ChemWiki - User contributions [en]

MRD:JFG17 01347171

2019-05-10T16:48:36Z

Jfg17: /* Question 6 - By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? Locate the appr...

== Molecular Reaction Dynamics Laboratory ==

=== Question 1 - On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===

A transition state is widely considered to be the point of maximum energy in a given reaction. As such, it is clear that the mathematical definition of a transition state should be inclusive of a description of a local maximum. However, to describe the true transition state on a potential surface, there must also be a local minimum at the very same point as the said local maximum; the transition state on a potential energy surface is a saddle point.

With respect to the systems under consideration:

If V''(r1) > 0 ; there is a local minimum wrt. r1.

If V''(r1) < 0 ; there is a local maximum wrt. r1.

However, a potential energy surface is a multi-variable function: a function of both r1 and r2. As such, the second derivative test should be used to confirm that the point under consideration (eg:(rx,ry)) is a true saddle point, not just a local max/min.

Defining the function D as the discriminant for this test:

D(r1,r2) = Vr1r1(r1,r2)Vr2r2(r1,r2)-(Vr1r2(r1,r2))2

If D < 0, the selected point (rx,ry) is a true saddle point - the transition state of the potential energy surface.

If D > 0, the selected point is only a local max/min.

=== Question 2 - 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. ===

The transition state of the (H-H, H) system is found at approximately r1 = r2 = 0.90775 Angstrom.

[[File:JFG TS LOCATION.png|Contour plot showing location of TS.]]

[[File:JFG TS LOCATION1.png|Internuclear distance vs time graph.]]

The internuclear distance vs time graph demonstrates that from the starting point of (0.90775, 0.90775), internuclear distance does not change with time. This indicates that the system is completely stable at this point. There is no propensity for the atoms to move because the gradient of the potential energy surface is zero, in all directions - this is a saddle point and as such can be determined as the TS. (Note: AB and BC) are overlayed in this graph.

=== Question 3 - Comment on how the mep and the trajectory you just calculated differ. ===

In a dynamic calculation, momentum is carried over from step to step. As such, vibrations in the bonded hydrogen atoms are visible because energy is carried forward. In the mep calculation, momentum is not carried forward from step to step. This causes any oscillations that would be present in the HH bond to become 'invisible' because each the calculation does not consider momentum. The mep calculation serves well to find energy minima because momentum is not carried forward by steps and only the gradient of the surface is considered - once a minimum is reached, there is no momentum to carry on past the well.

The visibility of oscillations in the two different calculation types can be seen in the images below - it can be seen that there is no oscillation in the mep calculated plot.

[[File:JFG MEP 1.png ]]
[[File:JFG MEP 2.png ]]

=== Question 4 - Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ===

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy!! Reactive? !! Description of dynamics!!
|-
| -1.25 || -2.5 || -99.119 || Reactive || [[File:JFG TABLE PLOT1.png]] || Overall reactive because the original AB distance of 0.74 is somewhat mirrored in the BC distance at the endpoint of the calculation, and the same for BC to AB. AB distance remains approx. constant whilst BC distance is reducing – collision. AB distance then increases (bond shown to be broken), BC distance fluctuates around 0.75 due to energy of collision.

|-
| -1.5|| -2.0 || -100.456 || Unreactive || [[File:JFG TABLE PLOT2.png]] || Overall unreactive because TS not overcome. AB distance remains approx. 0.75, indicating bond remains unbroken. Reduction is BC distance represents collision, but energy of the system not correct to overcome TS. Small amount of oscillation in the AB distance (HH bond) visible due to vibrations in the bond.
|-
| -1.5 || -2.5 || -98.956 || Reactive || [[File:JFG TABLE PLOT3.png]] || Overall reactive because the TS is overcome and calculation ends in product region. The product HF bond can be seen vibrating after the collision.
|-
| -2.5 || -5.0 || -84.956 || Unreactive || [[File:JFG TABLE PLOT4.png]] || Overall unreactive. Although the TS is passed, the oscillations shown in the plot show that the system returns to the reactant region. There is a high amount of vibration in the AB (HH) bond, perhaps indicating that rather than a minimum amount of energy to react, there is an optimum energy region that will allow for reaction.
|-
| -2.5 || -5.2 || -83.416 || Reactive || [[File:JFG TABLE PLOT5.png]] || Overall reactive. Despite a high amount of vibration in the AB (HH) bond as before, there is a slight increase in the amount of momentum of the BC distance. BC distance ends up being approx. 1 Angstrom, but there is a high amount of oscillation in the potential energy of the products.
|}

The data in the table demonstrates that energy is not the only factor that determines whether this reaction occurs. In the fourth example in the table for example, there are momenta values but an unreactive calculation.

=== Question 5 - 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? ===

A key assumption usually associated with TS theory is that once the TS has been reached/overcome, the reaction will necessarily result in products. However, as exhibited under the fourth set of conditions in the table above, the TS can be overcome but momenta distribution between atoms can take the system back to reactants.

A second assumption of TS theory is that the system of species (reactants/TS/products) follow the Boltzmann distribution. However, when a particularly shortlived TS species is involved, it cannot populate the Boltzmann distribution properly. This could cause a discrepancy in calculated and observed reaction rates.

=== Question 6 - By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? Locate the approximate position of the transition state. Report the activation energy for both reactions. ===

Bond energies:
HH bond: 436 kJmol-1
HF bond: 569 kJmol-1

Using these values, it is predicted that the reaction from the (H-H, F) system to the (H, H-F) system will be exothermic; breaking the HH bond and making the HF bond corresponds to a reaction energy of -133 kJmol-1. This prediction is confirmed by the shape of the potential energy surface for the reaction, shown below. There is a clear drop in potential energy from the (H-H, F) system to the (H, H-F) system.

[[File:JFG HHF SURFACE.png]]

The z axis of the surface, the energy axis, clearly exhibits a higher energy system whilst the AB distance is approx. 0.75 A. In this calculation, the AB distance corresponds to a bonded pair of Hydrogen atoms. When this HH bond is broken, AB (HH) distance increases and BC (HF) distance decreases, the product region of the graph is entered and this can be seen to be lower energy.

The transition state of the surface can be found at approximately (AB,BC) = (0.746 Angstrom ,1.789 Angstrom). Using Hammond's postulate helps to locate this TS as close to the (H-H, F) system because the reaction is exothermic and thus an early TS can be expected.

[[File:JFG HHF TS.png]]

By using the mep calculation type to find the energy minima for the reactants and products regions, the energy difference between these minima and the TS can be calculated. As such, the activation energy for the exothermic and endothermic reactions can be calculated.

Using this method, the following activation energies were calculated:

Exothermic activation energy: approx. 0.1 kcalmol-1

Endothermic activation energy: approx. 28.5 kcalmol-1

The following energy vs time plots exhibit these activation energies visually.

[[File:JFG HHF ACTIVATION ENERGY EXO.png]]

[[File:JFG HHF ACTIVATION ENERGY ENDO.png]]

(Bond energy reference: https://en.wikipedia.org/wiki/Bond-dissociation_energy.)

=== Question 7 - In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.===

By establishing a reactive pathway from (H-H, F) to (H, H-F), the mechanism of the release of energy can be examined. Because the reaction is exothermic, the potential energy of the products is lower than the reactants. However, energy must be conserved and so the energy released by the reaction must adopt another form. The energy vs time graph shown below confirms that total energy remains constant, and that potential energy is reduced in the reaction. It also shows that excess energy is released as kinetic energy. Examining the momentum vs time graph helps to confirm that the energy is released into large vibrations in the HF bond in the products - kinetic energy.

[[File:JFG HHF REACTIVE ENERGY.png]]

[[File:JFG HHF REACTIVE MOMENTUM.png]]

=== Question 8 - 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 theorise that for an early TS (an exothermic reaction), translational energy is effective for overcoming the TS and producing a reaction. Vibrational energy of a bond, for example in the AB (HH) bond, has very little effect on the ability of the system to overcome the TS barrier and complete the reaction. For a late TS (an endothermic reaction), the opposite set of ideas apply.

Considering Polanyi's rules for the exothermic reaction with a late TS, we can expect that increasing the vibrational energy of the AB (HH) bond will be ineffective for overcoming the TS and resulting in products. This is demonstrated in the plots below, which XXX.

In plot 1, AB momentum = -3 and BC momentum = -0.5.
In plot 2, AB momentum = -3 and BC momentum = -0.8. Comparing to plot 1 shows the effect of increasing the magnitude of momentum in the BC distance (translational energy; effective).
In plot 3, AB momentum = 0.1 and BC momentum = -0.8. Comparing with plot 2 shows the effect of decreasing the magnitude of the AB (decreased vibrational energy; little effect on reactivity).

[[File:JFG POL EXO 1.png ‎|| Plot 1 for exothermic reaction. ]]
[[File:JFG POL EXO 2.png |‎| Plot 2 for exothermic reaction.‎ ]]
[[File:JFG POL EXO 3.png ‎|‎| Plot 3 for exothermic reaction.]]

Polanyi's rules are certainly supported by the evidence provided for the exothermic reaction - that translational energy has a tremendously greater effect on the successful reactivity of am exothermic reaction than does vibrational energy.

Similarly considering the rules for the endothermic reaction with a late TS, we can expect that increasing the vibrational energy of the BC (HF) bond will be effective in producing a reactive pathway.

Reference for Polanyi's Rules: Zhang, Z., Zhou, Y., Zhang, D., Czakó, G. and Bowman, J. (2012). Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction. The Journal of Physical Chemistry Letters, 3(23), pp.3416-3419.

MRD:JFG17 01347171

2019-05-10T16:47:53Z

Jfg17: /* Question 4 - Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the ta...

== Molecular Reaction Dynamics Laboratory ==

=== Question 1 - On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===

A transition state is widely considered to be the point of maximum energy in a given reaction. As such, it is clear that the mathematical definition of a transition state should be inclusive of a description of a local maximum. However, to describe the true transition state on a potential surface, there must also be a local minimum at the very same point as the said local maximum; the transition state on a potential energy surface is a saddle point.

With respect to the systems under consideration:

If V''(r1) > 0 ; there is a local minimum wrt. r1.

If V''(r1) < 0 ; there is a local maximum wrt. r1.

However, a potential energy surface is a multi-variable function: a function of both r1 and r2. As such, the second derivative test should be used to confirm that the point under consideration (eg:(rx,ry)) is a true saddle point, not just a local max/min.

Defining the function D as the discriminant for this test:

D(r1,r2) = Vr1r1(r1,r2)Vr2r2(r1,r2)-(Vr1r2(r1,r2))2

If D < 0, the selected point (rx,ry) is a true saddle point - the transition state of the potential energy surface.

If D > 0, the selected point is only a local max/min.

=== Question 2 - 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. ===

The transition state of the (H-H, H) system is found at approximately r1 = r2 = 0.90775 Angstrom.

[[File:JFG TS LOCATION.png|Contour plot showing location of TS.]]

[[File:JFG TS LOCATION1.png|Internuclear distance vs time graph.]]

The internuclear distance vs time graph demonstrates that from the starting point of (0.90775, 0.90775), internuclear distance does not change with time. This indicates that the system is completely stable at this point. There is no propensity for the atoms to move because the gradient of the potential energy surface is zero, in all directions - this is a saddle point and as such can be determined as the TS. (Note: AB and BC) are overlayed in this graph.

=== Question 3 - Comment on how the mep and the trajectory you just calculated differ. ===

In a dynamic calculation, momentum is carried over from step to step. As such, vibrations in the bonded hydrogen atoms are visible because energy is carried forward. In the mep calculation, momentum is not carried forward from step to step. This causes any oscillations that would be present in the HH bond to become 'invisible' because each the calculation does not consider momentum. The mep calculation serves well to find energy minima because momentum is not carried forward by steps and only the gradient of the surface is considered - once a minimum is reached, there is no momentum to carry on past the well.

The visibility of oscillations in the two different calculation types can be seen in the images below - it can be seen that there is no oscillation in the mep calculated plot.

[[File:JFG MEP 1.png ]]
[[File:JFG MEP 2.png ]]

=== Question 4 - Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ===

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy!! Reactive? !! Description of dynamics!!
|-
| -1.25 || -2.5 || -99.119 || Reactive || [[File:JFG TABLE PLOT1.png]] || Overall reactive because the original AB distance of 0.74 is somewhat mirrored in the BC distance at the endpoint of the calculation, and the same for BC to AB. AB distance remains approx. constant whilst BC distance is reducing – collision. AB distance then increases (bond shown to be broken), BC distance fluctuates around 0.75 due to energy of collision.

|-
| -1.5|| -2.0 || -100.456 || Unreactive || [[File:JFG TABLE PLOT2.png]] || Overall unreactive because TS not overcome. AB distance remains approx. 0.75, indicating bond remains unbroken. Reduction is BC distance represents collision, but energy of the system not correct to overcome TS. Small amount of oscillation in the AB distance (HH bond) visible due to vibrations in the bond.
|-
| -1.5 || -2.5 || -98.956 || Reactive || [[File:JFG TABLE PLOT3.png]] || Overall reactive because the TS is overcome and calculation ends in product region. The product HF bond can be seen vibrating after the collision.
|-
| -2.5 || -5.0 || -84.956 || Unreactive || [[File:JFG TABLE PLOT4.png]] || Overall unreactive. Although the TS is passed, the oscillations shown in the plot show that the system returns to the reactant region. There is a high amount of vibration in the AB (HH) bond, perhaps indicating that rather than a minimum amount of energy to react, there is an optimum energy region that will allow for reaction.
|-
| -2.5 || -5.2 || -83.416 || Reactive || [[File:JFG TABLE PLOT5.png]] || Overall reactive. Despite a high amount of vibration in the AB (HH) bond as before, there is a slight increase in the amount of momentum of the BC distance. BC distance ends up being approx. 1 Angstrom, but there is a high amount of oscillation in the potential energy of the products.
|}

The data in the table demonstrates that energy is not the only factor that determines whether this reaction occurs. In the fourth example in the table for example, there are momenta values but an unreactive calculation.

=== Question 5 - 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? ===

A key assumption usually associated with TS theory is that once the TS has been reached/overcome, the reaction will necessarily result in products. However, as exhibited under the fourth set of conditions in the table above, the TS can be overcome but momenta distribution between atoms can take the system back to reactants.

A second assumption of TS theory is that the system of species (reactants/TS/products) follow the Boltzmann distribution. However, when a particularly shortlived TS species is involved, it cannot populate the Boltzmann distribution properly. This could cause a discrepancy in calculated and observed reaction rates.

=== Question 6 - By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? Locate the approximate position of the transition state. Report the activation energy for both reactions. ===

Bond energies:
HH bond: 436 kJmol-1
HF bond: 569 kJmol-1

Using these values, it is predicted that the reaction from the (H-H, F) system to the (H, H-F) system will be exothermic; breaking the HH bond and making the HF bond corresponds to a reaction energy of -133 kJmol-1. This prediction is confirmed by the shape of the potential energy surface for the reaction, shown below. There is a clear drop in potential energy from the (H-H, F) system to the (H, H-F) system.

[[File:JFG HHF SURFACE.png]]

The z axis of the surface, the energy axis, clearly exhibits a higher energy system whilst the AB distance is approx. 0.75 A. In this calculation, the AB distance corresponds to a bonded pair of Hydrogen atoms. When this HH bond is broken, AB (HH) distance increases and BC (HF) distance decreases, the product region of the graph is entered and this can be seen to be lower energy.

The transition state of the surface can be found at approximately (AB,BC) = (0.746 A,1.789 A). Using Hammonds postulate helps to locate this TS as close to the (H-H, F) system because the reaction is exothermic and thus an early TS can be expected.

[[File:JFG HHF TS.png]]

By using the mep calculation type to find the energy minima for the reactants and products regions, the energy difference between these minima and the TS can be calculated. As such, the activation energy for the exothermic and endothermic reactions can be calculated.

Using this method, the following activation energies were calculated:

Exothermic activation energy: approx. 0.1 kcalmol-1

Endothermic activation energy: approx. 28.5 kcalmol-1

The following energy vs time plots exhibit these activation energies visually.

[[File:JFG HHF ACTIVATION ENERGY EXO.png]]

[[File:JFG HHF ACTIVATION ENERGY ENDO.png]]

(Bond energy reference: https://en.wikipedia.org/wiki/Bond-dissociation_energy.)

=== Question 7 - In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.===

By establishing a reactive pathway from (H-H, F) to (H, H-F), the mechanism of the release of energy can be examined. Because the reaction is exothermic, the potential energy of the products is lower than the reactants. However, energy must be conserved and so the energy released by the reaction must adopt another form. The energy vs time graph shown below confirms that total energy remains constant, and that potential energy is reduced in the reaction. It also shows that excess energy is released as kinetic energy. Examining the momentum vs time graph helps to confirm that the energy is released into large vibrations in the HF bond in the products - kinetic energy.

[[File:JFG HHF REACTIVE ENERGY.png]]

[[File:JFG HHF REACTIVE MOMENTUM.png]]

=== Question 8 - 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 theorise that for an early TS (an exothermic reaction), translational energy is effective for overcoming the TS and producing a reaction. Vibrational energy of a bond, for example in the AB (HH) bond, has very little effect on the ability of the system to overcome the TS barrier and complete the reaction. For a late TS (an endothermic reaction), the opposite set of ideas apply.

Considering Polanyi's rules for the exothermic reaction with a late TS, we can expect that increasing the vibrational energy of the AB (HH) bond will be ineffective for overcoming the TS and resulting in products. This is demonstrated in the plots below, which XXX.

In plot 1, AB momentum = -3 and BC momentum = -0.5.
In plot 2, AB momentum = -3 and BC momentum = -0.8. Comparing to plot 1 shows the effect of increasing the magnitude of momentum in the BC distance (translational energy; effective).
In plot 3, AB momentum = 0.1 and BC momentum = -0.8. Comparing with plot 2 shows the effect of decreasing the magnitude of the AB (decreased vibrational energy; little effect on reactivity).

[[File:JFG POL EXO 1.png ‎|| Plot 1 for exothermic reaction. ]]
[[File:JFG POL EXO 2.png |‎| Plot 2 for exothermic reaction.‎ ]]
[[File:JFG POL EXO 3.png ‎|‎| Plot 3 for exothermic reaction.]]

Polanyi's rules are certainly supported by the evidence provided for the exothermic reaction - that translational energy has a tremendously greater effect on the successful reactivity of am exothermic reaction than does vibrational energy.

Similarly considering the rules for the endothermic reaction with a late TS, we can expect that increasing the vibrational energy of the BC (HF) bond will be effective in producing a reactive pathway.

Reference for Polanyi's Rules: Zhang, Z., Zhou, Y., Zhang, D., Czakó, G. and Bowman, J. (2012). Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction. The Journal of Physical Chemistry Letters, 3(23), pp.3416-3419.

MRD:JFG17 01347171

2019-05-10T16:45:00Z

== Molecular Reaction Dynamics Laboratory ==

=== Question 1 - On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===

A transition state is widely considered to be the point of maximum energy in a given reaction. As such, it is clear that the mathematical definition of a transition state should be inclusive of a description of a local maximum. However, to describe the true transition state on a potential surface, there must also be a local minimum at the very same point as the said local maximum; the transition state on a potential energy surface is a saddle point.

With respect to the systems under consideration:

If V''(r1) > 0 ; there is a local minimum wrt. r1.

If V''(r1) < 0 ; there is a local maximum wrt. r1.

However, a potential energy surface is a multi-variable function: a function of both r1 and r2. As such, the second derivative test should be used to confirm that the point under consideration (eg:(rx,ry)) is a true saddle point, not just a local max/min.

Defining the function D as the discriminant for this test:

D(r1,r2) = Vr1r1(r1,r2)Vr2r2(r1,r2)-(Vr1r2(r1,r2))2

If D < 0, the selected point (rx,ry) is a true saddle point - the transition state of the potential energy surface.

If D > 0, the selected point is only a local max/min.

=== Question 2 - 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. ===

The transition state of the (H-H, H) system is found at approximately r1 = r2 = 0.90775 Angstrom.

[[File:JFG TS LOCATION.png|Contour plot showing location of TS.]]

[[File:JFG TS LOCATION1.png|Internuclear distance vs time graph.]]

The internuclear distance vs time graph demonstrates that from the starting point of (0.90775, 0.90775), internuclear distance does not change with time. This indicates that the system is completely stable at this point. There is no propensity for the atoms to move because the gradient of the potential energy surface is zero, in all directions - this is a saddle point and as such can be determined as the TS. (Note: AB and BC) are overlayed in this graph.

=== Question 3 - Comment on how the mep and the trajectory you just calculated differ. ===

In a dynamic calculation, momentum is carried over from step to step. As such, vibrations in the bonded hydrogen atoms are visible because energy is carried forward. In the mep calculation, momentum is not carried forward from step to step. This causes any oscillations that would be present in the HH bond to become 'invisible' because each the calculation does not consider momentum. The mep calculation serves well to find energy minima because momentum is not carried forward by steps and only the gradient of the surface is considered - once a minimum is reached, there is no momentum to carry on past the well.

The visibility of oscillations in the two different calculation types can be seen in the images below - it can be seen that there is no oscillation in the mep calculated plot.

[[File:JFG MEP 1.png ]]
[[File:JFG MEP 2.png ]]

=== Question 4 - Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ===

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy!! Reactive? !! Description of dynamics!!
|-
| -1.25 || -2.5 || -99.119 || Reactive || [[File:JFG TABLE PLOT1.png]] || Overall reactive because the original AB distance of 0.74 is somewhat mirrored in the BC distance at the endpoint of the calculation, and the same for BC to AB. AB distance remains approx. constant whilst BC distance is reducing – collision. AB distance then increases (bond shown to be broken), BC distance fluctuates around 0.75 due to energy of collision.

|-
| -1.5|| -2.0 || -100.456 || Unreactive || [[File:JFG TABLE PLOT2.png]] || Overall unreactive because TS not overcome. AB distance remains approx. 0.75, indicating bond remains unbroken. Reduction is BC distance represents collision, but energy of the system not correct to overcome TS. Small amount of oscillation in the AB distance (HH bond) visible due to vibrations in the bond.
|-
| -1.5 || -2.5 || -98.956 || Reactive || [[File:JFG TABLE PLOT3.png]] || Overall reactive because the TS is overcome and calculation ends in product region.
|-
| -2.5 || -5.0 || -84.956 || Unreactive || [[File:JFG TABLE PLOT4.png]] || Overall unreactive. Although the TS is passed, the oscillations shown in the plot show that the system returns to the reactant region. There is a high amount of vibration in the AB (HH) bond, perhaps indicating that rather than a minimum amount of energy to react, there is an optimum energy region that will allow for reaction.
|-
| -2.5 || -5.2 || -83.416 || Reactive || [[File:JFG TABLE PLOT5.png]] || Overall reactive. Despite a high amount of vibration in the AB (HH) bond as before, there is a slight increase in the amount of momentum of the BC distance. BC distance ends up being approx. 1 Angstrom, but there is a high amount of oscillation in the potential energy of the products.
|}

The data in the table demonstrates that energy is not the only factor that determines whether this reaction occurs. In the fourth example in the table for example, there are momenta values but an unreactive calculation.

=== Question 5 - 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? ===

A key assumption usually associated with TS theory is that once the TS has been reached/overcome, the reaction will necessarily result in products. However, as exhibited under the fourth set of conditions in the table above, the TS can be overcome but momenta distribution between atoms can take the system back to reactants.

A second assumption of TS theory is that the system of species (reactants/TS/products) follow the Boltzmann distribution. However, when a particularly shortlived TS species is involved, it cannot populate the Boltzmann distribution properly. This could cause a discrepancy in calculated and observed reaction rates.

=== Question 6 - By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? Locate the approximate position of the transition state. Report the activation energy for both reactions. ===

Bond energies:
HH bond: 436 kJmol-1
HF bond: 569 kJmol-1

Using these values, it is predicted that the reaction from the (H-H, F) system to the (H, H-F) system will be exothermic; breaking the HH bond and making the HF bond corresponds to a reaction energy of -133 kJmol-1. This prediction is confirmed by the shape of the potential energy surface for the reaction, shown below. There is a clear drop in potential energy from the (H-H, F) system to the (H, H-F) system.

[[File:JFG HHF SURFACE.png]]

The z axis of the surface, the energy axis, clearly exhibits a higher energy system whilst the AB distance is approx. 0.75 A. In this calculation, the AB distance corresponds to a bonded pair of Hydrogen atoms. When this HH bond is broken, AB (HH) distance increases and BC (HF) distance decreases, the product region of the graph is entered and this can be seen to be lower energy.

The transition state of the surface can be found at approximately (AB,BC) = (0.746 A,1.789 A). Using Hammonds postulate helps to locate this TS as close to the (H-H, F) system because the reaction is exothermic and thus an early TS can be expected.

[[File:JFG HHF TS.png]]

By using the mep calculation type to find the energy minima for the reactants and products regions, the energy difference between these minima and the TS can be calculated. As such, the activation energy for the exothermic and endothermic reactions can be calculated.

Using this method, the following activation energies were calculated:

Exothermic activation energy: approx. 0.1 kcalmol-1

Endothermic activation energy: approx. 28.5 kcalmol-1

The following energy vs time plots exhibit these activation energies visually.

[[File:JFG HHF ACTIVATION ENERGY EXO.png]]

[[File:JFG HHF ACTIVATION ENERGY ENDO.png]]

(Bond energy reference: https://en.wikipedia.org/wiki/Bond-dissociation_energy.)

=== Question 7 - In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.===

By establishing a reactive pathway from (H-H, F) to (H, H-F), the mechanism of the release of energy can be examined. Because the reaction is exothermic, the potential energy of the products is lower than the reactants. However, energy must be conserved and so the energy released by the reaction must adopt another form. The energy vs time graph shown below confirms that total energy remains constant, and that potential energy is reduced in the reaction. It also shows that excess energy is released as kinetic energy. Examining the momentum vs time graph helps to confirm that the energy is released into large vibrations in the HF bond in the products - kinetic energy.

[[File:JFG HHF REACTIVE ENERGY.png]]

[[File:JFG HHF REACTIVE MOMENTUM.png]]

=== Question 8 - 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 theorise that for an early TS (an exothermic reaction), translational energy is effective for overcoming the TS and producing a reaction. Vibrational energy of a bond, for example in the AB (HH) bond, has very little effect on the ability of the system to overcome the TS barrier and complete the reaction. For a late TS (an endothermic reaction), the opposite set of ideas apply.

Considering Polanyi's rules for the exothermic reaction with a late TS, we can expect that increasing the vibrational energy of the AB (HH) bond will be ineffective for overcoming the TS and resulting in products. This is demonstrated in the plots below, which XXX.

In plot 1, AB momentum = -3 and BC momentum = -0.5.
In plot 2, AB momentum = -3 and BC momentum = -0.8. Comparing to plot 1 shows the effect of increasing the magnitude of momentum in the BC distance (translational energy; effective).
In plot 3, AB momentum = 0.1 and BC momentum = -0.8. Comparing with plot 2 shows the effect of decreasing the magnitude of the AB (decreased vibrational energy; little effect on reactivity).

[[File:JFG POL EXO 1.png ‎|| Plot 1 for exothermic reaction. ]]
[[File:JFG POL EXO 2.png |‎| Plot 2 for exothermic reaction.‎ ]]
[[File:JFG POL EXO 3.png ‎|‎| Plot 3 for exothermic reaction.]]

Polanyi's rules are certainly supported by the evidence provided for the exothermic reaction - that translational energy has a tremendously greater effect on the successful reactivity of am exothermic reaction than does vibrational energy.

Similarly considering the rules for the endothermic reaction with a late TS, we can expect that increasing the vibrational energy of the BC (HF) bond will be effective in producing a reactive pathway.

Reference for Polanyi's Rules: Zhang, Z., Zhou, Y., Zhang, D., Czakó, G. and Bowman, J. (2012). Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction. The Journal of Physical Chemistry Letters, 3(23), pp.3416-3419.

MRD:JFG17 01347171

2019-05-10T16:41:06Z

== Molecular Reaction Dynamics Laboratory ==

=== Question 1 - On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===

A transition state is widely considered to be the point of maximum energy in a given reaction. As such, it is clear that the mathematical definition of a transition state should be inclusive of a description of a local maximum. However, to describe the true transition state on a potential surface, there must also be a local minimum at the very same point as the said local maximum; the transition state on a potential energy surface is a saddle point.

With respect to the systems under consideration:

If V''(r1) > 0 ; there is a local minimum wrt. r1.

If V''(r1) < 0 ; there is a local maximum wrt. r1.

However, a potential energy surface is a multi-variable function: a function of both r1 and r2. As such, the second derivative test should be used to confirm that the point under consideration (eg:(rx,ry)) is a true saddle point, not just a local max/min.

Defining the function D as the discriminant for this test:

D(r1,r2) = Vr1r1(r1,r2)Vr2r2(r1,r2)-(Vr1r2(r1,r2))2

If D < 0, the selected point (rx,ry) is a true saddle point - the transition state of the potential energy surface.

If D > 0, the selected point is only a local max/min.

=== Question 2 - 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. ===

The transition state of the (H-H, H) system is found at approximately r1 = r2 = 0.90775 Angstrom.

[[File:JFG TS LOCATION.png|Contour plot showing location of TS.]]

[[File:JFG TS LOCATION1.png|Internuclear distance vs time graph.]]

The internuclear distance vs time graph demonstrates that from the starting point of (0.90775, 0.90775), internuclear distance does not change with time. This indicates that the system is completely stable at this point. There is no propensity for the atoms to move because the gradient of the potential energy surface is zero, in all directions - this is a saddle point and as such can be determined as the TS. (Note: AB and BC) are overlayed in this graph.

=== Question 3 - Comment on how the mep and the trajectory you just calculated differ. ===

In a dynamic calculation, momentum is carried over from step to step. As such, vibrations in the bonded hydrogen atoms are visible because energy is carried forward. In the mep calculation, momentum is not carried forward from step to step. This causes any oscillations that would be present in the HH bond to become 'invisible' because each the calculation does not consider momentum. The mep calculation serves well to find energy minima because momentum is not carried forward by steps and only the gradient of the surface is considered - once a minimum is reached, there is no momentum to carry on past the well.

The visibility of oscillations in the two different calculation types can be seen in the images below - it can be seen that there is no oscillation in the mep calculated plot.

[[File:JFG MEP 1.png ]]
[[File:JFG MEP 2.png ]]

=== Question 4 - Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ===

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy!! Reactive? !! Description of dynamics!!
|-
| -1.25 || -2.5 || -99.119 || Reactive || [[File:JFG TABLE PLOT1.png]] || Overall reactive because the original AB distance of 0.74 is somewhat mirrored in the BC distance at the endpoint of the calculation, and the same for BC to AB. AB distance remains approx. constant whilst BC distance is reducing – collision. AB distance then increases (bond shown to be broken), BC distance fluctuates around 0.75 due to energy of collision.

|-
| -1.5|| -2.0 || -100.456 || Unreactive || [[File:JFG TABLE PLOT2.png]] || Overall unreactive because TS not overcome. AB distance remains approx. 0.75, indicating bond remains unbroken. Reduction is BC distance represents collision, but energy of the system not correct to overcome TS. Small amount of oscillation in the AB distance (HH bond) visible due to vibrations in the bond.
|-
| -1.5 || -2.5 || -98.956 || Reactive || [[File:JFG TABLE PLOT3.png]] || Overall reactive because the TS is overcome and calculation ends in product region.
|-
| -2.5 || -5.0 || -84.956 || Unreactive || [[File:JFG TABLE PLOT4.png]] || Overall unreactive. Although the TS is passed, the oscillations shown in the plot show that the system returns to the reactant region. There is a high amount of vibration in the AB (HH) bond, perhaps indicating that rather than a minimum amount of energy to react, there is an optimum energy region that will allow for reaction.
|-
| -2.5 || -5.2 || -83.416 || Reactive || [[File:JFG TABLE PLOT5.png]] || Overall reactive. Despite a high amount of vibration in the AB (HH) bond as before, there is a slight increase in the amount of momentum of the BC distance. BC distance ends up being approx. 1 Angstrom, but there is a high amount of oscillation in the potential energy of the products.
|}

The data in the table demonstrates that energy is not the only factor that determines whether this reaction occurs. In the fourth example in the table for example, there are momenta values but an unreactive calculation.

=== Question 5 - 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? ===

A key assumption usually associated with TS theory is that once the TS has been reached/overcome, the reaction will necessarily result in products. However, as exhibited under the fourth set of conditions in the table above, the TS can be overcome but momenta distribution between atoms can take the system back to reactants.

A second assumption of TS theory is that the system of species (reactants/TS/products) follow the Boltzmann distribution. However, when a particularly shortlived TS species is involved, it cannot populate the Boltzmann distribution properly. This could cause a discrepancy in calculated and observed reaction rates.

=== Question 6 - By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? Locate the approximate position of the transition state. Report the activation energy for both reactions. ===

Bond energies:
HH bond: 436 kJmol-1
HF bond: 569 kJmol-1

Using these values, it is predicted that the reaction from the (H-H, F) system to the (H, H-F) system will be exothermic; breaking the HH bond and making the HF bond corresponds to a reaction energy of -133 kJmol-1. This prediction is confirmed by the shape of the potential energy surface for the reaction, shown below. There is a clear drop in potential energy from the (H-H, F) system to the (H, H-F) system.

[[File:JFG HHF SURFACE.png]]

The z axis of the surface, the energy axis, clearly exhibits a higher energy system whilst the AB distance is approx. 0.75 A. In this calculation, the AB distance corresponds to a bonded pair of Hydrogen atoms. When this HH bond is broken, AB (HH) distance increases and BC (HF) distance decreases, the product region of the graph is entered and this can be seen to be lower energy.

The transition state of the surface can be found at approximately (AB,BC) = (0.746 A,1.789 A).

[[File:JFG HHF TS.png]]

By using the mep calculation type to find the energy minima for the reactants and products regions, the energy difference between these minima and the TS can be calculated. As such, the activation energy for the exothermic and endothermic reactions can be calculated.

Using this method, the following activation energies were calculated:

Exothermic activation energy: approx. 0.1 kCalmol-1

Endothermic activation energy: approx. 28.5 kCalmol-1

The following energy vs time plots exhibit these activation energies visually.

[[File:JFG HHF ACTIVATION ENERGY EXO.png]]

[[File:JFG HHF ACTIVATION ENERGY ENDO.png]]

(Bond energy reference: https://en.wikipedia.org/wiki/Bond-dissociation_energy.)

=== Question 7 - In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.===

By establishing a reactive pathway from (H-H, F) to (H, H-F), the mechanism of the release of energy can be examined. Because the reaction is exothermic, the potential energy of the products is lower than the reactants. However, energy must be conserved and so the energy released by the reaction must adopt another form. The energy vs time graph shown below confirms that total energy remains constant, and that potential energy is reduced in the reaction. It also shows that excess energy is released as kinetic energy. Examining the momentum vs time graph helps to confirm that the energy is released into large vibrations in the HF bond in the products - kinetic energy.

[[File:JFG HHF REACTIVE ENERGY.png]]

[[File:JFG HHF REACTIVE MOMENTUM.png]]

=== Question 8 - 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 theorise that for an early TS (an exothermic reaction), translational energy is effective for overcoming the TS and producing a reaction. Vibrational energy of a bond, for example in the AB (HH) bond, has very little effect on the ability of the system to overcome the TS barrier and complete the reaction. For a late TS (an endothermic reaction), the opposite set of ideas apply.

Considering Polanyi's rules for the exothermic reaction with a late TS, we can expect that increasing the vibrational energy of the AB (HH) bond will be ineffective for overcoming the TS and resulting in products. This is demonstrated in the plots below, which XXX.

In plot 1, AB momentum = -3 and BC momentum = -0.5.
In plot 2, AB momentum = -3 and BC momentum = -0.8. Comparing to plot 1 shows the effect of increasing the magnitude of momentum in the BC distance (translational energy; effective).
In plot 3, AB momentum = 0.1 and BC momentum = -0.8. Comparing with plot 2 shows the effect of decreasing the magnitude of the AB (decreased vibrational energy; little effect on reactivity).

[[File:JFG POL EXO 1.png ‎|| Plot 1 for exothermic reaction. ]]
[[File:JFG POL EXO 2.png |‎| Plot 2 for exothermic reaction.‎ ]]
[[File:JFG POL EXO 3.png ‎|‎| Plot 3 for exothermic reaction.]]

Polanyi's rules are certainly supported by the evidence provided for the exothermic reaction - that translational energy has a tremendously greater effect on the successful reactivity of am exothermic reaction than does vibrational energy.

Similarly considering the rules for the endothermic reaction with a late TS, we can expect that increasing the vibrational energy of the BC (HF) bond will be effective in producing a reactive pathway.

Reference for Polanyi's Rules: Zhang, Z., Zhou, Y., Zhang, D., Czakó, G. and Bowman, J. (2012). Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction. The Journal of Physical Chemistry Letters, 3(23), pp.3416-3419.

MRD:JFG17 01347171

2019-05-10T16:37:48Z

Jfg17: /* Question 5 - 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? */

== Molecular Reaction Dynamics Laboratory ==

=== Question 1 - On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===

A transition state is widely considered to be the point of maximum energy in a given reaction. As such, it is clear that the mathematical definition of a transition state should be inclusive of a description of a local maximum. However, to describe the true transition state on a potential surface, there must also be a local minimum at the very same point as the said local maximum; the transition state on a potential energy surface is a saddle point.

With respect to the systems under consideration:

If V''(r1) > 0 ; there is a local minimum wrt. r1.

If V''(r1) < 0 ; there is a local maximum wrt. r1.

However, a potential energy surface is a multi-variable function: a function of both r1 and r2. As such, the second derivative test should be used to confirm that the point under consideration (eg:(rx,ry)) is a true saddle point, not just a local max/min.

Defining the function D as the discriminant for this test:

D(r1,r2) = Vr1r1(r1,r2)Vr2r2(r1,r2)-(Vr1r2(r1,r2))2

If D < 0, the selected point (rx,ry) is a true saddle point - the transition state of the potential energy surface.

If D > 0, the selected point is only a local max/min.

=== Question 2 - 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. ===

The transition state of the (H-H, H) system is found at approximately r1 = r2 = 0.90775 Angstrom.

[[File:JFG TS LOCATION.png|Contour plot showing location of TS.]]

[[File:JFG TS LOCATION1.png|Internuclear distance vs time graph.]]

The internuclear distance vs time graph demonstrates that from the starting point of (0.90775, 0.90775), internuclear distance does not change with time. This indicates that the system is completely stable at this point. There is no propensity for the atoms to move because the gradient of the potential energy surface is zero, in all directions - this is a saddle point and as such can be determined as the TS. (Note: AB and BC) are overlayed in this graph.

=== Question 3 - Comment on how the mep and the trajectory you just calculated differ. ===

In a dynamic calculation, momentum is carried over from step to step. As such, vibrations in the bonded hydrogen atoms are visible because energy is carried forward. In the mep calculation, momentum is not carried forward from step to step. This causes any oscillations that would be present in the HH bond to become 'invisible' because each the calculation does not consider momentum. The mep calculation serves well to find energy minima because momentum is not carried forward by steps and only the gradient of the surface is considered - once a minimum is reached, there is no momentum to carry on past the well.

The visibility of oscillations in the two different calculation types can be seen in the images below - it can be seen that there is no oscillation in the mep calculated plot.

[[File:JFG MEP 1.png ]]
[[File:JFG MEP 2.png ]]

=== Question 4 - Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ===

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy!! Reactive? !! Description of dynamics!!
|-
| -1.25 || -2.5 || -99.119 || Reactive || [[File:JFG TABLE PLOT1.png]] || Overall reactive because the original AB distance of 0.74 is somewhat mirrored in the BC distance at the endpoint of the calculation, and the same for BC to AB. AB distance remains approx. constant whilst BC distance is reducing – collision. AB distance then increases (bond shown to be broken), BC distance fluctuates around 0.75 due to energy of collision.

|-
| -1.5|| -2.0 || -100.456 || Unreactive || [[File:JFG TABLE PLOT2.png]] || Overall unreactive because TS not overcome. AB distance remains approx. 0.75, indicating bond remains unbroken. Reduction is BC distance represents collision, but energy of the system not correct to overcome TS. Small amount of oscillation in the AB distance (HH bond) visible due to vibrations in the bond.
|-
| -1.5 || -2.5 || -98.956 || Reactive || [[File:JFG TABLE PLOT3.png]] || Overall reactive because the TS is overcome and calculation ends in product region.
|-
| -2.5 || -5.0 || -84.956 || Unreactive || [[File:JFG TABLE PLOT4.png]] || Overall unreactive. Although the TS is passed, the oscillations shown in the plot show that the system returns to the reactant region. There is a high amount of vibration in the AB (HH) bond, perhaps indicating that rather than a minimum amount of energy to react, there is an optimum energy region that will allow for reaction.
|-
| -2.5 || -5.2 || -83.416 || Reactive || [[File:JFG TABLE PLOT5.png]] || Overall reactive. Despite a high amount of vibration in the AB (HH) bond as before, there is a slight increase in the amount of momentum of the BC distance. BC distance ends up being approx. 1 Angstrom, but there is a high amount of oscillation in the potential energy of the products.
|}

The data in the table demonstrates that energy is not the only factor that determines whether this reaction occurs. In the fourth example in the table for example, there are momenta values but an unreactive calculation.

=== Question 5 - 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? ===

A key assumption usually associated with TS theory is that once the TS has been reached/overcome, the reaction will necessarily result in products. However, as exhibited under the fourth set of conditions in the table above, the TS can be overcome but momenta distribution between atoms can take the system back to reactants.

A second assumption of TS theory is that the system of species (reactants/TS/products) follow the Boltzmann distribution. However, when a particularly shortlived TS species is involved, it cannot populate the Boltzmann distribution properly. This could cause a discrepancy in calculated and observed reaction rates.

=== Question 6 - By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? Locate the approximate position of the transition state. Report the activation energy for both reactions. ===

Bond energies:
HH bond: 436 kJmol-1
HF bond: 569 kJmol-1

Using these values, it is predicted that the reaction from the (H-H, F) system to the (H, H-F) system will be exothermic; breaking the HH bond and making the HF bond corresponds to a reaction energy of -133 kJmol-1. This prediction is confirmed by the shape of the potential energy surface for the reaction, shown below.

[[File:JFG HHF SURFACE.png]]

The z axis of the surface, the energy axis, clearly exhibits a higher energy system whilst the AB distance is approx. 0.75 A. In this calculation, the AB distance corresponds to a bonded pair of Hydrogen atoms. When this HH bond is broken, AB (HH) distance increases and BC (HF) distance decreases, the product region of the graph is entered and this can be seen to be lower energy.

The transition state of the surface can be found at approximately (AB,BC) = (0.746 A,1.789 A).

[[File:JFG HHF TS.png]]

By using the mep calculation type to find the energy minima for the reactants and products regions, the energy difference between these minima and the TS can be calculated. As such, the activation energy for the exothermic and endothermic reactions can be calculated.

Using this method, the following activation energies were calculated:

Exothermic activation energy: approx. 0.1 kCalmol-1

Endothermic activation energy: approx. 28.5 kCalmol-1

The following energy vs time plots exhibit these activation energies visually.

[[File:JFG HHF ACTIVATION ENERGY EXO.png]]

[[File:JFG HHF ACTIVATION ENERGY ENDO.png]]

=== Question 7 - In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.===

By establishing a reactive pathway from (H-H, F) to (H, H-F), the mechanism of the release of energy can be examined. Because the reaction is exothermic, the potential energy of the products is lower than the reactants. However, energy must be conserved and so the energy released by the reaction must adopt another form. The energy vs time graph shown below confirms that total energy remains constant, and that potential energy is reduced in the reaction. It also shows that excess energy is released as kinetic energy. Examining the momentum vs time graph helps to confirm that the energy is released into large vibrations in the HF bond in the products - kinetic energy.

[[File:JFG HHF REACTIVE ENERGY.png]]

[[File:JFG HHF REACTIVE MOMENTUM.png]]

=== Question 8 - 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 theorise that for an early TS (an exothermic reaction), translational energy is effective for overcoming the TS and producing a reaction. Vibrational energy of a bond, for example in the AB (HH) bond, has very little effect on the ability of the system to overcome the TS barrier and complete the reaction. For a late TS (an endothermic reaction), the opposite set of ideas apply.

Considering Polanyi's rules for the exothermic reaction with a late TS, we can expect that increasing the vibrational energy of the AB (HH) bond will be ineffective for overcoming the TS and resulting in products. This is demonstrated in the plots below, which XXX.

In plot 1, AB momentum = -3 and BC momentum = -0.5.
In plot 2, AB momentum = -3 and BC momentum = -0.8. Comparing to plot 1 shows the effect of increasing the magnitude of momentum in the BC distance (translational energy; effective).
In plot 3, AB momentum = 0.1 and BC momentum = -0.8. Comparing with plot 2 shows the effect of decreasing the magnitude of the AB (decreased vibrational energy; little effect on reactivity).

[[File:JFG POL EXO 1.png ‎|| Plot 1 for exothermic reaction. ]]
[[File:JFG POL EXO 2.png |‎| Plot 2 for exothermic reaction.‎ ]]
[[File:JFG POL EXO 3.png ‎|‎| Plot 3 for exothermic reaction.]]

Polanyi's rules are certainly supported by the evidence provided for the exothermic reaction - that translational energy has a tremendously greater effect on the successful reactivity of am exothermic reaction than does vibrational energy.

Similarly considering the rules for the endothermic reaction with a late TS, we can expect that increasing the vibrational energy of the BC (HF) bond will be effective in producing a reactive pathway.

Reference for Polanyi's Rules: Zhang, Z., Zhou, Y., Zhang, D., Czakó, G. and Bowman, J. (2012). Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction. The Journal of Physical Chemistry Letters, 3(23), pp.3416-3419.

MRD:JFG17 01347171

2019-05-10T16:16:53Z

Jfg17: /* Question 3 - Comment on how the mep and the trajectory you just calculated differ. */

== Molecular Reaction Dynamics Laboratory ==

=== Question 1 - On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===

A transition state is widely considered to be the point of maximum energy in a given reaction. As such, it is clear that the mathematical definition of a transition state should be inclusive of a description of a local maximum. However, to describe the true transition state on a potential surface, there must also be a local minimum at the very same point as the said local maximum; the transition state on a potential energy surface is a saddle point.

With respect to the systems under consideration:

If V''(r1) > 0 ; there is a local minimum wrt. r1.

If V''(r1) < 0 ; there is a local maximum wrt. r1.

However, a potential energy surface is a multi-variable function: a function of both r1 and r2. As such, the second derivative test should be used to confirm that the point under consideration (eg:(rx,ry)) is a true saddle point, not just a local max/min.

Defining the function D as the discriminant for this test:

D(r1,r2) = Vr1r1(r1,r2)Vr2r2(r1,r2)-(Vr1r2(r1,r2))2

If D < 0, the selected point (rx,ry) is a true saddle point - the transition state of the potential energy surface.

If D > 0, the selected point is only a local max/min.

=== Question 2 - 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. ===

The transition state of the (H-H, H) system is found at approximately r1 = r2 = 0.90775 Angstrom.

[[File:JFG TS LOCATION.png|Contour plot showing location of TS.]]

[[File:JFG TS LOCATION1.png|Internuclear distance vs time graph.]]

The internuclear distance vs time graph demonstrates that from the starting point of (0.90775, 0.90775), internuclear distance does not change with time. This indicates that the system is completely stable at this point. There is no propensity for the atoms to move because the gradient of the potential energy surface is zero, in all directions - this is a saddle point and as such can be determined as the TS. (Note: AB and BC) are overlayed in this graph.

=== Question 3 - Comment on how the mep and the trajectory you just calculated differ. ===

In a dynamic calculation, momentum is carried over from step to step. As such, vibrations in the bonded hydrogen atoms are visible because energy is carried forward. In the mep calculation, momentum is not carried forward from step to step. This causes any oscillations that would be present in the HH bond to become 'invisible' because each the calculation does not consider momentum. The mep calculation serves well to find energy minima because momentum is not carried forward by steps and only the gradient of the surface is considered - once a minimum is reached, there is no momentum to carry on past the well.

The visibility of oscillations in the two different calculation types can be seen in the images below - it can be seen that there is no oscillation in the mep calculated plot.

[[File:JFG MEP 1.png ]]
[[File:JFG MEP 2.png ]]

=== Question 4 - Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ===

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy!! Reactive? !! Description of dynamics!!
|-
| -1.25 || -2.5 || -99.119 || Reactive || [[File:JFG TABLE PLOT1.png]] || Overall reactive because the original AB distance of 0.74 is somewhat mirrored in the BC distance at the endpoint of the calculation, and the same for BC to AB. AB distance remains approx. constant whilst BC distance is reducing – collision. AB distance then increases (bond shown to be broken), BC distance fluctuates around 0.75 due to energy of collision.

|-
| -1.5|| -2.0 || -100.456 || Unreactive || [[File:JFG TABLE PLOT2.png]] || Overall unreactive because TS not overcome. AB distance remains approx. 0.75, indicating bond remains unbroken. Reduction is BC distance represents collision, but energy of the system not correct to overcome TS. Small amount of oscillation in the AB distance (HH bond) visible due to vibrations in the bond.
|-
| -1.5 || -2.5 || -98.956 || Reactive || [[File:JFG TABLE PLOT3.png]] || Overall reactive because the TS is overcome and calculation ends in product region.
|-
| -2.5 || -5.0 || -84.956 || Unreactive || [[File:JFG TABLE PLOT4.png]] || Overall unreactive. Although the TS is passed, the oscillations shown in the plot show that the system returns to the reactant region. There is a high amount of vibration in the AB (HH) bond, perhaps indicating that rather than a minimum amount of energy to react, there is an optimum energy region that will allow for reaction.
|-
| -2.5 || -5.2 || -83.416 || Reactive || [[File:JFG TABLE PLOT5.png]] || Overall reactive. Despite a high amount of vibration in the AB (HH) bond as before, there is a slight increase in the amount of momentum of the BC distance. BC distance ends up being approx. 1 Angstrom, but there is a high amount of oscillation in the potential energy of the products.
|}

The data in the table demonstrates that energy is not the only factor that determines whether this reaction occurs. In the fourth example in the table for example, there are momenta values but an unreactive calculation.

=== Question 5 - 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? ===

=== Question 6 - By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? Locate the approximate position of the transition state. Report the activation energy for both reactions. ===

Bond energies:
HH bond: 436 kJmol-1
HF bond: 569 kJmol-1

Using these values, it is predicted that the reaction from the (H-H, F) system to the (H, H-F) system will be exothermic; breaking the HH bond and making the HF bond corresponds to a reaction energy of -133 kJmol-1. This prediction is confirmed by the shape of the potential energy surface for the reaction, shown below.

[[File:JFG HHF SURFACE.png]]

The z axis of the surface, the energy axis, clearly exhibits a higher energy system whilst the AB distance is approx. 0.75 A. In this calculation, the AB distance corresponds to a bonded pair of Hydrogen atoms. When this HH bond is broken, AB (HH) distance increases and BC (HF) distance decreases, the product region of the graph is entered and this can be seen to be lower energy.

The transition state of the surface can be found at approximately (AB,BC) = (0.746 A,1.789 A).

[[File:JFG HHF TS.png]]

By using the mep calculation type to find the energy minima for the reactants and products regions, the energy difference between these minima and the TS can be calculated. As such, the activation energy for the exothermic and endothermic reactions can be calculated.

Using this method, the following activation energies were calculated:

Exothermic activation energy: approx. 0.1 kCalmol-1

Endothermic activation energy: approx. 28.5 kCalmol-1

The following energy vs time plots exhibit these activation energies visually.

[[File:JFG HHF ACTIVATION ENERGY EXO.png]]

[[File:JFG HHF ACTIVATION ENERGY ENDO.png]]

=== Question 7 - In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.===

By establishing a reactive pathway from (H-H, F) to (H, H-F), the mechanism of the release of energy can be examined. Because the reaction is exothermic, the potential energy of the products is lower than the reactants. However, energy must be conserved and so the energy released by the reaction must adopt another form. The energy vs time graph shown below confirms that total energy remains constant, and that potential energy is reduced in the reaction. It also shows that excess energy is released as kinetic energy. Examining the momentum vs time graph helps to confirm that the energy is released into large vibrations in the HF bond in the products - kinetic energy.

[[File:JFG HHF REACTIVE ENERGY.png]]

[[File:JFG HHF REACTIVE MOMENTUM.png]]

=== Question 8 - 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 theorise that for an early TS (an exothermic reaction), translational energy is effective for overcoming the TS and producing a reaction. Vibrational energy of a bond, for example in the AB (HH) bond, has very little effect on the ability of the system to overcome the TS barrier and complete the reaction. For a late TS (an endothermic reaction), the opposite set of ideas apply.

Considering Polanyi's rules for the exothermic reaction with a late TS, we can expect that increasing the vibrational energy of the AB (HH) bond will be ineffective for overcoming the TS and resulting in products. This is demonstrated in the plots below, which XXX.

In plot 1, AB momentum = -3 and BC momentum = -0.5.
In plot 2, AB momentum = -3 and BC momentum = -0.8. Comparing to plot 1 shows the effect of increasing the magnitude of momentum in the BC distance (translational energy; effective).
In plot 3, AB momentum = 0.1 and BC momentum = -0.8. Comparing with plot 2 shows the effect of decreasing the magnitude of the AB (decreased vibrational energy; little effect on reactivity).

[[File:JFG POL EXO 1.png ‎|| Plot 1 for exothermic reaction. ]]
[[File:JFG POL EXO 2.png |‎| Plot 2 for exothermic reaction.‎ ]]
[[File:JFG POL EXO 3.png ‎|‎| Plot 3 for exothermic reaction.]]

Polanyi's rules are certainly supported by the evidence provided for the exothermic reaction - that translational energy has a tremendously greater effect on the successful reactivity of am exothermic reaction than does vibrational energy.

Similarly considering the rules for the endothermic reaction with a late TS, we can expect that increasing the vibrational energy of the BC (HF) bond will be effective in producing a reactive pathway.

Reference for Polanyi's Rules: Zhang, Z., Zhou, Y., Zhang, D., Czakó, G. and Bowman, J. (2012). Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction. The Journal of Physical Chemistry Letters, 3(23), pp.3416-3419.

MRD:JFG17 01347171

2019-05-10T16:16:39Z

Jfg17: /* Question 3 - Comment on how the mep and the trajectory you just calculated differ. */

== Molecular Reaction Dynamics Laboratory ==

=== Question 1 - On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===

A transition state is widely considered to be the point of maximum energy in a given reaction. As such, it is clear that the mathematical definition of a transition state should be inclusive of a description of a local maximum. However, to describe the true transition state on a potential surface, there must also be a local minimum at the very same point as the said local maximum; the transition state on a potential energy surface is a saddle point.

With respect to the systems under consideration:

If V''(r1) > 0 ; there is a local minimum wrt. r1.

If V''(r1) < 0 ; there is a local maximum wrt. r1.

However, a potential energy surface is a multi-variable function: a function of both r1 and r2. As such, the second derivative test should be used to confirm that the point under consideration (eg:(rx,ry)) is a true saddle point, not just a local max/min.

Defining the function D as the discriminant for this test:

D(r1,r2) = Vr1r1(r1,r2)Vr2r2(r1,r2)-(Vr1r2(r1,r2))2

If D < 0, the selected point (rx,ry) is a true saddle point - the transition state of the potential energy surface.

If D > 0, the selected point is only a local max/min.

=== Question 2 - 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. ===

The transition state of the (H-H, H) system is found at approximately r1 = r2 = 0.90775 Angstrom.

[[File:JFG TS LOCATION.png|Contour plot showing location of TS.]]

[[File:JFG TS LOCATION1.png|Internuclear distance vs time graph.]]

The internuclear distance vs time graph demonstrates that from the starting point of (0.90775, 0.90775), internuclear distance does not change with time. This indicates that the system is completely stable at this point. There is no propensity for the atoms to move because the gradient of the potential energy surface is zero, in all directions - this is a saddle point and as such can be determined as the TS. (Note: AB and BC) are overlayed in this graph.

=== Question 3 - Comment on how the mep and the trajectory you just calculated differ. ===

In a dynamic calculation, momentum is carried over from step to step. As such, vibrations in the bonded hydrogen atoms are visible because energy is carried forward. In the mep calculation, momentum is not carried forward from step to step. This causes any oscillations that would be present in the HH bond to become 'invisible' because each the calculation does not consider momentum. The mep calculation serves well to find energy minima because momentum is not carried forward by steps and only the gradient of the surface is considered - once a minimum is reached, there is no momentum to carry on past the well.

The visibility of oscillations in the two different calculation types can be seen in the images below - it can be seen that there is no oscillation in the mep calculated plot.

[[File:File:JFG MEP 1.png ]]
[[File:JFG MEP 2.png ]]

=== Question 4 - Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ===

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy!! Reactive? !! Description of dynamics!!
|-
| -1.25 || -2.5 || -99.119 || Reactive || [[File:JFG TABLE PLOT1.png]] || Overall reactive because the original AB distance of 0.74 is somewhat mirrored in the BC distance at the endpoint of the calculation, and the same for BC to AB. AB distance remains approx. constant whilst BC distance is reducing – collision. AB distance then increases (bond shown to be broken), BC distance fluctuates around 0.75 due to energy of collision.

|-
| -1.5|| -2.0 || -100.456 || Unreactive || [[File:JFG TABLE PLOT2.png]] || Overall unreactive because TS not overcome. AB distance remains approx. 0.75, indicating bond remains unbroken. Reduction is BC distance represents collision, but energy of the system not correct to overcome TS. Small amount of oscillation in the AB distance (HH bond) visible due to vibrations in the bond.
|-
| -1.5 || -2.5 || -98.956 || Reactive || [[File:JFG TABLE PLOT3.png]] || Overall reactive because the TS is overcome and calculation ends in product region.
|-
| -2.5 || -5.0 || -84.956 || Unreactive || [[File:JFG TABLE PLOT4.png]] || Overall unreactive. Although the TS is passed, the oscillations shown in the plot show that the system returns to the reactant region. There is a high amount of vibration in the AB (HH) bond, perhaps indicating that rather than a minimum amount of energy to react, there is an optimum energy region that will allow for reaction.
|-
| -2.5 || -5.2 || -83.416 || Reactive || [[File:JFG TABLE PLOT5.png]] || Overall reactive. Despite a high amount of vibration in the AB (HH) bond as before, there is a slight increase in the amount of momentum of the BC distance. BC distance ends up being approx. 1 Angstrom, but there is a high amount of oscillation in the potential energy of the products.
|}

The data in the table demonstrates that energy is not the only factor that determines whether this reaction occurs. In the fourth example in the table for example, there are momenta values but an unreactive calculation.

=== Question 5 - 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? ===

=== Question 6 - By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? Locate the approximate position of the transition state. Report the activation energy for both reactions. ===

Bond energies:
HH bond: 436 kJmol-1
HF bond: 569 kJmol-1

Using these values, it is predicted that the reaction from the (H-H, F) system to the (H, H-F) system will be exothermic; breaking the HH bond and making the HF bond corresponds to a reaction energy of -133 kJmol-1. This prediction is confirmed by the shape of the potential energy surface for the reaction, shown below.

[[File:JFG HHF SURFACE.png]]

The z axis of the surface, the energy axis, clearly exhibits a higher energy system whilst the AB distance is approx. 0.75 A. In this calculation, the AB distance corresponds to a bonded pair of Hydrogen atoms. When this HH bond is broken, AB (HH) distance increases and BC (HF) distance decreases, the product region of the graph is entered and this can be seen to be lower energy.

The transition state of the surface can be found at approximately (AB,BC) = (0.746 A,1.789 A).

[[File:JFG HHF TS.png]]

By using the mep calculation type to find the energy minima for the reactants and products regions, the energy difference between these minima and the TS can be calculated. As such, the activation energy for the exothermic and endothermic reactions can be calculated.

Using this method, the following activation energies were calculated:

Exothermic activation energy: approx. 0.1 kCalmol-1

Endothermic activation energy: approx. 28.5 kCalmol-1

The following energy vs time plots exhibit these activation energies visually.

[[File:JFG HHF ACTIVATION ENERGY EXO.png]]

[[File:JFG HHF ACTIVATION ENERGY ENDO.png]]

=== Question 7 - In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.===

By establishing a reactive pathway from (H-H, F) to (H, H-F), the mechanism of the release of energy can be examined. Because the reaction is exothermic, the potential energy of the products is lower than the reactants. However, energy must be conserved and so the energy released by the reaction must adopt another form. The energy vs time graph shown below confirms that total energy remains constant, and that potential energy is reduced in the reaction. It also shows that excess energy is released as kinetic energy. Examining the momentum vs time graph helps to confirm that the energy is released into large vibrations in the HF bond in the products - kinetic energy.

[[File:JFG HHF REACTIVE ENERGY.png]]

[[File:JFG HHF REACTIVE MOMENTUM.png]]

=== Question 8 - 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 theorise that for an early TS (an exothermic reaction), translational energy is effective for overcoming the TS and producing a reaction. Vibrational energy of a bond, for example in the AB (HH) bond, has very little effect on the ability of the system to overcome the TS barrier and complete the reaction. For a late TS (an endothermic reaction), the opposite set of ideas apply.

Considering Polanyi's rules for the exothermic reaction with a late TS, we can expect that increasing the vibrational energy of the AB (HH) bond will be ineffective for overcoming the TS and resulting in products. This is demonstrated in the plots below, which XXX.

In plot 1, AB momentum = -3 and BC momentum = -0.5.
In plot 2, AB momentum = -3 and BC momentum = -0.8. Comparing to plot 1 shows the effect of increasing the magnitude of momentum in the BC distance (translational energy; effective).
In plot 3, AB momentum = 0.1 and BC momentum = -0.8. Comparing with plot 2 shows the effect of decreasing the magnitude of the AB (decreased vibrational energy; little effect on reactivity).

[[File:JFG POL EXO 1.png ‎|| Plot 1 for exothermic reaction. ]]
[[File:JFG POL EXO 2.png |‎| Plot 2 for exothermic reaction.‎ ]]
[[File:JFG POL EXO 3.png ‎|‎| Plot 3 for exothermic reaction.]]

Polanyi's rules are certainly supported by the evidence provided for the exothermic reaction - that translational energy has a tremendously greater effect on the successful reactivity of am exothermic reaction than does vibrational energy.

Similarly considering the rules for the endothermic reaction with a late TS, we can expect that increasing the vibrational energy of the BC (HF) bond will be effective in producing a reactive pathway.

Reference for Polanyi's Rules: Zhang, Z., Zhou, Y., Zhang, D., Czakó, G. and Bowman, J. (2012). Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction. The Journal of Physical Chemistry Letters, 3(23), pp.3416-3419.

File:JFG MEP 2.png

2019-05-10T16:10:38Z

Jfg17:

File:JFG MEP 1.png

2019-05-10T16:10:27Z

Jfg17:

MRD:JFG17 01347171

2019-05-10T16:04:26Z

Jfg17: /* Question 2 - 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. */

== Molecular Reaction Dynamics Laboratory ==

=== Question 1 - On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===

A transition state is widely considered to be the point of maximum energy in a given reaction. As such, it is clear that the mathematical definition of a transition state should be inclusive of a description of a local maximum. However, to describe the true transition state on a potential surface, there must also be a local minimum at the very same point as the said local maximum; the transition state on a potential energy surface is a saddle point.

With respect to the systems under consideration:

If V''(r1) > 0 ; there is a local minimum wrt. r1.

If V''(r1) < 0 ; there is a local maximum wrt. r1.

However, a potential energy surface is a multi-variable function: a function of both r1 and r2. As such, the second derivative test should be used to confirm that the point under consideration (eg:(rx,ry)) is a true saddle point, not just a local max/min.

Defining the function D as the discriminant for this test:

D(r1,r2) = Vr1r1(r1,r2)Vr2r2(r1,r2)-(Vr1r2(r1,r2))2

If D < 0, the selected point (rx,ry) is a true saddle point - the transition state of the potential energy surface.

If D > 0, the selected point is only a local max/min.

=== Question 2 - 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. ===

The transition state of the (H-H, H) system is found at approximately r1 = r2 = 0.90775 Angstrom.

[[File:JFG TS LOCATION.png|Contour plot showing location of TS.]]

[[File:JFG TS LOCATION1.png|Internuclear distance vs time graph.]]

The internuclear distance vs time graph demonstrates that from the starting point of (0.90775, 0.90775), internuclear distance does not change with time. This indicates that the system is completely stable at this point. There is no propensity for the atoms to move because the gradient of the potential energy surface is zero, in all directions - this is a saddle point and as such can be determined as the TS. (Note: AB and BC) are overlayed in this graph.

=== Question 3 - Comment on how the mep and the trajectory you just calculated differ. ===

In a dynamic calculation, momentum is carried over from step to step. As such, vibrations in the bonded hydrogen atoms are visible because energy

=== Question 4 - Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ===

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy!! Reactive? !! Description of dynamics!!
|-
| -1.25 || -2.5 || -99.119 || Reactive || [[File:JFG TABLE PLOT1.png]] || Overall reactive because the original AB distance of 0.74 is somewhat mirrored in the BC distance at the endpoint of the calculation, and the same for BC to AB. AB distance remains approx. constant whilst BC distance is reducing – collision. AB distance then increases (bond shown to be broken), BC distance fluctuates around 0.75 due to energy of collision.

|-
| -1.5|| -2.0 || -100.456 || Unreactive || [[File:JFG TABLE PLOT2.png]] || Overall unreactive because TS not overcome. AB distance remains approx. 0.75, indicating bond remains unbroken. Reduction is BC distance represents collision, but energy of the system not correct to overcome TS. Small amount of oscillation in the AB distance (HH bond) visible due to vibrations in the bond.
|-
| -1.5 || -2.5 || -98.956 || Reactive || [[File:JFG TABLE PLOT3.png]] || Overall reactive because the TS is overcome and calculation ends in product region.
|-
| -2.5 || -5.0 || -84.956 || Unreactive || [[File:JFG TABLE PLOT4.png]] || Overall unreactive. Although the TS is passed, the oscillations shown in the plot show that the system returns to the reactant region. There is a high amount of vibration in the AB (HH) bond, perhaps indicating that rather than a minimum amount of energy to react, there is an optimum energy region that will allow for reaction.
|-
| -2.5 || -5.2 || -83.416 || Reactive || [[File:JFG TABLE PLOT5.png]] || Overall reactive. Despite a high amount of vibration in the AB (HH) bond as before, there is a slight increase in the amount of momentum of the BC distance. BC distance ends up being approx. 1 Angstrom, but there is a high amount of oscillation in the potential energy of the products.
|}

The data in the table demonstrates that energy is not the only factor that determines whether this reaction occurs. In the fourth example in the table for example, there are momenta values but an unreactive calculation.

=== Question 5 - 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? ===

=== Question 6 - By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? Locate the approximate position of the transition state. Report the activation energy for both reactions. ===

Bond energies:
HH bond: 436 kJmol-1
HF bond: 569 kJmol-1

Using these values, it is predicted that the reaction from the (H-H, F) system to the (H, H-F) system will be exothermic; breaking the HH bond and making the HF bond corresponds to a reaction energy of -133 kJmol-1. This prediction is confirmed by the shape of the potential energy surface for the reaction, shown below.

[[File:JFG HHF SURFACE.png]]

The z axis of the surface, the energy axis, clearly exhibits a higher energy system whilst the AB distance is approx. 0.75 A. In this calculation, the AB distance corresponds to a bonded pair of Hydrogen atoms. When this HH bond is broken, AB (HH) distance increases and BC (HF) distance decreases, the product region of the graph is entered and this can be seen to be lower energy.

The transition state of the surface can be found at approximately (AB,BC) = (0.746 A,1.789 A).

[[File:JFG HHF TS.png]]

By using the mep calculation type to find the energy minima for the reactants and products regions, the energy difference between these minima and the TS can be calculated. As such, the activation energy for the exothermic and endothermic reactions can be calculated.

Using this method, the following activation energies were calculated:

Exothermic activation energy: approx. 0.1 kCalmol-1

Endothermic activation energy: approx. 28.5 kCalmol-1

The following energy vs time plots exhibit these activation energies visually.

[[File:JFG HHF ACTIVATION ENERGY EXO.png]]

[[File:JFG HHF ACTIVATION ENERGY ENDO.png]]

=== Question 7 - In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.===

By establishing a reactive pathway from (H-H, F) to (H, H-F), the mechanism of the release of energy can be examined. Because the reaction is exothermic, the potential energy of the products is lower than the reactants. However, energy must be conserved and so the energy released by the reaction must adopt another form. The energy vs time graph shown below confirms that total energy remains constant, and that potential energy is reduced in the reaction. It also shows that excess energy is released as kinetic energy. Examining the momentum vs time graph helps to confirm that the energy is released into large vibrations in the HF bond in the products - kinetic energy.

[[File:JFG HHF REACTIVE ENERGY.png]]

[[File:JFG HHF REACTIVE MOMENTUM.png]]

=== Question 8 - 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 theorise that for an early TS (an exothermic reaction), translational energy is effective for overcoming the TS and producing a reaction. Vibrational energy of a bond, for example in the AB (HH) bond, has very little effect on the ability of the system to overcome the TS barrier and complete the reaction. For a late TS (an endothermic reaction), the opposite set of ideas apply.

Considering Polanyi's rules for the exothermic reaction with a late TS, we can expect that increasing the vibrational energy of the AB (HH) bond will be ineffective for overcoming the TS and resulting in products. This is demonstrated in the plots below, which XXX.

In plot 1, AB momentum = -3 and BC momentum = -0.5.
In plot 2, AB momentum = -3 and BC momentum = -0.8. Comparing to plot 1 shows the effect of increasing the magnitude of momentum in the BC distance (translational energy; effective).
In plot 3, AB momentum = 0.1 and BC momentum = -0.8. Comparing with plot 2 shows the effect of decreasing the magnitude of the AB (decreased vibrational energy; little effect on reactivity).

[[File:JFG POL EXO 1.png ‎|| Plot 1 for exothermic reaction. ]]
[[File:JFG POL EXO 2.png |‎| Plot 2 for exothermic reaction.‎ ]]
[[File:JFG POL EXO 3.png ‎|‎| Plot 3 for exothermic reaction.]]

Polanyi's rules are certainly supported by the evidence provided for the exothermic reaction - that translational energy has a tremendously greater effect on the successful reactivity of am exothermic reaction than does vibrational energy.

Similarly considering the rules for the endothermic reaction with a late TS, we can expect that increasing the vibrational energy of the BC (HF) bond will be effective in producing a reactive pathway.

Reference for Polanyi's Rules: Zhang, Z., Zhou, Y., Zhang, D., Czakó, G. and Bowman, J. (2012). Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction. The Journal of Physical Chemistry Letters, 3(23), pp.3416-3419.

MRD:JFG17 01347171

2019-05-10T16:02:54Z

Jfg17: /* Question 8 - 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. */

== Molecular Reaction Dynamics Laboratory ==

=== Question 1 - On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===

A transition state is widely considered to be the point of maximum energy in a given reaction. As such, it is clear that the mathematical definition of a transition state should be inclusive of a description of a local maximum. However, to describe the true transition state on a potential surface, there must also be a local minimum at the very same point as the said local maximum; the transition state on a potential energy surface is a saddle point.

With respect to the systems under consideration:

If V''(r1) > 0 ; there is a local minimum wrt. r1.

If V''(r1) < 0 ; there is a local maximum wrt. r1.

However, a potential energy surface is a multi-variable function: a function of both r1 and r2. As such, the second derivative test should be used to confirm that the point under consideration (eg:(rx,ry)) is a true saddle point, not just a local max/min.

Defining the function D as the discriminant for this test:

D(r1,r2) = Vr1r1(r1,r2)Vr2r2(r1,r2)-(Vr1r2(r1,r2))2

If D < 0, the selected point (rx,ry) is a true saddle point - the transition state of the potential energy surface.

If D > 0, the selected point is only a local max/min.

=== Question 2 - 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. ===

The transition state of the (H-H, H) system is found at approximately r1=r2=0.90775/A.

[[File:JFG TS LOCATION.png|thumb|Contour plot showing location of TS.]]

[[File:JFG TS LOCATION1.png|thumb|Internuclear distance vs time graph.]]

The internuclear distance vs time graph demonstrates that from the starting point of (0.90775, 0.90775), internuclear distance does not change with time. This indicates that the system is completely stable at this point. There is no propensity for the atoms to move because the gradient of the potential energy surface is zero, in all directions. (Note: AB and BC) are overlayed in this graph.

=== Question 3 - Comment on how the mep and the trajectory you just calculated differ. ===

In a dynamic calculation, momentum is carried over from step to step. As such, vibrations in the bonded hydrogen atoms are visible because energy

=== Question 4 - Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ===

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy!! Reactive? !! Description of dynamics!!
|-
| -1.25 || -2.5 || -99.119 || Reactive || [[File:JFG TABLE PLOT1.png]] || Overall reactive because the original AB distance of 0.74 is somewhat mirrored in the BC distance at the endpoint of the calculation, and the same for BC to AB. AB distance remains approx. constant whilst BC distance is reducing – collision. AB distance then increases (bond shown to be broken), BC distance fluctuates around 0.75 due to energy of collision.

|-
| -1.5|| -2.0 || -100.456 || Unreactive || [[File:JFG TABLE PLOT2.png]] || Overall unreactive because TS not overcome. AB distance remains approx. 0.75, indicating bond remains unbroken. Reduction is BC distance represents collision, but energy of the system not correct to overcome TS. Small amount of oscillation in the AB distance (HH bond) visible due to vibrations in the bond.
|-
| -1.5 || -2.5 || -98.956 || Reactive || [[File:JFG TABLE PLOT3.png]] || Overall reactive because the TS is overcome and calculation ends in product region.
|-
| -2.5 || -5.0 || -84.956 || Unreactive || [[File:JFG TABLE PLOT4.png]] || Overall unreactive. Although the TS is passed, the oscillations shown in the plot show that the system returns to the reactant region. There is a high amount of vibration in the AB (HH) bond, perhaps indicating that rather than a minimum amount of energy to react, there is an optimum energy region that will allow for reaction.
|-
| -2.5 || -5.2 || -83.416 || Reactive || [[File:JFG TABLE PLOT5.png]] || Overall reactive. Despite a high amount of vibration in the AB (HH) bond as before, there is a slight increase in the amount of momentum of the BC distance. BC distance ends up being approx. 1 Angstrom, but there is a high amount of oscillation in the potential energy of the products.
|}

The data in the table demonstrates that energy is not the only factor that determines whether this reaction occurs. In the fourth example in the table for example, there are momenta values but an unreactive calculation.

=== Question 5 - 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? ===

=== Question 6 - By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? Locate the approximate position of the transition state. Report the activation energy for both reactions. ===

Bond energies:
HH bond: 436 kJmol-1
HF bond: 569 kJmol-1

Using these values, it is predicted that the reaction from the (H-H, F) system to the (H, H-F) system will be exothermic; breaking the HH bond and making the HF bond corresponds to a reaction energy of -133 kJmol-1. This prediction is confirmed by the shape of the potential energy surface for the reaction, shown below.

[[File:JFG HHF SURFACE.png]]

The z axis of the surface, the energy axis, clearly exhibits a higher energy system whilst the AB distance is approx. 0.75 A. In this calculation, the AB distance corresponds to a bonded pair of Hydrogen atoms. When this HH bond is broken, AB (HH) distance increases and BC (HF) distance decreases, the product region of the graph is entered and this can be seen to be lower energy.

The transition state of the surface can be found at approximately (AB,BC) = (0.746 A,1.789 A).

[[File:JFG HHF TS.png]]

By using the mep calculation type to find the energy minima for the reactants and products regions, the energy difference between these minima and the TS can be calculated. As such, the activation energy for the exothermic and endothermic reactions can be calculated.

Using this method, the following activation energies were calculated:

Exothermic activation energy: approx. 0.1 kCalmol-1

Endothermic activation energy: approx. 28.5 kCalmol-1

The following energy vs time plots exhibit these activation energies visually.

[[File:JFG HHF ACTIVATION ENERGY EXO.png]]

[[File:JFG HHF ACTIVATION ENERGY ENDO.png]]

=== Question 7 - In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.===

By establishing a reactive pathway from (H-H, F) to (H, H-F), the mechanism of the release of energy can be examined. Because the reaction is exothermic, the potential energy of the products is lower than the reactants. However, energy must be conserved and so the energy released by the reaction must adopt another form. The energy vs time graph shown below confirms that total energy remains constant, and that potential energy is reduced in the reaction. It also shows that excess energy is released as kinetic energy. Examining the momentum vs time graph helps to confirm that the energy is released into large vibrations in the HF bond in the products - kinetic energy.

[[File:JFG HHF REACTIVE ENERGY.png]]

[[File:JFG HHF REACTIVE MOMENTUM.png]]

=== Question 8 - 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 theorise that for an early TS (an exothermic reaction), translational energy is effective for overcoming the TS and producing a reaction. Vibrational energy of a bond, for example in the AB (HH) bond, has very little effect on the ability of the system to overcome the TS barrier and complete the reaction. For a late TS (an endothermic reaction), the opposite set of ideas apply.

Considering Polanyi's rules for the exothermic reaction with a late TS, we can expect that increasing the vibrational energy of the AB (HH) bond will be ineffective for overcoming the TS and resulting in products. This is demonstrated in the plots below, which XXX.

In plot 1, AB momentum = -3 and BC momentum = -0.5.
In plot 2, AB momentum = -3 and BC momentum = -0.8. Comparing to plot 1 shows the effect of increasing the magnitude of momentum in the BC distance (translational energy; effective).
In plot 3, AB momentum = 0.1 and BC momentum = -0.8. Comparing with plot 2 shows the effect of decreasing the magnitude of the AB (decreased vibrational energy; little effect on reactivity).

[[File:JFG POL EXO 1.png ‎|| Plot 1 for exothermic reaction. ]]
[[File:JFG POL EXO 2.png |‎| Plot 2 for exothermic reaction.‎ ]]
[[File:JFG POL EXO 3.png ‎|‎| Plot 3 for exothermic reaction.]]

Polanyi's rules are certainly supported by the evidence provided for the exothermic reaction - that translational energy has a tremendously greater effect on the successful reactivity of am exothermic reaction than does vibrational energy.

Similarly considering the rules for the endothermic reaction with a late TS, we can expect that increasing the vibrational energy of the BC (HF) bond will be effective in producing a reactive pathway.

Reference for Polanyi's Rules: Zhang, Z., Zhou, Y., Zhang, D., Czakó, G. and Bowman, J. (2012). Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction. The Journal of Physical Chemistry Letters, 3(23), pp.3416-3419.

MRD:JFG17 01347171

2019-05-10T15:57:33Z

== Molecular Reaction Dynamics Laboratory ==

=== Question 1 - On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===

A transition state is widely considered to be the point of maximum energy in a given reaction. As such, it is clear that the mathematical definition of a transition state should be inclusive of a description of a local maximum. However, to describe the true transition state on a potential surface, there must also be a local minimum at the very same point as the said local maximum; the transition state on a potential energy surface is a saddle point.

With respect to the systems under consideration:

If V''(r1) > 0 ; there is a local minimum wrt. r1.

If V''(r1) < 0 ; there is a local maximum wrt. r1.

However, a potential energy surface is a multi-variable function: a function of both r1 and r2. As such, the second derivative test should be used to confirm that the point under consideration (eg:(rx,ry)) is a true saddle point, not just a local max/min.

Defining the function D as the discriminant for this test:

D(r1,r2) = Vr1r1(r1,r2)Vr2r2(r1,r2)-(Vr1r2(r1,r2))2

If D < 0, the selected point (rx,ry) is a true saddle point - the transition state of the potential energy surface.

If D > 0, the selected point is only a local max/min.

=== Question 2 - 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. ===

The transition state of the (H-H, H) system is found at approximately r1=r2=0.90775/A.

[[File:JFG TS LOCATION.png|thumb|Contour plot showing location of TS.]]

[[File:JFG TS LOCATION1.png|thumb|Internuclear distance vs time graph.]]

The internuclear distance vs time graph demonstrates that from the starting point of (0.90775, 0.90775), internuclear distance does not change with time. This indicates that the system is completely stable at this point. There is no propensity for the atoms to move because the gradient of the potential energy surface is zero, in all directions. (Note: AB and BC) are overlayed in this graph.

=== Question 3 - Comment on how the mep and the trajectory you just calculated differ. ===

In a dynamic calculation, momentum is carried over from step to step. As such, vibrations in the bonded hydrogen atoms are visible because energy

=== Question 4 - Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ===

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy!! Reactive? !! Description of dynamics!!
|-
| -1.25 || -2.5 || -99.119 || Reactive || [[File:JFG TABLE PLOT1.png]] || Overall reactive because the original AB distance of 0.74 is somewhat mirrored in the BC distance at the endpoint of the calculation, and the same for BC to AB. AB distance remains approx. constant whilst BC distance is reducing – collision. AB distance then increases (bond shown to be broken), BC distance fluctuates around 0.75 due to energy of collision.

|-
| -1.5|| -2.0 || -100.456 || Unreactive || [[File:JFG TABLE PLOT2.png]] || Overall unreactive because TS not overcome. AB distance remains approx. 0.75, indicating bond remains unbroken. Reduction is BC distance represents collision, but energy of the system not correct to overcome TS. Small amount of oscillation in the AB distance (HH bond) visible due to vibrations in the bond.
|-
| -1.5 || -2.5 || -98.956 || Reactive || [[File:JFG TABLE PLOT3.png]] || Overall reactive because the TS is overcome and calculation ends in product region.
|-
| -2.5 || -5.0 || -84.956 || Unreactive || [[File:JFG TABLE PLOT4.png]] || Overall unreactive. Although the TS is passed, the oscillations shown in the plot show that the system returns to the reactant region. There is a high amount of vibration in the AB (HH) bond, perhaps indicating that rather than a minimum amount of energy to react, there is an optimum energy region that will allow for reaction.
|-
| -2.5 || -5.2 || -83.416 || Reactive || [[File:JFG TABLE PLOT5.png]] || Overall reactive. Despite a high amount of vibration in the AB (HH) bond as before, there is a slight increase in the amount of momentum of the BC distance. BC distance ends up being approx. 1 Angstrom, but there is a high amount of oscillation in the potential energy of the products.
|}

The data in the table demonstrates that energy is not the only factor that determines whether this reaction occurs. In the fourth example in the table for example, there are momenta values but an unreactive calculation.

=== Question 5 - 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? ===

=== Question 6 - By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? Locate the approximate position of the transition state. Report the activation energy for both reactions. ===

Bond energies:
HH bond: 436 kJmol-1
HF bond: 569 kJmol-1

Using these values, it is predicted that the reaction from the (H-H, F) system to the (H, H-F) system will be exothermic; breaking the HH bond and making the HF bond corresponds to a reaction energy of -133 kJmol-1. This prediction is confirmed by the shape of the potential energy surface for the reaction, shown below.

[[File:JFG HHF SURFACE.png]]

The z axis of the surface, the energy axis, clearly exhibits a higher energy system whilst the AB distance is approx. 0.75 A. In this calculation, the AB distance corresponds to a bonded pair of Hydrogen atoms. When this HH bond is broken, AB (HH) distance increases and BC (HF) distance decreases, the product region of the graph is entered and this can be seen to be lower energy.

The transition state of the surface can be found at approximately (AB,BC) = (0.746 A,1.789 A).

[[File:JFG HHF TS.png]]

By using the mep calculation type to find the energy minima for the reactants and products regions, the energy difference between these minima and the TS can be calculated. As such, the activation energy for the exothermic and endothermic reactions can be calculated.

Using this method, the following activation energies were calculated:

Exothermic activation energy: approx. 0.1 kCalmol-1

Endothermic activation energy: approx. 28.5 kCalmol-1

The following energy vs time plots exhibit these activation energies visually.

[[File:JFG HHF ACTIVATION ENERGY EXO.png]]

[[File:JFG HHF ACTIVATION ENERGY ENDO.png]]

=== Question 7 - In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.===

By establishing a reactive pathway from (H-H, F) to (H, H-F), the mechanism of the release of energy can be examined. Because the reaction is exothermic, the potential energy of the products is lower than the reactants. However, energy must be conserved and so the energy released by the reaction must adopt another form. The energy vs time graph shown below confirms that total energy remains constant, and that potential energy is reduced in the reaction. It also shows that excess energy is released as kinetic energy. Examining the momentum vs time graph helps to confirm that the energy is released into large vibrations in the HF bond in the products - kinetic energy.

[[File:JFG HHF REACTIVE ENERGY.png]]

[[File:JFG HHF REACTIVE MOMENTUM.png]]

=== Question 8 - 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 theorise that for an early TS (an exothermic reaction), translational energy is effective for overcoming the TS and producing a reaction. Vibrational energy of a bond, for example in the AB (HH) bond, has very little effect on the ability of the system to overcome the TS barrier and complete the reaction. For a late TS (an endothermic reaction), the opposite set of ideas apply.

Considering Polanyi's rules for the exothermic reaction with a late TS, we can expect that increasing the vibrational energy of the AB (HH) bond will be ineffective for overcoming the TS and resulting in products. This is demonstrated in the plots below, which XXX.

In plot 1, AB momentum = -3 and BC momentum = -0.5.
In plot 2, AB momentum = -3 and BC momentum = -0.8. Comparing to plot 1 shows the effect of increasing the magnitude of momentum in the BC distance (translational energy; effective).
In plot 3, AB momentum = 0.1 and BC momentum = -0.8. Comparing with plot 2 shows the effect of decreasing the magnitude of the AB (decreased vibrational energy; little effect on reactivity).

[[File:JFG POL EXO 1.png ‎|| Plot 1 for exothermic reaction. ]]
[[File:JFG POL EXO 2.png |‎| Plot 2 for exothermic reaction.‎ ]]
[[File:JFG POL EXO 3.png ‎|‎| Plot 3 for exothermic reaction.]]

Polanyi's rules are certainly supported by the evidence provided for the exothermic reaction - that translational energy has a tremendously greater effect on the successful reactivity of am exothermic reaction than does vibrational energy.

Similarly considering the rules for the endothermic reaction with a late TS, we can expect that increasing the vibrational energy of the BC (HF) bond will be effective in producing a reactive pathway.

MRD:JFG17 01347171

2019-05-10T15:55:09Z

== Molecular Reaction Dynamics Laboratory ==

=== Question 1 - On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===

A transition state is widely considered to be the point of maximum energy in a given reaction. As such, it is clear that the mathematical definition of a transition state should be inclusive of a description of a local maximum. However, to describe the true transition state on a potential surface, there must also be a local minimum at the very same point as the said local maximum; the transition state on a potential energy surface is a saddle point.

With respect to the systems under consideration:

If V''(r1) > 0 ; there is a local minimum wrt. r1.

If V''(r1) < 0 ; there is a local maximum wrt. r1.

However, a potential energy surface is a multi-variable function: a function of both r1 and r2. As such, the second derivative test should be used to confirm that the point under consideration (eg:(rx,ry)) is a true saddle point, not just a local max/min.

Defining the function D as the discriminant for this test:

D(r1,r2) = Vr1r1(r1,r2)Vr2r2(r1,r2)-(Vr1r2(r1,r2))2

If D < 0, the selected point (rx,ry) is a true saddle point - the transition state of the potential energy surface.

If D > 0, the selected point is only a local max/min.

=== Question 2 - 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. ===

The transition state of the (H-H, H) system is found at approximately r1=r2=0.90775/A.

[[File:JFG TS LOCATION.png|thumb|Contour plot showing location of TS.]]

[[File:JFG TS LOCATION1.png|thumb|Internuclear distance vs time graph.]]

The internuclear distance vs time graph demonstrates that from the starting point of (0.90775, 0.90775), internuclear distance does not change with time. This indicates that the system is completely stable at this point. There is no propensity for the atoms to move because the gradient of the potential energy surface is zero, in all directions. (Note: AB and BC) are overlayed in this graph.

=== Question 3 - Comment on how the mep and the trajectory you just calculated differ. ===

In a dynamic calculation, momentum is carried over from step to step. As such, vibrations in the bonded hydrogen atoms are visible because energy

=== Question 4 - Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ===

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy!! Reactive? !! Description of dynamics!!
|-
| -1.25 || -2.5 || -99.119 || Reactive || [[File:JFG TABLE PLOT1.png]] || Overall reactive because the original AB distance of 0.74 is somewhat mirrored in the BC distance at the endpoint of the calculation, and the same for BC to AB. AB distance remains approx. constant whilst BC distance is reducing – collision. AB distance then increases (bond shown to be broken), BC distance fluctuates around 0.75 due to energy of collision.

|-
| -1.5|| -2.0 || -100.456 || Unreactive || [[File:JFG TABLE PLOT2.png]] || Overall unreactive because TS not overcome. AB distance remains approx. 0.75, indicating bond remains unbroken. Reduction is BC distance represents collision, but energy of the system not correct to overcome TS. Small amount of oscillation in the AB distance (HH bond) visible due to vibrations in the bond.
|-
| -1.5 || -2.5 || -98.956 || Reactive || [[File:JFG TABLE PLOT3.png]] || Overall reactive because the TS is overcome and calculation ends in product region.
|-
| -2.5 || -5.0 || -84.956 || Unreactive || [[File:JFG TABLE PLOT4.png]] || Overall unreactive. Although the TS is passed, the oscillations shown in the plot show that the system returns to the reactant region. There is a high amount of vibration in the AB (HH) bond, perhaps indicating that rather than a minimum amount of energy to react, there is an optimum energy region that will allow for reaction.
|-
| -2.5 || -5.2 || -83.416 || Reactive || [[File:JFG TABLE PLOT5.png]] || Overall reactive. Despite a high amount of vibration in the AB (HH) bond as before, there is a slight increase in the amount of momentum of the BC distance. BC distance ends up being approx. 1 Angstrom, but there is a high amount of oscillation in the potential energy of the products.
|}

The data in the table demonstrates that energy is not the only factor that determines whether this reaction occurs. In the fourth example in the table for example, there are momenta values but an unreactive calculation.

=== Question 5 - 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? ===

=== Question 6 - By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? Locate the approximate position of the transition state. Report the activation energy for both reactions. ===

Bond energies:
HH bond: 436 kJmol-1
HF bond: 569 kJmol-1

Using these values, it is predicted that the reaction from the (H-H, F) system to the (H, H-F) system will be exothermic; breaking the HH bond and making the HF bond corresponds to a reaction energy of -133 kJmol-1. This prediction is confirmed by the shape of the potential energy surface for the reaction, shown below.

[[File:JFG HHF SURFACE.png]]

The z axis of the surface, the energy axis, clearly exhibits a higher energy system whilst the AB distance is approx. 0.75 A. In this calculation, the AB distance corresponds to a bonded pair of Hydrogen atoms. When this HH bond is broken, AB (HH) distance increases and BC (HF) distance decreases, the product region of the graph is entered and this can be seen to be lower energy.

The transition state of the surface can be found at approximately (AB,BC) = (0.746 A,1.789 A).

[[File:JFG HHF TS.png]]

By using the mep calculation type to find the energy minima for the reactants and products regions, the energy difference between these minima and the TS can be calculated. As such, the activation energy for the exothermic and endothermic reactions can be calculated.

Using this method, the following activation energies were calculated:

Exothermic activation energy: approx. 0.1 kCalmol-1

Endothermic activation energy: approx. 28.5 kCalmol-1

The following energy vs time plots exhibit these activation energies visually.

[[File:JFG HHF ACTIVATION ENERGY EXO.png]]

[[File:JFG HHF ACTIVATION ENERGY ENDO.png]]

=== Question 7 - In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.===

By establishing a reactive pathway from (H-H, F) to (H, H-F), the mechanism of the release of energy can be examined. Because the reaction is exothermic, the potential energy of the products is lower than the reactants. However, energy must be conserved and so the energy released by the reaction must adopt another form. The energy vs time graph shown below confirms that total energy remains constant, and that potential energy is reduced in the reaction. It also shows that excess energy is released as kinetic energy. Examining the momentum vs time graph helps to confirm that the energy is released into large vibrations in the HF bond in the products - kinetic energy.

[[File:JFG HHF REACTIVE ENERGY.png]]

[[File:JFG HHF REACTIVE MOMENTUM.png]]

=== Question 8 - 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 theorise that for an early TS (an exothermic reaction), translational energy is effective for overcoming the TS and producing a reaction. Vibrational energy of a bond, for example in the AB (HH) bond, has very little effect on the ability of the system to overcome the TS barrier and complete the reaction. For a late TS (an endothermic reaction), the opposite set of ideas apply.

Considering Polanyi's rules for the exothermic reaction with a late TS, we can expect that increasing the vibrational energy of the AB (HH) bond will be ineffective for overcoming the TS and resulting in products. This is demonstrated in the plots below, which XXX.

In plot 1, AB momentum = -3 and BC momentum = -0.5.
In plot 2, AB momentum = -3 and BC momentum = -0.8. Comparing to plot 1 shows the effect of increasing the magnitude of momentum in the BC distance (translational energy; effective).
In plot 3, AB momentum = 0.1 and BC momentum = -0.8. Comparing with plot 2 shows the effect of decreasing the magnitude of the AB (decreased vibrational energy; little effect on reactivity).

[[File:JFG POL EXO 1.png ‎| Plot 1 for exothermic reaction. ]]
[[File:JFG POL EXO 2.png |‎| Plot 2 for exothermic reaction.‎ ]]
[[File:JFG POL EXO 3.png ‎|‎| Plot 3 for exothermic reaction.]]

Similarly considering the rules for the endothermic reaction with a late TS, we can expect that increasing the vibrational energy of the BC (HF) bond will be effective in producing a reactive pathway.

MRD:JFG17 01347171

2019-05-10T15:54:34Z

== Molecular Reaction Dynamics Laboratory ==

=== Question 1 - On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===

A transition state is widely considered to be the point of maximum energy in a given reaction. As such, it is clear that the mathematical definition of a transition state should be inclusive of a description of a local maximum. However, to describe the true transition state on a potential surface, there must also be a local minimum at the very same point as the said local maximum; the transition state on a potential energy surface is a saddle point.

With respect to the systems under consideration:

If V''(r1) > 0 ; there is a local minimum wrt. r1.

If V''(r1) < 0 ; there is a local maximum wrt. r1.

However, a potential energy surface is a multi-variable function: a function of both r1 and r2. As such, the second derivative test should be used to confirm that the point under consideration (eg:(rx,ry)) is a true saddle point, not just a local max/min.

Defining the function D as the discriminant for this test:

D(r1,r2) = Vr1r1(r1,r2)Vr2r2(r1,r2)-(Vr1r2(r1,r2))2

If D < 0, the selected point (rx,ry) is a true saddle point - the transition state of the potential energy surface.

If D > 0, the selected point is only a local max/min.

=== Question 2 - 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. ===

The transition state of the (H-H, H) system is found at approximately r1=r2=0.90775/A.

[[File:JFG TS LOCATION.png|thumb|Contour plot showing location of TS.]]

[[File:JFG TS LOCATION1.png|thumb|Internuclear distance vs time graph.]]

The internuclear distance vs time graph demonstrates that from the starting point of (0.90775, 0.90775), internuclear distance does not change with time. This indicates that the system is completely stable at this point. There is no propensity for the atoms to move because the gradient of the potential energy surface is zero, in all directions. (Note: AB and BC) are overlayed in this graph.

=== Question 3 - Comment on how the mep and the trajectory you just calculated differ. ===

In a dynamic calculation, momentum is carried over from step to step. As such, vibrations in the bonded hydrogen atoms are visible because energy

=== Question 4 - Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ===

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy!! Reactive? !! Description of dynamics!!
|-
| -1.25 || -2.5 || -99.119 || Reactive || [[File:JFG TABLE PLOT1.png]] || Overall reactive because the original AB distance of 0.74 is somewhat mirrored in the BC distance at the endpoint of the calculation, and the same for BC to AB. AB distance remains approx. constant whilst BC distance is reducing – collision. AB distance then increases (bond shown to be broken), BC distance fluctuates around 0.75 due to energy of collision.

|-
| -1.5|| -2.0 || -100.456 || Unreactive || [[File:JFG TABLE PLOT2.png]] || Overall unreactive because TS not overcome. AB distance remains approx. 0.75, indicating bond remains unbroken. Reduction is BC distance represents collision, but energy of the system not correct to overcome TS. Small amount of oscillation in the AB distance (HH bond) visible due to vibrations in the bond.
|-
| -1.5 || -2.5 || -98.956 || Reactive || [[File:JFG TABLE PLOT3.png]] || Overall reactive because the TS is overcome and calculation ends in product region.
|-
| -2.5 || -5.0 || -84.956 || Unreactive || [[File:JFG TABLE PLOT4.png]] || Overall unreactive. Although the TS is passed, the oscillations shown in the plot show that the system returns to the reactant region. There is a high amount of vibration in the AB (HH) bond, perhaps indicating that rather than a minimum amount of energy to react, there is an optimum energy region that will allow for reaction.
|-
| -2.5 || -5.2 || -83.416 || Reactive || [[File:JFG TABLE PLOT5.png]] || Overall reactive. Despite a high amount of vibration in the AB (HH) bond as before, there is a slight increase in the amount of momentum of the BC distance. BC distance ends up being approx. 1 Angstrom, but there is a high amount of oscillation in the potential energy of the products.
|}

The data in the table demonstrates that energy is not the only factor that determines whether this reaction occurs. In the fourth example in the table for example, there are momenta values but an unreactive calculation.

=== Question 5 - 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? ===

=== Question 6 - By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? Locate the approximate position of the transition state. Report the activation energy for both reactions. ===

Bond energies:
HH bond: 436 kJmol-1
HF bond: 569 kJmol-1

Using these values, it is predicted that the reaction from the (H-H, F) system to the (H, H-F) system will be exothermic; breaking the HH bond and making the HF bond corresponds to a reaction energy of -133 kJmol-1. This prediction is confirmed by the shape of the potential energy surface for the reaction, shown below.

[[File:JFG HHF SURFACE.png]]

The z axis of the surface, the energy axis, clearly exhibits a higher energy system whilst the AB distance is approx. 0.75 A. In this calculation, the AB distance corresponds to a bonded pair of Hydrogen atoms. When this HH bond is broken, AB (HH) distance increases and BC (HF) distance decreases, the product region of the graph is entered and this can be seen to be lower energy.

The transition state of the surface can be found at approximately (AB,BC) = (0.746 A,1.789 A).

[[File:JFG HHF TS.png]]

By using the mep calculation type to find the energy minima for the reactants and products regions, the energy difference between these minima and the TS can be calculated. As such, the activation energy for the exothermic and endothermic reactions can be calculated.

Using this method, the following activation energies were calculated:

Exothermic activation energy: approx. 0.1 kCalmol-1

Endothermic activation energy: approx. 28.5 kCalmol-1

The following energy vs time plots exhibit these activation energies visually.

[[File:JFG HHF ACTIVATION ENERGY EXO.png]]

[[File:JFG HHF ACTIVATION ENERGY ENDO.png]]

=== Question 7 - In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.===

By establishing a reactive pathway from (H-H, F) to (H, H-F), the mechanism of the release of energy can be examined. Because the reaction is exothermic, the potential energy of the products is lower than the reactants. However, energy must be conserved and so the energy released by the reaction must adopt another form. The energy vs time graph shown below confirms that total energy remains constant, and that potential energy is reduced in the reaction. It also shows that excess energy is released as kinetic energy. Examining the momentum vs time graph helps to confirm that the energy is released into large vibrations in the HF bond in the products - kinetic energy.

[[File:JFG HHF REACTIVE ENERGY.png]]

[[File:JFG HHF REACTIVE MOMENTUM.png]]

=== Question 8 - 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 theorise that for an early TS (an exothermic reaction), translational energy is effective for overcoming the TS and producing a reaction. Vibrational energy of a bond, for example in the AB (HH) bond, has very little effect on the ability of the system to overcome the TS barrier and complete the reaction. For a late TS (an endothermic reaction), the opposite set of ideas apply.

Considering Polanyi's rules for the exothermic reaction with a late TS, we can expect that increasing the vibrational energy of the AB (HH) bond will be ineffective for overcoming the TS and resulting in products. This is demonstrated in the plots below, which XXX.

In plot 1, AB momentum = -3 and BC momentum = -0.5.
In plot 2, AB momentum = -3 and BC momentum = -0.8. Comparing to plot 1 shows the effect of increasing the magnitude of momentum in the BC distance (translational energy; effective).
In plot 3, AB momentum = 0.1 and BC momentum = -0.8. Comparing with plot 2 shows the effect of decreasing the magnitude of the AB (decreased vibrational energy; little effect on reactivity).

[[File:JFG POL EXO 1.png |‎| Plot 1 for exothermic reaction. ]]
[[File:JFG POL EXO 2.png |‎| Plot 2 for exothermic reaction.‎ ]]
[[File:JFG POL EXO 3.png ‎|‎| Plot 3 for exothermic reaction.]]

Similarly considering the rules for the endothermic reaction with a late TS, we can expect that increasing the vibrational energy of the BC (HF) bond will be effective in producing a reactive pathway.

MRD:JFG17 01347171

2019-05-10T15:54:13Z

== Molecular Reaction Dynamics Laboratory ==

=== Question 1 - On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===

A transition state is widely considered to be the point of maximum energy in a given reaction. As such, it is clear that the mathematical definition of a transition state should be inclusive of a description of a local maximum. However, to describe the true transition state on a potential surface, there must also be a local minimum at the very same point as the said local maximum; the transition state on a potential energy surface is a saddle point.

With respect to the systems under consideration:

If V''(r1) > 0 ; there is a local minimum wrt. r1.

If V''(r1) < 0 ; there is a local maximum wrt. r1.

However, a potential energy surface is a multi-variable function: a function of both r1 and r2. As such, the second derivative test should be used to confirm that the point under consideration (eg:(rx,ry)) is a true saddle point, not just a local max/min.

Defining the function D as the discriminant for this test:

D(r1,r2) = Vr1r1(r1,r2)Vr2r2(r1,r2)-(Vr1r2(r1,r2))2

If D < 0, the selected point (rx,ry) is a true saddle point - the transition state of the potential energy surface.

If D > 0, the selected point is only a local max/min.

=== Question 2 - 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. ===

The transition state of the (H-H, H) system is found at approximately r1=r2=0.90775/A.

[[File:JFG TS LOCATION.png|thumb|Contour plot showing location of TS.]]

[[File:JFG TS LOCATION1.png|thumb|Internuclear distance vs time graph.]]

The internuclear distance vs time graph demonstrates that from the starting point of (0.90775, 0.90775), internuclear distance does not change with time. This indicates that the system is completely stable at this point. There is no propensity for the atoms to move because the gradient of the potential energy surface is zero, in all directions. (Note: AB and BC) are overlayed in this graph.

=== Question 3 - Comment on how the mep and the trajectory you just calculated differ. ===

In a dynamic calculation, momentum is carried over from step to step. As such, vibrations in the bonded hydrogen atoms are visible because energy

=== Question 4 - Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ===

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy!! Reactive? !! Description of dynamics!!
|-
| -1.25 || -2.5 || -99.119 || Reactive || [[File:JFG TABLE PLOT1.png]] || Overall reactive because the original AB distance of 0.74 is somewhat mirrored in the BC distance at the endpoint of the calculation, and the same for BC to AB. AB distance remains approx. constant whilst BC distance is reducing – collision. AB distance then increases (bond shown to be broken), BC distance fluctuates around 0.75 due to energy of collision.

|-
| -1.5|| -2.0 || -100.456 || Unreactive || [[File:JFG TABLE PLOT2.png]] || Overall unreactive because TS not overcome. AB distance remains approx. 0.75, indicating bond remains unbroken. Reduction is BC distance represents collision, but energy of the system not correct to overcome TS. Small amount of oscillation in the AB distance (HH bond) visible due to vibrations in the bond.
|-
| -1.5 || -2.5 || -98.956 || Reactive || [[File:JFG TABLE PLOT3.png]] || Overall reactive because the TS is overcome and calculation ends in product region.
|-
| -2.5 || -5.0 || -84.956 || Unreactive || [[File:JFG TABLE PLOT4.png]] || Overall unreactive. Although the TS is passed, the oscillations shown in the plot show that the system returns to the reactant region. There is a high amount of vibration in the AB (HH) bond, perhaps indicating that rather than a minimum amount of energy to react, there is an optimum energy region that will allow for reaction.
|-
| -2.5 || -5.2 || -83.416 || Reactive || [[File:JFG TABLE PLOT5.png]] || Overall reactive. Despite a high amount of vibration in the AB (HH) bond as before, there is a slight increase in the amount of momentum of the BC distance. BC distance ends up being approx. 1 Angstrom, but there is a high amount of oscillation in the potential energy of the products.
|}

The data in the table demonstrates that energy is not the only factor that determines whether this reaction occurs. In the fourth example in the table for example, there are momenta values but an unreactive calculation.

=== Question 5 - 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? ===

=== Question 6 - By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? Locate the approximate position of the transition state. Report the activation energy for both reactions. ===

Bond energies:
HH bond: 436 kJmol-1
HF bond: 569 kJmol-1

Using these values, it is predicted that the reaction from the (H-H, F) system to the (H, H-F) system will be exothermic; breaking the HH bond and making the HF bond corresponds to a reaction energy of -133 kJmol-1. This prediction is confirmed by the shape of the potential energy surface for the reaction, shown below.

[[File:JFG HHF SURFACE.png]]

The z axis of the surface, the energy axis, clearly exhibits a higher energy system whilst the AB distance is approx. 0.75 A. In this calculation, the AB distance corresponds to a bonded pair of Hydrogen atoms. When this HH bond is broken, AB (HH) distance increases and BC (HF) distance decreases, the product region of the graph is entered and this can be seen to be lower energy.

The transition state of the surface can be found at approximately (AB,BC) = (0.746 A,1.789 A).

[[File:JFG HHF TS.png]]

By using the mep calculation type to find the energy minima for the reactants and products regions, the energy difference between these minima and the TS can be calculated. As such, the activation energy for the exothermic and endothermic reactions can be calculated.

Using this method, the following activation energies were calculated:

Exothermic activation energy: approx. 0.1 kCalmol-1

Endothermic activation energy: approx. 28.5 kCalmol-1

The following energy vs time plots exhibit these activation energies visually.

[[File:JFG HHF ACTIVATION ENERGY EXO.png]]

[[File:JFG HHF ACTIVATION ENERGY ENDO.png]]

=== Question 7 - In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.===

By establishing a reactive pathway from (H-H, F) to (H, H-F), the mechanism of the release of energy can be examined. Because the reaction is exothermic, the potential energy of the products is lower than the reactants. However, energy must be conserved and so the energy released by the reaction must adopt another form. The energy vs time graph shown below confirms that total energy remains constant, and that potential energy is reduced in the reaction. It also shows that excess energy is released as kinetic energy. Examining the momentum vs time graph helps to confirm that the energy is released into large vibrations in the HF bond in the products - kinetic energy.

[[File:JFG HHF REACTIVE ENERGY.png]]

[[File:JFG HHF REACTIVE MOMENTUM.png]]

=== Question 8 - 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 theorise that for an early TS (an exothermic reaction), translational energy is effective for overcoming the TS and producing a reaction. Vibrational energy of a bond, for example in the AB (HH) bond, has very little effect on the ability of the system to overcome the TS barrier and complete the reaction. For a late TS (an endothermic reaction), the opposite set of ideas apply.

Considering Polanyi's rules for the exothermic reaction with a late TS, we can expect that increasing the vibrational energy of the AB (HH) bond will be ineffective for overcoming the TS and resulting in products. This is demonstrated in the plots below, which XXX.

In plot 1, AB momentum = -3 and BC momentum = -0.5.
In plot 2, AB momentum = -3 and BC momentum = -0.8. Comparing to plot 1 shows the effect of increasing the magnitude of momentum in the BC distance (translational energy; effective).
In plot 3, AB momentum = 0.1 and BC momentum = -0.8. Comparing with plot 2 shows the effect of decreasing the magnitude of the AB (decreased vibrational energy; little effect on reactivity).

[[File:JFG POL EXO 1.png |‎| Plot 1 for exothermic reaction. ]]
[[File:JFG POL EXO 2.png |‎| Plot 2 for exothermic reaction.‎ ]]
[[File:JFG POL EXO 3.png ‎ |‎| Plot 3 for exothermic reaction.]]

Similarly considering the rules for the endothermic reaction with a late TS, we can expect that increasing the vibrational energy of the BC (HF) bond will be effective in producing a reactive pathway.

File:JFG POL EXO 3.png

2019-05-10T15:52:20Z

Jfg17:

File:JFG POL EXO 2.png

2019-05-10T15:52:11Z

Jfg17:

File:JFG POL EXO 1.png

2019-05-10T15:51:51Z

Jfg17:

MRD:JFG17 01347171

2019-05-10T15:26:29Z

== Molecular Reaction Dynamics Laboratory ==

=== Question 1 - On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===

A transition state is widely considered to be the point of maximum energy in a given reaction. As such, it is clear that the mathematical definition of a transition state should be inclusive of a description of a local maximum. However, to describe the true transition state on a potential surface, there must also be a local minimum at the very same point as the said local maximum; the transition state on a potential energy surface is a saddle point.

With respect to the systems under consideration:

If V''(r1) > 0 ; there is a local minimum wrt. r1.

If V''(r1) < 0 ; there is a local maximum wrt. r1.

However, a potential energy surface is a multi-variable function: a function of both r1 and r2. As such, the second derivative test should be used to confirm that the point under consideration (eg:(rx,ry)) is a true saddle point, not just a local max/min.

Defining the function D as the discriminant for this test:

D(r1,r2) = Vr1r1(r1,r2)Vr2r2(r1,r2)-(Vr1r2(r1,r2))2

If D < 0, the selected point (rx,ry) is a true saddle point - the transition state of the potential energy surface.

If D > 0, the selected point is only a local max/min.

=== Question 2 - 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. ===

The transition state of the (H-H, H) system is found at approximately r1=r2=0.90775/A.

[[File:JFG TS LOCATION.png|thumb|Contour plot showing location of TS.]]

[[File:JFG TS LOCATION1.png|thumb|Internuclear distance vs time graph.]]

The internuclear distance vs time graph demonstrates that from the starting point of (0.90775, 0.90775), internuclear distance does not change with time. This indicates that the system is completely stable at this point. There is no propensity for the atoms to move because the gradient of the potential energy surface is zero, in all directions. (Note: AB and BC) are overlayed in this graph.

=== Question 3 - Comment on how the mep and the trajectory you just calculated differ. ===

In a dynamic calculation, momentum is carried over from step to step. As such, vibrations in the bonded hydrogen atoms are visible because energy

=== Question 4 - Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ===

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy!! Reactive? !! Description of dynamics!!
|-
| -1.25 || -2.5 || -99.119 || Reactive || [[File:JFG TABLE PLOT1.png]] || Overall reactive because the original AB distance of 0.74 is somewhat mirrored in the BC distance at the endpoint of the calculation, and the same for BC to AB. AB distance remains approx. constant whilst BC distance is reducing – collision. AB distance then increases (bond shown to be broken), BC distance fluctuates around 0.75 due to energy of collision.

|-
| -1.5|| -2.0 || -100.456 || Unreactive || [[File:JFG TABLE PLOT2.png]] || Overall unreactive because TS not overcome. AB distance remains approx. 0.75, indicating bond remains unbroken. Reduction is BC distance represents collision, but energy of the system not correct to overcome TS. Small amount of oscillation in the AB distance (HH bond) visible due to vibrations in the bond.
|-
| -1.5 || -2.5 || -98.956 || Reactive || [[File:JFG TABLE PLOT3.png]] || Overall reactive because the TS is overcome and calculation ends in product region.
|-
| -2.5 || -5.0 || -84.956 || Unreactive || [[File:JFG TABLE PLOT4.png]] || Overall unreactive. Although the TS is passed, the oscillations shown in the plot show that the system returns to the reactant region. There is a high amount of vibration in the AB (HH) bond, perhaps indicating that rather than a minimum amount of energy to react, there is an optimum energy region that will allow for reaction.
|-
| -2.5 || -5.2 || -83.416 || Reactive || [[File:JFG TABLE PLOT5.png]] || Overall reactive. Despite a high amount of vibration in the AB (HH) bond as before, there is a slight increase in the amount of momentum of the BC distance. BC distance ends up being approx. 1 Angstrom, but there is a high amount of oscillation in the potential energy of the products.
|}

The data in the table demonstrates that energy is not the only factor that determines whether this reaction occurs. In the fourth example in the table for example, there are momenta values but an unreactive calculation.

=== Question 5 - 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? ===

=== Question 6 - By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? Locate the approximate position of the transition state. Report the activation energy for both reactions. ===

Bond energies:
HH bond: 436 kJmol-1
HF bond: 569 kJmol-1

Using these values, it is predicted that the reaction from the (H-H, F) system to the (H, H-F) system will be exothermic; breaking the HH bond and making the HF bond corresponds to a reaction energy of -133 kJmol-1. This prediction is confirmed by the shape of the potential energy surface for the reaction, shown below.

[[File:JFG HHF SURFACE.png]]

The z axis of the surface, the energy axis, clearly exhibits a higher energy system whilst the AB distance is approx. 0.75 A. In this calculation, the AB distance corresponds to a bonded pair of Hydrogen atoms. When this HH bond is broken, AB (HH) distance increases and BC (HF) distance decreases, the product region of the graph is entered and this can be seen to be lower energy.

The transition state of the surface can be found at approximately (AB,BC) = (0.746 A,1.789 A).

[[File:JFG HHF TS.png]]

By using the mep calculation type to find the energy minima for the reactants and products regions, the energy difference between these minima and the TS can be calculated. As such, the activation energy for the exothermic and endothermic reactions can be calculated.

Using this method, the following activation energies were calculated:

Exothermic activation energy: approx. 0.1 kCalmol-1

Endothermic activation energy: approx. 28.5 kCalmol-1

The following energy vs time plots exhibit these activation energies visually.

[[File:JFG HHF ACTIVATION ENERGY EXO.png]]

[[File:JFG HHF ACTIVATION ENERGY ENDO.png]]

=== Question 7 - In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.===

By establishing a reactive pathway from (H-H, F) to (H, H-F), the mechanism of the release of energy can be examined. Because the reaction is exothermic, the potential energy of the products is lower than the reactants. However, energy must be conserved and so the energy released by the reaction must adopt another form. The energy vs time graph shown below confirms that total energy remains constant, and that potential energy is reduced in the reaction. It also shows that excess energy is released as kinetic energy. Examining the momentum vs time graph helps to confirm that the energy is released into large vibrations in the HF bond in the products - kinetic energy.

[[File:JFG HHF REACTIVE ENERGY.png]]

[[File:JFG HHF REACTIVE MOMENTUM.png]]

=== Question 8 - 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 theorise that for an early TS (an exothermic reaction), translational energy is effective for overcoming the TS and producing a reaction. Vibrational energy of a bond, for example in the AB (HH) bond, has very little effect on the ability of the system to overcome the TS barrier and complete the reaction. For a late TS (an endothermic reaction), the opposite set of ideas apply.

Considering Polanyi's rules for the exothermic reaction with a late TS, we can expect that increasing the vibrational energy of the AB (HH) bond will be ineffective for overcoming the TS and resulting in products. This is demonstrated in the plots below, which XXX.

Similarly considering the rules for the endothermic reaction with a late TS, we can expect that increasing the vibrational energy of the BC (HF) bond will be effective in producing a reactive pathway.

MRD:JFG17 01347171

2019-05-10T15:10:13Z

MRD:JFG17 01347171

2019-05-10T15:06:14Z

MRD:JFG17 01347171

2019-05-10T15:02:57Z

Jfg17: /* Question 7 - In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally. */

MRD:JFG17 01347171

2019-05-10T14:59:12Z

File:JFG HHF ACTIVATION ENERGY ENDO.png

2019-05-10T14:58:37Z

Jfg17:

File:JFG HHF ACTIVATION ENERGY EXO.png

2019-05-10T14:58:21Z

Jfg17:

MRD:JFG17 01347171

2019-05-10T14:39:15Z

MRD:JFG17 01347171

2019-05-10T14:31:01Z

MRD:JFG17 01347171

2019-05-10T14:29:49Z

MRD:JFG17 01347171

2019-05-10T14:25:24Z

Jfg17: /* Question 3 - Comment on how the mep and the trajectory you just calculated differ. */

MRD:JFG17 01347171

2019-05-10T14:20:48Z

Jfg17: /* Molecular Reaction Dynamics Laboratory */

MRD:JFG17 01347171

2019-05-10T14:19:30Z

Jfg17: /* Molecular Reaction Dynamics Laboratory */

MRD:JFG17 01347171

2019-05-10T13:59:14Z

Jfg17: /* Question 7 */

== Molecular Reaction Dynamics Laboratory ==

=== Question 1 ===
==== On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ====

A transition state is widely considered to be the point of maximum energy in a given reaction. As such, it is clear that the mathematical definition of a transition state should be inclusive of a description of a local maximum. However, to describe the true transition state on a potential surface, there must also be a local minimum at the very same point as the said local maximum; the transition state on a potential energy surface is a saddle point.

With respect to the systems under consideration:

If V''(r1) > 0 ; there is a local minimum wrt. r1.

If V''(r1) < 0 ; there is a local maximum wrt. r1.

However, a potential energy surface is a multi-variable function: a function of both r1 and r2. As such, the second derivative test should be used to confirm that the point under consideration (eg:(rx,ry)) is a true saddle point, not just a local max/min.

Defining the function D as the discriminant for this test:

D(r1,r2) = Vr1r1(r1,r2)Vr2r2(r1,r2)-(Vr1r2(r1,r2))2

If D < 0, the selected point (rx,ry) is a true saddle point - the transition state of the potential energy surface.

If D > 0, the selected point is only a local max/min.

=== Question 2 ===
==== 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. ====

The transition state of the (H-H, H) system is found at approximately r1=r2=0.90775/A.

[[File:JFG TS LOCATION.png]]

[[File:JFG TS LOCATION1.png]]

The internuclear distance vs time graph demonstrates that from the starting point of (0.90775, 0.90775), internuclear distance does not change with time. This indicates that the system is completely stable at this point. There is no propensity for the atoms to move because the gradient of the potential energy surface is zero, in all directions. (Note: AB and BC) are overlayed in this graph.

=== Question 3 ===
==== Comment on how the mep and the trajectory you just calculated differ. ====
=== Question 4 ===
==== Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ====

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy!! Reactive? !! Description of dynamics!!
|-
| -1.25 || -2.5 || -99.119 || Reactive || [[File:JFG TABLE PLOT1.png]] || cell
|-
| -1.5|| -2.0 || -100.456 || Unreactive || [[File:JFG TABLE PLOT2.png]] || cell
|-
| -1.5 || -2.5 || -98.956 || Reactive || [[File:JFG TABLE PLOT3.png]] || cell
|-
| -2.5 || -5.0 || -84.956 || Unreactive || [[File:JFG TABLE PLOT4.png]] || cell
|-
| -2.5 || -5.2 || -83.416 || Reactive || [[File:JFG TABLE PLOT5.png]] || cell
|}

=== Question 5 ===
==== 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? ====
=== Question 6 ===
==== By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? Locate the approximate position of the transition state. Report the activation energy for both reactions. ====

Bond energies:
HH bond: 436 kJmol-1
HF bond: 569 kJmol-1

Using these values, it is predicted that the reaction from the (H-H, F) system to the (H, H-F) system will be exothermic; breaking the HH bond and making the HF bond corresponds to a reaction energy of -133 kJmol-1. This prediction is confirmed by the shape of the potential energy surface for the reaction, shown below.

[[File:JFG HHF SURFACE.png]]

The z axis of the surface, the energy axis, clearly exhibits a higher energy system whilst the AB distance is approx. 0.75 A. In this calculation, the AB distance corresponds to a bonded pair of Hydrogen atoms. When this HH bond is broken, AB (HH) distance increases and BC (HF) distance decreases, the product region of the graph is entered and this can be seen to be lower energy.

The transition state of the surface can be found at approximately (AB,BC) = (0.746 A,1.789 A).

[[File:JFG HHF TS.png]]

=== Question 7 ===
==== In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally. ====

By establishing a reactive pathway from (H-H, F) to (H, H-F), the mechanism of the release of energy can be examined. Because the reaction is exothermic, the potential energy of the products is lower than the reactants. However, energy must be conserved and so the energy released by the reaction must adopt another form. The energy vs time graph shown below confirms that total energy remains constant, and that potential energy is reduced in the reaction. It also shows that excess energy is released as kinetic energy. Examining the momentum vs time graph helps to confirm that the kinetic energy is released into large vibrations in the HF bond in the products.

[[File:JFG HHF REACTIVE ENERGY.png]]

[[File:JFG HHF REACTIVE MOMENTUM.png]]

=== Question 8 ===
==== 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:JFG HHF REACTIVE MOMENTUM.png

2019-05-10T13:56:50Z

Jfg17:

File:JFG HHF REACTIVE ENERGY.png

2019-05-10T13:56:38Z

Jfg17:

File:JFG HHF REACTIVE.png

2019-05-10T13:42:50Z

Jfg17:

MRD:JFG17 01347171

2019-05-10T13:25:06Z

Jfg17: /* By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? Locate the approximate posit...

== Molecular Reaction Dynamics Laboratory ==

=== Question 1 ===
==== On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ====

A transition state is widely considered to be the point of maximum energy in a given reaction. As such, it is clear that the mathematical definition of a transition state should be inclusive of a description of a local maximum. However, to describe the true transition state on a potential surface, there must also be a local minimum at the very same point as the said local maximum; the transition state on a potential energy surface is a saddle point.

With respect to the systems under consideration:

If V''(r1) > 0 ; there is a local minimum wrt. r1.

If V''(r1) < 0 ; there is a local maximum wrt. r1.

However, a potential energy surface is a multi-variable function: a function of both r1 and r2. As such, the second derivative test should be used to confirm that the point under consideration (eg:(rx,ry)) is a true saddle point, not just a local max/min.

Defining the function D as the discriminant for this test:

D(r1,r2) = Vr1r1(r1,r2)Vr2r2(r1,r2)-(Vr1r2(r1,r2))2

If D < 0, the selected point (rx,ry) is a true saddle point - the transition state of the potential energy surface.

If D > 0, the selected point is only a local max/min.

=== Question 2 ===
==== 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. ====

The transition state of the (H-H, H) system is found at approximately r1=r2=0.90775/A.

[[File:JFG TS LOCATION.png]]

[[File:JFG TS LOCATION1.png]]

The internuclear distance vs time graph demonstrates that from the starting point of (0.90775, 0.90775), internuclear distance does not change with time. This indicates that the system is completely stable at this point. There is no propensity for the atoms to move because the gradient of the potential energy surface is zero, in all directions. (Note: AB and BC) are overlayed in this graph.

=== Question 3 ===
==== Comment on how the mep and the trajectory you just calculated differ. ====
=== Question 4 ===
==== Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ====

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy!! Reactive? !! Description of dynamics!!
|-
| -1.25 || -2.5 || -99.119 || Reactive || [[File:JFG TABLE PLOT1.png]] || cell
|-
| -1.5|| -2.0 || -100.456 || Unreactive || [[File:JFG TABLE PLOT2.png]] || cell
|-
| -1.5 || -2.5 || -98.956 || Reactive || [[File:JFG TABLE PLOT3.png]] || cell
|-
| -2.5 || -5.0 || -84.956 || Unreactive || [[File:JFG TABLE PLOT4.png]] || cell
|-
| -2.5 || -5.2 || -83.416 || Reactive || [[File:JFG TABLE PLOT5.png]] || cell
|}

=== Question 5 ===
==== 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? ====
=== Question 6 ===
==== By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? Locate the approximate position of the transition state. Report the activation energy for both reactions. ====

Bond energies:
HH bond: 436 kJmol-1
HF bond: 569 kJmol-1

Using these values, it is predicted that the reaction from the (H-H, F) system to the (H, H-F) system will be exothermic; breaking the HH bond and making the HF bond corresponds to a reaction energy of -133 kJmol-1. This prediction is confirmed by the shape of the potential energy surface for the reaction, shown below.

[[File:JFG HHF SURFACE.png]]

The z axis of the surface, the energy axis, clearly exhibits a higher energy system whilst the AB distance is approx. 0.75 A. In this calculation, the AB distance corresponds to a bonded pair of Hydrogen atoms. When this HH bond is broken, AB (HH) distance increases and BC (HF) distance decreases, the product region of the graph is entered and this can be seen to be lower energy.

The transition state of the surface can be found at approximately (AB,BC) = (0.746 A,1.789 A).

[[File:JFG HHF TS.png]]

=== Question 7 ===
==== In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally. ====
=== Question 8 ===
==== 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:JFG HHF TS.png

2019-05-10T13:20:28Z

Jfg17:

MRD:JFG17 01347171

2019-05-10T13:03:38Z

Jfg17: /* Question 6 */

== Molecular Reaction Dynamics Laboratory ==

=== Question 1 ===
==== On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ====

A transition state is widely considered to be the point of maximum energy in a given reaction. As such, it is clear that the mathematical definition of a transition state should be inclusive of a description of a local maximum. However, to describe the true transition state on a potential surface, there must also be a local minimum at the very same point as the said local maximum; the transition state on a potential energy surface is a saddle point.

With respect to the systems under consideration:

If V''(r1) > 0 ; there is a local minimum wrt. r1.

If V''(r1) < 0 ; there is a local maximum wrt. r1.

However, a potential energy surface is a multi-variable function: a function of both r1 and r2. As such, the second derivative test should be used to confirm that the point under consideration (eg:(rx,ry)) is a true saddle point, not just a local max/min.

Defining the function D as the discriminant for this test:

D(r1,r2) = Vr1r1(r1,r2)Vr2r2(r1,r2)-(Vr1r2(r1,r2))2

If D < 0, the selected point (rx,ry) is a true saddle point - the transition state of the potential energy surface.

If D > 0, the selected point is only a local max/min.

=== Question 2 ===
==== 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. ====

The transition state of the (H-H, H) system is found at approximately r1=r2=0.90775/A.

[[File:JFG TS LOCATION.png]]

[[File:JFG TS LOCATION1.png]]

The internuclear distance vs time graph demonstrates that from the starting point of (0.90775, 0.90775), internuclear distance does not change with time. This indicates that the system is completely stable at this point. There is no propensity for the atoms to move because the gradient of the potential energy surface is zero, in all directions. (Note: AB and BC) are overlayed in this graph.

=== Question 3 ===
==== Comment on how the mep and the trajectory you just calculated differ. ====
=== Question 4 ===
==== Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ====

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy!! Reactive? !! Description of dynamics!!
|-
| -1.25 || -2.5 || -99.119 || Reactive || [[File:JFG TABLE PLOT1.png]] || cell
|-
| -1.5|| -2.0 || -100.456 || Unreactive || [[File:JFG TABLE PLOT2.png]] || cell
|-
| -1.5 || -2.5 || -98.956 || Reactive || [[File:JFG TABLE PLOT3.png]] || cell
|-
| -2.5 || -5.0 || -84.956 || Unreactive || [[File:JFG TABLE PLOT4.png]] || cell
|-
| -2.5 || -5.2 || -83.416 || Reactive || [[File:JFG TABLE PLOT5.png]] || cell
|}

=== Question 5 ===
==== 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? ====
=== Question 6 ===
==== By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? Locate the approximate position of the transition state. Report the activation energy for both reactions. ====

Bond energies:
HH bond: 436 kJmol-1
HF bond: 569 kJmol-1

Using these values, it is predicted that the reaction from the (H-H, F) system to the (H, H-F) system will be exothermic; breaking the HH bond and making the HF bond corresponds to a reaction energy of -133 kJmol-1. This prediction is confirmed by the shape of the potential energy surface for the reaction, shown below.

[[File:JFG HHF SURFACE.png]]

The z axis of the surface, the energy axis, clearly exhibits a higher energy system whilst the AB distance is approx. 0.75 A. In this calculation, the AB distance corresponds to a bonded pair of Hydrogen atoms. When this HH bond is broken, AB (HH) distance increases and BC (HF) distance decreases, the product region of the graph is entered and this can be seen to be lower energy.

=== Question 7 ===
==== In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally. ====
=== Question 8 ===
==== 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:JFG17 01347171

2019-05-10T13:03:06Z

Jfg17: /* Question 6 */

== Molecular Reaction Dynamics Laboratory ==

=== Question 1 ===
==== On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ====

A transition state is widely considered to be the point of maximum energy in a given reaction. As such, it is clear that the mathematical definition of a transition state should be inclusive of a description of a local maximum. However, to describe the true transition state on a potential surface, there must also be a local minimum at the very same point as the said local maximum; the transition state on a potential energy surface is a saddle point.

With respect to the systems under consideration:

If V''(r1) > 0 ; there is a local minimum wrt. r1.

If V''(r1) < 0 ; there is a local maximum wrt. r1.

However, a potential energy surface is a multi-variable function: a function of both r1 and r2. As such, the second derivative test should be used to confirm that the point under consideration (eg:(rx,ry)) is a true saddle point, not just a local max/min.

Defining the function D as the discriminant for this test:

D(r1,r2) = Vr1r1(r1,r2)Vr2r2(r1,r2)-(Vr1r2(r1,r2))2

If D < 0, the selected point (rx,ry) is a true saddle point - the transition state of the potential energy surface.

If D > 0, the selected point is only a local max/min.

=== Question 2 ===
==== 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. ====

The transition state of the (H-H, H) system is found at approximately r1=r2=0.90775/A.

[[File:JFG TS LOCATION.png]]

[[File:JFG TS LOCATION1.png]]

The internuclear distance vs time graph demonstrates that from the starting point of (0.90775, 0.90775), internuclear distance does not change with time. This indicates that the system is completely stable at this point. There is no propensity for the atoms to move because the gradient of the potential energy surface is zero, in all directions. (Note: AB and BC) are overlayed in this graph.

=== Question 3 ===
==== Comment on how the mep and the trajectory you just calculated differ. ====
=== Question 4 ===
==== Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ====

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy!! Reactive? !! Description of dynamics!!
|-
| -1.25 || -2.5 || -99.119 || Reactive || [[File:JFG TABLE PLOT1.png]] || cell
|-
| -1.5|| -2.0 || -100.456 || Unreactive || [[File:JFG TABLE PLOT2.png]] || cell
|-
| -1.5 || -2.5 || -98.956 || Reactive || [[File:JFG TABLE PLOT3.png]] || cell
|-
| -2.5 || -5.0 || -84.956 || Unreactive || [[File:JFG TABLE PLOT4.png]] || cell
|-
| -2.5 || -5.2 || -83.416 || Reactive || [[File:JFG TABLE PLOT5.png]] || cell
|}

=== Question 5 ===
==== 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? ====
=== Question 6 ===
==== By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

Bond energies:
HH bond: 436 kJmol-1
HF bond: 569 kJmol-1

Using these values, it is predicted that the reaction from the (H-H, F) system to the (H, H-F) system will be exothermic; breaking the HH bond and making the HF bond corresponds to a reaction energy of -133 kJmol-1. This prediction is confirmed by the shape of the potential energy surface for the reaction, shown below.

[[File:JFG HHF SURFACE.png]]

The z axis of the surface, the energy axis, clearly exhibits a higher energy system whilst the AB distance is approx. 0.75 A. In this calculation, the AB distance corresponds to a bonded pair of Hydrogen atoms. When this HH bond is broken, AB (HH) distance increases and BC (HF) distance decreases, the product region of the graph is entered and this can be seen to be lower energy.

Locate the approximate position of the transition state.

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

=== Question 7 ===
==== In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally. ====
=== Question 8 ===
==== 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:JFG HHF SURFACE.png

2019-05-10T12:58:09Z

Jfg17:

MRD:JFG17 01347171

2019-05-10T12:41:32Z

Jfg17: /* Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? */

== Molecular Reaction Dynamics Laboratory ==

=== Question 1 ===
==== On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ====

A transition state is widely considered to be the point of maximum energy in a given reaction. As such, it is clear that the mathematical definition of a transition state should be inclusive of a description of a local maximum. However, to describe the true transition state on a potential surface, there must also be a local minimum at the very same point as the said local maximum; the transition state on a potential energy surface is a saddle point.

With respect to the systems under consideration:

If V''(r1) > 0 ; there is a local minimum wrt. r1.

If V''(r1) < 0 ; there is a local maximum wrt. r1.

However, a potential energy surface is a multi-variable function: a function of both r1 and r2. As such, the second derivative test should be used to confirm that the point under consideration (eg:(rx,ry)) is a true saddle point, not just a local max/min.

Defining the function D as the discriminant for this test:

D(r1,r2) = Vr1r1(r1,r2)Vr2r2(r1,r2)-(Vr1r2(r1,r2))2

If D < 0, the selected point (rx,ry) is a true saddle point - the transition state of the potential energy surface.

If D > 0, the selected point is only a local max/min.

=== Question 2 ===
==== 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. ====

The transition state of the (H-H, H) system is found at approximately r1=r2=0.90775/A.

[[File:JFG TS LOCATION.png]]

[[File:JFG TS LOCATION1.png]]

The internuclear distance vs time graph demonstrates that from the starting point of (0.90775, 0.90775), internuclear distance does not change with time. This indicates that the system is completely stable at this point. There is no propensity for the atoms to move because the gradient of the potential energy surface is zero, in all directions. (Note: AB and BC) are overlayed in this graph.

=== Question 3 ===
==== Comment on how the mep and the trajectory you just calculated differ. ====
=== Question 4 ===
==== Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ====

{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Total Energy!! Reactive? !! Description of dynamics!!
|-
| -1.25 || -2.5 || -99.119 || Reactive || [[File:JFG TABLE PLOT1.png]] || cell
|-
| -1.5|| -2.0 || -100.456 || Unreactive || [[File:JFG TABLE PLOT2.png]] || cell
|-
| -1.5 || -2.5 || -98.956 || Reactive || [[File:JFG TABLE PLOT3.png]] || cell
|-
| -2.5 || -5.0 || -84.956 || Unreactive || [[File:JFG TABLE PLOT4.png]] || cell
|-
| -2.5 || -5.2 || -83.416 || Reactive || [[File:JFG TABLE PLOT5.png]] || cell
|}

=== Question 5 ===
==== 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? ====
=== Question 6 ===
==== By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

Locate the approximate position of the transition state.

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

=== Question 7 ===
==== In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally. ====
=== Question 8 ===
==== 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:JFG TABLE PLOT5.png

2019-05-10T12:40:11Z

Jfg17:

File:JFG TABLE PLOT4.png

2019-05-10T12:40:00Z

Jfg17:

File:JFG TABLE PLOT3.png

2019-05-10T12:39:49Z

Jfg17:

File:JFG TABLE PLOT2.png

2019-05-10T12:39:34Z

Jfg17:

File:JFG TABLE PLOT1.png

2019-05-10T12:39:23Z

Jfg17:

MRD:JFG17 01347171

2019-05-10T12:33:20Z

MRD:JFG17 01347171

2019-05-10T12:28:13Z

Jfg17: /* Question 2 */

MRD:JFG17 01347171

2019-05-10T12:23:27Z

Jfg17: /* Question 2 */

File:JFG TS LOCATION1.png

2019-05-10T12:20:54Z

Jfg17: