier:
Kinetic energy effective in overcoming barrier
Vibrational energy ineffective

Late barrier:
Vibrational energy effective in overcoming barrier
Kinetic energy ineffective

MRD:sc3916

2018-05-24T16:59:45Z

Sc3916: /* Reaction dynamics */

=Molecular Reaction Dynamics Report=

==Potential Energy Surfaces==

'''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.'''

The transition structure is at a saddle point - the intersection of the minimum of a curve in one plane, and the maximum of a curve in the other. In this case, the minimum curve is on the plane corresponding to RAB=RBC, and the maximum is on the perpendicular plane. The gradient of both these curves is equal to zero.

This is different from when the potential energy surface is at a minimum - in one plane the gradient = 0, but not in the other. This allows us to distinguish between the transition structure and minima by looking at the curvature in '''both''' planes.

[[File:Saddle_Point_Visualisation_sc3916.png|thumb|none|500px|Saddle Point Visualisation]]

'''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.'''

'''rts= 0.9075 angstroms =rAB=rBC'''

To estimate this, a rough estimate was found by looking at the saddle point. At the saddle point, gradient = 0 in both planes - so if the system is at exactly this point, the atoms will remain stationary if it has no initial momentum. If it isn't at this point, it will start to oscillate - and the further away from the saddle point, the greater the oscillation. Therefore by plotting this internuclear distance vs time, and adjusting the distance by small amounts until there was very little oscillation seen, a more precise value was found.

[[File: Sc3916_1.PNG|thumb|300px|none|Internuclear Distances vs Time]]

==Reaction trajectories==

[[File:Sc3916_mep.PNG|thumb|400px|left|Minimum Energy Path from from r1 = rts+δ, r2 = rts]]

[[File:Sc3916_dynamics.PNG|thumb|400px|center|Dynamic Energy Path from r1 = rts+δ, r2 = rts]]

'''Comment on how the MEP and the trajectory you just calculated differ.'''

The minimum energy path represents an infinitely slow reaction trajectory. We start off a very small distance from the transition structure, and then the trajectory "rolls" down the potential energy surface, following where the gradient is most negative. Since velocity resets to 0 every time interval, so does momentum (p=mv). This means at every infinitesimally small step, this pathway will minimise its energy. This is not how it works in reality.

The dynamic trajectory is much closer to reality. As the trajectory starts to move down the potential energy surface it gains momentum. This means it will carry on rolling up the curve on the other side of the minimum (which increases its potential energy), leading to oscillations. This reflects the vibrational motion of atoms, which MEP does not include.

Dynamics Internuclear distance vs time plot: 
Final AB distance = 0.8 
Final BC distance = 9.0 

Dynamics Internuclear Momenta vs time plot: 
Final average momentum for AB = 1.25 
Final average momentum for BC = 2.5 

This shows the reaction path from the above calculation leads to A-B + C

If we switch around initial values, so now rAB=rts+δ and rBC=rts, the reaction path will output the same values, except the products are B-C + A.

'''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.'''

The reaction proceeds from left to right across the below plot, A-B + C -> A + B-C. This reaction is essentially the reverse of the previous trajectory (allowing the pathway to "roll" down from the transition state), however it crosses over, to form the products.

[[File:Sc3916_dynamics2.PNG|thumb|400px|none| initial conditions: 
rAB = 0.8 
rBC = 9.0 
pAB = -1.25 
pBC = -2.5]]

==Reactive and unreactive trajectories==
The following table is for the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, with varying momentum.

{| class="wikitable" border=1
! p1 !! p2 !! E(kcal/mol) !! Reactive?
|-
| -1.25 || -2.5 || -99.119 || Yes
|-
| -1.5 || -2.0 || -100.456 || No
|-
| -1.5 || -2.5 || -98.956 || Yes
|-
| -2.5 || -5.0 || -84.956 || No
|-
| -2.5 || -5.2 || -83.416 || Yes
|}

[[File:Sc3916_rxn1.PNG|thumb|300px|none|Initial p1=-1.25, r2=-2.5 
the trajectory goes smoothly up to the transition state, then starts oscillating as it rolls down to the products]]

[[File:Sc3916_rxn2.PNG|thumb|300px|none|Initial p1=-1.5, r2=-2.0 
the trajectory goes up towards the transition state smoothly (but not very far), and then rolls back down the way it came, with oscillating]]

[[File:Sc3916_rxn3.PNG|thumb|300px|none|Initial p1=-1.5, r2=-2.5 
the trajectory goes smoothly up to the transition state, then starts oscillating as it rolls down to the products]]

[[File:Sc3916_rxn4b.PNG|thumb|300px|none|Initial p1=-2.5, r2=-5.0 
the trajectory goes smoothly up to the transition state, crosses it, but recrosses it and reverts back to the reactants, with very large oscillations]]

[[File:Sc3916_rxn5.PNG|thumb|300px|none|Initial p1=-2.5, r2=-5.2 
the trajectory crosses the transition state, forming the products, then recrosses it to the reactant side, and then recrosses it again, forming the products finally]]

'''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?'''

4 major assumptions are used. Firstly, the Born-Oppenheimer approximation is used, and that the effects of quantum tunneling are negligible (there is a non-zero probability that reactants can convert to products without reaching the activation energy). The reactants are assumed to have a Boltzmann distribution - which is perfectly valid at low temperatures. The final assumption is that "once the system attains the transition state, with a velocity towards the product configuration, it will not reenter the initial state region again."[https://doi.org/10.1016/B978-044452837-7.50005-8 1]

This last assumption does not fit with our simulations. Both the 4th, and 5th plots contain the trajectory crossing over the transition state and back again. Since there will be more trajectories (like the 4th) which involve crossing over and are still unreactive - which are not predicted by Transition State Theory - experimental rate values will be slower than those predicted by the theory.

==EXERCISE 2: F - H - H system==

'''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 is exothermic - the products have a more negative energy than the reactants. Because the enthalpy change only involves breaking a H-H bond and making a H-F bond, ΔH = +H(H-H) - H(H-F).

Since ΔH < 0:

H(H-F)>H(H-H)

This means HF has a greater bond enthalpy (and therefore strength) than H2. It can also be seen that H + HF is endothermic, which leads to the same conclusion.

'''Locate the approximate position of the transition state.'''

rts: 
RHH = 0.745 angstroms 
RHF = 1.811 angstroms

(0.7448778, 1.810754) is the super accurate one

'''Report the activation energy for both reactions.'''

By performing an MEP from next to the transition structure, the path from the TS to the reactants, or products, was found. By subtracting the final energy from the TS, the Ea was obtained. Since the activation energy for H2 + F is very small, the MEP has a very small gradient, and as such the calculation required an order of magnitude more steps in the calculation (than HF + H) to reach the reactants. Conversely, in HF + H a step change is seen.

[[File:Mep_energyvstime_sc3916.PNG|thumb|left|initial conditions: 
RHH=0.745 
RHF=1.83]]

[[File:3916_energyvstime_hf2.PNG|thumb|center|initial conditions: 
RHH=0.745 
RHF=1.8]]

Ea(H2 + F Ea) = 0.256 kcal/mol 
Ea(HF + H Ea) = 30.202 kcal/mol 

=== Reaction dynamics ===
* Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Internuclear Momenta vs Time”. {{fontcolor1|blue|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

[[File:H2+f_surface__3916.PNG|thumb|none|F + H2, initial conditions: 
RHH (AB) = 0.745 angstroms 
RHF (BC) = 2 angstroms 
AB momentum = +3 
BC momentum = -5]]

[[File:H2+F_MOMENTAVTIME3916.PNG|thumb|none|F + H2 Momentum vs Time Plot]]

The animation of this reaction clearly shows the velocity of the lone H atom formed has increased, and vibrations occurring in HF. This can be seen in the momentum vs time plot - with the increased AB and AC momentum reflecting the increase in velocity, and BC momentum oscillation reflecting the vibrations of the H-F bond. Since translational and vibrational energy contribute to kinetic energy, it is safe to assume that the energy released by this exothermic reaction (a decrease in enthalpy) is conserved in the form of increased kinetic energy. This can be seen in the below graph. An increase in average kinetic energy results in increased temperature - which is easy to measure experimentally.

[[File:H2+F_ENERGYVSTIME3916.PNG|thumb|none|F + H2 Energy vs Time Plot]]

* Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 0.74, with a momentum pFH = -0.5, and explore several values of pHH in the range -3 to 3 (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.

[[File:SC3916-3CONTOUR.PNG|thumb|none|F + H2, initial conditions: 
rHH=0.745 rHF=1.83 
pHH=-3.0 pHF=-0.5 ]]

[[File:SC3916-2CONTOUR.PNG|thumb|none|F + H2, initial conditions: 
rHH=0.745 rHF=1.83 
pHH=-2.0 pHF=-0.5 ]]

[[File:SC3916-1.5CONTOUR.PNG|thumb|none|F + H2, initial conditions: 
rHH=0.745 rHF=1.83 
pHH=-1 pHF=-0.5 ]]

[[File:SC39160CONTOUR.PNG|thumb|none|F + H2, initial conditions: 
rHH=0.745 rHF=1.83 
pHH=0.0 pHF=-0.5 ]]

[[File:SC3916+1.PNG|thumb|none|F + H2, initial conditions: 
rHH=0.745 rHF=1.83 
pHH=+1.0 pHF=-0.5 ]]

[[File:SC3916+3.PNG|thumb|none|F + H2, initial conditions: 
rHH=0.745 rHF=1.83 
pHH=+3.0 pHF=-0.5 ]]

* For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now?

[[File:SC3916_-0.8+0.1.PNG|thumb|none|F + H2, initial conditions: 
rHH=0.745 rHF=1.83 
pHH=+0.1 pHF=-0.8 ]]

*Let us now focus on the reverse reaction, H + HF (an H atom with high kinetic energy colliding with HF).

[[File:Sc3916_hf_contour.PNG|thumb|none|F + H2, initial conditions: 
rHH=0.745 rHF=1.83 
pHH=-0.5 pHF=-10 ]]

The cases studied are an illustration of Polanyi's empirical rules. {{fontcolor|blue|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:SC3916-3CONTOUR.PNG

2018-05-24T16:54:43Z

Sc3916:

MRD:sc3916

2018-05-24T16:52:02Z

Sc3916: /* Reaction dynamics */

=Molecular Reaction Dynamics Report=

==Potential Energy Surfaces==

'''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.'''

The transition structure is at a saddle point - the intersection of the minimum of a curve in one plane, and the maximum of a curve in the other. In this case, the minimum curve is on the plane corresponding to RAB=RBC, and the maximum is on the perpendicular plane. The gradient of both these curves is equal to zero.

This is different from when the potential energy surface is at a minimum - in one plane the gradient = 0, but not in the other. This allows us to distinguish between the transition structure and minima by looking at the curvature in '''both''' planes.

[[File:Saddle_Point_Visualisation_sc3916.png|thumb|none|500px|Saddle Point Visualisation]]

'''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.'''

'''rts= 0.9075 angstroms =rAB=rBC'''

To estimate this, a rough estimate was found by looking at the saddle point. At the saddle point, gradient = 0 in both planes - so if the system is at exactly this point, the atoms will remain stationary if it has no initial momentum. If it isn't at this point, it will start to oscillate - and the further away from the saddle point, the greater the oscillation. Therefore by plotting this internuclear distance vs time, and adjusting the distance by small amounts until there was very little oscillation seen, a more precise value was found.

[[File: Sc3916_1.PNG|thumb|300px|none|Internuclear Distances vs Time]]

==Reaction trajectories==

[[File:Sc3916_mep.PNG|thumb|400px|left|Minimum Energy Path from from r1 = rts+δ, r2 = rts]]

[[File:Sc3916_dynamics.PNG|thumb|400px|center|Dynamic Energy Path from r1 = rts+δ, r2 = rts]]

'''Comment on how the MEP and the trajectory you just calculated differ.'''

The minimum energy path represents an infinitely slow reaction trajectory. We start off a very small distance from the transition structure, and then the trajectory "rolls" down the potential energy surface, following where the gradient is most negative. Since velocity resets to 0 every time interval, so does momentum (p=mv). This means at every infinitesimally small step, this pathway will minimise its energy. This is not how it works in reality.

The dynamic trajectory is much closer to reality. As the trajectory starts to move down the potential energy surface it gains momentum. This means it will carry on rolling up the curve on the other side of the minimum (which increases its potential energy), leading to oscillations. This reflects the vibrational motion of atoms, which MEP does not include.

Dynamics Internuclear distance vs time plot: 
Final AB distance = 0.8 
Final BC distance = 9.0 

Dynamics Internuclear Momenta vs time plot: 
Final average momentum for AB = 1.25 
Final average momentum for BC = 2.5 

This shows the reaction path from the above calculation leads to A-B + C

If we switch around initial values, so now rAB=rts+δ and rBC=rts, the reaction path will output the same values, except the products are B-C + A.

'''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.'''

The reaction proceeds from left to right across the below plot, A-B + C -> A + B-C. This reaction is essentially the reverse of the previous trajectory (allowing the pathway to "roll" down from the transition state), however it crosses over, to form the products.

[[File:Sc3916_dynamics2.PNG|thumb|400px|none| initial conditions: 
rAB = 0.8 
rBC = 9.0 
pAB = -1.25 
pBC = -2.5]]

==Reactive and unreactive trajectories==
The following table is for the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, with varying momentum.

{| class="wikitable" border=1
! p1 !! p2 !! E(kcal/mol) !! Reactive?
|-
| -1.25 || -2.5 || -99.119 || Yes
|-
| -1.5 || -2.0 || -100.456 || No
|-
| -1.5 || -2.5 || -98.956 || Yes
|-
| -2.5 || -5.0 || -84.956 || No
|-
| -2.5 || -5.2 || -83.416 || Yes
|}

[[File:Sc3916_rxn1.PNG|thumb|300px|none|Initial p1=-1.25, r2=-2.5 
the trajectory goes smoothly up to the transition state, then starts oscillating as it rolls down to the products]]

[[File:Sc3916_rxn2.PNG|thumb|300px|none|Initial p1=-1.5, r2=-2.0 
the trajectory goes up towards the transition state smoothly (but not very far), and then rolls back down the way it came, with oscillating]]

[[File:Sc3916_rxn3.PNG|thumb|300px|none|Initial p1=-1.5, r2=-2.5 
the trajectory goes smoothly up to the transition state, then starts oscillating as it rolls down to the products]]

[[File:Sc3916_rxn4b.PNG|thumb|300px|none|Initial p1=-2.5, r2=-5.0 
the trajectory goes smoothly up to the transition state, crosses it, but recrosses it and reverts back to the reactants, with very large oscillations]]

[[File:Sc3916_rxn5.PNG|thumb|300px|none|Initial p1=-2.5, r2=-5.2 
the trajectory crosses the transition state, forming the products, then recrosses it to the reactant side, and then recrosses it again, forming the products finally]]

'''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?'''

4 major assumptions are used. Firstly, the Born-Oppenheimer approximation is used, and that the effects of quantum tunneling are negligible (there is a non-zero probability that reactants can convert to products without reaching the activation energy). The reactants are assumed to have a Boltzmann distribution - which is perfectly valid at low temperatures. The final assumption is that "once the system attains the transition state, with a velocity towards the product configuration, it will not reenter the initial state region again."[https://doi.org/10.1016/B978-044452837-7.50005-8 1]

This last assumption does not fit with our simulations. Both the 4th, and 5th plots contain the trajectory crossing over the transition state and back again. Since there will be more trajectories (like the 4th) which involve crossing over and are still unreactive - which are not predicted by Transition State Theory - experimental rate values will be slower than those predicted by the theory.

==EXERCISE 2: F - H - H system==

'''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 is exothermic - the products have a more negative energy than the reactants. Because the enthalpy change only involves breaking a H-H bond and making a H-F bond, ΔH = +H(H-H) - H(H-F).

Since ΔH < 0:

H(H-F)>H(H-H)

This means HF has a greater bond enthalpy (and therefore strength) than H2. It can also be seen that H + HF is endothermic, which leads to the same conclusion.

'''Locate the approximate position of the transition state.'''

rts: 
RHH = 0.745 angstroms 
RHF = 1.811 angstroms

(0.7448778, 1.810754) is the super accurate one

'''Report the activation energy for both reactions.'''

By performing an MEP from next to the transition structure, the path from the TS to the reactants, or products, was found. By subtracting the final energy from the TS, the Ea was obtained. Since the activation energy for H2 + F is very small, the MEP has a very small gradient, and as such the calculation required an order of magnitude more steps in the calculation (than HF + H) to reach the reactants. Conversely, in HF + H a step change is seen.

[[File:Mep_energyvstime_sc3916.PNG|thumb|left|initial conditions: 
RHH=0.745 
RHF=1.83]]

[[File:3916_energyvstime_hf2.PNG|thumb|center|initial conditions: 
RHH=0.745 
RHF=1.8]]

Ea(H2 + F Ea) = 0.256 kcal/mol 
Ea(HF + H Ea) = 30.202 kcal/mol 

=== Reaction dynamics ===
* Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Internuclear Momenta vs Time”. {{fontcolor1|blue|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

[[File:H2+f_surface__3916.PNG|thumb|none|F + H2, initial conditions: 
RHH (AB) = 0.745 angstroms 
RHF (BC) = 2 angstroms 
AB momentum = +3 
BC momentum = -5]]

[[File:H2+F_MOMENTAVTIME3916.PNG|thumb|none|F + H2 Momentum vs Time Plot]]

The animation of this reaction clearly shows the velocity of the lone H atom formed has increased, and vibrations occurring in HF. This can be seen in the momentum vs time plot - with the increased AB and AC momentum reflecting the increase in velocity, and BC momentum oscillation reflecting the vibrations of the H-F bond. Since translational and vibrational energy contribute to kinetic energy, it is safe to assume that the energy released by this exothermic reaction (a decrease in enthalpy) is conserved in the form of increased kinetic energy. This can be seen in the below graph. An increase in average kinetic energy results in increased temperature - which is easy to measure experimentally.

[[File:H2+F_ENERGYVSTIME3916.PNG|thumb|none|F + H2 Energy vs Time Plot]]

* Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 0.74, with a momentum pFH = -0.5, and explore several values of pHH in the range -3 to 3 (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.

[[File:SC3916-3CONTOUR.PNG|thumb|none|F + H2, initial conditions: 
rHH=0.745 rHF=1.83 
pHH=-3.0 pHF=-0.5 ]]

[[File:SC3916-2CONTOUR.PNG|thumb|none|F + H2, initial conditions: 
rHH=0.745 rHF=1.83 
pHH=-2.0 pHF=-0.5 ]]

[[File:SC3916-1CONTOUR.PNG|thumb|none|F + H2, initial conditions: 
rHH=0.745 rHF=1.83 
pHH=-1 pHF=-0.5 ]]

[[File:SC39160CONTOUR.PNG|thumb|none|F + H2, initial conditions: 
rHH=0.745 rHF=1.83 
pHH=0.0 pHF=-0.5 ]]

[[File:SC3916+1.PNG|thumb|none|F + H2, initial conditions: 
rHH=0.745 rHF=1.83 
pHH=+1.0 pHF=-0.5 ]]

[[File:SC3916+3.PNG|thumb|none|F + H2, initial conditions: 
rHH=0.745 rHF=1.83 
pHH=+3.0 pHF=-0.5 ]]

SC3916-3CONTOUR.PNG - RECROSSING, UNREACTIVE
SC3916-2CONTOUR.PNG - RECROSSING, UNREACTIVE
SC3916-2CONTOUR.PNG - REACTIVE, SOME RETRACING
SC39160CONTOUR.PNG - REACTIVE, SOME RETRACING
SC3916+1.PNG - REACTIVE, NO RETRACING
SC3916+3.PNG - RECROSSING, UNREACTIVE

* For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now?

SC3916_-0.8+0.1.PNG REACTIVE, NO RETRACING

Let us now focus on the reverse reaction, H + HF.

(an H atom colliding with a high kinetic energy).

Sc3916_hf_contour.PNG

The cases studied are an illustration of Polanyi's empirical rules. {{fontcolor|blue|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.}}

MRD:sc3916

2018-05-24T16:46:28Z

Sc3916: /* EXERCISE 2: F - H - H system */

=Molecular Reaction Dynamics Report=

==Potential Energy Surfaces==

'''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.'''

The transition structure is at a saddle point - the intersection of the minimum of a curve in one plane, and the maximum of a curve in the other. In this case, the minimum curve is on the plane corresponding to RAB=RBC, and the maximum is on the perpendicular plane. The gradient of both these curves is equal to zero.

This is different from when the potential energy surface is at a minimum - in one plane the gradient = 0, but not in the other. This allows us to distinguish between the transition structure and minima by looking at the curvature in '''both''' planes.

[[File:Saddle_Point_Visualisation_sc3916.png|thumb|none|500px|Saddle Point Visualisation]]

'''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.'''

'''rts= 0.9075 angstroms =rAB=rBC'''

To estimate this, a rough estimate was found by looking at the saddle point. At the saddle point, gradient = 0 in both planes - so if the system is at exactly this point, the atoms will remain stationary if it has no initial momentum. If it isn't at this point, it will start to oscillate - and the further away from the saddle point, the greater the oscillation. Therefore by plotting this internuclear distance vs time, and adjusting the distance by small amounts until there was very little oscillation seen, a more precise value was found.

[[File: Sc3916_1.PNG|thumb|300px|none|Internuclear Distances vs Time]]

==Reaction trajectories==

[[File:Sc3916_mep.PNG|thumb|400px|left|Minimum Energy Path from from r1 = rts+δ, r2 = rts]]

[[File:Sc3916_dynamics.PNG|thumb|400px|center|Dynamic Energy Path from r1 = rts+δ, r2 = rts]]

'''Comment on how the MEP and the trajectory you just calculated differ.'''

The minimum energy path represents an infinitely slow reaction trajectory. We start off a very small distance from the transition structure, and then the trajectory "rolls" down the potential energy surface, following where the gradient is most negative. Since velocity resets to 0 every time interval, so does momentum (p=mv). This means at every infinitesimally small step, this pathway will minimise its energy. This is not how it works in reality.

The dynamic trajectory is much closer to reality. As the trajectory starts to move down the potential energy surface it gains momentum. This means it will carry on rolling up the curve on the other side of the minimum (which increases its potential energy), leading to oscillations. This reflects the vibrational motion of atoms, which MEP does not include.

Dynamics Internuclear distance vs time plot: 
Final AB distance = 0.8 
Final BC distance = 9.0 

Dynamics Internuclear Momenta vs time plot: 
Final average momentum for AB = 1.25 
Final average momentum for BC = 2.5 

This shows the reaction path from the above calculation leads to A-B + C

If we switch around initial values, so now rAB=rts+δ and rBC=rts, the reaction path will output the same values, except the products are B-C + A.

'''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.'''

The reaction proceeds from left to right across the below plot, A-B + C -> A + B-C. This reaction is essentially the reverse of the previous trajectory (allowing the pathway to "roll" down from the transition state), however it crosses over, to form the products.

[[File:Sc3916_dynamics2.PNG|thumb|400px|none| initial conditions: 
rAB = 0.8 
rBC = 9.0 
pAB = -1.25 
pBC = -2.5]]

==Reactive and unreactive trajectories==
The following table is for the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, with varying momentum.

{| class="wikitable" border=1
! p1 !! p2 !! E(kcal/mol) !! Reactive?
|-
| -1.25 || -2.5 || -99.119 || Yes
|-
| -1.5 || -2.0 || -100.456 || No
|-
| -1.5 || -2.5 || -98.956 || Yes
|-
| -2.5 || -5.0 || -84.956 || No
|-
| -2.5 || -5.2 || -83.416 || Yes
|}

[[File:Sc3916_rxn1.PNG|thumb|300px|none|Initial p1=-1.25, r2=-2.5 
the trajectory goes smoothly up to the transition state, then starts oscillating as it rolls down to the products]]

[[File:Sc3916_rxn2.PNG|thumb|300px|none|Initial p1=-1.5, r2=-2.0 
the trajectory goes up towards the transition state smoothly (but not very far), and then rolls back down the way it came, with oscillating]]

[[File:Sc3916_rxn3.PNG|thumb|300px|none|Initial p1=-1.5, r2=-2.5 
the trajectory goes smoothly up to the transition state, then starts oscillating as it rolls down to the products]]

[[File:Sc3916_rxn4b.PNG|thumb|300px|none|Initial p1=-2.5, r2=-5.0 
the trajectory goes smoothly up to the transition state, crosses it, but recrosses it and reverts back to the reactants, with very large oscillations]]

[[File:Sc3916_rxn5.PNG|thumb|300px|none|Initial p1=-2.5, r2=-5.2 
the trajectory crosses the transition state, forming the products, then recrosses it to the reactant side, and then recrosses it again, forming the products finally]]

'''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?'''

4 major assumptions are used. Firstly, the Born-Oppenheimer approximation is used, and that the effects of quantum tunneling are negligible (there is a non-zero probability that reactants can convert to products without reaching the activation energy). The reactants are assumed to have a Boltzmann distribution - which is perfectly valid at low temperatures. The final assumption is that "once the system attains the transition state, with a velocity towards the product configuration, it will not reenter the initial state region again."[https://doi.org/10.1016/B978-044452837-7.50005-8 1]

This last assumption does not fit with our simulations. Both the 4th, and 5th plots contain the trajectory crossing over the transition state and back again. Since there will be more trajectories (like the 4th) which involve crossing over and are still unreactive - which are not predicted by Transition State Theory - experimental rate values will be slower than those predicted by the theory.

==EXERCISE 2: F - H - H system==

'''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 is exothermic - the products have a more negative energy than the reactants. Because the enthalpy change only involves breaking a H-H bond and making a H-F bond, ΔH = +H(H-H) - H(H-F).

Since ΔH < 0:

H(H-F)>H(H-H)

This means HF has a greater bond enthalpy (and therefore strength) than H2. It can also be seen that H + HF is endothermic, which leads to the same conclusion.

'''Locate the approximate position of the transition state.'''

rts: 
RHH = 0.745 angstroms 
RHF = 1.811 angstroms

(0.7448778, 1.810754) is the super accurate one

'''Report the activation energy for both reactions.'''

By performing an MEP from next to the transition structure, the path from the TS to the reactants, or products, was found. By subtracting the final energy from the TS, the Ea was obtained. Since the activation energy for H2 + F is very small, the MEP has a very small gradient, and as such the calculation required an order of magnitude more steps in the calculation (than HF + H) to reach the reactants. Conversely, in HF + H a step change is seen.

[[File:Mep_energyvstime_sc3916.PNG|thumb|left|initial conditions: 
RHH=0.745 
RHF=1.83]]

[[File:3916_energyvstime_hf2.PNG|thumb|center|initial conditions: 
RHH=0.745 
RHF=1.8]]

Ea(H2 + F Ea) = 0.256 kcal/mol 
Ea(HF + H Ea) = 30.202 kcal/mol 

=== Reaction dynamics ===
* Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Internuclear Momenta vs Time”. {{fontcolor1|blue|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

[[File:H2+f_surface__3916.PNG|thumb|none|F + H2, initial conditions: 
RHH (AB) = 0.745 angstroms 
RHF (BC) = 2 angstroms 
AB momentum = +3 
BC momentum = -5]]

[[File:H2+F_MOMENTAVTIME3916.PNG|thumb|none|F + H2 Momentum vs Time Plot]]

The animation of this reaction clearly shows the velocity of the lone H atom formed has increased, and vibrations occurring in HF. This can be seen in the momentum vs time plot - with the increased AB and AC momentum reflecting the increase in velocity, and BC momentum oscillation reflecting the vibrations of the H-F bond. Since translational and vibrational energy contribute to kinetic energy, it is safe to assume that the energy released by this exothermic reaction (a decrease in enthalpy) is conserved in the form of increased kinetic energy. This can be seen in the below graph. An increase in average kinetic energy results in increased temperature - which is easy to measure experimentally.

[[File:H2+F_ENERGYVSTIME3916.PNG|thumb|none|F + H2 Energy vs Time Plot]]

* Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 0.74, with a momentum pFH = -0.5, and explore several values of pHH in the range -3 to 3 (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.

SC3916-3CONTOUR.PNG - RECROSSING, UNREACTIVE
SC3916-2CONTOUR.PNG - RECROSSING, UNREACTIVE
SC3916-2CONTOUR.PNG - REACTIVE, SOME RETRACING
SC39160CONTOUR.PNG - REACTIVE, SOME RETRACING
SC3916+1.PNG - REACTIVE, NO RETRACING
SC3916+3.PNG - RECROSSING, UNREACTIVE

* For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now?

SC3916_-0.8+0.1.PNG REACTIVE, NO RETRACING

Let us now focus on the reverse reaction, H + HF.

(an H atom colliding with a high kinetic energy).

Sc3916_hf_contour.PNG

The cases studied are an illustration of Polanyi's empirical rules. {{fontcolor|blue|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:Sc3916 hf contour.PNG

2018-05-24T16:40:12Z

Sc3916:

MRD:sc3916

2018-05-24T16:28:58Z

Sc3916: /* Reaction trajectories */

=Molecular Reaction Dynamics Report=

==Potential Energy Surfaces==

'''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.'''

The transition structure is at a saddle point - the intersection of the minimum of a curve in one plane, and the maximum of a curve in the other. In this case, the minimum curve is on the plane corresponding to RAB=RBC, and the maximum is on the perpendicular plane. The gradient of both these curves is equal to zero.

This is different from when the potential energy surface is at a minimum - in one plane the gradient = 0, but not in the other. This allows us to distinguish between the transition structure and minima by looking at the curvature in '''both''' planes.

[[File:Saddle_Point_Visualisation_sc3916.png|thumb|none|500px|Saddle Point Visualisation]]

'''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.'''

'''rts= 0.9075 angstroms =rAB=rBC'''

To estimate this, a rough estimate was found by looking at the saddle point. At the saddle point, gradient = 0 in both planes - so if the system is at exactly this point, the atoms will remain stationary if it has no initial momentum. If it isn't at this point, it will start to oscillate - and the further away from the saddle point, the greater the oscillation. Therefore by plotting this internuclear distance vs time, and adjusting the distance by small amounts until there was very little oscillation seen, a more precise value was found.

[[File: Sc3916_1.PNG|thumb|300px|none|Internuclear Distances vs Time]]

==Reaction trajectories==

[[File:Sc3916_mep.PNG|thumb|400px|left|Minimum Energy Path from from r1 = rts+δ, r2 = rts]]

[[File:Sc3916_dynamics.PNG|thumb|400px|center|Dynamic Energy Path from r1 = rts+δ, r2 = rts]]

'''Comment on how the MEP and the trajectory you just calculated differ.'''

The minimum energy path represents an infinitely slow reaction trajectory. We start off a very small distance from the transition structure, and then the trajectory "rolls" down the potential energy surface, following where the gradient is most negative. Since velocity resets to 0 every time interval, so does momentum (p=mv). This means at every infinitesimally small step, this pathway will minimise its energy. This is not how it works in reality.

The dynamic trajectory is much closer to reality. As the trajectory starts to move down the potential energy surface it gains momentum. This means it will carry on rolling up the curve on the other side of the minimum (which increases its potential energy), leading to oscillations. This reflects the vibrational motion of atoms, which MEP does not include.

Dynamics Internuclear distance vs time plot: 
Final AB distance = 0.8 
Final BC distance = 9.0 

Dynamics Internuclear Momenta vs time plot: 
Final average momentum for AB = 1.25 
Final average momentum for BC = 2.5 

This shows the reaction path from the above calculation leads to A-B + C

If we switch around initial values, so now rAB=rts+δ and rBC=rts, the reaction path will output the same values, except the products are B-C + A.

'''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.'''

The reaction proceeds from left to right across the below plot, A-B + C -> A + B-C. This reaction is essentially the reverse of the previous trajectory (allowing the pathway to "roll" down from the transition state), however it crosses over, to form the products.

[[File:Sc3916_dynamics2.PNG|thumb|400px|none| initial conditions: 
rAB = 0.8 
rBC = 9.0 
pAB = -1.25 
pBC = -2.5]]

==Reactive and unreactive trajectories==
The following table is for the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, with varying momentum.

{| class="wikitable" border=1
! p1 !! p2 !! E(kcal/mol) !! Reactive?
|-
| -1.25 || -2.5 || -99.119 || Yes
|-
| -1.5 || -2.0 || -100.456 || No
|-
| -1.5 || -2.5 || -98.956 || Yes
|-
| -2.5 || -5.0 || -84.956 || No
|-
| -2.5 || -5.2 || -83.416 || Yes
|}

[[File:Sc3916_rxn1.PNG|thumb|300px|none|Initial p1=-1.25, r2=-2.5 
the trajectory goes smoothly up to the transition state, then starts oscillating as it rolls down to the products]]

[[File:Sc3916_rxn2.PNG|thumb|300px|none|Initial p1=-1.5, r2=-2.0 
the trajectory goes up towards the transition state smoothly (but not very far), and then rolls back down the way it came, with oscillating]]

[[File:Sc3916_rxn3.PNG|thumb|300px|none|Initial p1=-1.5, r2=-2.5 
the trajectory goes smoothly up to the transition state, then starts oscillating as it rolls down to the products]]

[[File:Sc3916_rxn4b.PNG|thumb|300px|none|Initial p1=-2.5, r2=-5.0 
the trajectory goes smoothly up to the transition state, crosses it, but recrosses it and reverts back to the reactants, with very large oscillations]]

[[File:Sc3916_rxn5.PNG|thumb|300px|none|Initial p1=-2.5, r2=-5.2 
the trajectory crosses the transition state, forming the products, then recrosses it to the reactant side, and then recrosses it again, forming the products finally]]

'''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?'''

4 major assumptions are used. Firstly, the Born-Oppenheimer approximation is used, and that the effects of quantum tunneling are negligible (there is a non-zero probability that reactants can convert to products without reaching the activation energy). The reactants are assumed to have a Boltzmann distribution - which is perfectly valid at low temperatures. The final assumption is that "once the system attains the transition state, with a velocity towards the product configuration, it will not reenter the initial state region again."[https://doi.org/10.1016/B978-044452837-7.50005-8 1]

This last assumption does not fit with our simulations. Both the 4th, and 5th plots contain the trajectory crossing over the transition state and back again. Since there will be more trajectories (like the 4th) which involve crossing over and are still unreactive - which are not predicted by Transition State Theory - experimental rate values will be slower than those predicted by the theory.

==EXERCISE 2: F - H - H system==

'''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 is exothermic - the products have a more negative energy than the reactants. Because the enthalpy change only involves breaking a H-H bond and making a H-F bond, ΔH = +H(H-H) - H(H-F).

Since ΔH < 0:

H(H-F)>H(H-H)

This means HF has a greater bond enthalpy (and therefore strength) than H2. It can also be seen that H + HF is endothermic, which leads to the same conclusion.

'''Locate the approximate position of the transition state.'''

rts: 
RHH = 0.745 angstroms 
RHF = 1.811 angstroms

(0.7448778, 1.810754) is the super accurate one

'''Report the activation energy for both reactions.'''

By performing an MEP from next to the transition structure, the path from the TS to the reactants, or products, was found. By subtracting the final energy from the TS, the Ea was obtained. Since the activation energy for H2 + F is very small, the MEP has a very small gradient, and as such the calculation required an order of magnitude more steps in the calculation (than HF + H) to reach the reactants. Conversely, in HF + H a step change is seen.

[[File:Mep_energyvstime_sc3916.PNG|thumb|left|initial conditions: 
RHH=0.745 
RHF=1.83]]

[[File:3916_energyvstime_hf2.PNG|thumb|center|initial conditions: 
RHH=0.745 
RHF=1.8]]

Ea(H2 + F Ea) = 0.256 kcal/mol 
Ea(HF + H Ea) = 30.202 kcal/mol 

=== Reaction dynamics ===
* Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Internuclear Momenta vs Time”. {{fontcolor1|blue|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

[[File:H2+f_surface__3916.PNG|thumb|none|F + H2, initial conditions: 
RHH (AB) = 0.745 angstroms 
RHF (BC) = 2 angstroms 
AB momentum = +3 
BC momentum = -5]]

[[File:H2+F_MOMENTAVTIME3916.PNG|thumb|none|F + H2 Momentum vs Time Plot]]

The animation of this reaction clearly shows the velocity of the lone H atom formed has increased, and vibrations occurring in HF. This can be seen in the momentum vs time plot - with the increased AB and AC momentum reflecting the increase in velocity, and BC momentum oscillation reflecting the vibrations of the H-F bond. Since translational and vibrational energy contribute to kinetic energy, it is safe to assume that the energy released by this exothermic reaction (a decrease in enthalpy) is conserved in the form of increased kinetic energy. This can be seen in the below graph. An increase in average kinetic energy results in increased temperature - which is easy to measure experimentally.

[[File:H2+F_ENERGYVSTIME3916.PNG|thumb|none|F + H2 Energy vs Time Plot]]

* Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 0.74, with a momentum pFH = -0.5, and explore several values of pHH in the range -3 to 3 (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.

SC3916-3CONTOUR.PNG - RECROSSING, UNREACTIVE
SC3916-2CONTOUR.PNG - RECROSSING, UNREACTIVE
SC3916-2CONTOUR.PNG - REACTIVE, SOME RETRACING
SC39160CONTOUR.PNG - REACTIVE, SOME RETRACING
SC3916+1.PNG - REACTIVE, NO RETRACING
SC3916+3.PNG - RECROSSING, UNREACTIVE

* For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now?

SC3916_-0.8+0.1.PNG REACTIVE, NO RETRACING

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 [[#invert momenta|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.)

The cases studied are an illustration of Polanyi's empirical rules. {{fontcolor|blue|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.}}

MRD:sc3916

2018-05-24T16:27:43Z

Sc3916: /* Reaction trajectories */

=Molecular Reaction Dynamics Report=

==Potential Energy Surfaces==

'''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.'''

The transition structure is at a saddle point - the intersection of the minimum of a curve in one plane, and the maximum of a curve in the other. In this case, the minimum curve is on the plane corresponding to RAB=RBC, and the maximum is on the perpendicular plane. The gradient of both these curves is equal to zero.

This is different from when the potential energy surface is at a minimum - in one plane the gradient = 0, but not in the other. This allows us to distinguish between the transition structure and minima by looking at the curvature in '''both''' planes.

[[File:Saddle_Point_Visualisation_sc3916.png|thumb|none|500px|Saddle Point Visualisation]]

'''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.'''

'''rts= 0.9075 angstroms =rAB=rBC'''

To estimate this, a rough estimate was found by looking at the saddle point. At the saddle point, gradient = 0 in both planes - so if the system is at exactly this point, the atoms will remain stationary if it has no initial momentum. If it isn't at this point, it will start to oscillate - and the further away from the saddle point, the greater the oscillation. Therefore by plotting this internuclear distance vs time, and adjusting the distance by small amounts until there was very little oscillation seen, a more precise value was found.

[[File: Sc3916_1.PNG|thumb|300px|none|Internuclear Distances vs Time]]

==Reaction trajectories==

[[File:Sc3916_mep.PNG|thumb|400px|left|Minimum Energy Path from from r1 = rts+δ, r2 = rts]]

[[File:Sc3916_dynamics.PNG|thumb|400px|center|Dynamic Energy Path from r1 = rts+δ, r2 = rts]]

'''Comment on how the MEP and the trajectory you just calculated differ.'''

The minimum energy path represents an infinitely slow reaction trajectory. We start off a very small distance from the transition structure, and then the trajectory "rolls" down the potential energy surface, following where the gradient is most negative. Since velocity resets to 0 every time interval, so does momentum (p=mv). This means at every infinitesimally small step, this pathway will minimise its energy. This is not how it works in reality.

The dynamic trajectory is much closer to reality. As the trajectory starts to move down the potential energy surface it gains momentum. This means it will carry on rolling up the curve on the other side of the minimum (which increases its potential energy), leading to oscillations. This reflects the vibrational motion of atoms, which MEP does not include.

Dynamics Internuclear distance vs time plot: 
Final AB distance = 0.8 
Final BC distance = 9.0 

Dynamics Internuclear Momenta vs time plot: 
Final average momentum for AB = 1.25 
Final average momentum for BC = 2.5 

This shows the reaction path from the above calculation leads to A-B + C

If we switch around initial values, so now rAB=rts+δ and rBC=rts, the reaction path will output the same values, except the products are B-C + A.

'''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.'''

The reaction proceeds from left to right across the below plot, A-B + C -> A + B-C. This reaction is essentially the reverse of the previous trajectory (allowing the pathway to "roll" down from the transition state).

[[File:Sc3916_dynamics2.PNG|thumb|400px|none| initial conditions: 
rAB = 0.8 
rBC = 9.0 
pAB = -1.25 
pBC = -2.5]]

==Reactive and unreactive trajectories==
The following table is for the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, with varying momentum.

{| class="wikitable" border=1
! p1 !! p2 !! E(kcal/mol) !! Reactive?
|-
| -1.25 || -2.5 || -99.119 || Yes
|-
| -1.5 || -2.0 || -100.456 || No
|-
| -1.5 || -2.5 || -98.956 || Yes
|-
| -2.5 || -5.0 || -84.956 || No
|-
| -2.5 || -5.2 || -83.416 || Yes
|}

[[File:Sc3916_rxn1.PNG|thumb|300px|none|Initial p1=-1.25, r2=-2.5 
the trajectory goes smoothly up to the transition state, then starts oscillating as it rolls down to the products]]

[[File:Sc3916_rxn2.PNG|thumb|300px|none|Initial p1=-1.5, r2=-2.0 
the trajectory goes up towards the transition state smoothly (but not very far), and then rolls back down the way it came, with oscillating]]

[[File:Sc3916_rxn3.PNG|thumb|300px|none|Initial p1=-1.5, r2=-2.5 
the trajectory goes smoothly up to the transition state, then starts oscillating as it rolls down to the products]]

[[File:Sc3916_rxn4b.PNG|thumb|300px|none|Initial p1=-2.5, r2=-5.0 
the trajectory goes smoothly up to the transition state, crosses it, but recrosses it and reverts back to the reactants, with very large oscillations]]

[[File:Sc3916_rxn5.PNG|thumb|300px|none|Initial p1=-2.5, r2=-5.2 
the trajectory crosses the transition state, forming the products, then recrosses it to the reactant side, and then recrosses it again, forming the products finally]]

'''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?'''

4 major assumptions are used. Firstly, the Born-Oppenheimer approximation is used, and that the effects of quantum tunneling are negligible (there is a non-zero probability that reactants can convert to products without reaching the activation energy). The reactants are assumed to have a Boltzmann distribution - which is perfectly valid at low temperatures. The final assumption is that "once the system attains the transition state, with a velocity towards the product configuration, it will not reenter the initial state region again."[https://doi.org/10.1016/B978-044452837-7.50005-8 1]

This last assumption does not fit with our simulations. Both the 4th, and 5th plots contain the trajectory crossing over the transition state and back again. Since there will be more trajectories (like the 4th) which involve crossing over and are still unreactive - which are not predicted by Transition State Theory - experimental rate values will be slower than those predicted by the theory.

==EXERCISE 2: F - H - H system==

'''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 is exothermic - the products have a more negative energy than the reactants. Because the enthalpy change only involves breaking a H-H bond and making a H-F bond, ΔH = +H(H-H) - H(H-F).

Since ΔH < 0:

H(H-F)>H(H-H)

This means HF has a greater bond enthalpy (and therefore strength) than H2. It can also be seen that H + HF is endothermic, which leads to the same conclusion.

'''Locate the approximate position of the transition state.'''

rts: 
RHH = 0.745 angstroms 
RHF = 1.811 angstroms

(0.7448778, 1.810754) is the super accurate one

'''Report the activation energy for both reactions.'''

By performing an MEP from next to the transition structure, the path from the TS to the reactants, or products, was found. By subtracting the final energy from the TS, the Ea was obtained. Since the activation energy for H2 + F is very small, the MEP has a very small gradient, and as such the calculation required an order of magnitude more steps in the calculation (than HF + H) to reach the reactants. Conversely, in HF + H a step change is seen.

[[File:Mep_energyvstime_sc3916.PNG|thumb|left|initial conditions: 
RHH=0.745 
RHF=1.83]]

[[File:3916_energyvstime_hf2.PNG|thumb|center|initial conditions: 
RHH=0.745 
RHF=1.8]]

Ea(H2 + F Ea) = 0.256 kcal/mol 
Ea(HF + H Ea) = 30.202 kcal/mol 

=== Reaction dynamics ===
* Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Internuclear Momenta vs Time”. {{fontcolor1|blue|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

[[File:H2+f_surface__3916.PNG|thumb|none|F + H2, initial conditions: 
RHH (AB) = 0.745 angstroms 
RHF (BC) = 2 angstroms 
AB momentum = +3 
BC momentum = -5]]

[[File:H2+F_MOMENTAVTIME3916.PNG|thumb|none|F + H2 Momentum vs Time Plot]]

The animation of this reaction clearly shows the velocity of the lone H atom formed has increased, and vibrations occurring in HF. This can be seen in the momentum vs time plot - with the increased AB and AC momentum reflecting the increase in velocity, and BC momentum oscillation reflecting the vibrations of the H-F bond. Since translational and vibrational energy contribute to kinetic energy, it is safe to assume that the energy released by this exothermic reaction (a decrease in enthalpy) is conserved in the form of increased kinetic energy. This can be seen in the below graph. An increase in average kinetic energy results in increased temperature - which is easy to measure experimentally.

[[File:H2+F_ENERGYVSTIME3916.PNG|thumb|none|F + H2 Energy vs Time Plot]]

* Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 0.74, with a momentum pFH = -0.5, and explore several values of pHH in the range -3 to 3 (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.

SC3916-3CONTOUR.PNG - RECROSSING, UNREACTIVE
SC3916-2CONTOUR.PNG - RECROSSING, UNREACTIVE
SC3916-2CONTOUR.PNG - REACTIVE, SOME RETRACING
SC39160CONTOUR.PNG - REACTIVE, SOME RETRACING
SC3916+1.PNG - REACTIVE, NO RETRACING
SC3916+3.PNG - RECROSSING, UNREACTIVE

* For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now?

SC3916_-0.8+0.1.PNG REACTIVE, NO RETRACING

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 [[#invert momenta|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.)

The cases studied are an illustration of Polanyi's empirical rules. {{fontcolor|blue|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.}}

MRD:sc3916

2018-05-24T16:12:25Z

Sc3916: /* EXERCISE 2: F - H - H system */

=Molecular Reaction Dynamics Report=

==Potential Energy Surfaces==

'''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.'''

The transition structure is at a saddle point - the intersection of the minimum of a curve in one plane, and the maximum of a curve in the other. In this case, the minimum curve is on the plane corresponding to RAB=RBC, and the maximum is on the perpendicular plane. The gradient of both these curves is equal to zero.

This is different from when the potential energy surface is at a minimum - in one plane the gradient = 0, but not in the other. This allows us to distinguish between the transition structure and minima by looking at the curvature in '''both''' planes.

[[File:Saddle_Point_Visualisation_sc3916.png|thumb|none|500px|Saddle Point Visualisation]]

'''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.'''

'''rts= 0.9075 angstroms =rAB=rBC'''

To estimate this, a rough estimate was found by looking at the saddle point. At the saddle point, gradient = 0 in both planes - so if the system is at exactly this point, the atoms will remain stationary if it has no initial momentum. If it isn't at this point, it will start to oscillate - and the further away from the saddle point, the greater the oscillation. Therefore by plotting this internuclear distance vs time, and adjusting the distance by small amounts until there was very little oscillation seen, a more precise value was found.

[[File: Sc3916_1.PNG|thumb|300px|none|Internuclear Distances vs Time]]

==Reaction trajectories==

[[File:Sc3916_mep.PNG|thumb|400px|left|Minimum Energy Path from from r1 = rts+δ, r2 = rts]]

[[File:Sc3916_dynamics.PNG|thumb|400px|center|Dynamic Energy Path from r1 = rts+δ, r2 = rts]]

'''Comment on how the MEP and the trajectory you just calculated differ.'''

The minimum energy path represents an infinitely slow reaction trajectory. We start off a very small distance from the transition structure, and then the trajectory "rolls" down the potential energy surface, following where the gradient is most negative. Since velocity resets to 0 every time interval, so does momentum (p=mv). This means at every infinitesimally small step, this pathway will minimise its energy. This is not how it works in reality.

The dynamic trajectory is much closer to reality. As the trajectory starts to move down the potential energy surface it gains momentum. This means it will carry on rolling up the curve on the other side of the minimum (which increases its potential energy), leading to oscillations. This reflects the vibrational motion of atoms, which MEP does not include.

Dynamics Internuclear distance vs time plot: 
Final AB distance = 0.8 
Final BC distance = 9.0 

Dynamics Internuclear Momenta vs time plot: 
Final average momentum for BC = 2.5 
Final average momentum for AB = 1.25 

This shows the reaction path from the above calculation leads to A-B + C

If we switch around initial values, so now rAB=rts+δ and rBC=rts, the reaction path will output the same values, except the products are B-C + A.

'''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.'''

The reaction proceeds from left to right across the below plot, A-B + C -> A + B-C.
Obviously the reaction starts with enough momentum to proceed all the way past the transition state and to the products.
COMMENT ON THIS USING INSIGHT GLEANED FROM BELOW SECTION AND REDO GRAPH WITH FILLED LINE

AND LABEL GRAPH

[[File:Sc3916_dynamics2.PNG|thumb|400px|none]]

==Reactive and unreactive trajectories==
The following table is for the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, with varying momentum.

{| class="wikitable" border=1
! p1 !! p2 !! E(kcal/mol) !! Reactive?
|-
| -1.25 || -2.5 || -99.119 || Yes
|-
| -1.5 || -2.0 || -100.456 || No
|-
| -1.5 || -2.5 || -98.956 || Yes
|-
| -2.5 || -5.0 || -84.956 || No
|-
| -2.5 || -5.2 || -83.416 || Yes
|}

[[File:Sc3916_rxn1.PNG|thumb|300px|none|Initial p1=-1.25, r2=-2.5 
the trajectory goes smoothly up to the transition state, then starts oscillating as it rolls down to the products]]

[[File:Sc3916_rxn2.PNG|thumb|300px|none|Initial p1=-1.5, r2=-2.0 
the trajectory goes up towards the transition state smoothly (but not very far), and then rolls back down the way it came, with oscillating]]

[[File:Sc3916_rxn3.PNG|thumb|300px|none|Initial p1=-1.5, r2=-2.5 
the trajectory goes smoothly up to the transition state, then starts oscillating as it rolls down to the products]]

[[File:Sc3916_rxn4b.PNG|thumb|300px|none|Initial p1=-2.5, r2=-5.0 
the trajectory goes smoothly up to the transition state, crosses it, but recrosses it and reverts back to the reactants, with very large oscillations]]

[[File:Sc3916_rxn5.PNG|thumb|300px|none|Initial p1=-2.5, r2=-5.2 
the trajectory crosses the transition state, forming the products, then recrosses it to the reactant side, and then recrosses it again, forming the products finally]]

'''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?'''

4 major assumptions are used. Firstly, the Born-Oppenheimer approximation is used, and that the effects of quantum tunneling are negligible (there is a non-zero probability that reactants can convert to products without reaching the activation energy). The reactants are assumed to have a Boltzmann distribution - which is perfectly valid at low temperatures. The final assumption is that "once the system attains the transition state, with a velocity towards the product configuration, it will not reenter the initial state region again."[https://doi.org/10.1016/B978-044452837-7.50005-8 1]

This last assumption does not fit with our simulations. Both the 4th, and 5th plots contain the trajectory crossing over the transition state and back again. Since there will be more trajectories (like the 4th) which involve crossing over and are still unreactive - which are not predicted by Transition State Theory - experimental rate values will be slower than those predicted by the theory.

==EXERCISE 2: F - H - H system==

'''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 is exothermic - the products have a more negative energy than the reactants. Because the enthalpy change only involves breaking a H-H bond and making a H-F bond, ΔH = +H(H-H) - H(H-F).

Since ΔH < 0:

H(H-F)>H(H-H)

This means HF has a greater bond enthalpy (and therefore strength) than H2. It can also be seen that H + HF is endothermic, which leads to the same conclusion.

'''Locate the approximate position of the transition state.'''

rts: 
RHH = 0.745 angstroms 
RHF = 1.811 angstroms

(0.7448778, 1.810754) is the super accurate one

'''Report the activation energy for both reactions.'''

By performing an MEP from next to the transition structure, the path from the TS to the reactants, or products, was found. By subtracting the final energy from the TS, the Ea was obtained. Since the activation energy for H2 + F is very small, the MEP has a very small gradient, and as such the calculation required an order of magnitude more steps in the calculation (than HF + H) to reach the reactants. Conversely, in HF + H a step change is seen.

[[File:Mep_energyvstime_sc3916.PNG|thumb|left|initial conditions: 
RHH=0.745 
RHF=1.83]]

[[File:3916_energyvstime_hf2.PNG|thumb|center|initial conditions: 
RHH=0.745 
RHF=1.8]]

Ea(H2 + F Ea) = 0.256 kcal/mol 
Ea(HF + H Ea) = 30.202 kcal/mol 

=== Reaction dynamics ===
* Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Internuclear Momenta vs Time”. {{fontcolor1|blue|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

[[File:H2+f_surface__3916.PNG|thumb|none|F + H2, initial conditions: 
RHH (AB) = 0.745 angstroms 
RHF (BC) = 2 angstroms 
AB momentum = +3 
BC momentum = -5]]

[[File:H2+F_MOMENTAVTIME3916.PNG|thumb|none|F + H2 Momentum vs Time Plot]]

The animation of this reaction clearly shows the velocity of the lone H atom formed has increased, and vibrations occurring in HF. This can be seen in the momentum vs time plot - with the increased AB and AC momentum reflecting the increase in velocity, and BC momentum oscillation reflecting the vibrations of the H-F bond. Since translational and vibrational energy contribute to kinetic energy, it is safe to assume that the energy released by this exothermic reaction (a decrease in enthalpy) is conserved in the form of increased kinetic energy. This can be seen in the below graph. An increase in average kinetic energy results in increased temperature - which is easy to measure experimentally.

[[File:H2+F_ENERGYVSTIME3916.PNG|thumb|none|F + H2 Energy vs Time Plot]]

* Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 0.74, with a momentum pFH = -0.5, and explore several values of pHH in the range -3 to 3 (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.

SC3916-3CONTOUR.PNG - RECROSSING, UNREACTIVE
SC3916-2CONTOUR.PNG - RECROSSING, UNREACTIVE
SC3916-2CONTOUR.PNG - REACTIVE, SOME RETRACING
SC39160CONTOUR.PNG - REACTIVE, SOME RETRACING
SC3916+1.PNG - REACTIVE, NO RETRACING
SC3916+3.PNG - RECROSSING, UNREACTIVE

* For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now?

SC3916_-0.8+0.1.PNG REACTIVE, NO RETRACING

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 [[#invert momenta|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.)

The cases studied are an illustration of Polanyi's empirical rules. {{fontcolor|blue|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.}}

MRD:sc3916

2018-05-24T16:10:06Z

Sc3916: /* Potential Energy Surfaces */

=Molecular Reaction Dynamics Report=

==Potential Energy Surfaces==

'''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.'''

The transition structure is at a saddle point - the intersection of the minimum of a curve in one plane, and the maximum of a curve in the other. In this case, the minimum curve is on the plane corresponding to RAB=RBC, and the maximum is on the perpendicular plane. The gradient of both these curves is equal to zero.

This is different from when the potential energy surface is at a minimum - in one plane the gradient = 0, but not in the other. This allows us to distinguish between the transition structure and minima by looking at the curvature in '''both''' planes.

[[File:Saddle_Point_Visualisation_sc3916.png|thumb|none|500px|Saddle Point Visualisation]]

'''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.'''

'''rts= 0.9075 angstroms =rAB=rBC'''

To estimate this, a rough estimate was found by looking at the saddle point. At the saddle point, gradient = 0 in both planes - so if the system is at exactly this point, the atoms will remain stationary if it has no initial momentum. If it isn't at this point, it will start to oscillate - and the further away from the saddle point, the greater the oscillation. Therefore by plotting this internuclear distance vs time, and adjusting the distance by small amounts until there was very little oscillation seen, a more precise value was found.

[[File: Sc3916_1.PNG|thumb|300px|none|Internuclear Distances vs Time]]

==Reaction trajectories==

[[File:Sc3916_mep.PNG|thumb|400px|left|Minimum Energy Path from from r1 = rts+δ, r2 = rts]]

[[File:Sc3916_dynamics.PNG|thumb|400px|center|Dynamic Energy Path from r1 = rts+δ, r2 = rts]]

'''Comment on how the MEP and the trajectory you just calculated differ.'''

The minimum energy path represents an infinitely slow reaction trajectory. We start off a very small distance from the transition structure, and then the trajectory "rolls" down the potential energy surface, following where the gradient is most negative. Since velocity resets to 0 every time interval, so does momentum (p=mv). This means at every infinitesimally small step, this pathway will minimise its energy. This is not how it works in reality.

The dynamic trajectory is much closer to reality. As the trajectory starts to move down the potential energy surface it gains momentum. This means it will carry on rolling up the curve on the other side of the minimum (which increases its potential energy), leading to oscillations. This reflects the vibrational motion of atoms, which MEP does not include.

Dynamics Internuclear distance vs time plot: 
Final AB distance = 0.8 
Final BC distance = 9.0 

Dynamics Internuclear Momenta vs time plot: 
Final average momentum for BC = 2.5 
Final average momentum for AB = 1.25 

This shows the reaction path from the above calculation leads to A-B + C

If we switch around initial values, so now rAB=rts+δ and rBC=rts, the reaction path will output the same values, except the products are B-C + A.

'''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.'''

The reaction proceeds from left to right across the below plot, A-B + C -> A + B-C.
Obviously the reaction starts with enough momentum to proceed all the way past the transition state and to the products.
COMMENT ON THIS USING INSIGHT GLEANED FROM BELOW SECTION AND REDO GRAPH WITH FILLED LINE

AND LABEL GRAPH

[[File:Sc3916_dynamics2.PNG|thumb|400px|none]]

==Reactive and unreactive trajectories==
The following table is for the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, with varying momentum.

{| class="wikitable" border=1
! p1 !! p2 !! E(kcal/mol) !! Reactive?
|-
| -1.25 || -2.5 || -99.119 || Yes
|-
| -1.5 || -2.0 || -100.456 || No
|-
| -1.5 || -2.5 || -98.956 || Yes
|-
| -2.5 || -5.0 || -84.956 || No
|-
| -2.5 || -5.2 || -83.416 || Yes
|}

[[File:Sc3916_rxn1.PNG|thumb|300px|none|Initial p1=-1.25, r2=-2.5 
the trajectory goes smoothly up to the transition state, then starts oscillating as it rolls down to the products]]

[[File:Sc3916_rxn2.PNG|thumb|300px|none|Initial p1=-1.5, r2=-2.0 
the trajectory goes up towards the transition state smoothly (but not very far), and then rolls back down the way it came, with oscillating]]

[[File:Sc3916_rxn3.PNG|thumb|300px|none|Initial p1=-1.5, r2=-2.5 
the trajectory goes smoothly up to the transition state, then starts oscillating as it rolls down to the products]]

[[File:Sc3916_rxn4b.PNG|thumb|300px|none|Initial p1=-2.5, r2=-5.0 
the trajectory goes smoothly up to the transition state, crosses it, but recrosses it and reverts back to the reactants, with very large oscillations]]

[[File:Sc3916_rxn5.PNG|thumb|300px|none|Initial p1=-2.5, r2=-5.2 
the trajectory crosses the transition state, forming the products, then recrosses it to the reactant side, and then recrosses it again, forming the products finally]]

'''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?'''

4 major assumptions are used. Firstly, the Born-Oppenheimer approximation is used, and that the effects of quantum tunneling are negligible (there is a non-zero probability that reactants can convert to products without reaching the activation energy). The reactants are assumed to have a Boltzmann distribution - which is perfectly valid at low temperatures. The final assumption is that "once the system attains the transition state, with a velocity towards the product configuration, it will not reenter the initial state region again."[https://doi.org/10.1016/B978-044452837-7.50005-8 1]

This last assumption does not fit with our simulations. Both the 4th, and 5th plots contain the trajectory crossing over the transition state and back again. Since there will be more trajectories (like the 4th) which involve crossing over and are still unreactive - which are not predicted by Transition State Theory - experimental rate values will be slower than those predicted by the theory.

==EXERCISE 2: F - H - H system==

'''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 is exothermic - the products have a more negative energy than the reactants. Because the enthalpy change only involves breaking a H-H bond and making a H-F bond, ΔH = +H(H-H) - H(H-F).

Since ΔH < 0:

H(H-F)>H(H-H)

This means HF has a greater bond enthalpy (and therefore strength) than H2. It can also be seen that H + HF is endothermic, which leads to the same conclusion.

'''Locate the approximate position of the transition state.'''

rts: 
RHH = 0.745 angstroms 
RHF = 1.811 angstroms

(0.7448778, 1.810754) is the super accurate one

'''Report the activation energy for both reactions.'''

By performing an MEP from next to the transition structure, the path from the TS to the reactants, or products, was found. By subtracting the final energy from the TS, the Ea was obtained. Since the activation energy for H2 + F is very small, the MEP has a very small gradient, and as such the calculation required an order of magnitude more steps in the calculation (than HF + H) to reach the reactants.

[[File:Mep_energyvstime_sc3916.PNG|thumb|left|initial conditions: 
RHH=0.745 
RHF=1.83]]

[[File:3916_energyvstime_hf2.PNG|thumb|center|initial conditions: 
RHH=0.745 
RHF=1.8]]

Ea(H2 + F Ea) = 0.256 kcal/mol 
Ea(HF + H Ea) = 30.202 kcal/mol 

=== Reaction dynamics ===
* Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Internuclear Momenta vs Time”. {{fontcolor1|blue|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

[[File:H2+f_surface__3916.PNG|thumb|none|F + H2, initial conditions: 
RHH (AB) = 0.745 angstroms 
RHF (BC) = 2 angstroms 
AB momentum = +3 
BC momentum = -5]]

[[File:H2+F_MOMENTAVTIME3916.PNG|thumb|none|F + H2 Momentum vs Time Plot]]

The animation of this reaction clearly shows the velocity of the lone H atom formed has increased, and vibrations occurring in HF. This can be seen in the momentum vs time plot - with the increased AB and AC momentum reflecting the increase in velocity, and BC momentum oscillation reflecting the vibrations of the H-F bond. Since translational and vibrational energy contribute to kinetic energy, it is safe to assume that the energy released by this exothermic reaction (a decrease in enthalpy) is conserved in the form of increased kinetic energy. This can be seen in the below graph. An increase in average kinetic energy results in increased temperature - which is easy to measure experimentally.

[[File:H2+F_ENERGYVSTIME3916.PNG|thumb|none|F + H2 Energy vs Time Plot]]

* Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 0.74, with a momentum pFH = -0.5, and explore several values of pHH in the range -3 to 3 (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.

SC3916-3CONTOUR.PNG - RECROSSING, UNREACTIVE
SC3916-2CONTOUR.PNG - RECROSSING, UNREACTIVE
SC3916-2CONTOUR.PNG - REACTIVE, SOME RETRACING
SC39160CONTOUR.PNG - REACTIVE, SOME RETRACING
SC3916+1.PNG - REACTIVE, NO RETRACING
SC3916+3.PNG - RECROSSING, UNREACTIVE

* For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now?

SC3916_-0.8+0.1.PNG REACTIVE, NO RETRACING

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 [[#invert momenta|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.)

The cases studied are an illustration of Polanyi's empirical rules. {{fontcolor|blue|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.}}

MRD:sc3916

2018-05-24T16:00:04Z

Sc3916:

=Molecular Reaction Dynamics Report=

==Potential Energy Surfaces==

'''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.'''

The transition structure is at a saddle point - the intersection of the minimum of a curve in one plane, and the maximum of a curve in the other. In this case, the minimum curve is on the plane corresponding to RAB=RBC, and the maximum is on the perpendicular plane. The gradient of both these curves is equal to zero.

This is different from when the potential energy surface is at a minimum - in one plane the gradient = 0, but not in the other. This allows us to distinguish between the transition structure and minima by looking at the curvature in '''both''' planes.

[[File:Saddle_Point_Visualisation_sc3916.png|thumb|none|500px|Saddle Point Visualisation]]

'''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.'''

'''rts= 0.9075 angstroms'''

To estimate this, a rough estimate was found by looking at the saddle point. At the saddle point, gradient = 0 in both planes - so if the system is at exactly this point, the atoms will remain stationary if it has no initial momentum. If it isn't at this point, it will start to oscillate - and the further away from the saddle point, the greater the oscillation. Therefore by plotting this internuclear distance vs time, and adjusting the distance by small amounts until there was very little oscillation seen, a more precise value was found.

[[File: Sc3916_1.PNG|thumb|300px|none|Internuclear Distances vs Time]]

==Reaction trajectories==

[[File:Sc3916_mep.PNG|thumb|400px|left|Minimum Energy Path from from r1 = rts+δ, r2 = rts]]

[[File:Sc3916_dynamics.PNG|thumb|400px|center|Dynamic Energy Path from r1 = rts+δ, r2 = rts]]

'''Comment on how the MEP and the trajectory you just calculated differ.'''

The minimum energy path represents an infinitely slow reaction trajectory. We start off a very small distance from the transition structure, and then the trajectory "rolls" down the potential energy surface, following where the gradient is most negative. Since velocity resets to 0 every time interval, so does momentum (p=mv). This means at every infinitesimally small step, this pathway will minimise its energy. This is not how it works in reality.

The dynamic trajectory is much closer to reality. As the trajectory starts to move down the potential energy surface it gains momentum. This means it will carry on rolling up the curve on the other side of the minimum (which increases its potential energy), leading to oscillations. This reflects the vibrational motion of atoms, which MEP does not include.

Dynamics Internuclear distance vs time plot: 
Final AB distance = 0.8 
Final BC distance = 9.0 

Dynamics Internuclear Momenta vs time plot: 
Final average momentum for BC = 2.5 
Final average momentum for AB = 1.25 

This shows the reaction path from the above calculation leads to A-B + C

If we switch around initial values, so now rAB=rts+δ and rBC=rts, the reaction path will output the same values, except the products are B-C + A.

'''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.'''

The reaction proceeds from left to right across the below plot, A-B + C -> A + B-C.
Obviously the reaction starts with enough momentum to proceed all the way past the transition state and to the products.
COMMENT ON THIS USING INSIGHT GLEANED FROM BELOW SECTION AND REDO GRAPH WITH FILLED LINE

AND LABEL GRAPH

[[File:Sc3916_dynamics2.PNG|thumb|400px|none]]

==Reactive and unreactive trajectories==
The following table is for the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, with varying momentum.

{| class="wikitable" border=1
! p1 !! p2 !! E(kcal/mol) !! Reactive?
|-
| -1.25 || -2.5 || -99.119 || Yes
|-
| -1.5 || -2.0 || -100.456 || No
|-
| -1.5 || -2.5 || -98.956 || Yes
|-
| -2.5 || -5.0 || -84.956 || No
|-
| -2.5 || -5.2 || -83.416 || Yes
|}

[[File:Sc3916_rxn1.PNG|thumb|300px|none|Initial p1=-1.25, r2=-2.5 
the trajectory goes smoothly up to the transition state, then starts oscillating as it rolls down to the products]]

[[File:Sc3916_rxn2.PNG|thumb|300px|none|Initial p1=-1.5, r2=-2.0 
the trajectory goes up towards the transition state smoothly (but not very far), and then rolls back down the way it came, with oscillating]]

[[File:Sc3916_rxn3.PNG|thumb|300px|none|Initial p1=-1.5, r2=-2.5 
the trajectory goes smoothly up to the transition state, then starts oscillating as it rolls down to the products]]

[[File:Sc3916_rxn4b.PNG|thumb|300px|none|Initial p1=-2.5, r2=-5.0 
the trajectory goes smoothly up to the transition state, crosses it, but recrosses it and reverts back to the reactants, with very large oscillations]]

[[File:Sc3916_rxn5.PNG|thumb|300px|none|Initial p1=-2.5, r2=-5.2 
the trajectory crosses the transition state, forming the products, then recrosses it to the reactant side, and then recrosses it again, forming the products finally]]

'''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?'''

4 major assumptions are used. Firstly, the Born-Oppenheimer approximation is used, and that the effects of quantum tunneling are negligible (there is a non-zero probability that reactants can convert to products without reaching the activation energy). The reactants are assumed to have a Boltzmann distribution - which is perfectly valid at low temperatures. The final assumption is that "once the system attains the transition state, with a velocity towards the product configuration, it will not reenter the initial state region again."[https://doi.org/10.1016/B978-044452837-7.50005-8 1]

This last assumption does not fit with our simulations. Both the 4th, and 5th plots contain the trajectory crossing over the transition state and back again. Since there will be more trajectories (like the 4th) which involve crossing over and are still unreactive - which are not predicted by Transition State Theory - experimental rate values will be slower than those predicted by the theory.

==EXERCISE 2: F - H - H system==

'''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 is exothermic - the products have a more negative energy than the reactants. Because the enthalpy change only involves breaking a H-H bond and making a H-F bond, ΔH = +H(H-H) - H(H-F).

Since ΔH < 0:

H(H-F)>H(H-H)

This means HF has a greater bond enthalpy (and therefore strength) than H2. It can also be seen that H + HF is endothermic, which leads to the same conclusion.

'''Locate the approximate position of the transition state.'''

rts: 
RHH = 0.745 angstroms 
RHF = 1.811 angstroms

(0.7448778, 1.810754) is the super accurate one

'''Report the activation energy for both reactions.'''

By performing an MEP from next to the transition structure, the path from the TS to the reactants, or products, was found. By subtracting the final energy from the TS, the Ea was obtained. Since the activation energy for H2 + F is very small, the MEP has a very small gradient, and as such the calculation required an order of magnitude more steps in the calculation (than HF + H) to reach the reactants.

[[File:Mep_energyvstime_sc3916.PNG|thumb|left|initial conditions: 
RHH=0.745 
RHF=1.83]]

[[File:3916_energyvstime_hf2.PNG|thumb|center|initial conditions: 
RHH=0.745 
RHF=1.8]]

Ea(H2 + F Ea) = 0.256 kcal/mol 
Ea(HF + H Ea) = 30.202 kcal/mol 

=== Reaction dynamics ===
* Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Internuclear Momenta vs Time”. {{fontcolor1|blue|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

[[File:H2+f_surface__3916.PNG|thumb|none|F + H2, initial conditions: 
RHH (AB) = 0.745 angstroms 
RHF (BC) = 2 angstroms 
AB momentum = +3 
BC momentum = -5]]

[[File:H2+F_MOMENTAVTIME3916.PNG|thumb|none|F + H2 Momentum vs Time Plot]]

The animation of this reaction clearly shows the velocity of the lone H atom formed has increased, and vibrations occurring in HF. This can be seen in the momentum vs time plot - with the increased AB and AC momentum reflecting the increase in velocity, and BC momentum oscillation reflecting the vibrations of the H-F bond. Since translational and vibrational energy contribute to kinetic energy, it is safe to assume that the energy released by this exothermic reaction (a decrease in enthalpy) is conserved in the form of increased kinetic energy. This can be seen in the below graph. An increase in average kinetic energy results in increased temperature - which is easy to measure experimentally.

[[File:H2+F_ENERGYVSTIME3916.PNG|thumb|none|F + H2 Energy vs Time Plot]]

* Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 0.74, with a momentum pFH = -0.5, and explore several values of pHH in the range -3 to 3 (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.

SC3916-3CONTOUR.PNG - RECROSSING, UNREACTIVE
SC3916-2CONTOUR.PNG - RECROSSING, UNREACTIVE
SC3916-2CONTOUR.PNG - REACTIVE, SOME RETRACING
SC39160CONTOUR.PNG - REACTIVE, SOME RETRACING
SC3916+1.PNG - REACTIVE, NO RETRACING
SC3916+3.PNG - RECROSSING, UNREACTIVE

* For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now?

SC3916_-0.8+0.1.PNG REACTIVE, NO RETRACING

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 [[#invert momenta|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.)

The cases studied are an illustration of Polanyi's empirical rules. {{fontcolor|blue|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:SC3916 -0.8+0.1.PNG

2018-05-24T15:59:15Z

Sc3916:

File:SC3916+3.PNG

2018-05-24T15:57:32Z

Sc3916:

File:SC3916+1.PNG

2018-05-24T15:57:22Z

Sc3916:

File:SC3916+3CONTOUR.PNG

2018-05-24T15:53:29Z

Sc3916:

File:SC3916+2CONTOUR.PNG

2018-05-24T15:24:52Z

Sc3916:

MRD:sc3916

2018-05-24T15:23:38Z

Sc3916: /* Reaction dynamics */

=Molecular Reaction Dynamics Report=

==Potential Energy Surfaces==

'''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.'''

The transition structure is at a saddle point - the intersection of the minimum of a curve in one plane, and the maximum of a curve in the other. In this case, the minimum curve is on the plane corresponding to RAB=RBC, and the maximum is on the perpendicular plane. The gradient of both these curves is equal to zero.

This is different from when the potential energy surface is at a minimum - in one plane the gradient = 0, but not in the other. This allows us to distinguish between the transition structure and minima by looking at the curvature in '''both''' planes.

[[File:Saddle_Point_Visualisation_sc3916.png|thumb|none|500px|Saddle Point Visualisation]]

'''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.'''

'''rts= 0.9075 angstroms'''

To estimate this, a rough estimate was found by looking at the saddle point. At the saddle point, gradient = 0 in both planes - so if the system is at exactly this point, the atoms will remain stationary if it has no initial momentum. If it isn't at this point, it will start to oscillate - and the further away from the saddle point, the greater the oscillation. Therefore by plotting this internuclear distance vs time, and adjusting the distance by small amounts until there was very little oscillation seen, a more precise value was found.

[[File: Sc3916_1.PNG|thumb|300px|none|Internuclear Distances vs Time]]

==Reaction trajectories==

[[File:Sc3916_mep.PNG|thumb|400px|left|Minimum Energy Path from from r1 = rts+δ, r2 = rts]]

[[File:Sc3916_dynamics.PNG|thumb|400px|center|Dynamic Energy Path from r1 = rts+δ, r2 = rts]]

'''Comment on how the MEP and the trajectory you just calculated differ.'''

The minimum energy path represents an infinitely slow reaction trajectory. We start off a very small distance from the transition structure, and then the trajectory "rolls" down the potential energy surface, following where the gradient is most negative. Since velocity resets to 0 every time interval, so does momentum (p=mv). This means at every infinitesimally small step, this pathway will minimise its energy. This is not how it works in reality.

The dynamic trajectory is much closer to reality. As the trajectory starts to move down the potential energy surface it gains momentum. This means it will carry on rolling up the curve on the other side of the minimum (which increases its potential energy), leading to oscillations. This reflects the vibrational motion of atoms, which MEP does not include.

Dynamics Internuclear distance vs time plot: 
Final AB distance = 0.8 
Final BC distance = 9.0 

Dynamics Internuclear Momenta vs time plot: 
Final average momentum for BC = 2.5 
Final average momentum for AB = 1.25 

This shows the reaction path from the above calculation leads to A-B + C

If we switch around initial values, so now rAB=rts+δ and rBC=rts, the reaction path will output the same values, except the products are B-C + A.

'''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.'''

The reaction proceeds from left to right across the below plot, A-B + C -> A + B-C.
Obviously the reaction starts with enough momentum to proceed all the way past the transition state and to the products.
COMMENT ON THIS USING INSIGHT GLEANED FROM BELOW SECTION AND REDO GRAPH WITH FILLED LINE

AND LABEL GRAPH

[[File:Sc3916_dynamics2.PNG|thumb|400px|none]]

==Reactive and unreactive trajectories==
The following table is for the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, with varying momentum.

{| class="wikitable" border=1
! p1 !! p2 !! E(kcal/mol) !! Reactive?
|-
| -1.25 || -2.5 || -99.119 || Yes
|-
| -1.5 || -2.0 || -100.456 || No
|-
| -1.5 || -2.5 || -98.956 || Yes
|-
| -2.5 || -5.0 || -84.956 || No
|-
| -2.5 || -5.2 || -83.416 || Yes
|}

[[File:Sc3916_rxn1.PNG|thumb|300px|none|Initial p1=-1.25, r2=-2.5 
the trajectory goes smoothly up to the transition state, then starts oscillating as it rolls down to the products]]

[[File:Sc3916_rxn2.PNG|thumb|300px|none|Initial p1=-1.5, r2=-2.0 
the trajectory goes up towards the transition state smoothly (but not very far), and then rolls back down the way it came, with oscillating]]

[[File:Sc3916_rxn3.PNG|thumb|300px|none|Initial p1=-1.5, r2=-2.5 
the trajectory goes smoothly up to the transition state, then starts oscillating as it rolls down to the products]]

[[File:Sc3916_rxn4b.PNG|thumb|300px|none|Initial p1=-2.5, r2=-5.0 
the trajectory goes smoothly up to the transition state, crosses it, but recrosses it and reverts back to the reactants, with very large oscillations]]

[[File:Sc3916_rxn5.PNG|thumb|300px|none|Initial p1=-2.5, r2=-5.2 
the trajectory crosses the transition state, forming the products, then recrosses it to the reactant side, and then recrosses it again, forming the products finally]]

'''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?'''

4 major assumptions are used. Firstly, the Born-Oppenheimer approximation is used, and that the effects of quantum tunneling are negligible (there is a non-zero probability that reactants can convert to products without reaching the activation energy). The reactants are assumed to have a Boltzmann distribution - which is perfectly valid at low temperatures. The final assumption is that "once the system attains the transition state, with a velocity towards the product configuration, it will not reenter the initial state region again."[https://doi.org/10.1016/B978-044452837-7.50005-8 1]

This last assumption does not fit with our simulations. Both the 4th, and 5th plots contain the trajectory crossing over the transition state and back again. Since there will be more trajectories (like the 4th) which involve crossing over and are still unreactive - which are not predicted by Transition State Theory - experimental rate values will be slower than those predicted by the theory.

==EXERCISE 2: F - H - H system==

'''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 is exothermic - the products have a more negative energy than the reactants. Because the enthalpy change only involves breaking a H-H bond and making a H-F bond, ΔH = +H(H-H) - H(H-F).

Since ΔH < 0:

H(H-F)>H(H-H)

This means HF has a greater bond enthalpy (and therefore strength) than H2. It can also be seen that H + HF is endothermic, which leads to the same conclusion.

'''Locate the approximate position of the transition state.'''

rts: 
RHH = 0.745 angstroms 
RHF = 1.811 angstroms

(0.7448778, 1.810754) is the super accurate one

'''Report the activation energy for both reactions.'''

By performing an MEP from next to the transition structure, the path from the TS to the reactants, or products, was found. By subtracting the final energy from the TS, the Ea was obtained. Since the activation energy for H2 + F is very small, the MEP has a very small gradient, and as such the calculation required an order of magnitude more steps in the calculation (than HF + H) to reach the reactants.

[[File:Mep_energyvstime_sc3916.PNG|thumb|left|initial conditions: 
RHH=0.745 
RHF=1.83]]

[[File:3916_energyvstime_hf2.PNG|thumb|center|initial conditions: 
RHH=0.745 
RHF=1.8]]

Ea(H2 + F Ea) = 0.256 kcal/mol 
Ea(HF + H Ea) = 30.202 kcal/mol 

=== Reaction dynamics ===
* Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Internuclear Momenta vs Time”. {{fontcolor1|blue|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

[[File:H2+f_surface__3916.PNG|thumb|none|F + H2, initial conditions: 
RHH (AB) = 0.745 angstroms 
RHF (BC) = 2 angstroms 
AB momentum = +3 
BC momentum = -5]]

[[File:H2+F_MOMENTAVTIME3916.PNG|thumb|none|F + H2 Momentum vs Time Plot]]

The animation of this reaction clearly shows the velocity of the lone H atom formed has increased, and vibrations occurring in HF. This can be seen in the momentum vs time plot - with the increased AB and AC momentum reflecting the increase in velocity, and BC momentum oscillation reflecting the vibrations of the H-F bond. Since translational and vibrational energy contribute to kinetic energy, it is safe to assume that the energy released by this exothermic reaction (a decrease in enthalpy) is conserved in the form of increased kinetic energy. This can be seen in the below graph. An increase in average kinetic energy results in increased temperature - which is easy to measure experimentally.

[[File:H2+F_ENERGYVSTIME3916.PNG|thumb|none|F + H2 Energy vs Time Plot]]

* Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 0.74, with a momentum pFH = -0.5, and explore several values of pHH in the range -3 to 3 (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.

SC3916-3CONTOUR.PNG - RECROSSING, UNREACTIVE
SC3916-2CONTOUR.PNG - RECROSSING, UNREACTIVE
SC3916-2CONTOUR.PNG - REACTIVE, SOME RETRACING
SC39160CONTOUR.PNG - REACTIVE, SOME RETRACING
SC3916+1CONTOUR.PNG - REACTIVE, NO RETRACING

* For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now?

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 [[#invert momenta|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.)

The cases studied are an illustration of Polanyi's empirical rules. {{fontcolor|blue|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:SC3916+1CONTOUR.PNG

2018-05-24T15:22:38Z

Sc3916:

File:SC39160CONTOUR.PNG

2018-05-24T15:20:36Z

Sc3916:

File:SC3916-1.5CONTOUR.PNG

2018-05-24T15:18:00Z

Sc3916:

File:SC3916-2CONTOUR.PNG

2018-05-24T15:17:07Z

Sc3916:

File:SC3916-2.5CONTOUR.PNG

2018-05-24T15:14:52Z

Sc3916:

MRD:sc3916

2018-05-24T15:11:28Z

Sc3916: /* Reaction trajectories */

=Molecular Reaction Dynamics Report=

==Potential Energy Surfaces==

'''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.'''

The transition structure is at a saddle point - the intersection of the minimum of a curve in one plane, and the maximum of a curve in the other. In this case, the minimum curve is on the plane corresponding to RAB=RBC, and the maximum is on the perpendicular plane. The gradient of both these curves is equal to zero.

This is different from when the potential energy surface is at a minimum - in one plane the gradient = 0, but not in the other. This allows us to distinguish between the transition structure and minima by looking at the curvature in '''both''' planes.

[[File:Saddle_Point_Visualisation_sc3916.png|thumb|none|500px|Saddle Point Visualisation]]

'''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.'''

'''rts= 0.9075 angstroms'''

To estimate this, a rough estimate was found by looking at the saddle point. At the saddle point, gradient = 0 in both planes - so if the system is at exactly this point, the atoms will remain stationary if it has no initial momentum. If it isn't at this point, it will start to oscillate - and the further away from the saddle point, the greater the oscillation. Therefore by plotting this internuclear distance vs time, and adjusting the distance by small amounts until there was very little oscillation seen, a more precise value was found.

[[File: Sc3916_1.PNG|thumb|300px|none|Internuclear Distances vs Time]]

==Reaction trajectories==

[[File:Sc3916_mep.PNG|thumb|400px|left|Minimum Energy Path from from r1 = rts+δ, r2 = rts]]

[[File:Sc3916_dynamics.PNG|thumb|400px|center|Dynamic Energy Path from r1 = rts+δ, r2 = rts]]

'''Comment on how the MEP and the trajectory you just calculated differ.'''

The minimum energy path represents an infinitely slow reaction trajectory. We start off a very small distance from the transition structure, and then the trajectory "rolls" down the potential energy surface, following where the gradient is most negative. Since velocity resets to 0 every time interval, so does momentum (p=mv). This means at every infinitesimally small step, this pathway will minimise its energy. This is not how it works in reality.

The dynamic trajectory is much closer to reality. As the trajectory starts to move down the potential energy surface it gains momentum. This means it will carry on rolling up the curve on the other side of the minimum (which increases its potential energy), leading to oscillations. This reflects the vibrational motion of atoms, which MEP does not include.

Dynamics Internuclear distance vs time plot: 
Final AB distance = 0.8 
Final BC distance = 9.0 

Dynamics Internuclear Momenta vs time plot: 
Final average momentum for BC = 2.5 
Final average momentum for AB = 1.25 

This shows the reaction path from the above calculation leads to A-B + C

If we switch around initial values, so now rAB=rts+δ and rBC=rts, the reaction path will output the same values, except the products are B-C + A.

'''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.'''

The reaction proceeds from left to right across the below plot, A-B + C -> A + B-C.
Obviously the reaction starts with enough momentum to proceed all the way past the transition state and to the products.
COMMENT ON THIS USING INSIGHT GLEANED FROM BELOW SECTION AND REDO GRAPH WITH FILLED LINE

AND LABEL GRAPH

[[File:Sc3916_dynamics2.PNG|thumb|400px|none]]

==Reactive and unreactive trajectories==
The following table is for the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, with varying momentum.

{| class="wikitable" border=1
! p1 !! p2 !! E(kcal/mol) !! Reactive?
|-
| -1.25 || -2.5 || -99.119 || Yes
|-
| -1.5 || -2.0 || -100.456 || No
|-
| -1.5 || -2.5 || -98.956 || Yes
|-
| -2.5 || -5.0 || -84.956 || No
|-
| -2.5 || -5.2 || -83.416 || Yes
|}

[[File:Sc3916_rxn1.PNG|thumb|300px|none|Initial p1=-1.25, r2=-2.5 
the trajectory goes smoothly up to the transition state, then starts oscillating as it rolls down to the products]]

[[File:Sc3916_rxn2.PNG|thumb|300px|none|Initial p1=-1.5, r2=-2.0 
the trajectory goes up towards the transition state smoothly (but not very far), and then rolls back down the way it came, with oscillating]]

[[File:Sc3916_rxn3.PNG|thumb|300px|none|Initial p1=-1.5, r2=-2.5 
the trajectory goes smoothly up to the transition state, then starts oscillating as it rolls down to the products]]

[[File:Sc3916_rxn4b.PNG|thumb|300px|none|Initial p1=-2.5, r2=-5.0 
the trajectory goes smoothly up to the transition state, crosses it, but recrosses it and reverts back to the reactants, with very large oscillations]]

[[File:Sc3916_rxn5.PNG|thumb|300px|none|Initial p1=-2.5, r2=-5.2 
the trajectory crosses the transition state, forming the products, then recrosses it to the reactant side, and then recrosses it again, forming the products finally]]

'''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?'''

4 major assumptions are used. Firstly, the Born-Oppenheimer approximation is used, and that the effects of quantum tunneling are negligible (there is a non-zero probability that reactants can convert to products without reaching the activation energy). The reactants are assumed to have a Boltzmann distribution - which is perfectly valid at low temperatures. The final assumption is that "once the system attains the transition state, with a velocity towards the product configuration, it will not reenter the initial state region again."[https://doi.org/10.1016/B978-044452837-7.50005-8 1]

This last assumption does not fit with our simulations. Both the 4th, and 5th plots contain the trajectory crossing over the transition state and back again. Since there will be more trajectories (like the 4th) which involve crossing over and are still unreactive - which are not predicted by Transition State Theory - experimental rate values will be slower than those predicted by the theory.

==EXERCISE 2: F - H - H system==

'''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 is exothermic - the products have a more negative energy than the reactants. Because the enthalpy change only involves breaking a H-H bond and making a H-F bond, ΔH = +H(H-H) - H(H-F).

Since ΔH < 0:

H(H-F)>H(H-H)

This means HF has a greater bond enthalpy (and therefore strength) than H2. It can also be seen that H + HF is endothermic, which leads to the same conclusion.

'''Locate the approximate position of the transition state.'''

rts: 
RHH = 0.745 angstroms 
RHF = 1.811 angstroms

(0.7448778, 1.810754) is the super accurate one

'''Report the activation energy for both reactions.'''

By performing an MEP from next to the transition structure, the path from the TS to the reactants, or products, was found. By subtracting the final energy from the TS, the Ea was obtained. Since the activation energy for H2 + F is very small, the MEP has a very small gradient, and as such the calculation required an order of magnitude more steps in the calculation (than HF + H) to reach the reactants.

[[File:Mep_energyvstime_sc3916.PNG|thumb|left|initial conditions: 
RHH=0.745 
RHF=1.83]]

[[File:3916_energyvstime_hf2.PNG|thumb|center|initial conditions: 
RHH=0.745 
RHF=1.8]]

Ea(H2 + F Ea) = 0.256 kcal/mol 
Ea(HF + H Ea) = 30.202 kcal/mol 

=== Reaction dynamics ===
* Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Internuclear Momenta vs Time”. {{fontcolor1|blue|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

[[File:H2+f_surface__3916.PNG|thumb|none|F + H2, initial conditions: 
RHH (AB) = 0.745 angstroms 
RHF (BC) = 2 angstroms 
AB momentum = +3 
BC momentum = -5]]

[[File:H2+F_MOMENTAVTIME3916.PNG|thumb|none|F + H2 Momentum vs Time Plot]]

The animation of this reaction clearly shows the velocity of the lone H atom formed has increased, and vibrations occurring in HF. This can be seen in the momentum vs time plot - with the increased AB and AC momentum reflecting the increase in velocity, and BC momentum oscillation reflecting the vibrations of the H-F bond. Since translational and vibrational energy contribute to kinetic energy, it is safe to assume that the energy released by this exothermic reaction (a decrease in enthalpy) is conserved in the form of increased kinetic energy. This can be seen in the below graph. An increase in average kinetic energy results in increased temperature - which is easy to measure experimentally.

[[File:H2+F_ENERGYVSTIME3916.PNG|thumb|none|F + H2 Energy vs Time Plot]]

* Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 0.74, with a momentum pFH = -0.5, and explore several values of pHH in the range -3 to 3 (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.

* For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now?

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 [[#invert momenta|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.)

The cases studied are an illustration of Polanyi's empirical rules. {{fontcolor|blue|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.}}

MRD:sc3916

2018-05-24T15:09:47Z

Sc3916: /* EXERCISE 2: F - H - H system */

=Molecular Reaction Dynamics Report=

==Potential Energy Surfaces==

'''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.'''

The transition structure is at a saddle point - the intersection of the minimum of a curve in one plane, and the maximum of a curve in the other. In this case, the minimum curve is on the plane corresponding to RAB=RBC, and the maximum is on the perpendicular plane. The gradient of both these curves is equal to zero.

This is different from when the potential energy surface is at a minimum - in one plane the gradient = 0, but not in the other. This allows us to distinguish between the transition structure and minima by looking at the curvature in '''both''' planes.

[[File:Saddle_Point_Visualisation_sc3916.png|thumb|none|500px|Saddle Point Visualisation]]

'''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.'''

'''rts= 0.9075 angstroms'''

To estimate this, a rough estimate was found by looking at the saddle point. At the saddle point, gradient = 0 in both planes - so if the system is at exactly this point, the atoms will remain stationary if it has no initial momentum. If it isn't at this point, it will start to oscillate - and the further away from the saddle point, the greater the oscillation. Therefore by plotting this internuclear distance vs time, and adjusting the distance by small amounts until there was very little oscillation seen, a more precise value was found.

[[File: Sc3916_1.PNG|thumb|300px|none|Internuclear Distances vs Time]]

==Reaction trajectories==

[[File:Sc3916_mep.PNG|thumb|400px|left|Minimum Energy Path from from r1 = rts+δ, r2 = rts]]

[[File:Sc3916_dynamics.PNG|thumb|400px|center|Dynamic Energy Path from r1 = rts+δ, r2 = rts]]

'''Comment on how the MEP and the trajectory you just calculated differ.'''

The minimum energy path represents an infinitely slow reaction trajectory. We start off a very small distance from the transition structure, and then the trajectory "rolls" down the potential energy surface, following where the gradient is most negative. Since velocity resets to 0 every time interval, so does momentum (p=mv). This means at every infinitesimally small step, this pathway will minimise its energy. This is not how it works in reality.

The dynamic trajectory is much closer to reality. As the trajectory starts to move down the potential energy surface it gains momentum. This means it will carry on rolling up the curve on the other side of the minimum (which increases its potential energy), leading to oscillations. This reflects the vibrational motion of atoms, which MEP does not include.

Dynamics Internuclear distance vs time plot: 
Final AB distance = 0.8 
Final BC distance = 9.0 

Dynamics Internuclear Momenta vs time plot: 
Final average momentum for BC = 2.5 
Final average momentum for AB = 1.25 

This shows the reaction path from the above calculation leads to A-B + C

If we switch around initial values, so now rAB=rts+δ and rBC=rts, the reaction path will output the same values, except the products are B-C + A.

'''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.'''

The reaction proceeds from left to right across the below plot, A-B + C -> A + B-C.
Obviously the reaction starts with enough momentum to proceed all the way past the transition state and to the products.
COMMENT ON THIS USING INSIGHT GLEANED FROM BELOW SECTION AND REDO GRAPH WITH FILLED LINE

[[File:Sc3916_dynamics2.PNG|thumb|400px|none]]

==Reactive and unreactive trajectories==
The following table is for the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, with varying momentum.

{| class="wikitable" border=1
! p1 !! p2 !! E(kcal/mol) !! Reactive?
|-
| -1.25 || -2.5 || -99.119 || Yes
|-
| -1.5 || -2.0 || -100.456 || No
|-
| -1.5 || -2.5 || -98.956 || Yes
|-
| -2.5 || -5.0 || -84.956 || No
|-
| -2.5 || -5.2 || -83.416 || Yes
|}

[[File:Sc3916_rxn1.PNG|thumb|300px|none|Initial p1=-1.25, r2=-2.5 
the trajectory goes smoothly up to the transition state, then starts oscillating as it rolls down to the products]]

[[File:Sc3916_rxn2.PNG|thumb|300px|none|Initial p1=-1.5, r2=-2.0 
the trajectory goes up towards the transition state smoothly (but not very far), and then rolls back down the way it came, with oscillating]]

[[File:Sc3916_rxn3.PNG|thumb|300px|none|Initial p1=-1.5, r2=-2.5 
the trajectory goes smoothly up to the transition state, then starts oscillating as it rolls down to the products]]

[[File:Sc3916_rxn4b.PNG|thumb|300px|none|Initial p1=-2.5, r2=-5.0 
the trajectory goes smoothly up to the transition state, crosses it, but recrosses it and reverts back to the reactants, with very large oscillations]]

[[File:Sc3916_rxn5.PNG|thumb|300px|none|Initial p1=-2.5, r2=-5.2 
the trajectory crosses the transition state, forming the products, then recrosses it to the reactant side, and then recrosses it again, forming the products finally]]

'''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?'''

4 major assumptions are used. Firstly, the Born-Oppenheimer approximation is used, and that the effects of quantum tunneling are negligible (there is a non-zero probability that reactants can convert to products without reaching the activation energy). The reactants are assumed to have a Boltzmann distribution - which is perfectly valid at low temperatures. The final assumption is that "once the system attains the transition state, with a velocity towards the product configuration, it will not reenter the initial state region again."[https://doi.org/10.1016/B978-044452837-7.50005-8 1]

This last assumption does not fit with our simulations. Both the 4th, and 5th plots contain the trajectory crossing over the transition state and back again. Since there will be more trajectories (like the 4th) which involve crossing over and are still unreactive - which are not predicted by Transition State Theory - experimental rate values will be slower than those predicted by the theory.

==EXERCISE 2: F - H - H system==

'''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 is exothermic - the products have a more negative energy than the reactants. Because the enthalpy change only involves breaking a H-H bond and making a H-F bond, ΔH = +H(H-H) - H(H-F).

Since ΔH < 0:

H(H-F)>H(H-H)

This means HF has a greater bond enthalpy (and therefore strength) than H2. It can also be seen that H + HF is endothermic, which leads to the same conclusion.

'''Locate the approximate position of the transition state.'''

rts: 
RHH = 0.745 angstroms 
RHF = 1.811 angstroms

(0.7448778, 1.810754) is the super accurate one

'''Report the activation energy for both reactions.'''

By performing an MEP from next to the transition structure, the path from the TS to the reactants, or products, was found. By subtracting the final energy from the TS, the Ea was obtained. Since the activation energy for H2 + F is very small, the MEP has a very small gradient, and as such the calculation required an order of magnitude more steps in the calculation (than HF + H) to reach the reactants.

[[File:Mep_energyvstime_sc3916.PNG|thumb|left|initial conditions: 
RHH=0.745 
RHF=1.83]]

[[File:3916_energyvstime_hf2.PNG|thumb|center|initial conditions: 
RHH=0.745 
RHF=1.8]]

Ea(H2 + F Ea) = 0.256 kcal/mol 
Ea(HF + H Ea) = 30.202 kcal/mol 

=== Reaction dynamics ===
* Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Internuclear Momenta vs Time”. {{fontcolor1|blue|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

[[File:H2+f_surface__3916.PNG|thumb|none|F + H2, initial conditions: 
RHH (AB) = 0.745 angstroms 
RHF (BC) = 2 angstroms 
AB momentum = +3 
BC momentum = -5]]

[[File:H2+F_MOMENTAVTIME3916.PNG|thumb|none|F + H2 Momentum vs Time Plot]]

The animation of this reaction clearly shows the velocity of the lone H atom formed has increased, and vibrations occurring in HF. This can be seen in the momentum vs time plot - with the increased AB and AC momentum reflecting the increase in velocity, and BC momentum oscillation reflecting the vibrations of the H-F bond. Since translational and vibrational energy contribute to kinetic energy, it is safe to assume that the energy released by this exothermic reaction (a decrease in enthalpy) is conserved in the form of increased kinetic energy. This can be seen in the below graph. An increase in average kinetic energy results in increased temperature - which is easy to measure experimentally.

[[File:H2+F_ENERGYVSTIME3916.PNG|thumb|none|F + H2 Energy vs Time Plot]]

* Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 0.74, with a momentum pFH = -0.5, and explore several values of pHH in the range -3 to 3 (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.

* For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now?

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 [[#invert momenta|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.)

The cases studied are an illustration of Polanyi's empirical rules. {{fontcolor|blue|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:3916 energyvstime hf2.PNG

2018-05-24T15:08:48Z

Sc3916:

MRD:sc3916

2018-05-24T15:07:42Z

Sc3916: /* EXERCISE 2: F - H - H system */

=Molecular Reaction Dynamics Report=

==Potential Energy Surfaces==

'''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.'''

The transition structure is at a saddle point - the intersection of the minimum of a curve in one plane, and the maximum of a curve in the other. In this case, the minimum curve is on the plane corresponding to RAB=RBC, and the maximum is on the perpendicular plane. The gradient of both these curves is equal to zero.

This is different from when the potential energy surface is at a minimum - in one plane the gradient = 0, but not in the other. This allows us to distinguish between the transition structure and minima by looking at the curvature in '''both''' planes.

[[File:Saddle_Point_Visualisation_sc3916.png|thumb|none|500px|Saddle Point Visualisation]]

'''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.'''

'''rts= 0.9075 angstroms'''

To estimate this, a rough estimate was found by looking at the saddle point. At the saddle point, gradient = 0 in both planes - so if the system is at exactly this point, the atoms will remain stationary if it has no initial momentum. If it isn't at this point, it will start to oscillate - and the further away from the saddle point, the greater the oscillation. Therefore by plotting this internuclear distance vs time, and adjusting the distance by small amounts until there was very little oscillation seen, a more precise value was found.

[[File: Sc3916_1.PNG|thumb|300px|none|Internuclear Distances vs Time]]

==Reaction trajectories==

[[File:Sc3916_mep.PNG|thumb|400px|left|Minimum Energy Path from from r1 = rts+δ, r2 = rts]]

[[File:Sc3916_dynamics.PNG|thumb|400px|center|Dynamic Energy Path from r1 = rts+δ, r2 = rts]]

'''Comment on how the MEP and the trajectory you just calculated differ.'''

The minimum energy path represents an infinitely slow reaction trajectory. We start off a very small distance from the transition structure, and then the trajectory "rolls" down the potential energy surface, following where the gradient is most negative. Since velocity resets to 0 every time interval, so does momentum (p=mv). This means at every infinitesimally small step, this pathway will minimise its energy. This is not how it works in reality.

The dynamic trajectory is much closer to reality. As the trajectory starts to move down the potential energy surface it gains momentum. This means it will carry on rolling up the curve on the other side of the minimum (which increases its potential energy), leading to oscillations. This reflects the vibrational motion of atoms, which MEP does not include.

Dynamics Internuclear distance vs time plot: 
Final AB distance = 0.8 
Final BC distance = 9.0 

Dynamics Internuclear Momenta vs time plot: 
Final average momentum for BC = 2.5 
Final average momentum for AB = 1.25 

This shows the reaction path from the above calculation leads to A-B + C

If we switch around initial values, so now rAB=rts+δ and rBC=rts, the reaction path will output the same values, except the products are B-C + A.

'''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.'''

The reaction proceeds from left to right across the below plot, A-B + C -> A + B-C.
Obviously the reaction starts with enough momentum to proceed all the way past the transition state and to the products.
COMMENT ON THIS USING INSIGHT GLEANED FROM BELOW SECTION AND REDO GRAPH WITH FILLED LINE

[[File:Sc3916_dynamics2.PNG|thumb|400px|none]]

==Reactive and unreactive trajectories==
The following table is for the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, with varying momentum.

{| class="wikitable" border=1
! p1 !! p2 !! E(kcal/mol) !! Reactive?
|-
| -1.25 || -2.5 || -99.119 || Yes
|-
| -1.5 || -2.0 || -100.456 || No
|-
| -1.5 || -2.5 || -98.956 || Yes
|-
| -2.5 || -5.0 || -84.956 || No
|-
| -2.5 || -5.2 || -83.416 || Yes
|}

[[File:Sc3916_rxn1.PNG|thumb|300px|none|Initial p1=-1.25, r2=-2.5 
the trajectory goes smoothly up to the transition state, then starts oscillating as it rolls down to the products]]

[[File:Sc3916_rxn2.PNG|thumb|300px|none|Initial p1=-1.5, r2=-2.0 
the trajectory goes up towards the transition state smoothly (but not very far), and then rolls back down the way it came, with oscillating]]

[[File:Sc3916_rxn3.PNG|thumb|300px|none|Initial p1=-1.5, r2=-2.5 
the trajectory goes smoothly up to the transition state, then starts oscillating as it rolls down to the products]]

[[File:Sc3916_rxn4b.PNG|thumb|300px|none|Initial p1=-2.5, r2=-5.0 
the trajectory goes smoothly up to the transition state, crosses it, but recrosses it and reverts back to the reactants, with very large oscillations]]

[[File:Sc3916_rxn5.PNG|thumb|300px|none|Initial p1=-2.5, r2=-5.2 
the trajectory crosses the transition state, forming the products, then recrosses it to the reactant side, and then recrosses it again, forming the products finally]]

'''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?'''

4 major assumptions are used. Firstly, the Born-Oppenheimer approximation is used, and that the effects of quantum tunneling are negligible (there is a non-zero probability that reactants can convert to products without reaching the activation energy). The reactants are assumed to have a Boltzmann distribution - which is perfectly valid at low temperatures. The final assumption is that "once the system attains the transition state, with a velocity towards the product configuration, it will not reenter the initial state region again."[https://doi.org/10.1016/B978-044452837-7.50005-8 1]

This last assumption does not fit with our simulations. Both the 4th, and 5th plots contain the trajectory crossing over the transition state and back again. Since there will be more trajectories (like the 4th) which involve crossing over and are still unreactive - which are not predicted by Transition State Theory - experimental rate values will be slower than those predicted by the theory.

==EXERCISE 2: F - H - H system==

'''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 is exothermic - the products have a more negative energy than the reactants. Because the enthalpy change only involves breaking a H-H bond and making a H-F bond, ΔH = +H(H-H) - H(H-F).

Since ΔH < 0:

H(H-F)>H(H-H)

This means HF has a greater bond enthalpy (and therefore strength) than H2. It can also be seen that H + HF is endothermic, which leads to the same conclusion.

'''Locate the approximate position of the transition state.'''

rts: 
RHH = 0.745 angstroms 
RHF = 1.811 angstroms

(0.7448778, 1.810754) is the super accurate one

'''Report the activation energy for both reactions.'''

By performing an MEP from next to the transition structure, the path from the TS to the reactants, or products, was found. By subtracting the final energy from the TS, the Ea was obtained. Since the activation energy for H2 + F is very small, the MEP has a very small gradient, and as such the calculation required an order of magnitude more steps in the calculation (than HF + H) to reach the reactants.

-103.752 TS

[[File:Mep_energyvstime_sc3916.PNG|thumb|left|initial conditions: 
RHH=0.745 
RHF=1.83]]

[[3916_energyvstime_hf.PNG|thumb|center|initial conditions: 
RHH=0.745 
RHF=1.8]]

Ea(H2 + F Ea) = 0.256 kcal/mol 
Ea(HF + H Ea) = 30.202 kcal/mol 

=== Reaction dynamics ===
* Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Internuclear Momenta vs Time”. {{fontcolor1|blue|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

[[File:H2+f_surface__3916.PNG|thumb|none|F + H2, initial conditions: 
RHH (AB) = 0.745 angstroms 
RHF (BC) = 2 angstroms 
AB momentum = +3 
BC momentum = -5]]

[[File:H2+F_MOMENTAVTIME3916.PNG|thumb|none|F + H2 Momentum vs Time Plot]]

The animation of this reaction clearly shows the velocity of the lone H atom formed has increased, and vibrations occurring in HF. This can be seen in the momentum vs time plot - with the increased AB and AC momentum reflecting the increase in velocity, and BC momentum oscillation reflecting the vibrations of the H-F bond. Since translational and vibrational energy contribute to kinetic energy, it is safe to assume that the energy released by this exothermic reaction (a decrease in enthalpy) is conserved in the form of increased kinetic energy. This can be seen in the below graph. An increase in average kinetic energy results in increased temperature - which is easy to measure experimentally.

[[File:H2+F_ENERGYVSTIME3916.PNG|thumb|none|F + H2 Energy vs Time Plot]]

* Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 0.74, with a momentum pFH = -0.5, and explore several values of pHH in the range -3 to 3 (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.

* For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now?

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 [[#invert momenta|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.)

The cases studied are an illustration of Polanyi's empirical rules. {{fontcolor|blue|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.}}

MRD:sc3916

2018-05-24T15:03:51Z

Sc3916: /* Reaction trajectories */

=Molecular Reaction Dynamics Report=

==Potential Energy Surfaces==

'''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.'''

The transition structure is at a saddle point - the intersection of the minimum of a curve in one plane, and the maximum of a curve in the other. In this case, the minimum curve is on the plane corresponding to RAB=RBC, and the maximum is on the perpendicular plane. The gradient of both these curves is equal to zero.

This is different from when the potential energy surface is at a minimum - in one plane the gradient = 0, but not in the other. This allows us to distinguish between the transition structure and minima by looking at the curvature in '''both''' planes.

[[File:Saddle_Point_Visualisation_sc3916.png|thumb|none|500px|Saddle Point Visualisation]]

'''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.'''

'''rts= 0.9075 angstroms'''

To estimate this, a rough estimate was found by looking at the saddle point. At the saddle point, gradient = 0 in both planes - so if the system is at exactly this point, the atoms will remain stationary if it has no initial momentum. If it isn't at this point, it will start to oscillate - and the further away from the saddle point, the greater the oscillation. Therefore by plotting this internuclear distance vs time, and adjusting the distance by small amounts until there was very little oscillation seen, a more precise value was found.

[[File: Sc3916_1.PNG|thumb|300px|none|Internuclear Distances vs Time]]

==Reaction trajectories==

[[File:Sc3916_mep.PNG|thumb|400px|left|Minimum Energy Path from from r1 = rts+δ, r2 = rts]]

[[File:Sc3916_dynamics.PNG|thumb|400px|center|Dynamic Energy Path from r1 = rts+δ, r2 = rts]]

'''Comment on how the MEP and the trajectory you just calculated differ.'''

The minimum energy path represents an infinitely slow reaction trajectory. We start off a very small distance from the transition structure, and then the trajectory "rolls" down the potential energy surface, following where the gradient is most negative. Since velocity resets to 0 every time interval, so does momentum (p=mv). This means at every infinitesimally small step, this pathway will minimise its energy. This is not how it works in reality.

The dynamic trajectory is much closer to reality. As the trajectory starts to move down the potential energy surface it gains momentum. This means it will carry on rolling up the curve on the other side of the minimum (which increases its potential energy), leading to oscillations. This reflects the vibrational motion of atoms, which MEP does not include.

Dynamics Internuclear distance vs time plot: 
Final AB distance = 0.8 
Final BC distance = 9.0 

Dynamics Internuclear Momenta vs time plot: 
Final average momentum for BC = 2.5 
Final average momentum for AB = 1.25 

This shows the reaction path from the above calculation leads to A-B + C

If we switch around initial values, so now rAB=rts+δ and rBC=rts, the reaction path will output the same values, except the products are B-C + A.

'''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.'''

The reaction proceeds from left to right across the below plot, A-B + C -> A + B-C.
Obviously the reaction starts with enough momentum to proceed all the way past the transition state and to the products.
COMMENT ON THIS USING INSIGHT GLEANED FROM BELOW SECTION AND REDO GRAPH WITH FILLED LINE

[[File:Sc3916_dynamics2.PNG|thumb|400px|none]]

==Reactive and unreactive trajectories==
The following table is for the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, with varying momentum.

{| class="wikitable" border=1
! p1 !! p2 !! E(kcal/mol) !! Reactive?
|-
| -1.25 || -2.5 || -99.119 || Yes
|-
| -1.5 || -2.0 || -100.456 || No
|-
| -1.5 || -2.5 || -98.956 || Yes
|-
| -2.5 || -5.0 || -84.956 || No
|-
| -2.5 || -5.2 || -83.416 || Yes
|}

[[File:Sc3916_rxn1.PNG|thumb|300px|none|Initial p1=-1.25, r2=-2.5 
the trajectory goes smoothly up to the transition state, then starts oscillating as it rolls down to the products]]

[[File:Sc3916_rxn2.PNG|thumb|300px|none|Initial p1=-1.5, r2=-2.0 
the trajectory goes up towards the transition state smoothly (but not very far), and then rolls back down the way it came, with oscillating]]

[[File:Sc3916_rxn3.PNG|thumb|300px|none|Initial p1=-1.5, r2=-2.5 
the trajectory goes smoothly up to the transition state, then starts oscillating as it rolls down to the products]]

[[File:Sc3916_rxn4b.PNG|thumb|300px|none|Initial p1=-2.5, r2=-5.0 
the trajectory goes smoothly up to the transition state, crosses it, but recrosses it and reverts back to the reactants, with very large oscillations]]

[[File:Sc3916_rxn5.PNG|thumb|300px|none|Initial p1=-2.5, r2=-5.2 
the trajectory crosses the transition state, forming the products, then recrosses it to the reactant side, and then recrosses it again, forming the products finally]]

'''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?'''

4 major assumptions are used. Firstly, the Born-Oppenheimer approximation is used, and that the effects of quantum tunneling are negligible (there is a non-zero probability that reactants can convert to products without reaching the activation energy). The reactants are assumed to have a Boltzmann distribution - which is perfectly valid at low temperatures. The final assumption is that "once the system attains the transition state, with a velocity towards the product configuration, it will not reenter the initial state region again."[https://doi.org/10.1016/B978-044452837-7.50005-8 1]

This last assumption does not fit with our simulations. Both the 4th, and 5th plots contain the trajectory crossing over the transition state and back again. Since there will be more trajectories (like the 4th) which involve crossing over and are still unreactive - which are not predicted by Transition State Theory - experimental rate values will be slower than those predicted by the theory.

==EXERCISE 2: F - H - H system==

'''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 is exothermic - the products have a more negative energy than the reactants. Because the enthalpy change only involves breaking a H-H bond and making a H-F bond, ΔH = +H(H-H) - H(H-F).

Since ΔH < 0:

H(H-F)>H(H-H)

This means HF has a greater bond enthalpy (and therefore strength) than H2. It can also be seen that H + HF is endothermic, which leads to the same conclusion.

'''Locate the approximate position of the transition state.'''

rts: 
RHH = 0.745 angstroms 
RHF = 1.811 angstroms

(0.7448778, 1.810754) is the super accurate one

'''Report the activation energy for both reactions.'''

By performing an MEP from next to the transition structure, the path from the TS to the reactants, or products, was found. By subtracting the final energy from the TS, the Ea was obtained. Since the activation energy for H2 + F is very small, the MEP has a very small gradient, and as such the calculation required an order of magnitude more steps in the calculation (than HF + H) to reach the reactants.

-103.752 TS

[[File:Mep_energyvstime_sc3916.PNG|thumb|left|initial conditions: 
RHH=0.745 
RHF=1.83]]

[[3916_energyvstime_hf.PNG|thumb|center|initial conditions: 
RHH=0.745 
RHF=1.8]]

Ea(H2 + F Ea) = kcal/mol 
Ea(HF + H Ea) = 30.20 kcal/mol 

=== Reaction dynamics ===
* Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Internuclear Momenta vs Time”. {{fontcolor1|blue|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

[[File:H2+f_surface__3916.PNG|thumb|none|F + H2, initial conditions: 
RHH (AB) = 0.745 angstroms 
RHF (BC) = 2 angstroms 
AB momentum = +3 
BC momentum = -5]]

[[File:H2+F_MOMENTAVTIME3916.PNG|thumb|none|F + H2 Momentum vs Time Plot]]

The animation of this reaction clearly shows the velocity of the lone H atom formed has increased, and vibrations occurring in HF. This can be seen in the momentum vs time plot - with the increased AB and AC momentum reflecting the increase in velocity, and BC momentum oscillation reflecting the vibrations of the H-F bond. Since translational and vibrational energy contribute to kinetic energy, it is safe to assume that the energy released by this exothermic reaction (a decrease in enthalpy) is conserved in the form of increased kinetic energy. This can be seen in the below graph. An increase in average kinetic energy results in increased temperature - which is easy to measure experimentally.

[[File:H2+F_ENERGYVSTIME3916.PNG|thumb|none|F + H2 Energy vs Time Plot]]

* Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 0.74, with a momentum pFH = -0.5, and explore several values of pHH in the range -3 to 3 (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.

* For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now?

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 [[#invert momenta|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.)

The cases studied are an illustration of Polanyi's empirical rules. {{fontcolor|blue|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.}}

MRD:sc3916

2018-05-24T15:02:44Z

Sc3916:

=Molecular Reaction Dynamics Report=

==Potential Energy Surfaces==

'''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.'''

The transition structure is at a saddle point - the intersection of the minimum of a curve in one plane, and the maximum of a curve in the other. In this case, the minimum curve is on the plane corresponding to RAB=RBC, and the maximum is on the perpendicular plane. The gradient of both these curves is equal to zero.

This is different from when the potential energy surface is at a minimum - in one plane the gradient = 0, but not in the other. This allows us to distinguish between the transition structure and minima by looking at the curvature in '''both''' planes.

[[File:Saddle_Point_Visualisation_sc3916.png|thumb|none|500px|Saddle Point Visualisation]]

'''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.'''

'''rts= 0.9075 angstroms'''

To estimate this, a rough estimate was found by looking at the saddle point. At the saddle point, gradient = 0 in both planes - so if the system is at exactly this point, the atoms will remain stationary if it has no initial momentum. If it isn't at this point, it will start to oscillate - and the further away from the saddle point, the greater the oscillation. Therefore by plotting this internuclear distance vs time, and adjusting the distance by small amounts until there was very little oscillation seen, a more precise value was found.

[[File: Sc3916_1.PNG|thumb|300px|none|Internuclear Distances vs Time]]

==Reaction trajectories==

[[File:Sc3916_mep.PNG|thumb|400px|left|Minimum Energy Path from from r1 = rts+δ, r2 = rts]]

[[File:Sc3916_dynamics.PNG|thumb|400px|center|Dynamic Energy Path from r1 = rts+δ, r2 = rts]]

'''Comment on how the MEP and the trajectory you just calculated differ.'''

The minimum energy path represents an infinitely slow reaction trajectory. We start off a very small distance from the transition structure, and then the trajectory "rolls" down the potential energy surface, following where the gradient is most negative. Since velocity resets to 0 every time interval, so does momentum (p=mv). This means at every infinitesimally small step, this pathway will minimise its energy. This is not how it works in reality.

The dynamic trajectory is much closer to reality. As the trajectory starts to move down the potential energy surface it gains momentum. This means it will carry on rolling up the curve on the other side of the minimum (which increases its potential energy), leading to oscillations. This reflects the vibrational motion of atoms, which MEP does not include.

ASK IF YOU HAVE MISSED ANYTHING OUT

Dynamics Internuclear distance vs time plot: 
Final AB distance = 0.8 
Final BC distance = 9.0 

Dynamics Internuclear Momenta vs time plot: 
Final average momentum for BC = 2.5 
Final average momentum for AB = 1.25 

This shows the reaction path from the above calculation leads to A-B + C

If we switch around initial values, so now rAB=rts+δ and rBC=rts, the reaction path will output the same values, except the products are B-C + A.

'''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.'''

The reaction proceeds from left to right across the below plot, A-B + C -> A + B-C.
Obviously the reaction starts with enough momentum to proceed all the way past the transition state and to the products.
COMMENT ON THIS USING INSIGHT GLEANED FROM BELOW SECTION AND REDO GRAPH WITH FILLED LINE

[[File:Sc3916_dynamics2.PNG|thumb|400px|none]]

==Reactive and unreactive trajectories==
The following table is for the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, with varying momentum.

{| class="wikitable" border=1
! p1 !! p2 !! E(kcal/mol) !! Reactive?
|-
| -1.25 || -2.5 || -99.119 || Yes
|-
| -1.5 || -2.0 || -100.456 || No
|-
| -1.5 || -2.5 || -98.956 || Yes
|-
| -2.5 || -5.0 || -84.956 || No
|-
| -2.5 || -5.2 || -83.416 || Yes
|}

[[File:Sc3916_rxn1.PNG|thumb|300px|none|Initial p1=-1.25, r2=-2.5 
the trajectory goes smoothly up to the transition state, then starts oscillating as it rolls down to the products]]

[[File:Sc3916_rxn2.PNG|thumb|300px|none|Initial p1=-1.5, r2=-2.0 
the trajectory goes up towards the transition state smoothly (but not very far), and then rolls back down the way it came, with oscillating]]

[[File:Sc3916_rxn3.PNG|thumb|300px|none|Initial p1=-1.5, r2=-2.5 
the trajectory goes smoothly up to the transition state, then starts oscillating as it rolls down to the products]]

[[File:Sc3916_rxn4b.PNG|thumb|300px|none|Initial p1=-2.5, r2=-5.0 
the trajectory goes smoothly up to the transition state, crosses it, but recrosses it and reverts back to the reactants, with very large oscillations]]

[[File:Sc3916_rxn5.PNG|thumb|300px|none|Initial p1=-2.5, r2=-5.2 
the trajectory crosses the transition state, forming the products, then recrosses it to the reactant side, and then recrosses it again, forming the products finally]]

'''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?'''

4 major assumptions are used. Firstly, the Born-Oppenheimer approximation is used, and that the effects of quantum tunneling are negligible (there is a non-zero probability that reactants can convert to products without reaching the activation energy). The reactants are assumed to have a Boltzmann distribution - which is perfectly valid at low temperatures. The final assumption is that "once the system attains the transition state, with a velocity towards the product configuration, it will not reenter the initial state region again."[https://doi.org/10.1016/B978-044452837-7.50005-8 1]

This last assumption does not fit with our simulations. Both the 4th, and 5th plots contain the trajectory crossing over the transition state and back again. Since there will be more trajectories (like the 4th) which involve crossing over and are still unreactive - which are not predicted by Transition State Theory - experimental rate values will be slower than those predicted by the theory.

==EXERCISE 2: F - H - H system==

'''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 is exothermic - the products have a more negative energy than the reactants. Because the enthalpy change only involves breaking a H-H bond and making a H-F bond, ΔH = +H(H-H) - H(H-F).

Since ΔH < 0:

H(H-F)>H(H-H)

This means HF has a greater bond enthalpy (and therefore strength) than H2. It can also be seen that H + HF is endothermic, which leads to the same conclusion.

'''Locate the approximate position of the transition state.'''

rts: 
RHH = 0.745 angstroms 
RHF = 1.811 angstroms

(0.7448778, 1.810754) is the super accurate one

'''Report the activation energy for both reactions.'''

By performing an MEP from next to the transition structure, the path from the TS to the reactants, or products, was found. By subtracting the final energy from the TS, the Ea was obtained. Since the activation energy for H2 + F is very small, the MEP has a very small gradient, and as such the calculation required an order of magnitude more steps in the calculation (than HF + H) to reach the reactants.

-103.752 TS

[[File:Mep_energyvstime_sc3916.PNG|thumb|left|initial conditions: 
RHH=0.745 
RHF=1.83]]

[[3916_energyvstime_hf.PNG|thumb|center|initial conditions: 
RHH=0.745 
RHF=1.8]]

Ea(H2 + F Ea) = kcal/mol 
Ea(HF + H Ea) = 30.20 kcal/mol 

=== Reaction dynamics ===
* Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Internuclear Momenta vs Time”. {{fontcolor1|blue|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

[[File:H2+f_surface__3916.PNG|thumb|none|F + H2, initial conditions: 
RHH (AB) = 0.745 angstroms 
RHF (BC) = 2 angstroms 
AB momentum = +3 
BC momentum = -5]]

[[File:H2+F_MOMENTAVTIME3916.PNG|thumb|none|F + H2 Momentum vs Time Plot]]

The animation of this reaction clearly shows the velocity of the lone H atom formed has increased, and vibrations occurring in HF. This can be seen in the momentum vs time plot - with the increased AB and AC momentum reflecting the increase in velocity, and BC momentum oscillation reflecting the vibrations of the H-F bond. Since translational and vibrational energy contribute to kinetic energy, it is safe to assume that the energy released by this exothermic reaction (a decrease in enthalpy) is conserved in the form of increased kinetic energy. This can be seen in the below graph. An increase in average kinetic energy results in increased temperature - which is easy to measure experimentally.

[[File:H2+F_ENERGYVSTIME3916.PNG|thumb|none|F + H2 Energy vs Time Plot]]

* Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 0.74, with a momentum pFH = -0.5, and explore several values of pHH in the range -3 to 3 (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.

* For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now?

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 [[#invert momenta|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.)

The cases studied are an illustration of Polanyi's empirical rules. {{fontcolor|blue|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.}}

MRD:sc3916

2018-05-24T15:01:07Z

Sc3916: /* EXERCISE 2: F - H - H system */

=Molecular Reaction Dynamics Report=

==Potential Energy Surfaces==

'''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.'''

The transition structure is at a saddle point - the intersection of the minimum of a curve in one plane, and the maximum of a curve in the other. In this case, the minimum curve is on the plane corresponding to RAB=RBC, and the maximum is on the perpendicular plane. The gradient of both these curves is equal to zero.

This is different from when the potential energy surface is at a minimum - in one plane the gradient = 0, but not in the other. This allows us to distinguish between the transition structure and minima by looking at the curvature in '''both''' planes.

[[File:Saddle_Point_Visualisation_sc3916.png|thumb|none|500px|Saddle Point Visualisation]]

'''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.'''

'''rts= 0.9075 angstroms'''

To estimate this, a rough estimate was found by looking at the saddle point. At the saddle point, gradient = 0 in both planes - so if the system is at exactly this point, the atoms will remain stationary if it has no initial momentum. If it isn't at this point, it will start to oscillate - and the further away from the saddle point, the greater the oscillation. Therefore by plotting this internuclear distance vs time, and adjusting the distance by small amounts until there was very little oscillation seen, a more precise value was found.

[[File: Sc3916_1.PNG|thumb|300px|none|Internuclear Distances vs Time]]

==Reaction trajectories==

[[File:Sc3916_mep.PNG|thumb|400px|left|Minimum Energy Path from from r1 = rts+δ, r2 = rts]]

[[File:Sc3916_dynamics.PNG|thumb|400px|center|Dynamic Energy Path from r1 = rts+δ, r2 = rts]]

'''Comment on how the MEP and the trajectory you just calculated differ.'''

The minimum energy path represents an infinitely slow reaction trajectory. We start off a very small distance from the transition structure, and then the trajectory "rolls" down the potential energy surface, following where the gradient is most negative. Since velocity resets to 0 every time interval, so does momentum (p=mv). This means at every infinitesimally small step, this pathway will minimise its energy. This is not how it works in reality.

The dynamic trajectory is much closer to reality. As the trajectory starts to move down the potential energy surface it gains momentum. This means it will carry on rolling up the curve on the other side of the minimum (which increases its potential energy), leading to oscillations. This reflects the vibrational motion of atoms, which MEP does not include.

ASK IF YOU HAVE MISSED ANYTHING OUT

Dynamics Internuclear distance vs time plot: 
Final AB distance = 0.8 
Final BC distance = 9.0 

Dynamics Internuclear Momenta vs time plot: 
Final average momentum for BC = 2.5 
Final average momentum for AB = 1.25 

This shows the reaction path from the above calculation leads to A-B + C

If we switch around initial values, so now rAB=rts+δ and rBC=rts, the reaction path will output the same values, except the products are B-C + A.

'''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.'''

The reaction proceeds from left to right across the below plot, A-B + C -> A + B-C.
Obviously the reaction starts with enough momentum to proceed all the way past the transition state and to the products.
COMMENT ON THIS USING INSIGHT GLEANED FROM BELOW SECTION AND REDO GRAPH WITH FILLED LINE

[[File:Sc3916_dynamics2.PNG|thumb|400px|none]]

==Reactive and unreactive trajectories==
The following table is for the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, with varying momentum.

{| class="wikitable" border=1
! p1 !! p2 !! E(kcal/mol) !! Reactive?
|-
| -1.25 || -2.5 || -99.119 || Yes
|-
| -1.5 || -2.0 || -100.456 || No
|-
| -1.5 || -2.5 || -98.956 || Yes
|-
| -2.5 || -5.0 || -84.956 || No
|-
| -2.5 || -5.2 || -83.416 || Yes
|}

[[File:Sc3916_rxn1.PNG|thumb|300px|none|Initial p1=-1.25, r2=-2.5 
the trajectory goes smoothly up to the transition state, then starts oscillating as it rolls down to the products]]

[[File:Sc3916_rxn2.PNG|thumb|300px|none|Initial p1=-1.5, r2=-2.0 
the trajectory goes up towards the transition state smoothly (but not very far), and then rolls back down the way it came, with oscillating]]

[[File:Sc3916_rxn3.PNG|thumb|300px|none|Initial p1=-1.5, r2=-2.5 
the trajectory goes smoothly up to the transition state, then starts oscillating as it rolls down to the products]]

[[File:Sc3916_rxn4b.PNG|thumb|300px|none|Initial p1=-2.5, r2=-5.0 
the trajectory goes smoothly up to the transition state, crosses it, but recrosses it and reverts back to the reactants, with very large oscillations]]

[[File:Sc3916_rxn5.PNG|thumb|300px|none|Initial p1=-2.5, r2=-5.2 
the trajectory crosses the transition state, forming the products, then recrosses it to the reactant side, and then recrosses it again, forming the products finally]]

'''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?'''

4 major assumptions are used. Firstly, the Born-Oppenheimer approximation is used, and that the effects of quantum tunneling are negligible (there is a non-zero probability that reactants can convert to products without reaching the activation energy). The reactants are assumed to have a Boltzmann distribution - which is perfectly valid at low temperatures. The final assumption is that "once the system attains the transition state, with a velocity towards the product configuration, it will not reenter the initial state region again."[https://doi.org/10.1016/B978-044452837-7.50005-8 1]

This last assumption does not fit with our simulations. Both the 4th, and 5th plots contain the trajectory crossing over the transition state and back again. Since there will be more trajectories (like the 4th) which involve crossing over and are still unreactive - which are not predicted by Transition State Theory - experimental rate values will be slower than those predicted by the theory.

==EXERCISE 2: F - H - H system==

'''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 is exothermic - the products have a more negative energy than the reactants. Because the enthalpy change only involves breaking a H-H bond and making a H-F bond, ΔH = +H(H-H) - H(H-F).

Since ΔH < 0:

H(H-F)>H(H-H)

This means HF has a greater bond enthalpy (and therefore strength) than H2. It can also be seen that H + HF is endothermic, which leads to the same conclusion.

'''Locate the approximate position of the transition state.'''

rts: 
RHH = 0.745 angstroms 
RHF = 1.811 angstroms

(0.7448778, 1.810754) is the super accurate one

'''Report the activation energy for both reactions.'''

By performing an MEP from next to the transition structure, the path from the TS to the reactants, or products, was found. By subtracting the final energy from the TS, the Ea was obtained. Since the activation energy for H2 + F is very small, the MEP has a very small gradient, and as such the calculation required an order of magnitude more steps in the calculation (than HF + H) to reach the reactants.

-103.752 TS

133.954

[[File:Mep_energyvstime_sc3916.PNG|thumb|left|initial conditions: 
RHH=0.745 
RHF=1.83]]

[[3916_energyvstime_hf.PNG|thumb|right|initial conditions: 
RHH=0.745 
RHF=1.8]]

Ea(H2 + F Ea) = 
Ea(HF + H Ea) = 30.20 kcal/mol 

=== Reaction dynamics ===
* Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Internuclear Momenta vs Time”. {{fontcolor1|blue|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

[[File:H2+f_surface__3916.PNG|thumb|none|F + H2, initial conditions: 
RHH (AB) = 0.745 angstroms 
RHF (BC) = 2 angstroms 
AB momentum = +3 
BC momentum = -5]]

[[File:H2+F_MOMENTAVTIME3916.PNG|thumb|none|F + H2 Momentum vs Time Plot]]

The animation of this reaction clearly shows the velocity of the lone H atom formed has increased, and vibrations occurring in HF. This can be seen in the momentum vs time plot - with the increased AB and AC momentum reflecting the increase in velocity, and BC momentum oscillation reflecting the vibrations of the H-F bond. Since translational and vibrational energy contribute to kinetic energy, it is safe to assume that the energy released by this exothermic reaction (a decrease in enthalpy) is conserved in the form of increased kinetic energy. This can be seen in the below graph. An increase in average kinetic energy results in increased temperature - which is easy to measure experimentally.

[[File:H2+F_ENERGYVSTIME3916.PNG|thumb|none|F + H2 Energy vs Time Plot]]

* Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 0.74, with a momentum pFH = -0.5, and explore several values of pHH in the range -3 to 3 (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.

* For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now?

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 [[#invert momenta|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.)

The cases studied are an illustration of Polanyi's empirical rules. {{fontcolor|blue|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:3916 energyvstime hf.PNG

2018-05-24T14:48:52Z

Sc3916: