MRD:YX8818CX

2020-05-21T14:11:22Z

Yx8818: /* EXERCISE 2: F - H - H system */

Explicitly address questions highlighted in blue in the script and backup your answers with results and illustrations of calculations you perform using the program.

== EXERCISE 1: H + H2 system ==

=== Dynamics from the transition state region ===
'''In this report r1 = rAB, r2 = rBC.'''

{{fontcolor|blue|Q1: 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?}}

[[File:Saddle point.svg|thumb|right|300px|A saddle point (in red) on the graph of z=x2−y2 Reference: Wikipedia: Saddle point https://https://en.wikipedia.org/wiki/Saddle_point]]

On a potential energy (PE) surface, the transition state is a saddle point. At that point the gradient of the potential is zero. Mathematically speaking, a saddle point is a point on a surface of a function graph where the slopes (derivatives) in orthogonal directions are all zero, but which is not a local maximum or minimum of the function.

The saddle point can be determined mathematically by calculating the Hessian matrix for the function at that point. For a function f(x,y) of two variables, a point is a saddle point if:

If fx = 0, fy = 0, and fxxfyy - f2xy < 0.
Where fx and fxx are the first and second derivative of f with respect to x respectively.

The transition state (the saddle point) can be identified in a three-dimensional PE plot by looking at the surface at both direcions (r1 and r2 ). A saddle point will be a maximum in one direction and a minimum in the other direction. It can be easily distinguished from a local minimum, as a local minimum will be a minimum in all directions.

Take the matrix along the AB and BC. Consider plotting the eigenvectors.

==== Trajectories from r1 = r2: locating the transition state ====
{{fontcolor|blue|Q2: 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.}}

My best estimate of the TS position rts is 90.8 pm.

As p1 = p2 = 0, there is no initial momentum, hence no kinetic energy. So if the H + H2 surface start at the transition state, there will be no oscillation seen in the Distance vs Time plot, thus rAB and rBC will be constant. When r1 = r2 = 90.8 pm, there is negligible oscillatory behaviour displayed in the Distance vs Time plot, as shown in Figure. This indicates the transition state position ('''rts''') is very close to this value.
{| class="wikitable"
|[[File:TS_force_zero_YX8818.PNG|thumb|center|500px|Settings for rAB and rBC when forces is zero]]
|[[File:Q2_Distance_vs_time_plot_TS_YX8818.png|thumb|center|500px|Internuclear Distances vs Time Plot]]
|}

==== Calculating the reaction path ====

==== Trajectories from r1 = rts+δ, r2 = rts ====
{{fontcolor1|blue|Q3: Comment on how the ''mep'' and the trajectory you just calculated differ.}}

Using r1 = rts+1 = 91.8 pm, r2 = rts = 90.8 pm

As shown from the surface plots below, The minimum energy path (''mep'') corresponded to infinitely slow motions, with velocities / momenta reset to zero for each step. Thus, the trajectory corresponds to infinitely slow motion, and simply follows the valley floor. While the mep plot is useful in characterising the reaction, it does not give an realistic account of the atomic motions, i.e no oscillations are observed.

In comparison, the dynamics surface plot gave a more accurate description of atomic motions (inertial) during the reaction. Oscillations of the potential for the H atoms was observed for this plot.

{| class="wikitable"
|-
! Header 1
! Header 2
! Header 3
|-
| row1
| [[File:Q3_contour_plot_dynamics_YX8818.png|thumb|center|300px|Contour plot using Dynamics calculation type]]
| [[File:Q3_surface_plot_dynamics_YX8818.png|thumb|center|300px|Surface plot using Dynamics calculation type]]
|-
| row2
| [[File:Q3_contour_plot_MEP_YX8818.png|thumb|center|300px|Contour plot using MEP calculation type]]
| [[File:Q3_surface_plot_MEP_YX8818.png|thumb|center|300px|Surface plot using MEP calculation type]]
|}

==== Additional Questions: ====
'''A1'': 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?'''''

As shown in the “Internuclear Distances vs Time” and “Momenta vs Time” plots below, when r1 = rts+1, r2 = rts, the reaction proceed to form the products, and H2 molecule BC and H atom A were formed. This is because the transition state was displaced slightly towards the products.

On the other hand, when r1 = rts, r2 = rts+1, the reaction proceed to form the reactants, and molecule AB and atom C were formed.This is because the transition state was displaced slightly towards the reactants.

{| class="wikitable"
|-
!initial conditions
!r1 = rts+1, r2 = rts
!r1 = rts, r2 = rts+1
|-
|Internuclear Distances vs Time Plot
| [[File:internuclear_vs_time_r2TS_YX8818.png|300px]]
|[[File:internuclear_vs_time_r1TS_YX8818.png|300px]]
|-
|Momenta vs Time Plot
| [[File:Momenta_vs_time_r2TS_YX8818.png|300px]]
| [[File:Momenta_vs_time_r1TS_YX8818.png|300px]]
|}

'''A2: Note final values of the positions r1(t) r2(t) and p1(t) p2(t) for your trajectory for large enough t .'''

'''t = 100 fs '''was chosen, and for r1 = rts, r2 = rts+1, ''' r1'''(t) = 731 pm, '''r2'''(t) = 75.5 pm, '''p1'''(t) = 5.1 g.mol-1.pm.fs-1, '''p2'''(t) =2.5 g.mol-1.pm.fs-1.

'''A3'': 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. ''What do you observe?'''

When momentum was reversed, the trajectory traced back to the original position (slightly towards the product from the TS). But as it does not have enough energy to overcome the energy barrier at the TS, it traced back to the product, H2 molecules BC and H atoms A.

[[File:A3_momentum_reversed_YX8818.png|thumb|center|350px|Contour plot when monenta were reversed]]
''<nowiki/>''

=== Reactive and unreactive trajectories ===
{{fontcolor1|blue|Q4: 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?}}
* Initial positions '''r1''' = 74 pm and '''r2''' = 200 pm.
* Hypthesis: For a reactive trajectory, all trajectories starting with the same positions but with higher values of momenta (higher kinetic energy) would be reactive, as they have enough kinetic energy to overcome the activation barrier.
* From the simulations of the trajectory, it is found that the hypothesis is false. This is because simulation 4, which have momenta higher than some reactive trajectories, was not reactive. This shows that not all all trajectories starting with the same positions but with higher values of momenta (higher kinetic energy) would be reactive.

{| class="wikitable"
!No
!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:
contour plot
!Illustration of the trajectory:

surface plot
|-
|'''1'''
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|The distance rBC decreases as the molecule AB approaches H atom C and rBC reaches minium as the system reaches the transition state (TS). As the system has sufficient energy to overcome the energy barrier, the products BC molecules is formed (indicated by relatively constant small rBC) and increasing rAB as the H atom A moves away.
|[[File:Q4_1_Contour_YX8818.png|250px]]
|[[File:Q4_1_Surface_YX8818.png|300px]]
|-
|'''2'''
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|The distance rBC decreases as the molecule AB approaches H atom C, but upon collision there is insufficient energy to overcome the energy barrier at TS. Hence, the molecule AB remains, and atom C moves away (rBC increases)
|[[File:Q4_2_Contour_YX8818.png|250px]]
|[[File:Q4_2_Surface_YX8818.png|300px]]
|-
|'''3'''
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|The distance rBC decreases as the molecule AB approaches H atom C and rBC reaches minium as the system reaches the transition state (TS). As the system has sufficient energy to overcome the energy barrier, the products BC molecules is formed (indicated by relatively constant small rBC) and increasing rAB as the H atom A moves away.
|[[File:Q4_3_Contour_YX8818.png|250px]]
|[[File:Q4_3_Surface_YX8818.png|300px]]
|-
|'''4'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|The distance rBC decreases at first when the molecule AB approaches H atom C and reaches the transition state (TS). As the system has sufficient energy, it was able to cross the TS and from the product BC molecule and atom A. However, due to the excess kinetic energy / momentum the system possess, the system recrossed the energy barrier at the TS again, and refromed the reactants molecule AB, and atom C.
|[[File:Q4_4_Contour_YX8818.png|250px]]
|[[File:Q4_4_Surface_YX8818.png|300px]]
|-
|'''5'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|The excess kinetic energy / momentum of the system allows it to cross the energy barrier at the TS to the product side, recrossed to the reactant side, and then crossed the energy barrier again to finallay form the refromed the products, molecule AB, and atom C.
|[[File:Q4_5_Contour_YX8818.png|250px]]
|[[File:Q4_5_Surface_YX8818.png|300px]]
|}

==== Transition State Theory ====
{{fontcolor1|blue|Q5: Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}

During a molecular collision, we must consider two molecules to form a single quantum-mechanical entity, which is called a supermolecule.

The Transition State Theory (TST) chooses a boundary surface located between the reactant and product regions and assumes that all supermolecules that cross this boundary surface become products. The boundary surface, called the (critical) dividing surface, is taken to pass through the saddle point of the potential-energy surface.

TST has the following three assumptions:

1. All supermolecules that cross the critical dividing surface from the reactant side becomes product.

2. During the reaction, the Boltzmann distribution of energy is maintained for the reactant molecules.

3. The supermolecules crossing the critical surface from the reactant side have a Boltzmann distribution of energy corresponding to the temperature of the reacting system.

It is found that majority of the results obtained (Set 1, 2, 3, and 5) agrees with the Transition State Theory, as the supermolecules with sufficient momenta to cross the critical dividing surface from the reactant side becomes product, and those that did not crossed the critical dividing surface remain as reactants. However, experimental values does not always agree with experimental values. The results from Set 4, where the supermolecule recrossed the boundary, contradicted Assumption 1 of TST, showing that some supermolecules that crossed the critical dividing surface containing the transition state, can rebound, and reformed the reactants. This shows that TST prediction for reaction rate values might not accurately agrees with experimental values for some cases.

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

=== PES inspection ===
{{fontcolor1|blue|Q6: 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?}}

The potential energy surface for F + H2 → HF + H prcess is shown below:
[[File:Q6_PE_surface_F_H2_YX8818.png|thumb|right|350px|Potential energy surface for F + H2 → H + HF prcess ]]

F + H2 → HF + H : Exothermic.

This can be shown from the lowering of potential energy going from reactant to products.

HF + H → F + H2: Endothermic.

This can be shown from the increase in energy going from reactant to product.

Bond strength of HF = 565 kJ.mol-1

Bond strength of H2 = 432 kJ.mol-1

F + H2 → HF + H, ΔH = -565 + 432 = -133 kJ.mol-1, which supports it being an exothermic process as energy is being released overall. This is because the energy required to break the H-H bond is less than the energy released in the formation of H-F bond.

HF + H → F + H2 , ΔH = 565 - 432 = 133 kJ.mol-1, which supports it being an enothermic process as energy is being absorbed overall. This is because the energy required to break the H-F bond is more than the energy released in the formation of H-H bond.

{{fontcolor1|blue|Q7: Locate the approximate position of the transition state.}}

By Hammond postulate, an exothermic reaction of F + H2 to HF + Hhas an early transition state that resemble the reactant.

Thus, the distance rAB between F and H2 molecule should be large, while the H2 molecule distance rBC should be close the H2 bond distance.

Using the same method in Q2, the transition state position was found to be as follows:

F-H distance, rAB = 181.13 pm.

H-H distance, rBC = 74.45 pm.

{| class="wikitable"
|[[File:Q7_TS_zero_force_YX8818.png|thumb|center|500px|Settings for rAB and rBC when forces is zero]]
|[[File:Q7_TS_distance_vs_Time_YX8818.png|thumb|center|400px|Internuclear Distances vs Time Plot]]
|[[File:Q7_TS_contour_plot_YX8818.png|thumb|400px|Contour plot]]
|}

{{fontcolor1|blue|Q8: Report the activation energy for both reactions.}}
{| class="wikitable"
!system
!rHF / pm
!rHH / pm
!Total Energy / kJ.mol-1
!Activation EnergyTotal Energy / kJ.mol-1
!Energy vs time plot (full scale)
!Energy vs time plot (zoomed)
!Contour plot
|-
|transition state: F-H-H
|181.13
|74.45
|<nowiki>-433.978</nowiki>
|<nowiki>---</nowiki>
|
|
|
|-
|reactant to product:
F + H2 → HF + H
|180.13
|74.45
|<nowiki>-560.328</nowiki>
|<nowiki>-433.978 - (-560.328) = 126.35</nowiki>
|[[File:Q8_forming_HF_Activation_energy_E_vs_time_YX8818.png|400px]]
|[[File:Q8_zoomed_HF_Activation_energy_E_vs_time_YX8818.png|400px]]
|[[File:Q8_forming_HF_Contour_Plot_YX8818.png|400px]]
|-
|product to reactant:
HF + H → F + H2
|182.95
|74.45
|<nowiki>-435.015</nowiki>
|<nowiki>-433.978 - (-435.015) = 1.037</nowiki>
|[[File:Q8_original_forming_reactant_H2_Activation_energy_E_vs_time_YX8818.png|400px]]
|[[File:Q8_forming_reactant_H2_Activation_energy_E_vs_time_YX8818.png|400px]]
|[[File:Q8_forming_reactant_H2_contour_YX8818.png|400px]]
|}

=== Reaction dynamics ===
{{fontcolor1|blue|Q9: 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.}}

{{fontcolor1|blue|Q10: 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.}}

* For the initial positions '''r1''' = rHF = pm and '''r2''' = rHH = 74 pm, p1 = 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?
{| 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>-1.0 </nowiki>
|<nowiki>-5.9</nowiki>
|
|
|
|
|-
|<nowiki>-1.0 </nowiki>
|2.5
|
|
|
|
|-
|<nowiki>-1.0</nowiki>
|0
|
|
|
|
|-
|<nowiki>-1.0</nowiki>
|2.5
|
|
|
|
|-
|<nowiki>-1.0</nowiki>
|5.9
|
|
|
|
|}
* 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?
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-1.6</nowiki>
|0.2
|
|
|
|
|}

Let us now focus on the reverse reaction, H + HF.
* Setup initial conditions starting at the bottom of the entry channel, with very low vibrational motion on on the H - F bond, and an arbitrarily high value of pHH above the activation energy (an H atom colliding with a high kinetic energy).

* Try to obtain a reactive trajectory by decreasing the momentum of the incoming H atom and increasing the energy of the H - F vibration. (It may be difficult to find the initial conditions that generate a reactive trajectory for this reaction. Using the inversion of momentum procedure for a trajectory starting on the transition state can be useful in this case. Working on the Skew Plot framework could also be helpful.)
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|
|
|
|
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|
|
|
|
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|
|
|
|
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|
|
|
|
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|
|
|
|
|}

Polanyi's empirical rules

= References =
The quick brown fox jumps over the lazy dog.<nowiki><ref name="LazyDog" />
<references>
<ref name="LazyDog">This is the lazy dog reference.</ref>
</references></nowiki>

H-F bond distance NIST Diatomic Spectral Database (www.physics.nist.gov/PhysRefData/MolSpec/Diatomic/index.html)

MRD:YX8818CX

2020-05-21T14:09:37Z

Yx8818: /* EXERCISE 2: F - H - H system */

Explicitly address questions highlighted in blue in the script and backup your answers with results and illustrations of calculations you perform using the program.

== EXERCISE 1: H + H2 system ==

=== Dynamics from the transition state region ===
'''In this report r1 = rAB, r2 = rBC.'''

{{fontcolor|blue|Q1: 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?}}

[[File:Saddle point.svg|thumb|right|300px|A saddle point (in red) on the graph of z=x2−y2 Reference: Wikipedia: Saddle point https://https://en.wikipedia.org/wiki/Saddle_point]]

On a potential energy (PE) surface, the transition state is a saddle point. At that point the gradient of the potential is zero. Mathematically speaking, a saddle point is a point on a surface of a function graph where the slopes (derivatives) in orthogonal directions are all zero, but which is not a local maximum or minimum of the function.

The saddle point can be determined mathematically by calculating the Hessian matrix for the function at that point. For a function f(x,y) of two variables, a point is a saddle point if:

If fx = 0, fy = 0, and fxxfyy - f2xy < 0.
Where fx and fxx are the first and second derivative of f with respect to x respectively.

The transition state (the saddle point) can be identified in a three-dimensional PE plot by looking at the surface at both direcions (r1 and r2 ). A saddle point will be a maximum in one direction and a minimum in the other direction. It can be easily distinguished from a local minimum, as a local minimum will be a minimum in all directions.

Take the matrix along the AB and BC. Consider plotting the eigenvectors.

==== Trajectories from r1 = r2: locating the transition state ====
{{fontcolor|blue|Q2: 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.}}

My best estimate of the TS position rts is 90.8 pm.

As p1 = p2 = 0, there is no initial momentum, hence no kinetic energy. So if the H + H2 surface start at the transition state, there will be no oscillation seen in the Distance vs Time plot, thus rAB and rBC will be constant. When r1 = r2 = 90.8 pm, there is negligible oscillatory behaviour displayed in the Distance vs Time plot, as shown in Figure. This indicates the transition state position ('''rts''') is very close to this value.
{| class="wikitable"
|[[File:TS_force_zero_YX8818.PNG|thumb|center|500px|Settings for rAB and rBC when forces is zero]]
|[[File:Q2_Distance_vs_time_plot_TS_YX8818.png|thumb|center|500px|Internuclear Distances vs Time Plot]]
|}

==== Calculating the reaction path ====

==== Trajectories from r1 = rts+δ, r2 = rts ====
{{fontcolor1|blue|Q3: Comment on how the ''mep'' and the trajectory you just calculated differ.}}

Using r1 = rts+1 = 91.8 pm, r2 = rts = 90.8 pm

As shown from the surface plots below, The minimum energy path (''mep'') corresponded to infinitely slow motions, with velocities / momenta reset to zero for each step. Thus, the trajectory corresponds to infinitely slow motion, and simply follows the valley floor. While the mep plot is useful in characterising the reaction, it does not give an realistic account of the atomic motions, i.e no oscillations are observed.

In comparison, the dynamics surface plot gave a more accurate description of atomic motions (inertial) during the reaction. Oscillations of the potential for the H atoms was observed for this plot.

{| class="wikitable"
|-
! Header 1
! Header 2
! Header 3
|-
| row1
| [[File:Q3_contour_plot_dynamics_YX8818.png|thumb|center|300px|Contour plot using Dynamics calculation type]]
| [[File:Q3_surface_plot_dynamics_YX8818.png|thumb|center|300px|Surface plot using Dynamics calculation type]]
|-
| row2
| [[File:Q3_contour_plot_MEP_YX8818.png|thumb|center|300px|Contour plot using MEP calculation type]]
| [[File:Q3_surface_plot_MEP_YX8818.png|thumb|center|300px|Surface plot using MEP calculation type]]
|}

==== Additional Questions: ====
'''A1'': 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?'''''

As shown in the “Internuclear Distances vs Time” and “Momenta vs Time” plots below, when r1 = rts+1, r2 = rts, the reaction proceed to form the products, and H2 molecule BC and H atom A were formed. This is because the transition state was displaced slightly towards the products.

On the other hand, when r1 = rts, r2 = rts+1, the reaction proceed to form the reactants, and molecule AB and atom C were formed.This is because the transition state was displaced slightly towards the reactants.

{| class="wikitable"
|-
!initial conditions
!r1 = rts+1, r2 = rts
!r1 = rts, r2 = rts+1
|-
|Internuclear Distances vs Time Plot
| [[File:internuclear_vs_time_r2TS_YX8818.png|300px]]
|[[File:internuclear_vs_time_r1TS_YX8818.png|300px]]
|-
|Momenta vs Time Plot
| [[File:Momenta_vs_time_r2TS_YX8818.png|300px]]
| [[File:Momenta_vs_time_r1TS_YX8818.png|300px]]
|}

'''A2: Note final values of the positions r1(t) r2(t) and p1(t) p2(t) for your trajectory for large enough t .'''

'''t = 100 fs '''was chosen, and for r1 = rts, r2 = rts+1, ''' r1'''(t) = 731 pm, '''r2'''(t) = 75.5 pm, '''p1'''(t) = 5.1 g.mol-1.pm.fs-1, '''p2'''(t) =2.5 g.mol-1.pm.fs-1.

'''A3'': 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. ''What do you observe?'''

When momentum was reversed, the trajectory traced back to the original position (slightly towards the product from the TS). But as it does not have enough energy to overcome the energy barrier at the TS, it traced back to the product, H2 molecules BC and H atoms A.

[[File:A3_momentum_reversed_YX8818.png|thumb|center|350px|Contour plot when monenta were reversed]]
''<nowiki/>''

=== Reactive and unreactive trajectories ===
{{fontcolor1|blue|Q4: 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?}}
* Initial positions '''r1''' = 74 pm and '''r2''' = 200 pm.
* Hypthesis: For a reactive trajectory, all trajectories starting with the same positions but with higher values of momenta (higher kinetic energy) would be reactive, as they have enough kinetic energy to overcome the activation barrier.
* From the simulations of the trajectory, it is found that the hypothesis is false. This is because simulation 4, which have momenta higher than some reactive trajectories, was not reactive. This shows that not all all trajectories starting with the same positions but with higher values of momenta (higher kinetic energy) would be reactive.

{| class="wikitable"
!No
!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:
contour plot
!Illustration of the trajectory:

surface plot
|-
|'''1'''
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|The distance rBC decreases as the molecule AB approaches H atom C and rBC reaches minium as the system reaches the transition state (TS). As the system has sufficient energy to overcome the energy barrier, the products BC molecules is formed (indicated by relatively constant small rBC) and increasing rAB as the H atom A moves away.
|[[File:Q4_1_Contour_YX8818.png|250px]]
|[[File:Q4_1_Surface_YX8818.png|300px]]
|-
|'''2'''
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|The distance rBC decreases as the molecule AB approaches H atom C, but upon collision there is insufficient energy to overcome the energy barrier at TS. Hence, the molecule AB remains, and atom C moves away (rBC increases)
|[[File:Q4_2_Contour_YX8818.png|250px]]
|[[File:Q4_2_Surface_YX8818.png|300px]]
|-
|'''3'''
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|The distance rBC decreases as the molecule AB approaches H atom C and rBC reaches minium as the system reaches the transition state (TS). As the system has sufficient energy to overcome the energy barrier, the products BC molecules is formed (indicated by relatively constant small rBC) and increasing rAB as the H atom A moves away.
|[[File:Q4_3_Contour_YX8818.png|250px]]
|[[File:Q4_3_Surface_YX8818.png|300px]]
|-
|'''4'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|The distance rBC decreases at first when the molecule AB approaches H atom C and reaches the transition state (TS). As the system has sufficient energy, it was able to cross the TS and from the product BC molecule and atom A. However, due to the excess kinetic energy / momentum the system possess, the system recrossed the energy barrier at the TS again, and refromed the reactants molecule AB, and atom C.
|[[File:Q4_4_Contour_YX8818.png|250px]]
|[[File:Q4_4_Surface_YX8818.png|300px]]
|-
|'''5'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|The excess kinetic energy / momentum of the system allows it to cross the energy barrier at the TS to the product side, recrossed to the reactant side, and then crossed the energy barrier again to finallay form the refromed the products, molecule AB, and atom C.
|[[File:Q4_5_Contour_YX8818.png|250px]]
|[[File:Q4_5_Surface_YX8818.png|300px]]
|}

==== Transition State Theory ====
{{fontcolor1|blue|Q5: Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}

During a molecular collision, we must consider two molecules to form a single quantum-mechanical entity, which is called a supermolecule.

The Transition State Theory (TST) chooses a boundary surface located between the reactant and product regions and assumes that all supermolecules that cross this boundary surface become products. The boundary surface, called the (critical) dividing surface, is taken to pass through the saddle point of the potential-energy surface.

TST has the following three assumptions:

1. All supermolecules that cross the critical dividing surface from the reactant side becomes product.

2. During the reaction, the Boltzmann distribution of energy is maintained for the reactant molecules.

3. The supermolecules crossing the critical surface from the reactant side have a Boltzmann distribution of energy corresponding to the temperature of the reacting system.

It is found that majority of the results obtained (Set 1, 2, 3, and 5) agrees with the Transition State Theory, as the supermolecules with sufficient momenta to cross the critical dividing surface from the reactant side becomes product, and those that did not crossed the critical dividing surface remain as reactants. However, experimental values does not always agree with experimental values. The results from Set 4, where the supermolecule recrossed the boundary, contradicted Assumption 1 of TST, showing that some supermolecules that crossed the critical dividing surface containing the transition state, can rebound, and reformed the reactants. This shows that TST prediction for reaction rate values might not accurately agrees with experimental values for some cases.

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

=== PES inspection ===
{{fontcolor1|blue|Q6: 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?}}

The potential energy surface for F + H2 → HF + H prcess is shown below:
[[File:Q6_PE_surface_F_H2_YX8818.png|thumb|right|350px|Potential energy surface for F + H2 → H + HF prcess ]]

F + H2 → HF + H : Exothermic.

This can be shown from the lowering of potential energy going from reactant to products.

HF + H → F + H2: Endothermic.

This can be shown from the increase in energy going from reactant to product.

Bond strength of HF = 565 kJ.mol-1

Bond strength of H2 = 432 kJ.mol-1

F + H2 → HF + H, ΔH = -565 + 432 = -133 kJ.mol-1, which supports it being an exothermic process as energy is being released overall. This is because the energy required to break the H-H bond is less than the energy released in the formation of H-F bond.

HF + H → F + H2 , ΔH = 565 - 432 = 133 kJ.mol-1, which supports it being an enothermic process as energy is being absorbed overall. This is because the energy required to break the H-F bond is more than the energy released in the formation of H-H bond.

{{fontcolor1|blue|Q7: Locate the approximate position of the transition state.}}

By Hammond postulate, an exothermic reaction of F + H2 to HF + Hhas an early transition state that resemble the reactant.

Thus, the distance rAB between F and H2 molecule should be large, while the H2 molecule distance rBC should be close the H2 bond distance.

Using the same method in Q2, the transition state position was found to be as follows:

F-H distance, rAB = 181.13 pm.

H-H distance, rBC = 74.45 pm.

{| class="wikitable"
|[[File:Q7_TS_zero_force_YX8818.png|thumb|center|500px|Settings for rAB and rBC when forces is zero]]
|[[File:Q7_TS_distance_vs_Time_YX8818.png|thumb|center|400px|Internuclear Distances vs Time Plot]]
|[[File:Q7_TS_contour_plot_YX8818.png|thumb|400px|Contour plot]]
|}

{{fontcolor1|blue|Q8: Report the activation energy for both reactions.}}
{| class="wikitable"
!system
!rHF / pm
!rHH / pm
!Total Energy / kJ.mol-1
!Activation EnergyTotal Energy / kJ.mol-1
!Energy vs time plot (full scale)
!Energy vs time plot (zoomed)
!Contour plot
|-
|transition state: F-H-H
|181.13
|74.45
|<nowiki>-433.978</nowiki>
|<nowiki>---</nowiki>
|
|
|
|-
|reactant to product:
F + H2 → HF + H
|180.13
|74.45
|<nowiki>-560.328</nowiki>
|<nowiki>-433.978 - (-560.328) = 126.35</nowiki>
|Q8_forming_HF_Activation_energy_E_vs_time_YX8818.png
|Q8_zoomed_HF_Activation_energy_E_vs_time_YX8818.png
|Q8_forming_HF_Contour_Plot_YX8818.png
|-
|product to reactant:
HF + H → F + H2
|182.95
|74.45
|<nowiki>-435.015</nowiki>
|<nowiki>-433.978 - (-435.015) = 1.037</nowiki>
|Q8_original_forming_reactant_H2_Activation_energy_E_vs_time_YX8818.png
|Q8_forming_reactant_H2_Activation_energy_E_vs_time_YX8818.png
|Q8_forming_reactant_H2_contour_YX8818.png
|}

=== Reaction dynamics ===
{{fontcolor1|blue|Q9: 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.}}

{{fontcolor1|blue|Q10: 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.}}

* For the initial positions '''r1''' = rHF = pm and '''r2''' = rHH = 74 pm, p1 = 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?
{| 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>-1.0 </nowiki>
|<nowiki>-5.9</nowiki>
|
|
|
|
|-
|<nowiki>-1.0 </nowiki>
|2.5
|
|
|
|
|-
|<nowiki>-1.0</nowiki>
|0
|
|
|
|
|-
|<nowiki>-1.0</nowiki>
|2.5
|
|
|
|
|-
|<nowiki>-1.0</nowiki>
|5.9
|
|
|
|
|}
* 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?
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-1.6</nowiki>
|0.2
|
|
|
|
|}

Let us now focus on the reverse reaction, H + HF.
* Setup initial conditions starting at the bottom of the entry channel, with very low vibrational motion on on the H - F bond, and an arbitrarily high value of pHH above the activation energy (an H atom colliding with a high kinetic energy).

* Try to obtain a reactive trajectory by decreasing the momentum of the incoming H atom and increasing the energy of the H - F vibration. (It may be difficult to find the initial conditions that generate a reactive trajectory for this reaction. Using the inversion of momentum procedure for a trajectory starting on the transition state can be useful in this case. Working on the Skew Plot framework could also be helpful.)
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|
|
|
|
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|
|
|
|
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|
|
|
|
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|
|
|
|
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|
|
|
|
|}

Polanyi's empirical rules

= References =
The quick brown fox jumps over the lazy dog.<nowiki><ref name="LazyDog" />
<references>
<ref name="LazyDog">This is the lazy dog reference.</ref>
</references></nowiki>

H-F bond distance NIST Diatomic Spectral Database (www.physics.nist.gov/PhysRefData/MolSpec/Diatomic/index.html)

MRD:YX8818CX

2020-05-21T14:08:14Z

2020-05-20T00:11:22Z

Yx8818:

MRD:YX8818CX

2020-05-20T00:00:57Z

Yx8818: /* Reactive and unreactive trajectories */