ChemWiki - User contributions [en]

MRD:01524831.molecularreactiondynamics

2020-05-15T22:18:24Z

Hk3918: /* Report the activation energy for both reactions. */

== Exercise 1: H + H2 system ==

[[File:Surface_Plot_first_trajcectory_hk3918.png|thumb|Figure 1. A surface plot showing the transition state of a H + H2 system]]
[[File:Surface_Plot_transtion_point-h3_hk3918.png |thumb|Figure 2. A contour plot showing the transition state of a H + H2 system at 91 pm]]

[[File:Animation mep hk3918.png ‎|thumb|Figure 3. A surface plot showing the mep of a H + H2 system when slightly displacing BC distance]]

=== 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? ===
On a potential energy surface diagram , the transition state is mathematically defined as the saddle point.<ref name=":0">Steinfeld, Jeffrey I, Chemical kinetics and dynamics, Prentice Hall, 1999, ch. 10, pp. 287-321.</ref> The saddle point is a stationary point on the plot, where along the reaction path it is a maxima and perpendicular to the reaction path the saddle point is perceived to be a minima. The transition point can be identified in this simple example when the distances between the 3 hydrogen atoms are of equal distance. As shown in figure 1, the reaction path is oscillating at the saddle point. This depicts the vibrational freedom of the system at the transition state. A saddle point is mathematically found as shown below.

D = fxx.fyy - fxy2

<nowiki> </nowiki>If D > 0, fxx < 0 , local minima.

If D< 0 , then it is a saddle point.

=== 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 estimated '''rts '''is found to be around 91 pm. The transition state in this H + H2 system is defined to be the position where internuclear distances are equal, and where it is at a potential energy maxima in relation to the reaction path trajectory. From figure 1 it is evident that there are vibrations within the system, depicted by the oscillations on the ridge. The transition state position was found by determining at which position the internuclear distance was equal and constant, and where the potential energy was a maximum, and thus the stable point of the transition state.<ref name=":0" /> The saddle point was found as shown in figure 2.

=== Comment on how the ''mep'' and the trajectory you just calculated differ. ===
When calculating under mep the reaction path is depicted different to when the calculation type is set to dynamics. What is absent in the map calculation is the motion of the atoms. The vibrational motion of the three hydrogen atoms are not depicted as shown in figure 3. The initial condition is set so that the distance between AB is '''rts''' and the distance between BC deviates from '''rts '''by 1 pm, resulting in the reaction pathway following through the trajectory of minimum energy, and advances towards the HA + HB-HC product. The flat BC line indicates the fact that the mep calculation does not consider vibrational motion of atoms, and solely is helps depict the favoured trajectory when slightly displacing said atoms away from the transition state, as it chooses the minimum energy path. This could be useful to understand energy diagrams and help calculate activation energies.

==== Look at the “Internuclear Distances vs Time” and “Momenta vs Time”. What would change if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead? ====
The change of initial conditions to those above will change final product from HA + HB-HC to HA-HB + HC as it is the HB-HC bond distance that deviates away from the '''rts. '''As a result the internuclear distances vs time plot will now show AB to be constant over time. Momentum remains unchanged as initial momentum is 0 for both scenarios.

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

The 'Get last geometry' option was selected, and the momenta were then reversed under the dynamic calculation type. The plot remains unchanged but with reversed directions as expected. The reason for this is because the initial argument resulted in a trajectory towards a larger BC distance. Using the momenta of this trajectory but reversing the sign simply results in the hydrogen and hydrogen molecule to start in the final position, but now approach each other due to the negative momentum value.
=== 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"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot/kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Hydrogen atoms AB and C approach
eachother. Notably, the vibrational

motion present after the transition state is

not present before.
|[[File:Surface_Plot_question4_1_hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-4.1 </nowiki>
|<nowiki>-420.077</nowiki>
|No
|

Hydrogen atoms AB approaches C, however

before approaching the saddle point to AB

swiftly moves away from C

|[[File:Animation exercise2 p2 -4.1 hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|Hydrogen atom C approaches AB with a large
amount of momentum, resulting in a new BC Hydrogen

molecule forming, releasing the A hydrogen atom in the

reaction.
|[[File:Animation exercise2 p2 -5.1 p1 -3.1 hk3918.png|300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.1 </nowiki>
|<nowiki>-357.277</nowiki>
|No
|Hydrogen atoms AB and C approach each other very
quickly. Upon impact, large internuclear distance and velocities

are depicted, showing the B atom being tugged by both A and C.

In the end, B forms a molecule AB and moves away from C.
|[[File:Animation exercise2 p2 -10.1 hk3918.png |300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.6 </nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|Upon impact, the B atom is tugged between the two atoms A and
C, and is finally bonded to to atom C. A then moves away from

BC.
|[[File:Animation p2 -10.6 hk3918.png |300px]]
|}
The table above shows that a reaction isn't only reliant on the amount of kinetic energy in the system. As shown in the table above, there are instances where kinetic energy is higher than the activation energy, however no reaction takes place. One therefore needs to consider that successful collisions are not only reliant on the system having enough energy to break and reform bonds.

=== Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===
Transition State Theory has a fundamental assumption which will result in its predictions being different to experimental values. The assumption that all trajectories possessing enough kinetic energy to circumvent the activation barrier will be reactive will result in a larger ''Rcl '' and thus a larger rate prediction than the experimental outcome. As seen in the table above, not all collisions are reactive even if there is sufficient kinetic energy - the case of barrier recrossing perfectly illustrates this.

== Exercise 2 ==

=== 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? ===
[[File:Surface_Plot_HF_hk3918.png|thumb|Figure 4. A surface plot showing the trajectory of F + H2 to H + HF reaction]|300px]]The reaction is exothermic as shown in figure 4, there is a decrease in potential energy when passing the transition state and forming the new HF bond. This relates to bond-strength. H-H has a bond strength of 432 kJ/mol whilst H-F has a bond strength of 565 kJ/mol.<ref>Wired Chemist,http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html, (accessed May 2020).</ref> As HF is a polar bond as a result of the very electronegative F, there is an ionic element to the bond, resulting in a much greater attraction between the two atoms, thus releasing a lot of energy on formation. As the energy released is greater than that absorbed, the reaction is exothermic as depicted in the trajectory shown in figure 4.<ref>Laidler, Keith J, Chemical kinetics, HarperCollins, 1989, ch. 12.3, pp. 460-471.</ref> There is an energy well adjacent to the transition state, indicating the difference in energy between the reactant and product.<ref>Niels E. Henriksen and Flemming Y. Hansen, Theories of Molecular Reaction Dynamics: The Microscopic Foundation of Chemical Kinetics, Oxford University Press, Oxford, United Kingdom ; New York, NY, 2019, ch. 3, pp. 39-54 </ref>

[[File:Surface_Plot_hf-h_hk3918.png|thumb|Figure 5. A surface plot showing the trajectory of the FH + H to F + H2 reaction|300px]]

Figure 5 on the other hand shows an endothermic reaction, illustrated by the increase in potential energy along the trajectory, indicating that the energy absorbed to break the stronger H-F bond is greater than the energy released when forming the weaker H-H bond.

[[File:Plot small activation.png|thumb|Figure 6. Activation energy for the F +H2 reaction |300px]]

[[File:Plot activation big.png|thumb|Figure 7. Activation energy for the HF + H reaction |300px]]

[[File:Animation_momenta_hk3918.png |thumb|Figure 8. Momenta vs time plot for F + H2 reaction |300px]]

[[File:Animation_-74_-6.1.png ‎ |thumb|Figure 9. Contour plot for F + H2 reaction. Parameters: AB=200, BC=74, Pab=-1, Pbc=-6.9 |300px]]

[[File: Animation_low_momentum_hk3918.png ‎ |thumb|Figure 10. Contour plot for F + H2 reaction. Parameters: AB=200, BC=74, Pab=-1.6, Pbc=0.2 |300px]]

[[File:Animation low vibrational.png ‎ |thumb|Figure 11. Contour plot for the HF + H reaction where H has high momenta and HF has low initial momenta |300px]]

[[File:Animation highvirabtional.png‎ |thumb|Figure 12. Skew plot for the HF + H reaction where H has low momenta and HF has high initial momenta |300px]]

=== Locate the approximate position of the transition state ===
The transition state position was calculated to be when the H-F bond was 181.8 pm, and the H-H bond was 73.1 pm as shown in figure 5. Analoguos to the calculation when approximating the transition state position of the H + H2 system, in the F + H2 reaction, simulations were tested in order to acquire a constant bond length, at the maximum potential energy, and where the trajectory remained stable. As the F + H2 reaction was exothermic, adopting Hammonds postulate, the assumption was made that the reaction had an early transition state thus the region near the reactants was analysed. The mep calculation was used to remove the vibrational noise as it made it difficult to read the plots. As HF + H is a late transition state, resembling its product F + H2, it will have the same transition state positions. To confirm the transition state position, under mep calculations, incremental increases to the transition state bond length were made to make sure the trajectory rolled over the transition state to either the reactant or products.

=== Report the activation energy for both reactions. ===
In order to find the activation energy for HF + H, the bond distance of H-F transition state was first decreased by 1 pm. This then resulted in the energy falling to form the products H-F and H. The difference in this energy as shown in figure 7 was used to calculate the activation energy of the H-F + H endothermic reaction. The activation energy was estimated to be 120 kJ/mol. For the exothermic F + H2 reaction, the activation energy was calculated by trying to shift the trajectory towards to the reactants. This was done by thus increasing the F-H distance, forming the F + H2 reactant. The difference in this energy was very small as shown in figure 8 and was used to calculate the activation energy for the F + H2 reaction to be around 1.2 kJ/mol.

=== 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. ===
Figure 8 illustrates the transition between translational and vibrational energy in the F + H2 reaction. Initially, the H-H bond (B-C) is vibrationally excited shown by the oscillation, whilst the H-F(A-B) bond isn't as is indicated by the absence of an oscillation. The H-F bond is at the vibrational ground state . After the collision we see a transition. The H-F bond now is vibrationally excited illustrated by the oscillation in the momenta vs time plot, and the H-H bonds loses this vibrationally energy and now only has translational energy due to the breaking of the bond, depicted by the plateau after 30s. Experimentally IR could be used to see the progress of the reaction. Only molecules with a dipole are IR active, thus after the formation of a H-F bond, IR absorption absent in the reactant will start to appear with overtones suggesting higher vibrational energies reached.

=== 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. ===
Below, various examples of both reactions are used to illustrate Polanyl's rules.<ref>Steinfeld, Jeffrey I, Chemical kinetics and dynamics, Prentice Hall, 1999, ch. 9, pp. 272-274.</ref> Reactions with late transition states are more efficient when the initial trajectory possesses vibrational energy whereas reactions with early transition states are more efficient when the initial trajectory possesses translational energy. The following examples illustrate these rules.

==== Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, with a momentum pFH = -1.0 g.mol-1.pm.fs-1, and explore several values of pHH in the range -6.1 to 6.1 g.mol-1.pm.fs-1 (explore values also close to these limits). What do you observe? ====
When using the parameters above and setting the F-H value to 200 pm, at lower momentum values vibrational frequencies were obversed. The reactions did not therefore take place, satisfying Polyanyi's rules, where possessing translational energy is more efficient when there is an early barrier rather than vibrational energy. Even at higher momentum extremes shown in figure 9, the reaction does not go to completion, indicating even though a lot of energy is being put into the system, surpassing the activation energy , it is not an efficient reaction as there is not enough translational energy present, satisfying the rules.

==== For the same initial position, increase slightly the momentum pFH = -1.6 g.mol-1.pm.fs-1, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.2 g.mol-1.pm.fs-1. What do you observe now? ====
Figure 10 shows that, even at lower momentum values, resulting in lower vibrational energies the reaction still does not proceed to completion, as there is not enough translational energy in the reaction, further satisfying Polyanyi's rules.

==== Focus on the reverse reaction, H + HF. ====
This reaction has a late transition state, thus according to Polanyi's rules, a vibrational energy trajectory is more efficient. This was tested firstly by using a low H-F vibrational energy and a high H translational energy as shown in figure 11. As predicted the reaction did not proceed to completion as there was not enough vibrational energy in the reaction. Figure 12 shows an example where the reaction has a larger vibrational energy, and therefore is more efficient, resulting in a successful reaction, satisfying Polanyi's rules.

== References ==

MRD:01524831.molecularreactiondynamics

2020-05-15T22:16:28Z

Hk3918: /* Focus on the reverse reaction, H + HF. */

== Exercise 1: H + H2 system ==

[[File:Surface_Plot_first_trajcectory_hk3918.png|thumb|Figure 1. A surface plot showing the transition state of a H + H2 system]]
[[File:Surface_Plot_transtion_point-h3_hk3918.png |thumb|Figure 2. A contour plot showing the transition state of a H + H2 system at 91 pm]]

[[File:Animation mep hk3918.png ‎|thumb|Figure 3. A surface plot showing the mep of a H + H2 system when slightly displacing BC distance]]

=== 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? ===
On a potential energy surface diagram , the transition state is mathematically defined as the saddle point.<ref name=":0">Steinfeld, Jeffrey I, Chemical kinetics and dynamics, Prentice Hall, 1999, ch. 10, pp. 287-321.</ref> The saddle point is a stationary point on the plot, where along the reaction path it is a maxima and perpendicular to the reaction path the saddle point is perceived to be a minima. The transition point can be identified in this simple example when the distances between the 3 hydrogen atoms are of equal distance. As shown in figure 1, the reaction path is oscillating at the saddle point. This depicts the vibrational freedom of the system at the transition state. A saddle point is mathematically found as shown below.

D = fxx.fyy - fxy2

<nowiki> </nowiki>If D > 0, fxx < 0 , local minima.

If D< 0 , then it is a saddle point.

=== 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 estimated '''rts '''is found to be around 91 pm. The transition state in this H + H2 system is defined to be the position where internuclear distances are equal, and where it is at a potential energy maxima in relation to the reaction path trajectory. From figure 1 it is evident that there are vibrations within the system, depicted by the oscillations on the ridge. The transition state position was found by determining at which position the internuclear distance was equal and constant, and where the potential energy was a maximum, and thus the stable point of the transition state.<ref name=":0" /> The saddle point was found as shown in figure 2.

=== Comment on how the ''mep'' and the trajectory you just calculated differ. ===
When calculating under mep the reaction path is depicted different to when the calculation type is set to dynamics. What is absent in the map calculation is the motion of the atoms. The vibrational motion of the three hydrogen atoms are not depicted as shown in figure 3. The initial condition is set so that the distance between AB is '''rts''' and the distance between BC deviates from '''rts '''by 1 pm, resulting in the reaction pathway following through the trajectory of minimum energy, and advances towards the HA + HB-HC product. The flat BC line indicates the fact that the mep calculation does not consider vibrational motion of atoms, and solely is helps depict the favoured trajectory when slightly displacing said atoms away from the transition state, as it chooses the minimum energy path. This could be useful to understand energy diagrams and help calculate activation energies.

==== Look at the “Internuclear Distances vs Time” and “Momenta vs Time”. What would change if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead? ====
The change of initial conditions to those above will change final product from HA + HB-HC to HA-HB + HC as it is the HB-HC bond distance that deviates away from the '''rts. '''As a result the internuclear distances vs time plot will now show AB to be constant over time. Momentum remains unchanged as initial momentum is 0 for both scenarios.

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

The 'Get last geometry' option was selected, and the momenta were then reversed under the dynamic calculation type. The plot remains unchanged but with reversed directions as expected. The reason for this is because the initial argument resulted in a trajectory towards a larger BC distance. Using the momenta of this trajectory but reversing the sign simply results in the hydrogen and hydrogen molecule to start in the final position, but now approach each other due to the negative momentum value.
=== 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"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot/kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Hydrogen atoms AB and C approach
eachother. Notably, the vibrational

motion present after the transition state is

not present before.
|[[File:Surface_Plot_question4_1_hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-4.1 </nowiki>
|<nowiki>-420.077</nowiki>
|No
|

Hydrogen atoms AB approaches C, however

before approaching the saddle point to AB

swiftly moves away from C

|[[File:Animation exercise2 p2 -4.1 hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|Hydrogen atom C approaches AB with a large
amount of momentum, resulting in a new BC Hydrogen

molecule forming, releasing the A hydrogen atom in the

reaction.
|[[File:Animation exercise2 p2 -5.1 p1 -3.1 hk3918.png|300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.1 </nowiki>
|<nowiki>-357.277</nowiki>
|No
|Hydrogen atoms AB and C approach each other very
quickly. Upon impact, large internuclear distance and velocities

are depicted, showing the B atom being tugged by both A and C.

In the end, B forms a molecule AB and moves away from C.
|[[File:Animation exercise2 p2 -10.1 hk3918.png |300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.6 </nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|Upon impact, the B atom is tugged between the two atoms A and
C, and is finally bonded to to atom C. A then moves away from

BC.
|[[File:Animation p2 -10.6 hk3918.png |300px]]
|}
The table above shows that a reaction isn't only reliant on the amount of kinetic energy in the system. As shown in the table above, there are instances where kinetic energy is higher than the activation energy, however no reaction takes place. One therefore needs to consider that successful collisions are not only reliant on the system having enough energy to break and reform bonds.

=== Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===
Transition State Theory has a fundamental assumption which will result in its predictions being different to experimental values. The assumption that all trajectories possessing enough kinetic energy to circumvent the activation barrier will be reactive will result in a larger ''Rcl '' and thus a larger rate prediction than the experimental outcome. As seen in the table above, not all collisions are reactive even if there is sufficient kinetic energy - the case of barrier recrossing perfectly illustrates this.

== Exercise 2 ==

=== 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? ===
[[File:Surface_Plot_HF_hk3918.png|thumb|Figure 4. A surface plot showing the trajectory of F + H2 to H + HF reaction]|300px]]The reaction is exothermic as shown in figure 4, there is a decrease in potential energy when passing the transition state and forming the new HF bond. This relates to bond-strength. H-H has a bond strength of 432 kJ/mol whilst H-F has a bond strength of 565 kJ/mol.<ref>Wired Chemist,http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html, (accessed May 2020).</ref> As HF is a polar bond as a result of the very electronegative F, there is an ionic element to the bond, resulting in a much greater attraction between the two atoms, thus releasing a lot of energy on formation. As the energy released is greater than that absorbed, the reaction is exothermic as depicted in the trajectory shown in figure 4.<ref>Laidler, Keith J, Chemical kinetics, HarperCollins, 1989, ch. 12.3, pp. 460-471.</ref> There is an energy well adjacent to the transition state, indicating the difference in energy between the reactant and product.<ref>Niels E. Henriksen and Flemming Y. Hansen, Theories of Molecular Reaction Dynamics: The Microscopic Foundation of Chemical Kinetics, Oxford University Press, Oxford, United Kingdom ; New York, NY, 2019, ch. 3, pp. 39-54 </ref>

[[File:Surface_Plot_hf-h_hk3918.png|thumb|Figure 5. A surface plot showing the trajectory of the FH + H to F + H2 reaction|300px]]

Figure 5 on the other hand shows an endothermic reaction, illustrated by the increase in potential energy along the trajectory, indicating that the energy absorbed to break the stronger H-F bond is greater than the energy released when forming the weaker H-H bond.

[[File:Plot small activation.png|thumb|Figure 6. Activation energy for the F +H2 reaction |300px]]

[[File:Plot activation big.png|thumb|Figure 7. Activation energy for the HF + H reaction |300px]]

[[File:Animation_momenta_hk3918.png |thumb|Figure 8. Momenta vs time plot for F + H2 reaction |300px]]

[[File:Animation_-74_-6.1.png ‎ |thumb|Figure 9. Contour plot for F + H2 reaction. Parameters: AB=200, BC=74, Pab=-1, Pbc=-6.9 |300px]]

[[File: Animation_low_momentum_hk3918.png ‎ |thumb|Figure 10. Contour plot for F + H2 reaction. Parameters: AB=200, BC=74, Pab=-1.6, Pbc=0.2 |300px]]

[[File:Animation low vibrational.png ‎ |thumb|Figure 11. Contour plot for the HF + H reaction where H has high momenta and HF has low initial momenta |300px]]

[[File:Animation highvirabtional.png‎ |thumb|Figure 12. Skew plot for the HF + H reaction where H has low momenta and HF has high initial momenta |300px]]

=== Locate the approximate position of the transition state ===
The transition state position was calculated to be when the H-F bond was 181.8 pm, and the H-H bond was 73.1 pm as shown in figure 5. Analoguos to the calculation when approximating the transition state position of the H + H2 system, in the F + H2 reaction, simulations were tested in order to acquire a constant bond length, at the maximum potential energy, and where the trajectory remained stable. As the F + H2 reaction was exothermic, adopting Hammonds postulate, the assumption was made that the reaction had an early transition state thus the region near the reactants was analysed. The mep calculation was used to remove the vibrational noise as it made it difficult to read the plots. As HF + H is a late transition state, resembling its product F + H2, it will have the same transition state positions. To confirm the transition state position, under mep calculations, incremental increases to the transition state bond length were made to make sure the trajectory rolled over the transition state to either the reactant or products.

=== Report the activation energy for both reactions. ===
In order to find the activation energy for HF + H, the bond distance of H-F transition state was first decreased by 1 pm. This then resulted in the energy falling to form the products H-F and H. The difference in this energy as shown in figure 7 was used to calculate the activation energy of the H-F + H endothermic reaction. The activation energy was estimated to be 120 kJ/mol. For the exothermic F + H2 reaction, the activation energy was calculated by trying to shift the trajectory towards to the reactants. This was done by thus increasing the F-H distance forming the F + H2 reactant. The difference in this energy was very small as shown in figure 8 and was used to calculate the activation energy for the F + H2 reaction to be around 1.2 kJ/mol.

=== 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. ===
Figure 8 illustrates the transition between translational and vibrational energy in the F + H2 reaction. Initially, the H-H bond (B-C) is vibrationally excited shown by the oscillation, whilst the H-F(A-B) bond isn't as is indicated by the absence of an oscillation. The H-F bond is at the vibrational ground state . After the collision we see a transition. The H-F bond now is vibrationally excited illustrated by the oscillation in the momenta vs time plot, and the H-H bonds loses this vibrationally energy and now only has translational energy due to the breaking of the bond, depicted by the plateau after 30s. Experimentally IR could be used to see the progress of the reaction. Only molecules with a dipole are IR active, thus after the formation of a H-F bond, IR absorption absent in the reactant will start to appear with overtones suggesting higher vibrational energies reached.

=== 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. ===
Below, various examples of both reactions are used to illustrate Polanyl's rules.<ref>Steinfeld, Jeffrey I, Chemical kinetics and dynamics, Prentice Hall, 1999, ch. 9, pp. 272-274.</ref> Reactions with late transition states are more efficient when the initial trajectory possesses vibrational energy whereas reactions with early transition states are more efficient when the initial trajectory possesses translational energy. The following examples illustrate these rules.

==== Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, with a momentum pFH = -1.0 g.mol-1.pm.fs-1, and explore several values of pHH in the range -6.1 to 6.1 g.mol-1.pm.fs-1 (explore values also close to these limits). What do you observe? ====
When using the parameters above and setting the F-H value to 200 pm, at lower momentum values vibrational frequencies were obversed. The reactions did not therefore take place, satisfying Polyanyi's rules, where possessing translational energy is more efficient when there is an early barrier rather than vibrational energy. Even at higher momentum extremes shown in figure 9, the reaction does not go to completion, indicating even though a lot of energy is being put into the system, surpassing the activation energy , it is not an efficient reaction as there is not enough translational energy present, satisfying the rules.

==== For the same initial position, increase slightly the momentum pFH = -1.6 g.mol-1.pm.fs-1, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.2 g.mol-1.pm.fs-1. What do you observe now? ====
Figure 10 shows that, even at lower momentum values, resulting in lower vibrational energies the reaction still does not proceed to completion, as there is not enough translational energy in the reaction, further satisfying Polyanyi's rules.

==== Focus on the reverse reaction, H + HF. ====
This reaction has a late transition state, thus according to Polanyi's rules, a vibrational energy trajectory is more efficient. This was tested firstly by using a low H-F vibrational energy and a high H translational energy as shown in figure 11. As predicted the reaction did not proceed to completion as there was not enough vibrational energy in the reaction. Figure 12 shows an example where the reaction has a larger vibrational energy, and therefore is more efficient, resulting in a successful reaction, satisfying Polanyi's rules.

== References ==

MRD:01524831.molecularreactiondynamics

2020-05-15T22:15:39Z

Hk3918: /* Locate the approximate position of the transition state */

== Exercise 1: H + H2 system ==

[[File:Surface_Plot_first_trajcectory_hk3918.png|thumb|Figure 1. A surface plot showing the transition state of a H + H2 system]]
[[File:Surface_Plot_transtion_point-h3_hk3918.png |thumb|Figure 2. A contour plot showing the transition state of a H + H2 system at 91 pm]]

[[File:Animation mep hk3918.png ‎|thumb|Figure 3. A surface plot showing the mep of a H + H2 system when slightly displacing BC distance]]

=== 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? ===
On a potential energy surface diagram , the transition state is mathematically defined as the saddle point.<ref name=":0">Steinfeld, Jeffrey I, Chemical kinetics and dynamics, Prentice Hall, 1999, ch. 10, pp. 287-321.</ref> The saddle point is a stationary point on the plot, where along the reaction path it is a maxima and perpendicular to the reaction path the saddle point is perceived to be a minima. The transition point can be identified in this simple example when the distances between the 3 hydrogen atoms are of equal distance. As shown in figure 1, the reaction path is oscillating at the saddle point. This depicts the vibrational freedom of the system at the transition state. A saddle point is mathematically found as shown below.

D = fxx.fyy - fxy2

<nowiki> </nowiki>If D > 0, fxx < 0 , local minima.

If D< 0 , then it is a saddle point.

=== 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 estimated '''rts '''is found to be around 91 pm. The transition state in this H + H2 system is defined to be the position where internuclear distances are equal, and where it is at a potential energy maxima in relation to the reaction path trajectory. From figure 1 it is evident that there are vibrations within the system, depicted by the oscillations on the ridge. The transition state position was found by determining at which position the internuclear distance was equal and constant, and where the potential energy was a maximum, and thus the stable point of the transition state.<ref name=":0" /> The saddle point was found as shown in figure 2.

=== Comment on how the ''mep'' and the trajectory you just calculated differ. ===
When calculating under mep the reaction path is depicted different to when the calculation type is set to dynamics. What is absent in the map calculation is the motion of the atoms. The vibrational motion of the three hydrogen atoms are not depicted as shown in figure 3. The initial condition is set so that the distance between AB is '''rts''' and the distance between BC deviates from '''rts '''by 1 pm, resulting in the reaction pathway following through the trajectory of minimum energy, and advances towards the HA + HB-HC product. The flat BC line indicates the fact that the mep calculation does not consider vibrational motion of atoms, and solely is helps depict the favoured trajectory when slightly displacing said atoms away from the transition state, as it chooses the minimum energy path. This could be useful to understand energy diagrams and help calculate activation energies.

==== Look at the “Internuclear Distances vs Time” and “Momenta vs Time”. What would change if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead? ====
The change of initial conditions to those above will change final product from HA + HB-HC to HA-HB + HC as it is the HB-HC bond distance that deviates away from the '''rts. '''As a result the internuclear distances vs time plot will now show AB to be constant over time. Momentum remains unchanged as initial momentum is 0 for both scenarios.

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

The 'Get last geometry' option was selected, and the momenta were then reversed under the dynamic calculation type. The plot remains unchanged but with reversed directions as expected. The reason for this is because the initial argument resulted in a trajectory towards a larger BC distance. Using the momenta of this trajectory but reversing the sign simply results in the hydrogen and hydrogen molecule to start in the final position, but now approach each other due to the negative momentum value.
=== 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"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot/kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Hydrogen atoms AB and C approach
eachother. Notably, the vibrational

motion present after the transition state is

not present before.
|[[File:Surface_Plot_question4_1_hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-4.1 </nowiki>
|<nowiki>-420.077</nowiki>
|No
|

Hydrogen atoms AB approaches C, however

before approaching the saddle point to AB

swiftly moves away from C

|[[File:Animation exercise2 p2 -4.1 hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|Hydrogen atom C approaches AB with a large
amount of momentum, resulting in a new BC Hydrogen

molecule forming, releasing the A hydrogen atom in the

reaction.
|[[File:Animation exercise2 p2 -5.1 p1 -3.1 hk3918.png|300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.1 </nowiki>
|<nowiki>-357.277</nowiki>
|No
|Hydrogen atoms AB and C approach each other very
quickly. Upon impact, large internuclear distance and velocities

are depicted, showing the B atom being tugged by both A and C.

In the end, B forms a molecule AB and moves away from C.
|[[File:Animation exercise2 p2 -10.1 hk3918.png |300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.6 </nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|Upon impact, the B atom is tugged between the two atoms A and
C, and is finally bonded to to atom C. A then moves away from

BC.
|[[File:Animation p2 -10.6 hk3918.png |300px]]
|}
The table above shows that a reaction isn't only reliant on the amount of kinetic energy in the system. As shown in the table above, there are instances where kinetic energy is higher than the activation energy, however no reaction takes place. One therefore needs to consider that successful collisions are not only reliant on the system having enough energy to break and reform bonds.

=== Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===
Transition State Theory has a fundamental assumption which will result in its predictions being different to experimental values. The assumption that all trajectories possessing enough kinetic energy to circumvent the activation barrier will be reactive will result in a larger ''Rcl '' and thus a larger rate prediction than the experimental outcome. As seen in the table above, not all collisions are reactive even if there is sufficient kinetic energy - the case of barrier recrossing perfectly illustrates this.

== Exercise 2 ==

=== 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? ===
[[File:Surface_Plot_HF_hk3918.png|thumb|Figure 4. A surface plot showing the trajectory of F + H2 to H + HF reaction]|300px]]The reaction is exothermic as shown in figure 4, there is a decrease in potential energy when passing the transition state and forming the new HF bond. This relates to bond-strength. H-H has a bond strength of 432 kJ/mol whilst H-F has a bond strength of 565 kJ/mol.<ref>Wired Chemist,http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html, (accessed May 2020).</ref> As HF is a polar bond as a result of the very electronegative F, there is an ionic element to the bond, resulting in a much greater attraction between the two atoms, thus releasing a lot of energy on formation. As the energy released is greater than that absorbed, the reaction is exothermic as depicted in the trajectory shown in figure 4.<ref>Laidler, Keith J, Chemical kinetics, HarperCollins, 1989, ch. 12.3, pp. 460-471.</ref> There is an energy well adjacent to the transition state, indicating the difference in energy between the reactant and product.<ref>Niels E. Henriksen and Flemming Y. Hansen, Theories of Molecular Reaction Dynamics: The Microscopic Foundation of Chemical Kinetics, Oxford University Press, Oxford, United Kingdom ; New York, NY, 2019, ch. 3, pp. 39-54 </ref>

[[File:Surface_Plot_hf-h_hk3918.png|thumb|Figure 5. A surface plot showing the trajectory of the FH + H to F + H2 reaction|300px]]

Figure 5 on the other hand shows an endothermic reaction, illustrated by the increase in potential energy along the trajectory, indicating that the energy absorbed to break the stronger H-F bond is greater than the energy released when forming the weaker H-H bond.

[[File:Plot small activation.png|thumb|Figure 6. Activation energy for the F +H2 reaction |300px]]

[[File:Plot activation big.png|thumb|Figure 7. Activation energy for the HF + H reaction |300px]]

[[File:Animation_momenta_hk3918.png |thumb|Figure 8. Momenta vs time plot for F + H2 reaction |300px]]

[[File:Animation_-74_-6.1.png ‎ |thumb|Figure 9. Contour plot for F + H2 reaction. Parameters: AB=200, BC=74, Pab=-1, Pbc=-6.9 |300px]]

[[File: Animation_low_momentum_hk3918.png ‎ |thumb|Figure 10. Contour plot for F + H2 reaction. Parameters: AB=200, BC=74, Pab=-1.6, Pbc=0.2 |300px]]

[[File:Animation low vibrational.png ‎ |thumb|Figure 11. Contour plot for the HF + H reaction where H has high momenta and HF has low initial momenta |300px]]

[[File:Animation highvirabtional.png‎ |thumb|Figure 12. Skew plot for the HF + H reaction where H has low momenta and HF has high initial momenta |300px]]

=== Locate the approximate position of the transition state ===
The transition state position was calculated to be when the H-F bond was 181.8 pm, and the H-H bond was 73.1 pm as shown in figure 5. Analoguos to the calculation when approximating the transition state position of the H + H2 system, in the F + H2 reaction, simulations were tested in order to acquire a constant bond length, at the maximum potential energy, and where the trajectory remained stable. As the F + H2 reaction was exothermic, adopting Hammonds postulate, the assumption was made that the reaction had an early transition state thus the region near the reactants was analysed. The mep calculation was used to remove the vibrational noise as it made it difficult to read the plots. As HF + H is a late transition state, resembling its product F + H2, it will have the same transition state positions. To confirm the transition state position, under mep calculations, incremental increases to the transition state bond length were made to make sure the trajectory rolled over the transition state to either the reactant or products.

=== Report the activation energy for both reactions. ===
In order to find the activation energy for HF + H, the bond distance of H-F transition state was first decreased by 1 pm. This then resulted in the energy falling to form the products H-F and H. The difference in this energy as shown in figure 7 was used to calculate the activation energy of the H-F + H endothermic reaction. The activation energy was estimated to be 120 kJ/mol. For the exothermic F + H2 reaction, the activation energy was calculated by trying to shift the trajectory towards to the reactants. This was done by thus increasing the F-H distance forming the F + H2 reactant. The difference in this energy was very small as shown in figure 8 and was used to calculate the activation energy for the F + H2 reaction to be around 1.2 kJ/mol.

=== 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. ===
Figure 8 illustrates the transition between translational and vibrational energy in the F + H2 reaction. Initially, the H-H bond (B-C) is vibrationally excited shown by the oscillation, whilst the H-F(A-B) bond isn't as is indicated by the absence of an oscillation. The H-F bond is at the vibrational ground state . After the collision we see a transition. The H-F bond now is vibrationally excited illustrated by the oscillation in the momenta vs time plot, and the H-H bonds loses this vibrationally energy and now only has translational energy due to the breaking of the bond, depicted by the plateau after 30s. Experimentally IR could be used to see the progress of the reaction. Only molecules with a dipole are IR active, thus after the formation of a H-F bond, IR absorption absent in the reactant will start to appear with overtones suggesting higher vibrational energies reached.

=== 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. ===
Below, various examples of both reactions are used to illustrate Polanyl's rules.<ref>Steinfeld, Jeffrey I, Chemical kinetics and dynamics, Prentice Hall, 1999, ch. 9, pp. 272-274.</ref> Reactions with late transition states are more efficient when the initial trajectory possesses vibrational energy whereas reactions with early transition states are more efficient when the initial trajectory possesses translational energy. The following examples illustrate these rules.

==== Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, with a momentum pFH = -1.0 g.mol-1.pm.fs-1, and explore several values of pHH in the range -6.1 to 6.1 g.mol-1.pm.fs-1 (explore values also close to these limits). What do you observe? ====
When using the parameters above and setting the F-H value to 200 pm, at lower momentum values vibrational frequencies were obversed. The reactions did not therefore take place, satisfying Polyanyi's rules, where possessing translational energy is more efficient when there is an early barrier rather than vibrational energy. Even at higher momentum extremes shown in figure 9, the reaction does not go to completion, indicating even though a lot of energy is being put into the system, surpassing the activation energy , it is not an efficient reaction as there is not enough translational energy present, satisfying the rules.

==== For the same initial position, increase slightly the momentum pFH = -1.6 g.mol-1.pm.fs-1, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.2 g.mol-1.pm.fs-1. What do you observe now? ====
Figure 10 shows that, even at lower momentum values, resulting in lower vibrational energies the reaction still does not proceed to completion, as there is not enough translational energy in the reaction, further satisfying Polyanyi's rules.

==== Focus on the reverse reaction, H + HF. ====
This reaction has a late transition state, thus according to Polanyi's rules, vibrational energy trajectory is more efficient. This was tested firstly by using a low H-F vibrational energy and a high H translational energy as shown in figure 11. As predicted the reaction did not proceed to completion as there was not enough vibrational energy in the reaction. Figure 12 shows an example where the reaction has a larger vibrational energy, and therefore is more efficient, resulting in a successful reaction.

== References ==

MRD:01524831.molecularreactiondynamics

2020-05-15T22:07:21Z

Hk3918:

== Exercise 1: H + H2 system ==

[[File:Surface_Plot_first_trajcectory_hk3918.png|thumb|Figure 1. A surface plot showing the transition state of a H + H2 system]]
[[File:Surface_Plot_transtion_point-h3_hk3918.png |thumb|Figure 2. A contour plot showing the transition state of a H + H2 system at 91 pm]]

[[File:Animation mep hk3918.png ‎|thumb|Figure 3. A surface plot showing the mep of a H + H2 system when slightly displacing BC distance]]

=== 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? ===
On a potential energy surface diagram , the transition state is mathematically defined as the saddle point.<ref name=":0">Steinfeld, Jeffrey I, Chemical kinetics and dynamics, Prentice Hall, 1999, ch. 10, pp. 287-321.</ref> The saddle point is a stationary point on the plot, where along the reaction path it is a maxima and perpendicular to the reaction path the saddle point is perceived to be a minima. The transition point can be identified in this simple example when the distances between the 3 hydrogen atoms are of equal distance. As shown in figure 1, the reaction path is oscillating at the saddle point. This depicts the vibrational freedom of the system at the transition state. A saddle point is mathematically found as shown below.

D = fxx.fyy - fxy2

<nowiki> </nowiki>If D > 0, fxx < 0 , local minima.

If D< 0 , then it is a saddle point.

=== 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 estimated '''rts '''is found to be around 91 pm. The transition state in this H + H2 system is defined to be the position where internuclear distances are equal, and where it is at a potential energy maxima in relation to the reaction path trajectory. From figure 1 it is evident that there are vibrations within the system, depicted by the oscillations on the ridge. The transition state position was found by determining at which position the internuclear distance was equal and constant, and where the potential energy was a maximum, and thus the stable point of the transition state.<ref name=":0" /> The saddle point was found as shown in figure 2.

=== Comment on how the ''mep'' and the trajectory you just calculated differ. ===
When calculating under mep the reaction path is depicted different to when the calculation type is set to dynamics. What is absent in the map calculation is the motion of the atoms. The vibrational motion of the three hydrogen atoms are not depicted as shown in figure 3. The initial condition is set so that the distance between AB is '''rts''' and the distance between BC deviates from '''rts '''by 1 pm, resulting in the reaction pathway following through the trajectory of minimum energy, and advances towards the HA + HB-HC product. The flat BC line indicates the fact that the mep calculation does not consider vibrational motion of atoms, and solely is helps depict the favoured trajectory when slightly displacing said atoms away from the transition state, as it chooses the minimum energy path. This could be useful to understand energy diagrams and help calculate activation energies.

==== Look at the “Internuclear Distances vs Time” and “Momenta vs Time”. What would change if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead? ====
The change of initial conditions to those above will change final product from HA + HB-HC to HA-HB + HC as it is the HB-HC bond distance that deviates away from the '''rts. '''As a result the internuclear distances vs time plot will now show AB to be constant over time. Momentum remains unchanged as initial momentum is 0 for both scenarios.

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

The 'Get last geometry' option was selected, and the momenta were then reversed under the dynamic calculation type. The plot remains unchanged but with reversed directions as expected. The reason for this is because the initial argument resulted in a trajectory towards a larger BC distance. Using the momenta of this trajectory but reversing the sign simply results in the hydrogen and hydrogen molecule to start in the final position, but now approach each other due to the negative momentum value.
=== 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"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot/kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Hydrogen atoms AB and C approach
eachother. Notably, the vibrational

motion present after the transition state is

not present before.
|[[File:Surface_Plot_question4_1_hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-4.1 </nowiki>
|<nowiki>-420.077</nowiki>
|No
|

Hydrogen atoms AB approaches C, however

before approaching the saddle point to AB

swiftly moves away from C

|[[File:Animation exercise2 p2 -4.1 hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|Hydrogen atom C approaches AB with a large
amount of momentum, resulting in a new BC Hydrogen

molecule forming, releasing the A hydrogen atom in the

reaction.
|[[File:Animation exercise2 p2 -5.1 p1 -3.1 hk3918.png|300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.1 </nowiki>
|<nowiki>-357.277</nowiki>
|No
|Hydrogen atoms AB and C approach each other very
quickly. Upon impact, large internuclear distance and velocities

are depicted, showing the B atom being tugged by both A and C.

In the end, B forms a molecule AB and moves away from C.
|[[File:Animation exercise2 p2 -10.1 hk3918.png |300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.6 </nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|Upon impact, the B atom is tugged between the two atoms A and
C, and is finally bonded to to atom C. A then moves away from

BC.
|[[File:Animation p2 -10.6 hk3918.png |300px]]
|}
The table above shows that a reaction isn't only reliant on the amount of kinetic energy in the system. As shown in the table above, there are instances where kinetic energy is higher than the activation energy, however no reaction takes place. One therefore needs to consider that successful collisions are not only reliant on the system having enough energy to break and reform bonds.

=== Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===
Transition State Theory has a fundamental assumption which will result in its predictions being different to experimental values. The assumption that all trajectories possessing enough kinetic energy to circumvent the activation barrier will be reactive will result in a larger ''Rcl '' and thus a larger rate prediction than the experimental outcome. As seen in the table above, not all collisions are reactive even if there is sufficient kinetic energy - the case of barrier recrossing perfectly illustrates this.

== Exercise 2 ==

=== 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? ===
[[File:Surface_Plot_HF_hk3918.png|thumb|Figure 4. A surface plot showing the trajectory of F + H2 to H + HF reaction]|300px]]The reaction is exothermic as shown in figure 4, there is a decrease in potential energy when passing the transition state and forming the new HF bond. This relates to bond-strength. H-H has a bond strength of 432 kJ/mol whilst H-F has a bond strength of 565 kJ/mol.<ref>Wired Chemist,http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html, (accessed May 2020).</ref> As HF is a polar bond as a result of the very electronegative F, there is an ionic element to the bond, resulting in a much greater attraction between the two atoms, thus releasing a lot of energy on formation. As the energy released is greater than that absorbed, the reaction is exothermic as depicted in the trajectory shown in figure 4.<ref>Laidler, Keith J, Chemical kinetics, HarperCollins, 1989, ch. 12.3, pp. 460-471.</ref> There is an energy well adjacent to the transition state, indicating the difference in energy between the reactant and product.<ref>Niels E. Henriksen and Flemming Y. Hansen, Theories of Molecular Reaction Dynamics: The Microscopic Foundation of Chemical Kinetics, Oxford University Press, Oxford, United Kingdom ; New York, NY, 2019, ch. 3, pp. 39-54 </ref>

[[File:Surface_Plot_hf-h_hk3918.png|thumb|Figure 5. A surface plot showing the trajectory of the FH + H to F + H2 reaction|300px]]

Figure 5 on the other hand shows an endothermic reaction, illustrated by the increase in potential energy along the trajectory, indicating that the energy absorbed to break the stronger H-F bond is greater than the energy released when forming the weaker H-H bond.

[[File:Plot small activation.png|thumb|Figure 6. Activation energy for the F +H2 reaction |300px]]

[[File:Plot activation big.png|thumb|Figure 7. Activation energy for the HF + H reaction |300px]]

[[File:Animation_momenta_hk3918.png |thumb|Figure 8. Momenta vs time plot for F + H2 reaction |300px]]

[[File:Animation_-74_-6.1.png ‎ |thumb|Figure 9. Contour plot for F + H2 reaction. Parameters: AB=200, BC=74, Pab=-1, Pbc=-6.9 |300px]]

[[File: Animation_low_momentum_hk3918.png ‎ |thumb|Figure 10. Contour plot for F + H2 reaction. Parameters: AB=200, BC=74, Pab=-1.6, Pbc=0.2 |300px]]

[[File:Animation low vibrational.png ‎ |thumb|Figure 11. Contour plot for the HF + H reaction where H has high momenta and HF has low initial momenta |300px]]

[[File:Animation highvirabtional.png‎ |thumb|Figure 12. Skew plot for the HF + H reaction where H has low momenta and HF has high initial momenta |300px]]

=== Locate the approximate position of the transition state ===
The transition state position was calculated to be when the H-F bond had a bond length of 181.8 pm, and the H-H bond was 73.1 pm as shown in figure 5. Analoguos to the calculation when approximating the transition state position of the H + H2 system, in the F + H2 reaction simulations were tested in order to acquire a constant bond length, at the maximum potential energy, and where the trajectory remained stable. As the F + H2 reaction was exothermic, adopting Hammonds postulate, the assumption was made that the reaction had an early transition state thus the region near the reactants was analysed. The mep calculation was used to remove the vibrational noise as it made it difficult to read the plots. As HF + H is a late transition state, resembling its product F + H2, it will have the same transition state positions. To confirm the transition state position, under mep calculations, incremental increases to the transition state bond length were made to make sure the trajectory rolled over the transition state to either the reactant or products.

=== Report the activation energy for both reactions. ===
In order to find the activation energy for HF + H, the bond distance of H-F transition state was first decreased by 1 pm. This then resulted in the energy falling to form the products H-F and H. The difference in this energy as shown in figure 7 was used to calculate the activation energy of the H-F + H endothermic reaction. The activation energy was estimated to be 120 kJ/mol. For the exothermic F + H2 reaction, the activation energy was calculated by trying to shift the trajectory towards to the reactants. This was done by thus increasing the F-H distance forming the F + H2 reactant. The difference in this energy was very small as shown in figure 8 and was used to calculate the activation energy for the F + H2 reaction to be around 1.2 kJ/mol.

=== 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. ===
Figure 8 illustrates the transition between translational and vibrational energy in the F + H2 reaction. Initially, the H-H bond (B-C) is vibrationally excited shown by the oscillation, whilst the H-F(A-B) bond isn't indicated by the absence of an oscillation. The H-F bond is at the vibrational ground state . After the collision we see a transition. The H-F bond now is vibrationally excited illustrated by the oscillation in the momenta vs time plot, and the H-H bonds loses this vibrationally energy and now only has translational energy due to the breaking of the bond, depicted by the plateau after 30s. Experimentally IR could be used to see the progress of the reaction. Only molecules with a dipole are IR active, thus after the formation of a H-F bond, IR absorption absent in the reactant will start to appear with overtones suggesting higher vibrational energies reached.

=== 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. ===
Below various example of both reactions are used to illustrate Polanyl's rules.<ref>Steinfeld, Jeffrey I, Chemical kinetics and dynamics, Prentice Hall, 1999, ch. 9, pp. 272-274.</ref> Reactions with late transition states are more efficient when the initial trajectory possesses vibrational energy whereas reactions with early transition states are more efficient when the initial trajectory possesses translational energy. The following examples illustrate these rules.

==== Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, with a momentum pFH = -1.0 g.mol-1.pm.fs-1, and explore several values of pHH in the range -6.1 to 6.1 g.mol-1.pm.fs-1 (explore values also close to these limits). What do you observe? ====
When using the parameters above and setting the F-H value to 200 pm, at lower momentum values larger vibrational frequencies were obversed. The reactions did not therefore take place, satisfying Polyanyi's rules, where the having translational energy is more efficient when there is an early barrier rather than vibrational energy. Even at higher momentum extremes shown in figure 9, the reaction does not go to completion, indicating even though a lot of energy is being put into the system , it is not an efficient reaction, satisfying the rules.

==== For the same initial position, increase slightly the momentum pFH = -1.6 g.mol-1.pm.fs-1, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.2 g.mol-1.pm.fs-1. What do you observe now? ====
Figure 10 shows that, even at lower momentum values, resulting in lower vibrational energies the reaction still does not proceed to completion. As there is not enough translational energy in the reaction, further satisfying Polyanyi's rules.

==== Focus on the reverse reaction, H + HF. ====
This reaction has a late transition state, thus according to Polanyi's rules, vibrational energy trajectory is more efficient. This was tested firstly by using a low H-F vibrational energy and a high H translational energy as shown in figure 11. As predicted the reaction did not proceed to completion as there was not enough vibrational energy in the reaction. Figure 12 shows an example where the reaction has a larger vibrational energy, and therefore is more efficient, resulting in a successful reaction.

== References ==

MRD:01524831.molecularreactiondynamics

2020-05-15T22:05:11Z

Hk3918:

== Exercise 1: H + H2 system ==

[[File:Surface_Plot_first_trajcectory_hk3918.png|thumb|Figure 1. A surface plot showing the transition state of a H + H2 system]]
[[File:Surface_Plot_transtion_point-h3_hk3918.png |thumb|Figure 2. A contour plot showing the transition state of a H + H2 system at 91 pm]]

[[File:Animation mep hk3918.png ‎|thumb|Figure 3. A surface plot showing the mep of a H + H2 system when slightly displacing BC distance]]

=== 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? ===
On a potential energy surface diagram , the transition state is mathematically defined as the saddle point.<ref name=":0">Steinfeld, Jeffrey I, Chemical kinetics and dynamics, Prentice Hall, 1999, ch. 10, pp. 287-321.</ref> The saddle point is a stationary point on the plot, where along the reaction path it is a maxima and perpendicular to the reaction path the saddle point is perceived to be a minima. The transition point can be identified in this simple example when the distances between the 3 hydrogen atoms are of equal distance. As shown in figure 1, the reaction path is oscillating at the saddle point. This depicts the vibrational freedom of the system at the transition state. A saddle point is mathematically found as shown below.

D = fxx.fyy - fxy2

<nowiki> </nowiki>If D > 0, fxx < 0 , local minima.

If D< 0 , then it is a saddle point.

=== 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 estimated '''rts '''is found to be around 91 pm. The transition state in this H + H2 system is defined to be the position where internuclear distances are equal, and where it is at a potential energy maxima in relation to the reaction path trajectory. From figure 1 it is evident that there are vibrations within the system, depicted by the oscillations on the ridge. The transition state position was found by determining at which position the internuclear distance was equal and constant, and where the potential energy was a maximum, and thus the stable point of the transition state.<ref name=":0" /> The saddle point was found as shown in figure 2.

=== Comment on how the ''mep'' and the trajectory you just calculated differ. ===
When calculating under mep the reaction path is depicted different to when the calculation type is set to dynamics. What is absent in the map calculation is the motion of the atoms. The vibrational motion of the three hydrogen atoms are not depicted as shown in figure 3. The initial condition is set so that the distance between AB is '''rts''' and the distance between BC deviates from '''rts '''by 1 pm, resulting in the reaction pathway following through the trajectory of minimum energy, and advances towards the HA + HB-HC product. The flat BC line indicates the fact that the mep calculation does not consider vibrational motion of atoms, and solely is helps depict the favoured trajectory when slightly displacing said atoms away from the transition state, as it chooses the minimum energy path. This could be useful to understand energy diagrams and help calculate activation energies.

==== Look at the “Internuclear Distances vs Time” and “Momenta vs Time”. What would change if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead? ====
The change of initial conditions to those above will change final product from HA + HB-HC to HA-HB + HC as it is the HB-HC bond distance that deviates away from the '''rts. '''As a result the internuclear distances vs time plot will now show AB to be constant over time. Momentum remains unchanged as initial momentum is 0 for both scenarios.

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

The 'Get last geometry' option was selected, and the momenta were then reversed under the dynamic calculation type. The plot remains unchanged but with reversed directions as expected. The reason for this is because the initial argument resulted in a trajectory towards a larger BC distance. Using the momenta of this trajectory but reversing the sign simply results in the hydrogen and hydrogen molecule to start in the final position, but now approach each other due to the negative momentum value.
=== 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"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot/kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Hydrogen atoms AB and C approach
eachother. Notably, the vibrational

motion present after the transition state is

not present before.
|[[File:Surface_Plot_question4_1_hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-4.1 </nowiki>
|<nowiki>-420.077</nowiki>
|No
|

Hydrogen atoms AB approaches C, however

before approaching the saddle point to AB

swiftly moves away from C

|[[File:Animation exercise2 p2 -4.1 hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|Hydrogen atom C approaches AB with a large
amount of momentum, resulting in a new BC Hydrogen

molecule forming, releasing the A hydrogen atom in the

reaction.
|[[File:Animation exercise2 p2 -5.1 p1 -3.1 hk3918.png|300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.1 </nowiki>
|<nowiki>-357.277</nowiki>
|No
|Hydrogen atoms AB and C approach each other very
quickly. Upon impact, large internuclear distance and velocities

are depicted, showing the B atom being tugged by both A and C.

In the end, B forms a molecule AB and moves away from C.
|[[File:Animation exercise2 p2 -10.1 hk3918.png |300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.6 </nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|Upon impact, the B atom is tugged between the two atoms A and
C, and is finally bonded to to atom C. A then moves away from

BC.
|[[File:Animation p2 -10.6 hk3918.png |300px]]
|}
The table above shows that a reaction isn't only reliant on the amount of kinetic energy in the system. As shown in the table above, there are instances where kinetic energy is higher than the activation energy, however no reaction takes place. One therefore needs to consider that successful collisions are not only reliant on the system having enough energy to break and reform bonds.

=== Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===
Transition State Theory has a fundamental assumption which will result in its predictions being different to experimental values. The assumption that all trajectories possessing enough kinetic energy to circumvent the activation barrier will be reactive will result in a larger ''Rcl '' and thus a larger rate prediction than the experimental outcome. As seen in the table above, not all collisions are reactive even if there is sufficient kinetic energy - the case of barrier recrossing perfectly illustrates this.

== Exercise 2 ==

=== 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? ===
[[File:Surface_Plot_HF_hk3918.png|thumb|Figure 4. A surface plot showing the trajectory of F + H2 to H + HF reaction]|300px]]The reaction is exothermic as shown in figure 4, there is a decrease in potential energy when passing the transition state and forming the new HF bond. This relates to bond-strength. H-H has a bond strength of 432 kJ/mol whilst H-F has a bond strength of 565 kJ/mol.<ref>Wired Chemist,http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html, (accessed May 2020).</ref> As HF is a polar bond as a result of the very electronegative F, there is an ionic element to the bond, resulting in a much greater attraction between the two atoms, thus releasing a lot of energy on formation. As the energy released is greater than that absorbed, the reaction is exothermic as depicted in the trajectory shown in figure 4.<ref>Laidler, Keith J, Chemical kinetics, HarperCollins, 1989, ch. 12.3, pp. 460-471.</ref> There is an energy well adjacent to the transition state, indicating the difference in energy between the reactant and product.<ref>Niels E. Henriksen and Flemming Y. Hansen, Theories of Molecular Reaction Dynamics: The Microscopic Foundation of Chemical Kinetics, Oxford University Press, Oxford, United Kingdom ; New York, NY, 2019, ch. 3, pp. 39-54 </ref>

[[File:Surface_Plot_hf-h_hk3918.png|thumb|Figure 5. A surface plot showing the trajectory of the FH + H to F + H2 reaction|300px]]

Figure 5 on the other hand shows an endothermic reaction, illustrated by the increase in potential energy along the trajectory, indicating that the energy absorbed to break the stronger H-F bond is greater than the energy released when forming the weaker H-H bond.

[[File:Plot small activation.png|thumb|Figure 6. Activation energy for the F +H2 reaction |300px]]

[[File:Plot activation big.png|thumb|Figure 7. Activation energy for the HF + H reaction |300px]]

[[File:Animation_momenta_hk3918.png |thumb|Figure 8. Momenta vs time plot for F + H2 reaction |300px]]

[[File:Animation_-74_-6.1.png ‎ |thumb|Figure 9. Contour plot for F + H2 reaction. Parameters: AB=200, BC=74, Pab=-1, Pbc=-6.9 |300px]]

[[File: Animation_low_momentum_hk3918.png ‎ |thumb|Figure 10. Contour plot for F + H2 reaction. Parameters: AB=200, BC=74, Pab=-1.6, Pbc=0.2 |300px]]

[[File:Animation low vibrational.png ‎ |thumb|Figure 11. Contour plot for the HF + H reaction where H has high momenta and HF has low initial momenta |300px]]

[[File:Animation highvirabtional.png‎ |thumb|Figure 12. Skew plot for the HF + H reaction where H has low momenta and HF has high initial momenta |300px]]

=== Locate the approximate position of the transition state ===
The transition state position was calculated to be when the H-F bond had a bond length of 181.8 pm, and the H-H bond was 73.1 pm as shown in figure 5. Analoguos to the calculation when approximating the transition state position of the H + H2 system, in the F + H2 reaction simulations were tested in order to acquire a constant bond length, at the maximum potential energy, and where the trajectory remained stable. As the F + H2 reaction was exothermic, adopting Hammonds postulate, the assumption was made that the reaction had an early transition state thus the region near the reactants was analysed. The mep calculation was used to remove the vibrational noise as it made it difficult to read the plots. As HF + H is a late transition state, resembling its product F + H2, it will have the same transition state positions. To confirm the transition state position, under mep calculations, incremental increases to the transition state bond length were made to make sure the trajectory rolled over the transition state to either the reactant or products.

=== Report the activation energy for both reactions. ===
In order to find the activation energy for HF + H, the bond distance of H-F transition state was first decreased by 1 pm. This then resulted in the energy falling to form the products H-F and H. The difference in this energy as shown in figure 7 was used to calculate the activation energy of the H-F + H endothermic reaction. The activation energy was estimated to be 120 kJ/mol. For the exothermic F + H2 reaction, the activation energy was calculated by trying to shift the trajectory towards to the reactants. This was done by thus increasing the F-H distance forming the F + H2 reactant. The difference in this energy was very small as shown in figure 8 and was used to calculate the activation energy for the F + H2 reaction to be around 1.2 kJ/mol.

=== 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. ===
Figure 8 illustrates the transition between translational and vibrational energy in the F + H2 reaction. Initially, the H-H bond (B-C) is vibrationally excited shown by the oscillation, whilst the H-F(A-B) bond isn't indicated by the absence of an oscillation. The H-F bond is at the vibrational ground state . After the collision we see a transition. The H-F bond now is vibrationally excited illustrated by the oscillation in the momenta vs time plot, and the H-H bonds loses this vibrationally energy and now only has translational energy due to the breaking of the bond, depicted by the plateau after 30s. Experimentally IR could be used to see the progress of the reaction. Only molecules with a dipole are IR active, thus after the formation of a H-F bond, IR absorption absent in the reactant will start to appear with overtones suggesting higher vibrational energies reached.

=== 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. ===
Below various example of both reactions are used to illustrate Polanyl's rules. Reactions with late transition states are more efficient when the initial trajectory possesses vibrational energy whereas reactions with early transition states are more efficient when the initial trajectory possesses translational energy. The following examples illustrate these rules.

==== Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, with a momentum pFH = -1.0 g.mol-1.pm.fs-1, and explore several values of pHH in the range -6.1 to 6.1 g.mol-1.pm.fs-1 (explore values also close to these limits). What do you observe? ====
When using the parameters above and setting the F-H value to 200 pm, at lower momentum values larger vibrational frequencies were obversed. The reactions did not therefore take place, satisfying Polyanyi's rules, where the having translational energy is more efficient when there is an early barrier rather than vibrational energy. Even at higher momentum extremes shown in figure 9, the reaction does not go to completion, indicating even though a lot of energy is being put into the system , it is not an efficient reaction, satisfying the rules.

==== For the same initial position, increase slightly the momentum pFH = -1.6 g.mol-1.pm.fs-1, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.2 g.mol-1.pm.fs-1. What do you observe now? ====
Figure 10 shows that, even at lower momentum values, resulting in lower vibrational energies the reaction still does not proceed to completion. As there is not enough translational energy in the reaction, further satisfying Polyanyi's rules.

==== Focus on the reverse reaction, H + HF. ====
This reaction has a late transition state, thus according to Polanyi's rules, vibrational energy trajectory is more efficient. This was tested firstly by using a low H-F vibrational energy and a high H translational energy as shown in figure 11. As predicted the reaction did not proceed to completion as there was not enough vibrational energy in the reaction. Figure 12 shows an example where the reaction has a larger vibrational energy, and therefore is more efficient, resulting in a successful reaction.

MRD:01524831.molecularreactiondynamics

2020-05-15T21:31:59Z

Hk3918: /* 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? */

== Exercise 1: H + H2 system ==

[[File:Surface_Plot_first_trajcectory_hk3918.png|thumb|Figure 1. A surface plot showing the transition state of a H + H2 system]]
[[File:Surface_Plot_transtion_point-h3_hk3918.png |thumb|Figure 2. A contour plot showing the transition state of a H + H2 system at 91 pm]]

[[File:Animation mep hk3918.png ‎|thumb|Figure 3. A surface plot showing the mep of a H + H2 system when slightly displacing BC distance]]

=== 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? ===
On a potential energy surface diagram , the transition state is mathematically defined as the saddle point. The saddle point is a stationary point on the plot, where along the reaction path it is a maxima and perpendicular to the reaction path the saddle point is perceived to be a minima. The transition point can be identified in this simple example when the distances between the 3 hydrogen atoms are of equal distance. As shown in figure 1, the reaction path is oscillating at the saddle point. This depicts the vibrational freedom of the system at the transition state. A saddle point is mathematically found as shown below.

D = fxx.fyy - fxy2

<nowiki> </nowiki>If D > 0, fxx < 0 , local minima.

If D< 0 , then it is a saddle point.

=== 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 estimated '''rts '''is found to be around 91 pm. The transition state in this H + H2 system is defined to be the position where internuclear distances are equal, and where it is at a potential energy maxima in relation to the reaction path trajectory. From figure 1 it is evident that there are vibrations within the system, depicted by the oscillations on the ridge. The transition state position was found by determining at which position the internuclear distance was equal and constant, and where the potential energy was a maximum, and thus the stable point of the transition state. The saddle point was found as shown in figure 2.

=== Comment on how the ''mep'' and the trajectory you just calculated differ. ===
When calculating under mep the reaction path is depicted different to when the calculation type is set to dynamics. What is absent in the map calculation is the motion of the atoms. The vibrational motion of the three hydrogen atoms are not depicted as shown in figure 3. The initial condition is set so that the distance between AB is '''rts''' and the distance between BC deviates from '''rts '''by 1 pm, resulting in the reaction pathway following through the trajectory of minimum energy, and advances towards the HA + HB-HC product. The flat BC line indicates the fact that the mep calculation does not consider vibrational motion of atoms, and solely is helps depict the favoured trajectory when slightly displacing said atoms away from the transition state, as it chooses the minimum energy path. This could be useful to understand energy diagrams and help calculate activation energies.

==== Look at the “Internuclear Distances vs Time” and “Momenta vs Time”. What would change if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead? ====
The change of initial conditions to those above will change final product from HA + HB-HC to HA-HB + HC as it is the HB-HC bond distance that deviates away from the '''rts. '''As a result the internuclear distances vs time plot will now show AB to be constant over time. Momentum remains unchanged as initial momentum is 0 for both scenarios

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

The 'Get last geometry' option was selected, and the momenta were then reversed under the dynamic calculation type. The plot remains unchanged but with reversed directions as expected. The reason for this is because the initial argument resulted in a trajectory towards a larger BC distance. Using the momenta of this trajectory but reversing the sign simply results in the hydrogen and hydrogen molecule to start in the final position, but now approach each other due to the negative momentum value.
=== 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"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot/kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Hydrogen atoms AB and C approach
eachother. Notably, the vibrational

motion present after the transition state is

not present before.
|[[File:Surface_Plot_question4_1_hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-4.1 </nowiki>
|<nowiki>-420.077</nowiki>
|No
|

Hydrogen atoms AB approaches C, however

before approaching the saddle point to AB

swiftly moves away from C

|[[File:Animation exercise2 p2 -4.1 hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|Hydrogen atom C approaches AB with a large
amount of momentum, resulting in a new BC Hydrogen

molecule forming, releasing the A hydrogen atom in the

reaction.
|[[File:Animation exercise2 p2 -5.1 p1 -3.1 hk3918.png|300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.1 </nowiki>
|<nowiki>-357.277</nowiki>
|No
|Hydrogen atoms AB and C approach each other very
quickly. Upon impact, large internuclear distance and velocities

are depicted, showing the B atom being tugged by both A and C.

In the end, B forms a molecule AB and moves away from C.
|[[File:Animation exercise2 p2 -10.1 hk3918.png |300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.6 </nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|Upon impact, the B atom is tugged between the two atoms A and
C, and is finally bonded to to atom C. A then moves away from

BC.
|[[File:Animation p2 -10.6 hk3918.png |300px]]
|}
The table above shows that a reaction isn't only reliant on the amount of kinetic energy in the system. As shown in the table above, there are instances where kinetic energy is higher than the activation energy, however no reaction takes place. One therefore needs to consider that successful collisions are not only reliant on the system having enough energy to break and reform bonds.

=== Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===
Transition State Theory has a fundamental assumption which will result in its predictions being different to experimental values. The assumption that all trajectories possessing enough kinetic energy to circumvent the activation barrier will be reactive will result in a larger ''Rcl '' and thus a larger rate prediction than the experimental outcome. As seen in the table above, not all collisions are reactive even if there is sufficient kinetic energy - the case of barrier recrossing perfectly illustrates this.

== Exercise 2 ==

=== 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? ===
[[File:Surface_Plot_HF_hk3918.png|thumb|Figure 4. A surface plot showing the trajectory of F + H2 to H + HF reaction]|300px]]The reaction is exothermic as shown in figure 4, there is a decrease in potential energy when passing the transition state and forming the new HF bond. This relates to bond-strength. H-H has a bond strength of 432 kJ/mol whilst H-F has a bond strength of 565 kJ/mol.(reference). As HF is a polar bond as a result of the very electronegative F, there is an ionic element to the bond, resulting in a much greater attraction between the two atoms, thus releasing a lot of energy on formation. As the energy released is greater than that absorbed, the reaction is exothermic as depicted in the trajectory shown in figure 4.

[[File:Surface_Plot_hf-h_hk3918.png|thumb|Figure 5. A surface plot showing the trajectory of the FH + H to F + H2 reaction|300px]]

Figure 5 on the other hand shows an endothermic reaction, illustrated by the increase in potential energy along the trajectory, indicating that the energy absorbed to break the stronger H-F bond is greater than the energy released when forming the weaker H-H bond.

[[File:Plot small activation.png|thumb|Figure 6. Activation energy for the F +H2 reaction |300px]]

[[File:Plot activation big.png|thumb|Figure 7. Activation energy for the HF + H reaction |300px]]

[[File:Animation_momenta_hk3918.png |thumb|Figure 8. Momenta vs time plot for F + H2 reaction |300px]]

[[File:Animation_-74_-6.1.png ‎ |thumb|Figure 9. Contour plot for F + H2 reaction. Parameters: AB=200, BC=74, Pab=-1, Pbc=-6.9 |300px]]

[[File: Animation_low_momentum_hk3918.png ‎ |thumb|Figure 10. Contour plot for F + H2 reaction. Parameters: AB=200, BC=74, Pab=-1.6, Pbc=0.2 |300px]]

[[File:Animation low vibrational.png ‎ |thumb|Figure 11. Contour plot for the HF + H reaction where H has high momenta and HF has low initial momenta |300px]]

[[File:Animation highvirabtional.png‎ |thumb|Figure 12. Skew plot for the HF + H reaction where H has low momenta and HF has high initial momenta |300px]]

=== Locate the approximate position of the transition state ===
The transition state position was calculated to be when the H-F bond had a bond length of 181.8 pm, and the H-H bond was 73.1 pm as shown in figure 5. Analoguos to the calculation when approximating the transition state position of the H + H2 system, in the F + H2 reaction simulations were tested in order to acquire a constant bond length, at the maximum potential energy, and where the trajectory remained stable. As the F + H2 reaction was exothermic, adopting Hammonds postulate, the assumption was made that the reaction had an early transition state thus the region near the reactants was analysed. The mep calculation was used to remove the vibrational noise as it made it difficult to read the plots. As HF + H is a late transition state, resembling its product F + H2, it will have the same transition state positions. To confirm the transition state position, under mep calculations, incremental increases to the transition state bond length were made to make sure the trajectory rolled over the transition state to either the reactant or products.

=== Report the activation energy for both reactions. ===
In order to find the activation energy for HF + H, the bond distance of H-F transition state was first decreased by 1 pm. This then resulted in the energy falling to form the products H-F and H. The difference in this energy as shown in figure 7 was used to calculate the activation energy of the H-F + H endothermic reaction. The activation energy was estimated to be 120 kJ/mol. For the exothermic F + H2 reaction, the activation energy was calculated by trying to shift the trajectory towards to the reactants. This was done by thus increasing the F-H distance forming the F + H2 reactant. The difference in this energy was very small as shown in figure 8 and was used to calculate the activation energy for the F + H2 reaction to be around 1.2 kJ/mol.

=== 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. ===
Figure 8 illustrates the transition between translational and vibrational energy in the F + H2 reaction. Initially, the H-H bond (B-C) is vibrationally excited shown by the oscillation, whilst the H-F(A-B) bond isn't indicated by the absence of an oscillation. The H-F bond is at the vibrational ground state . After the collision we see a transition. The H-F bond now is vibrationally excited illustrated by the oscillation in the momenta vs time plot, and the H-H bonds loses this vibrationally energy and now only has translational energy due to the breaking of the bond, depicted by the plateau after 30s. Experimentally IR could be used to see the progress of the reaction. Only molecules with a dipole are IR active, thus after the formation of a H-F bond, IR absorption absent in the reactant will start to appear with overtones suggesting higher vibrational energies reached.

=== 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. ===
Below various example of both reactions are used to illustrate Polanyl's rules. Reactions with late transition states are more efficient when the initial trajectory possesses vibrational energy whereas reactions with early transition states are more efficient when the initial trajectory possesses translational energy. The following examples illustrate these rules.

==== Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, with a momentum pFH = -1.0 g.mol-1.pm.fs-1, and explore several values of pHH in the range -6.1 to 6.1 g.mol-1.pm.fs-1 (explore values also close to these limits). What do you observe? ====
When using the parameters above and setting the F-H value to 200 pm, at lower momentum values larger vibrational frequencies were obversed. The reactions did not therefore take place, satisfying Polyanyi's rules, where the having translational energy is more efficient when there is an early barrier rather than vibrational energy. Even at higher momentum extremes shown in figure 9, the reaction does not go to completion, indicating even though a lot of energy is being put into the system , it is not an efficient reaction, satisfying the rules.

==== For the same initial position, increase slightly the momentum pFH = -1.6 g.mol-1.pm.fs-1, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.2 g.mol-1.pm.fs-1. What do you observe now? ====
Figure 10 shows that, even at lower momentum values, resulting in lower vibrational energies the reaction still does not proceed to completion. As there is not enough translational energy in the reaction, further satisfying Polyanyi's rules.

==== Focus on the reverse reaction, H + HF. ====
This reaction has a late transition state, thus according to Polanyi's rules, vibrational energy trajectory is more efficient. This was tested firstly by using a low H-F vibrational energy and a high H translational energy as shown in figure 11. As predicted the reaction did not proceed to completion as there was not enough vibrational energy in the reaction. Figure 12 shows an example where the reaction has a larger vibrational energy, and therefore is more efficient, resulting in a successful reaction.

MRD:01524831.molecularreactiondynamics

2020-05-15T21:21:46Z

Hk3918: /* 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? */

== Exercise 1: H + H2 system ==

[[File:Surface_Plot_first_trajcectory_hk3918.png|thumb|Figure 1. A surface plot showing the transition state of a H + H2 system]]
[[File:Surface_Plot_transtion_point-h3_hk3918.png |thumb|Figure 2. A contour plot showing the transition state of a H + H2 system at 91 pm]]

[[File:Animation mep hk3918.png ‎|thumb|Figure 3. A surface plot showing the mep of a H + H2 system when slightly displacing BC distance]]

=== 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? ===
On a potential energy surface diagram , the transition state is mathematically defined as the saddle point. The saddle point is a stationary point on the plot, where along the reaction path it is a maxima and perpendicular to the reaction path the saddle point is perceived to be a minima. The transition point can be identified in this simple example when the distances between the 3 hydrogen atoms are of equal distance. As shown in figure 1, the reaction path is oscillating at the saddle point. This depicts the vibrational freedom of the system at the transition state. A saddle point is mathematically found as shown below.

D = fxx.fyy - fxy2

<nowiki> </nowiki>If D > 0, fxx < 0 , local minima.

If D< 0 , then it is a saddle point.

=== 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 estimated '''rts '''is found to be around 91 pm. The transition state in this H + H2 system is defined to be the position where internuclear distances are equal, and where it is at a potential energy maxima in relation to the reaction path trajectory. From figure 1 it is evident that there are vibrations within the system, depicted by the oscillations on the ridge. The transition state position was found by determining at which position the internuclear distance was equal and constant, and where the potential energy was a maximum, and thus the stable point of the transition state. The saddle point was found as shown in figure 2.

=== Comment on how the ''mep'' and the trajectory you just calculated differ. ===
When calculating under mep the reaction path is depicted different to when the calculation type is set to dynamics. What is absent in the map calculation is the motion of the atoms. The vibrational motion of the three hydrogen atoms are not depicted as shown in figure 3. The initial condition is set so that the distance between AB is '''rts''' and the distance between BC deviates from '''rts '''by 1 pm, resulting in the reaction pathway following through the trajectory of minimum energy, and advances towards the HA + HB-HC product. The flat BC line indicates the fact that the mep calculation does not consider vibrational motion of atoms, and solely is helps depict the favoured trajectory when slightly displacing said atoms away from the transition state, as it chooses the minimum energy path. This could be useful to understand energy diagrams and help calculate activation energies.

==== Look at the “Internuclear Distances vs Time” and “Momenta vs Time”. What would change if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead? ====
The change of initial conditions to those above will change final product from HA + HB-HC to HA-HB + HC as it is the HB-HC bond distance that deviates away from the '''rts. '''As a result the internuclear distances vs time plot will now show AB to be constant over time. Momentum remains unchanged as initial momentum is 0 for both scenarios

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

The 'Get last geometry' option was selected, and the momenta were then reversed under the dynamic calculation type. The plot remains unchanged but with reversed directions as expected. The reason for this is because the initial argument resulted in a trajectory towards a larger BC distance. Using the momenta of this trajectory but reversing the sign simply results in the hydrogen and hydrogen molecule to start in the final position, but now approach each other due to the negative momentum value.
=== 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"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot/kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Hydrogen atoms AB and C approach
eachother. Notably, the vibrational

motion present after the transition state is

not present before.
|[[File:Surface_Plot_question4_1_hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-4.1 </nowiki>
|<nowiki>-420.077</nowiki>
|No
|

Hydrogen atoms AB approaches C, however

before approaching the saddle point to AB

swiftly moves away from C

|[[File:Animation exercise2 p2 -4.1 hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|Hydrogen atom C approaches AB with a large
amount of momentum, resulting in a new BC Hydrogen

molecule forming, releasing the A hydrogen atom in the

reaction.
|[[File:Animation exercise2 p2 -5.1 p1 -3.1 hk3918.png|300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.1 </nowiki>
|<nowiki>-357.277</nowiki>
|No
|Hydrogen atoms AB and C approach each other very
quickly. Upon impact, large internuclear distance and velocities

are depicted, showing the B atom being tugged by both A and C.

In the end, B forms a molecule AB and moves away from C.
|[[File:Animation exercise2 p2 -10.1 hk3918.png |300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.6 </nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|Upon impact, the B atom is tugged between the two atoms A and
C, and is finally bonded to to atom C. A then moves away from

BC.
|[[File:Animation p2 -10.6 hk3918.png |300px]]
|}
The table above shows that a reaction isn't only reliant on the amount of kinetic energy in the system. As shown in the table above, there are instances where kinetic energy is higher than the activation energy, however no reaction takes place. One therefore needs to consider that successful collisions are not only reliant on the system having enough energy to break and reform bonds.

=== Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===
Transition State Theory has a fundamental assumption which will result in its predictions being different to experimental values. The assumption that all trajectories possessing enough kinetic energy to circumvent the activation barrier will be reactive will result in a larger ''Rcl '' and thus a larger rate prediction than the experimental outcome. As seen in the table above, not all collisions are reactive even if there is sufficient kinetic energy - the case of barrier recrossing perfectly illustrates this.

== Exercise 2 ==

=== 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? ===
[[File:Surface_Plot_HF_hk3918.png|thumb|Figure 4. A surface plot showing the trajectory of F + H2 to H + HF reaction]|300px]]The reaction is exothermic as shown in figure 4, there is a decrease in potential energy when passing the transition state and forming the new HF bond. This relates to bond-strength, H-H has a bond strength of 432 kJ/mol whilst H-F has a bond strength of 565 kJ/mol.(reference). As HF is a polar bond as a result of the very electronegative F, there is an ionic element to the bond, resulting in a much greater attraction between the two atoms, thus releasing a lot of energy on formation. As the energy released is greater than that absorbed, the reaction is exothermic as depicted in the trajectory shown in figure 4.

[[File:Surface_Plot_hf-h_hk3918.png|thumb|Figure 5. A surface plot showing the trajectory of the FH + H to F + H2 reaction|300px]]

Figure 5 on the other hand shows an endothermic reaction, illustrated by the increase in potential energy along the trajectory, indicating that the energy absorbed to break the stronger H-F bond is greater than the energy released when forming the weaker H-H bond.

[[File:Plot small activation.png|thumb|Figure 6. Activation energy for the F +H2 reaction |300px]]

[[File:Plot activation big.png|thumb|Figure 7. Activation energy for the HF + H reaction |300px]]

[[File:Animation_momenta_hk3918.png |thumb|Figure 8. Momenta vs time plot for F + H2 reaction |300px]]

[[File:Animation_-74_-6.1.png ‎ |thumb|Figure 9. Contour plot for F + H2 reaction. Parameters: AB=200, BC=74, Pab=-1, Pbc=-6.9 |300px]]

[[File: Animation_low_momentum_hk3918.png ‎ |thumb|Figure 10. Contour plot for F + H2 reaction. Parameters: AB=200, BC=74, Pab=-1.6, Pbc=0.2 |300px]]

[[File:Animation low vibrational.png ‎ |thumb|Figure 11. Contour plot for the HF + H reaction where H has high momenta and HF has low initial momenta |300px]]

[[File:Animation highvirabtional.png‎ |thumb|Figure 12. Skew plot for the HF + H reaction where H has low momenta and HF has high initial momenta |300px]]

=== Locate the approximate position of the transition state ===
The transition state position was calculated to be when the H-F bond had a bond length of 181.8 pm, and the H-H bond was 73.1 pm as shown in figure 5. Analoguos to the calculation when approximating the transition state position of the H + H2 system, in the F + H2 reaction simulations were tested in order to acquire a constant bond length, at the maximum potential energy. As the first part of this reaction was exothermic, adopting Hammonds postulate, the assumption was made that the reaction had an early transition state thus the region near the reactants was analysed. The MEP calculation was used to remove the vibrational noise as it made it difficult to read the plots. As HF + H is a late transition state, resembling the its product F + H2, it will have the same transition state positions.

=== Report the activation energy for both reactions. ===
In order to find the activation energy the bond distance of H-F was first decreased. This then resulted in the energy falling to form the products H-F and H. The difference in this energy as shown in figure 7 was used to calculate the activation energy of the H-F + H endothermic reaction. The activation energy was estimated to be 120 kJ/mol. For the exothermic F + H2 reaction, the activation energy was calculated by trying to shift the trajectory towards to the reactants. This was done by thus increasing the F-H distance forming the F + H2 reactant. The difference in this energy was very small as shown in figure 8 and was used to calculate the activation energy for the F + H2 reaction.

=== 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. ===
Figure 8 illustrates the transition between translational and vibrational energy in the F + H2 reaction. Initially, the H-H bond (B-C) is vibrationally excited shown by the oscillation, whilst the H-F(A-B) bond isn't indicated by the absence of an oscillation. The H-F bond is at the vibrational ground state . After the collision we see a transition. The H-F bond now is vibrationally excited illustrated by the oscillation in the momenta vs time plot, and the H-H bonds loses this vibrationally energy and now only has translational energy due to the breaking of the bond, depicted by the plateau after 30s. Experimentally IR could be used to see the progress of the reaction. Only molecules with a dipole are IR active, thus after the formation of a H-F bond, IR absorption absent in the reactant will start to appear with overtones suggesting higher vibrational energies reached.

=== 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. ===
Below various example of both reactions are used to illustrate Polanyl's rules. Reactions with late transition states are more efficient when the initial trajectory possesses vibrational energy whereas reactions with early transition states are more efficient when the initial trajectory possesses translational energy. The following examples illustrate these rules.

==== Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, with a momentum pFH = -1.0 g.mol-1.pm.fs-1, and explore several values of pHH in the range -6.1 to 6.1 g.mol-1.pm.fs-1 (explore values also close to these limits). What do you observe? ====
When using the parameters above and setting the F-H value to 200 pm, at lower momentum values larger vibrational frequencies were obversed. The reactions did not therefore take place, satisfying Polyanyi's rules, where the having translational energy is more efficient when there is an early barrier rather than vibrational energy. Even at higher momentum extremes shown in figure 9, the reaction does not go to completion, indicating even though a lot of energy is being put into the system , it is not an efficient reaction, satisfying the rules.

==== For the same initial position, increase slightly the momentum pFH = -1.6 g.mol-1.pm.fs-1, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.2 g.mol-1.pm.fs-1. What do you observe now? ====
Figure 10 shows that, even at lower momentum values, resulting in lower vibrational energies the reaction still does not proceed to completion. As there is not enough translational energy in the reaction, further satisfying Polyanyi's rules.

==== Focus on the reverse reaction, H + HF. ====
This reaction has a late transition state, thus according to Polanyi's rules, vibrational energy trajectory is more efficient. This was tested firstly by using a low H-F vibrational energy and a high H translational energy as shown in figure 11. As predicted the reaction did not proceed to completion as there was not enough vibrational energy in the reaction. Figure 12 shows an example where the reaction has a larger vibrational energy, and therefore is more efficient, resulting in a successful reaction.

MRD:01524831.molecularreactiondynamics

2020-05-15T21:20:30Z

Hk3918: /* Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed. */

== Exercise 1: H + H2 system ==

[[File:Surface_Plot_first_trajcectory_hk3918.png|thumb|Figure 1. A surface plot showing the transition state of a H + H2 system]]
[[File:Surface_Plot_transtion_point-h3_hk3918.png |thumb|Figure 2. A contour plot showing the transition state of a H + H2 system at 91 pm]]

[[File:Animation mep hk3918.png ‎|thumb|Figure 3. A surface plot showing the mep of a H + H2 system when slightly displacing BC distance]]

=== 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? ===
On a potential energy surface diagram , the transition state is mathematically defined as the saddle point. The saddle point is a stationary point on the plot, where along the reaction path it is a maxima and perpendicular to the reaction path the saddle point is perceived to be a minima. The transition point can be identified in this simple example when the distances between the 3 hydrogen atoms are of equal distance. As shown in figure 1, the reaction path is oscillating at the saddle point. This depicts the vibrational freedom of the system at the transition state. A saddle point is mathematically found as shown below.

D = fxx.fyy - fxy2

<nowiki> </nowiki>If D > 0, fxx < 0 , local minima.

If D< 0 , then it is a saddle point.

=== 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 estimated '''rts '''is found to be around 91 pm. The transition state in this H + H2 system is defined to be the position where internuclear distances are equal, and where it is at a potential energy maxima in relation to the reaction path trajectory. From figure 1 it is evident that there are vibrations within the system, depicted by the oscillations on the ridge. The transition state position was found by determining at which position the internuclear distance was equal and constant, and where the potential energy was a maximum, and thus the stable point of the transition state. The saddle point was found as shown in figure 2.

=== Comment on how the ''mep'' and the trajectory you just calculated differ. ===
When calculating under mep the reaction path is depicted different to when the calculation type is set to dynamics. What is absent in the map calculation is the motion of the atoms. The vibrational motion of the three hydrogen atoms are not depicted as shown in figure 3. The initial condition is set so that the distance between AB is '''rts''' and the distance between BC deviates from '''rts '''by 1 pm, resulting in the reaction pathway following through the trajectory of minimum energy, and advances towards the HA + HB-HC product. The flat BC line indicates the fact that the mep calculation does not consider vibrational motion of atoms, and solely is helps depict the favoured trajectory when slightly displacing said atoms away from the transition state, as it chooses the minimum energy path. This could be useful to understand energy diagrams and help calculate activation energies.

==== Look at the “Internuclear Distances vs Time” and “Momenta vs Time”. What would change if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead? ====
The change of initial conditions to those above will change final product from HA + HB-HC to HA-HB + HC as it is the HB-HC bond distance that deviates away from the '''rts. '''As a result the internuclear distances vs time plot will now show AB to be constant over time. Momentum remains unchanged as initial momentum is 0 for both scenarios

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

The 'Get last geometry' option was selected, and the momenta were then reversed under the dynamic calculation type. The plot remains unchanged but with reversed directions as expected. The reason for this is because the initial argument resulted in a trajectory towards a larger BC distance. Using the momenta of this trajectory but reversing the sign simply results in the hydrogen and hydrogen molecule to start in the final position, but now approach each other due to the negative momentum value.
=== 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"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot/kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Hydrogen atoms AB and C approach
eachother. Notably, the vibrational

motion present after the transition state is

not present before.
|[[File:Surface_Plot_question4_1_hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-4.1 </nowiki>
|<nowiki>-420.077</nowiki>
|No
|

Hydrogen atoms AB approaches C, however

before approaching the saddle point to AB

swiftly moves away from C

|[[File:Animation exercise2 p2 -4.1 hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|Hydrogen atom C approaches AB with a large
amount of momentum, resulting in a new BC Hydrogen

molecule forming, releasing the A hydrogen atom in the

reaction.
|[[File:Animation exercise2 p2 -5.1 p1 -3.1 hk3918.png|300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.1 </nowiki>
|<nowiki>-357.277</nowiki>
|No
|Hydrogen atoms AB and C approach each other very
quickly. Upon impact, large internuclear distance and velocities

are depicted, showing the B atom being tugged by both A and C.

In the end, B forms a molecule AB and moves away from C.
|[[File:Animation exercise2 p2 -10.1 hk3918.png |300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.6 </nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|Upon impact, the B atom is tugged between the two atoms A and
C, and is finally bonded to to atom C. A then moves away from

BC.
|[[File:Animation p2 -10.6 hk3918.png |300px]]
|}
The table above shows that a reaction isn't only reliant on the amount of kinetic energy in the system. As shown in the table above, there are instances where kinetic energy is higher than the activation energy, however no reaction takes place. One therefore needs to consider that successful collision are reliant on the system having enough energy to break and reform bonds, and also the orientation at which atoms collide.

=== Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===
Transition State Theory has a fundamental assumption which will result in its predictions being different to experimental values. The assumption that all trajectories possessing enough kinetic energy to circumvent the activation barrier will be reactive will result in a larger ''Rcl '' and thus a larger rate prediction than the experimental outcome. As seen in the table above, not all collisions are reactive even if there is sufficient kinetic energy - the case of barrier recrossing perfectly illustrates this.

== Exercise 2 ==

=== 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? ===
[[File:Surface_Plot_HF_hk3918.png|thumb|Figure 4. A surface plot showing the trajectory of F + H2 to H + HF reaction]|300px]]The reaction is exothermic as shown in figure 4, there is a decrease in potential energy when passing the transition state and forming the new HF bond. This relates to bond-strength, H-H has a bond strength of 432 kJ/mol whilst H-F has a bond strength of 565 kJ/mol.(reference). As HF is a polar bond as a result of the very electronegative F, there is an ionic element to the bond, resulting in a much greater attraction between the two atoms, thus releasing a lot of energy on formation. As the energy released is greater than that absorbed, the reaction is exothermic as depicted in the trajectory shown in figure 4.

[[File:Surface_Plot_hf-h_hk3918.png|thumb|Figure 5. A surface plot showing the trajectory of the FH + H to F + H2 reaction|300px]]

Figure 5 on the other hand shows an endothermic reaction, illustrated by the increase in potential energy along the trajectory, indicating that the energy absorbed to break the stronger H-F bond is greater than the energy released when forming the weaker H-H bond.

[[File:Plot small activation.png|thumb|Figure 6. Activation energy for the F +H2 reaction |300px]]

[[File:Plot activation big.png|thumb|Figure 7. Activation energy for the HF + H reaction |300px]]

[[File:Animation_momenta_hk3918.png |thumb|Figure 8. Momenta vs time plot for F + H2 reaction |300px]]

[[File:Animation_-74_-6.1.png ‎ |thumb|Figure 9. Contour plot for F + H2 reaction. Parameters: AB=200, BC=74, Pab=-1, Pbc=-6.9 |300px]]

[[File: Animation_low_momentum_hk3918.png ‎ |thumb|Figure 10. Contour plot for F + H2 reaction. Parameters: AB=200, BC=74, Pab=-1.6, Pbc=0.2 |300px]]

[[File:Animation low vibrational.png ‎ |thumb|Figure 11. Contour plot for the HF + H reaction where H has high momenta and HF has low initial momenta |300px]]

[[File:Animation highvirabtional.png‎ |thumb|Figure 12. Skew plot for the HF + H reaction where H has low momenta and HF has high initial momenta |300px]]

=== Locate the approximate position of the transition state ===
The transition state position was calculated to be when the H-F bond had a bond length of 181.8 pm, and the H-H bond was 73.1 pm as shown in figure 5. Analoguos to the calculation when approximating the transition state position of the H + H2 system, in the F + H2 reaction simulations were tested in order to acquire a constant bond length, at the maximum potential energy. As the first part of this reaction was exothermic, adopting Hammonds postulate, the assumption was made that the reaction had an early transition state thus the region near the reactants was analysed. The MEP calculation was used to remove the vibrational noise as it made it difficult to read the plots. As HF + H is a late transition state, resembling the its product F + H2, it will have the same transition state positions.

=== Report the activation energy for both reactions. ===
In order to find the activation energy the bond distance of H-F was first decreased. This then resulted in the energy falling to form the products H-F and H. The difference in this energy as shown in figure 7 was used to calculate the activation energy of the H-F + H endothermic reaction. The activation energy was estimated to be 120 kJ/mol. For the exothermic F + H2 reaction, the activation energy was calculated by trying to shift the trajectory towards to the reactants. This was done by thus increasing the F-H distance forming the F + H2 reactant. The difference in this energy was very small as shown in figure 8 and was used to calculate the activation energy for the F + H2 reaction.

=== 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. ===
Figure 8 illustrates the transition between translational and vibrational energy in the F + H2 reaction. Initially, the H-H bond (B-C) is vibrationally excited shown by the oscillation, whilst the H-F(A-B) bond isn't indicated by the absence of an oscillation. The H-F bond is at the vibrational ground state . After the collision we see a transition. The H-F bond now is vibrationally excited illustrated by the oscillation in the momenta vs time plot, and the H-H bonds loses this vibrationally energy and now only has translational energy due to the breaking of the bond, depicted by the plateau after 30s. Experimentally IR could be used to see the progress of the reaction. Only molecules with a dipole are IR active, thus after the formation of a H-F bond, IR absorption absent in the reactant will start to appear with overtones suggesting higher vibrational energies reached.

=== 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. ===
Below various example of both reactions are used to illustrate Polanyl's rules. Reactions with late transition states are more efficient when the initial trajectory possesses vibrational energy whereas reactions with early transition states are more efficient when the initial trajectory possesses translational energy. The following examples illustrate these rules.

==== Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, with a momentum pFH = -1.0 g.mol-1.pm.fs-1, and explore several values of pHH in the range -6.1 to 6.1 g.mol-1.pm.fs-1 (explore values also close to these limits). What do you observe? ====
When using the parameters above and setting the F-H value to 200 pm, at lower momentum values larger vibrational frequencies were obversed. The reactions did not therefore take place, satisfying Polyanyi's rules, where the having translational energy is more efficient when there is an early barrier rather than vibrational energy. Even at higher momentum extremes shown in figure 9, the reaction does not go to completion, indicating even though a lot of energy is being put into the system , it is not an efficient reaction, satisfying the rules.

==== For the same initial position, increase slightly the momentum pFH = -1.6 g.mol-1.pm.fs-1, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.2 g.mol-1.pm.fs-1. What do you observe now? ====
Figure 10 shows that, even at lower momentum values, resulting in lower vibrational energies the reaction still does not proceed to completion. As there is not enough translational energy in the reaction, further satisfying Polyanyi's rules.

==== Focus on the reverse reaction, H + HF. ====
This reaction has a late transition state, thus according to Polanyi's rules, vibrational energy trajectory is more efficient. This was tested firstly by using a low H-F vibrational energy and a high H translational energy as shown in figure 11. As predicted the reaction did not proceed to completion as there was not enough vibrational energy in the reaction. Figure 12 shows an example where the reaction has a larger vibrational energy, and therefore is more efficient, resulting in a successful reaction.

MRD:01524831.molecularreactiondynamics

2020-05-15T21:18:49Z

Hk3918: /* Exercise 1: H + H2 system */

== Exercise 1: H + H2 system ==

[[File:Surface_Plot_first_trajcectory_hk3918.png|thumb|Figure 1. A surface plot showing the transition state of a H + H2 system]]
[[File:Surface_Plot_transtion_point-h3_hk3918.png |thumb|Figure 2. A contour plot showing the transition state of a H + H2 system at 91 pm]]

[[File:Animation mep hk3918.png ‎|thumb|Figure 3. A surface plot showing the mep of a H + H2 system when slightly displacing BC distance]]

=== 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? ===
On a potential energy surface diagram , the transition state is mathematically defined as the saddle point. The saddle point is a stationary point on the plot, where along the reaction path it is a maxima and perpendicular to the reaction path the saddle point is perceived to be a minima. The transition point can be identified in this simple example when the distances between the 3 hydrogen atoms are of equal distance. As shown in figure 1, the reaction path is oscillating at the saddle point. This depicts the vibrational freedom of the system at the transition state. A saddle point is mathematically found as shown below.

D = fxx.fyy - fxy2

<nowiki> </nowiki>If D > 0, fxx < 0 , local minima.

If D< 0 , then it is a saddle point.

=== 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 estimated '''rts '''is found to be around 91 pm. The transition state in this H + H2 system is defined to be the position where internuclear distances are equal, and where it is at a potential energy maxima in relation to the reaction path trajectory. From figure 1 it is evident that there are vibrations within the system, depicted by the oscillations on the ridge. The transition state position was found by determining at which position the internuclear distance was equal and constant, and where the potential energy was a maximum, and thus the stable point of the transition state. The saddle point was found as shown in figure 2.

=== Comment on how the ''mep'' and the trajectory you just calculated differ. ===
When calculating under mep the reaction path is depicted different to when the calculation type is set to dynamics. What is absent in the map calculation is the motion of the atoms. The vibrational motion of the three hydrogen atoms are not depicted as shown in figure 3. The initial condition is set so that the distance between AB is '''rts''' and the distance between BC deviates from '''rts '''by 1 pm, resulting in the reaction pathway following through the trajectory of minimum energy, and advances towards the HA + HB-HC product. The flat BC line indicates the fact that the mep calculation does not consider vibrational motion of atoms, and solely is helps depict the favoured trajectory when slightly displacing said atoms away from the transition state, as it chooses the minimum energy path. This could be useful to understand energy diagrams and help calculate activation energies.

==== Look at the “Internuclear Distances vs Time” and “Momenta vs Time”. What would change if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead? ====
The change of initial conditions to those above will change final product from HA + HB-HC to HA-HB + HC as it is the HB-HC bond distance that deviates away from the '''rts. '''As a result the internuclear distances vs time plot will now show AB to be constant over time. Momentum remains unchanged as initial momentum is 0 for both scenarios

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

The 'Get last geometry' option was selected, and the momenta were then reversed under the dynamic calculation type. The plot remains unchanged as expected. The reason for this is because the initial argument resulted in a trajectory towards a larger BC distance. Using the momenta of this trajectory but reversing the sign simply results in the hydrogen and hydrogen molecule to start in the final position, but now approach each other due to the negative momentum value.
=== 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"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot/kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Hydrogen atoms AB and C approach
eachother. Notably, the vibrational

motion present after the transition state is

not present before.
|[[File:Surface_Plot_question4_1_hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-4.1 </nowiki>
|<nowiki>-420.077</nowiki>
|No
|

Hydrogen atoms AB approaches C, however

before approaching the saddle point to AB

swiftly moves away from C

|[[File:Animation exercise2 p2 -4.1 hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|Hydrogen atom C approaches AB with a large
amount of momentum, resulting in a new BC Hydrogen

molecule forming, releasing the A hydrogen atom in the

reaction.
|[[File:Animation exercise2 p2 -5.1 p1 -3.1 hk3918.png|300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.1 </nowiki>
|<nowiki>-357.277</nowiki>
|No
|Hydrogen atoms AB and C approach each other very
quickly. Upon impact, large internuclear distance and velocities

are depicted, showing the B atom being tugged by both A and C.

In the end, B forms a molecule AB and moves away from C.
|[[File:Animation exercise2 p2 -10.1 hk3918.png |300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.6 </nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|Upon impact, the B atom is tugged between the two atoms A and
C, and is finally bonded to to atom C. A then moves away from

BC.
|[[File:Animation p2 -10.6 hk3918.png |300px]]
|}
The table above shows that a reaction isn't only reliant on the amount of kinetic energy in the system. As shown in the table above, there are instances where kinetic energy is higher than the activation energy, however no reaction takes place. One therefore needs to consider that successful collision are reliant on the system having enough energy to break and reform bonds, and also the orientation at which atoms collide.

=== Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===
Transition State Theory has a fundamental assumption which will result in its predictions being different to experimental values. The assumption that all trajectories possessing enough kinetic energy to circumvent the activation barrier will be reactive will result in a larger ''Rcl '' and thus a larger rate prediction than the experimental outcome. As seen in the table above, not all collisions are reactive even if there is sufficient kinetic energy - the case of barrier recrossing perfectly illustrates this.

== Exercise 2 ==

=== 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? ===
[[File:Surface_Plot_HF_hk3918.png|thumb|Figure 4. A surface plot showing the trajectory of F + H2 to H + HF reaction]|300px]]The reaction is exothermic as shown in figure 4, there is a decrease in potential energy when passing the transition state and forming the new HF bond. This relates to bond-strength, H-H has a bond strength of 432 kJ/mol whilst H-F has a bond strength of 565 kJ/mol.(reference). As HF is a polar bond as a result of the very electronegative F, there is an ionic element to the bond, resulting in a much greater attraction between the two atoms, thus releasing a lot of energy on formation. As the energy released is greater than that absorbed, the reaction is exothermic as depicted in the trajectory shown in figure 4.

[[File:Surface_Plot_hf-h_hk3918.png|thumb|Figure 5. A surface plot showing the trajectory of the FH + H to F + H2 reaction|300px]]

Figure 5 on the other hand shows an endothermic reaction, illustrated by the increase in potential energy along the trajectory, indicating that the energy absorbed to break the stronger H-F bond is greater than the energy released when forming the weaker H-H bond.

[[File:Plot small activation.png|thumb|Figure 6. Activation energy for the F +H2 reaction |300px]]

[[File:Plot activation big.png|thumb|Figure 7. Activation energy for the HF + H reaction |300px]]

[[File:Animation_momenta_hk3918.png |thumb|Figure 8. Momenta vs time plot for F + H2 reaction |300px]]

[[File:Animation_-74_-6.1.png ‎ |thumb|Figure 9. Contour plot for F + H2 reaction. Parameters: AB=200, BC=74, Pab=-1, Pbc=-6.9 |300px]]

[[File: Animation_low_momentum_hk3918.png ‎ |thumb|Figure 10. Contour plot for F + H2 reaction. Parameters: AB=200, BC=74, Pab=-1.6, Pbc=0.2 |300px]]

[[File:Animation low vibrational.png ‎ |thumb|Figure 11. Contour plot for the HF + H reaction where H has high momenta and HF has low initial momenta |300px]]

[[File:Animation highvirabtional.png‎ |thumb|Figure 12. Skew plot for the HF + H reaction where H has low momenta and HF has high initial momenta |300px]]

=== Locate the approximate position of the transition state ===
The transition state position was calculated to be when the H-F bond had a bond length of 181.8 pm, and the H-H bond was 73.1 pm as shown in figure 5. Analoguos to the calculation when approximating the transition state position of the H + H2 system, in the F + H2 reaction simulations were tested in order to acquire a constant bond length, at the maximum potential energy. As the first part of this reaction was exothermic, adopting Hammonds postulate, the assumption was made that the reaction had an early transition state thus the region near the reactants was analysed. The MEP calculation was used to remove the vibrational noise as it made it difficult to read the plots. As HF + H is a late transition state, resembling the its product F + H2, it will have the same transition state positions.

=== Report the activation energy for both reactions. ===
In order to find the activation energy the bond distance of H-F was first decreased. This then resulted in the energy falling to form the products H-F and H. The difference in this energy as shown in figure 7 was used to calculate the activation energy of the H-F + H endothermic reaction. The activation energy was estimated to be 120 kJ/mol. For the exothermic F + H2 reaction, the activation energy was calculated by trying to shift the trajectory towards to the reactants. This was done by thus increasing the F-H distance forming the F + H2 reactant. The difference in this energy was very small as shown in figure 8 and was used to calculate the activation energy for the F + H2 reaction.

=== 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. ===
Figure 8 illustrates the transition between translational and vibrational energy in the F + H2 reaction. Initially, the H-H bond (B-C) is vibrationally excited shown by the oscillation, whilst the H-F(A-B) bond isn't indicated by the absence of an oscillation. The H-F bond is at the vibrational ground state . After the collision we see a transition. The H-F bond now is vibrationally excited illustrated by the oscillation in the momenta vs time plot, and the H-H bonds loses this vibrationally energy and now only has translational energy due to the breaking of the bond, depicted by the plateau after 30s. Experimentally IR could be used to see the progress of the reaction. Only molecules with a dipole are IR active, thus after the formation of a H-F bond, IR absorption absent in the reactant will start to appear with overtones suggesting higher vibrational energies reached.

=== 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. ===
Below various example of both reactions are used to illustrate Polanyl's rules. Reactions with late transition states are more efficient when the initial trajectory possesses vibrational energy whereas reactions with early transition states are more efficient when the initial trajectory possesses translational energy. The following examples illustrate these rules.

==== Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, with a momentum pFH = -1.0 g.mol-1.pm.fs-1, and explore several values of pHH in the range -6.1 to 6.1 g.mol-1.pm.fs-1 (explore values also close to these limits). What do you observe? ====
When using the parameters above and setting the F-H value to 200 pm, at lower momentum values larger vibrational frequencies were obversed. The reactions did not therefore take place, satisfying Polyanyi's rules, where the having translational energy is more efficient when there is an early barrier rather than vibrational energy. Even at higher momentum extremes shown in figure 9, the reaction does not go to completion, indicating even though a lot of energy is being put into the system , it is not an efficient reaction, satisfying the rules.

==== For the same initial position, increase slightly the momentum pFH = -1.6 g.mol-1.pm.fs-1, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.2 g.mol-1.pm.fs-1. What do you observe now? ====
Figure 10 shows that, even at lower momentum values, resulting in lower vibrational energies the reaction still does not proceed to completion. As there is not enough translational energy in the reaction, further satisfying Polyanyi's rules.

==== Focus on the reverse reaction, H + HF. ====
This reaction has a late transition state, thus according to Polanyi's rules, vibrational energy trajectory is more efficient. This was tested firstly by using a low H-F vibrational energy and a high H translational energy as shown in figure 11. As predicted the reaction did not proceed to completion as there was not enough vibrational energy in the reaction. Figure 12 shows an example where the reaction has a larger vibrational energy, and therefore is more efficient, resulting in a successful reaction.

MRD:01524831.molecularreactiondynamics

2020-05-15T21:11:42Z

Hk3918: /* Exercise 1: H + H2 system */

== Exercise 1: H + H2 system ==

[[File:Surface_Plot_first_trajcectory_hk3918.png|thumb|Figure 1. A surface plot showing the transition state of a H + H2 system]]
[[File:Surface_Plot_transtion_point-h3_hk3918.png |thumb|Figure 2. A contour plot showing the transition state of a H + H2 system at 91 pm]]

[[File:Animation_newmep_hk3918.png ‎|thumb|Figure 3. A surface plot showing the mep of a H + H2 system when slightly displacing BC distance]]

=== 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? ===
On a potential energy surface diagram , the transition state is mathematically defined as the saddle point. The saddle point is a stationary point on the plot, where along the reaction path it is a maxima and perpendicular to the reaction path the saddle point is perceived to be a minima. The transition point can be identified in this simple example when the distances between the 3 hydrogen atoms are of equal distance. As shown in figure 1, the reaction path is oscillating at the saddle point. This depicts the vibrational freedom of the system at the transition state. A saddle point is mathematically found as shown below.

D = fxx.fyy - fxy2

<nowiki> </nowiki>If D > 0, fxx < 0 , local minima.

If D< 0 , then it is a saddle point.

=== 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 estimated '''rts '''is found to be around 91 pm. The transition state in this H + H2 system is defined to be the position where internuclear distances are equal, and where it is at a potential energy maxima in relation to the reaction path trajectory. From figure 1 it is evident that there are vibrations within the system, depicted by the oscillations on the ridge. The transition state position was found by determining at which position the internuclear distance was equal and constant, and where the potential energy was a maximum, and thus the stable point of the transition state. The saddle point was found as shown in figure 2.

=== Comment on how the ''mep'' and the trajectory you just calculated differ. ===
When calculating under mep the reaction path is depicted different to when the calculation type is set to dynamics. What is absent in the map calculation is the motion of the atoms. The vibrational motion of the three hydrogen atoms are not depicted as shown in figure 3. The initial condition is set so that the distance between AB is '''rts''' and the distance between BC deviates from '''rts '''by 1 pm, resulting in the reaction pathway following through the trajectory of minimum energy, and advances towards the HA + HB-HC product. The flat BC line indicates the fact that the mep calculation does not consider vibrational motion of atoms, and solely is helps depict the favoured trajectory when slightly displacing said atoms away from the transition state, as it chooses the minimum energy path. This could be useful to understand energy diagrams and help calculate activation energies.

==== Look at the “Internuclear Distances vs Time” and “Momenta vs Time”. What would change if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead? ====
The change of initial conditions to those above will change final product from HA + HB-HC to HA-HB + HC as it is the HB-HC bond distance that deviates away from the '''rts. '''As a result the internuclear distances vs time plot will now show AB to be constant over time. Momentum remains unchanged as initial momentum is 0 for both scenarios

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

The 'Get last geometry' option was selected, and the momenta were then reversed under the dynamic calculation type. The plot remains unchanged as expected. The reason for this is because the initial argument resulted in a trajectory towards a larger BC distance. Using the momenta of this trajectory but reversing the sign simply results in the hydrogen and hydrogen molecule to start in the final position, but now approach each other due to the negative momentum value.
=== 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"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot/kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Hydrogen atoms AB and C approach
eachother. Notably, the vibrational

motion present after the transition state is

not present before.
|[[File:Surface_Plot_question4_1_hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-4.1 </nowiki>
|<nowiki>-420.077</nowiki>
|No
|

Hydrogen atoms AB approaches C, however

before approaching the saddle point to AB

swiftly moves away from C

|[[File:Animation exercise2 p2 -4.1 hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|Hydrogen atom C approaches AB with a large
amount of momentum, resulting in a new BC Hydrogen

molecule forming, releasing the A hydrogen atom in the

reaction.
|[[File:Animation exercise2 p2 -5.1 p1 -3.1 hk3918.png|300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.1 </nowiki>
|<nowiki>-357.277</nowiki>
|No
|Hydrogen atoms AB and C approach each other very
quickly. Upon impact, large internuclear distance and velocities

are depicted, showing the B atom being tugged by both A and C.

In the end, B forms a molecule AB and moves away from C.
|[[File:Animation exercise2 p2 -10.1 hk3918.png |300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.6 </nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|Upon impact, the B atom is tugged between the two atoms A and
C, and is finally bonded to to atom C. A then moves away from

BC.
|[[File:Animation p2 -10.6 hk3918.png |300px]]
|}
The table above shows that a reaction isn't only reliant on the amount of kinetic energy in the system. As shown in the table above, there are instances where kinetic energy is higher than the activation energy, however no reaction takes place. One therefore needs to consider that successful collision are reliant on the system having enough energy to break and reform bonds, and also the orientation at which atoms collide.

=== Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===
Transition State Theory has a fundamental assumption which will result in its predictions being different to experimental values. The assumption that all trajectories possessing enough kinetic energy to circumvent the activation barrier will be reactive will result in a larger ''Rcl '' and thus a larger rate prediction than the experimental outcome. As seen in the table above, not all collisions are reactive even if there is sufficient kinetic energy - the case of barrier recrossing perfectly illustrates this.

== Exercise 2 ==

=== 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? ===
[[File:Surface_Plot_HF_hk3918.png|thumb|Figure 4. A surface plot showing the trajectory of F + H2 to H + HF reaction]|300px]]The reaction is exothermic as shown in figure 4, there is a decrease in potential energy when passing the transition state and forming the new HF bond. This relates to bond-strength, H-H has a bond strength of 432 kJ/mol whilst H-F has a bond strength of 565 kJ/mol.(reference). As HF is a polar bond as a result of the very electronegative F, there is an ionic element to the bond, resulting in a much greater attraction between the two atoms, thus releasing a lot of energy on formation. As the energy released is greater than that absorbed, the reaction is exothermic as depicted in the trajectory shown in figure 4.

[[File:Surface_Plot_hf-h_hk3918.png|thumb|Figure 5. A surface plot showing the trajectory of the FH + H to F + H2 reaction|300px]]

Figure 5 on the other hand shows an endothermic reaction, illustrated by the increase in potential energy along the trajectory, indicating that the energy absorbed to break the stronger H-F bond is greater than the energy released when forming the weaker H-H bond.

[[File:Plot small activation.png|thumb|Figure 6. Activation energy for the F +H2 reaction |300px]]

[[File:Plot activation big.png|thumb|Figure 7. Activation energy for the HF + H reaction |300px]]

[[File:Animation_momenta_hk3918.png |thumb|Figure 8. Momenta vs time plot for F + H2 reaction |300px]]

[[File:Animation_-74_-6.1.png ‎ |thumb|Figure 9. Contour plot for F + H2 reaction. Parameters: AB=200, BC=74, Pab=-1, Pbc=-6.9 |300px]]

[[File: Animation_low_momentum_hk3918.png ‎ |thumb|Figure 10. Contour plot for F + H2 reaction. Parameters: AB=200, BC=74, Pab=-1.6, Pbc=0.2 |300px]]

[[File:Animation low vibrational.png ‎ |thumb|Figure 11. Contour plot for the HF + H reaction where H has high momenta and HF has low initial momenta |300px]]

[[File:Animation highvirabtional.png‎ |thumb|Figure 12. Skew plot for the HF + H reaction where H has low momenta and HF has high initial momenta |300px]]

=== Locate the approximate position of the transition state ===
The transition state position was calculated to be when the H-F bond had a bond length of 181.8 pm, and the H-H bond was 73.1 pm as shown in figure 5. Analoguos to the calculation when approximating the transition state position of the H + H2 system, in the F + H2 reaction simulations were tested in order to acquire a constant bond length, at the maximum potential energy. As the first part of this reaction was exothermic, adopting Hammonds postulate, the assumption was made that the reaction had an early transition state thus the region near the reactants was analysed. The MEP calculation was used to remove the vibrational noise as it made it difficult to read the plots. As HF + H is a late transition state, resembling the its product F + H2, it will have the same transition state positions.

=== Report the activation energy for both reactions. ===
In order to find the activation energy the bond distance of H-F was first decreased. This then resulted in the energy falling to form the products H-F and H. The difference in this energy as shown in figure 7 was used to calculate the activation energy of the H-F + H endothermic reaction. The activation energy was estimated to be 120 kJ/mol. For the exothermic F + H2 reaction, the activation energy was calculated by trying to shift the trajectory towards to the reactants. This was done by thus increasing the F-H distance forming the F + H2 reactant. The difference in this energy was very small as shown in figure 8 and was used to calculate the activation energy for the F + H2 reaction.

=== 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. ===
Figure 8 illustrates the transition between translational and vibrational energy in the F + H2 reaction. Initially, the H-H bond (B-C) is vibrationally excited shown by the oscillation, whilst the H-F(A-B) bond isn't indicated by the absence of an oscillation. The H-F bond is at the vibrational ground state . After the collision we see a transition. The H-F bond now is vibrationally excited illustrated by the oscillation in the momenta vs time plot, and the H-H bonds loses this vibrationally energy and now only has translational energy due to the breaking of the bond, depicted by the plateau after 30s. Experimentally IR could be used to see the progress of the reaction. Only molecules with a dipole are IR active, thus after the formation of a H-F bond, IR absorption absent in the reactant will start to appear with overtones suggesting higher vibrational energies reached.

=== 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. ===
Below various example of both reactions are used to illustrate Polanyl's rules. Reactions with late transition states are more efficient when the initial trajectory possesses vibrational energy whereas reactions with early transition states are more efficient when the initial trajectory possesses translational energy. The following examples illustrate these rules.

==== Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, with a momentum pFH = -1.0 g.mol-1.pm.fs-1, and explore several values of pHH in the range -6.1 to 6.1 g.mol-1.pm.fs-1 (explore values also close to these limits). What do you observe? ====
When using the parameters above and setting the F-H value to 200 pm, at lower momentum values larger vibrational frequencies were obversed. The reactions did not therefore take place, satisfying Polyanyi's rules, where the having translational energy is more efficient when there is an early barrier rather than vibrational energy. Even at higher momentum extremes shown in figure 9, the reaction does not go to completion, indicating even though a lot of energy is being put into the system , it is not an efficient reaction, satisfying the rules.

==== For the same initial position, increase slightly the momentum pFH = -1.6 g.mol-1.pm.fs-1, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.2 g.mol-1.pm.fs-1. What do you observe now? ====
Figure 10 shows that, even at lower momentum values, resulting in lower vibrational energies the reaction still does not proceed to completion. As there is not enough translational energy in the reaction, further satisfying Polyanyi's rules.

==== Focus on the reverse reaction, H + HF. ====
This reaction has a late transition state, thus according to Polanyi's rules, vibrational energy trajectory is more efficient. This was tested firstly by using a low H-F vibrational energy and a high H translational energy as shown in figure 11. As predicted the reaction did not proceed to completion as there was not enough vibrational energy in the reaction. Figure 12 shows an example where the reaction has a larger vibrational energy, and therefore is more efficient, resulting in a successful reaction.

File:Animation newmep hk3918.png

2020-05-15T21:11:11Z

Hk3918:

MRD:01524831.molecularreactiondynamics

2020-05-15T21:09:11Z

Hk3918: /* Comment on how the mep and the trajectory you just calculated differ. */

== Exercise 1: H + H2 system ==

[[File:Surface_Plot_first_trajcectory_hk3918.png|thumb|Figure 1. A surface plot showing the transition state of a H + H2 system]]
[[File:Surface_Plot_transtion_point-h3_hk3918.png |thumb|Figure 2. A contour plot showing the transition state of a H + H2 system at 91 pm]]

[[File:Animation_mep_hk3918.png ‎|thumb|Figure 3. A surface plot showing the mep of a H + H2 system]]

=== 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? ===
On a potential energy surface diagram , the transition state is mathematically defined as the saddle point. The saddle point is a stationary point on the plot, where along the reaction path it is a maxima and perpendicular to the reaction path the saddle point is perceived to be a minima. The transition point can be identified in this simple example when the distances between the 3 hydrogen atoms are of equal distance. As shown in figure 1, the reaction path is oscillating at the saddle point. This depicts the vibrational freedom of the system at the transition state. A saddle point is mathematically found as shown below.

D = fxx.fyy - fxy2

<nowiki> </nowiki>If D > 0, fxx < 0 , local minima.

If D< 0 , then it is a saddle point.

=== 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 estimated '''rts '''is found to be around 91 pm. The transition state in this H + H2 system is defined to be the position where internuclear distances are equal, and where it is at a potential energy maxima in relation to the reaction path trajectory. From figure 1 it is evident that there are vibrations within the system, depicted by the oscillations on the ridge. The transition state position was found by determining at which position the internuclear distance was equal and constant, and where the potential energy was a maximum, and thus the stable point of the transition state. The saddle point was found as shown in figure 2.

=== Comment on how the ''mep'' and the trajectory you just calculated differ. ===
When calculating under mep the reaction path is depicted different to when the calculation type is set to dynamics. What is absent in the map calculation is the motion of the atoms. The vibrational motion of the three hydrogen atoms are not depicted as shown in figure 3. The initial condition is set so that the distance between AB is '''rts''' and the distance between BC deviates from '''rts '''by 1 pm, resulting in the reaction pathway following through the trajectory of minimum energy, and advances towards the HA + HB-HC product. The flat BC line indicates the fact that the mep calculation does not consider vibrational motion of atoms, and solely is helps depict the favoured trajectory when slightly displacing said atoms away from the transition state, as it chooses the minimum energy path. This could be useful to understand energy diagrams and help calculate activation energies.

==== Look at the “Internuclear Distances vs Time” and “Momenta vs Time”. What would change if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead? ====
The change of initial conditions to those above will change final product from HA + HB-HC to HA-HB + HC as it is the HB-HC bond distance that deviates away from the '''rts. '''As a result the internuclear distances vs time plot will now show AB to be constant over time. Momentum remains unchanged as initial momentum is 0 for both scenarios

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

The 'Get last geometry' option was selected, and the momenta were then reversed under the dynamic calculation type. The plot remains unchanged as expected. The reason for this is because the initial argument resulted in a trajectory towards a larger BC distance. Using the momenta of this trajectory but reversing the sign simply results in the hydrogen and hydrogen molecule to start in the final position, but now approach each other due to the negative momentum value.
=== 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"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot/kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Hydrogen atoms AB and C approach
eachother. Notably, the vibrational

motion present after the transition state is

not present before.
|[[File:Surface_Plot_question4_1_hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-4.1 </nowiki>
|<nowiki>-420.077</nowiki>
|No
|

Hydrogen atoms AB approaches C, however

before approaching the saddle point to AB

swiftly moves away from C

|[[File:Animation exercise2 p2 -4.1 hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|Hydrogen atom C approaches AB with a large
amount of momentum, resulting in a new BC Hydrogen

molecule forming, releasing the A hydrogen atom in the

reaction.
|[[File:Animation exercise2 p2 -5.1 p1 -3.1 hk3918.png|300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.1 </nowiki>
|<nowiki>-357.277</nowiki>
|No
|Hydrogen atoms AB and C approach each other very
quickly. Upon impact, large internuclear distance and velocities

are depicted, showing the B atom being tugged by both A and C.

In the end, B forms a molecule AB and moves away from C.
|[[File:Animation exercise2 p2 -10.1 hk3918.png |300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.6 </nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|Upon impact, the B atom is tugged between the two atoms A and
C, and is finally bonded to to atom C. A then moves away from

BC.
|[[File:Animation p2 -10.6 hk3918.png |300px]]
|}
The table above shows that a reaction isn't only reliant on the amount of kinetic energy in the system. As shown in the table above, there are instances where kinetic energy is higher than the activation energy, however no reaction takes place. One therefore needs to consider that successful collision are reliant on the system having enough energy to break and reform bonds, and also the orientation at which atoms collide.

=== Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===
Transition State Theory has a fundamental assumption which will result in its predictions being different to experimental values. The assumption that all trajectories possessing enough kinetic energy to circumvent the activation barrier will be reactive will result in a larger ''Rcl '' and thus a larger rate prediction than the experimental outcome. As seen in the table above, not all collisions are reactive even if there is sufficient kinetic energy - the case of barrier recrossing perfectly illustrates this.

== Exercise 2 ==

=== 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? ===
[[File:Surface_Plot_HF_hk3918.png|thumb|Figure 4. A surface plot showing the trajectory of F + H2 to H + HF reaction]|300px]]The reaction is exothermic as shown in figure 4, there is a decrease in potential energy when passing the transition state and forming the new HF bond. This relates to bond-strength, H-H has a bond strength of 432 kJ/mol whilst H-F has a bond strength of 565 kJ/mol.(reference). As HF is a polar bond as a result of the very electronegative F, there is an ionic element to the bond, resulting in a much greater attraction between the two atoms, thus releasing a lot of energy on formation. As the energy released is greater than that absorbed, the reaction is exothermic as depicted in the trajectory shown in figure 4.

[[File:Surface_Plot_hf-h_hk3918.png|thumb|Figure 5. A surface plot showing the trajectory of the FH + H to F + H2 reaction|300px]]

Figure 5 on the other hand shows an endothermic reaction, illustrated by the increase in potential energy along the trajectory, indicating that the energy absorbed to break the stronger H-F bond is greater than the energy released when forming the weaker H-H bond.

[[File:Plot small activation.png|thumb|Figure 6. Activation energy for the F +H2 reaction |300px]]

[[File:Plot activation big.png|thumb|Figure 7. Activation energy for the HF + H reaction |300px]]

[[File:Animation_momenta_hk3918.png |thumb|Figure 8. Momenta vs time plot for F + H2 reaction |300px]]

[[File:Animation_-74_-6.1.png ‎ |thumb|Figure 9. Contour plot for F + H2 reaction. Parameters: AB=200, BC=74, Pab=-1, Pbc=-6.9 |300px]]

[[File: Animation_low_momentum_hk3918.png ‎ |thumb|Figure 10. Contour plot for F + H2 reaction. Parameters: AB=200, BC=74, Pab=-1.6, Pbc=0.2 |300px]]

[[File:Animation low vibrational.png ‎ |thumb|Figure 11. Contour plot for the HF + H reaction where H has high momenta and HF has low initial momenta |300px]]

[[File:Animation highvirabtional.png‎ |thumb|Figure 12. Skew plot for the HF + H reaction where H has low momenta and HF has high initial momenta |300px]]

=== Locate the approximate position of the transition state ===
The transition state position was calculated to be when the H-F bond had a bond length of 181.8 pm, and the H-H bond was 73.1 pm as shown in figure 5. Analoguos to the calculation when approximating the transition state position of the H + H2 system, in the F + H2 reaction simulations were tested in order to acquire a constant bond length, at the maximum potential energy. As the first part of this reaction was exothermic, adopting Hammonds postulate, the assumption was made that the reaction had an early transition state thus the region near the reactants was analysed. The MEP calculation was used to remove the vibrational noise as it made it difficult to read the plots. As HF + H is a late transition state, resembling the its product F + H2, it will have the same transition state positions.

=== Report the activation energy for both reactions. ===
In order to find the activation energy the bond distance of H-F was first decreased. This then resulted in the energy falling to form the products H-F and H. The difference in this energy as shown in figure 7 was used to calculate the activation energy of the H-F + H endothermic reaction. The activation energy was estimated to be 120 kJ/mol. For the exothermic F + H2 reaction, the activation energy was calculated by trying to shift the trajectory towards to the reactants. This was done by thus increasing the F-H distance forming the F + H2 reactant. The difference in this energy was very small as shown in figure 8 and was used to calculate the activation energy for the F + H2 reaction.

=== 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. ===
Figure 8 illustrates the transition between translational and vibrational energy in the F + H2 reaction. Initially, the H-H bond (B-C) is vibrationally excited shown by the oscillation, whilst the H-F(A-B) bond isn't indicated by the absence of an oscillation. The H-F bond is at the vibrational ground state . After the collision we see a transition. The H-F bond now is vibrationally excited illustrated by the oscillation in the momenta vs time plot, and the H-H bonds loses this vibrationally energy and now only has translational energy due to the breaking of the bond, depicted by the plateau after 30s. Experimentally IR could be used to see the progress of the reaction. Only molecules with a dipole are IR active, thus after the formation of a H-F bond, IR absorption absent in the reactant will start to appear with overtones suggesting higher vibrational energies reached.

=== 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. ===
Below various example of both reactions are used to illustrate Polanyl's rules. Reactions with late transition states are more efficient when the initial trajectory possesses vibrational energy whereas reactions with early transition states are more efficient when the initial trajectory possesses translational energy. The following examples illustrate these rules.

==== Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, with a momentum pFH = -1.0 g.mol-1.pm.fs-1, and explore several values of pHH in the range -6.1 to 6.1 g.mol-1.pm.fs-1 (explore values also close to these limits). What do you observe? ====
When using the parameters above and setting the F-H value to 200 pm, at lower momentum values larger vibrational frequencies were obversed. The reactions did not therefore take place, satisfying Polyanyi's rules, where the having translational energy is more efficient when there is an early barrier rather than vibrational energy. Even at higher momentum extremes shown in figure 9, the reaction does not go to completion, indicating even though a lot of energy is being put into the system , it is not an efficient reaction, satisfying the rules.

==== For the same initial position, increase slightly the momentum pFH = -1.6 g.mol-1.pm.fs-1, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.2 g.mol-1.pm.fs-1. What do you observe now? ====
Figure 10 shows that, even at lower momentum values, resulting in lower vibrational energies the reaction still does not proceed to completion. As there is not enough translational energy in the reaction, further satisfying Polyanyi's rules.

==== Focus on the reverse reaction, H + HF. ====
This reaction has a late transition state, thus according to Polanyi's rules, vibrational energy trajectory is more efficient. This was tested firstly by using a low H-F vibrational energy and a high H translational energy as shown in figure 11. As predicted the reaction did not proceed to completion as there was not enough vibrational energy in the reaction. Figure 12 shows an example where the reaction has a larger vibrational energy, and therefore is more efficient, resulting in a successful reaction.

MRD:01524831.molecularreactiondynamics

2020-05-15T21:06:28Z

Hk3918: /* Exercise 1: H + H2 system */

== Exercise 1: H + H2 system ==

[[File:Surface_Plot_first_trajcectory_hk3918.png|thumb|Figure 1. A surface plot showing the transition state of a H + H2 system]]
[[File:Surface_Plot_transtion_point-h3_hk3918.png |thumb|Figure 2. A contour plot showing the transition state of a H + H2 system at 91 pm]]

[[File:Animation_mep_hk3918.png ‎|thumb|Figure 3. A surface plot showing the mep of a H + H2 system]]

=== 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? ===
On a potential energy surface diagram , the transition state is mathematically defined as the saddle point. The saddle point is a stationary point on the plot, where along the reaction path it is a maxima and perpendicular to the reaction path the saddle point is perceived to be a minima. The transition point can be identified in this simple example when the distances between the 3 hydrogen atoms are of equal distance. As shown in figure 1, the reaction path is oscillating at the saddle point. This depicts the vibrational freedom of the system at the transition state. A saddle point is mathematically found as shown below.

D = fxx.fyy - fxy2

<nowiki> </nowiki>If D > 0, fxx < 0 , local minima.

If D< 0 , then it is a saddle point.

=== 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 estimated '''rts '''is found to be around 91 pm. The transition state in this H + H2 system is defined to be the position where internuclear distances are equal, and where it is at a potential energy maxima in relation to the reaction path trajectory. From figure 1 it is evident that there are vibrations within the system, depicted by the oscillations on the ridge. The transition state position was found by determining at which position the internuclear distance was equal and constant, and where the potential energy was a maximum, and thus the stable point of the transition state. The saddle point was found as shown in figure 2.

=== Comment on how the ''mep'' and the trajectory you just calculated differ. ===
When calculating under mep the reaction path is depicted different to when the calculation type is set to dynamics. What is absent in the map calculation is the motion of the atoms. The vibrational motion of the three hydrogen atoms are not depicted as shown in figure .... The initial condition is set so that the distance between AB is '''rts''' and the distance between BC deviates from '''rts '''by 1 pm, resulting in the reaction pathway following through, and advance towards the HA + HB-HC product. The flat BC line indicates the fact that the mep calculation does not consider vibrational motion of atoms.

==== Look at the “Internuclear Distances vs Time” and “Momenta vs Time”. What would change if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead? ====
The change of initial conditions to those above will change final product from HA + HB-HC to HA-HB + HC as it is the HB-HC bond distance that deviates away from the '''rts. '''As a result the internuclear distances vs time plot will now show AB to be constant over time. Momentum remains unchanged as initial momentum is 0 for both scenarios

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

The 'Get last geometry' option was selected, and the momenta were then reversed under the dynamic calculation type. The plot remains unchanged as expected. The reason for this is because the initial argument resulted in a trajectory towards a larger BC distance. Using the momenta of this trajectory but reversing the sign simply results in the hydrogen and hydrogen molecule to start in the final position, but now approach each other due to the negative momentum value.
=== 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"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot/kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Hydrogen atoms AB and C approach
eachother. Notably, the vibrational

motion present after the transition state is

not present before.
|[[File:Surface_Plot_question4_1_hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-4.1 </nowiki>
|<nowiki>-420.077</nowiki>
|No
|

Hydrogen atoms AB approaches C, however

before approaching the saddle point to AB

swiftly moves away from C

|[[File:Animation exercise2 p2 -4.1 hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|Hydrogen atom C approaches AB with a large
amount of momentum, resulting in a new BC Hydrogen

molecule forming, releasing the A hydrogen atom in the

reaction.
|[[File:Animation exercise2 p2 -5.1 p1 -3.1 hk3918.png|300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.1 </nowiki>
|<nowiki>-357.277</nowiki>
|No
|Hydrogen atoms AB and C approach each other very
quickly. Upon impact, large internuclear distance and velocities

are depicted, showing the B atom being tugged by both A and C.

In the end, B forms a molecule AB and moves away from C.
|[[File:Animation exercise2 p2 -10.1 hk3918.png |300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.6 </nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|Upon impact, the B atom is tugged between the two atoms A and
C, and is finally bonded to to atom C. A then moves away from

BC.
|[[File:Animation p2 -10.6 hk3918.png |300px]]
|}
The table above shows that a reaction isn't only reliant on the amount of kinetic energy in the system. As shown in the table above, there are instances where kinetic energy is higher than the activation energy, however no reaction takes place. One therefore needs to consider that successful collision are reliant on the system having enough energy to break and reform bonds, and also the orientation at which atoms collide.

=== Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===
Transition State Theory has a fundamental assumption which will result in its predictions being different to experimental values. The assumption that all trajectories possessing enough kinetic energy to circumvent the activation barrier will be reactive will result in a larger ''Rcl '' and thus a larger rate prediction than the experimental outcome. As seen in the table above, not all collisions are reactive even if there is sufficient kinetic energy - the case of barrier recrossing perfectly illustrates this.

== Exercise 2 ==

=== 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? ===
[[File:Surface_Plot_HF_hk3918.png|thumb|Figure 4. A surface plot showing the trajectory of F + H2 to H + HF reaction]|300px]]The reaction is exothermic as shown in figure 4, there is a decrease in potential energy when passing the transition state and forming the new HF bond. This relates to bond-strength, H-H has a bond strength of 432 kJ/mol whilst H-F has a bond strength of 565 kJ/mol.(reference). As HF is a polar bond as a result of the very electronegative F, there is an ionic element to the bond, resulting in a much greater attraction between the two atoms, thus releasing a lot of energy on formation. As the energy released is greater than that absorbed, the reaction is exothermic as depicted in the trajectory shown in figure 4.

[[File:Surface_Plot_hf-h_hk3918.png|thumb|Figure 5. A surface plot showing the trajectory of the FH + H to F + H2 reaction|300px]]

Figure 5 on the other hand shows an endothermic reaction, illustrated by the increase in potential energy along the trajectory, indicating that the energy absorbed to break the stronger H-F bond is greater than the energy released when forming the weaker H-H bond.

[[File:Plot small activation.png|thumb|Figure 6. Activation energy for the F +H2 reaction |300px]]

[[File:Plot activation big.png|thumb|Figure 7. Activation energy for the HF + H reaction |300px]]

[[File:Animation_momenta_hk3918.png |thumb|Figure 8. Momenta vs time plot for F + H2 reaction |300px]]

[[File:Animation_-74_-6.1.png ‎ |thumb|Figure 9. Contour plot for F + H2 reaction. Parameters: AB=200, BC=74, Pab=-1, Pbc=-6.9 |300px]]

[[File: Animation_low_momentum_hk3918.png ‎ |thumb|Figure 10. Contour plot for F + H2 reaction. Parameters: AB=200, BC=74, Pab=-1.6, Pbc=0.2 |300px]]

[[File:Animation low vibrational.png ‎ |thumb|Figure 11. Contour plot for the HF + H reaction where H has high momenta and HF has low initial momenta |300px]]

[[File:Animation highvirabtional.png‎ |thumb|Figure 12. Skew plot for the HF + H reaction where H has low momenta and HF has high initial momenta |300px]]

=== Locate the approximate position of the transition state ===
The transition state position was calculated to be when the H-F bond had a bond length of 181.8 pm, and the H-H bond was 73.1 pm as shown in figure 5. Analoguos to the calculation when approximating the transition state position of the H + H2 system, in the F + H2 reaction simulations were tested in order to acquire a constant bond length, at the maximum potential energy. As the first part of this reaction was exothermic, adopting Hammonds postulate, the assumption was made that the reaction had an early transition state thus the region near the reactants was analysed. The MEP calculation was used to remove the vibrational noise as it made it difficult to read the plots. As HF + H is a late transition state, resembling the its product F + H2, it will have the same transition state positions.

=== Report the activation energy for both reactions. ===
In order to find the activation energy the bond distance of H-F was first decreased. This then resulted in the energy falling to form the products H-F and H. The difference in this energy as shown in figure 7 was used to calculate the activation energy of the H-F + H endothermic reaction. The activation energy was estimated to be 120 kJ/mol. For the exothermic F + H2 reaction, the activation energy was calculated by trying to shift the trajectory towards to the reactants. This was done by thus increasing the F-H distance forming the F + H2 reactant. The difference in this energy was very small as shown in figure 8 and was used to calculate the activation energy for the F + H2 reaction.

=== 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. ===
Figure 8 illustrates the transition between translational and vibrational energy in the F + H2 reaction. Initially, the H-H bond (B-C) is vibrationally excited shown by the oscillation, whilst the H-F(A-B) bond isn't indicated by the absence of an oscillation. The H-F bond is at the vibrational ground state . After the collision we see a transition. The H-F bond now is vibrationally excited illustrated by the oscillation in the momenta vs time plot, and the H-H bonds loses this vibrationally energy and now only has translational energy due to the breaking of the bond, depicted by the plateau after 30s. Experimentally IR could be used to see the progress of the reaction. Only molecules with a dipole are IR active, thus after the formation of a H-F bond, IR absorption absent in the reactant will start to appear with overtones suggesting higher vibrational energies reached.

=== 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. ===
Below various example of both reactions are used to illustrate Polanyl's rules. Reactions with late transition states are more efficient when the initial trajectory possesses vibrational energy whereas reactions with early transition states are more efficient when the initial trajectory possesses translational energy. The following examples illustrate these rules.

==== Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, with a momentum pFH = -1.0 g.mol-1.pm.fs-1, and explore several values of pHH in the range -6.1 to 6.1 g.mol-1.pm.fs-1 (explore values also close to these limits). What do you observe? ====
When using the parameters above and setting the F-H value to 200 pm, at lower momentum values larger vibrational frequencies were obversed. The reactions did not therefore take place, satisfying Polyanyi's rules, where the having translational energy is more efficient when there is an early barrier rather than vibrational energy. Even at higher momentum extremes shown in figure 9, the reaction does not go to completion, indicating even though a lot of energy is being put into the system , it is not an efficient reaction, satisfying the rules.

==== For the same initial position, increase slightly the momentum pFH = -1.6 g.mol-1.pm.fs-1, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.2 g.mol-1.pm.fs-1. What do you observe now? ====
Figure 10 shows that, even at lower momentum values, resulting in lower vibrational energies the reaction still does not proceed to completion. As there is not enough translational energy in the reaction, further satisfying Polyanyi's rules.

==== Focus on the reverse reaction, H + HF. ====
This reaction has a late transition state, thus according to Polanyi's rules, vibrational energy trajectory is more efficient. This was tested firstly by using a low H-F vibrational energy and a high H translational energy as shown in figure 11. As predicted the reaction did not proceed to completion as there was not enough vibrational energy in the reaction. Figure 12 shows an example where the reaction has a larger vibrational energy, and therefore is more efficient, resulting in a successful reaction.

File:Animation mep hk3918.png

2020-05-15T21:05:31Z

Hk3918:

MRD:01524831.molecularreactiondynamics

2020-05-15T20:58:10Z

Hk3918: /* Exercise 1: H + H2 system */

== Exercise 1: H + H2 system ==

[[File:Surface_Plot_first_trajcectory_hk3918.png|thumb|Figure 1. A surface plot showing the transition state of a H + H2 system]]
[[File:Surface_Plot_transtion_point-h3_hk3918.png |thumb|Figure 2. A contour plot showing the transition state of a H + H2 system at 91 pm]]

=== 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? ===
On a potential energy surface diagram , the transition state is mathematically defined as the saddle point. The saddle point is a stationary point on the plot, where along the reaction path it is a maxima and perpendicular to the reaction path the saddle point is perceived to be a minima. The transition point can be identified in this simple example when the distances between the 3 hydrogen atoms are of equal distance. As shown in figure 1, the reaction path is oscillating at the saddle point. This depicts the vibrational freedom of the system at the transition state. A saddle point is mathematically found as shown below.

D = fxx.fyy - fxy2

<nowiki> </nowiki>If D > 0, fxx < 0 , local minima.

If D< 0 , then it is a saddle point.

=== 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 estimated '''rts '''is found to be around 91 pm. The transition state in this H + H2 system is defined to be the position where internuclear distances are equal, and where it is at a potential energy maxima in relation to the reaction path trajectory. From figure 1 it is evident that there are vibrations within the system, depicted by the oscillations on the ridge. The transition state position was found by determining at which position the internuclear distance was equal and constant, and where the potential energy was a maximum, and thus the stable point of the transition state. The saddle point was found as shown in figure 2.

=== Comment on how the ''mep'' and the trajectory you just calculated differ. ===
When calculating under mep the reaction path is depicted different to when the calculation type is set to dynamics. What is absent in the map calculation is the motion of the atoms. The vibrational motion of the three hydrogen atoms are not depicted as shown in figure .... The initial condition is set so that the distance between AB is '''rts''' and the distance between BC deviates from '''rts '''by 1 pm, resulting in the reaction pathway following through, and advance towards the HA + HB-HC product. The flat BC line indicates the fact that the mep calculation does not consider vibrational motion of atoms.

==== Look at the “Internuclear Distances vs Time” and “Momenta vs Time”. What would change if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead? ====
The change of initial conditions to those above will change final product from HA + HB-HC to HA-HB + HC as it is the HB-HC bond distance that deviates away from the '''rts. '''As a result the internuclear distances vs time plot will now show AB to be constant over time. Momentum remains unchanged as initial momentum is 0 for both scenarios

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

The 'Get last geometry' option was selected, and the momenta were then reversed under the dynamic calculation type. The plot remains unchanged as expected. The reason for this is because the initial argument resulted in a trajectory towards a larger BC distance. Using the momenta of this trajectory but reversing the sign simply results in the hydrogen and hydrogen molecule to start in the final position, but now approach each other due to the negative momentum value.
=== 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"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot/kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Hydrogen atoms AB and C approach
eachother. Notably, the vibrational

motion present after the transition state is

not present before.
|[[File:Surface_Plot_question4_1_hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-4.1 </nowiki>
|<nowiki>-420.077</nowiki>
|No
|

Hydrogen atoms AB approaches C, however

before approaching the saddle point to AB

swiftly moves away from C

|[[File:Animation exercise2 p2 -4.1 hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|Hydrogen atom C approaches AB with a large
amount of momentum, resulting in a new BC Hydrogen

molecule forming, releasing the A hydrogen atom in the

reaction.
|[[File:Animation exercise2 p2 -5.1 p1 -3.1 hk3918.png|300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.1 </nowiki>
|<nowiki>-357.277</nowiki>
|No
|Hydrogen atoms AB and C approach each other very
quickly. Upon impact, large internuclear distance and velocities

are depicted, showing the B atom being tugged by both A and C.

In the end, B forms a molecule AB and moves away from C.
|[[File:Animation exercise2 p2 -10.1 hk3918.png |300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.6 </nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|Upon impact, the B atom is tugged between the two atoms A and
C, and is finally bonded to to atom C. A then moves away from

BC.
|[[File:Animation p2 -10.6 hk3918.png |300px]]
|}
The table above shows that a reaction isn't only reliant on the amount of kinetic energy in the system. As shown in the table above, there are instances where kinetic energy is higher than the activation energy, however no reaction takes place. One therefore needs to consider that successful collision are reliant on the system having enough energy to break and reform bonds, and also the orientation at which atoms collide.

=== Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===
Transition State Theory has a fundamental assumption which will result in its predictions being different to experimental values. The assumption that all trajectories possessing enough kinetic energy to circumvent the activation barrier will be reactive will result in a larger ''Rcl '' and thus a larger rate prediction than the experimental outcome. As seen in the table above, not all collisions are reactive even if there is sufficient kinetic energy - the case of barrier recrossing perfectly illustrates this.

== Exercise 2 ==

=== 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? ===
[[File:Surface_Plot_HF_hk3918.png|thumb|Figure 4. A surface plot showing the trajectory of F + H2 to H + HF reaction]|300px]]The reaction is exothermic as shown in figure 4, there is a decrease in potential energy when passing the transition state and forming the new HF bond. This relates to bond-strength, H-H has a bond strength of 432 kJ/mol whilst H-F has a bond strength of 565 kJ/mol.(reference). As HF is a polar bond as a result of the very electronegative F, there is an ionic element to the bond, resulting in a much greater attraction between the two atoms, thus releasing a lot of energy on formation. As the energy released is greater than that absorbed, the reaction is exothermic as depicted in the trajectory shown in figure 4.

[[File:Surface_Plot_hf-h_hk3918.png|thumb|Figure 5. A surface plot showing the trajectory of the FH + H to F + H2 reaction|300px]]

Figure 5 on the other hand shows an endothermic reaction, illustrated by the increase in potential energy along the trajectory, indicating that the energy absorbed to break the stronger H-F bond is greater than the energy released when forming the weaker H-H bond.

[[File:Plot small activation.png|thumb|Figure 6. Activation energy for the F +H2 reaction |300px]]

[[File:Plot activation big.png|thumb|Figure 7. Activation energy for the HF + H reaction |300px]]

[[File:Animation_momenta_hk3918.png |thumb|Figure 8. Momenta vs time plot for F + H2 reaction |300px]]

[[File:Animation_-74_-6.1.png ‎ |thumb|Figure 9. Contour plot for F + H2 reaction. Parameters: AB=200, BC=74, Pab=-1, Pbc=-6.9 |300px]]

[[File: Animation_low_momentum_hk3918.png ‎ |thumb|Figure 10. Contour plot for F + H2 reaction. Parameters: AB=200, BC=74, Pab=-1.6, Pbc=0.2 |300px]]

[[File:Animation low vibrational.png ‎ |thumb|Figure 11. Contour plot for the HF + H reaction where H has high momenta and HF has low initial momenta |300px]]

[[File:Animation highvirabtional.png‎ |thumb|Figure 12. Skew plot for the HF + H reaction where H has low momenta and HF has high initial momenta |300px]]

=== Locate the approximate position of the transition state ===
The transition state position was calculated to be when the H-F bond had a bond length of 181.8 pm, and the H-H bond was 73.1 pm as shown in figure 5. Analoguos to the calculation when approximating the transition state position of the H + H2 system, in the F + H2 reaction simulations were tested in order to acquire a constant bond length, at the maximum potential energy. As the first part of this reaction was exothermic, adopting Hammonds postulate, the assumption was made that the reaction had an early transition state thus the region near the reactants was analysed. The MEP calculation was used to remove the vibrational noise as it made it difficult to read the plots. As HF + H is a late transition state, resembling the its product F + H2, it will have the same transition state positions.

=== Report the activation energy for both reactions. ===
In order to find the activation energy the bond distance of H-F was first decreased. This then resulted in the energy falling to form the products H-F and H. The difference in this energy as shown in figure 7 was used to calculate the activation energy of the H-F + H endothermic reaction. The activation energy was estimated to be 120 kJ/mol. For the exothermic F + H2 reaction, the activation energy was calculated by trying to shift the trajectory towards to the reactants. This was done by thus increasing the F-H distance forming the F + H2 reactant. The difference in this energy was very small as shown in figure 8 and was used to calculate the activation energy for the F + H2 reaction.

=== 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. ===
Figure 8 illustrates the transition between translational and vibrational energy in the F + H2 reaction. Initially, the H-H bond (B-C) is vibrationally excited shown by the oscillation, whilst the H-F(A-B) bond isn't indicated by the absence of an oscillation. The H-F bond is at the vibrational ground state . After the collision we see a transition. The H-F bond now is vibrationally excited illustrated by the oscillation in the momenta vs time plot, and the H-H bonds loses this vibrationally energy and now only has translational energy due to the breaking of the bond, depicted by the plateau after 30s. Experimentally IR could be used to see the progress of the reaction. Only molecules with a dipole are IR active, thus after the formation of a H-F bond, IR absorption absent in the reactant will start to appear with overtones suggesting higher vibrational energies reached.

=== 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. ===
Below various example of both reactions are used to illustrate Polanyl's rules. Reactions with late transition states are more efficient when the initial trajectory possesses vibrational energy whereas reactions with early transition states are more efficient when the initial trajectory possesses translational energy. The following examples illustrate these rules.

==== Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, with a momentum pFH = -1.0 g.mol-1.pm.fs-1, and explore several values of pHH in the range -6.1 to 6.1 g.mol-1.pm.fs-1 (explore values also close to these limits). What do you observe? ====
When using the parameters above and setting the F-H value to 200 pm, at lower momentum values larger vibrational frequencies were obversed. The reactions did not therefore take place, satisfying Polyanyi's rules, where the having translational energy is more efficient when there is an early barrier rather than vibrational energy. Even at higher momentum extremes shown in figure 9, the reaction does not go to completion, indicating even though a lot of energy is being put into the system , it is not an efficient reaction, satisfying the rules.

==== For the same initial position, increase slightly the momentum pFH = -1.6 g.mol-1.pm.fs-1, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.2 g.mol-1.pm.fs-1. What do you observe now? ====
Figure 10 shows that, even at lower momentum values, resulting in lower vibrational energies the reaction still does not proceed to completion. As there is not enough translational energy in the reaction, further satisfying Polyanyi's rules.

==== Focus on the reverse reaction, H + HF. ====
This reaction has a late transition state, thus according to Polanyi's rules, vibrational energy trajectory is more efficient. This was tested firstly by using a low H-F vibrational energy and a high H translational energy as shown in figure 11. As predicted the reaction did not proceed to completion as there was not enough vibrational energy in the reaction. Figure 12 shows an example where the reaction has a larger vibrational energy, and therefore is more efficient, resulting in a successful reaction.

File:Surface Plot transtion point-h3 hk3918.png

2020-05-15T20:57:22Z

Hk3918:

MRD:01524831.molecularreactiondynamics

2020-05-15T20:55:38Z

Hk3918: /* 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. */

== Exercise 1: H + H2 system ==

[[File:Surface_Plot_first_trajcectory_hk3918.png|thumb|Figure 1. A surface plot showing the transition state of a H + H2 system]]

=== 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? ===
On a potential energy surface diagram , the transition state is mathematically defined as the saddle point. The saddle point is a stationary point on the plot, where along the reaction path it is a maxima and perpendicular to the reaction path the saddle point is perceived to be a minima. The transition point can be identified in this simple example when the distances between the 3 hydrogen atoms are of equal distance. As shown in figure 1, the reaction path is oscillating at the saddle point. This depicts the vibrational freedom of the system at the transition state. A saddle point is mathematically found as shown below.

D = fxx.fyy - fxy2

<nowiki> </nowiki>If D > 0, fxx < 0 , local minima.

If D< 0 , then it is a saddle point.

=== 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 estimated '''rts '''is found to be around 91 pm. The transition state in this H + H2 system is defined to be the position where internuclear distances are equal, and where it is at a potential energy maxima in relation to the reaction path trajectory. From figure 1 it is evident that there are vibrations within the system, depicted by the oscillations on the ridge. The transition state position was found by determining at which position the internuclear distance was equal and constant, and where the potential energy was a maximum, and thus the stable point of the transition state. The saddle point was found as shown in figure 2.

=== Comment on how the ''mep'' and the trajectory you just calculated differ. ===
When calculating under mep the reaction path is depicted different to when the calculation type is set to dynamics. What is absent in the map calculation is the motion of the atoms. The vibrational motion of the three hydrogen atoms are not depicted as shown in figure .... The initial condition is set so that the distance between AB is '''rts''' and the distance between BC deviates from '''rts '''by 1 pm, resulting in the reaction pathway following through, and advance towards the HA + HB-HC product. The flat BC line indicates the fact that the mep calculation does not consider vibrational motion of atoms.

==== Look at the “Internuclear Distances vs Time” and “Momenta vs Time”. What would change if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead? ====
The change of initial conditions to those above will change final product from HA + HB-HC to HA-HB + HC as it is the HB-HC bond distance that deviates away from the '''rts. '''As a result the internuclear distances vs time plot will now show AB to be constant over time. Momentum remains unchanged as initial momentum is 0 for both scenarios

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

The 'Get last geometry' option was selected, and the momenta were then reversed under the dynamic calculation type. The plot remains unchanged as expected. The reason for this is because the initial argument resulted in a trajectory towards a larger BC distance. Using the momenta of this trajectory but reversing the sign simply results in the hydrogen and hydrogen molecule to start in the final position, but now approach each other due to the negative momentum value.
=== 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"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot/kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Hydrogen atoms AB and C approach
eachother. Notably, the vibrational

motion present after the transition state is

not present before.
|[[File:Surface_Plot_question4_1_hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-4.1 </nowiki>
|<nowiki>-420.077</nowiki>
|No
|

Hydrogen atoms AB approaches C, however

before approaching the saddle point to AB

swiftly moves away from C

|[[File:Animation exercise2 p2 -4.1 hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|Hydrogen atom C approaches AB with a large
amount of momentum, resulting in a new BC Hydrogen

molecule forming, releasing the A hydrogen atom in the

reaction.
|[[File:Animation exercise2 p2 -5.1 p1 -3.1 hk3918.png|300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.1 </nowiki>
|<nowiki>-357.277</nowiki>
|No
|Hydrogen atoms AB and C approach each other very
quickly. Upon impact, large internuclear distance and velocities

are depicted, showing the B atom being tugged by both A and C.

In the end, B forms a molecule AB and moves away from C.
|[[File:Animation exercise2 p2 -10.1 hk3918.png |300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.6 </nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|Upon impact, the B atom is tugged between the two atoms A and
C, and is finally bonded to to atom C. A then moves away from

BC.
|[[File:Animation p2 -10.6 hk3918.png |300px]]
|}
The table above shows that a reaction isn't only reliant on the amount of kinetic energy in the system. As shown in the table above, there are instances where kinetic energy is higher than the activation energy, however no reaction takes place. One therefore needs to consider that successful collision are reliant on the system having enough energy to break and reform bonds, and also the orientation at which atoms collide.

=== Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===
Transition State Theory has a fundamental assumption which will result in its predictions being different to experimental values. The assumption that all trajectories possessing enough kinetic energy to circumvent the activation barrier will be reactive will result in a larger ''Rcl '' and thus a larger rate prediction than the experimental outcome. As seen in the table above, not all collisions are reactive even if there is sufficient kinetic energy - the case of barrier recrossing perfectly illustrates this.

== Exercise 2 ==

=== 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? ===
[[File:Surface_Plot_HF_hk3918.png|thumb|Figure 4. A surface plot showing the trajectory of F + H2 to H + HF reaction]|300px]]The reaction is exothermic as shown in figure 4, there is a decrease in potential energy when passing the transition state and forming the new HF bond. This relates to bond-strength, H-H has a bond strength of 432 kJ/mol whilst H-F has a bond strength of 565 kJ/mol.(reference). As HF is a polar bond as a result of the very electronegative F, there is an ionic element to the bond, resulting in a much greater attraction between the two atoms, thus releasing a lot of energy on formation. As the energy released is greater than that absorbed, the reaction is exothermic as depicted in the trajectory shown in figure 4.

[[File:Surface_Plot_hf-h_hk3918.png|thumb|Figure 5. A surface plot showing the trajectory of the FH + H to F + H2 reaction|300px]]

Figure 5 on the other hand shows an endothermic reaction, illustrated by the increase in potential energy along the trajectory, indicating that the energy absorbed to break the stronger H-F bond is greater than the energy released when forming the weaker H-H bond.

[[File:Plot small activation.png|thumb|Figure 6. Activation energy for the F +H2 reaction |300px]]

[[File:Plot activation big.png|thumb|Figure 7. Activation energy for the HF + H reaction |300px]]

[[File:Animation_momenta_hk3918.png |thumb|Figure 8. Momenta vs time plot for F + H2 reaction |300px]]

[[File:Animation_-74_-6.1.png ‎ |thumb|Figure 9. Contour plot for F + H2 reaction. Parameters: AB=200, BC=74, Pab=-1, Pbc=-6.9 |300px]]

[[File: Animation_low_momentum_hk3918.png ‎ |thumb|Figure 10. Contour plot for F + H2 reaction. Parameters: AB=200, BC=74, Pab=-1.6, Pbc=0.2 |300px]]

[[File:Animation low vibrational.png ‎ |thumb|Figure 11. Contour plot for the HF + H reaction where H has high momenta and HF has low initial momenta |300px]]

[[File:Animation highvirabtional.png‎ |thumb|Figure 12. Skew plot for the HF + H reaction where H has low momenta and HF has high initial momenta |300px]]

=== Locate the approximate position of the transition state ===
The transition state position was calculated to be when the H-F bond had a bond length of 181.8 pm, and the H-H bond was 73.1 pm as shown in figure 5. Analoguos to the calculation when approximating the transition state position of the H + H2 system, in the F + H2 reaction simulations were tested in order to acquire a constant bond length, at the maximum potential energy. As the first part of this reaction was exothermic, adopting Hammonds postulate, the assumption was made that the reaction had an early transition state thus the region near the reactants was analysed. The MEP calculation was used to remove the vibrational noise as it made it difficult to read the plots. As HF + H is a late transition state, resembling the its product F + H2, it will have the same transition state positions.

=== Report the activation energy for both reactions. ===
In order to find the activation energy the bond distance of H-F was first decreased. This then resulted in the energy falling to form the products H-F and H. The difference in this energy as shown in figure 7 was used to calculate the activation energy of the H-F + H endothermic reaction. The activation energy was estimated to be 120 kJ/mol. For the exothermic F + H2 reaction, the activation energy was calculated by trying to shift the trajectory towards to the reactants. This was done by thus increasing the F-H distance forming the F + H2 reactant. The difference in this energy was very small as shown in figure 8 and was used to calculate the activation energy for the F + H2 reaction.

=== 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. ===
Figure 8 illustrates the transition between translational and vibrational energy in the F + H2 reaction. Initially, the H-H bond (B-C) is vibrationally excited shown by the oscillation, whilst the H-F(A-B) bond isn't indicated by the absence of an oscillation. The H-F bond is at the vibrational ground state . After the collision we see a transition. The H-F bond now is vibrationally excited illustrated by the oscillation in the momenta vs time plot, and the H-H bonds loses this vibrationally energy and now only has translational energy due to the breaking of the bond, depicted by the plateau after 30s. Experimentally IR could be used to see the progress of the reaction. Only molecules with a dipole are IR active, thus after the formation of a H-F bond, IR absorption absent in the reactant will start to appear with overtones suggesting higher vibrational energies reached.

=== 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. ===
Below various example of both reactions are used to illustrate Polanyl's rules. Reactions with late transition states are more efficient when the initial trajectory possesses vibrational energy whereas reactions with early transition states are more efficient when the initial trajectory possesses translational energy. The following examples illustrate these rules.

==== Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, with a momentum pFH = -1.0 g.mol-1.pm.fs-1, and explore several values of pHH in the range -6.1 to 6.1 g.mol-1.pm.fs-1 (explore values also close to these limits). What do you observe? ====
When using the parameters above and setting the F-H value to 200 pm, at lower momentum values larger vibrational frequencies were obversed. The reactions did not therefore take place, satisfying Polyanyi's rules, where the having translational energy is more efficient when there is an early barrier rather than vibrational energy. Even at higher momentum extremes shown in figure 9, the reaction does not go to completion, indicating even though a lot of energy is being put into the system , it is not an efficient reaction, satisfying the rules.

==== For the same initial position, increase slightly the momentum pFH = -1.6 g.mol-1.pm.fs-1, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.2 g.mol-1.pm.fs-1. What do you observe now? ====
Figure 10 shows that, even at lower momentum values, resulting in lower vibrational energies the reaction still does not proceed to completion. As there is not enough translational energy in the reaction, further satisfying Polyanyi's rules.

==== Focus on the reverse reaction, H + HF. ====
This reaction has a late transition state, thus according to Polanyi's rules, vibrational energy trajectory is more efficient. This was tested firstly by using a low H-F vibrational energy and a high H translational energy as shown in figure 11. As predicted the reaction did not proceed to completion as there was not enough vibrational energy in the reaction. Figure 12 shows an example where the reaction has a larger vibrational energy, and therefore is more efficient, resulting in a successful reaction.

MRD:01524831.molecularreactiondynamics

2020-05-15T20:53:41Z

Hk3918: /* 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. */

== Exercise 1: H + H2 system ==

[[File:Surface_Plot_first_trajcectory_hk3918.png|thumb|Figure 1. A surface plot showing the transition state of a H + H2 system]]

=== 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? ===
On a potential energy surface diagram , the transition state is mathematically defined as the saddle point. The saddle point is a stationary point on the plot, where along the reaction path it is a maxima and perpendicular to the reaction path the saddle point is perceived to be a minima. The transition point can be identified in this simple example when the distances between the 3 hydrogen atoms are of equal distance. As shown in figure 1, the reaction path is oscillating at the saddle point. This depicts the vibrational freedom of the system at the transition state. A saddle point is mathematically found as shown below.

D = fxx.fyy - fxy2

<nowiki> </nowiki>If D > 0, fxx < 0 , local minima.

If D< 0 , then it is a saddle point.

=== 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 estimated '''rts '''is found to be around 91 pm. The transition state in this H + H2 system is defined to be the position where internuclear distances are equal, and where it is at a potential energy maxima in relation to the reaction path trajectory. From figure 1 it is evident that there are vibrations within the system, depicted by the oscillations on the ridge. The transition state position was found by determining which position the system stops oscillating, where the potential energy was a maximum, and thus the stable point of the transition state, the saddle point was found as shown in figure 2 and 3.

=== Comment on how the ''mep'' and the trajectory you just calculated differ. ===
When calculating under mep the reaction path is depicted different to when the calculation type is set to dynamics. What is absent in the map calculation is the motion of the atoms. The vibrational motion of the three hydrogen atoms are not depicted as shown in figure .... The initial condition is set so that the distance between AB is '''rts''' and the distance between BC deviates from '''rts '''by 1 pm, resulting in the reaction pathway following through, and advance towards the HA + HB-HC product. The flat BC line indicates the fact that the mep calculation does not consider vibrational motion of atoms.

==== Look at the “Internuclear Distances vs Time” and “Momenta vs Time”. What would change if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead? ====
The change of initial conditions to those above will change final product from HA + HB-HC to HA-HB + HC as it is the HB-HC bond distance that deviates away from the '''rts. '''As a result the internuclear distances vs time plot will now show AB to be constant over time. Momentum remains unchanged as initial momentum is 0 for both scenarios

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

The 'Get last geometry' option was selected, and the momenta were then reversed under the dynamic calculation type. The plot remains unchanged as expected. The reason for this is because the initial argument resulted in a trajectory towards a larger BC distance. Using the momenta of this trajectory but reversing the sign simply results in the hydrogen and hydrogen molecule to start in the final position, but now approach each other due to the negative momentum value.
=== 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"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot/kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Hydrogen atoms AB and C approach
eachother. Notably, the vibrational

motion present after the transition state is

not present before.
|[[File:Surface_Plot_question4_1_hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-4.1 </nowiki>
|<nowiki>-420.077</nowiki>
|No
|

Hydrogen atoms AB approaches C, however

before approaching the saddle point to AB

swiftly moves away from C

|[[File:Animation exercise2 p2 -4.1 hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|Hydrogen atom C approaches AB with a large
amount of momentum, resulting in a new BC Hydrogen

molecule forming, releasing the A hydrogen atom in the

reaction.
|[[File:Animation exercise2 p2 -5.1 p1 -3.1 hk3918.png|300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.1 </nowiki>
|<nowiki>-357.277</nowiki>
|No
|Hydrogen atoms AB and C approach each other very
quickly. Upon impact, large internuclear distance and velocities

are depicted, showing the B atom being tugged by both A and C.

In the end, B forms a molecule AB and moves away from C.
|[[File:Animation exercise2 p2 -10.1 hk3918.png |300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.6 </nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|Upon impact, the B atom is tugged between the two atoms A and
C, and is finally bonded to to atom C. A then moves away from

BC.
|[[File:Animation p2 -10.6 hk3918.png |300px]]
|}
The table above shows that a reaction isn't only reliant on the amount of kinetic energy in the system. As shown in the table above, there are instances where kinetic energy is higher than the activation energy, however no reaction takes place. One therefore needs to consider that successful collision are reliant on the system having enough energy to break and reform bonds, and also the orientation at which atoms collide.

=== Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===
Transition State Theory has a fundamental assumption which will result in its predictions being different to experimental values. The assumption that all trajectories possessing enough kinetic energy to circumvent the activation barrier will be reactive will result in a larger ''Rcl '' and thus a larger rate prediction than the experimental outcome. As seen in the table above, not all collisions are reactive even if there is sufficient kinetic energy - the case of barrier recrossing perfectly illustrates this.

== Exercise 2 ==

=== 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? ===
[[File:Surface_Plot_HF_hk3918.png|thumb|Figure 4. A surface plot showing the trajectory of F + H2 to H + HF reaction]|300px]]The reaction is exothermic as shown in figure 4, there is a decrease in potential energy when passing the transition state and forming the new HF bond. This relates to bond-strength, H-H has a bond strength of 432 kJ/mol whilst H-F has a bond strength of 565 kJ/mol.(reference). As HF is a polar bond as a result of the very electronegative F, there is an ionic element to the bond, resulting in a much greater attraction between the two atoms, thus releasing a lot of energy on formation. As the energy released is greater than that absorbed, the reaction is exothermic as depicted in the trajectory shown in figure 4.

[[File:Surface_Plot_hf-h_hk3918.png|thumb|Figure 5. A surface plot showing the trajectory of the FH + H to F + H2 reaction|300px]]

Figure 5 on the other hand shows an endothermic reaction, illustrated by the increase in potential energy along the trajectory, indicating that the energy absorbed to break the stronger H-F bond is greater than the energy released when forming the weaker H-H bond.

[[File:Plot small activation.png|thumb|Figure 6. Activation energy for the F +H2 reaction |300px]]

[[File:Plot activation big.png|thumb|Figure 7. Activation energy for the HF + H reaction |300px]]

[[File:Animation_momenta_hk3918.png |thumb|Figure 8. Momenta vs time plot for F + H2 reaction |300px]]

[[File:Animation_-74_-6.1.png ‎ |thumb|Figure 9. Contour plot for F + H2 reaction. Parameters: AB=200, BC=74, Pab=-1, Pbc=-6.9 |300px]]

[[File: Animation_low_momentum_hk3918.png ‎ |thumb|Figure 10. Contour plot for F + H2 reaction. Parameters: AB=200, BC=74, Pab=-1.6, Pbc=0.2 |300px]]

[[File:Animation low vibrational.png ‎ |thumb|Figure 11. Contour plot for the HF + H reaction where H has high momenta and HF has low initial momenta |300px]]

[[File:Animation highvirabtional.png‎ |thumb|Figure 12. Skew plot for the HF + H reaction where H has low momenta and HF has high initial momenta |300px]]

=== Locate the approximate position of the transition state ===
The transition state position was calculated to be when the H-F bond had a bond length of 181.8 pm, and the H-H bond was 73.1 pm as shown in figure 5. Analoguos to the calculation when approximating the transition state position of the H + H2 system, in the F + H2 reaction simulations were tested in order to acquire a constant bond length, at the maximum potential energy. As the first part of this reaction was exothermic, adopting Hammonds postulate, the assumption was made that the reaction had an early transition state thus the region near the reactants was analysed. The MEP calculation was used to remove the vibrational noise as it made it difficult to read the plots. As HF + H is a late transition state, resembling the its product F + H2, it will have the same transition state positions.

=== Report the activation energy for both reactions. ===
In order to find the activation energy the bond distance of H-F was first decreased. This then resulted in the energy falling to form the products H-F and H. The difference in this energy as shown in figure 7 was used to calculate the activation energy of the H-F + H endothermic reaction. The activation energy was estimated to be 120 kJ/mol. For the exothermic F + H2 reaction, the activation energy was calculated by trying to shift the trajectory towards to the reactants. This was done by thus increasing the F-H distance forming the F + H2 reactant. The difference in this energy was very small as shown in figure 8 and was used to calculate the activation energy for the F + H2 reaction.

=== 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. ===
Figure 8 illustrates the transition between translational and vibrational energy in the F + H2 reaction. Initially, the H-H bond (B-C) is vibrationally excited shown by the oscillation, whilst the H-F(A-B) bond isn't indicated by the absence of an oscillation. The H-F bond is at the vibrational ground state . After the collision we see a transition. The H-F bond now is vibrationally excited illustrated by the oscillation in the momenta vs time plot, and the H-H bonds loses this vibrationally energy and now only has translational energy due to the breaking of the bond, depicted by the plateau after 30s. Experimentally IR could be used to see the progress of the reaction. Only molecules with a dipole are IR active, thus after the formation of a H-F bond, IR absorption absent in the reactant will start to appear with overtones suggesting higher vibrational energies reached.

=== 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. ===
Below various example of both reactions are used to illustrate Polanyl's rules. Reactions with late transition states are more efficient when the initial trajectory possesses vibrational energy whereas reactions with early transition states are more efficient when the initial trajectory possesses translational energy. The following examples illustrate these rules.

==== Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, with a momentum pFH = -1.0 g.mol-1.pm.fs-1, and explore several values of pHH in the range -6.1 to 6.1 g.mol-1.pm.fs-1 (explore values also close to these limits). What do you observe? ====
When using the parameters above and setting the F-H value to 200 pm, at lower momentum values larger vibrational frequencies were obversed. The reactions did not therefore take place, satisfying Polyanyi's rules, where the having translational energy is more efficient when there is an early barrier rather than vibrational energy. Even at higher momentum extremes shown in figure 9, the reaction does not go to completion, indicating even though a lot of energy is being put into the system , it is not an efficient reaction, satisfying the rules.

==== For the same initial position, increase slightly the momentum pFH = -1.6 g.mol-1.pm.fs-1, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.2 g.mol-1.pm.fs-1. What do you observe now? ====
Figure 10 shows that, even at lower momentum values, resulting in lower vibrational energies the reaction still does not proceed to completion. As there is not enough translational energy in the reaction, further satisfying Polyanyi's rules.

==== Focus on the reverse reaction, H + HF. ====
This reaction has a late transition state, thus according to Polanyi's rules, vibrational energy trajectory is more efficient. This was tested firstly by using a low H-F vibrational energy and a high H translational energy as shown in figure 11. As predicted the reaction did not proceed to completion as there was not enough vibrational energy in the reaction. Figure 12 shows an example where the reaction has a larger vibrational energy, and therefore is more efficient, resulting in a successful reaction.

MRD:01524831.molecularreactiondynamics

2020-05-15T20:51:30Z

Hk3918: /* 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? */

== Exercise 1: H + H2 system ==

[[File:Surface_Plot_first_trajcectory_hk3918.png|thumb|Figure 1. A surface plot showing the transition state of a H + H2 system]]

=== 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? ===
On a potential energy surface diagram , the transition state is mathematically defined as the saddle point. The saddle point is a stationary point on the plot, where along the reaction path it is a maxima and perpendicular to the reaction path the saddle point is perceived to be a minima. The transition point can be identified in this simple example when the distances between the 3 hydrogen atoms are of equal distance. As shown in figure 1, the reaction path is oscillating at the saddle point. This depicts the vibrational freedom of the system at the transition state. A saddle point is mathematically found as shown below.

D = fxx.fyy - fxy2

<nowiki> </nowiki>If D > 0, fxx < 0 , local minima.

If D< 0 , then it is a saddle point.

=== 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 estimated '''rts '''is found to be around 91 pm. The transition state is defined as the position where internuclear distances are equal, and at a energy maxima in relation to the reaction path. From figure 1 it is evident that there are vibrations within the system, depicted by the oscillations on the ridge. The transition state position was found by determining which position the system stops oscillating, where the potential energy was a maximum, and thus the stable point of the transition state, the saddle point was found as shown in figure 2 and 3.

=== Comment on how the ''mep'' and the trajectory you just calculated differ. ===
When calculating under mep the reaction path is depicted different to when the calculation type is set to dynamics. What is absent in the map calculation is the motion of the atoms. The vibrational motion of the three hydrogen atoms are not depicted as shown in figure .... The initial condition is set so that the distance between AB is '''rts''' and the distance between BC deviates from '''rts '''by 1 pm, resulting in the reaction pathway following through, and advance towards the HA + HB-HC product. The flat BC line indicates the fact that the mep calculation does not consider vibrational motion of atoms.

==== Look at the “Internuclear Distances vs Time” and “Momenta vs Time”. What would change if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead? ====
The change of initial conditions to those above will change final product from HA + HB-HC to HA-HB + HC as it is the HB-HC bond distance that deviates away from the '''rts. '''As a result the internuclear distances vs time plot will now show AB to be constant over time. Momentum remains unchanged as initial momentum is 0 for both scenarios

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

The 'Get last geometry' option was selected, and the momenta were then reversed under the dynamic calculation type. The plot remains unchanged as expected. The reason for this is because the initial argument resulted in a trajectory towards a larger BC distance. Using the momenta of this trajectory but reversing the sign simply results in the hydrogen and hydrogen molecule to start in the final position, but now approach each other due to the negative momentum value.
=== 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"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot/kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Hydrogen atoms AB and C approach
eachother. Notably, the vibrational

motion present after the transition state is

not present before.
|[[File:Surface_Plot_question4_1_hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-4.1 </nowiki>
|<nowiki>-420.077</nowiki>
|No
|

Hydrogen atoms AB approaches C, however

before approaching the saddle point to AB

swiftly moves away from C

|[[File:Animation exercise2 p2 -4.1 hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|Hydrogen atom C approaches AB with a large
amount of momentum, resulting in a new BC Hydrogen

molecule forming, releasing the A hydrogen atom in the

reaction.
|[[File:Animation exercise2 p2 -5.1 p1 -3.1 hk3918.png|300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.1 </nowiki>
|<nowiki>-357.277</nowiki>
|No
|Hydrogen atoms AB and C approach each other very
quickly. Upon impact, large internuclear distance and velocities

are depicted, showing the B atom being tugged by both A and C.

In the end, B forms a molecule AB and moves away from C.
|[[File:Animation exercise2 p2 -10.1 hk3918.png |300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.6 </nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|Upon impact, the B atom is tugged between the two atoms A and
C, and is finally bonded to to atom C. A then moves away from

BC.
|[[File:Animation p2 -10.6 hk3918.png |300px]]
|}
The table above shows that a reaction isn't only reliant on the amount of kinetic energy in the system. As shown in the table above, there are instances where kinetic energy is higher than the activation energy, however no reaction takes place. One therefore needs to consider that successful collision are reliant on the system having enough energy to break and reform bonds, and also the orientation at which atoms collide.

=== Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===
Transition State Theory has a fundamental assumption which will result in its predictions being different to experimental values. The assumption that all trajectories possessing enough kinetic energy to circumvent the activation barrier will be reactive will result in a larger ''Rcl '' and thus a larger rate prediction than the experimental outcome. As seen in the table above, not all collisions are reactive even if there is sufficient kinetic energy - the case of barrier recrossing perfectly illustrates this.

== Exercise 2 ==

=== 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? ===
[[File:Surface_Plot_HF_hk3918.png|thumb|Figure 4. A surface plot showing the trajectory of F + H2 to H + HF reaction]|300px]]The reaction is exothermic as shown in figure 4, there is a decrease in potential energy when passing the transition state and forming the new HF bond. This relates to bond-strength, H-H has a bond strength of 432 kJ/mol whilst H-F has a bond strength of 565 kJ/mol.(reference). As HF is a polar bond as a result of the very electronegative F, there is an ionic element to the bond, resulting in a much greater attraction between the two atoms, thus releasing a lot of energy on formation. As the energy released is greater than that absorbed, the reaction is exothermic as depicted in the trajectory shown in figure 4.

[[File:Surface_Plot_hf-h_hk3918.png|thumb|Figure 5. A surface plot showing the trajectory of the FH + H to F + H2 reaction|300px]]

Figure 5 on the other hand shows an endothermic reaction, illustrated by the increase in potential energy along the trajectory, indicating that the energy absorbed to break the stronger H-F bond is greater than the energy released when forming the weaker H-H bond.

[[File:Plot small activation.png|thumb|Figure 6. Activation energy for the F +H2 reaction |300px]]

[[File:Plot activation big.png|thumb|Figure 7. Activation energy for the HF + H reaction |300px]]

[[File:Animation_momenta_hk3918.png |thumb|Figure 8. Momenta vs time plot for F + H2 reaction |300px]]

[[File:Animation_-74_-6.1.png ‎ |thumb|Figure 9. Contour plot for F + H2 reaction. Parameters: AB=200, BC=74, Pab=-1, Pbc=-6.9 |300px]]

[[File: Animation_low_momentum_hk3918.png ‎ |thumb|Figure 10. Contour plot for F + H2 reaction. Parameters: AB=200, BC=74, Pab=-1.6, Pbc=0.2 |300px]]

[[File:Animation low vibrational.png ‎ |thumb|Figure 11. Contour plot for the HF + H reaction where H has high momenta and HF has low initial momenta |300px]]

[[File:Animation highvirabtional.png‎ |thumb|Figure 12. Skew plot for the HF + H reaction where H has low momenta and HF has high initial momenta |300px]]

=== Locate the approximate position of the transition state ===
The transition state position was calculated to be when the H-F bond had a bond length of 181.8 pm, and the H-H bond was 73.1 pm as shown in figure 5. Analoguos to the calculation when approximating the transition state position of the H + H2 system, in the F + H2 reaction simulations were tested in order to acquire a constant bond length, at the maximum potential energy. As the first part of this reaction was exothermic, adopting Hammonds postulate, the assumption was made that the reaction had an early transition state thus the region near the reactants was analysed. The MEP calculation was used to remove the vibrational noise as it made it difficult to read the plots. As HF + H is a late transition state, resembling the its product F + H2, it will have the same transition state positions.

=== Report the activation energy for both reactions. ===
In order to find the activation energy the bond distance of H-F was first decreased. This then resulted in the energy falling to form the products H-F and H. The difference in this energy as shown in figure 7 was used to calculate the activation energy of the H-F + H endothermic reaction. The activation energy was estimated to be 120 kJ/mol. For the exothermic F + H2 reaction, the activation energy was calculated by trying to shift the trajectory towards to the reactants. This was done by thus increasing the F-H distance forming the F + H2 reactant. The difference in this energy was very small as shown in figure 8 and was used to calculate the activation energy for the F + H2 reaction.

=== 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. ===
Figure 8 illustrates the transition between translational and vibrational energy in the F + H2 reaction. Initially, the H-H bond (B-C) is vibrationally excited shown by the oscillation, whilst the H-F(A-B) bond isn't indicated by the absence of an oscillation. The H-F bond is at the vibrational ground state . After the collision we see a transition. The H-F bond now is vibrationally excited illustrated by the oscillation in the momenta vs time plot, and the H-H bonds loses this vibrationally energy and now only has translational energy due to the breaking of the bond, depicted by the plateau after 30s. Experimentally IR could be used to see the progress of the reaction. Only molecules with a dipole are IR active, thus after the formation of a H-F bond, IR absorption absent in the reactant will start to appear with overtones suggesting higher vibrational energies reached.

=== 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. ===
Below various example of both reactions are used to illustrate Polanyl's rules. Reactions with late transition states are more efficient when the initial trajectory possesses vibrational energy whereas reactions with early transition states are more efficient when the initial trajectory possesses translational energy. The following examples illustrate these rules.

==== Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, with a momentum pFH = -1.0 g.mol-1.pm.fs-1, and explore several values of pHH in the range -6.1 to 6.1 g.mol-1.pm.fs-1 (explore values also close to these limits). What do you observe? ====
When using the parameters above and setting the F-H value to 200 pm, at lower momentum values larger vibrational frequencies were obversed. The reactions did not therefore take place, satisfying Polyanyi's rules, where the having translational energy is more efficient when there is an early barrier rather than vibrational energy. Even at higher momentum extremes shown in figure 9, the reaction does not go to completion, indicating even though a lot of energy is being put into the system , it is not an efficient reaction, satisfying the rules.

==== For the same initial position, increase slightly the momentum pFH = -1.6 g.mol-1.pm.fs-1, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.2 g.mol-1.pm.fs-1. What do you observe now? ====
Figure 10 shows that, even at lower momentum values, resulting in lower vibrational energies the reaction still does not proceed to completion. As there is not enough translational energy in the reaction, further satisfying Polyanyi's rules.

==== Focus on the reverse reaction, H + HF. ====
This reaction has a late transition state, thus according to Polanyi's rules, vibrational energy trajectory is more efficient. This was tested firstly by using a low H-F vibrational energy and a high H translational energy as shown in figure 11. As predicted the reaction did not proceed to completion as there was not enough vibrational energy in the reaction. Figure 12 shows an example where the reaction has a larger vibrational energy, and therefore is more efficient, resulting in a successful reaction.

MRD:01524831.molecularreactiondynamics

2020-05-15T20:46:43Z

Hk3918: /* Exercise 1: H + H2 system */

== Exercise 1: H + H2 system ==

[[File:Surface_Plot_first_trajcectory_hk3918.png|thumb|Figure 1. A surface plot showing the transition state of a H + H2 system]]

=== 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? ===
On a potential energy surface diagram , the transition state is mathematically defined as the saddle point. The saddle point is a stationary point on the plot, where along the reaction path it is a maxima and perpendicular to the reaction path the saddle point is perceived to be a minima. The transition point can be identified in this simple example when the distances between the 3 hydrogen atoms are of equal distance. As shown in figure 1, the reaction path is oscillating at the saddle point. This depicts the vibrational freedom of the system at the transition state. A saddle point is found by...

=== 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 estimated '''rts '''is found to be around 91 pm. The transition state is defined as the position where internuclear distances are equal, and at a energy maxima in relation to the reaction path. From figure 1 it is evident that there are vibrations within the system, depicted by the oscillations on the ridge. The transition state position was found by determining which position the system stops oscillating, where the potential energy was a maximum, and thus the stable point of the transition state, the saddle point was found as shown in figure 2 and 3.

=== Comment on how the ''mep'' and the trajectory you just calculated differ. ===
When calculating under mep the reaction path is depicted different to when the calculation type is set to dynamics. What is absent in the map calculation is the motion of the atoms. The vibrational motion of the three hydrogen atoms are not depicted as shown in figure .... The initial condition is set so that the distance between AB is '''rts''' and the distance between BC deviates from '''rts '''by 1 pm, resulting in the reaction pathway following through, and advance towards the HA + HB-HC product. The flat BC line indicates the fact that the mep calculation does not consider vibrational motion of atoms.

==== Look at the “Internuclear Distances vs Time” and “Momenta vs Time”. What would change if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead? ====
The change of initial conditions to those above will change final product from HA + HB-HC to HA-HB + HC as it is the HB-HC bond distance that deviates away from the '''rts. '''As a result the internuclear distances vs time plot will now show AB to be constant over time. Momentum remains unchanged as initial momentum is 0 for both scenarios

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

The 'Get last geometry' option was selected, and the momenta were then reversed under the dynamic calculation type. The plot remains unchanged as expected. The reason for this is because the initial argument resulted in a trajectory towards a larger BC distance. Using the momenta of this trajectory but reversing the sign simply results in the hydrogen and hydrogen molecule to start in the final position, but now approach each other due to the negative momentum value.
=== 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"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot/kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Hydrogen atoms AB and C approach
eachother. Notably, the vibrational

motion present after the transition state is

not present before.
|[[File:Surface_Plot_question4_1_hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-4.1 </nowiki>
|<nowiki>-420.077</nowiki>
|No
|

Hydrogen atoms AB approaches C, however

before approaching the saddle point to AB

swiftly moves away from C

|[[File:Animation exercise2 p2 -4.1 hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|Hydrogen atom C approaches AB with a large
amount of momentum, resulting in a new BC Hydrogen

molecule forming, releasing the A hydrogen atom in the

reaction.
|[[File:Animation exercise2 p2 -5.1 p1 -3.1 hk3918.png|300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.1 </nowiki>
|<nowiki>-357.277</nowiki>
|No
|Hydrogen atoms AB and C approach each other very
quickly. Upon impact, large internuclear distance and velocities

are depicted, showing the B atom being tugged by both A and C.

In the end, B forms a molecule AB and moves away from C.
|[[File:Animation exercise2 p2 -10.1 hk3918.png |300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.6 </nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|Upon impact, the B atom is tugged between the two atoms A and
C, and is finally bonded to to atom C. A then moves away from

BC.
|[[File:Animation p2 -10.6 hk3918.png |300px]]
|}
The table above shows that a reaction isn't only reliant on the amount of kinetic energy in the system. As shown in the table above, there are instances where kinetic energy is higher than the activation energy, however no reaction takes place. One therefore needs to consider that successful collision are reliant on the system having enough energy to break and reform bonds, and also the orientation at which atoms collide.

=== Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===
Transition State Theory has a fundamental assumption which will result in its predictions being different to experimental values. The assumption that all trajectories possessing enough kinetic energy to circumvent the activation barrier will be reactive will result in a larger ''Rcl '' and thus a larger rate prediction than the experimental outcome. As seen in the table above, not all collisions are reactive even if there is sufficient kinetic energy - the case of barrier recrossing perfectly illustrates this.

== Exercise 2 ==

=== 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? ===
[[File:Surface_Plot_HF_hk3918.png|thumb|Figure 4. A surface plot showing the trajectory of F + H2 to H + HF reaction]|300px]]The reaction is exothermic as shown in figure 4, there is a decrease in potential energy when passing the transition state and forming the new HF bond. This relates to bond-strength, H-H has a bond strength of 432 kJ/mol whilst H-F has a bond strength of 565 kJ/mol.(reference). As HF is a polar bond as a result of the very electronegative F, there is an ionic element to the bond, resulting in a much greater attraction between the two atoms, thus releasing a lot of energy on formation. As the energy released is greater than that absorbed, the reaction is exothermic as depicted in the trajectory shown in figure 4.

[[File:Surface_Plot_hf-h_hk3918.png|thumb|Figure 5. A surface plot showing the trajectory of the FH + H to F + H2 reaction|300px]]

Figure 5 on the other hand shows an endothermic reaction, illustrated by the increase in potential energy along the trajectory, indicating that the energy absorbed to break the stronger H-F bond is greater than the energy released when forming the weaker H-H bond.

[[File:Plot small activation.png|thumb|Figure 6. Activation energy for the F +H2 reaction |300px]]

[[File:Plot activation big.png|thumb|Figure 7. Activation energy for the HF + H reaction |300px]]

[[File:Animation_momenta_hk3918.png |thumb|Figure 8. Momenta vs time plot for F + H2 reaction |300px]]

[[File:Animation_-74_-6.1.png ‎ |thumb|Figure 9. Contour plot for F + H2 reaction. Parameters: AB=200, BC=74, Pab=-1, Pbc=-6.9 |300px]]

[[File: Animation_low_momentum_hk3918.png ‎ |thumb|Figure 10. Contour plot for F + H2 reaction. Parameters: AB=200, BC=74, Pab=-1.6, Pbc=0.2 |300px]]

[[File:Animation low vibrational.png ‎ |thumb|Figure 11. Contour plot for the HF + H reaction where H has high momenta and HF has low initial momenta |300px]]

[[File:Animation highvirabtional.png‎ |thumb|Figure 12. Skew plot for the HF + H reaction where H has low momenta and HF has high initial momenta |300px]]

=== Locate the approximate position of the transition state ===
The transition state position was calculated to be when the H-F bond had a bond length of 181.8 pm, and the H-H bond was 73.1 pm as shown in figure 5. Analoguos to the calculation when approximating the transition state position of the H + H2 system, in the F + H2 reaction simulations were tested in order to acquire a constant bond length, at the maximum potential energy. As the first part of this reaction was exothermic, adopting Hammonds postulate, the assumption was made that the reaction had an early transition state thus the region near the reactants was analysed. The MEP calculation was used to remove the vibrational noise as it made it difficult to read the plots. As HF + H is a late transition state, resembling the its product F + H2, it will have the same transition state positions.

=== Report the activation energy for both reactions. ===
In order to find the activation energy the bond distance of H-F was first decreased. This then resulted in the energy falling to form the products H-F and H. The difference in this energy as shown in figure 7 was used to calculate the activation energy of the H-F + H endothermic reaction. The activation energy was estimated to be 120 kJ/mol. For the exothermic F + H2 reaction, the activation energy was calculated by trying to shift the trajectory towards to the reactants. This was done by thus increasing the F-H distance forming the F + H2 reactant. The difference in this energy was very small as shown in figure 8 and was used to calculate the activation energy for the F + H2 reaction.

=== 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. ===
Figure 8 illustrates the transition between translational and vibrational energy in the F + H2 reaction. Initially, the H-H bond (B-C) is vibrationally excited shown by the oscillation, whilst the H-F(A-B) bond isn't indicated by the absence of an oscillation. The H-F bond is at the vibrational ground state . After the collision we see a transition. The H-F bond now is vibrationally excited illustrated by the oscillation in the momenta vs time plot, and the H-H bonds loses this vibrationally energy and now only has translational energy due to the breaking of the bond, depicted by the plateau after 30s. Experimentally IR could be used to see the progress of the reaction. Only molecules with a dipole are IR active, thus after the formation of a H-F bond, IR absorption absent in the reactant will start to appear with overtones suggesting higher vibrational energies reached.

=== 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. ===
Below various example of both reactions are used to illustrate Polanyl's rules. Reactions with late transition states are more efficient when the initial trajectory possesses vibrational energy whereas reactions with early transition states are more efficient when the initial trajectory possesses translational energy. The following examples illustrate these rules.

==== Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, with a momentum pFH = -1.0 g.mol-1.pm.fs-1, and explore several values of pHH in the range -6.1 to 6.1 g.mol-1.pm.fs-1 (explore values also close to these limits). What do you observe? ====
When using the parameters above and setting the F-H value to 200 pm, at lower momentum values larger vibrational frequencies were obversed. The reactions did not therefore take place, satisfying Polyanyi's rules, where the having translational energy is more efficient when there is an early barrier rather than vibrational energy. Even at higher momentum extremes shown in figure 9, the reaction does not go to completion, indicating even though a lot of energy is being put into the system , it is not an efficient reaction, satisfying the rules.

==== For the same initial position, increase slightly the momentum pFH = -1.6 g.mol-1.pm.fs-1, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.2 g.mol-1.pm.fs-1. What do you observe now? ====
Figure 10 shows that, even at lower momentum values, resulting in lower vibrational energies the reaction still does not proceed to completion. As there is not enough translational energy in the reaction, further satisfying Polyanyi's rules.

==== Focus on the reverse reaction, H + HF. ====
This reaction has a late transition state, thus according to Polanyi's rules, vibrational energy trajectory is more efficient. This was tested firstly by using a low H-F vibrational energy and a high H translational energy as shown in figure 11. As predicted the reaction did not proceed to completion as there was not enough vibrational energy in the reaction. Figure 12 shows an example where the reaction has a larger vibrational energy, and therefore is more efficient, resulting in a successful reaction.

File:Surface Plot first trajcectory hk3918.png

2020-05-15T20:46:24Z

Hk3918:

MRD:01524831.molecularreactiondynamics

2020-05-15T19:30:10Z

== Exercise 1: H + H2 system ==

[[File:Surface_Plot.png|thumb|Figure 1. A surface plot showing the transition state of a H + H2 system]]

=== 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? ===
On a potential energy surface diagram , the transition state is mathematically defined as the saddle point. The saddle point is a stationary point on the plot, where along the reaction path it is a maxima and perpendicular to the reaction path the saddle point is perceived to be a minima. The transition point can be identified in this simple example when the distances between the 3 hydrogen atoms are of equal distance. As shown in figure 1, the reaction path is oscillating at the saddle point. This depicts the vibrational freedom of the system at the transition state. A saddle point is found by...

=== 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 estimated '''rts '''is found to be around 91 pm. The transition state is defined as the position where internuclear distances are equal, and at a energy maxima in relation to the reaction path. From figure 1 it is evident that there are vibrations within the system, depicted by the oscillations on the ridge. The transition state position was found by determining which position the system stops oscillating, where the potential energy was a maximum, and thus the stable point of the transition state, the saddle point was found as shown in figure 2 and 3.

=== Comment on how the ''mep'' and the trajectory you just calculated differ. ===
When calculating under mep the reaction path is depicted different to when the calculation type is set to dynamics. What is absent in the map calculation is the motion of the atoms. The vibrational motion of the three hydrogen atoms are not depicted as shown in figure .... The initial condition is set so that the distance between AB is '''rts''' and the distance between BC deviates from '''rts '''by 1 pm, resulting in the reaction pathway following through, and advance towards the HA + HB-HC product. The flat BC line indicates the fact that the mep calculation does not consider vibrational motion of atoms.

==== Look at the “Internuclear Distances vs Time” and “Momenta vs Time”. What would change if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead? ====
The change of initial conditions to those above will change final product from HA + HB-HC to HA-HB + HC as it is the HB-HC bond distance that deviates away from the '''rts. '''As a result the internuclear distances vs time plot will now show AB to be constant over time. Momentum remains unchanged as initial momentum is 0 for both scenarios

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

The 'Get last geometry' option was selected, and the momenta were then reversed under the dynamic calculation type. The plot remains unchanged as expected. The reason for this is because the initial argument resulted in a trajectory towards a larger BC distance. Using the momenta of this trajectory but reversing the sign simply results in the hydrogen and hydrogen molecule to start in the final position, but now approach each other due to the negative momentum value.
=== 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"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot/kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Hydrogen atoms AB and C approach
eachother. Notably, the vibrational

motion present after the transition state is

not present before.
|[[File:Surface_Plot_question4_1_hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-4.1 </nowiki>
|<nowiki>-420.077</nowiki>
|No
|

Hydrogen atoms AB approaches C, however

before approaching the saddle point to AB

swiftly moves away from C

|[[File:Animation exercise2 p2 -4.1 hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|Hydrogen atom C approaches AB with a large
amount of momentum, resulting in a new BC Hydrogen

molecule forming, releasing the A hydrogen atom in the

reaction.
|[[File:Animation exercise2 p2 -5.1 p1 -3.1 hk3918.png|300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.1 </nowiki>
|<nowiki>-357.277</nowiki>
|No
|Hydrogen atoms AB and C approach each other very
quickly. Upon impact, large internuclear distance and velocities

are depicted, showing the B atom being tugged by both A and C.

In the end, B forms a molecule AB and moves away from C.
|[[File:Animation exercise2 p2 -10.1 hk3918.png |300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.6 </nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|Upon impact, the B atom is tugged between the two atoms A and
C, and is finally bonded to to atom C. A then moves away from

BC.
|[[File:Animation p2 -10.6 hk3918.png |300px]]
|}
The table above shows that a reaction isn't only reliant on the amount of kinetic energy in the system. As shown in the table above, there are instances where kinetic energy is higher than the activation energy, however no reaction takes place. One therefore needs to consider that successful collision are reliant on the system having enough energy to break and reform bonds, and also the orientation at which atoms collide.

=== Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===
Transition State Theory has a fundamental assumption which will result in its predictions being different to experimental values. The assumption that all trajectories possessing enough kinetic energy to circumvent the activation barrier will be reactive will result in a larger ''Rcl '' and thus a larger rate prediction than the experimental outcome. As seen in the table above, not all collisions are reactive even if there is sufficient kinetic energy - the case of barrier recrossing perfectly illustrates this.

== Exercise 2 ==

=== 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? ===
[[File:Surface_Plot_HF_hk3918.png|thumb|Figure 4. A surface plot showing the trajectory of F + H2 to H + HF reaction]|300px]]The reaction is exothermic as shown in figure 4, there is a decrease in potential energy when passing the transition state and forming the new HF bond. This relates to bond-strength, H-H has a bond strength of 432 kJ/mol whilst H-F has a bond strength of 565 kJ/mol.(reference). As HF is a polar bond as a result of the very electronegative F, there is an ionic element to the bond, resulting in a much greater attraction between the two atoms, thus releasing a lot of energy on formation. As the energy released is greater than that absorbed, the reaction is exothermic as depicted in the trajectory shown in figure 4.

[[File:Surface_Plot_hf-h_hk3918.png|thumb|Figure 5. A surface plot showing the trajectory of the FH + H to F + H2 reaction|300px]]

Figure 5 on the other hand shows an endothermic reaction, illustrated by the increase in potential energy along the trajectory, indicating that the energy absorbed to break the stronger H-F bond is greater than the energy released when forming the weaker H-H bond.

[[File:Plot small activation.png|thumb|Figure 6. Activation energy for the F +H2 reaction |300px]]

[[File:Plot activation big.png|thumb|Figure 7. Activation energy for the HF + H reaction |300px]]

[[File:Animation_momenta_hk3918.png |thumb|Figure 8. Momenta vs time plot for F + H2 reaction |300px]]

[[File:Animation_-74_-6.1.png ‎ |thumb|Figure 9. Contour plot for F + H2 reaction. Parameters: AB=200, BC=74, Pab=-1, Pbc=-6.9 |300px]]

[[File: Animation_low_momentum_hk3918.png ‎ |thumb|Figure 10. Contour plot for F + H2 reaction. Parameters: AB=200, BC=74, Pab=-1.6, Pbc=0.2 |300px]]

[[File:Animation low vibrational.png ‎ |thumb|Figure 11. Contour plot for the HF + H reaction where H has high momenta and HF has low initial momenta |300px]]

[[File:Animation highvirabtional.png‎ |thumb|Figure 12. Skew plot for the HF + H reaction where H has low momenta and HF has high initial momenta |300px]]

=== Locate the approximate position of the transition state ===
The transition state position was calculated to be when the H-F bond had a bond length of 181.8 pm, and the H-H bond was 73.1 pm as shown in figure 5. Analoguos to the calculation when approximating the transition state position of the H + H2 system, in the F + H2 reaction simulations were tested in order to acquire a constant bond length, at the maximum potential energy. As the first part of this reaction was exothermic, adopting Hammonds postulate, the assumption was made that the reaction had an early transition state thus the region near the reactants was analysed. The MEP calculation was used to remove the vibrational noise as it made it difficult to read the plots. As HF + H is a late transition state, resembling the its product F + H2, it will have the same transition state positions.

=== Report the activation energy for both reactions. ===
In order to find the activation energy the bond distance of H-F was first decreased. This then resulted in the energy falling to form the products H-F and H. The difference in this energy as shown in figure 7 was used to calculate the activation energy of the H-F + H endothermic reaction. The activation energy was estimated to be 120 kJ/mol. For the exothermic F + H2 reaction, the activation energy was calculated by trying to shift the trajectory towards to the reactants. This was done by thus increasing the F-H distance forming the F + H2 reactant. The difference in this energy was very small as shown in figure 8 and was used to calculate the activation energy for the F + H2 reaction.

=== 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. ===
Figure 8 illustrates the transition between translational and vibrational energy in the F + H2 reaction. Initially, the H-H bond (B-C) is vibrationally excited shown by the oscillation, whilst the H-F(A-B) bond isn't indicated by the absence of an oscillation. The H-F bond is at the vibrational ground state . After the collision we see a transition. The H-F bond now is vibrationally excited illustrated by the oscillation in the momenta vs time plot, and the H-H bonds loses this vibrationally energy and now only has translational energy due to the breaking of the bond, depicted by the plateau after 30s. Experimentally IR could be used to see the progress of the reaction. Only molecules with a dipole are IR active, thus after the formation of a H-F bond, IR absorption absent in the reactant will start to appear with overtones suggesting higher vibrational energies reached.

=== 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. ===
Below various example of both reactions are used to illustrate Polanyl's rules. Reactions with late transition states are more efficient when the initial trajectory possesses vibrational energy whereas reactions with early transition states are more efficient when the initial trajectory possesses translational energy. The following examples illustrate these rules.

==== Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, with a momentum pFH = -1.0 g.mol-1.pm.fs-1, and explore several values of pHH in the range -6.1 to 6.1 g.mol-1.pm.fs-1 (explore values also close to these limits). What do you observe? ====
When using the parameters above and setting the F-H value to 200 pm, at lower momentum values larger vibrational frequencies were obversed. The reactions did not therefore take place, satisfying Polyanyi's rules, where the having translational energy is more efficient when there is an early barrier rather than vibrational energy. Even at higher momentum extremes shown in figure 9, the reaction does not go to completion, indicating even though a lot of energy is being put into the system , it is not an efficient reaction, satisfying the rules.

==== For the same initial position, increase slightly the momentum pFH = -1.6 g.mol-1.pm.fs-1, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.2 g.mol-1.pm.fs-1. What do you observe now? ====
Figure 10 shows that, even at lower momentum values, resulting in lower vibrational energies the reaction still does not proceed to completion. As there is not enough translational energy in the reaction, further satisfying Polyanyi's rules.

==== Focus on the reverse reaction, H + HF. ====
This reaction has a late transition state, thus according to Polanyi's rules, vibrational energy trajectory is more efficient. This was tested firstly by using a low H-F vibrational energy and a high H translational energy as shown in figure 11. As predicted the reaction did not proceed to completion as there was not enough vibrational energy in the reaction. Figure 12 shows an example where the reaction has a larger vibrational energy, and therefore is more efficient, resulting in a successful reaction.

MRD:01524831.molecularreactiondynamics

2020-05-15T19:27:14Z

Hk3918: /* 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. */

== Exercise 1: H + H2 system ==

[[File:Surface_Plot.png|thumb|Figure 1. A surface plot showing the transition state of a H + H2 system]]

=== 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? ===
On a potential energy surface diagram , the transition state is mathematically defined as the saddle point. The saddle point is a stationary point on the plot, where along the reaction path it is a maxima and perpendicular to the reaction path the saddle point is perceived to be a minima. The transition point can be identified in this simple example when the distances between the 3 hydrogen atoms are of equal distance. As shown in figure 1, the reaction path is oscillating at the saddle point. This depicts the vibrational freedom of the system at the transition state. A saddle point is found by...

=== 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 estimated '''rts '''is found to be around 91 pm. The transition state is defined as the position where internuclear distances are equal, and at a energy maxima in relation to the reaction path. From figure 1 it is evident that there are vibrations within the system, depicted by the oscillations on the ridge. The transition state position was found by determining which position the system stops oscillating, where the potential energy was a maximum, and thus the stable point of the transition state, the saddle point was found as shown in figure 2 and 3.

=== Comment on how the ''mep'' and the trajectory you just calculated differ. ===
When calculating under mep the reaction path is depicted different to when the calculation type is set to dynamics. What is absent in the map calculation is the motion of the atoms. The vibrational motion of the three hydrogen atoms are not depicted as shown in figure .... The initial condition is set so that the distance between AB is '''rts''' and the distance between BC deviates from '''rts '''by 1 pm, resulting in the reaction pathway following through, and advance towards the HA + HB-HC product. The flat BC line indicates the fact that the mep calculation does not consider vibrational motion of atoms.

==== Look at the “Internuclear Distances vs Time” and “Momenta vs Time”. What would change if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead? ====
The change of initial conditions to those above will change final product from HA + HB-HC to HA-HB + HC as it is the HB-HC bond distance that deviates away from the '''rts. '''As a result the internuclear distances vs time plot will now show AB to be constant over time. Momentum remains unchanged as initial momentum is 0 for both scenarios

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

The 'Get last geometry' option was selected, and the momenta were then reversed under the dynamic calculation type. The plot remains unchanged as expected. The reason for this is because the initial argument resulted in a trajectory towards a larger BC distance. Using the momenta of this trajectory but reversing the sign simply results in the hydrogen and hydrogen molecule to start in the final position, but now approach each other due to the negative momentum value.
=== 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"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot/kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Hydrogen atoms AB and C approach
eachother. Notably, the vibrational

motion present after the transition state is

not present before.
|[[File:Surface_Plot_question4_1_hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-4.1 </nowiki>
|<nowiki>-420.077</nowiki>
|No
|

Hydrogen atoms AB approaches C, however

before approaching the saddle point to AB

swiftly moves away from C

|[[File:Animation exercise2 p2 -4.1 hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|Hydrogen atom C approaches AB with a large
amount of momentum, resulting in a new BC Hydrogen

molecule forming, releasing the A hydrogen atom in the

reaction.
|[[File:Animation exercise2 p2 -5.1 p1 -3.1 hk3918.png|300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.1 </nowiki>
|<nowiki>-357.277</nowiki>
|No
|Hydrogen atoms AB and C approach each other very
quickly. Upon impact, large internuclear distance and velocities

are depicted, showing the B atom being tugged by both A and C.

In the end, B forms a molecule AB and moves away from C.
|[[File:Animation exercise2 p2 -10.1 hk3918.png |300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.6 </nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|Upon impact, the B atom is tugged between the two atoms A and
C, and is finally bonded to to atom C. A then moves away from

BC.
|[[File:Animation p2 -10.6 hk3918.png |300px]]
|}
The table above shows that a reaction isn't only reliant on the amount of kinetic energy in the system. As shown in the table above, there are instances where kinetic energy is higher than the activation energy, however no reaction takes place. One therefore needs to consider that successful collision are reliant on the system having enough energy to break and reform bonds, and also the orientation at which atoms collide.

=== Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===
Transition State Theory has a fundamental assumption which will result in its predictions being different to experimental values. The assumption that all trajectories possessing enough kinetic energy to circumvent the activation barrier will be reactive will result in a larger ''Rcl '' and thus a larger rate prediction than the experimental outcome. As seen in the table above, not all collisions are reactive even if there is sufficient kinetic energy - the case of barrier recrossing perfectly illustrates this.

== Exercise 2 ==

=== 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? ===
[[File:Surface_Plot_HF_hk3918.png|thumb|Figure 4. A surface plot showing the trajectory of F + H2 to H + HF reaction]|300px]]The reaction is exothermic as shown in figure 4, there is a decrease in potential energy when passing the transition state and forming the new HF bond. This relates to bond-strength, H-H has a bond strength of 432 kJ/mol whilst H-F has a bond strength of 565 kJ/mol.(reference). As HF is a polar bond as a result of the very electronegative F, there is an ionic element to the bond, resulting in a much greater attraction between the two atoms, thus releasing a lot of energy on formation. As the energy released is greater than that absorbed, the reaction is exothermic as depicted in the trajectory shown in figure 4.

[[File:Surface_Plot_hf-h_hk3918.png|thumb|Figure 5. A surface plot showing the trajectory of the FH + H to F + H2 reaction|300px]]

Figure 5 on the other hand shows an endothermic reaction, illustrated by the increase in potential energy along the trajectory, indicating that the energy absorbed to break the stronger H-F bond is greater than the energy released when forming the weaker H-H bond.

[[File:Plot small activation.png|thumb|Figure 6. Activation energy for the F +H2 reaction |300px]]

[[File:Plot activation big.png|thumb|Figure 7. Activation energy for the HF + H reaction |300px]]

[[File:Animation_momenta_hk3918.png |thumb|Figure 8. Momenta vs time plot for F + H2 reaction |300px]]

[[File:Animation_-74_-6.1.png ‎ |thumb|Figure 9. Contour plot for F + H2 reaction. Parameters: AB=200, BC=74, Pab=-1, Pbc=-6.9 |300px]]

[[File: Animation_low_momentum_hk3918.png ‎ |thumb|Figure 10. Contour plot for F + H2 reaction. Parameters: AB=200, BC=74, Pab=-1.6, Pbc=0.2 |300px]]

=== Locate the approximate position of the transition state ===
The transition state position was calculated to be when the H-F bond had a bond length of 181.8 pm, and the H-H bond was 73.1 pm as shown in figure 5. Analoguos to the calculation when approximating the transition state position of the H + H2 system, in the F + H2 reaction simulations were tested in order to acquire a constant bond length, at the maximum potential energy. As the first part of this reaction was exothermic, adopting Hammonds postulate, the assumption was made that the reaction had an early transition state thus the region near the reactants was analysed. The MEP calculation was used to remove the vibrational noise as it made it difficult to read the plots. As HF + H is a late transition state, resembling the its product F + H2, it will have the same transition state positions.

=== Report the activation energy for both reactions. ===
In order to find the activation energy the bond distance of H-F was first decreased. This then resulted in the energy falling to form the products H-F and H. The difference in this energy as shown in figure 7 was used to calculate the activation energy of the H-F + H endothermic reaction. The activation energy was estimated to be 120 kJ/mol. For the exothermic F + H2 reaction, the activation energy was calculated by trying to shift the trajectory towards to the reactants. This was done by thus increasing the F-H distance forming the F + H2 reactant. The difference in this energy was very small as shown in figure 8 and was used to calculate the activation energy for the F + H2 reaction.

=== 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. ===
Figure 8 illustrates the transition between translational and vibrational energy in the F + H2 reaction. Initially, the H-H bond (B-C) is vibrationally excited shown by the oscillation, whilst the H-F(A-B) bond isn't indicated by the absence of an oscillation. The H-F bond is at the vibrational ground state . After the collision we see a transition. The H-F bond now is vibrationally excited illustrated by the oscillation in the momenta vs time plot, and the H-H bonds loses this vibrationally energy and now only has translational energy due to the breaking of the bond, depicted by the plateau after 30s. Experimentally IR could be used to see the progress of the reaction. Only molecules with a dipole are IR active, thus after the formation of a H-F bond, IR absorption absent in the reactant will start to appear with overtones suggesting higher vibrational energies reached.

=== 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. ===
Below various example of both reactions are used to illustrate Polanyl's rules. Reactions with late transition states are more efficient when the initial trajectory possesses vibrational energy whereas reactions with early transition states are more efficient when the initial trajectory possesses translational energy. The following examples illustrate these rules.

==== Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, with a momentum pFH = -1.0 g.mol-1.pm.fs-1, and explore several values of pHH in the range -6.1 to 6.1 g.mol-1.pm.fs-1 (explore values also close to these limits). What do you observe? ====
When using the parameters above and setting the F-H value to 200 pm, at lower momentum values larger vibrational frequencies were obversed. The reactions did not therefore take place, satisfying Polyanyi's rules, where the having translational energy is more efficient when there is an early barrier rather than vibrational energy. Even at higher momentum extremes shown in figure 9, the reaction does not go to completion, indicating even though a lot of energy is being put into the system , it is not an efficient reaction, satisfying the rules.

==== For the same initial position, increase slightly the momentum pFH = -1.6 g.mol-1.pm.fs-1, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.2 g.mol-1.pm.fs-1. What do you observe now? ====
Figure 10 shows that, even at lower momentum values, resulting in lower vibrational energies the reaction still does not proceed to completion. As there is not enough translational energy in the reaction, further satisfying Polyanyi's rules.

==== Focus on the reverse reaction, H + HF. ====
This reaction has a late transition state, thus according to Polanyi's rules, vibrational energy trajectory is more efficient. This was tested firstly by using a low H-F vibrational energy and a high H translational energy as shown in figure 11. As predicted the reaction did not proceed to completion as there was not enough vibrational energy in the reaction. Figure 12 shows an example where the reaction has a larger vibrational energy, and therefore is more efficient, resulting in a successful reaction.

MRD:01524831.molecularreactiondynamics

2020-05-15T19:21:04Z

Hk3918: /* 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. */

== Exercise 1: H + H2 system ==

[[File:Surface_Plot.png|thumb|Figure 1. A surface plot showing the transition state of a H + H2 system]]

=== 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? ===
On a potential energy surface diagram , the transition state is mathematically defined as the saddle point. The saddle point is a stationary point on the plot, where along the reaction path it is a maxima and perpendicular to the reaction path the saddle point is perceived to be a minima. The transition point can be identified in this simple example when the distances between the 3 hydrogen atoms are of equal distance. As shown in figure 1, the reaction path is oscillating at the saddle point. This depicts the vibrational freedom of the system at the transition state. A saddle point is found by...

=== 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 estimated '''rts '''is found to be around 91 pm. The transition state is defined as the position where internuclear distances are equal, and at a energy maxima in relation to the reaction path. From figure 1 it is evident that there are vibrations within the system, depicted by the oscillations on the ridge. The transition state position was found by determining which position the system stops oscillating, where the potential energy was a maximum, and thus the stable point of the transition state, the saddle point was found as shown in figure 2 and 3.

=== Comment on how the ''mep'' and the trajectory you just calculated differ. ===
When calculating under mep the reaction path is depicted different to when the calculation type is set to dynamics. What is absent in the map calculation is the motion of the atoms. The vibrational motion of the three hydrogen atoms are not depicted as shown in figure .... The initial condition is set so that the distance between AB is '''rts''' and the distance between BC deviates from '''rts '''by 1 pm, resulting in the reaction pathway following through, and advance towards the HA + HB-HC product. The flat BC line indicates the fact that the mep calculation does not consider vibrational motion of atoms.

==== Look at the “Internuclear Distances vs Time” and “Momenta vs Time”. What would change if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead? ====
The change of initial conditions to those above will change final product from HA + HB-HC to HA-HB + HC as it is the HB-HC bond distance that deviates away from the '''rts. '''As a result the internuclear distances vs time plot will now show AB to be constant over time. Momentum remains unchanged as initial momentum is 0 for both scenarios

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

The 'Get last geometry' option was selected, and the momenta were then reversed under the dynamic calculation type. The plot remains unchanged as expected. The reason for this is because the initial argument resulted in a trajectory towards a larger BC distance. Using the momenta of this trajectory but reversing the sign simply results in the hydrogen and hydrogen molecule to start in the final position, but now approach each other due to the negative momentum value.
=== 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"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot/kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Hydrogen atoms AB and C approach
eachother. Notably, the vibrational

motion present after the transition state is

not present before.
|[[File:Surface_Plot_question4_1_hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-4.1 </nowiki>
|<nowiki>-420.077</nowiki>
|No
|

Hydrogen atoms AB approaches C, however

before approaching the saddle point to AB

swiftly moves away from C

|[[File:Animation exercise2 p2 -4.1 hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|Hydrogen atom C approaches AB with a large
amount of momentum, resulting in a new BC Hydrogen

molecule forming, releasing the A hydrogen atom in the

reaction.
|[[File:Animation exercise2 p2 -5.1 p1 -3.1 hk3918.png|300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.1 </nowiki>
|<nowiki>-357.277</nowiki>
|No
|Hydrogen atoms AB and C approach each other very
quickly. Upon impact, large internuclear distance and velocities

are depicted, showing the B atom being tugged by both A and C.

In the end, B forms a molecule AB and moves away from C.
|[[File:Animation exercise2 p2 -10.1 hk3918.png |300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.6 </nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|Upon impact, the B atom is tugged between the two atoms A and
C, and is finally bonded to to atom C. A then moves away from

BC.
|[[File:Animation p2 -10.6 hk3918.png |300px]]
|}
The table above shows that a reaction isn't only reliant on the amount of kinetic energy in the system. As shown in the table above, there are instances where kinetic energy is higher than the activation energy, however no reaction takes place. One therefore needs to consider that successful collision are reliant on the system having enough energy to break and reform bonds, and also the orientation at which atoms collide.

=== Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===
Transition State Theory has a fundamental assumption which will result in its predictions being different to experimental values. The assumption that all trajectories possessing enough kinetic energy to circumvent the activation barrier will be reactive will result in a larger ''Rcl '' and thus a larger rate prediction than the experimental outcome. As seen in the table above, not all collisions are reactive even if there is sufficient kinetic energy - the case of barrier recrossing perfectly illustrates this.

== Exercise 2 ==

=== 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? ===
[[File:Surface_Plot_HF_hk3918.png|thumb|Figure 4. A surface plot showing the trajectory of F + H2 to H + HF reaction]|300px]]The reaction is exothermic as shown in figure 4, there is a decrease in potential energy when passing the transition state and forming the new HF bond. This relates to bond-strength, H-H has a bond strength of 432 kJ/mol whilst H-F has a bond strength of 565 kJ/mol.(reference). As HF is a polar bond as a result of the very electronegative F, there is an ionic element to the bond, resulting in a much greater attraction between the two atoms, thus releasing a lot of energy on formation. As the energy released is greater than that absorbed, the reaction is exothermic as depicted in the trajectory shown in figure 4.

[[File:Surface_Plot_hf-h_hk3918.png|thumb|Figure 5. A surface plot showing the trajectory of the FH + H to F + H2 reaction|300px]]

Figure 5 on the other hand shows an endothermic reaction, illustrated by the increase in potential energy along the trajectory, indicating that the energy absorbed to break the stronger H-F bond is greater than the energy released when forming the weaker H-H bond.

[[File:Plot small activation.png|thumb|Figure 6. Activation energy for the F +H2 reaction |300px]]

[[File:Plot activation big.png|thumb|Figure 7. Activation energy for the HF + H reaction |300px]]

[[File:Animation_momenta_hk3918.png |thumb|Figure 8. Momenta vs time plot for F + H2 reaction |300px]]

[[File:Animation_-74_-6.1.png ‎ |thumb|Figure 9. Contour plot for F + H2 reaction. Parameters: AB=200, BC=74, Pab=-1, Pbc=-6.9 |300px]]

[[File: Animation_low_momentum_hk3918.png ‎ |thumb|Figure 10. Contour plot for F + H2 reaction. Parameters: AB=200, BC=74, Pab=-1.6, Pbc=0.2 |300px]]

=== Locate the approximate position of the transition state ===
The transition state position was calculated to be when the H-F bond had a bond length of 181.8 pm, and the H-H bond was 73.1 pm as shown in figure 5. Analoguos to the calculation when approximating the transition state position of the H + H2 system, in the F + H2 reaction simulations were tested in order to acquire a constant bond length, at the maximum potential energy. As the first part of this reaction was exothermic, adopting Hammonds postulate, the assumption was made that the reaction had an early transition state thus the region near the reactants was analysed. The MEP calculation was used to remove the vibrational noise as it made it difficult to read the plots. As HF + H is a late transition state, resembling the its product F + H2, it will have the same transition state positions.

=== Report the activation energy for both reactions. ===
In order to find the activation energy the bond distance of H-F was first decreased. This then resulted in the energy falling to form the products H-F and H. The difference in this energy as shown in figure 7 was used to calculate the activation energy of the H-F + H endothermic reaction. The activation energy was estimated to be 120 kJ/mol. For the exothermic F + H2 reaction, the activation energy was calculated by trying to shift the trajectory towards to the reactants. This was done by thus increasing the F-H distance forming the F + H2 reactant. The difference in this energy was very small as shown in figure 8 and was used to calculate the activation energy for the F + H2 reaction.

=== 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. ===
Figure 8 illustrates the transition between translational and vibrational energy in the F + H2 reaction. Initially, the H-H bond (B-C) is vibrationally excited shown by the oscillation, whilst the H-F(A-B) bond isn't indicated by the absence of an oscillation. The H-F bond is at the vibrational ground state . After the collision we see a transition. The H-F bond now is vibrationally excited illustrated by the oscillation in the momenta vs time plot, and the H-H bonds loses this vibrationally energy and now only has translational energy due to the breaking of the bond, depicted by the plateau after 30s. Experimentally IR could be used to see the progress of the reaction. Only molecules with a dipole are IR active, thus after the formation of a H-F bond, IR absorption absent in the reactant will start to appear with overtones suggesting higher vibrational energies reached.

==== Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, with a momentum pFH = -1.0 g.mol-1.pm.fs-1, and explore several values of pHH in the range -6.1 to 6.1 g.mol-1.pm.fs-1 (explore values also close to these limits). What do you observe? Note that we are putting a significant amount of energy (much more than the activation energy) into the system on the H - H vibration. ====
When using the parameters above and setting the F-H value to 200 pm, at lower momentum values larger vibrational frequencies were obversed. The reactions did not therefore take place, satisfying Polyanyi's rules, where the having translational energy is more efficient when there is an early barrier rather than vibrational energy. Even at higher momentum extremes shown in figure 9, the reaction does not go to completion, indicating even though a lot of energy is being put into the system , it is not an efficient reaction, satisfying the rules.

==== For the same initial position, increase slightly the momentum pFH = -1.6 g.mol-1.pm.fs-1, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.2 g.mol-1.pm.fs-1. What do you observe now? ====
Figure 10 shows that, even at lower momentum values, resulting in lower vibrational energies the reaction still does not proceed to completion. As there is not enough translational energy in the reaction, further satisfying Polyanyi's rules.

==== Focus on the reverse reaction, H + HF. ====
This reaction has a late transition state, thus according to Polanyi's rules, vibrational energy trajectory is more efficient. This was tested firstly by using a low H-F vibrational energy and a high H translational energy as shown in figure 11. As predicted the reaction did not proceed to completion as there was not enough vibrational energy in the reaction. Figure 12 shows an example where the reaction has a larger vibrational energy, and therefore is more efficient, resulting in a successful reaction.

MRD:01524831.molecularreactiondynamics

2020-05-15T19:19:17Z

Hk3918: /* For the same initial position, increase slightly the momentum pFH = -1.6 g.mol-1.pm.fs-1, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.2 g.mol-1.pm.fs-1. What do you observe now? */

== Exercise 1: H + H2 system ==

[[File:Surface_Plot.png|thumb|Figure 1. A surface plot showing the transition state of a H + H2 system]]

=== 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? ===
On a potential energy surface diagram , the transition state is mathematically defined as the saddle point. The saddle point is a stationary point on the plot, where along the reaction path it is a maxima and perpendicular to the reaction path the saddle point is perceived to be a minima. The transition point can be identified in this simple example when the distances between the 3 hydrogen atoms are of equal distance. As shown in figure 1, the reaction path is oscillating at the saddle point. This depicts the vibrational freedom of the system at the transition state. A saddle point is found by...

=== 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 estimated '''rts '''is found to be around 91 pm. The transition state is defined as the position where internuclear distances are equal, and at a energy maxima in relation to the reaction path. From figure 1 it is evident that there are vibrations within the system, depicted by the oscillations on the ridge. The transition state position was found by determining which position the system stops oscillating, where the potential energy was a maximum, and thus the stable point of the transition state, the saddle point was found as shown in figure 2 and 3.

=== Comment on how the ''mep'' and the trajectory you just calculated differ. ===
When calculating under mep the reaction path is depicted different to when the calculation type is set to dynamics. What is absent in the map calculation is the motion of the atoms. The vibrational motion of the three hydrogen atoms are not depicted as shown in figure .... The initial condition is set so that the distance between AB is '''rts''' and the distance between BC deviates from '''rts '''by 1 pm, resulting in the reaction pathway following through, and advance towards the HA + HB-HC product. The flat BC line indicates the fact that the mep calculation does not consider vibrational motion of atoms.

==== Look at the “Internuclear Distances vs Time” and “Momenta vs Time”. What would change if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead? ====
The change of initial conditions to those above will change final product from HA + HB-HC to HA-HB + HC as it is the HB-HC bond distance that deviates away from the '''rts. '''As a result the internuclear distances vs time plot will now show AB to be constant over time. Momentum remains unchanged as initial momentum is 0 for both scenarios

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

The 'Get last geometry' option was selected, and the momenta were then reversed under the dynamic calculation type. The plot remains unchanged as expected. The reason for this is because the initial argument resulted in a trajectory towards a larger BC distance. Using the momenta of this trajectory but reversing the sign simply results in the hydrogen and hydrogen molecule to start in the final position, but now approach each other due to the negative momentum value.
=== 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"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot/kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Hydrogen atoms AB and C approach
eachother. Notably, the vibrational

motion present after the transition state is

not present before.
|[[File:Surface_Plot_question4_1_hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-4.1 </nowiki>
|<nowiki>-420.077</nowiki>
|No
|

Hydrogen atoms AB approaches C, however

before approaching the saddle point to AB

swiftly moves away from C

|[[File:Animation exercise2 p2 -4.1 hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|Hydrogen atom C approaches AB with a large
amount of momentum, resulting in a new BC Hydrogen

molecule forming, releasing the A hydrogen atom in the

reaction.
|[[File:Animation exercise2 p2 -5.1 p1 -3.1 hk3918.png|300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.1 </nowiki>
|<nowiki>-357.277</nowiki>
|No
|Hydrogen atoms AB and C approach each other very
quickly. Upon impact, large internuclear distance and velocities

are depicted, showing the B atom being tugged by both A and C.

In the end, B forms a molecule AB and moves away from C.
|[[File:Animation exercise2 p2 -10.1 hk3918.png |300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.6 </nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|Upon impact, the B atom is tugged between the two atoms A and
C, and is finally bonded to to atom C. A then moves away from

BC.
|[[File:Animation p2 -10.6 hk3918.png |300px]]
|}
The table above shows that a reaction isn't only reliant on the amount of kinetic energy in the system. As shown in the table above, there are instances where kinetic energy is higher than the activation energy, however no reaction takes place. One therefore needs to consider that successful collision are reliant on the system having enough energy to break and reform bonds, and also the orientation at which atoms collide.

=== Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===
Transition State Theory has a fundamental assumption which will result in its predictions being different to experimental values. The assumption that all trajectories possessing enough kinetic energy to circumvent the activation barrier will be reactive will result in a larger ''Rcl '' and thus a larger rate prediction than the experimental outcome. As seen in the table above, not all collisions are reactive even if there is sufficient kinetic energy - the case of barrier recrossing perfectly illustrates this.

== Exercise 2 ==

=== 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? ===
[[File:Surface_Plot_HF_hk3918.png|thumb|Figure 4. A surface plot showing the trajectory of F + H2 to H + HF reaction]|300px]]The reaction is exothermic as shown in figure 4, there is a decrease in potential energy when passing the transition state and forming the new HF bond. This relates to bond-strength, H-H has a bond strength of 432 kJ/mol whilst H-F has a bond strength of 565 kJ/mol.(reference). As HF is a polar bond as a result of the very electronegative F, there is an ionic element to the bond, resulting in a much greater attraction between the two atoms, thus releasing a lot of energy on formation. As the energy released is greater than that absorbed, the reaction is exothermic as depicted in the trajectory shown in figure 4.

[[File:Surface_Plot_hf-h_hk3918.png|thumb|Figure 5. A surface plot showing the trajectory of the FH + H to F + H2 reaction|300px]]

Figure 5 on the other hand shows an endothermic reaction, illustrated by the increase in potential energy along the trajectory, indicating that the energy absorbed to break the stronger H-F bond is greater than the energy released when forming the weaker H-H bond.

[[File:Plot small activation.png|thumb|Figure 6. Activation energy for the F +H2 reaction |300px]]

[[File:Plot activation big.png|thumb|Figure 7. Activation energy for the HF + H reaction |300px]]

[[File:Animation_momenta_hk3918.png |thumb|Figure 8. Momenta vs time plot for F + H2 reaction |300px]]

[[File:Animation_-74_-6.1.png ‎ |thumb|Figure 9. Contour plot for F + H2 reaction. Parameters: AB=200, BC=74, Pab=-1, Pbc=-6.9 |300px]]

[[File: Animation_low_momentum_hk3918.png ‎ |thumb|Figure 10. Contour plot for F + H2 reaction. Parameters: AB=200, BC=74, Pab=-1.6, Pbc=0.2 |300px]]

[[File: Animation low vibrational.png ‎ |thumb|Figure 11. Contour plot for HF + H reaction with high H momenta and low H-F momenta. Parameters: AB=200, BC=74, Pab=-1.6, Pbc=0.2 |300px]]

[[File: Animation highvirabtional.png|thumb|Figure 12. Skew plot for HF + H reaction with momentum of F-H set at -20 and momentum of H set below 1. Parameters: AB=200, BC=74, Pab=-1.6, Pbc=0.2 |300px]]

=== Locate the approximate position of the transition state ===
The transition state position was calculated to be when the H-F bond had a bond length of 181.8 pm, and the H-H bond was 73.1 pm as shown in figure 5. Analoguos to the calculation when approximating the transition state position of the H + H2 system, in the F + H2 reaction simulations were tested in order to acquire a constant bond length, at the maximum potential energy. As the first part of this reaction was exothermic, adopting Hammonds postulate, the assumption was made that the reaction had an early transition state thus the region near the reactants was analysed. The MEP calculation was used to remove the vibrational noise as it made it difficult to read the plots. As HF + H is a late transition state, resembling the its product F + H2, it will have the same transition state positions.

=== Report the activation energy for both reactions. ===
In order to find the activation energy the bond distance of H-F was first decreased. This then resulted in the energy falling to form the products H-F and H. The difference in this energy as shown in figure 7 was used to calculate the activation energy of the H-F + H endothermic reaction. The activation energy was estimated to be 120 kJ/mol. For the exothermic F + H2 reaction, the activation energy was calculated by trying to shift the trajectory towards to the reactants. This was done by thus increasing the F-H distance forming the F + H2 reactant. The difference in this energy was very small as shown in figure 8 and was used to calculate the activation energy for the F + H2 reaction.

=== 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. ===
Figure 8 illustrates the transition between translational and vibrational energy in the F + H2 reaction. Initially, the H-H bond (B-C) is vibrationally excited shown by the oscillation, whilst the H-F(A-B) bond isn't indicated by the absence of an oscillation. The H-F bond is at the vibrational ground state . After the collision we see a transition. The H-F bond now is vibrationally excited illustrated by the oscillation in the momenta vs time plot, and the H-H bonds loses this vibrationally energy and now only has translational energy due to the breaking of the bond, depicted by the plateau after 30s. Experimentally IR could be used to see the progress of the reaction. Only molecules with a dipole are IR active, thus after the formation of a H-F bond, IR absorption absent in the reactant will start to appear with overtones suggesting higher vibrational energies reached.

==== Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, with a momentum pFH = -1.0 g.mol-1.pm.fs-1, and explore several values of pHH in the range -6.1 to 6.1 g.mol-1.pm.fs-1 (explore values also close to these limits). What do you observe? Note that we are putting a significant amount of energy (much more than the activation energy) into the system on the H - H vibration. ====
When using the parameters above and setting the F-H value to 200 pm, at lower momentum values larger vibrational frequencies were obversed. The reactions did not therefore take place, satisfying Polyanyi's rules, where the having translational energy is more efficient when there is an early barrier rather than vibrational energy. Even at higher momentum extremes shown in figure 9, the reaction does not go to completion, indicating even though a lot of energy is being put into the system , it is not an efficient reaction, satisfying the rules.

==== For the same initial position, increase slightly the momentum pFH = -1.6 g.mol-1.pm.fs-1, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.2 g.mol-1.pm.fs-1. What do you observe now? ====
Figure 10 shows that, even at lower momentum values, resulting in lower vibrational energies the reaction still does not proceed to completion. As there is not enough translational energy in the reaction, further satisfying Polyanyi's rules.

==== Focus on the reverse reaction, H + HF. ====
This reaction has a late transition state, thus according to Polanyi's rules, vibrational energy trajectory is more efficient. This was tested firstly by using a low H-F vibrational energy and a high H translational energy as shown in figure 11. As predicted the reaction did not proceed to completion as there was not enough vibrational energy in the reaction. Figure 12 shows an example where the reaction has a larger vibrational energy, and therefore is more efficient, resulting in a successful reaction.

MRD:01524831.molecularreactiondynamics

2020-05-15T19:14:53Z

== Exercise 1: H + H2 system ==

[[File:Surface_Plot.png|thumb|Figure 1. A surface plot showing the transition state of a H + H2 system]]

=== 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? ===
On a potential energy surface diagram , the transition state is mathematically defined as the saddle point. The saddle point is a stationary point on the plot, where along the reaction path it is a maxima and perpendicular to the reaction path the saddle point is perceived to be a minima. The transition point can be identified in this simple example when the distances between the 3 hydrogen atoms are of equal distance. As shown in figure 1, the reaction path is oscillating at the saddle point. This depicts the vibrational freedom of the system at the transition state. A saddle point is found by...

=== 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 estimated '''rts '''is found to be around 91 pm. The transition state is defined as the position where internuclear distances are equal, and at a energy maxima in relation to the reaction path. From figure 1 it is evident that there are vibrations within the system, depicted by the oscillations on the ridge. The transition state position was found by determining which position the system stops oscillating, where the potential energy was a maximum, and thus the stable point of the transition state, the saddle point was found as shown in figure 2 and 3.

=== Comment on how the ''mep'' and the trajectory you just calculated differ. ===
When calculating under mep the reaction path is depicted different to when the calculation type is set to dynamics. What is absent in the map calculation is the motion of the atoms. The vibrational motion of the three hydrogen atoms are not depicted as shown in figure .... The initial condition is set so that the distance between AB is '''rts''' and the distance between BC deviates from '''rts '''by 1 pm, resulting in the reaction pathway following through, and advance towards the HA + HB-HC product. The flat BC line indicates the fact that the mep calculation does not consider vibrational motion of atoms.

==== Look at the “Internuclear Distances vs Time” and “Momenta vs Time”. What would change if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead? ====
The change of initial conditions to those above will change final product from HA + HB-HC to HA-HB + HC as it is the HB-HC bond distance that deviates away from the '''rts. '''As a result the internuclear distances vs time plot will now show AB to be constant over time. Momentum remains unchanged as initial momentum is 0 for both scenarios

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

The 'Get last geometry' option was selected, and the momenta were then reversed under the dynamic calculation type. The plot remains unchanged as expected. The reason for this is because the initial argument resulted in a trajectory towards a larger BC distance. Using the momenta of this trajectory but reversing the sign simply results in the hydrogen and hydrogen molecule to start in the final position, but now approach each other due to the negative momentum value.
=== 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"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot/kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Hydrogen atoms AB and C approach
eachother. Notably, the vibrational

motion present after the transition state is

not present before.
|[[File:Surface_Plot_question4_1_hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-4.1 </nowiki>
|<nowiki>-420.077</nowiki>
|No
|

Hydrogen atoms AB approaches C, however

before approaching the saddle point to AB

swiftly moves away from C

|[[File:Animation exercise2 p2 -4.1 hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|Hydrogen atom C approaches AB with a large
amount of momentum, resulting in a new BC Hydrogen

molecule forming, releasing the A hydrogen atom in the

reaction.
|[[File:Animation exercise2 p2 -5.1 p1 -3.1 hk3918.png|300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.1 </nowiki>
|<nowiki>-357.277</nowiki>
|No
|Hydrogen atoms AB and C approach each other very
quickly. Upon impact, large internuclear distance and velocities

are depicted, showing the B atom being tugged by both A and C.

In the end, B forms a molecule AB and moves away from C.
|[[File:Animation exercise2 p2 -10.1 hk3918.png |300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.6 </nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|Upon impact, the B atom is tugged between the two atoms A and
C, and is finally bonded to to atom C. A then moves away from

BC.
|[[File:Animation p2 -10.6 hk3918.png |300px]]
|}
The table above shows that a reaction isn't only reliant on the amount of kinetic energy in the system. As shown in the table above, there are instances where kinetic energy is higher than the activation energy, however no reaction takes place. One therefore needs to consider that successful collision are reliant on the system having enough energy to break and reform bonds, and also the orientation at which atoms collide.

=== Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===
Transition State Theory has a fundamental assumption which will result in its predictions being different to experimental values. The assumption that all trajectories possessing enough kinetic energy to circumvent the activation barrier will be reactive will result in a larger ''Rcl '' and thus a larger rate prediction than the experimental outcome. As seen in the table above, not all collisions are reactive even if there is sufficient kinetic energy - the case of barrier recrossing perfectly illustrates this.

== Exercise 2 ==

=== 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? ===
[[File:Surface_Plot_HF_hk3918.png|thumb|Figure 4. A surface plot showing the trajectory of F + H2 to H + HF reaction]|300px]]The reaction is exothermic as shown in figure 4, there is a decrease in potential energy when passing the transition state and forming the new HF bond. This relates to bond-strength, H-H has a bond strength of 432 kJ/mol whilst H-F has a bond strength of 565 kJ/mol.(reference). As HF is a polar bond as a result of the very electronegative F, there is an ionic element to the bond, resulting in a much greater attraction between the two atoms, thus releasing a lot of energy on formation. As the energy released is greater than that absorbed, the reaction is exothermic as depicted in the trajectory shown in figure 4.

[[File:Surface_Plot_hf-h_hk3918.png|thumb|Figure 5. A surface plot showing the trajectory of the FH + H to F + H2 reaction|300px]]

Figure 5 on the other hand shows an endothermic reaction, illustrated by the increase in potential energy along the trajectory, indicating that the energy absorbed to break the stronger H-F bond is greater than the energy released when forming the weaker H-H bond.

[[File:Plot small activation.png|thumb|Figure 6. Activation energy for the F +H2 reaction |300px]]

[[File:Plot activation big.png|thumb|Figure 7. Activation energy for the HF + H reaction |300px]]

[[File:Animation_momenta_hk3918.png |thumb|Figure 8. Momenta vs time plot for F + H2 reaction |300px]]

[[File:Animation_-74_-6.1.png ‎ |thumb|Figure 9. Contour plot for F + H2 reaction. Parameters: AB=200, BC=74, Pab=-1, Pbc=-6.9 |300px]]

[[File: Animation_low_momentum_hk3918.png ‎ |thumb|Figure 10. Contour plot for F + H2 reaction. Parameters: AB=200, BC=74, Pab=-1.6, Pbc=0.2 |300px]]

[[File: Animation low vibrational.png ‎ |thumb|Figure 11. Contour plot for HF + H reaction with high H momenta and low H-F momenta. Parameters: AB=200, BC=74, Pab=-1.6, Pbc=0.2 |300px]]

[[File: Animation highvirabtional.png|thumb|Figure 12. Skew plot for HF + H reaction with momentum of F-H set at -20 and momentum of H set below 1. Parameters: AB=200, BC=74, Pab=-1.6, Pbc=0.2 |300px]]

=== Locate the approximate position of the transition state ===
The transition state position was calculated to be when the H-F bond had a bond length of 181.8 pm, and the H-H bond was 73.1 pm as shown in figure 5. Analoguos to the calculation when approximating the transition state position of the H + H2 system, in the F + H2 reaction simulations were tested in order to acquire a constant bond length, at the maximum potential energy. As the first part of this reaction was exothermic, adopting Hammonds postulate, the assumption was made that the reaction had an early transition state thus the region near the reactants was analysed. The MEP calculation was used to remove the vibrational noise as it made it difficult to read the plots. As HF + H is a late transition state, resembling the its product F + H2, it will have the same transition state positions.

=== Report the activation energy for both reactions. ===
In order to find the activation energy the bond distance of H-F was first decreased. This then resulted in the energy falling to form the products H-F and H. The difference in this energy as shown in figure 7 was used to calculate the activation energy of the H-F + H endothermic reaction. The activation energy was estimated to be 120 kJ/mol. For the exothermic F + H2 reaction, the activation energy was calculated by trying to shift the trajectory towards to the reactants. This was done by thus increasing the F-H distance forming the F + H2 reactant. The difference in this energy was very small as shown in figure 8 and was used to calculate the activation energy for the F + H2 reaction.

=== 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. ===
Figure 8 illustrates the transition between translational and vibrational energy in the F + H2 reaction. Initially, the H-H bond (B-C) is vibrationally excited shown by the oscillation, whilst the H-F(A-B) bond isn't indicated by the absence of an oscillation. The H-F bond is at the vibrational ground state . After the collision we see a transition. The H-F bond now is vibrationally excited illustrated by the oscillation in the momenta vs time plot, and the H-H bonds loses this vibrationally energy and now only has translational energy due to the breaking of the bond, depicted by the plateau after 30s. Experimentally IR could be used to see the progress of the reaction. Only molecules with a dipole are IR active, thus after the formation of a H-F bond, IR absorption absent in the reactant will start to appear with overtones suggesting higher vibrational energies reached.

==== Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, with a momentum pFH = -1.0 g.mol-1.pm.fs-1, and explore several values of pHH in the range -6.1 to 6.1 g.mol-1.pm.fs-1 (explore values also close to these limits). What do you observe? Note that we are putting a significant amount of energy (much more than the activation energy) into the system on the H - H vibration. ====
When using the parameters above and setting the F-H value to 200 pm, at lower momentum values larger vibrational frequencies were obversed. The reactions did not therefore take place, satisfying Polyanyi's rules, where the having translational energy is more efficient when there is an early barrier rather than vibrational energy. Even at higher momentum extremes shown in figure 9, the reaction does not go to completion, indicating even though a lot of energy is being put into the system , it is not an efficient reaction, satisfying the rules.

==== For the same initial position, increase slightly the momentum pFH = -1.6 g.mol-1.pm.fs-1, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.2 g.mol-1.pm.fs-1. What do you observe now? ====
Figure 10 shows that, even at lower momentum values, resulting in lower vibrational energies the reaction still does not proceed to completion. As there is not enough translational energy in the reaction, further satisfying Polyanyi's rules.

File:Animation highvirabtional.png

2020-05-15T19:10:54Z

Hk3918: high vibrational AB-74 pm, Pab= -0.2 BC-200 pm, Pbc = -20

high vibrational
AB-74 pm, Pab= -0.2
BC-200 pm, Pbc = -20

File:Animation low vibrational.png

2020-05-15T19:09:23Z

Hk3918:

MRD:01524831.molecularreactiondynamics

2020-05-15T18:59:22Z

Hk3918: /* Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, with a momentum pFH = -1.0 g.mol-1.pm.fs-1, and explore several values of pHH in the range -6.1 to 6.1 g.mol-1.pm.fs-1 (explore values

== Exercise 1: H + H2 system ==

[[File:Surface_Plot.png|thumb|Figure 1. A surface plot showing the transition state of a H + H2 system]]

=== 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? ===
On a potential energy surface diagram , the transition state is mathematically defined as the saddle point. The saddle point is a stationary point on the plot, where along the reaction path it is a maxima and perpendicular to the reaction path the saddle point is perceived to be a minima. The transition point can be identified in this simple example when the distances between the 3 hydrogen atoms are of equal distance. As shown in figure 1, the reaction path is oscillating at the saddle point. This depicts the vibrational freedom of the system at the transition state. A saddle point is found by...

=== 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 estimated '''rts '''is found to be around 91 pm. The transition state is defined as the position where internuclear distances are equal, and at a energy maxima in relation to the reaction path. From figure 1 it is evident that there are vibrations within the system, depicted by the oscillations on the ridge. The transition state position was found by determining which position the system stops oscillating, where the potential energy was a maximum, and thus the stable point of the transition state, the saddle point was found as shown in figure 2 and 3.

=== Comment on how the ''mep'' and the trajectory you just calculated differ. ===
When calculating under mep the reaction path is depicted different to when the calculation type is set to dynamics. What is absent in the map calculation is the motion of the atoms. The vibrational motion of the three hydrogen atoms are not depicted as shown in figure .... The initial condition is set so that the distance between AB is '''rts''' and the distance between BC deviates from '''rts '''by 1 pm, resulting in the reaction pathway following through, and advance towards the HA + HB-HC product. The flat BC line indicates the fact that the mep calculation does not consider vibrational motion of atoms.

==== Look at the “Internuclear Distances vs Time” and “Momenta vs Time”. What would change if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead? ====
The change of initial conditions to those above will change final product from HA + HB-HC to HA-HB + HC as it is the HB-HC bond distance that deviates away from the '''rts. '''As a result the internuclear distances vs time plot will now show AB to be constant over time. Momentum remains unchanged as initial momentum is 0 for both scenarios

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

The 'Get last geometry' option was selected, and the momenta were then reversed under the dynamic calculation type. The plot remains unchanged as expected. The reason for this is because the initial argument resulted in a trajectory towards a larger BC distance. Using the momenta of this trajectory but reversing the sign simply results in the hydrogen and hydrogen molecule to start in the final position, but now approach each other due to the negative momentum value.
=== 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"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot/kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Hydrogen atoms AB and C approach
eachother. Notably, the vibrational

motion present after the transition state is

not present before.
|[[File:Surface_Plot_question4_1_hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-4.1 </nowiki>
|<nowiki>-420.077</nowiki>
|No
|

Hydrogen atoms AB approaches C, however

before approaching the saddle point to AB

swiftly moves away from C

|[[File:Animation exercise2 p2 -4.1 hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|Hydrogen atom C approaches AB with a large
amount of momentum, resulting in a new BC Hydrogen

molecule forming, releasing the A hydrogen atom in the

reaction.
|[[File:Animation exercise2 p2 -5.1 p1 -3.1 hk3918.png|300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.1 </nowiki>
|<nowiki>-357.277</nowiki>
|No
|Hydrogen atoms AB and C approach each other very
quickly. Upon impact, large internuclear distance and velocities

are depicted, showing the B atom being tugged by both A and C.

In the end, B forms a molecule AB and moves away from C.
|[[File:Animation exercise2 p2 -10.1 hk3918.png |300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.6 </nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|Upon impact, the B atom is tugged between the two atoms A and
C, and is finally bonded to to atom C. A then moves away from

BC.
|[[File:Animation p2 -10.6 hk3918.png |300px]]
|}
The table above shows that a reaction isn't only reliant on the amount of kinetic energy in the system. As shown in the table above, there are instances where kinetic energy is higher than the activation energy, however no reaction takes place. One therefore needs to consider that successful collision are reliant on the system having enough energy to break and reform bonds, and also the orientation at which atoms collide.

=== Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===
Transition State Theory has a fundamental assumption which will result in its predictions being different to experimental values. The assumption that all trajectories possessing enough kinetic energy to circumvent the activation barrier will be reactive will result in a larger ''Rcl '' and thus a larger rate prediction than the experimental outcome. As seen in the table above, not all collisions are reactive even if there is sufficient kinetic energy - the case of barrier recrossing perfectly illustrates this.

== Exercise 2 ==

=== 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? ===
[[File:Surface_Plot_HF_hk3918.png|thumb|Figure 4. A surface plot showing the trajectory of F + H2 to H + HF reaction]|300px]]The reaction is exothermic as shown in figure 4, there is a decrease in potential energy when passing the transition state and forming the new HF bond. This relates to bond-strength, H-H has a bond strength of 432 kJ/mol whilst H-F has a bond strength of 565 kJ/mol.(reference). As HF is a polar bond as a result of the very electronegative F, there is an ionic element to the bond, resulting in a much greater attraction between the two atoms, thus releasing a lot of energy on formation. As the energy released is greater than that absorbed, the reaction is exothermic as depicted in the trajectory shown in figure 4.

[[File:Surface_Plot_hf-h_hk3918.png|thumb|Figure 5. A surface plot showing the trajectory of the FH + H to F + H2 reaction|300px]]

Figure 5 on the other hand shows an endothermic reaction, illustrated by the increase in potential energy along the trajectory, indicating that the energy absorbed to break the stronger H-F bond is greater than the energy released when forming the weaker H-H bond.

[[File:Plot small activation.png|thumb|Figure 6. Activation energy for the F +H2 reaction |300px]]

[[File:Plot activation big.png|thumb|Figure 7. Activation energy for the HF + H reaction |300px]]

[[File:Animation_momenta_hk3918.png |thumb|Figure 8. Momenta vs time plot for F + H2 reaction |300px]]

[[File:Animation_-74_-6.1.png ‎ |thumb|Figure 9. Contour plot for F + H2 reaction. Parameters: AB=200, BC=74, Pab=-1, Pbc=-6.9 |300px]]

[[File: Animation_low_momentum_hk3918.png ‎ |thumb|Figure 10. Contour plot for F + H2 reaction. Parameters: AB=200, BC=74, Pab=-1.6, Pbc=0.2 |300px]]

=== Locate the approximate position of the transition state ===
The transition state position was calculated to be when the H-F bond had a bond length of 181.8 pm, and the H-H bond was 73.1 pm as shown in figure 5. Analoguos to the calculation when approximating the transition state position of the H + H2 system, in the F + H2 reaction simulations were tested in order to acquire a constant bond length, at the maximum potential energy. As the first part of this reaction was exothermic, adopting Hammonds postulate, the assumption was made that the reaction had an early transition state thus the region near the reactants was analysed. The MEP calculation was used to remove the vibrational noise as it made it difficult to read the plots. As HF + H is a late transition state, resembling the its product F + H2, it will have the same transition state positions.

=== Report the activation energy for both reactions. ===
In order to find the activation energy the bond distance of H-F was first decreased. This then resulted in the energy falling to form the products H-F and H. The difference in this energy as shown in figure 7 was used to calculate the activation energy of the H-F + H endothermic reaction. The activation energy was estimated to be 120 kJ/mol. For the exothermic F + H2 reaction, the activation energy was calculated by trying to shift the trajectory towards to the reactants. This was done by thus increasing the F-H distance forming the F + H2 reactant. The difference in this energy was very small as shown in figure 8 and was used to calculate the activation energy for the F + H2 reaction.

=== 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. ===
Figure 8 illustrates the transition between translational and vibrational energy in the F + H2 reaction. Initially, the H-H bond (B-C) is vibrationally excited shown by the oscillation, whilst the H-F(A-B) bond isn't indicated by the absence of an oscillation. The H-F bond is at the vibrational ground state . After the collision we see a transition. The H-F bond now is vibrationally excited illustrated by the oscillation in the momenta vs time plot, and the H-H bonds loses this vibrationally energy and now only has translational energy due to the breaking of the bond, depicted by the plateau after 30s. Experimentally IR could be used to see the progress of the reaction. Only molecules with a dipole are IR active, thus after the formation of a H-F bond, IR absorption absent in the reactant will start to appear with overtones suggesting higher vibrational energies reached.

==== Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, with a momentum pFH = -1.0 g.mol-1.pm.fs-1, and explore several values of pHH in the range -6.1 to 6.1 g.mol-1.pm.fs-1 (explore values also close to these limits). What do you observe? Note that we are putting a significant amount of energy (much more than the activation energy) into the system on the H - H vibration. ====
When using the parameters above and setting the F-H value to 200 pm, at lower momentum values larger vibrational frequencies were obversed. The reactions did not therefore take place, satisfying Polyanyi's rules, where the having translational energy is more efficient when there is an early barrier rather than vibrational energy. Even at higher momentum extremes shown in figure 9, the reaction does not go to completion, indicating even though a lot of energy is being put into the system , it is not an efficient reaction, satisfying the rules.

==== For the same initial position, increase slightly the momentum pFH = -1.6 g.mol-1.pm.fs-1, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.2 g.mol-1.pm.fs-1. What do you observe now? ====
Figure 10 shows that, even at lower momentum values, resulting in lower vibrational energies the reaction still does not proceed to completion. As there is not enough translational energy in the reaction, further satisfying Polyanyi's rules.

MRD:01524831.molecularreactiondynamics

2020-05-15T18:56:33Z

== Exercise 1: H + H2 system ==

[[File:Surface_Plot.png|thumb|Figure 1. A surface plot showing the transition state of a H + H2 system]]

=== 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? ===
On a potential energy surface diagram , the transition state is mathematically defined as the saddle point. The saddle point is a stationary point on the plot, where along the reaction path it is a maxima and perpendicular to the reaction path the saddle point is perceived to be a minima. The transition point can be identified in this simple example when the distances between the 3 hydrogen atoms are of equal distance. As shown in figure 1, the reaction path is oscillating at the saddle point. This depicts the vibrational freedom of the system at the transition state. A saddle point is found by...

=== 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 estimated '''rts '''is found to be around 91 pm. The transition state is defined as the position where internuclear distances are equal, and at a energy maxima in relation to the reaction path. From figure 1 it is evident that there are vibrations within the system, depicted by the oscillations on the ridge. The transition state position was found by determining which position the system stops oscillating, where the potential energy was a maximum, and thus the stable point of the transition state, the saddle point was found as shown in figure 2 and 3.

=== Comment on how the ''mep'' and the trajectory you just calculated differ. ===
When calculating under mep the reaction path is depicted different to when the calculation type is set to dynamics. What is absent in the map calculation is the motion of the atoms. The vibrational motion of the three hydrogen atoms are not depicted as shown in figure .... The initial condition is set so that the distance between AB is '''rts''' and the distance between BC deviates from '''rts '''by 1 pm, resulting in the reaction pathway following through, and advance towards the HA + HB-HC product. The flat BC line indicates the fact that the mep calculation does not consider vibrational motion of atoms.

==== Look at the “Internuclear Distances vs Time” and “Momenta vs Time”. What would change if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead? ====
The change of initial conditions to those above will change final product from HA + HB-HC to HA-HB + HC as it is the HB-HC bond distance that deviates away from the '''rts. '''As a result the internuclear distances vs time plot will now show AB to be constant over time. Momentum remains unchanged as initial momentum is 0 for both scenarios

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

The 'Get last geometry' option was selected, and the momenta were then reversed under the dynamic calculation type. The plot remains unchanged as expected. The reason for this is because the initial argument resulted in a trajectory towards a larger BC distance. Using the momenta of this trajectory but reversing the sign simply results in the hydrogen and hydrogen molecule to start in the final position, but now approach each other due to the negative momentum value.
=== 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"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot/kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Hydrogen atoms AB and C approach
eachother. Notably, the vibrational

motion present after the transition state is

not present before.
|[[File:Surface_Plot_question4_1_hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-4.1 </nowiki>
|<nowiki>-420.077</nowiki>
|No
|

Hydrogen atoms AB approaches C, however

before approaching the saddle point to AB

swiftly moves away from C

|[[File:Animation exercise2 p2 -4.1 hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|Hydrogen atom C approaches AB with a large
amount of momentum, resulting in a new BC Hydrogen

molecule forming, releasing the A hydrogen atom in the

reaction.
|[[File:Animation exercise2 p2 -5.1 p1 -3.1 hk3918.png|300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.1 </nowiki>
|<nowiki>-357.277</nowiki>
|No
|Hydrogen atoms AB and C approach each other very
quickly. Upon impact, large internuclear distance and velocities

are depicted, showing the B atom being tugged by both A and C.

In the end, B forms a molecule AB and moves away from C.
|[[File:Animation exercise2 p2 -10.1 hk3918.png |300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.6 </nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|Upon impact, the B atom is tugged between the two atoms A and
C, and is finally bonded to to atom C. A then moves away from

BC.
|[[File:Animation p2 -10.6 hk3918.png |300px]]
|}
The table above shows that a reaction isn't only reliant on the amount of kinetic energy in the system. As shown in the table above, there are instances where kinetic energy is higher than the activation energy, however no reaction takes place. One therefore needs to consider that successful collision are reliant on the system having enough energy to break and reform bonds, and also the orientation at which atoms collide.

=== Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===
Transition State Theory has a fundamental assumption which will result in its predictions being different to experimental values. The assumption that all trajectories possessing enough kinetic energy to circumvent the activation barrier will be reactive will result in a larger ''Rcl '' and thus a larger rate prediction than the experimental outcome. As seen in the table above, not all collisions are reactive even if there is sufficient kinetic energy - the case of barrier recrossing perfectly illustrates this.

== Exercise 2 ==

=== 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? ===
[[File:Surface_Plot_HF_hk3918.png|thumb|Figure 4. A surface plot showing the trajectory of F + H2 to H + HF reaction]|300px]]The reaction is exothermic as shown in figure 4, there is a decrease in potential energy when passing the transition state and forming the new HF bond. This relates to bond-strength, H-H has a bond strength of 432 kJ/mol whilst H-F has a bond strength of 565 kJ/mol.(reference). As HF is a polar bond as a result of the very electronegative F, there is an ionic element to the bond, resulting in a much greater attraction between the two atoms, thus releasing a lot of energy on formation. As the energy released is greater than that absorbed, the reaction is exothermic as depicted in the trajectory shown in figure 4.

[[File:Surface_Plot_hf-h_hk3918.png|thumb|Figure 5. A surface plot showing the trajectory of the FH + H to F + H2 reaction|300px]]

Figure 5 on the other hand shows an endothermic reaction, illustrated by the increase in potential energy along the trajectory, indicating that the energy absorbed to break the stronger H-F bond is greater than the energy released when forming the weaker H-H bond.

[[File:Plot small activation.png|thumb|Figure 6. Activation energy for the F +H2 reaction |300px]]

[[File:Plot activation big.png|thumb|Figure 7. Activation energy for the HF + H reaction |300px]]

[[File:Animation_momenta_hk3918.png |thumb|Figure 8. Momenta vs time plot for F + H2 reaction |300px]]

[[File:Animation_-74_-6.1.png ‎ |thumb|Figure 9. Contour plot for F + H2 reaction. Parameters: AB=200, BC=74, Pab=-1, Pbc=-6.9 |300px]]

[[File: Animation_low_momentum_hk3918.png ‎ |thumb|Figure 10. Contour plot for F + H2 reaction. Parameters: AB=200, BC=74, Pab=-1.6, Pbc=0.2 |300px]]

=== Locate the approximate position of the transition state ===
The transition state position was calculated to be when the H-F bond had a bond length of 181.8 pm, and the H-H bond was 73.1 pm as shown in figure 5. Analoguos to the calculation when approximating the transition state position of the H + H2 system, in the F + H2 reaction simulations were tested in order to acquire a constant bond length, at the maximum potential energy. As the first part of this reaction was exothermic, adopting Hammonds postulate, the assumption was made that the reaction had an early transition state thus the region near the reactants was analysed. The MEP calculation was used to remove the vibrational noise as it made it difficult to read the plots. As HF + H is a late transition state, resembling the its product F + H2, it will have the same transition state positions.

=== Report the activation energy for both reactions. ===
In order to find the activation energy the bond distance of H-F was first decreased. This then resulted in the energy falling to form the products H-F and H. The difference in this energy as shown in figure 7 was used to calculate the activation energy of the H-F + H endothermic reaction. The activation energy was estimated to be 120 kJ/mol. For the exothermic F + H2 reaction, the activation energy was calculated by trying to shift the trajectory towards to the reactants. This was done by thus increasing the F-H distance forming the F + H2 reactant. The difference in this energy was very small as shown in figure 8 and was used to calculate the activation energy for the F + H2 reaction.

=== 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. ===
Figure 8 illustrates the transition between translational and vibrational energy in the F + H2 reaction. Initially, the H-H bond (B-C) is vibrationally excited shown by the oscillation, whilst the H-F(A-B) bond isn't indicated by the absence of an oscillation. The H-F bond is at the vibrational ground state . After the collision we see a transition. The H-F bond now is vibrationally excited illustrated by the oscillation in the momenta vs time plot, and the H-H bonds loses this vibrationally energy and now only has translational energy due to the breaking of the bond, depicted by the plateau after 30s. Experimentally IR could be used to see the progress of the reaction. Only molecules with a dipole are IR active, thus after the formation of a H-F bond, IR absorption absent in the reactant will start to appear with overtones suggesting higher vibrational energies reached.

==== Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, with a momentum pFH = -1.0 g.mol-1.pm.fs-1, and explore several values of pHH in the range -6.1 to 6.1 g.mol-1.pm.fs-1 (explore values also close to these limits). What do you observe? Note that we are putting a significant amount of energy (much more than the activation energy) into the system on the H - H vibration. ====
When using the parameters above and setting the F-H value to 200 pm, at lower momentum values larger vibrational frequencies were obversed. The reactions did not therefore take place, satisfying Polyanyi's rules, where the having translational energy is more efficient when there is an early barrier rather than vibrational energy. Even at higher momentum extremes shown in figure 9, the reaction does not go to completion, indicating even though a lot of energy is being put into the system , it is not an efficient reaction, satisfying the rules.

File:Animation low momentum hk3918.png

2020-05-15T18:55:12Z

Hk3918:

MRD:01524831.molecularreactiondynamics

2020-05-15T18:54:10Z

== Exercise 1: H + H2 system ==

[[File:Surface_Plot.png|thumb|Figure 1. A surface plot showing the transition state of a H + H2 system]]

=== 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? ===
On a potential energy surface diagram , the transition state is mathematically defined as the saddle point. The saddle point is a stationary point on the plot, where along the reaction path it is a maxima and perpendicular to the reaction path the saddle point is perceived to be a minima. The transition point can be identified in this simple example when the distances between the 3 hydrogen atoms are of equal distance. As shown in figure 1, the reaction path is oscillating at the saddle point. This depicts the vibrational freedom of the system at the transition state. A saddle point is found by...

=== 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 estimated '''rts '''is found to be around 91 pm. The transition state is defined as the position where internuclear distances are equal, and at a energy maxima in relation to the reaction path. From figure 1 it is evident that there are vibrations within the system, depicted by the oscillations on the ridge. The transition state position was found by determining which position the system stops oscillating, where the potential energy was a maximum, and thus the stable point of the transition state, the saddle point was found as shown in figure 2 and 3.

=== Comment on how the ''mep'' and the trajectory you just calculated differ. ===
When calculating under mep the reaction path is depicted different to when the calculation type is set to dynamics. What is absent in the map calculation is the motion of the atoms. The vibrational motion of the three hydrogen atoms are not depicted as shown in figure .... The initial condition is set so that the distance between AB is '''rts''' and the distance between BC deviates from '''rts '''by 1 pm, resulting in the reaction pathway following through, and advance towards the HA + HB-HC product. The flat BC line indicates the fact that the mep calculation does not consider vibrational motion of atoms.

==== Look at the “Internuclear Distances vs Time” and “Momenta vs Time”. What would change if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead? ====
The change of initial conditions to those above will change final product from HA + HB-HC to HA-HB + HC as it is the HB-HC bond distance that deviates away from the '''rts. '''As a result the internuclear distances vs time plot will now show AB to be constant over time. Momentum remains unchanged as initial momentum is 0 for both scenarios

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

The 'Get last geometry' option was selected, and the momenta were then reversed under the dynamic calculation type. The plot remains unchanged as expected. The reason for this is because the initial argument resulted in a trajectory towards a larger BC distance. Using the momenta of this trajectory but reversing the sign simply results in the hydrogen and hydrogen molecule to start in the final position, but now approach each other due to the negative momentum value.
=== 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"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot/kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Hydrogen atoms AB and C approach
eachother. Notably, the vibrational

motion present after the transition state is

not present before.
|[[File:Surface_Plot_question4_1_hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-4.1 </nowiki>
|<nowiki>-420.077</nowiki>
|No
|

Hydrogen atoms AB approaches C, however

before approaching the saddle point to AB

swiftly moves away from C

|[[File:Animation exercise2 p2 -4.1 hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|Hydrogen atom C approaches AB with a large
amount of momentum, resulting in a new BC Hydrogen

molecule forming, releasing the A hydrogen atom in the

reaction.
|[[File:Animation exercise2 p2 -5.1 p1 -3.1 hk3918.png|300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.1 </nowiki>
|<nowiki>-357.277</nowiki>
|No
|Hydrogen atoms AB and C approach each other very
quickly. Upon impact, large internuclear distance and velocities

are depicted, showing the B atom being tugged by both A and C.

In the end, B forms a molecule AB and moves away from C.
|[[File:Animation exercise2 p2 -10.1 hk3918.png |300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.6 </nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|Upon impact, the B atom is tugged between the two atoms A and
C, and is finally bonded to to atom C. A then moves away from

BC.
|[[File:Animation p2 -10.6 hk3918.png |300px]]
|}
The table above shows that a reaction isn't only reliant on the amount of kinetic energy in the system. As shown in the table above, there are instances where kinetic energy is higher than the activation energy, however no reaction takes place. One therefore needs to consider that successful collision are reliant on the system having enough energy to break and reform bonds, and also the orientation at which atoms collide.

=== Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===
Transition State Theory has a fundamental assumption which will result in its predictions being different to experimental values. The assumption that all trajectories possessing enough kinetic energy to circumvent the activation barrier will be reactive will result in a larger ''Rcl '' and thus a larger rate prediction than the experimental outcome. As seen in the table above, not all collisions are reactive even if there is sufficient kinetic energy - the case of barrier recrossing perfectly illustrates this.

== Exercise 2 ==

=== 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? ===
[[File:Surface_Plot_HF_hk3918.png|thumb|Figure 4. A surface plot showing the trajectory of F + H2 to H + HF reaction]|300px]]The reaction is exothermic as shown in figure 4, there is a decrease in potential energy when passing the transition state and forming the new HF bond. This relates to bond-strength, H-H has a bond strength of 432 kJ/mol whilst H-F has a bond strength of 565 kJ/mol.(reference). As HF is a polar bond as a result of the very electronegative F, there is an ionic element to the bond, resulting in a much greater attraction between the two atoms, thus releasing a lot of energy on formation. As the energy released is greater than that absorbed, the reaction is exothermic as depicted in the trajectory shown in figure 4.

[[File:Surface_Plot_hf-h_hk3918.png|thumb|Figure 5. A surface plot showing the trajectory of the FH + H to F + H2 reaction|300px]]

Figure 5 on the other hand shows an endothermic reaction, illustrated by the increase in potential energy along the trajectory, indicating that the energy absorbed to break the stronger H-F bond is greater than the energy released when forming the weaker H-H bond.

[[File:Plot small activation.png|thumb|Figure 6. Activation energy for the F +H2 reaction |300px]]

[[File:Plot activation big.png|thumb|Figure 7. Activation energy for the HF + H reaction |300px]]

[[File:Animation_momenta_hk3918.png |thumb|Figure 8. Momenta vs time plot for F + H2 reaction |300px]]

[[File:Animation_-74_-6.1.png ‎ |thumb|Figure 9. Contour plot for F + H2 reaction. Parameters: AB=200, BC=74, Pab=-1, Pbc=-6.9 |300px]]

=== Locate the approximate position of the transition state ===
The transition state position was calculated to be when the H-F bond had a bond length of 181.8 pm, and the H-H bond was 73.1 pm as shown in figure 5. Analoguos to the calculation when approximating the transition state position of the H + H2 system, in the F + H2 reaction simulations were tested in order to acquire a constant bond length, at the maximum potential energy. As the first part of this reaction was exothermic, adopting Hammonds postulate, the assumption was made that the reaction had an early transition state thus the region near the reactants was analysed. The MEP calculation was used to remove the vibrational noise as it made it difficult to read the plots. As HF + H is a late transition state, resembling the its product F + H2, it will have the same transition state positions.

=== Report the activation energy for both reactions. ===
In order to find the activation energy the bond distance of H-F was first decreased. This then resulted in the energy falling to form the products H-F and H. The difference in this energy as shown in figure 7 was used to calculate the activation energy of the H-F + H endothermic reaction. The activation energy was estimated to be 120 kJ/mol. For the exothermic F + H2 reaction, the activation energy was calculated by trying to shift the trajectory towards to the reactants. This was done by thus increasing the F-H distance forming the F + H2 reactant. The difference in this energy was very small as shown in figure 8 and was used to calculate the activation energy for the F + H2 reaction.

=== 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. ===
Figure 8 illustrates the transition between translational and vibrational energy in the F + H2 reaction. Initially, the H-H bond (B-C) is vibrationally excited shown by the oscillation, whilst the H-F(A-B) bond isn't indicated by the absence of an oscillation. The H-F bond is at the vibrational ground state . After the collision we see a transition. The H-F bond now is vibrationally excited illustrated by the oscillation in the momenta vs time plot, and the H-H bonds loses this vibrationally energy and now only has translational energy due to the breaking of the bond, depicted by the plateau after 30s. Experimentally IR could be used to see the progress of the reaction. Only molecules with a dipole are IR active, thus after the formation of a H-F bond, IR absorption absent in the reactant will start to appear with overtones suggesting higher vibrational energies reached.

==== Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, with a momentum pFH = -1.0 g.mol-1.pm.fs-1, and explore several values of pHH in the range -6.1 to 6.1 g.mol-1.pm.fs-1 (explore values also close to these limits). What do you observe? Note that we are putting a significant amount of energy (much more than the activation energy) into the system on the H - H vibration. ====
When using the parameters above and setting the F-H value to 200 pm, at lower momentum values larger vibrational frequencies were obversed. The reactions did not therefore take place, satisfying Polyanyi's rules, where the having translational energy is more efficient when there is an early barrier rather than vibrational energy. Even at higher momentum extremes shown in figure 9, the reaction does not go to completion, indicating even though a lot of energy is being put into the system , it is not an efficient reaction, satisfying the rules.

MRD:01524831.molecularreactiondynamics

2020-05-15T18:43:45Z

== Exercise 1: H + H2 system ==

[[File:Surface_Plot.png|thumb|Figure 1. A surface plot showing the transition state of a H + H2 system]]

=== 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? ===
On a potential energy surface diagram , the transition state is mathematically defined as the saddle point. The saddle point is a stationary point on the plot, where along the reaction path it is a maxima and perpendicular to the reaction path the saddle point is perceived to be a minima. The transition point can be identified in this simple example when the distances between the 3 hydrogen atoms are of equal distance. As shown in figure 1, the reaction path is oscillating at the saddle point. This depicts the vibrational freedom of the system at the transition state. A saddle point is found by...

=== 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 estimated '''rts '''is found to be around 91 pm. The transition state is defined as the position where internuclear distances are equal, and at a energy maxima in relation to the reaction path. From figure 1 it is evident that there are vibrations within the system, depicted by the oscillations on the ridge. The transition state position was found by determining which position the system stops oscillating, where the potential energy was a maximum, and thus the stable point of the transition state, the saddle point was found as shown in figure 2 and 3.

=== Comment on how the ''mep'' and the trajectory you just calculated differ. ===
When calculating under mep the reaction path is depicted different to when the calculation type is set to dynamics. What is absent in the map calculation is the motion of the atoms. The vibrational motion of the three hydrogen atoms are not depicted as shown in figure .... The initial condition is set so that the distance between AB is '''rts''' and the distance between BC deviates from '''rts '''by 1 pm, resulting in the reaction pathway following through, and advance towards the HA + HB-HC product. The flat BC line indicates the fact that the mep calculation does not consider vibrational motion of atoms.

==== Look at the “Internuclear Distances vs Time” and “Momenta vs Time”. What would change if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead? ====
The change of initial conditions to those above will change final product from HA + HB-HC to HA-HB + HC as it is the HB-HC bond distance that deviates away from the '''rts. '''As a result the internuclear distances vs time plot will now show AB to be constant over time. Momentum remains unchanged as initial momentum is 0 for both scenarios

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

The 'Get last geometry' option was selected, and the momenta were then reversed under the dynamic calculation type. The plot remains unchanged as expected. The reason for this is because the initial argument resulted in a trajectory towards a larger BC distance. Using the momenta of this trajectory but reversing the sign simply results in the hydrogen and hydrogen molecule to start in the final position, but now approach each other due to the negative momentum value.
=== 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"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot/kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Hydrogen atoms AB and C approach
eachother. Notably, the vibrational

motion present after the transition state is

not present before.
|[[File:Surface_Plot_question4_1_hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-4.1 </nowiki>
|<nowiki>-420.077</nowiki>
|No
|

Hydrogen atoms AB approaches C, however

before approaching the saddle point to AB

swiftly moves away from C

|[[File:Animation exercise2 p2 -4.1 hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|Hydrogen atom C approaches AB with a large
amount of momentum, resulting in a new BC Hydrogen

molecule forming, releasing the A hydrogen atom in the

reaction.
|[[File:Animation exercise2 p2 -5.1 p1 -3.1 hk3918.png|300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.1 </nowiki>
|<nowiki>-357.277</nowiki>
|No
|Hydrogen atoms AB and C approach each other very
quickly. Upon impact, large internuclear distance and velocities

are depicted, showing the B atom being tugged by both A and C.

In the end, B forms a molecule AB and moves away from C.
|[[File:Animation exercise2 p2 -10.1 hk3918.png |300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.6 </nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|Upon impact, the B atom is tugged between the two atoms A and
C, and is finally bonded to to atom C. A then moves away from

BC.
|[[File:Animation p2 -10.6 hk3918.png |300px]]
|}
The table above shows that a reaction isn't only reliant on the amount of kinetic energy in the system. As shown in the table above, there are instances where kinetic energy is higher than the activation energy, however no reaction takes place. One therefore needs to consider that successful collision are reliant on the system having enough energy to break and reform bonds, and also the orientation at which atoms collide.

=== Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===
Transition State Theory has a fundamental assumption which will result in its predictions being different to experimental values. The assumption that all trajectories possessing enough kinetic energy to circumvent the activation barrier will be reactive will result in a larger ''Rcl '' and thus a larger rate prediction than the experimental outcome. As seen in the table above, not all collisions are reactive even if there is sufficient kinetic energy - the case of barrier recrossing perfectly illustrates this.

== Exercise 2 ==

=== 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? ===
[[File:Surface_Plot_HF_hk3918.png|thumb|Figure 4. A surface plot showing the trajectory of F + H2 to H + HF reaction]|300px]]The reaction is exothermic as shown in figure 4, there is a decrease in potential energy when passing the transition state and forming the new HF bond. This relates to bond-strength, H-H has a bond strength of 432 kJ/mol whilst H-F has a bond strength of 565 kJ/mol.(reference). As HF is a polar bond as a result of the very electronegative F, there is an ionic element to the bond, resulting in a much greater attraction between the two atoms, thus releasing a lot of energy on formation. As the energy released is greater than that absorbed, the reaction is exothermic as depicted in the trajectory shown in figure 4.

[[File:Surface_Plot_hf-h_hk3918.png|thumb|Figure 5. A surface plot showing the trajectory of the FH + H to F + H2 reaction|300px]]

Figure 5 on the other hand shows an endothermic reaction, illustrated by the increase in potential energy along the trajectory, indicating that the energy absorbed to break the stronger H-F bond is greater than the energy released when forming the weaker H-H bond.

[[File:Plot small activation.png|thumb|Figure 6. Activation energy for the F +H2 reaction |300px]]

[[File:Plot activation big.png|thumb|Figure 7. Activation energy for the HF + H reaction |300px]]

[[File:Animation_momenta_hk3918.png |thumb|Figure 8. Momenta vs time plot for F + H2 reaction |300px]]

[[File:Animation_-74_-6.1.png ‎ |thumb|Figure 9. Contour plot for F + H2 reaction. Parameters: AB=200, BC=74, Pab=-1, Pbc=-6.9 |300px]]

=== Locate the approximate position of the transition state ===
The transition state position was calculated to be when the H-F bond had a bond length of 181.8 pm, and the H-H bond was 73.1 pm as shown in figure 5. Analoguos to the calculation when approximating the transition state position of the H + H2 system, in the F + H2 reaction simulations were tested in order to acquire a constant bond length, at the maximum potential energy. As the first part of this reaction was exothermic, adopting Hammonds postulate, the assumption was made that the reaction had an early transition state thus the region near the reactants was analysed. The MEP calculation was used to remove the vibrational noise as it made it difficult to read the plots. As HF + H is a late transition state, resembling the its product F + H2, it will have the same transition state positions.

=== Report the activation energy for both reactions. ===
In order to find the activation energy the bond distance of H-F was first decreased. This then resulted in the energy falling to form the products H-F and H. The difference in this energy as shown in figure 7 was used to calculate the activation energy of the H-F + H endothermic reaction. The activation energy was estimated to be 120 kJ/mol. For the exothermic F + H2 reaction, the activation energy was calculated by trying to shift the trajectory towards to the reactants. This was done by thus increasing the F-H distance forming the F + H2 reactant. The difference in this energy was very small as shown in figure 8 and was used to calculate the activation energy for the F + H2 reaction.

=== 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. ===
Figure 8 illustrates the transition between translational and vibrational energy in the F + H2 reaction. Initially, the H-H bond (B-C) is vibrationally excited shown by the oscillation, whilst the H-F(A-B) bond isn't indicated by the absence of an oscillation. The H-F bond is at the vibrational ground state . After the collision we see a transition. The H-F bond now is vibrationally excited illustrated by the oscillation in the momenta vs time plot, and the H-H bonds loses this vibrationally energy and now only has translational energy due to the breaking of the bond, depicted by the plateau after 30s. Experimentally IR could be used to see the progress of the reaction. Only molecules with a dipole are IR active, thus after the formation of a H-F bond, IR absorption absent in the reactant will start to appear with overtones suggesting higher vibrational energies reached.

==== Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, with a momentum pFH = -1.0 g.mol-1.pm.fs-1, and explore several values of pHH in the range -6.1 to 6.1 g.mol-1.pm.fs-1 (explore values also close to these limits). What do you observe? Note that we are putting a significant amount of energy (much more than the activation energy) into the system on the H - H vibration. ====
When using the parameters above and setting the F-H value to 200 pm, at lower momentum values larger vibrational frequencies were obversed. The reactions did not therefore take place, satisfying Polyanyi's rules, where the having translational energy is more efficient when there is an early barrier rather than vibrational energy. What was interesting to see was that at higher momentum extremes, the vibrational energies were less pronounced, less frequent and the reaction proceeded further, however still did not proceed to completion (see figure 9). This further satisfies the Polyanyi's rules.

MRD:01524831.molecularreactiondynamics

2020-05-15T18:42:18Z

== Exercise 1: H + H2 system ==

[[File:Surface_Plot.png|thumb|Figure 1. A surface plot showing the transition state of a H + H2 system]]

=== 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? ===
On a potential energy surface diagram , the transition state is mathematically defined as the saddle point. The saddle point is a stationary point on the plot, where along the reaction path it is a maxima and perpendicular to the reaction path the saddle point is perceived to be a minima. The transition point can be identified in this simple example when the distances between the 3 hydrogen atoms are of equal distance. As shown in figure 1, the reaction path is oscillating at the saddle point. This depicts the vibrational freedom of the system at the transition state. A saddle point is found by...

=== 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 estimated '''rts '''is found to be around 91 pm. The transition state is defined as the position where internuclear distances are equal, and at a energy maxima in relation to the reaction path. From figure 1 it is evident that there are vibrations within the system, depicted by the oscillations on the ridge. The transition state position was found by determining which position the system stops oscillating, where the potential energy was a maximum, and thus the stable point of the transition state, the saddle point was found as shown in figure 2 and 3.

=== Comment on how the ''mep'' and the trajectory you just calculated differ. ===
When calculating under mep the reaction path is depicted different to when the calculation type is set to dynamics. What is absent in the map calculation is the motion of the atoms. The vibrational motion of the three hydrogen atoms are not depicted as shown in figure .... The initial condition is set so that the distance between AB is '''rts''' and the distance between BC deviates from '''rts '''by 1 pm, resulting in the reaction pathway following through, and advance towards the HA + HB-HC product. The flat BC line indicates the fact that the mep calculation does not consider vibrational motion of atoms.

==== Look at the “Internuclear Distances vs Time” and “Momenta vs Time”. What would change if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead? ====
The change of initial conditions to those above will change final product from HA + HB-HC to HA-HB + HC as it is the HB-HC bond distance that deviates away from the '''rts. '''As a result the internuclear distances vs time plot will now show AB to be constant over time. Momentum remains unchanged as initial momentum is 0 for both scenarios

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

The 'Get last geometry' option was selected, and the momenta were then reversed under the dynamic calculation type. The plot remains unchanged as expected. The reason for this is because the initial argument resulted in a trajectory towards a larger BC distance. Using the momenta of this trajectory but reversing the sign simply results in the hydrogen and hydrogen molecule to start in the final position, but now approach each other due to the negative momentum value.
=== 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"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot/kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Hydrogen atoms AB and C approach
eachother. Notably, the vibrational

motion present after the transition state is

not present before.
|[[File:Surface_Plot_question4_1_hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-4.1 </nowiki>
|<nowiki>-420.077</nowiki>
|No
|

Hydrogen atoms AB approaches C, however

before approaching the saddle point to AB

swiftly moves away from C

|[[File:Animation exercise2 p2 -4.1 hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|Hydrogen atom C approaches AB with a large
amount of momentum, resulting in a new BC Hydrogen

molecule forming, releasing the A hydrogen atom in the

reaction.
|[[File:Animation exercise2 p2 -5.1 p1 -3.1 hk3918.png|300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.1 </nowiki>
|<nowiki>-357.277</nowiki>
|No
|Hydrogen atoms AB and C approach each other very
quickly. Upon impact, large internuclear distance and velocities

are depicted, showing the B atom being tugged by both A and C.

In the end, B forms a molecule AB and moves away from C.
|[[File:Animation exercise2 p2 -10.1 hk3918.png |300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.6 </nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|Upon impact, the B atom is tugged between the two atoms A and
C, and is finally bonded to to atom C. A then moves away from

BC.
|[[File:Animation p2 -10.6 hk3918.png |300px]]
|}
The table above shows that a reaction isn't only reliant on the amount of kinetic energy in the system. As shown in the table above, there are instances where kinetic energy is higher than the activation energy, however no reaction takes place. One therefore needs to consider that successful collision are reliant on the system having enough energy to break and reform bonds, and also the orientation at which atoms collide.

=== Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===
Transition State Theory has a fundamental assumption which will result in its predictions being different to experimental values. The assumption that all trajectories possessing enough kinetic energy to circumvent the activation barrier will be reactive will result in a larger ''Rcl '' and thus a larger rate prediction than the experimental outcome. As seen in the table above, not all collisions are reactive even if there is sufficient kinetic energy - the case of barrier recrossing perfectly illustrates this.

== Exercise 2 ==

=== 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? ===
[[File:Surface_Plot_HF_hk3918.png|thumb|Figure 4. A surface plot showing the trajectory of F + H2 to H + HF reaction]|300px]]The reaction is exothermic as shown in figure 4, there is a decrease in potential energy when passing the transition state and forming the new HF bond. This relates to bond-strength, H-H has a bond strength of 432 kJ/mol whilst H-F has a bond strength of 565 kJ/mol.(reference). As HF is a polar bond as a result of the very electronegative F, there is an ionic element to the bond, resulting in a much greater attraction between the two atoms, thus releasing a lot of energy on formation. As the energy released is greater than that absorbed, the reaction is exothermic as depicted in the trajectory shown in figure 4.

[[File:Surface_Plot_hf-h_hk3918.png|thumb|Figure 5. A surface plot showing the trajectory of the FH + H to F + H2 reaction|300px]]

Figure 5 on the other hand shows an endothermic reaction, illustrated by the increase in potential energy along the trajectory, indicating that the energy absorbed to break the stronger H-F bond is greater than the energy released when forming the weaker H-H bond.

[[File:Plot small activation.png|thumb|Figure 6. Activation energy for the F +H2 reaction |300px]]

[[File:Plot activation big.png|thumb|Figure 7. Activation energy for the HF + H reaction |300px]]

[[File:Animation_momenta_hk3918.png |thumb|Figure 8. Momenta vs time plot for F + H2 reaction |300px]]

=== Locate the approximate position of the transition state ===
The transition state position was calculated to be when the H-F bond had a bond length of 181.8 pm, and the H-H bond was 73.1 pm as shown in figure 5. Analoguos to the calculation when approximating the transition state position of the H + H2 system, in the F + H2 reaction simulations were tested in order to acquire a constant bond length, at the maximum potential energy. As the first part of this reaction was exothermic, adopting Hammonds postulate, the assumption was made that the reaction had an early transition state thus the region near the reactants was analysed. The MEP calculation was used to remove the vibrational noise as it made it difficult to read the plots. As HF + H is a late transition state, resembling the its product F + H2, it will have the same transition state positions.

=== Report the activation energy for both reactions. ===
In order to find the activation energy the bond distance of H-F was first decreased. This then resulted in the energy falling to form the products H-F and H. The difference in this energy as shown in figure 7 was used to calculate the activation energy of the H-F + H endothermic reaction. The activation energy was estimated to be 120 kJ/mol. For the exothermic F + H2 reaction, the activation energy was calculated by trying to shift the trajectory towards to the reactants. This was done by thus increasing the F-H distance forming the F + H2 reactant. The difference in this energy was very small as shown in figure 8 and was used to calculate the activation energy for the F + H2 reaction.

=== 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. ===
Figure 8 illustrates the transition between translational and vibrational energy in the F + H2 reaction. Initially, the H-H bond (B-C) is vibrationally excited shown by the oscillation, whilst the H-F(A-B) bond isn't indicated by the absence of an oscillation. The H-F bond is at the vibrational ground state . After the collision we see a transition. The H-F bond now is vibrationally excited illustrated by the oscillation in the momenta vs time plot, and the H-H bonds loses this vibrationally energy and now only has translational energy due to the breaking of the bond, depicted by the plateau after 30s. Experimentally IR could be used to see the progress of the reaction. Only molecules with a dipole are IR active, thus after the formation of a H-F bond, IR absorption absent in the reactant will start to appear with overtones suggesting higher vibrational energies reached.

==== Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, with a momentum pFH = -1.0 g.mol-1.pm.fs-1, and explore several values of pHH in the range -6.1 to 6.1 g.mol-1.pm.fs-1 (explore values also close to these limits). What do you observe? Note that we are putting a significant amount of energy (much more than the activation energy) into the system on the H - H vibration. ====
When using the parameters above and setting the F-H value to 200 pm, at lower momentum values larger vibrational frequencies were obversed. The reactions did not therefore take place, satisfying Polyanyi's rules, where the having translational energy is more efficient when there is an early barrier rather than vibrational energy. What was interesting to see was that at higher momentum extremes, the vibrational energies were less pronounced, less frequent and the reaction proceeded further, however still did not proceed to completion (see figure 9). This further satisfies the Polyanyi's rules.

File:Animation -74 -6.1.png

2020-05-15T18:41:22Z

Hk3918: contour plot of F + H2

contour plot of F + H2

MRD:01524831.molecularreactiondynamics

2020-05-15T18:40:00Z

Hk3918:

== Exercise 1: H + H2 system ==

[[File:Surface_Plot.png|thumb|Figure 1. A surface plot showing the transition state of a H + H2 system]]

=== 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? ===
On a potential energy surface diagram , the transition state is mathematically defined as the saddle point. The saddle point is a stationary point on the plot, where along the reaction path it is a maxima and perpendicular to the reaction path the saddle point is perceived to be a minima. The transition point can be identified in this simple example when the distances between the 3 hydrogen atoms are of equal distance. As shown in figure 1, the reaction path is oscillating at the saddle point. This depicts the vibrational freedom of the system at the transition state. A saddle point is found by...

=== 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 estimated '''rts '''is found to be around 91 pm. The transition state is defined as the position where internuclear distances are equal, and at a energy maxima in relation to the reaction path. From figure 1 it is evident that there are vibrations within the system, depicted by the oscillations on the ridge. The transition state position was found by determining which position the system stops oscillating, where the potential energy was a maximum, and thus the stable point of the transition state, the saddle point was found as shown in figure 2 and 3.

=== Comment on how the ''mep'' and the trajectory you just calculated differ. ===
When calculating under mep the reaction path is depicted different to when the calculation type is set to dynamics. What is absent in the map calculation is the motion of the atoms. The vibrational motion of the three hydrogen atoms are not depicted as shown in figure .... The initial condition is set so that the distance between AB is '''rts''' and the distance between BC deviates from '''rts '''by 1 pm, resulting in the reaction pathway following through, and advance towards the HA + HB-HC product. The flat BC line indicates the fact that the mep calculation does not consider vibrational motion of atoms.

==== Look at the “Internuclear Distances vs Time” and “Momenta vs Time”. What would change if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead? ====
The change of initial conditions to those above will change final product from HA + HB-HC to HA-HB + HC as it is the HB-HC bond distance that deviates away from the '''rts. '''As a result the internuclear distances vs time plot will now show AB to be constant over time. Momentum remains unchanged as initial momentum is 0 for both scenarios

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

The 'Get last geometry' option was selected, and the momenta were then reversed under the dynamic calculation type. The plot remains unchanged as expected. The reason for this is because the initial argument resulted in a trajectory towards a larger BC distance. Using the momenta of this trajectory but reversing the sign simply results in the hydrogen and hydrogen molecule to start in the final position, but now approach each other due to the negative momentum value.
=== 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"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot/kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Hydrogen atoms AB and C approach
eachother. Notably, the vibrational

motion present after the transition state is

not present before.
|[[File:Surface_Plot_question4_1_hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-4.1 </nowiki>
|<nowiki>-420.077</nowiki>
|No
|

Hydrogen atoms AB approaches C, however

before approaching the saddle point to AB

swiftly moves away from C

|[[File:Animation exercise2 p2 -4.1 hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|Hydrogen atom C approaches AB with a large
amount of momentum, resulting in a new BC Hydrogen

molecule forming, releasing the A hydrogen atom in the

reaction.
|[[File:Animation exercise2 p2 -5.1 p1 -3.1 hk3918.png|300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.1 </nowiki>
|<nowiki>-357.277</nowiki>
|No
|Hydrogen atoms AB and C approach each other very
quickly. Upon impact, large internuclear distance and velocities

are depicted, showing the B atom being tugged by both A and C.

In the end, B forms a molecule AB and moves away from C.
|[[File:Animation exercise2 p2 -10.1 hk3918.png |300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.6 </nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|Upon impact, the B atom is tugged between the two atoms A and
C, and is finally bonded to to atom C. A then moves away from

BC.
|[[File:Animation p2 -10.6 hk3918.png |300px]]
|}
The table above shows that a reaction isn't only reliant on the amount of kinetic energy in the system. As shown in the table above, there are instances where kinetic energy is higher than the activation energy, however no reaction takes place. One therefore needs to consider that successful collision are reliant on the system having enough energy to break and reform bonds, and also the orientation at which atoms collide.

=== Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===
Transition State Theory has a fundamental assumption which will result in its predictions being different to experimental values. The assumption that all trajectories possessing enough kinetic energy to circumvent the activation barrier will be reactive will result in a larger ''Rcl '' and thus a larger rate prediction than the experimental outcome. As seen in the table above, not all collisions are reactive even if there is sufficient kinetic energy - the case of barrier recrossing perfectly illustrates this.

== Exercise 2 ==

=== 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? ===
[[File:Surface_Plot_HF_hk3918.png|thumb|Figure 4. A surface plot showing the trajectory of F + H2 to H + HF reaction]|300px]]The reaction is exothermic as shown in figure 4, there is a decrease in potential energy when passing the transition state and forming the new HF bond. This relates to bond-strength, H-H has a bond strength of 432 kJ/mol whilst H-F has a bond strength of 565 kJ/mol.(reference). As HF is a polar bond as a result of the very electronegative F, there is an ionic element to the bond, resulting in a much greater attraction between the two atoms, thus releasing a lot of energy on formation. As the energy released is greater than that absorbed, the reaction is exothermic as depicted in the trajectory shown in figure 4.

[[File:Surface_Plot_hf-h_hk3918.png|thumb|Figure 5. A surface plot showing the trajectory of the FH + H to F + H2 reaction|300px]]

Figure 5 on the other hand shows an endothermic reaction, illustrated by the increase in potential energy along the trajectory, indicating that the energy absorbed to break the stronger H-F bond is greater than the energy released when forming the weaker H-H bond.

[[File:Plot small activation.png|thumb|Figure 6. Activation energy for the F +H2 reaction |300px]]

[[File:Plot activation big.png|thumb|Figure 7. Activation energy for the HF + H reaction |300px]]

[[File:Animation_momenta_hk3918.png |thumb|Figure 8. Momenta vs time plot for F + H2 reaction |300px]]

=== Locate the approximate position of the transition state ===
The transition state position was calculated to be when the H-F bond had a bond length of 181.8 pm, and the H-H bond was 73.1 pm as shown in figure 5. Analoguos to the calculation when approximating the transition state position of the H + H2 system, in the F + H2 reaction simulations were tested in order to acquire a constant bond length, at the maximum potential energy. As the first part of this reaction was exothermic, adopting Hammonds postulate, the assumption was made that the reaction had an early transition state thus the region near the reactants was analysed. The MEP calculation was used to remove the vibrational noise as it made it difficult to read the plots. As HF + H is a late transition state, resembling the its product F + H2, it will have the same transition state positions.

=== Report the activation energy for both reactions. ===
In order to find the activation energy the bond distance of H-F was first decreased. This then resulted in the energy falling to form the products H-F and H. The difference in this energy as shown in figure 7 was used to calculate the activation energy of the H-F + H endothermic reaction. The activation energy was estimated to be 120 kJ/mol. For the exothermic F + H2 reaction, the activation energy was calculated by trying to shift the trajectory towards to the reactants. This was done by thus increasing the F-H distance forming the F + H2 reactant. The difference in this energy was very small as shown in figure 8 and was used to calculate the activation energy for the F + H2 reaction.

=== 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. ===
Figure 8 illustrates the transition between translational and vibrational energy in the F + H2 reaction. Initially, the H-H bond (B-C) is vibrationally excited shown by the oscillation, whilst the H-F(A-B) bond isn't indicated by the absence of an oscillation. The H-F bond is at the vibrational ground state . After the collision we see a transition. The H-F bond now is vibrationally excited illustrated by the oscillation in the momenta vs time plot, and the H-H bonds loses this vibrationally energy and now only has translational energy due to the breaking of the bond, depicted by the plateau after 30s. Experimentally IR could be used to see the progress of the reaction. Only molecules with a dipole are IR active, thus after the formation of a H-F bond, IR absorption absent in the reactant will start to appear with overtones suggesting higher vibrational energies reached.

==== Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, with a momentum pFH = -1.0 g.mol-1.pm.fs-1, and explore several values of pHH in the range -6.1 to 6.1 g.mol-1.pm.fs-1 (explore values also close to these limits). What do you observe? Note that we are putting a significant amount of energy (much more than the activation energy) into the system on the H - H vibration. ====
When using the parameters above and setting the F-H value to 200 pm, at lower momentum values larger vibrational frequencies were obversed. The reactions did not therefore take place, satisfying Polyanyi's rules, where the having translational energy is more efficient when there is an early barrier rather than vibrational energy. What was interesting to see was that at higher momentum extremes, the vibrational energies were less pronounced, less frequent and the reaction proceeded further, however did not proceed to completion. This further satisfies the Polyanyi's rules.

MRD:01524831.molecularreactiondynamics

2020-05-15T17:56:40Z

Hk3918: /* 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. */

== Exercise 1: H + H2 system ==

[[File:Surface_Plot.png|thumb|Figure 1. A surface plot showing the transition state of a H + H2 system]]

=== 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? ===
On a potential energy surface diagram , the transition state is mathematically defined as the saddle point. The saddle point is a stationary point on the plot, where along the reaction path it is a maxima and perpendicular to the reaction path the saddle point is perceived to be a minima. The transition point can be identified in this simple example when the distances between the 3 hydrogen atoms are of equal distance. As shown in figure 1, the reaction path is oscillating at the saddle point. This depicts the vibrational freedom of the system at the transition state. A saddle point is found by...

=== 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 estimated '''rts '''is found to be around 91 pm. The transition state is defined as the position where internuclear distances are equal, and at a energy maxima in relation to the reaction path. From figure 1 it is evident that there are vibrations within the system, depicted by the oscillations on the ridge. The transition state position was found by determining which position the system stops oscillating, where the potential energy was a maximum, and thus the stable point of the transition state, the saddle point was found as shown in figure 2 and 3.

=== Comment on how the ''mep'' and the trajectory you just calculated differ. ===
When calculating under mep the reaction path is depicted different to when the calculation type is set to dynamics. What is absent in the map calculation is the motion of the atoms. The vibrational motion of the three hydrogen atoms are not depicted as shown in figure .... The initial condition is set so that the distance between AB is '''rts''' and the distance between BC deviates from '''rts '''by 1 pm, resulting in the reaction pathway following through, and advance towards the HA + HB-HC product. The flat BC line indicates the fact that the mep calculation does not consider vibrational motion of atoms.

==== Look at the “Internuclear Distances vs Time” and “Momenta vs Time”. What would change if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead? ====
The change of initial conditions to those above will change final product from HA + HB-HC to HA-HB + HC as it is the HB-HC bond distance that deviates away from the '''rts. '''As a result the internuclear distances vs time plot will now show AB to be constant over time. Momentum remains unchanged as initial momentum is 0 for both scenarios

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

The 'Get last geometry' option was selected, and the momenta were then reversed under the dynamic calculation type. The plot remains unchanged as expected. The reason for this is because the initial argument resulted in a trajectory towards a larger BC distance. Using the momenta of this trajectory but reversing the sign simply results in the hydrogen and hydrogen molecule to start in the final position, but now approach each other due to the negative momentum value.
=== 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"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot/kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Hydrogen atoms AB and C approach
eachother. Notably, the vibrational

motion present after the transition state is

not present before.
|[[File:Surface_Plot_question4_1_hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-4.1 </nowiki>
|<nowiki>-420.077</nowiki>
|No
|

Hydrogen atoms AB approaches C, however

before approaching the saddle point to AB

swiftly moves away from C

|[[File:Animation exercise2 p2 -4.1 hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|Hydrogen atom C approaches AB with a large
amount of momentum, resulting in a new BC Hydrogen

molecule forming, releasing the A hydrogen atom in the

reaction.
|[[File:Animation exercise2 p2 -5.1 p1 -3.1 hk3918.png|300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.1 </nowiki>
|<nowiki>-357.277</nowiki>
|No
|Hydrogen atoms AB and C approach each other very
quickly. Upon impact, large internuclear distance and velocities

are depicted, showing the B atom being tugged by both A and C.

In the end, B forms a molecule AB and moves away from C.
|[[File:Animation exercise2 p2 -10.1 hk3918.png |300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.6 </nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|Upon impact, the B atom is tugged between the two atoms A and
C, and is finally bonded to to atom C. A then moves away from

BC.
|[[File:Animation p2 -10.6 hk3918.png |300px]]
|}
The table above shows that a reaction isn't only reliant on the amount of kinetic energy in the system. As shown in the table above, there are instances where kinetic energy is higher than the activation energy, however no reaction takes place. One therefore needs to consider that successful collision are reliant on the system having enough energy to break and reform bonds, and also the orientation at which atoms collide.

=== Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===
Transition State Theory has a fundamental assumption which will result in its predictions being different to experimental values. The assumption that all trajectories possessing enough kinetic energy to circumvent the activation barrier will be reactive will result in a larger ''Rcl '' and thus a larger rate prediction than the experimental outcome. As seen in the table above, not all collisions are reactive even if there is sufficient kinetic energy - the case of barrier recrossing perfectly illustrates this.

== Exercise 2 ==

=== 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? ===
[[File:Surface_Plot_HF_hk3918.png|thumb|Figure 4. A surface plot showing the trajectory of F + H2 to H + HF reaction]|300px]]The reaction is exothermic as shown in figure 4, there is a decrease in potential energy when passing the transition state and forming the new HF bond. This relates to bond-strength, H-H has a bond strength of 432 kJ/mol whilst H-F has a bond strength of 565 kJ/mol.(reference). As HF is a polar bond as a result of the very electronegative F, there is an ionic element to the bond, resulting in a much greater attraction between the two atoms, thus releasing a lot of energy on formation. As the energy released is greater than that absorbed, the reaction is exothermic as depicted in the trajectory shown in figure 4.

[[File:Surface_Plot_hf-h_hk3918.png|thumb|Figure 5. A surface plot showing the trajectory of the FH + H to F + H2 reaction|300px]]

Figure 5 on the other hand shows an endothermic reaction, illustrated by the increase in potential energy along the trajectory, indicating that the energy absorbed to break the stronger H-F bond is greater than the energy released when forming the weaker H-H bond.

[[File:Plot small activation.png|thumb|Figure 6. Activation energy for the F +H2 reaction |300px]]

[[File:Plot activation big.png|thumb|Figure 7. Activation energy for the HF + H reaction |300px]]

[[File:Animation_momenta_hk3918.png |thumb|Figure 8. Momenta vs time plot for F + H2 reaction |300px]]

=== Locate the approximate position of the transition state ===
The transition state position was calculated to be when the H-F bond had a bond length of 181.8 pm, and the H-H bond was 73.1 pm as shown in figure 5. Analoguos to the calculation when approximating the transition state position of the H + H2 system, in the F + H2 reaction simulations were tested in order to acquire a constant bond length, at the maximum potential energy. As the first part of this reaction was exothermic, adopting Hammonds postulate, the assumption was made that the reaction had an early transition state thus the region near the reactants was analysed. The MEP calculation was used to remove the vibrational noise as it made it difficult to read the plots. As HF + H is a late transition state, resembling the its product F + H2, it will have the same transition state positions.

=== Report the activation energy for both reactions. ===
In order to find the activation energy the bond distance of H-F was first decreased. This then resulted in the energy falling to form the products H-F and H. The difference in this energy as shown in figure 7 was used to calculate the activation energy of the H-F + H endothermic reaction. The activation energy was estimated to be 120 kJ/mol. For the exothermic F + H2 reaction, the activation energy was calculated by trying to shift the trajectory towards to the reactants. This was done by thus increasing the F-H distance forming the F + H2 reactant. The difference in this energy was very small as shown in figure 8 and was used to calculate the activation energy for the F + H2 reaction.

=== 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. ===
Figure 8 illustrates the transition between translational and vibrational energy in the F + H2 reaction. Initially, the H-H bond (B-C) is vibrationally excited shown by the oscillation, whilst the H-F(A-B) bond isn't indicated by the absence of an oscillation. The H-F bond is at the vibrational ground state . After the collision we see a transition. The H-F bond now is vibrationally excited illustrated by the oscillation in the momenta vs time plot, and the H-H bonds loses this vibrationally energy and now only has translational energy due to the breaking of the bond, depicted by the plateau after 30s. Experimentally IR could be used to see the progress of the reaction. Only molecules with a dipole are IR active, thus after the formation of a H-F bond, IR absorption absent in the reactant will start to appear with overtones suggesting higher vibrational energies reached.

MRD:01524831.molecularreactiondynamics

2020-05-15T17:54:20Z

== Exercise 1: H + H2 system ==

[[File:Surface_Plot.png|thumb|Figure 1. A surface plot showing the transition state of a H + H2 system]]

=== 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? ===
On a potential energy surface diagram , the transition state is mathematically defined as the saddle point. The saddle point is a stationary point on the plot, where along the reaction path it is a maxima and perpendicular to the reaction path the saddle point is perceived to be a minima. The transition point can be identified in this simple example when the distances between the 3 hydrogen atoms are of equal distance. As shown in figure 1, the reaction path is oscillating at the saddle point. This depicts the vibrational freedom of the system at the transition state. A saddle point is found by...

=== 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 estimated '''rts '''is found to be around 91 pm. The transition state is defined as the position where internuclear distances are equal, and at a energy maxima in relation to the reaction path. From figure 1 it is evident that there are vibrations within the system, depicted by the oscillations on the ridge. The transition state position was found by determining which position the system stops oscillating, where the potential energy was a maximum, and thus the stable point of the transition state, the saddle point was found as shown in figure 2 and 3.

=== Comment on how the ''mep'' and the trajectory you just calculated differ. ===
When calculating under mep the reaction path is depicted different to when the calculation type is set to dynamics. What is absent in the map calculation is the motion of the atoms. The vibrational motion of the three hydrogen atoms are not depicted as shown in figure .... The initial condition is set so that the distance between AB is '''rts''' and the distance between BC deviates from '''rts '''by 1 pm, resulting in the reaction pathway following through, and advance towards the HA + HB-HC product. The flat BC line indicates the fact that the mep calculation does not consider vibrational motion of atoms.

==== Look at the “Internuclear Distances vs Time” and “Momenta vs Time”. What would change if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead? ====
The change of initial conditions to those above will change final product from HA + HB-HC to HA-HB + HC as it is the HB-HC bond distance that deviates away from the '''rts. '''As a result the internuclear distances vs time plot will now show AB to be constant over time. Momentum remains unchanged as initial momentum is 0 for both scenarios

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

The 'Get last geometry' option was selected, and the momenta were then reversed under the dynamic calculation type. The plot remains unchanged as expected. The reason for this is because the initial argument resulted in a trajectory towards a larger BC distance. Using the momenta of this trajectory but reversing the sign simply results in the hydrogen and hydrogen molecule to start in the final position, but now approach each other due to the negative momentum value.
=== 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"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot/kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Hydrogen atoms AB and C approach
eachother. Notably, the vibrational

motion present after the transition state is

not present before.
|[[File:Surface_Plot_question4_1_hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-4.1 </nowiki>
|<nowiki>-420.077</nowiki>
|No
|

Hydrogen atoms AB approaches C, however

before approaching the saddle point to AB

swiftly moves away from C

|[[File:Animation exercise2 p2 -4.1 hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|Hydrogen atom C approaches AB with a large
amount of momentum, resulting in a new BC Hydrogen

molecule forming, releasing the A hydrogen atom in the

reaction.
|[[File:Animation exercise2 p2 -5.1 p1 -3.1 hk3918.png|300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.1 </nowiki>
|<nowiki>-357.277</nowiki>
|No
|Hydrogen atoms AB and C approach each other very
quickly. Upon impact, large internuclear distance and velocities

are depicted, showing the B atom being tugged by both A and C.

In the end, B forms a molecule AB and moves away from C.
|[[File:Animation exercise2 p2 -10.1 hk3918.png |300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.6 </nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|Upon impact, the B atom is tugged between the two atoms A and
C, and is finally bonded to to atom C. A then moves away from

BC.
|[[File:Animation p2 -10.6 hk3918.png |300px]]
|}
The table above shows that a reaction isn't only reliant on the amount of kinetic energy in the system. As shown in the table above, there are instances where kinetic energy is higher than the activation energy, however no reaction takes place. One therefore needs to consider that successful collision are reliant on the system having enough energy to break and reform bonds, and also the orientation at which atoms collide.

=== Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===
Transition State Theory has a fundamental assumption which will result in its predictions being different to experimental values. The assumption that all trajectories possessing enough kinetic energy to circumvent the activation barrier will be reactive will result in a larger ''Rcl '' and thus a larger rate prediction than the experimental outcome. As seen in the table above, not all collisions are reactive even if there is sufficient kinetic energy - the case of barrier recrossing perfectly illustrates this.

== Exercise 2 ==

=== 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? ===
[[File:Surface_Plot_HF_hk3918.png|thumb|Figure 4. A surface plot showing the trajectory of F + H2 to H + HF reaction]|300px]]The reaction is exothermic as shown in figure 4, there is a decrease in potential energy when passing the transition state and forming the new HF bond. This relates to bond-strength, H-H has a bond strength of 432 kJ/mol whilst H-F has a bond strength of 565 kJ/mol.(reference). As HF is a polar bond as a result of the very electronegative F, there is an ionic element to the bond, resulting in a much greater attraction between the two atoms, thus releasing a lot of energy on formation. As the energy released is greater than that absorbed, the reaction is exothermic as depicted in the trajectory shown in figure 4.

[[File:Surface_Plot_hf-h_hk3918.png|thumb|Figure 5. A surface plot showing the trajectory of the FH + H to F + H2 reaction|300px]]

Figure 5 on the other hand shows an endothermic reaction, illustrated by the increase in potential energy along the trajectory, indicating that the energy absorbed to break the stronger H-F bond is greater than the energy released when forming the weaker H-H bond.

[[File:Plot small activation.png|thumb|Figure 6. Activation energy for the F +H2 reaction |300px]]

[[File:Plot activation big.png|thumb|Figure 7. Activation energy for the HF + H reaction |300px]]

[[File:Animation_momenta_hk3918.png |thumb|Figure 8. Momenta vs time plot for F + H2 reaction |300px]]

=== Locate the approximate position of the transition state ===
The transition state position was calculated to be when the H-F bond had a bond length of 181.8 pm, and the H-H bond was 73.1 pm as shown in figure 5. Analoguos to the calculation when approximating the transition state position of the H + H2 system, in the F + H2 reaction simulations were tested in order to acquire a constant bond length, at the maximum potential energy. As the first part of this reaction was exothermic, adopting Hammonds postulate, the assumption was made that the reaction had an early transition state thus the region near the reactants was analysed. The MEP calculation was used to remove the vibrational noise as it made it difficult to read the plots. As HF + H is a late transition state, resembling the its product F + H2, it will have the same transition state positions.

=== Report the activation energy for both reactions. ===
In order to find the activation energy the bond distance of H-F was first decreased. This then resulted in the energy falling to form the products H-F and H. The difference in this energy as shown in figure 7 was used to calculate the activation energy of the H-F + H endothermic reaction. The activation energy was estimated to be 120 kJ/mol. For the exothermic F + H2 reaction, the activation energy was calculated by trying to shift the trajectory towards to the reactants. This was done by thus increasing the F-H distance forming the F + H2 reactant. The difference in this energy was very small as shown in figure 8 and was used to calculate the activation energy for the F + H2 reaction.

=== 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. ===
Figure 8 illustrates the transition between translational and vibrational energy in the F + H2 reaction. Initially, the H-H bond (B-C) is vibrationally excited shown by the oscillation, whilst the H-F(A-B) bond isn't indicated by the absence of an oscillation. The H-F bond is at the vibrational ground state . After the collision we see a transition. The H-F bond now is vibrationally excited illustrated by the oscillation in the momenta vs time plot, and the H-H bonds loses this vibrationally energy and now only has translational energy due to the breaking of the bond, depicted by the plateau after 30s. Experimentally IR could be used to see the progress of the reaction. Only molecules with a dipole are IR active, thus after the formation of a H-F bond, IR absorption absent in the reactant will start to appear.

MRD:01524831.molecularreactiondynamics

2020-05-15T17:53:56Z

== Exercise 1: H + H2 system ==

[[File:Surface_Plot.png|thumb|Figure 1. A surface plot showing the transition state of a H + H2 system]]

=== 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? ===
On a potential energy surface diagram , the transition state is mathematically defined as the saddle point. The saddle point is a stationary point on the plot, where along the reaction path it is a maxima and perpendicular to the reaction path the saddle point is perceived to be a minima. The transition point can be identified in this simple example when the distances between the 3 hydrogen atoms are of equal distance. As shown in figure 1, the reaction path is oscillating at the saddle point. This depicts the vibrational freedom of the system at the transition state. A saddle point is found by...

=== 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 estimated '''rts '''is found to be around 91 pm. The transition state is defined as the position where internuclear distances are equal, and at a energy maxima in relation to the reaction path. From figure 1 it is evident that there are vibrations within the system, depicted by the oscillations on the ridge. The transition state position was found by determining which position the system stops oscillating, where the potential energy was a maximum, and thus the stable point of the transition state, the saddle point was found as shown in figure 2 and 3.

=== Comment on how the ''mep'' and the trajectory you just calculated differ. ===
When calculating under mep the reaction path is depicted different to when the calculation type is set to dynamics. What is absent in the map calculation is the motion of the atoms. The vibrational motion of the three hydrogen atoms are not depicted as shown in figure .... The initial condition is set so that the distance between AB is '''rts''' and the distance between BC deviates from '''rts '''by 1 pm, resulting in the reaction pathway following through, and advance towards the HA + HB-HC product. The flat BC line indicates the fact that the mep calculation does not consider vibrational motion of atoms.

==== Look at the “Internuclear Distances vs Time” and “Momenta vs Time”. What would change if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead? ====
The change of initial conditions to those above will change final product from HA + HB-HC to HA-HB + HC as it is the HB-HC bond distance that deviates away from the '''rts. '''As a result the internuclear distances vs time plot will now show AB to be constant over time. Momentum remains unchanged as initial momentum is 0 for both scenarios

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

The 'Get last geometry' option was selected, and the momenta were then reversed under the dynamic calculation type. The plot remains unchanged as expected. The reason for this is because the initial argument resulted in a trajectory towards a larger BC distance. Using the momenta of this trajectory but reversing the sign simply results in the hydrogen and hydrogen molecule to start in the final position, but now approach each other due to the negative momentum value.
=== 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"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot/kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Hydrogen atoms AB and C approach
eachother. Notably, the vibrational

motion present after the transition state is

not present before.
|[[File:Surface_Plot_question4_1_hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-4.1 </nowiki>
|<nowiki>-420.077</nowiki>
|No
|

Hydrogen atoms AB approaches C, however

before approaching the saddle point to AB

swiftly moves away from C

|[[File:Animation exercise2 p2 -4.1 hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|Hydrogen atom C approaches AB with a large
amount of momentum, resulting in a new BC Hydrogen

molecule forming, releasing the A hydrogen atom in the

reaction.
|[[File:Animation exercise2 p2 -5.1 p1 -3.1 hk3918.png|300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.1 </nowiki>
|<nowiki>-357.277</nowiki>
|No
|Hydrogen atoms AB and C approach each other very
quickly. Upon impact, large internuclear distance and velocities

are depicted, showing the B atom being tugged by both A and C.

In the end, B forms a molecule AB and moves away from C.
|[[File:Animation exercise2 p2 -10.1 hk3918.png |300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.6 </nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|Upon impact, the B atom is tugged between the two atoms A and
C, and is finally bonded to to atom C. A then moves away from

BC.
|[[File:Animation p2 -10.6 hk3918.png |300px]]
|}
The table above shows that a reaction isn't only reliant on the amount of kinetic energy in the system. As shown in the table above, there are instances where kinetic energy is higher than the activation energy, however no reaction takes place. One therefore needs to consider that successful collision are reliant on the system having enough energy to break and reform bonds, and also the orientation at which atoms collide.

=== Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===
Transition State Theory has a fundamental assumption which will result in its predictions being different to experimental values. The assumption that all trajectories possessing enough kinetic energy to circumvent the activation barrier will be reactive will result in a larger ''Rcl '' and thus a larger rate prediction than the experimental outcome. As seen in the table above, not all collisions are reactive even if there is sufficient kinetic energy - the case of barrier recrossing perfectly illustrates this.

== Exercise 2 ==

=== 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? ===
[[File:Surface_Plot_HF_hk3918.png|thumb|Figure 4. A surface plot showing the trajectory of F + H2 to H + HF reaction]|300px]]The reaction is exothermic as shown in figure 4, there is a decrease in potential energy when passing the transition state and forming the new HF bond. This relates to bond-strength, H-H has a bond strength of 432 kJ/mol whilst H-F has a bond strength of 565 kJ/mol.(reference). As HF is a polar bond as a result of the very electronegative F, there is an ionic element to the bond, resulting in a much greater attraction between the two atoms, thus releasing a lot of energy on formation. As the energy released is greater than that absorbed, the reaction is exothermic as depicted in the trajectory shown in figure 4.

[[File:Surface_Plot_hf-h_hk3918.png|thumb|Figure 5. A surface plot showing the trajectory of the FH + H to F + H2 reaction|300px]]

Figure 5 on the other hand shows an endothermic reaction, illustrated by the increase in potential energy along the trajectory, indicating that the energy absorbed to break the stronger H-F bond is greater than the energy released when forming the weaker H-H bond.

[[File:Plot small activation.png|thumb|Figure 6. Activation energy for the F +H2 reaction |300px]]

[[File:Plot activation big.png|thumb|Figure 7. Activation energy for the HF + H reaction |300px]]

[[File:Plot activation big.png|thumb|Figure 8. Momenta vs time plot for F + H2 reaction |300px]]

=== Locate the approximate position of the transition state ===
The transition state position was calculated to be when the H-F bond had a bond length of 181.8 pm, and the H-H bond was 73.1 pm as shown in figure 5. Analoguos to the calculation when approximating the transition state position of the H + H2 system, in the F + H2 reaction simulations were tested in order to acquire a constant bond length, at the maximum potential energy. As the first part of this reaction was exothermic, adopting Hammonds postulate, the assumption was made that the reaction had an early transition state thus the region near the reactants was analysed. The MEP calculation was used to remove the vibrational noise as it made it difficult to read the plots. As HF + H is a late transition state, resembling the its product F + H2, it will have the same transition state positions.

=== Report the activation energy for both reactions. ===
In order to find the activation energy the bond distance of H-F was first decreased. This then resulted in the energy falling to form the products H-F and H. The difference in this energy as shown in figure 7 was used to calculate the activation energy of the H-F + H endothermic reaction. The activation energy was estimated to be 120 kJ/mol. For the exothermic F + H2 reaction, the activation energy was calculated by trying to shift the trajectory towards to the reactants. This was done by thus increasing the F-H distance forming the F + H2 reactant. The difference in this energy was very small as shown in figure 8 and was used to calculate the activation energy for the F + H2 reaction.

=== 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. ===
Figure 8 illustrates the transition between translational and vibrational energy in the F + H2 reaction. Initially, the H-H bond (B-C) is vibrationally excited shown by the oscillation, whilst the H-F(A-B) bond isn't indicated by the absence of an oscillation. The H-F bond is at the vibrational ground state . After the collision we see a transition. The H-F bond now is vibrationally excited illustrated by the oscillation in the momenta vs time plot, and the H-H bonds loses this vibrationally energy and now only has translational energy due to the breaking of the bond, depicted by the plateau after 30s. Experimentally IR could be used to see the progress of the reaction. Only molecules with a dipole are IR active, thus after the formation of a H-F bond, IR absorption absent in the reactant will start to appear.

MRD:01524831.molecularreactiondynamics

2020-05-15T17:53:12Z

Hk3918: /* Report the activation energy for both reactions. */

== Exercise 1: H + H2 system ==

[[File:Surface_Plot.png|thumb|Figure 1. A surface plot showing the transition state of a H + H2 system]]

=== 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? ===
On a potential energy surface diagram , the transition state is mathematically defined as the saddle point. The saddle point is a stationary point on the plot, where along the reaction path it is a maxima and perpendicular to the reaction path the saddle point is perceived to be a minima. The transition point can be identified in this simple example when the distances between the 3 hydrogen atoms are of equal distance. As shown in figure 1, the reaction path is oscillating at the saddle point. This depicts the vibrational freedom of the system at the transition state. A saddle point is found by...

=== 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 estimated '''rts '''is found to be around 91 pm. The transition state is defined as the position where internuclear distances are equal, and at a energy maxima in relation to the reaction path. From figure 1 it is evident that there are vibrations within the system, depicted by the oscillations on the ridge. The transition state position was found by determining which position the system stops oscillating, where the potential energy was a maximum, and thus the stable point of the transition state, the saddle point was found as shown in figure 2 and 3.

=== Comment on how the ''mep'' and the trajectory you just calculated differ. ===
When calculating under mep the reaction path is depicted different to when the calculation type is set to dynamics. What is absent in the map calculation is the motion of the atoms. The vibrational motion of the three hydrogen atoms are not depicted as shown in figure .... The initial condition is set so that the distance between AB is '''rts''' and the distance between BC deviates from '''rts '''by 1 pm, resulting in the reaction pathway following through, and advance towards the HA + HB-HC product. The flat BC line indicates the fact that the mep calculation does not consider vibrational motion of atoms.

==== Look at the “Internuclear Distances vs Time” and “Momenta vs Time”. What would change if we used the initial conditions '''r1''' = '''rts''' and '''r2''' = '''rts'''+1 pm instead? ====
The change of initial conditions to those above will change final product from HA + HB-HC to HA-HB + HC as it is the HB-HC bond distance that deviates away from the '''rts. '''As a result the internuclear distances vs time plot will now show AB to be constant over time. Momentum remains unchanged as initial momentum is 0 for both scenarios

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

The 'Get last geometry' option was selected, and the momenta were then reversed under the dynamic calculation type. The plot remains unchanged as expected. The reason for this is because the initial argument resulted in a trajectory towards a larger BC distance. Using the momenta of this trajectory but reversing the sign simply results in the hydrogen and hydrogen molecule to start in the final position, but now approach each other due to the negative momentum value.
=== 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"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot/kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Hydrogen atoms AB and C approach
eachother. Notably, the vibrational

motion present after the transition state is

not present before.
|[[File:Surface_Plot_question4_1_hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-4.1 </nowiki>
|<nowiki>-420.077</nowiki>
|No
|

Hydrogen atoms AB approaches C, however

before approaching the saddle point to AB

swiftly moves away from C

|[[File:Animation exercise2 p2 -4.1 hk3918.png|300px]]
|-
|<nowiki>-3.1 </nowiki>
|<nowiki>-5.1 </nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|Hydrogen atom C approaches AB with a large
amount of momentum, resulting in a new BC Hydrogen

molecule forming, releasing the A hydrogen atom in the

reaction.
|[[File:Animation exercise2 p2 -5.1 p1 -3.1 hk3918.png|300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.1 </nowiki>
|<nowiki>-357.277</nowiki>
|No
|Hydrogen atoms AB and C approach each other very
quickly. Upon impact, large internuclear distance and velocities

are depicted, showing the B atom being tugged by both A and C.

In the end, B forms a molecule AB and moves away from C.
|[[File:Animation exercise2 p2 -10.1 hk3918.png |300px]]
|-
|<nowiki>-5.1 </nowiki>
|<nowiki>-10.6 </nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|Upon impact, the B atom is tugged between the two atoms A and
C, and is finally bonded to to atom C. A then moves away from

BC.
|[[File:Animation p2 -10.6 hk3918.png |300px]]
|}
The table above shows that a reaction isn't only reliant on the amount of kinetic energy in the system. As shown in the table above, there are instances where kinetic energy is higher than the activation energy, however no reaction takes place. One therefore needs to consider that successful collision are reliant on the system having enough energy to break and reform bonds, and also the orientation at which atoms collide.

=== Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===
Transition State Theory has a fundamental assumption which will result in its predictions being different to experimental values. The assumption that all trajectories possessing enough kinetic energy to circumvent the activation barrier will be reactive will result in a larger ''Rcl '' and thus a larger rate prediction than the experimental outcome. As seen in the table above, not all collisions are reactive even if there is sufficient kinetic energy - the case of barrier recrossing perfectly illustrates this.

== Exercise 2 ==

=== 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? ===
[[File:Surface_Plot_HF_hk3918.png|thumb|Figure 4. A surface plot showing the trajectory of F + H2 to H + HF reaction]|300px]]The reaction is exothermic as shown in figure 4, there is a decrease in potential energy when passing the transition state and forming the new HF bond. This relates to bond-strength, H-H has a bond strength of 432 kJ/mol whilst H-F has a bond strength of 565 kJ/mol.(reference). As HF is a polar bond as a result of the very electronegative F, there is an ionic element to the bond, resulting in a much greater attraction between the two atoms, thus releasing a lot of energy on formation. As the energy released is greater than that absorbed, the reaction is exothermic as depicted in the trajectory shown in figure 4.

[[File:Surface_Plot_hf-h_hk3918.png|thumb|Figure 5. A surface plot showing the trajectory of the FH + H to F + H2 reaction|300px]]

Figure 5 on the other hand shows an endothermic reaction, illustrated by the increase in potential energy along the trajectory, indicating that the energy absorbed to break the stronger H-F bond is greater than the energy released when forming the weaker H-H bond.

[[File:Plot small activation.png|thumb|Figure 6. Activation energy for the F +H2 reaction |300px]]

[[File:Plot activation big.png|thumb|Figure 7. Activation energy for the HF + H reaction |300px]]

=== Locate the approximate position of the transition state ===
The transition state position was calculated to be when the H-F bond had a bond length of 181.8 pm, and the H-H bond was 73.1 pm as shown in figure 5. Analoguos to the calculation when approximating the transition state position of the H + H2 system, in the F + H2 reaction simulations were tested in order to acquire a constant bond length, at the maximum potential energy. As the first part of this reaction was exothermic, adopting Hammonds postulate, the assumption was made that the reaction had an early transition state thus the region near the reactants was analysed. The MEP calculation was used to remove the vibrational noise as it made it difficult to read the plots. As HF + H is a late transition state, resembling the its product F + H2, it will have the same transition state positions.

=== Report the activation energy for both reactions. ===
In order to find the activation energy the bond distance of H-F was first decreased. This then resulted in the energy falling to form the products H-F and H. The difference in this energy as shown in figure 7 was used to calculate the activation energy of the H-F + H endothermic reaction. The activation energy was estimated to be 120 kJ/mol. For the exothermic F + H2 reaction, the activation energy was calculated by trying to shift the trajectory towards to the reactants. This was done by thus increasing the F-H distance forming the F + H2 reactant. The difference in this energy was very small as shown in figure 8 and was used to calculate the activation energy for the F + H2 reaction.

=== 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. ===
Figure 8 illustrates the transition between translational and vibrational energy in the F + H2 reaction. Initially, the H-H bond (B-C) is vibrationally excited shown by the oscillation, whilst the H-F(A-B) bond isn't indicated by the absence of an oscillation. The H-F bond is at the vibrational ground state . After the collision we see a transition. The H-F bond now is vibrationally excited illustrated by the oscillation in the momenta vs time plot, and the H-H bonds loses this vibrationally energy and now only has translational energy due to the breaking of the bond, depicted by the plateau after 30s. Experimentally IR could be used to see the progress of the reaction. Only molecules with a dipole are IR active, thus after the formation of a H-F bond, IR absorption absent in the reactant will start to appear.

File:Animation momenta hk3918.png

2020-05-15T17:44:15Z

Hk3918: momenta vs time for F + H2

momenta vs time for F + H2

MRD:01524831.molecularreactiondynamics

2020-05-15T16:52:10Z

MRD:01524831.molecularreactiondynamics

2020-05-15T16:51:49Z

MRD:01524831.molecularreactiondynamics

2020-05-15T16:51:16Z

File:Plot small activation.png

2020-05-15T16:48:53Z

Hk3918: Activation energy for F + H2

Activation energy for F + H2

File:Plot activation big.png

2020-05-15T16:48:12Z

Hk3918: Activation energy for HF + H

Activation energy for HF + H

MRD:01524831.molecularreactiondynamics

2020-05-15T16:47:07Z

Hk3918: /* Report the activation energy for both reactions. */

MRD:01524831.molecularreactiondynamics

2020-05-15T16:41:12Z

Hk3918:

MRD:01524831.molecularreactiondynamics

2020-05-15T03:07:45Z

Hk3918: /* Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed. */

MRD:01524831.molecularreactiondynamics

2020-05-15T02:37:27Z

Hk3918: /* Look at the “Internuclear Distances vs Time” and “Momenta vs Time”. What would change if we used the initial conditions r1 = rts and r2 = rts+1 pm instead? */