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MRD:WXY0119

2018-05-18T17:01:01Z

Xw2816: /* Dynamics from the transition state region */

= Molecular Reaction Dynamics Report =

== EXERCISE 1ː H + H2 SYSTEM ==

[[File:HandH2wxy0119.png]]

In the H + H2 system, the distance between the first two H atoms (m1, m2) is r1 and the distance between the two H atoms in the H2 molecule (m2, m3) is r2.

=== Dynamics from the transition state region ===

{{fontcolor|red|Question 1 ː What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.}}

{| class="wikitable" border=1
|[[File:Q1TStransitionwxy0119.png|390px]]
|[[File:Q1TS2wxy0119.png|390px]]
|[[File:Q1TS3wxy0119.png|390px]]
|-
|Figure.1 Surface plot of reaction trajectory
|Figure.2 Transition state surface plot
|Figure.3 Transition state surface plot (different angle)
|}

In the reaction trajectory surface plot (Fig.1), AB is the distance r1 and BC is the distance r2. The gradient of the potential energy with regard to r1 and r2 are ∂V(r1)/∂r1 and ∂V(r2)/∂r2 respectively. The second derivatives of the potential energy with regard to r1 and r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 respectively.

At the two minimum structures which are at the two ends of the reaction pathway, ∂V(r1)/∂r1 = 0 and ∂V(r2)/∂r2 = 0 respectively. Since the two minimum structures are at two local minimum point , the second derivatives ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 are both greater than zero.

On the other hand, the components at the transition state structure (Fig.2) are different from that at minimum points. The gradient ∂V(r1)/∂r1 and ∂V(r2)/∂r2 are both equal to zero at the transition structure. However, ∂2V(r1)/∂r12 >0 and ∂2V(r2)/∂r22 <0. The transition state point is actually a saddle point which is observed more clearly in Fig.3. The Transition state point is thus distinguished from the minimum points as the potential energy surface curves inwards and downwards at the minimum points.

=== Locating the Transition State ===

{{fontcolor|red|Question 2ː 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.}}

{| class="wikitable" border=1
|[[File:TScontourwxy0119.png|390px]]
|[[File:TSsurfaceplwxy0119.png|390px]]
|[[File:TSestimatewxy0119.png|390px]]
|-
|Figure.4 Transition state contour plot
|Figure.5 Transition state surface plot
|Figure.6 Internuclear distance against time plot
|}

The best estimate of the transition state position is rts = 0.9078 Å.

When r1 = r2 = rts and momenta are set to zero, a graph of intermolecular distance against time is plotted. (Fig.6) In this graph, AB and BC lines are the same and overlap, the two lines on the graph are flat and horizontal, indicating that the atomic distances are the same at the position and with no momenta the state is at equilibrium, validating the position is the transition state position. The transition state is shown as a cross on the counter plot (Fig.4) and a dot on the surface plot (Fig.5).

=== Calculating and comparing the reaction path and trajectory ===

{{fontcolor1|red|Question 3ː Comment on how the ''mep'' and the trajectory you just calculated differ.}}

{| class="wikitable" border=1
|[[File:Mep1wxy0119.png|390px]]
|[[File:Mep2wxy0119.png|390px]]
|[[File:Mep3wxy0119.png|390px]]
|-
|Figure.7 mep calculation contour plot
|Figure.8 mep calculation surface plot
|Figure.9 mep calculation internuclear distance against time plot
|-
|[[File:Dynamic1wxy0119.png|390px]]
|[[File:Dynamic2wxy0119.png|390px]]
|[[File:Dynamic3wxy0119.png|390px]]
|-
|Figure.10 Dynamics calculation contour plot
|Figure.11 Dynamics calculation surface plot
|Figure.12 Dynamics calculation internuclear distance against time plot
|}
{| class="wikitable" border=1
|[[File:dynamics4wxy0119.png|600px]]
|[[File:Mep5wxy0119.png|600px]]
|-
|Figure.13 Dynamics calculation internuclear momenta against time plot
|Figure.14 mep calculation internuclear momenta against time plotsurface plot
|}

The mep (minimum energy path) trajectory is a smooth line on contour and surface plot (Fig.7 Fig.8) while the trajectory under dynamics calculation is oscillating along the pathway. (Fig.10 Fig.11) The mep trajectory takes more steps and longer time than the dynamics trajectory as shown in internuclear distance against time plots. (Fig. 9 Fig.12) The momentum is zero all the time under mep calculation (Fig.13) but under dynamics calculation, the momenta lines are oscillating against time.

The differences arise from mep and dynamics calculations can be explained by the the natures of the two calculations. Under mep calculation, the atoms are in extremely slow motion and the momentum is zero for each step, i.e. the atoms are "stop" after every step. The pathway under mep calculation is formed by connecting all the minimum energy points of each step and appears as a smooth line. On the other hand, atoms are in continuous motion and the momentum is accumulated after each step, atoms are then oscillating on the potential energy surface along the pathway. The steps are set to 5000 in the mep calculation but are only 500 in the dynamics calculatoin. Since the steps in mep are small, more time is needed to complete the same length on the trajectories.

=== Reactive and unreactive trajectories ===

{{fontcolor1|red|Question 4ː Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.}}

For the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, run trajectories with the following momenta p1 and p2 combinations:
{| class="wikitable" border=1
| Condition
| p1
| p2
| Total Energy/ kcal mol -1
| Reactivity
|-
|1.
| -1.25
| -2.5
| -99.018
| reactive
|-
|2.
| -1.5
| -2.0
| -100.456
| unreactive
|-
|3.
| -1.5
| -2.5
| -98.956
| reactive
|-
|4.
| -2.5
| -5.0
| -84.956
| unreactive
|-
|5.
| -2.5
| -5.2
| -83.416
| reactive
|}

{| class="wikitable" border=1
| Condition
| Surface Plot
| Contour Plot
| Description
|-
|1.
| [[File:condition1wxy0119.png|350px]]
| [[File:firstwxy0119.png|350px]]
| r2 (BC) decreases when HC approaches bonded HA and HB. The energy processed by the system is sufficient to overcome the activation barrier and crosses the transition state structure to break the HA and HB bond and form a new HB and HC bond. The new bond oscillate as r1 (AB) increases.
|-
|2.
| [[File:condition2wxy0119.png|350px]]
| [[File:secondwxy0119.png|350px]]
| HC approaches HB but the energy is insufficient to reach the transition state point, HC then moves further away from the bonded H2 molecule and no new bond is formed. The oscillation along r2 is a result of an increase in momentum p1.
|-
|3.
| [[File:condition3wxy0119.png|350px]]
| [[File:thirdwxy0119.png|350px]]
| Similarly to condition 1 but with a more negative p1, HA - HB bond oscillates more when r2 decreases. The reaction proceeds and crosses the transition state point to form the products.
|-
|4.
| [[File:condition4wxy0119.png|350px]]
| [[File:fourthwxy0119.png|350px]]
| Both p1 and p2 are more negative than the previous conditions. The HA - HB bond oscillates vigorously with larger amplitudes. The barrier recrossing happens after the system crosses the transition state region but eventually reverts back to the reactants.
|-
|5.
| [[File:condition5wxy0119.png|350px]]
| [[File:fifthwxy0119.png|350px]]
| With a slight more negative value of p1 as compared to that in condition 4, the system now processes appropriate energy to proceed to forming products. The vigorous oscillations are resulting from high momenta and barrier recrossing occurs.
|}

=== Transition State Theory ===

{{fontcolor|red|Question 5ː State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}

The main assumptions of Transition State Theory (TST) are<ref name="TS" />ː

1. The behaviours of the atomic nuclei follow the classical mechanics rules, i.e. the transition structure is formed as long as atoms or molecules collide with enough energy.

2. The intermediates have a long enough lifetime for the energies to be Boltzmann distributed preceding the next step.

3. The lowest energy saddle point on the potential energy surface is passed over in the reaction system.

The TST does not take into account of the quantum tunnelling effect which allows reactions to happen even when the energy carried by the system is lower than the activation energy barrier, especially for reactions with low energy barriers. Nor does the theory consider the intermediates with short lifetimes. When the energies are not fully distributed, the momentum of the reaction trajectory from the reactants to the intermediate can have effects on product selectivity. Moreover, the theory fails at high temperature because higher vibrational energy modes can be populated for a molecule at high temperature. The molecules can have complex motion and their collisions may result in a higher energy transition structure. The experimental results may thus deviate from that predicted by the TST. For example in condition 4, the kinetic energy of the reactants is much higher than the activation energy required. However, the momenta between atoms are very high and higher vibrational modes are populated that result in higher energy transition structures formed and the system does not pass though the lowest transition saddle point as shown in the diagram. Barrier recrossing occurs but the products are not formed as opposed to the predictions by the TST.

== EXERCISE 2: F - H - H SYSTEM ==

=== Reaction Energetics ===

{{fontcolor1|red|Question 6ː 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?}}

{| class="wikitable" border=1
|[[File:Q6FH2wxy0119.png|600px]]
|[[File:Q6HFHwxy0119.png|600px]]
|-
|Figure.15 F + H2 position on surface plot
|Figure.16 HF and H position on surface plot
|}

The surface plots of the potential energy surface of H + HF → F + H2 reaction is shown (Fig.15 Fig.16). Distance AB is the distance between H and F atoms and distance BC is the distance between two H atoms. The two minimum points shown in the graph are positions of F + H2 (Fig.15) and H + HF (Fig.16) respectively. If forward reaction is H + HF → F + H2, then the backward reaction is F + H2 → H + HF. As clearly shown in the diagram F + H2 are at a minimum point of higher potential energy than that of H + HF. Thus, by comparing the potential energy of reactants and products, the signs of changes in enthalpy (ΔH) and the energetics are determined. H + HF → F + H2 is endothermic and F + H2 → H + HF is exothermic.

The energetics of the two reactions are in accordance to the predictions by comparing bond strengths of chemical species. The bond energy of H-F bond (565 kJ/mol) is higher than the bond energy of H-H bond (432 kJ/mol). Thus, energy released by forming the H-H bond is not enough to compensate for energy required for breaking the H-F bond and the reaction H + HF → F + H2 is endothermic. On the contrary, the reaction F + H2 → H + HF is exothermic.

=== Transition State Approximation ===

{{fontcolor1|red|Question 7ː Locate the approximate position of the transition state.}}

{| class="wikitable" border=1
|[[File:Q7TStransitionwxy0119.png|390px]]
|[[File:Q7TS2wxy0119.png|390px]]
|[[File:Q7TS3wxy0119.png|390px]]
|-
|Figure.17 Transition state surface plot
|Figure.18 Transition state contour plot
|Figure.19 Internuclear distance against time plot (at TS point)
|}

The transition state (TS) is shown as a black dot on the potential energy surface plot (Fig.17) and a red cross on the contour plot. (Fig,18) The position of the TS is at the point where the distance between F and H (rFH) is 1.810Å and the distance between two H atoms (rHH) is 0.746Å. The TS is closer to F + H2 and further away from H + HF. At the transition state position, the distance between F and H, between two H atoms are constant with no momentum as shown in the internuclear distance against time plot as horizontal and flat lines, validating the position is the transition state point (Fig.19)

=== Activation Energies ===

{{fontcolor1|red|Question 8ː Report the activation energy for both reactions.}}

{| class="wikitable" border=1
|[[File:Q8Ea2wxy0119.png|600px]]
|[[File:Q8Ea1wxy0119.png|600px]]
|-
|Figure.21 Energy against time plot (HF + H)
|Figure.20 Energy against time plot (F + H2)
|}

By performing MEP calculation with slight increase and decrease of the rFH to 1.820Å to perform F + H2 → H + HF reaction and to 1.800Å to perform H + HF → F + H2 reaction, the activation energies (Ea) of both reactions can be calculated from the differences in their potential energies (Fig.20 Fig.21)ː

H + HF → F + H2ː Ea = -103.751 - (-133.624) = +29.873 kcal mol-1

F + H2 → H + HFː Ea = -103.751 - (-103.972) = + 0.221 kcal mol-1

=== Reaction Dynamics ===

{{fontcolor1|red|Question 9ː In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

Reactionː F + H2 → H + HF

Initial condition setː rHF = 2Å rHH = 0.74Å pHF = -0.5 pHH = 0.1

{| class="wikitable" border=1
|[[File:Q9plot1wxy0119.png|390px]]
|[[File:Q9plot2wxy0119.png|390px]]
|[[File:Q9plot3wxy0119.png|390px]]
|-
|Figure.23 Contour plot
|Figure.24 Surface plot
|Figure.25 Internuclear momentum against time plot
|-
|[[File:Q9ani1wxy0119.png|390px]]
|[[File:Q9ani2wxy0119.png|390px]]
|-
|Figure.26 Animation figure (at the start of the reaction)
|Figure.27 Animation figure (at the end of the reaction)
|}

As shown in the plots and animation figures, the H-H bond in the reactants is broken and H-F bond in the products is formed. There are small momenta between both H and F and H-H bond that provide the initial kinetic energy of the reactants and the initial vibrational energy of the H-H bond in the reactants. The initial energy is required to break the H-H bond and reduce the distance of H and F for bond forming and potential energy is released when H-F bond is formed. Since energy is conserved, the reaction energy released is converted to the vibrational energy of the H-F bond and the translational kinetic energies of the products. This is shown in the internuclear momentum against time graph as the initial vibrations of B-C (H-H bond) becomes flat as two H atoms are separated in the product and the A-B (H-F bond) formed in the product vibrates with a large amplitude. (Fig.25) The contour plot and the surface plot also clearly show there is a significant increase in the amplitude of the oscillation of H-F bond which is an indication of the large vibrational energy of H-F bond. (Fig.23 Fig.24)

The predictions can be proved experimentally. The vibrational energy of H-F bond can be determined by measuring the frequency and the intensity of the absorption band of the H-F bond with infrared spectroscopy. The conversion to the kinetic energies of the products can be observed by measuring the temperature of the reaction as the kinetic energy can be converted to thermal energy as products collide with solvent molecules. The temperature is expected to increase as this is an exothermic reaction.

=== Energy Distribution and Reactivity ===

{{fontcolor|red|Question 10ː 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.}}
{| class="wikitable" border=1
| Condition
| rFH /Å
| rHH /Å
| pFH
| pHH
|-
|1.
| 2
| 0.74
| -0.5
| -3
|-
|2.
| 2
| 0.74
| -0.5
| 1.9
|-
|3.
| 2.1
| 0.74
| -0.5
| 0.1
|-
|4.
| 2
| 0.74
| -0.8
| 0.1
|-
|5.
| 0.91
| 2
| 0.05
| -20
|-
|6.
| 0.91
| 2
| 0.8
| -7.5
|}
{| class="wikitable" border=1
|Condition
|Contour plot
|Observation
|-
|1.
|[[File:Q10plot1wxy0119.png|600px]]
| The products are not formed in the reaction between F and H2 even pHH is very large. Barrier recrossing occurs and the H-H bond oscillates significantly with a large amplitude in the diagram. It should be noted that while keeping pFH to be -0.5 and changing pHH from -3 to 3, the trend is not very obvious as there are many anomalous values of momentum that render the reaction unreactive while slight increase and decrease of momentum make the reaction reactive again. For example, pHH = -0.9 is reactive but pHH = -1.1 is unreactive.
|-
|2.
|[[File:Q10plot2wxy0119.png|600px]]
| There is reaction between F and H2 and products are formed. The F-H bond formed vibrates significantly along BC trajectory.
|-
|3.
|[[File:Q10plot9wxy0119.png|600px]]
| There is no reaction under this condition between F and H2. There is not enough energy to cross over the transition state point. Although investigations have shown that there is reaction when rFH is equal or less than 2Å, but there is no reaction when H and F atoms are further apart.
|-
|4.
|[[File:Q10plot8wxy0119.png|600px]]
| There is reaction between F and H2. By comparison between condition 4 and condition 3, a slight increase in the momentum of F and H which indicates a slight increase in kinetic translational energy result in the formation of products. Investigations have shown that even when rFH is slightly larger than 2Å, the products are still formed. Thus the change in reaction conditions illustrate that the changes in translational energy affect the reactivity the reaction more than the vibrational energy.
|-
|5.
|[[File:Q10plot5wxy0119.png|600px]]
| There is no reaction between H and HF even though the kinetic translational energy is very high and much higher than the activation energy. (pHF = -20) The trajectory reverts back before reaching the transition state structure.
|-
|6.
|[[File:Q10plot4wxy0119.png|600px]]
| There is reaction between H and HF. Compared to condition 5, the translational energy decreases significantly while vibrational energy increases slightly. Thus, the vibrational energy is more efficient in promoting the reaction than the translational energy.
|}

The transition state illustrated in Question 7 is closer to F and H2 than H and HF. From the above illustrations, it can be shown that the translational energy is more efficient in promoting the reaction between F and H2 which is exothermic, has a lower activation energy and has an earlier transition state. On the other hand, the vibrational energy is more efficient in the reactivity of the reaction between H and HF which is endothermic, has a higher activation energy and has a late transition state. These observations agree with the Polanyi's empirical rules.<ref name="PoRule" />

== References ==

<references>
<ref name="TS">Eyring, H. (1935). "The Activated Complex in Chemical Reactions". J. Chem. Phys. 3 (2): 107–115.</ref>
<ref name="PoRule">Polanyi J C. Concepts in reaction dynamics[J]. Accounts of Chemical Research, 1972, 5(5): 161-168.</ref>
</references>

MRD:WXY0119

2018-05-18T17:00:27Z

Xw2816: /* Dynamics from the transition state region */

= Molecular Reaction Dynamics Report =

== EXERCISE 1ː H + H2 SYSTEM ==

[[File:HandH2wxy0119.png]]

In the H + H2 system, the distance between the first two H atoms (m1, m2) is r1 and the distance between the two H atoms in the H2 molecule (m2, m3) is r2.

=== Dynamics from the transition state region ===

{{fontcolor|red|Question 1 ː What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.}}

{| class="wikitable" border=1
|[[File:Q1TStransitionwxy0119.png|390px]]
|[[File:Q1TS2wxy0119.png|390px]]
|[[File:Q1TS3wxy0119.png|390px]]
|-
|Figure.1 Surface plot of reaction trajectory
|Figure.2 Transition state surface plot
|Figure.3 Transition state surface plot (different angle)
|}

In the reaction trajectory surface plot (Fig.1), AB is the distance r1 and BC is the distance r2. The gradient of the potential energy with regard to r1 and r2 are ∂V(r1)/∂r1 and ∂V(r2)/∂r2 respectively. The second derivatives of the potential energy with regard to r1 and r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 respectively.

At the two minimum structures which are at the two ends of the reaction pathway, ∂V(r1)/∂r1 = 0 and ∂V(r2)/∂r2 = 0 respectively. Since the two minimum structures are at two local minimum point , the second derivatives r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 are both greater than zero.

On the other hand, the components at the transition state structure (Fig.2) are different from that at minimum points. The gradient ∂V(r1)/∂r1 and ∂V(r2)/∂r2 are both equal to zero at the transition structure. However, ∂2V(r1)/∂r12 >0 and ∂2V(r2)/∂r22 <0. The transition state point is actually a saddle point which is observed more clearly in Fig.3. The Transition state point is thus distinguished from the minimum points as the potential energy surface curves inwards and downwards at the minimum points.

=== Locating the Transition State ===

{{fontcolor|red|Question 2ː 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.}}

{| class="wikitable" border=1
|[[File:TScontourwxy0119.png|390px]]
|[[File:TSsurfaceplwxy0119.png|390px]]
|[[File:TSestimatewxy0119.png|390px]]
|-
|Figure.4 Transition state contour plot
|Figure.5 Transition state surface plot
|Figure.6 Internuclear distance against time plot
|}

The best estimate of the transition state position is rts = 0.9078 Å.

When r1 = r2 = rts and momenta are set to zero, a graph of intermolecular distance against time is plotted. (Fig.6) In this graph, AB and BC lines are the same and overlap, the two lines on the graph are flat and horizontal, indicating that the atomic distances are the same at the position and with no momenta the state is at equilibrium, validating the position is the transition state position. The transition state is shown as a cross on the counter plot (Fig.4) and a dot on the surface plot (Fig.5).

=== Calculating and comparing the reaction path and trajectory ===

{{fontcolor1|red|Question 3ː Comment on how the ''mep'' and the trajectory you just calculated differ.}}

{| class="wikitable" border=1
|[[File:Mep1wxy0119.png|390px]]
|[[File:Mep2wxy0119.png|390px]]
|[[File:Mep3wxy0119.png|390px]]
|-
|Figure.7 mep calculation contour plot
|Figure.8 mep calculation surface plot
|Figure.9 mep calculation internuclear distance against time plot
|-
|[[File:Dynamic1wxy0119.png|390px]]
|[[File:Dynamic2wxy0119.png|390px]]
|[[File:Dynamic3wxy0119.png|390px]]
|-
|Figure.10 Dynamics calculation contour plot
|Figure.11 Dynamics calculation surface plot
|Figure.12 Dynamics calculation internuclear distance against time plot
|}
{| class="wikitable" border=1
|[[File:dynamics4wxy0119.png|600px]]
|[[File:Mep5wxy0119.png|600px]]
|-
|Figure.13 Dynamics calculation internuclear momenta against time plot
|Figure.14 mep calculation internuclear momenta against time plotsurface plot
|}

The mep (minimum energy path) trajectory is a smooth line on contour and surface plot (Fig.7 Fig.8) while the trajectory under dynamics calculation is oscillating along the pathway. (Fig.10 Fig.11) The mep trajectory takes more steps and longer time than the dynamics trajectory as shown in internuclear distance against time plots. (Fig. 9 Fig.12) The momentum is zero all the time under mep calculation (Fig.13) but under dynamics calculation, the momenta lines are oscillating against time.

The differences arise from mep and dynamics calculations can be explained by the the natures of the two calculations. Under mep calculation, the atoms are in extremely slow motion and the momentum is zero for each step, i.e. the atoms are "stop" after every step. The pathway under mep calculation is formed by connecting all the minimum energy points of each step and appears as a smooth line. On the other hand, atoms are in continuous motion and the momentum is accumulated after each step, atoms are then oscillating on the potential energy surface along the pathway. The steps are set to 5000 in the mep calculation but are only 500 in the dynamics calculatoin. Since the steps in mep are small, more time is needed to complete the same length on the trajectories.

=== Reactive and unreactive trajectories ===

{{fontcolor1|red|Question 4ː Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.}}

For the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, run trajectories with the following momenta p1 and p2 combinations:
{| class="wikitable" border=1
| Condition
| p1
| p2
| Total Energy/ kcal mol -1
| Reactivity
|-
|1.
| -1.25
| -2.5
| -99.018
| reactive
|-
|2.
| -1.5
| -2.0
| -100.456
| unreactive
|-
|3.
| -1.5
| -2.5
| -98.956
| reactive
|-
|4.
| -2.5
| -5.0
| -84.956
| unreactive
|-
|5.
| -2.5
| -5.2
| -83.416
| reactive
|}

{| class="wikitable" border=1
| Condition
| Surface Plot
| Contour Plot
| Description
|-
|1.
| [[File:condition1wxy0119.png|350px]]
| [[File:firstwxy0119.png|350px]]
| r2 (BC) decreases when HC approaches bonded HA and HB. The energy processed by the system is sufficient to overcome the activation barrier and crosses the transition state structure to break the HA and HB bond and form a new HB and HC bond. The new bond oscillate as r1 (AB) increases.
|-
|2.
| [[File:condition2wxy0119.png|350px]]
| [[File:secondwxy0119.png|350px]]
| HC approaches HB but the energy is insufficient to reach the transition state point, HC then moves further away from the bonded H2 molecule and no new bond is formed. The oscillation along r2 is a result of an increase in momentum p1.
|-
|3.
| [[File:condition3wxy0119.png|350px]]
| [[File:thirdwxy0119.png|350px]]
| Similarly to condition 1 but with a more negative p1, HA - HB bond oscillates more when r2 decreases. The reaction proceeds and crosses the transition state point to form the products.
|-
|4.
| [[File:condition4wxy0119.png|350px]]
| [[File:fourthwxy0119.png|350px]]
| Both p1 and p2 are more negative than the previous conditions. The HA - HB bond oscillates vigorously with larger amplitudes. The barrier recrossing happens after the system crosses the transition state region but eventually reverts back to the reactants.
|-
|5.
| [[File:condition5wxy0119.png|350px]]
| [[File:fifthwxy0119.png|350px]]
| With a slight more negative value of p1 as compared to that in condition 4, the system now processes appropriate energy to proceed to forming products. The vigorous oscillations are resulting from high momenta and barrier recrossing occurs.
|}

=== Transition State Theory ===

{{fontcolor|red|Question 5ː State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}

The main assumptions of Transition State Theory (TST) are<ref name="TS" />ː

1. The behaviours of the atomic nuclei follow the classical mechanics rules, i.e. the transition structure is formed as long as atoms or molecules collide with enough energy.

2. The intermediates have a long enough lifetime for the energies to be Boltzmann distributed preceding the next step.

3. The lowest energy saddle point on the potential energy surface is passed over in the reaction system.

The TST does not take into account of the quantum tunnelling effect which allows reactions to happen even when the energy carried by the system is lower than the activation energy barrier, especially for reactions with low energy barriers. Nor does the theory consider the intermediates with short lifetimes. When the energies are not fully distributed, the momentum of the reaction trajectory from the reactants to the intermediate can have effects on product selectivity. Moreover, the theory fails at high temperature because higher vibrational energy modes can be populated for a molecule at high temperature. The molecules can have complex motion and their collisions may result in a higher energy transition structure. The experimental results may thus deviate from that predicted by the TST. For example in condition 4, the kinetic energy of the reactants is much higher than the activation energy required. However, the momenta between atoms are very high and higher vibrational modes are populated that result in higher energy transition structures formed and the system does not pass though the lowest transition saddle point as shown in the diagram. Barrier recrossing occurs but the products are not formed as opposed to the predictions by the TST.

== EXERCISE 2: F - H - H SYSTEM ==

=== Reaction Energetics ===

{{fontcolor1|red|Question 6ː 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?}}

{| class="wikitable" border=1
|[[File:Q6FH2wxy0119.png|600px]]
|[[File:Q6HFHwxy0119.png|600px]]
|-
|Figure.15 F + H2 position on surface plot
|Figure.16 HF and H position on surface plot
|}

The surface plots of the potential energy surface of H + HF → F + H2 reaction is shown (Fig.15 Fig.16). Distance AB is the distance between H and F atoms and distance BC is the distance between two H atoms. The two minimum points shown in the graph are positions of F + H2 (Fig.15) and H + HF (Fig.16) respectively. If forward reaction is H + HF → F + H2, then the backward reaction is F + H2 → H + HF. As clearly shown in the diagram F + H2 are at a minimum point of higher potential energy than that of H + HF. Thus, by comparing the potential energy of reactants and products, the signs of changes in enthalpy (ΔH) and the energetics are determined. H + HF → F + H2 is endothermic and F + H2 → H + HF is exothermic.

The energetics of the two reactions are in accordance to the predictions by comparing bond strengths of chemical species. The bond energy of H-F bond (565 kJ/mol) is higher than the bond energy of H-H bond (432 kJ/mol). Thus, energy released by forming the H-H bond is not enough to compensate for energy required for breaking the H-F bond and the reaction H + HF → F + H2 is endothermic. On the contrary, the reaction F + H2 → H + HF is exothermic.

=== Transition State Approximation ===

{{fontcolor1|red|Question 7ː Locate the approximate position of the transition state.}}

{| class="wikitable" border=1
|[[File:Q7TStransitionwxy0119.png|390px]]
|[[File:Q7TS2wxy0119.png|390px]]
|[[File:Q7TS3wxy0119.png|390px]]
|-
|Figure.17 Transition state surface plot
|Figure.18 Transition state contour plot
|Figure.19 Internuclear distance against time plot (at TS point)
|}

The transition state (TS) is shown as a black dot on the potential energy surface plot (Fig.17) and a red cross on the contour plot. (Fig,18) The position of the TS is at the point where the distance between F and H (rFH) is 1.810Å and the distance between two H atoms (rHH) is 0.746Å. The TS is closer to F + H2 and further away from H + HF. At the transition state position, the distance between F and H, between two H atoms are constant with no momentum as shown in the internuclear distance against time plot as horizontal and flat lines, validating the position is the transition state point (Fig.19)

=== Activation Energies ===

{{fontcolor1|red|Question 8ː Report the activation energy for both reactions.}}

{| class="wikitable" border=1
|[[File:Q8Ea2wxy0119.png|600px]]
|[[File:Q8Ea1wxy0119.png|600px]]
|-
|Figure.21 Energy against time plot (HF + H)
|Figure.20 Energy against time plot (F + H2)
|}

By performing MEP calculation with slight increase and decrease of the rFH to 1.820Å to perform F + H2 → H + HF reaction and to 1.800Å to perform H + HF → F + H2 reaction, the activation energies (Ea) of both reactions can be calculated from the differences in their potential energies (Fig.20 Fig.21)ː

H + HF → F + H2ː Ea = -103.751 - (-133.624) = +29.873 kcal mol-1

F + H2 → H + HFː Ea = -103.751 - (-103.972) = + 0.221 kcal mol-1

=== Reaction Dynamics ===

{{fontcolor1|red|Question 9ː In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

Reactionː F + H2 → H + HF

Initial condition setː rHF = 2Å rHH = 0.74Å pHF = -0.5 pHH = 0.1

{| class="wikitable" border=1
|[[File:Q9plot1wxy0119.png|390px]]
|[[File:Q9plot2wxy0119.png|390px]]
|[[File:Q9plot3wxy0119.png|390px]]
|-
|Figure.23 Contour plot
|Figure.24 Surface plot
|Figure.25 Internuclear momentum against time plot
|-
|[[File:Q9ani1wxy0119.png|390px]]
|[[File:Q9ani2wxy0119.png|390px]]
|-
|Figure.26 Animation figure (at the start of the reaction)
|Figure.27 Animation figure (at the end of the reaction)
|}

As shown in the plots and animation figures, the H-H bond in the reactants is broken and H-F bond in the products is formed. There are small momenta between both H and F and H-H bond that provide the initial kinetic energy of the reactants and the initial vibrational energy of the H-H bond in the reactants. The initial energy is required to break the H-H bond and reduce the distance of H and F for bond forming and potential energy is released when H-F bond is formed. Since energy is conserved, the reaction energy released is converted to the vibrational energy of the H-F bond and the translational kinetic energies of the products. This is shown in the internuclear momentum against time graph as the initial vibrations of B-C (H-H bond) becomes flat as two H atoms are separated in the product and the A-B (H-F bond) formed in the product vibrates with a large amplitude. (Fig.25) The contour plot and the surface plot also clearly show there is a significant increase in the amplitude of the oscillation of H-F bond which is an indication of the large vibrational energy of H-F bond. (Fig.23 Fig.24)

The predictions can be proved experimentally. The vibrational energy of H-F bond can be determined by measuring the frequency and the intensity of the absorption band of the H-F bond with infrared spectroscopy. The conversion to the kinetic energies of the products can be observed by measuring the temperature of the reaction as the kinetic energy can be converted to thermal energy as products collide with solvent molecules. The temperature is expected to increase as this is an exothermic reaction.

=== Energy Distribution and Reactivity ===

{{fontcolor|red|Question 10ː 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.}}
{| class="wikitable" border=1
| Condition
| rFH /Å
| rHH /Å
| pFH
| pHH
|-
|1.
| 2
| 0.74
| -0.5
| -3
|-
|2.
| 2
| 0.74
| -0.5
| 1.9
|-
|3.
| 2.1
| 0.74
| -0.5
| 0.1
|-
|4.
| 2
| 0.74
| -0.8
| 0.1
|-
|5.
| 0.91
| 2
| 0.05
| -20
|-
|6.
| 0.91
| 2
| 0.8
| -7.5
|}
{| class="wikitable" border=1
|Condition
|Contour plot
|Observation
|-
|1.
|[[File:Q10plot1wxy0119.png|600px]]
| The products are not formed in the reaction between F and H2 even pHH is very large. Barrier recrossing occurs and the H-H bond oscillates significantly with a large amplitude in the diagram. It should be noted that while keeping pFH to be -0.5 and changing pHH from -3 to 3, the trend is not very obvious as there are many anomalous values of momentum that render the reaction unreactive while slight increase and decrease of momentum make the reaction reactive again. For example, pHH = -0.9 is reactive but pHH = -1.1 is unreactive.
|-
|2.
|[[File:Q10plot2wxy0119.png|600px]]
| There is reaction between F and H2 and products are formed. The F-H bond formed vibrates significantly along BC trajectory.
|-
|3.
|[[File:Q10plot9wxy0119.png|600px]]
| There is no reaction under this condition between F and H2. There is not enough energy to cross over the transition state point. Although investigations have shown that there is reaction when rFH is equal or less than 2Å, but there is no reaction when H and F atoms are further apart.
|-
|4.
|[[File:Q10plot8wxy0119.png|600px]]
| There is reaction between F and H2. By comparison between condition 4 and condition 3, a slight increase in the momentum of F and H which indicates a slight increase in kinetic translational energy result in the formation of products. Investigations have shown that even when rFH is slightly larger than 2Å, the products are still formed. Thus the change in reaction conditions illustrate that the changes in translational energy affect the reactivity the reaction more than the vibrational energy.
|-
|5.
|[[File:Q10plot5wxy0119.png|600px]]
| There is no reaction between H and HF even though the kinetic translational energy is very high and much higher than the activation energy. (pHF = -20) The trajectory reverts back before reaching the transition state structure.
|-
|6.
|[[File:Q10plot4wxy0119.png|600px]]
| There is reaction between H and HF. Compared to condition 5, the translational energy decreases significantly while vibrational energy increases slightly. Thus, the vibrational energy is more efficient in promoting the reaction than the translational energy.
|}

The transition state illustrated in Question 7 is closer to F and H2 than H and HF. From the above illustrations, it can be shown that the translational energy is more efficient in promoting the reaction between F and H2 which is exothermic, has a lower activation energy and has an earlier transition state. On the other hand, the vibrational energy is more efficient in the reactivity of the reaction between H and HF which is endothermic, has a higher activation energy and has a late transition state. These observations agree with the Polanyi's empirical rules.<ref name="PoRule" />

== References ==

<references>
<ref name="TS">Eyring, H. (1935). "The Activated Complex in Chemical Reactions". J. Chem. Phys. 3 (2): 107–115.</ref>
<ref name="PoRule">Polanyi J C. Concepts in reaction dynamics[J]. Accounts of Chemical Research, 1972, 5(5): 161-168.</ref>
</references>

MRD:WXY0119

2018-05-18T16:57:07Z

Xw2816: /* Energy Distribution and Reactivity */

= Molecular Reaction Dynamics Report =

== EXERCISE 1ː H + H2 SYSTEM ==

[[File:HandH2wxy0119.png]]

In the H + H2 system, the distance between the first two H atoms (m1, m2) is r1 and the distance between the two H atoms in the H2 molecule (m2, m3) is r2.

=== Dynamics from the transition state region ===

{{fontcolor|red|Question 1 ː What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.}}

{| class="wikitable" border=1
|[[File:Q1TStransitionwxy0119.png|390px]]
|[[File:Q1TS2wxy0119.png|390px]]
|[[File:Q1TS3wxy0119.png|390px]]
|-
|Figure.1 Surface plot of reaction trajectory
|Figure.2 Transition state surface plot
|Figure.3 Transition state surface plot (different angle)
|}

In the reaction trajectory surface plot (Fig.1), AB is the distance r1 and BC is the distance r2. The gradient of the potential energy with regard to r1 and r2 are ∂V(r1)/∂r1 and ∂V(r2)/∂r2 respectively. The second derivatives of the potential energy with regard to r1 and r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 respectively.

At the two minimum structures which are at the two ends of the reaction pathway, ∂V(r1)/∂r1 = 0 and ∂V(r2)/∂r2 = 0 respectively. Since the two minimum structures are at two local minimum point , the second derivatives r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 are both greater than zero.

On the other hand, the components at the transition state structure (Fig.2) are different from that at minimum points. The gradient ∂V(r1)/∂r1 and ∂V(r2)/∂r2 are both equal to zero at the transition structure. However, r2 are ∂2V(r1)/∂r12 >0 and ∂2V(r2)/∂r22 <0. The transition state point is actually a saddle point which is observed more clearly in Fig.3. The Transition state point is thus distinguished from the minimum points as the potential energy surface curves inwards and downwards at the minimum points.

=== Locating the Transition State ===

{{fontcolor|red|Question 2ː 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.}}

{| class="wikitable" border=1
|[[File:TScontourwxy0119.png|390px]]
|[[File:TSsurfaceplwxy0119.png|390px]]
|[[File:TSestimatewxy0119.png|390px]]
|-
|Figure.4 Transition state contour plot
|Figure.5 Transition state surface plot
|Figure.6 Internuclear distance against time plot
|}

The best estimate of the transition state position is rts = 0.9078 Å.

When r1 = r2 = rts and momenta are set to zero, a graph of intermolecular distance against time is plotted. (Fig.6) In this graph, AB and BC lines are the same and overlap, the two lines on the graph are flat and horizontal, indicating that the atomic distances are the same at the position and with no momenta the state is at equilibrium, validating the position is the transition state position. The transition state is shown as a cross on the counter plot (Fig.4) and a dot on the surface plot (Fig.5).

=== Calculating and comparing the reaction path and trajectory ===

{{fontcolor1|red|Question 3ː Comment on how the ''mep'' and the trajectory you just calculated differ.}}

{| class="wikitable" border=1
|[[File:Mep1wxy0119.png|390px]]
|[[File:Mep2wxy0119.png|390px]]
|[[File:Mep3wxy0119.png|390px]]
|-
|Figure.7 mep calculation contour plot
|Figure.8 mep calculation surface plot
|Figure.9 mep calculation internuclear distance against time plot
|-
|[[File:Dynamic1wxy0119.png|390px]]
|[[File:Dynamic2wxy0119.png|390px]]
|[[File:Dynamic3wxy0119.png|390px]]
|-
|Figure.10 Dynamics calculation contour plot
|Figure.11 Dynamics calculation surface plot
|Figure.12 Dynamics calculation internuclear distance against time plot
|}
{| class="wikitable" border=1
|[[File:dynamics4wxy0119.png|600px]]
|[[File:Mep5wxy0119.png|600px]]
|-
|Figure.13 Dynamics calculation internuclear momenta against time plot
|Figure.14 mep calculation internuclear momenta against time plotsurface plot
|}

The mep (minimum energy path) trajectory is a smooth line on contour and surface plot (Fig.7 Fig.8) while the trajectory under dynamics calculation is oscillating along the pathway. (Fig.10 Fig.11) The mep trajectory takes more steps and longer time than the dynamics trajectory as shown in internuclear distance against time plots. (Fig. 9 Fig.12) The momentum is zero all the time under mep calculation (Fig.13) but under dynamics calculation, the momenta lines are oscillating against time.

The differences arise from mep and dynamics calculations can be explained by the the natures of the two calculations. Under mep calculation, the atoms are in extremely slow motion and the momentum is zero for each step, i.e. the atoms are "stop" after every step. The pathway under mep calculation is formed by connecting all the minimum energy points of each step and appears as a smooth line. On the other hand, atoms are in continuous motion and the momentum is accumulated after each step, atoms are then oscillating on the potential energy surface along the pathway. The steps are set to 5000 in the mep calculation but are only 500 in the dynamics calculatoin. Since the steps in mep are small, more time is needed to complete the same length on the trajectories.

=== Reactive and unreactive trajectories ===

{{fontcolor1|red|Question 4ː Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.}}

For the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, run trajectories with the following momenta p1 and p2 combinations:
{| class="wikitable" border=1
| Condition
| p1
| p2
| Total Energy/ kcal mol -1
| Reactivity
|-
|1.
| -1.25
| -2.5
| -99.018
| reactive
|-
|2.
| -1.5
| -2.0
| -100.456
| unreactive
|-
|3.
| -1.5
| -2.5
| -98.956
| reactive
|-
|4.
| -2.5
| -5.0
| -84.956
| unreactive
|-
|5.
| -2.5
| -5.2
| -83.416
| reactive
|}

{| class="wikitable" border=1
| Condition
| Surface Plot
| Contour Plot
| Description
|-
|1.
| [[File:condition1wxy0119.png|350px]]
| [[File:firstwxy0119.png|350px]]
| r2 (BC) decreases when HC approaches bonded HA and HB. The energy processed by the system is sufficient to overcome the activation barrier and crosses the transition state structure to break the HA and HB bond and form a new HB and HC bond. The new bond oscillate as r1 (AB) increases.
|-
|2.
| [[File:condition2wxy0119.png|350px]]
| [[File:secondwxy0119.png|350px]]
| HC approaches HB but the energy is insufficient to reach the transition state point, HC then moves further away from the bonded H2 molecule and no new bond is formed. The oscillation along r2 is a result of an increase in momentum p1.
|-
|3.
| [[File:condition3wxy0119.png|350px]]
| [[File:thirdwxy0119.png|350px]]
| Similarly to condition 1 but with a more negative p1, HA - HB bond oscillates more when r2 decreases. The reaction proceeds and crosses the transition state point to form the products.
|-
|4.
| [[File:condition4wxy0119.png|350px]]
| [[File:fourthwxy0119.png|350px]]
| Both p1 and p2 are more negative than the previous conditions. The HA - HB bond oscillates vigorously with larger amplitudes. The barrier recrossing happens after the system crosses the transition state region but eventually reverts back to the reactants.
|-
|5.
| [[File:condition5wxy0119.png|350px]]
| [[File:fifthwxy0119.png|350px]]
| With a slight more negative value of p1 as compared to that in condition 4, the system now processes appropriate energy to proceed to forming products. The vigorous oscillations are resulting from high momenta and barrier recrossing occurs.
|}

=== Transition State Theory ===

{{fontcolor|red|Question 5ː State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}

The main assumptions of Transition State Theory (TST) are<ref name="TS" />ː

1. The behaviours of the atomic nuclei follow the classical mechanics rules, i.e. the transition structure is formed as long as atoms or molecules collide with enough energy.

2. The intermediates have a long enough lifetime for the energies to be Boltzmann distributed preceding the next step.

3. The lowest energy saddle point on the potential energy surface is passed over in the reaction system.

The TST does not take into account of the quantum tunnelling effect which allows reactions to happen even when the energy carried by the system is lower than the activation energy barrier, especially for reactions with low energy barriers. Nor does the theory consider the intermediates with short lifetimes. When the energies are not fully distributed, the momentum of the reaction trajectory from the reactants to the intermediate can have effects on product selectivity. Moreover, the theory fails at high temperature because higher vibrational energy modes can be populated for a molecule at high temperature. The molecules can have complex motion and their collisions may result in a higher energy transition structure. The experimental results may thus deviate from that predicted by the TST. For example in condition 4, the kinetic energy of the reactants is much higher than the activation energy required. However, the momenta between atoms are very high and higher vibrational modes are populated that result in higher energy transition structures formed and the system does not pass though the lowest transition saddle point as shown in the diagram. Barrier recrossing occurs but the products are not formed as opposed to the predictions by the TST.

== EXERCISE 2: F - H - H SYSTEM ==

=== Reaction Energetics ===

{{fontcolor1|red|Question 6ː 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?}}

{| class="wikitable" border=1
|[[File:Q6FH2wxy0119.png|600px]]
|[[File:Q6HFHwxy0119.png|600px]]
|-
|Figure.15 F + H2 position on surface plot
|Figure.16 HF and H position on surface plot
|}

The surface plots of the potential energy surface of H + HF → F + H2 reaction is shown (Fig.15 Fig.16). Distance AB is the distance between H and F atoms and distance BC is the distance between two H atoms. The two minimum points shown in the graph are positions of F + H2 (Fig.15) and H + HF (Fig.16) respectively. If forward reaction is H + HF → F + H2, then the backward reaction is F + H2 → H + HF. As clearly shown in the diagram F + H2 are at a minimum point of higher potential energy than that of H + HF. Thus, by comparing the potential energy of reactants and products, the signs of changes in enthalpy (ΔH) and the energetics are determined. H + HF → F + H2 is endothermic and F + H2 → H + HF is exothermic.

The energetics of the two reactions are in accordance to the predictions by comparing bond strengths of chemical species. The bond energy of H-F bond (565 kJ/mol) is higher than the bond energy of H-H bond (432 kJ/mol). Thus, energy released by forming the H-H bond is not enough to compensate for energy required for breaking the H-F bond and the reaction H + HF → F + H2 is endothermic. On the contrary, the reaction F + H2 → H + HF is exothermic.

=== Transition State Approximation ===

{{fontcolor1|red|Question 7ː Locate the approximate position of the transition state.}}

{| class="wikitable" border=1
|[[File:Q7TStransitionwxy0119.png|390px]]
|[[File:Q7TS2wxy0119.png|390px]]
|[[File:Q7TS3wxy0119.png|390px]]
|-
|Figure.17 Transition state surface plot
|Figure.18 Transition state contour plot
|Figure.19 Internuclear distance against time plot (at TS point)
|}

The transition state (TS) is shown as a black dot on the potential energy surface plot (Fig.17) and a red cross on the contour plot. (Fig,18) The position of the TS is at the point where the distance between F and H (rFH) is 1.810Å and the distance between two H atoms (rHH) is 0.746Å. The TS is closer to F + H2 and further away from H + HF. At the transition state position, the distance between F and H, between two H atoms are constant with no momentum as shown in the internuclear distance against time plot as horizontal and flat lines, validating the position is the transition state point (Fig.19)

=== Activation Energies ===

{{fontcolor1|red|Question 8ː Report the activation energy for both reactions.}}

{| class="wikitable" border=1
|[[File:Q8Ea2wxy0119.png|600px]]
|[[File:Q8Ea1wxy0119.png|600px]]
|-
|Figure.21 Energy against time plot (HF + H)
|Figure.20 Energy against time plot (F + H2)
|}

By performing MEP calculation with slight increase and decrease of the rFH to 1.820Å to perform F + H2 → H + HF reaction and to 1.800Å to perform H + HF → F + H2 reaction, the activation energies (Ea) of both reactions can be calculated from the differences in their potential energies (Fig.20 Fig.21)ː

H + HF → F + H2ː Ea = -103.751 - (-133.624) = +29.873 kcal mol-1

F + H2 → H + HFː Ea = -103.751 - (-103.972) = + 0.221 kcal mol-1

=== Reaction Dynamics ===

{{fontcolor1|red|Question 9ː In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

Reactionː F + H2 → H + HF

Initial condition setː rHF = 2Å rHH = 0.74Å pHF = -0.5 pHH = 0.1

{| class="wikitable" border=1
|[[File:Q9plot1wxy0119.png|390px]]
|[[File:Q9plot2wxy0119.png|390px]]
|[[File:Q9plot3wxy0119.png|390px]]
|-
|Figure.23 Contour plot
|Figure.24 Surface plot
|Figure.25 Internuclear momentum against time plot
|-
|[[File:Q9ani1wxy0119.png|390px]]
|[[File:Q9ani2wxy0119.png|390px]]
|-
|Figure.26 Animation figure (at the start of the reaction)
|Figure.27 Animation figure (at the end of the reaction)
|}

As shown in the plots and animation figures, the H-H bond in the reactants is broken and H-F bond in the products is formed. There are small momenta between both H and F and H-H bond that provide the initial kinetic energy of the reactants and the initial vibrational energy of the H-H bond in the reactants. The initial energy is required to break the H-H bond and reduce the distance of H and F for bond forming and potential energy is released when H-F bond is formed. Since energy is conserved, the reaction energy released is converted to the vibrational energy of the H-F bond and the translational kinetic energies of the products. This is shown in the internuclear momentum against time graph as the initial vibrations of B-C (H-H bond) becomes flat as two H atoms are separated in the product and the A-B (H-F bond) formed in the product vibrates with a large amplitude. (Fig.25) The contour plot and the surface plot also clearly show there is a significant increase in the amplitude of the oscillation of H-F bond which is an indication of the large vibrational energy of H-F bond. (Fig.23 Fig.24)

The predictions can be proved experimentally. The vibrational energy of H-F bond can be determined by measuring the frequency and the intensity of the absorption band of the H-F bond with infrared spectroscopy. The conversion to the kinetic energies of the products can be observed by measuring the temperature of the reaction as the kinetic energy can be converted to thermal energy as products collide with solvent molecules. The temperature is expected to increase as this is an exothermic reaction.

=== Energy Distribution and Reactivity ===

{{fontcolor|red|Question 10ː 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.}}
{| class="wikitable" border=1
| Condition
| rFH /Å
| rHH /Å
| pFH
| pHH
|-
|1.
| 2
| 0.74
| -0.5
| -3
|-
|2.
| 2
| 0.74
| -0.5
| 1.9
|-
|3.
| 2.1
| 0.74
| -0.5
| 0.1
|-
|4.
| 2
| 0.74
| -0.8
| 0.1
|-
|5.
| 0.91
| 2
| 0.05
| -20
|-
|6.
| 0.91
| 2
| 0.8
| -7.5
|}
{| class="wikitable" border=1
|Condition
|Contour plot
|Observation
|-
|1.
|[[File:Q10plot1wxy0119.png|600px]]
| The products are not formed in the reaction between F and H2 even pHH is very large. Barrier recrossing occurs and the H-H bond oscillates significantly with a large amplitude in the diagram. It should be noted that while keeping pFH to be -0.5 and changing pHH from -3 to 3, the trend is not very obvious as there are many anomalous values of momentum that render the reaction unreactive while slight increase and decrease of momentum make the reaction reactive again. For example, pHH = -0.9 is reactive but pHH = -1.1 is unreactive.
|-
|2.
|[[File:Q10plot2wxy0119.png|600px]]
| There is reaction between F and H2 and products are formed. The F-H bond formed vibrates significantly along BC trajectory.
|-
|3.
|[[File:Q10plot9wxy0119.png|600px]]
| There is no reaction under this condition between F and H2. There is not enough energy to cross over the transition state point. Although investigations have shown that there is reaction when rFH is equal or less than 2Å, but there is no reaction when H and F atoms are further apart.
|-
|4.
|[[File:Q10plot8wxy0119.png|600px]]
| There is reaction between F and H2. By comparison between condition 4 and condition 3, a slight increase in the momentum of F and H which indicates a slight increase in kinetic translational energy result in the formation of products. Investigations have shown that even when rFH is slightly larger than 2Å, the products are still formed. Thus the change in reaction conditions illustrate that the changes in translational energy affect the reactivity the reaction more than the vibrational energy.
|-
|5.
|[[File:Q10plot5wxy0119.png|600px]]
| There is no reaction between H and HF even though the kinetic translational energy is very high and much higher than the activation energy. (pHF = -20) The trajectory reverts back before reaching the transition state structure.
|-
|6.
|[[File:Q10plot4wxy0119.png|600px]]
| There is reaction between H and HF. Compared to condition 5, the translational energy decreases significantly while vibrational energy increases slightly. Thus, the vibrational energy is more efficient in promoting the reaction than the translational energy.
|}

The transition state illustrated in Question 7 is closer to F and H2 than H and HF. From the above illustrations, it can be shown that the translational energy is more efficient in promoting the reaction between F and H2 which is exothermic, has a lower activation energy and has an earlier transition state. On the other hand, the vibrational energy is more efficient in the reactivity of the reaction between H and HF which is endothermic, has a higher activation energy and has a late transition state. These observations agree with the Polanyi's empirical rules.<ref name="PoRule" />

== References ==

<references>
<ref name="TS">Eyring, H. (1935). "The Activated Complex in Chemical Reactions". J. Chem. Phys. 3 (2): 107–115.</ref>
<ref name="PoRule">Polanyi J C. Concepts in reaction dynamics[J]. Accounts of Chemical Research, 1972, 5(5): 161-168.</ref>
</references>

MRD:WXY0119

2018-05-18T16:54:03Z

Xw2816: /* References */

= Molecular Reaction Dynamics Report =

== EXERCISE 1ː H + H2 SYSTEM ==

[[File:HandH2wxy0119.png]]

In the H + H2 system, the distance between the first two H atoms (m1, m2) is r1 and the distance between the two H atoms in the H2 molecule (m2, m3) is r2.

=== Dynamics from the transition state region ===

{{fontcolor|red|Question 1 ː What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.}}

{| class="wikitable" border=1
|[[File:Q1TStransitionwxy0119.png|390px]]
|[[File:Q1TS2wxy0119.png|390px]]
|[[File:Q1TS3wxy0119.png|390px]]
|-
|Figure.1 Surface plot of reaction trajectory
|Figure.2 Transition state surface plot
|Figure.3 Transition state surface plot (different angle)
|}

In the reaction trajectory surface plot (Fig.1), AB is the distance r1 and BC is the distance r2. The gradient of the potential energy with regard to r1 and r2 are ∂V(r1)/∂r1 and ∂V(r2)/∂r2 respectively. The second derivatives of the potential energy with regard to r1 and r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 respectively.

At the two minimum structures which are at the two ends of the reaction pathway, ∂V(r1)/∂r1 = 0 and ∂V(r2)/∂r2 = 0 respectively. Since the two minimum structures are at two local minimum point , the second derivatives r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 are both greater than zero.

On the other hand, the components at the transition state structure (Fig.2) are different from that at minimum points. The gradient ∂V(r1)/∂r1 and ∂V(r2)/∂r2 are both equal to zero at the transition structure. However, r2 are ∂2V(r1)/∂r12 >0 and ∂2V(r2)/∂r22 <0. The transition state point is actually a saddle point which is observed more clearly in Fig.3. The Transition state point is thus distinguished from the minimum points as the potential energy surface curves inwards and downwards at the minimum points.

=== Locating the Transition State ===

{{fontcolor|red|Question 2ː 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.}}

{| class="wikitable" border=1
|[[File:TScontourwxy0119.png|390px]]
|[[File:TSsurfaceplwxy0119.png|390px]]
|[[File:TSestimatewxy0119.png|390px]]
|-
|Figure.4 Transition state contour plot
|Figure.5 Transition state surface plot
|Figure.6 Internuclear distance against time plot
|}

The best estimate of the transition state position is rts = 0.9078 Å.

When r1 = r2 = rts and momenta are set to zero, a graph of intermolecular distance against time is plotted. (Fig.6) In this graph, AB and BC lines are the same and overlap, the two lines on the graph are flat and horizontal, indicating that the atomic distances are the same at the position and with no momenta the state is at equilibrium, validating the position is the transition state position. The transition state is shown as a cross on the counter plot (Fig.4) and a dot on the surface plot (Fig.5).

=== Calculating and comparing the reaction path and trajectory ===

{{fontcolor1|red|Question 3ː Comment on how the ''mep'' and the trajectory you just calculated differ.}}

{| class="wikitable" border=1
|[[File:Mep1wxy0119.png|390px]]
|[[File:Mep2wxy0119.png|390px]]
|[[File:Mep3wxy0119.png|390px]]
|-
|Figure.7 mep calculation contour plot
|Figure.8 mep calculation surface plot
|Figure.9 mep calculation internuclear distance against time plot
|-
|[[File:Dynamic1wxy0119.png|390px]]
|[[File:Dynamic2wxy0119.png|390px]]
|[[File:Dynamic3wxy0119.png|390px]]
|-
|Figure.10 Dynamics calculation contour plot
|Figure.11 Dynamics calculation surface plot
|Figure.12 Dynamics calculation internuclear distance against time plot
|}
{| class="wikitable" border=1
|[[File:dynamics4wxy0119.png|600px]]
|[[File:Mep5wxy0119.png|600px]]
|-
|Figure.13 Dynamics calculation internuclear momenta against time plot
|Figure.14 mep calculation internuclear momenta against time plotsurface plot
|}

The mep (minimum energy path) trajectory is a smooth line on contour and surface plot (Fig.7 Fig.8) while the trajectory under dynamics calculation is oscillating along the pathway. (Fig.10 Fig.11) The mep trajectory takes more steps and longer time than the dynamics trajectory as shown in internuclear distance against time plots. (Fig. 9 Fig.12) The momentum is zero all the time under mep calculation (Fig.13) but under dynamics calculation, the momenta lines are oscillating against time.

The differences arise from mep and dynamics calculations can be explained by the the natures of the two calculations. Under mep calculation, the atoms are in extremely slow motion and the momentum is zero for each step, i.e. the atoms are "stop" after every step. The pathway under mep calculation is formed by connecting all the minimum energy points of each step and appears as a smooth line. On the other hand, atoms are in continuous motion and the momentum is accumulated after each step, atoms are then oscillating on the potential energy surface along the pathway. The steps are set to 5000 in the mep calculation but are only 500 in the dynamics calculatoin. Since the steps in mep are small, more time is needed to complete the same length on the trajectories.

=== Reactive and unreactive trajectories ===

{{fontcolor1|red|Question 4ː Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.}}

For the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, run trajectories with the following momenta p1 and p2 combinations:
{| class="wikitable" border=1
| Condition
| p1
| p2
| Total Energy/ kcal mol -1
| Reactivity
|-
|1.
| -1.25
| -2.5
| -99.018
| reactive
|-
|2.
| -1.5
| -2.0
| -100.456
| unreactive
|-
|3.
| -1.5
| -2.5
| -98.956
| reactive
|-
|4.
| -2.5
| -5.0
| -84.956
| unreactive
|-
|5.
| -2.5
| -5.2
| -83.416
| reactive
|}

{| class="wikitable" border=1
| Condition
| Surface Plot
| Contour Plot
| Description
|-
|1.
| [[File:condition1wxy0119.png|350px]]
| [[File:firstwxy0119.png|350px]]
| r2 (BC) decreases when HC approaches bonded HA and HB. The energy processed by the system is sufficient to overcome the activation barrier and crosses the transition state structure to break the HA and HB bond and form a new HB and HC bond. The new bond oscillate as r1 (AB) increases.
|-
|2.
| [[File:condition2wxy0119.png|350px]]
| [[File:secondwxy0119.png|350px]]
| HC approaches HB but the energy is insufficient to reach the transition state point, HC then moves further away from the bonded H2 molecule and no new bond is formed. The oscillation along r2 is a result of an increase in momentum p1.
|-
|3.
| [[File:condition3wxy0119.png|350px]]
| [[File:thirdwxy0119.png|350px]]
| Similarly to condition 1 but with a more negative p1, HA - HB bond oscillates more when r2 decreases. The reaction proceeds and crosses the transition state point to form the products.
|-
|4.
| [[File:condition4wxy0119.png|350px]]
| [[File:fourthwxy0119.png|350px]]
| Both p1 and p2 are more negative than the previous conditions. The HA - HB bond oscillates vigorously with larger amplitudes. The barrier recrossing happens after the system crosses the transition state region but eventually reverts back to the reactants.
|-
|5.
| [[File:condition5wxy0119.png|350px]]
| [[File:fifthwxy0119.png|350px]]
| With a slight more negative value of p1 as compared to that in condition 4, the system now processes appropriate energy to proceed to forming products. The vigorous oscillations are resulting from high momenta and barrier recrossing occurs.
|}

=== Transition State Theory ===

{{fontcolor|red|Question 5ː State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}

The main assumptions of Transition State Theory (TST) are<ref name="TS" />ː

1. The behaviours of the atomic nuclei follow the classical mechanics rules, i.e. the transition structure is formed as long as atoms or molecules collide with enough energy.

2. The intermediates have a long enough lifetime for the energies to be Boltzmann distributed preceding the next step.

3. The lowest energy saddle point on the potential energy surface is passed over in the reaction system.

The TST does not take into account of the quantum tunnelling effect which allows reactions to happen even when the energy carried by the system is lower than the activation energy barrier, especially for reactions with low energy barriers. Nor does the theory consider the intermediates with short lifetimes. When the energies are not fully distributed, the momentum of the reaction trajectory from the reactants to the intermediate can have effects on product selectivity. Moreover, the theory fails at high temperature because higher vibrational energy modes can be populated for a molecule at high temperature. The molecules can have complex motion and their collisions may result in a higher energy transition structure. The experimental results may thus deviate from that predicted by the TST. For example in condition 4, the kinetic energy of the reactants is much higher than the activation energy required. However, the momenta between atoms are very high and higher vibrational modes are populated that result in higher energy transition structures formed and the system does not pass though the lowest transition saddle point as shown in the diagram. Barrier recrossing occurs but the products are not formed as opposed to the predictions by the TST.

== EXERCISE 2: F - H - H SYSTEM ==

=== Reaction Energetics ===

{{fontcolor1|red|Question 6ː 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?}}

{| class="wikitable" border=1
|[[File:Q6FH2wxy0119.png|600px]]
|[[File:Q6HFHwxy0119.png|600px]]
|-
|Figure.15 F + H2 position on surface plot
|Figure.16 HF and H position on surface plot
|}

The surface plots of the potential energy surface of H + HF → F + H2 reaction is shown (Fig.15 Fig.16). Distance AB is the distance between H and F atoms and distance BC is the distance between two H atoms. The two minimum points shown in the graph are positions of F + H2 (Fig.15) and H + HF (Fig.16) respectively. If forward reaction is H + HF → F + H2, then the backward reaction is F + H2 → H + HF. As clearly shown in the diagram F + H2 are at a minimum point of higher potential energy than that of H + HF. Thus, by comparing the potential energy of reactants and products, the signs of changes in enthalpy (ΔH) and the energetics are determined. H + HF → F + H2 is endothermic and F + H2 → H + HF is exothermic.

The energetics of the two reactions are in accordance to the predictions by comparing bond strengths of chemical species. The bond energy of H-F bond (565 kJ/mol) is higher than the bond energy of H-H bond (432 kJ/mol). Thus, energy released by forming the H-H bond is not enough to compensate for energy required for breaking the H-F bond and the reaction H + HF → F + H2 is endothermic. On the contrary, the reaction F + H2 → H + HF is exothermic.

=== Transition State Approximation ===

{{fontcolor1|red|Question 7ː Locate the approximate position of the transition state.}}

{| class="wikitable" border=1
|[[File:Q7TStransitionwxy0119.png|390px]]
|[[File:Q7TS2wxy0119.png|390px]]
|[[File:Q7TS3wxy0119.png|390px]]
|-
|Figure.17 Transition state surface plot
|Figure.18 Transition state contour plot
|Figure.19 Internuclear distance against time plot (at TS point)
|}

The transition state (TS) is shown as a black dot on the potential energy surface plot (Fig.17) and a red cross on the contour plot. (Fig,18) The position of the TS is at the point where the distance between F and H (rFH) is 1.810Å and the distance between two H atoms (rHH) is 0.746Å. The TS is closer to F + H2 and further away from H + HF. At the transition state position, the distance between F and H, between two H atoms are constant with no momentum as shown in the internuclear distance against time plot as horizontal and flat lines, validating the position is the transition state point (Fig.19)

=== Activation Energies ===

{{fontcolor1|red|Question 8ː Report the activation energy for both reactions.}}

{| class="wikitable" border=1
|[[File:Q8Ea2wxy0119.png|600px]]
|[[File:Q8Ea1wxy0119.png|600px]]
|-
|Figure.21 Energy against time plot (HF + H)
|Figure.20 Energy against time plot (F + H2)
|}

By performing MEP calculation with slight increase and decrease of the rFH to 1.820Å to perform F + H2 → H + HF reaction and to 1.800Å to perform H + HF → F + H2 reaction, the activation energies (Ea) of both reactions can be calculated from the differences in their potential energies (Fig.20 Fig.21)ː

H + HF → F + H2ː Ea = -103.751 - (-133.624) = +29.873 kcal mol-1

F + H2 → H + HFː Ea = -103.751 - (-103.972) = + 0.221 kcal mol-1

=== Reaction Dynamics ===

{{fontcolor1|red|Question 9ː In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

Reactionː F + H2 → H + HF

Initial condition setː rHF = 2Å rHH = 0.74Å pHF = -0.5 pHH = 0.1

{| class="wikitable" border=1
|[[File:Q9plot1wxy0119.png|390px]]
|[[File:Q9plot2wxy0119.png|390px]]
|[[File:Q9plot3wxy0119.png|390px]]
|-
|Figure.23 Contour plot
|Figure.24 Surface plot
|Figure.25 Internuclear momentum against time plot
|-
|[[File:Q9ani1wxy0119.png|390px]]
|[[File:Q9ani2wxy0119.png|390px]]
|-
|Figure.26 Animation figure (at the start of the reaction)
|Figure.27 Animation figure (at the end of the reaction)
|}

As shown in the plots and animation figures, the H-H bond in the reactants is broken and H-F bond in the products is formed. There are small momenta between both H and F and H-H bond that provide the initial kinetic energy of the reactants and the initial vibrational energy of the H-H bond in the reactants. The initial energy is required to break the H-H bond and reduce the distance of H and F for bond forming and potential energy is released when H-F bond is formed. Since energy is conserved, the reaction energy released is converted to the vibrational energy of the H-F bond and the translational kinetic energies of the products. This is shown in the internuclear momentum against time graph as the initial vibrations of B-C (H-H bond) becomes flat as two H atoms are separated in the product and the A-B (H-F bond) formed in the product vibrates with a large amplitude. (Fig.25) The contour plot and the surface plot also clearly show there is a significant increase in the amplitude of the oscillation of H-F bond which is an indication of the large vibrational energy of H-F bond. (Fig.23 Fig.24)

The predictions can be proved experimentally. The vibrational energy of H-F bond can be determined by measuring the frequency and the intensity of the absorption band of the H-F bond with infrared spectroscopy. The conversion to the kinetic energies of the products can be observed by measuring the temperature of the reaction as the kinetic energy can be converted to thermal energy as products collide with solvent molecules. The temperature is expected to increase as this is an exothermic reaction.

=== Energy Distribution and Reactivity ===

{{fontcolor|red|Question 10ː 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.}}
{| class="wikitable" border=1
| Condition
| rFH /Å
| rHH /Å
| pFH
| pHH
|-
|1.
| 2
| 0.74
| -0.5
| -3
|-
|2.
| 2
| 0.74
| -0.5
| 1.9
|-
|3.
| 2.1
| 0.74
| -0.5
| 0.1
|-
|4.
| 2
| 0.74
| -0.8
| 0.1
|-
|5.
| 0.91
| 2
| 0.05
| -20
|-
|6.
| 0.91
| 2
| 0.8
| -7.5
|}
{| class="wikitable" border=1
|Condition
|Contour plot
|Observation
|-
|1.
|[[File:Q10plot1wxy0119.png|600px]]
| The products are not formed in the reaction between F and H2 even pHH is very large. Barrier recrossing occurs and the H-H bond oscillates significantly with a large amplitude in the diagram. It should be noted that while keeping pFH to be -0.5 and changing pHH from -3 to 3, the trend is not very obvious as there are many anomalous values of momentum that render the reaction unreactive while slight increase and decrease of momentum make the reaction reactive again. For example, pHH = -0.9 is reactive but pHH = -1.1 is unreactive.
|-
|2.
|[[File:Q10plot2wxy0119.png|600px]]
| There is reaction between F and H2 and products are formed. The F-H bond formed vibrates significantly along BC trajectory.
|-
|3.
|[[File:Q10plot9wxy0119.png|600px]]
| There is no reaction under this condition between F and H2. There is not enough energy to cross over the transition state point. Although investigations have shown that there is reaction when rFH is equal or less than 2Å, but there is no reaction when H and F atoms are further apart.
|-
|4.
|[[File:Q10plot8wxy0119.png|600px]]
| There is reaction between F and H2. By comparison between condition 4 and condition 3, a slight increase in the momentum of F and H which indicates a slight increase in kinetic translational energy result in the formation of products. Investigations have shown that even when rFH is slightly larger than 2Å, the products are still formed. Thus the change in reaction conditions illustrate that the changes in translational energy affect the reactivity the reaction more than the vibrational energy.
|-
|5.
|[[File:Q10plot5wxy0119.png|600px]]
| There is no reaction between H and HF even though the kinetic translational energy is very high and much higher than the activation energy. (pHF = -20) The trajectory reverts back before reaching the transition state structure.
|-
|6.
|[[File:Q10plot4wxy0119.png|600px]]
| There is reaction between H and HF. Compared to condition 5, the translational energy is reduced significantly while vibrational energy increases slightly. Thus, the vibrational energy is more efficient in promoting the reaction than the translational energy.
|}

The transition state illustrated in Question 7 is closer to F and H2 than H and HF. From the above illustrations, it can be shown that the translational energy is more efficient in promoting the reaction between F and H2 which is exothermic, has a lower activation energy and has an earlier transition state. On the other hand, the vibrational energy is more efficient in the reactivity of the reaction between H and HF which is endothermic, has a higher activation energy and has a late transition state. These observations agree with the Polanyi's empirical rules.<ref name="PoRule" />

== References ==

<references>
<ref name="TS">Eyring, H. (1935). "The Activated Complex in Chemical Reactions". J. Chem. Phys. 3 (2): 107–115.</ref>
<ref name="PoRule">Polanyi J C. Concepts in reaction dynamics[J]. Accounts of Chemical Research, 1972, 5(5): 161-168.</ref>
</references>

MRD:WXY0119

2018-05-18T16:53:21Z

Xw2816: /* References */

= Molecular Reaction Dynamics Report =

== EXERCISE 1ː H + H2 SYSTEM ==

[[File:HandH2wxy0119.png]]

In the H + H2 system, the distance between the first two H atoms (m1, m2) is r1 and the distance between the two H atoms in the H2 molecule (m2, m3) is r2.

=== Dynamics from the transition state region ===

{{fontcolor|red|Question 1 ː What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.}}

{| class="wikitable" border=1
|[[File:Q1TStransitionwxy0119.png|390px]]
|[[File:Q1TS2wxy0119.png|390px]]
|[[File:Q1TS3wxy0119.png|390px]]
|-
|Figure.1 Surface plot of reaction trajectory
|Figure.2 Transition state surface plot
|Figure.3 Transition state surface plot (different angle)
|}

In the reaction trajectory surface plot (Fig.1), AB is the distance r1 and BC is the distance r2. The gradient of the potential energy with regard to r1 and r2 are ∂V(r1)/∂r1 and ∂V(r2)/∂r2 respectively. The second derivatives of the potential energy with regard to r1 and r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 respectively.

At the two minimum structures which are at the two ends of the reaction pathway, ∂V(r1)/∂r1 = 0 and ∂V(r2)/∂r2 = 0 respectively. Since the two minimum structures are at two local minimum point , the second derivatives r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 are both greater than zero.

On the other hand, the components at the transition state structure (Fig.2) are different from that at minimum points. The gradient ∂V(r1)/∂r1 and ∂V(r2)/∂r2 are both equal to zero at the transition structure. However, r2 are ∂2V(r1)/∂r12 >0 and ∂2V(r2)/∂r22 <0. The transition state point is actually a saddle point which is observed more clearly in Fig.3. The Transition state point is thus distinguished from the minimum points as the potential energy surface curves inwards and downwards at the minimum points.

=== Locating the Transition State ===

{{fontcolor|red|Question 2ː 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.}}

{| class="wikitable" border=1
|[[File:TScontourwxy0119.png|390px]]
|[[File:TSsurfaceplwxy0119.png|390px]]
|[[File:TSestimatewxy0119.png|390px]]
|-
|Figure.4 Transition state contour plot
|Figure.5 Transition state surface plot
|Figure.6 Internuclear distance against time plot
|}

The best estimate of the transition state position is rts = 0.9078 Å.

When r1 = r2 = rts and momenta are set to zero, a graph of intermolecular distance against time is plotted. (Fig.6) In this graph, AB and BC lines are the same and overlap, the two lines on the graph are flat and horizontal, indicating that the atomic distances are the same at the position and with no momenta the state is at equilibrium, validating the position is the transition state position. The transition state is shown as a cross on the counter plot (Fig.4) and a dot on the surface plot (Fig.5).

=== Calculating and comparing the reaction path and trajectory ===

{{fontcolor1|red|Question 3ː Comment on how the ''mep'' and the trajectory you just calculated differ.}}

{| class="wikitable" border=1
|[[File:Mep1wxy0119.png|390px]]
|[[File:Mep2wxy0119.png|390px]]
|[[File:Mep3wxy0119.png|390px]]
|-
|Figure.7 mep calculation contour plot
|Figure.8 mep calculation surface plot
|Figure.9 mep calculation internuclear distance against time plot
|-
|[[File:Dynamic1wxy0119.png|390px]]
|[[File:Dynamic2wxy0119.png|390px]]
|[[File:Dynamic3wxy0119.png|390px]]
|-
|Figure.10 Dynamics calculation contour plot
|Figure.11 Dynamics calculation surface plot
|Figure.12 Dynamics calculation internuclear distance against time plot
|}
{| class="wikitable" border=1
|[[File:dynamics4wxy0119.png|600px]]
|[[File:Mep5wxy0119.png|600px]]
|-
|Figure.13 Dynamics calculation internuclear momenta against time plot
|Figure.14 mep calculation internuclear momenta against time plotsurface plot
|}

The mep (minimum energy path) trajectory is a smooth line on contour and surface plot (Fig.7 Fig.8) while the trajectory under dynamics calculation is oscillating along the pathway. (Fig.10 Fig.11) The mep trajectory takes more steps and longer time than the dynamics trajectory as shown in internuclear distance against time plots. (Fig. 9 Fig.12) The momentum is zero all the time under mep calculation (Fig.13) but under dynamics calculation, the momenta lines are oscillating against time.

The differences arise from mep and dynamics calculations can be explained by the the natures of the two calculations. Under mep calculation, the atoms are in extremely slow motion and the momentum is zero for each step, i.e. the atoms are "stop" after every step. The pathway under mep calculation is formed by connecting all the minimum energy points of each step and appears as a smooth line. On the other hand, atoms are in continuous motion and the momentum is accumulated after each step, atoms are then oscillating on the potential energy surface along the pathway. The steps are set to 5000 in the mep calculation but are only 500 in the dynamics calculatoin. Since the steps in mep are small, more time is needed to complete the same length on the trajectories.

=== Reactive and unreactive trajectories ===

{{fontcolor1|red|Question 4ː Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.}}

For the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, run trajectories with the following momenta p1 and p2 combinations:
{| class="wikitable" border=1
| Condition
| p1
| p2
| Total Energy/ kcal mol -1
| Reactivity
|-
|1.
| -1.25
| -2.5
| -99.018
| reactive
|-
|2.
| -1.5
| -2.0
| -100.456
| unreactive
|-
|3.
| -1.5
| -2.5
| -98.956
| reactive
|-
|4.
| -2.5
| -5.0
| -84.956
| unreactive
|-
|5.
| -2.5
| -5.2
| -83.416
| reactive
|}

{| class="wikitable" border=1
| Condition
| Surface Plot
| Contour Plot
| Description
|-
|1.
| [[File:condition1wxy0119.png|350px]]
| [[File:firstwxy0119.png|350px]]
| r2 (BC) decreases when HC approaches bonded HA and HB. The energy processed by the system is sufficient to overcome the activation barrier and crosses the transition state structure to break the HA and HB bond and form a new HB and HC bond. The new bond oscillate as r1 (AB) increases.
|-
|2.
| [[File:condition2wxy0119.png|350px]]
| [[File:secondwxy0119.png|350px]]
| HC approaches HB but the energy is insufficient to reach the transition state point, HC then moves further away from the bonded H2 molecule and no new bond is formed. The oscillation along r2 is a result of an increase in momentum p1.
|-
|3.
| [[File:condition3wxy0119.png|350px]]
| [[File:thirdwxy0119.png|350px]]
| Similarly to condition 1 but with a more negative p1, HA - HB bond oscillates more when r2 decreases. The reaction proceeds and crosses the transition state point to form the products.
|-
|4.
| [[File:condition4wxy0119.png|350px]]
| [[File:fourthwxy0119.png|350px]]
| Both p1 and p2 are more negative than the previous conditions. The HA - HB bond oscillates vigorously with larger amplitudes. The barrier recrossing happens after the system crosses the transition state region but eventually reverts back to the reactants.
|-
|5.
| [[File:condition5wxy0119.png|350px]]
| [[File:fifthwxy0119.png|350px]]
| With a slight more negative value of p1 as compared to that in condition 4, the system now processes appropriate energy to proceed to forming products. The vigorous oscillations are resulting from high momenta and barrier recrossing occurs.
|}

=== Transition State Theory ===

{{fontcolor|red|Question 5ː State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}

The main assumptions of Transition State Theory (TST) are<ref name="TS" />ː

1. The behaviours of the atomic nuclei follow the classical mechanics rules, i.e. the transition structure is formed as long as atoms or molecules collide with enough energy.

2. The intermediates have a long enough lifetime for the energies to be Boltzmann distributed preceding the next step.

3. The lowest energy saddle point on the potential energy surface is passed over in the reaction system.

The TST does not take into account of the quantum tunnelling effect which allows reactions to happen even when the energy carried by the system is lower than the activation energy barrier, especially for reactions with low energy barriers. Nor does the theory consider the intermediates with short lifetimes. When the energies are not fully distributed, the momentum of the reaction trajectory from the reactants to the intermediate can have effects on product selectivity. Moreover, the theory fails at high temperature because higher vibrational energy modes can be populated for a molecule at high temperature. The molecules can have complex motion and their collisions may result in a higher energy transition structure. The experimental results may thus deviate from that predicted by the TST. For example in condition 4, the kinetic energy of the reactants is much higher than the activation energy required. However, the momenta between atoms are very high and higher vibrational modes are populated that result in higher energy transition structures formed and the system does not pass though the lowest transition saddle point as shown in the diagram. Barrier recrossing occurs but the products are not formed as opposed to the predictions by the TST.

== EXERCISE 2: F - H - H SYSTEM ==

=== Reaction Energetics ===

{{fontcolor1|red|Question 6ː 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?}}

{| class="wikitable" border=1
|[[File:Q6FH2wxy0119.png|600px]]
|[[File:Q6HFHwxy0119.png|600px]]
|-
|Figure.15 F + H2 position on surface plot
|Figure.16 HF and H position on surface plot
|}

The surface plots of the potential energy surface of H + HF → F + H2 reaction is shown (Fig.15 Fig.16). Distance AB is the distance between H and F atoms and distance BC is the distance between two H atoms. The two minimum points shown in the graph are positions of F + H2 (Fig.15) and H + HF (Fig.16) respectively. If forward reaction is H + HF → F + H2, then the backward reaction is F + H2 → H + HF. As clearly shown in the diagram F + H2 are at a minimum point of higher potential energy than that of H + HF. Thus, by comparing the potential energy of reactants and products, the signs of changes in enthalpy (ΔH) and the energetics are determined. H + HF → F + H2 is endothermic and F + H2 → H + HF is exothermic.

The energetics of the two reactions are in accordance to the predictions by comparing bond strengths of chemical species. The bond energy of H-F bond (565 kJ/mol) is higher than the bond energy of H-H bond (432 kJ/mol). Thus, energy released by forming the H-H bond is not enough to compensate for energy required for breaking the H-F bond and the reaction H + HF → F + H2 is endothermic. On the contrary, the reaction F + H2 → H + HF is exothermic.

=== Transition State Approximation ===

{{fontcolor1|red|Question 7ː Locate the approximate position of the transition state.}}

{| class="wikitable" border=1
|[[File:Q7TStransitionwxy0119.png|390px]]
|[[File:Q7TS2wxy0119.png|390px]]
|[[File:Q7TS3wxy0119.png|390px]]
|-
|Figure.17 Transition state surface plot
|Figure.18 Transition state contour plot
|Figure.19 Internuclear distance against time plot (at TS point)
|}

The transition state (TS) is shown as a black dot on the potential energy surface plot (Fig.17) and a red cross on the contour plot. (Fig,18) The position of the TS is at the point where the distance between F and H (rFH) is 1.810Å and the distance between two H atoms (rHH) is 0.746Å. The TS is closer to F + H2 and further away from H + HF. At the transition state position, the distance between F and H, between two H atoms are constant with no momentum as shown in the internuclear distance against time plot as horizontal and flat lines, validating the position is the transition state point (Fig.19)

=== Activation Energies ===

{{fontcolor1|red|Question 8ː Report the activation energy for both reactions.}}

{| class="wikitable" border=1
|[[File:Q8Ea2wxy0119.png|600px]]
|[[File:Q8Ea1wxy0119.png|600px]]
|-
|Figure.21 Energy against time plot (HF + H)
|Figure.20 Energy against time plot (F + H2)
|}

By performing MEP calculation with slight increase and decrease of the rFH to 1.820Å to perform F + H2 → H + HF reaction and to 1.800Å to perform H + HF → F + H2 reaction, the activation energies (Ea) of both reactions can be calculated from the differences in their potential energies (Fig.20 Fig.21)ː

H + HF → F + H2ː Ea = -103.751 - (-133.624) = +29.873 kcal mol-1

F + H2 → H + HFː Ea = -103.751 - (-103.972) = + 0.221 kcal mol-1

=== Reaction Dynamics ===

{{fontcolor1|red|Question 9ː In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

Reactionː F + H2 → H + HF

Initial condition setː rHF = 2Å rHH = 0.74Å pHF = -0.5 pHH = 0.1

{| class="wikitable" border=1
|[[File:Q9plot1wxy0119.png|390px]]
|[[File:Q9plot2wxy0119.png|390px]]
|[[File:Q9plot3wxy0119.png|390px]]
|-
|Figure.23 Contour plot
|Figure.24 Surface plot
|Figure.25 Internuclear momentum against time plot
|-
|[[File:Q9ani1wxy0119.png|390px]]
|[[File:Q9ani2wxy0119.png|390px]]
|-
|Figure.26 Animation figure (at the start of the reaction)
|Figure.27 Animation figure (at the end of the reaction)
|}

As shown in the plots and animation figures, the H-H bond in the reactants is broken and H-F bond in the products is formed. There are small momenta between both H and F and H-H bond that provide the initial kinetic energy of the reactants and the initial vibrational energy of the H-H bond in the reactants. The initial energy is required to break the H-H bond and reduce the distance of H and F for bond forming and potential energy is released when H-F bond is formed. Since energy is conserved, the reaction energy released is converted to the vibrational energy of the H-F bond and the translational kinetic energies of the products. This is shown in the internuclear momentum against time graph as the initial vibrations of B-C (H-H bond) becomes flat as two H atoms are separated in the product and the A-B (H-F bond) formed in the product vibrates with a large amplitude. (Fig.25) The contour plot and the surface plot also clearly show there is a significant increase in the amplitude of the oscillation of H-F bond which is an indication of the large vibrational energy of H-F bond. (Fig.23 Fig.24)

The predictions can be proved experimentally. The vibrational energy of H-F bond can be determined by measuring the frequency and the intensity of the absorption band of the H-F bond with infrared spectroscopy. The conversion to the kinetic energies of the products can be observed by measuring the temperature of the reaction as the kinetic energy can be converted to thermal energy as products collide with solvent molecules. The temperature is expected to increase as this is an exothermic reaction.

=== Energy Distribution and Reactivity ===

{{fontcolor|red|Question 10ː 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.}}
{| class="wikitable" border=1
| Condition
| rFH /Å
| rHH /Å
| pFH
| pHH
|-
|1.
| 2
| 0.74
| -0.5
| -3
|-
|2.
| 2
| 0.74
| -0.5
| 1.9
|-
|3.
| 2.1
| 0.74
| -0.5
| 0.1
|-
|4.
| 2
| 0.74
| -0.8
| 0.1
|-
|5.
| 0.91
| 2
| 0.05
| -20
|-
|6.
| 0.91
| 2
| 0.8
| -7.5
|}
{| class="wikitable" border=1
|Condition
|Contour plot
|Observation
|-
|1.
|[[File:Q10plot1wxy0119.png|600px]]
| The products are not formed in the reaction between F and H2 even pHH is very large. Barrier recrossing occurs and the H-H bond oscillates significantly with a large amplitude in the diagram. It should be noted that while keeping pFH to be -0.5 and changing pHH from -3 to 3, the trend is not very obvious as there are many anomalous values of momentum that render the reaction unreactive while slight increase and decrease of momentum make the reaction reactive again. For example, pHH = -0.9 is reactive but pHH = -1.1 is unreactive.
|-
|2.
|[[File:Q10plot2wxy0119.png|600px]]
| There is reaction between F and H2 and products are formed. The F-H bond formed vibrates significantly along BC trajectory.
|-
|3.
|[[File:Q10plot9wxy0119.png|600px]]
| There is no reaction under this condition between F and H2. There is not enough energy to cross over the transition state point. Although investigations have shown that there is reaction when rFH is equal or less than 2Å, but there is no reaction when H and F atoms are further apart.
|-
|4.
|[[File:Q10plot8wxy0119.png|600px]]
| There is reaction between F and H2. By comparison between condition 4 and condition 3, a slight increase in the momentum of F and H which indicates a slight increase in kinetic translational energy result in the formation of products. Investigations have shown that even when rFH is slightly larger than 2Å, the products are still formed. Thus the change in reaction conditions illustrate that the changes in translational energy affect the reactivity the reaction more than the vibrational energy.
|-
|5.
|[[File:Q10plot5wxy0119.png|600px]]
| There is no reaction between H and HF even though the kinetic translational energy is very high and much higher than the activation energy. (pHF = -20) The trajectory reverts back before reaching the transition state structure.
|-
|6.
|[[File:Q10plot4wxy0119.png|600px]]
| There is reaction between H and HF. Compared to condition 5, the translational energy is reduced significantly while vibrational energy increases slightly. Thus, the vibrational energy is more efficient in promoting the reaction than the translational energy.
|}

The transition state illustrated in Question 7 is closer to F and H2 than H and HF. From the above illustrations, it can be shown that the translational energy is more efficient in promoting the reaction between F and H2 which is exothermic, has a lower activation energy and has an earlier transition state. On the other hand, the vibrational energy is more efficient in the reactivity of the reaction between H and HF which is endothermic, has a higher activation energy and has a late transition state. These observations agree with the Polanyi's empirical rules.<ref name="PoRule" />

== References ==

<references>
<ref name="TS">Eyring, H. (1935). "The Activated Complex in Chemical Reactions". J. Chem. Phys. 3 (2): 107–115.</ref>
</references>

<references>
<ref name="PoRule">Polanyi J C. Concepts in reaction dynamics[J]. Accounts of Chemical Research, 1972, 5(5): 161-168.</ref>
</references>

MRD:WXY0119

2018-05-18T16:52:59Z

Xw2816: /* Energy Distribution and Reactivity */

= Molecular Reaction Dynamics Report =

== EXERCISE 1ː H + H2 SYSTEM ==

[[File:HandH2wxy0119.png]]

In the H + H2 system, the distance between the first two H atoms (m1, m2) is r1 and the distance between the two H atoms in the H2 molecule (m2, m3) is r2.

=== Dynamics from the transition state region ===

{{fontcolor|red|Question 1 ː What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.}}

{| class="wikitable" border=1
|[[File:Q1TStransitionwxy0119.png|390px]]
|[[File:Q1TS2wxy0119.png|390px]]
|[[File:Q1TS3wxy0119.png|390px]]
|-
|Figure.1 Surface plot of reaction trajectory
|Figure.2 Transition state surface plot
|Figure.3 Transition state surface plot (different angle)
|}

In the reaction trajectory surface plot (Fig.1), AB is the distance r1 and BC is the distance r2. The gradient of the potential energy with regard to r1 and r2 are ∂V(r1)/∂r1 and ∂V(r2)/∂r2 respectively. The second derivatives of the potential energy with regard to r1 and r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 respectively.

At the two minimum structures which are at the two ends of the reaction pathway, ∂V(r1)/∂r1 = 0 and ∂V(r2)/∂r2 = 0 respectively. Since the two minimum structures are at two local minimum point , the second derivatives r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 are both greater than zero.

On the other hand, the components at the transition state structure (Fig.2) are different from that at minimum points. The gradient ∂V(r1)/∂r1 and ∂V(r2)/∂r2 are both equal to zero at the transition structure. However, r2 are ∂2V(r1)/∂r12 >0 and ∂2V(r2)/∂r22 <0. The transition state point is actually a saddle point which is observed more clearly in Fig.3. The Transition state point is thus distinguished from the minimum points as the potential energy surface curves inwards and downwards at the minimum points.

=== Locating the Transition State ===

{{fontcolor|red|Question 2ː 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.}}

{| class="wikitable" border=1
|[[File:TScontourwxy0119.png|390px]]
|[[File:TSsurfaceplwxy0119.png|390px]]
|[[File:TSestimatewxy0119.png|390px]]
|-
|Figure.4 Transition state contour plot
|Figure.5 Transition state surface plot
|Figure.6 Internuclear distance against time plot
|}

The best estimate of the transition state position is rts = 0.9078 Å.

When r1 = r2 = rts and momenta are set to zero, a graph of intermolecular distance against time is plotted. (Fig.6) In this graph, AB and BC lines are the same and overlap, the two lines on the graph are flat and horizontal, indicating that the atomic distances are the same at the position and with no momenta the state is at equilibrium, validating the position is the transition state position. The transition state is shown as a cross on the counter plot (Fig.4) and a dot on the surface plot (Fig.5).

=== Calculating and comparing the reaction path and trajectory ===

{{fontcolor1|red|Question 3ː Comment on how the ''mep'' and the trajectory you just calculated differ.}}

{| class="wikitable" border=1
|[[File:Mep1wxy0119.png|390px]]
|[[File:Mep2wxy0119.png|390px]]
|[[File:Mep3wxy0119.png|390px]]
|-
|Figure.7 mep calculation contour plot
|Figure.8 mep calculation surface plot
|Figure.9 mep calculation internuclear distance against time plot
|-
|[[File:Dynamic1wxy0119.png|390px]]
|[[File:Dynamic2wxy0119.png|390px]]
|[[File:Dynamic3wxy0119.png|390px]]
|-
|Figure.10 Dynamics calculation contour plot
|Figure.11 Dynamics calculation surface plot
|Figure.12 Dynamics calculation internuclear distance against time plot
|}
{| class="wikitable" border=1
|[[File:dynamics4wxy0119.png|600px]]
|[[File:Mep5wxy0119.png|600px]]
|-
|Figure.13 Dynamics calculation internuclear momenta against time plot
|Figure.14 mep calculation internuclear momenta against time plotsurface plot
|}

The mep (minimum energy path) trajectory is a smooth line on contour and surface plot (Fig.7 Fig.8) while the trajectory under dynamics calculation is oscillating along the pathway. (Fig.10 Fig.11) The mep trajectory takes more steps and longer time than the dynamics trajectory as shown in internuclear distance against time plots. (Fig. 9 Fig.12) The momentum is zero all the time under mep calculation (Fig.13) but under dynamics calculation, the momenta lines are oscillating against time.

The differences arise from mep and dynamics calculations can be explained by the the natures of the two calculations. Under mep calculation, the atoms are in extremely slow motion and the momentum is zero for each step, i.e. the atoms are "stop" after every step. The pathway under mep calculation is formed by connecting all the minimum energy points of each step and appears as a smooth line. On the other hand, atoms are in continuous motion and the momentum is accumulated after each step, atoms are then oscillating on the potential energy surface along the pathway. The steps are set to 5000 in the mep calculation but are only 500 in the dynamics calculatoin. Since the steps in mep are small, more time is needed to complete the same length on the trajectories.

=== Reactive and unreactive trajectories ===

{{fontcolor1|red|Question 4ː Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.}}

For the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, run trajectories with the following momenta p1 and p2 combinations:
{| class="wikitable" border=1
| Condition
| p1
| p2
| Total Energy/ kcal mol -1
| Reactivity
|-
|1.
| -1.25
| -2.5
| -99.018
| reactive
|-
|2.
| -1.5
| -2.0
| -100.456
| unreactive
|-
|3.
| -1.5
| -2.5
| -98.956
| reactive
|-
|4.
| -2.5
| -5.0
| -84.956
| unreactive
|-
|5.
| -2.5
| -5.2
| -83.416
| reactive
|}

{| class="wikitable" border=1
| Condition
| Surface Plot
| Contour Plot
| Description
|-
|1.
| [[File:condition1wxy0119.png|350px]]
| [[File:firstwxy0119.png|350px]]
| r2 (BC) decreases when HC approaches bonded HA and HB. The energy processed by the system is sufficient to overcome the activation barrier and crosses the transition state structure to break the HA and HB bond and form a new HB and HC bond. The new bond oscillate as r1 (AB) increases.
|-
|2.
| [[File:condition2wxy0119.png|350px]]
| [[File:secondwxy0119.png|350px]]
| HC approaches HB but the energy is insufficient to reach the transition state point, HC then moves further away from the bonded H2 molecule and no new bond is formed. The oscillation along r2 is a result of an increase in momentum p1.
|-
|3.
| [[File:condition3wxy0119.png|350px]]
| [[File:thirdwxy0119.png|350px]]
| Similarly to condition 1 but with a more negative p1, HA - HB bond oscillates more when r2 decreases. The reaction proceeds and crosses the transition state point to form the products.
|-
|4.
| [[File:condition4wxy0119.png|350px]]
| [[File:fourthwxy0119.png|350px]]
| Both p1 and p2 are more negative than the previous conditions. The HA - HB bond oscillates vigorously with larger amplitudes. The barrier recrossing happens after the system crosses the transition state region but eventually reverts back to the reactants.
|-
|5.
| [[File:condition5wxy0119.png|350px]]
| [[File:fifthwxy0119.png|350px]]
| With a slight more negative value of p1 as compared to that in condition 4, the system now processes appropriate energy to proceed to forming products. The vigorous oscillations are resulting from high momenta and barrier recrossing occurs.
|}

=== Transition State Theory ===

{{fontcolor|red|Question 5ː State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}

The main assumptions of Transition State Theory (TST) are<ref name="TS" />ː

1. The behaviours of the atomic nuclei follow the classical mechanics rules, i.e. the transition structure is formed as long as atoms or molecules collide with enough energy.

2. The intermediates have a long enough lifetime for the energies to be Boltzmann distributed preceding the next step.

3. The lowest energy saddle point on the potential energy surface is passed over in the reaction system.

The TST does not take into account of the quantum tunnelling effect which allows reactions to happen even when the energy carried by the system is lower than the activation energy barrier, especially for reactions with low energy barriers. Nor does the theory consider the intermediates with short lifetimes. When the energies are not fully distributed, the momentum of the reaction trajectory from the reactants to the intermediate can have effects on product selectivity. Moreover, the theory fails at high temperature because higher vibrational energy modes can be populated for a molecule at high temperature. The molecules can have complex motion and their collisions may result in a higher energy transition structure. The experimental results may thus deviate from that predicted by the TST. For example in condition 4, the kinetic energy of the reactants is much higher than the activation energy required. However, the momenta between atoms are very high and higher vibrational modes are populated that result in higher energy transition structures formed and the system does not pass though the lowest transition saddle point as shown in the diagram. Barrier recrossing occurs but the products are not formed as opposed to the predictions by the TST.

== EXERCISE 2: F - H - H SYSTEM ==

=== Reaction Energetics ===

{{fontcolor1|red|Question 6ː 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?}}

{| class="wikitable" border=1
|[[File:Q6FH2wxy0119.png|600px]]
|[[File:Q6HFHwxy0119.png|600px]]
|-
|Figure.15 F + H2 position on surface plot
|Figure.16 HF and H position on surface plot
|}

The surface plots of the potential energy surface of H + HF → F + H2 reaction is shown (Fig.15 Fig.16). Distance AB is the distance between H and F atoms and distance BC is the distance between two H atoms. The two minimum points shown in the graph are positions of F + H2 (Fig.15) and H + HF (Fig.16) respectively. If forward reaction is H + HF → F + H2, then the backward reaction is F + H2 → H + HF. As clearly shown in the diagram F + H2 are at a minimum point of higher potential energy than that of H + HF. Thus, by comparing the potential energy of reactants and products, the signs of changes in enthalpy (ΔH) and the energetics are determined. H + HF → F + H2 is endothermic and F + H2 → H + HF is exothermic.

The energetics of the two reactions are in accordance to the predictions by comparing bond strengths of chemical species. The bond energy of H-F bond (565 kJ/mol) is higher than the bond energy of H-H bond (432 kJ/mol). Thus, energy released by forming the H-H bond is not enough to compensate for energy required for breaking the H-F bond and the reaction H + HF → F + H2 is endothermic. On the contrary, the reaction F + H2 → H + HF is exothermic.

=== Transition State Approximation ===

{{fontcolor1|red|Question 7ː Locate the approximate position of the transition state.}}

{| class="wikitable" border=1
|[[File:Q7TStransitionwxy0119.png|390px]]
|[[File:Q7TS2wxy0119.png|390px]]
|[[File:Q7TS3wxy0119.png|390px]]
|-
|Figure.17 Transition state surface plot
|Figure.18 Transition state contour plot
|Figure.19 Internuclear distance against time plot (at TS point)
|}

The transition state (TS) is shown as a black dot on the potential energy surface plot (Fig.17) and a red cross on the contour plot. (Fig,18) The position of the TS is at the point where the distance between F and H (rFH) is 1.810Å and the distance between two H atoms (rHH) is 0.746Å. The TS is closer to F + H2 and further away from H + HF. At the transition state position, the distance between F and H, between two H atoms are constant with no momentum as shown in the internuclear distance against time plot as horizontal and flat lines, validating the position is the transition state point (Fig.19)

=== Activation Energies ===

{{fontcolor1|red|Question 8ː Report the activation energy for both reactions.}}

{| class="wikitable" border=1
|[[File:Q8Ea2wxy0119.png|600px]]
|[[File:Q8Ea1wxy0119.png|600px]]
|-
|Figure.21 Energy against time plot (HF + H)
|Figure.20 Energy against time plot (F + H2)
|}

By performing MEP calculation with slight increase and decrease of the rFH to 1.820Å to perform F + H2 → H + HF reaction and to 1.800Å to perform H + HF → F + H2 reaction, the activation energies (Ea) of both reactions can be calculated from the differences in their potential energies (Fig.20 Fig.21)ː

H + HF → F + H2ː Ea = -103.751 - (-133.624) = +29.873 kcal mol-1

F + H2 → H + HFː Ea = -103.751 - (-103.972) = + 0.221 kcal mol-1

=== Reaction Dynamics ===

{{fontcolor1|red|Question 9ː In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

Reactionː F + H2 → H + HF

Initial condition setː rHF = 2Å rHH = 0.74Å pHF = -0.5 pHH = 0.1

{| class="wikitable" border=1
|[[File:Q9plot1wxy0119.png|390px]]
|[[File:Q9plot2wxy0119.png|390px]]
|[[File:Q9plot3wxy0119.png|390px]]
|-
|Figure.23 Contour plot
|Figure.24 Surface plot
|Figure.25 Internuclear momentum against time plot
|-
|[[File:Q9ani1wxy0119.png|390px]]
|[[File:Q9ani2wxy0119.png|390px]]
|-
|Figure.26 Animation figure (at the start of the reaction)
|Figure.27 Animation figure (at the end of the reaction)
|}

As shown in the plots and animation figures, the H-H bond in the reactants is broken and H-F bond in the products is formed. There are small momenta between both H and F and H-H bond that provide the initial kinetic energy of the reactants and the initial vibrational energy of the H-H bond in the reactants. The initial energy is required to break the H-H bond and reduce the distance of H and F for bond forming and potential energy is released when H-F bond is formed. Since energy is conserved, the reaction energy released is converted to the vibrational energy of the H-F bond and the translational kinetic energies of the products. This is shown in the internuclear momentum against time graph as the initial vibrations of B-C (H-H bond) becomes flat as two H atoms are separated in the product and the A-B (H-F bond) formed in the product vibrates with a large amplitude. (Fig.25) The contour plot and the surface plot also clearly show there is a significant increase in the amplitude of the oscillation of H-F bond which is an indication of the large vibrational energy of H-F bond. (Fig.23 Fig.24)

The predictions can be proved experimentally. The vibrational energy of H-F bond can be determined by measuring the frequency and the intensity of the absorption band of the H-F bond with infrared spectroscopy. The conversion to the kinetic energies of the products can be observed by measuring the temperature of the reaction as the kinetic energy can be converted to thermal energy as products collide with solvent molecules. The temperature is expected to increase as this is an exothermic reaction.

=== Energy Distribution and Reactivity ===

{{fontcolor|red|Question 10ː 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.}}
{| class="wikitable" border=1
| Condition
| rFH /Å
| rHH /Å
| pFH
| pHH
|-
|1.
| 2
| 0.74
| -0.5
| -3
|-
|2.
| 2
| 0.74
| -0.5
| 1.9
|-
|3.
| 2.1
| 0.74
| -0.5
| 0.1
|-
|4.
| 2
| 0.74
| -0.8
| 0.1
|-
|5.
| 0.91
| 2
| 0.05
| -20
|-
|6.
| 0.91
| 2
| 0.8
| -7.5
|}
{| class="wikitable" border=1
|Condition
|Contour plot
|Observation
|-
|1.
|[[File:Q10plot1wxy0119.png|600px]]
| The products are not formed in the reaction between F and H2 even pHH is very large. Barrier recrossing occurs and the H-H bond oscillates significantly with a large amplitude in the diagram. It should be noted that while keeping pFH to be -0.5 and changing pHH from -3 to 3, the trend is not very obvious as there are many anomalous values of momentum that render the reaction unreactive while slight increase and decrease of momentum make the reaction reactive again. For example, pHH = -0.9 is reactive but pHH = -1.1 is unreactive.
|-
|2.
|[[File:Q10plot2wxy0119.png|600px]]
| There is reaction between F and H2 and products are formed. The F-H bond formed vibrates significantly along BC trajectory.
|-
|3.
|[[File:Q10plot9wxy0119.png|600px]]
| There is no reaction under this condition between F and H2. There is not enough energy to cross over the transition state point. Although investigations have shown that there is reaction when rFH is equal or less than 2Å, but there is no reaction when H and F atoms are further apart.
|-
|4.
|[[File:Q10plot8wxy0119.png|600px]]
| There is reaction between F and H2. By comparison between condition 4 and condition 3, a slight increase in the momentum of F and H which indicates a slight increase in kinetic translational energy result in the formation of products. Investigations have shown that even when rFH is slightly larger than 2Å, the products are still formed. Thus the change in reaction conditions illustrate that the changes in translational energy affect the reactivity the reaction more than the vibrational energy.
|-
|5.
|[[File:Q10plot5wxy0119.png|600px]]
| There is no reaction between H and HF even though the kinetic translational energy is very high and much higher than the activation energy. (pHF = -20) The trajectory reverts back before reaching the transition state structure.
|-
|6.
|[[File:Q10plot4wxy0119.png|600px]]
| There is reaction between H and HF. Compared to condition 5, the translational energy is reduced significantly while vibrational energy increases slightly. Thus, the vibrational energy is more efficient in promoting the reaction than the translational energy.
|}

The transition state illustrated in Question 7 is closer to F and H2 than H and HF. From the above illustrations, it can be shown that the translational energy is more efficient in promoting the reaction between F and H2 which is exothermic, has a lower activation energy and has an earlier transition state. On the other hand, the vibrational energy is more efficient in the reactivity of the reaction between H and HF which is endothermic, has a higher activation energy and has a late transition state. These observations agree with the Polanyi's empirical rules.<ref name="PoRule" />

== References ==

<references>
<ref name="TS">Eyring, H. (1935). "The Activated Complex in Chemical Reactions". J. Chem. Phys. 3 (2): 107–115.</ref>
</references>

<references>
<ref name="Po Rule">Polanyi J C. Concepts in reaction dynamics[J]. Accounts of Chemical Research, 1972, 5(5): 161-168.</ref>
</references>

MRD:WXY0119

2018-05-18T16:52:31Z

Xw2816: /* References */

= Molecular Reaction Dynamics Report =

== EXERCISE 1ː H + H2 SYSTEM ==

[[File:HandH2wxy0119.png]]

In the H + H2 system, the distance between the first two H atoms (m1, m2) is r1 and the distance between the two H atoms in the H2 molecule (m2, m3) is r2.

=== Dynamics from the transition state region ===

{{fontcolor|red|Question 1 ː What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.}}

{| class="wikitable" border=1
|[[File:Q1TStransitionwxy0119.png|390px]]
|[[File:Q1TS2wxy0119.png|390px]]
|[[File:Q1TS3wxy0119.png|390px]]
|-
|Figure.1 Surface plot of reaction trajectory
|Figure.2 Transition state surface plot
|Figure.3 Transition state surface plot (different angle)
|}

In the reaction trajectory surface plot (Fig.1), AB is the distance r1 and BC is the distance r2. The gradient of the potential energy with regard to r1 and r2 are ∂V(r1)/∂r1 and ∂V(r2)/∂r2 respectively. The second derivatives of the potential energy with regard to r1 and r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 respectively.

At the two minimum structures which are at the two ends of the reaction pathway, ∂V(r1)/∂r1 = 0 and ∂V(r2)/∂r2 = 0 respectively. Since the two minimum structures are at two local minimum point , the second derivatives r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 are both greater than zero.

On the other hand, the components at the transition state structure (Fig.2) are different from that at minimum points. The gradient ∂V(r1)/∂r1 and ∂V(r2)/∂r2 are both equal to zero at the transition structure. However, r2 are ∂2V(r1)/∂r12 >0 and ∂2V(r2)/∂r22 <0. The transition state point is actually a saddle point which is observed more clearly in Fig.3. The Transition state point is thus distinguished from the minimum points as the potential energy surface curves inwards and downwards at the minimum points.

=== Locating the Transition State ===

{{fontcolor|red|Question 2ː 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.}}

{| class="wikitable" border=1
|[[File:TScontourwxy0119.png|390px]]
|[[File:TSsurfaceplwxy0119.png|390px]]
|[[File:TSestimatewxy0119.png|390px]]
|-
|Figure.4 Transition state contour plot
|Figure.5 Transition state surface plot
|Figure.6 Internuclear distance against time plot
|}

The best estimate of the transition state position is rts = 0.9078 Å.

When r1 = r2 = rts and momenta are set to zero, a graph of intermolecular distance against time is plotted. (Fig.6) In this graph, AB and BC lines are the same and overlap, the two lines on the graph are flat and horizontal, indicating that the atomic distances are the same at the position and with no momenta the state is at equilibrium, validating the position is the transition state position. The transition state is shown as a cross on the counter plot (Fig.4) and a dot on the surface plot (Fig.5).

=== Calculating and comparing the reaction path and trajectory ===

{{fontcolor1|red|Question 3ː Comment on how the ''mep'' and the trajectory you just calculated differ.}}

{| class="wikitable" border=1
|[[File:Mep1wxy0119.png|390px]]
|[[File:Mep2wxy0119.png|390px]]
|[[File:Mep3wxy0119.png|390px]]
|-
|Figure.7 mep calculation contour plot
|Figure.8 mep calculation surface plot
|Figure.9 mep calculation internuclear distance against time plot
|-
|[[File:Dynamic1wxy0119.png|390px]]
|[[File:Dynamic2wxy0119.png|390px]]
|[[File:Dynamic3wxy0119.png|390px]]
|-
|Figure.10 Dynamics calculation contour plot
|Figure.11 Dynamics calculation surface plot
|Figure.12 Dynamics calculation internuclear distance against time plot
|}
{| class="wikitable" border=1
|[[File:dynamics4wxy0119.png|600px]]
|[[File:Mep5wxy0119.png|600px]]
|-
|Figure.13 Dynamics calculation internuclear momenta against time plot
|Figure.14 mep calculation internuclear momenta against time plotsurface plot
|}

The mep (minimum energy path) trajectory is a smooth line on contour and surface plot (Fig.7 Fig.8) while the trajectory under dynamics calculation is oscillating along the pathway. (Fig.10 Fig.11) The mep trajectory takes more steps and longer time than the dynamics trajectory as shown in internuclear distance against time plots. (Fig. 9 Fig.12) The momentum is zero all the time under mep calculation (Fig.13) but under dynamics calculation, the momenta lines are oscillating against time.

The differences arise from mep and dynamics calculations can be explained by the the natures of the two calculations. Under mep calculation, the atoms are in extremely slow motion and the momentum is zero for each step, i.e. the atoms are "stop" after every step. The pathway under mep calculation is formed by connecting all the minimum energy points of each step and appears as a smooth line. On the other hand, atoms are in continuous motion and the momentum is accumulated after each step, atoms are then oscillating on the potential energy surface along the pathway. The steps are set to 5000 in the mep calculation but are only 500 in the dynamics calculatoin. Since the steps in mep are small, more time is needed to complete the same length on the trajectories.

=== Reactive and unreactive trajectories ===

{{fontcolor1|red|Question 4ː Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.}}

For the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, run trajectories with the following momenta p1 and p2 combinations:
{| class="wikitable" border=1
| Condition
| p1
| p2
| Total Energy/ kcal mol -1
| Reactivity
|-
|1.
| -1.25
| -2.5
| -99.018
| reactive
|-
|2.
| -1.5
| -2.0
| -100.456
| unreactive
|-
|3.
| -1.5
| -2.5
| -98.956
| reactive
|-
|4.
| -2.5
| -5.0
| -84.956
| unreactive
|-
|5.
| -2.5
| -5.2
| -83.416
| reactive
|}

{| class="wikitable" border=1
| Condition
| Surface Plot
| Contour Plot
| Description
|-
|1.
| [[File:condition1wxy0119.png|350px]]
| [[File:firstwxy0119.png|350px]]
| r2 (BC) decreases when HC approaches bonded HA and HB. The energy processed by the system is sufficient to overcome the activation barrier and crosses the transition state structure to break the HA and HB bond and form a new HB and HC bond. The new bond oscillate as r1 (AB) increases.
|-
|2.
| [[File:condition2wxy0119.png|350px]]
| [[File:secondwxy0119.png|350px]]
| HC approaches HB but the energy is insufficient to reach the transition state point, HC then moves further away from the bonded H2 molecule and no new bond is formed. The oscillation along r2 is a result of an increase in momentum p1.
|-
|3.
| [[File:condition3wxy0119.png|350px]]
| [[File:thirdwxy0119.png|350px]]
| Similarly to condition 1 but with a more negative p1, HA - HB bond oscillates more when r2 decreases. The reaction proceeds and crosses the transition state point to form the products.
|-
|4.
| [[File:condition4wxy0119.png|350px]]
| [[File:fourthwxy0119.png|350px]]
| Both p1 and p2 are more negative than the previous conditions. The HA - HB bond oscillates vigorously with larger amplitudes. The barrier recrossing happens after the system crosses the transition state region but eventually reverts back to the reactants.
|-
|5.
| [[File:condition5wxy0119.png|350px]]
| [[File:fifthwxy0119.png|350px]]
| With a slight more negative value of p1 as compared to that in condition 4, the system now processes appropriate energy to proceed to forming products. The vigorous oscillations are resulting from high momenta and barrier recrossing occurs.
|}

=== Transition State Theory ===

{{fontcolor|red|Question 5ː State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}

The main assumptions of Transition State Theory (TST) are<ref name="TS" />ː

1. The behaviours of the atomic nuclei follow the classical mechanics rules, i.e. the transition structure is formed as long as atoms or molecules collide with enough energy.

2. The intermediates have a long enough lifetime for the energies to be Boltzmann distributed preceding the next step.

3. The lowest energy saddle point on the potential energy surface is passed over in the reaction system.

The TST does not take into account of the quantum tunnelling effect which allows reactions to happen even when the energy carried by the system is lower than the activation energy barrier, especially for reactions with low energy barriers. Nor does the theory consider the intermediates with short lifetimes. When the energies are not fully distributed, the momentum of the reaction trajectory from the reactants to the intermediate can have effects on product selectivity. Moreover, the theory fails at high temperature because higher vibrational energy modes can be populated for a molecule at high temperature. The molecules can have complex motion and their collisions may result in a higher energy transition structure. The experimental results may thus deviate from that predicted by the TST. For example in condition 4, the kinetic energy of the reactants is much higher than the activation energy required. However, the momenta between atoms are very high and higher vibrational modes are populated that result in higher energy transition structures formed and the system does not pass though the lowest transition saddle point as shown in the diagram. Barrier recrossing occurs but the products are not formed as opposed to the predictions by the TST.

== EXERCISE 2: F - H - H SYSTEM ==

=== Reaction Energetics ===

{{fontcolor1|red|Question 6ː 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?}}

{| class="wikitable" border=1
|[[File:Q6FH2wxy0119.png|600px]]
|[[File:Q6HFHwxy0119.png|600px]]
|-
|Figure.15 F + H2 position on surface plot
|Figure.16 HF and H position on surface plot
|}

The surface plots of the potential energy surface of H + HF → F + H2 reaction is shown (Fig.15 Fig.16). Distance AB is the distance between H and F atoms and distance BC is the distance between two H atoms. The two minimum points shown in the graph are positions of F + H2 (Fig.15) and H + HF (Fig.16) respectively. If forward reaction is H + HF → F + H2, then the backward reaction is F + H2 → H + HF. As clearly shown in the diagram F + H2 are at a minimum point of higher potential energy than that of H + HF. Thus, by comparing the potential energy of reactants and products, the signs of changes in enthalpy (ΔH) and the energetics are determined. H + HF → F + H2 is endothermic and F + H2 → H + HF is exothermic.

The energetics of the two reactions are in accordance to the predictions by comparing bond strengths of chemical species. The bond energy of H-F bond (565 kJ/mol) is higher than the bond energy of H-H bond (432 kJ/mol). Thus, energy released by forming the H-H bond is not enough to compensate for energy required for breaking the H-F bond and the reaction H + HF → F + H2 is endothermic. On the contrary, the reaction F + H2 → H + HF is exothermic.

=== Transition State Approximation ===

{{fontcolor1|red|Question 7ː Locate the approximate position of the transition state.}}

{| class="wikitable" border=1
|[[File:Q7TStransitionwxy0119.png|390px]]
|[[File:Q7TS2wxy0119.png|390px]]
|[[File:Q7TS3wxy0119.png|390px]]
|-
|Figure.17 Transition state surface plot
|Figure.18 Transition state contour plot
|Figure.19 Internuclear distance against time plot (at TS point)
|}

The transition state (TS) is shown as a black dot on the potential energy surface plot (Fig.17) and a red cross on the contour plot. (Fig,18) The position of the TS is at the point where the distance between F and H (rFH) is 1.810Å and the distance between two H atoms (rHH) is 0.746Å. The TS is closer to F + H2 and further away from H + HF. At the transition state position, the distance between F and H, between two H atoms are constant with no momentum as shown in the internuclear distance against time plot as horizontal and flat lines, validating the position is the transition state point (Fig.19)

=== Activation Energies ===

{{fontcolor1|red|Question 8ː Report the activation energy for both reactions.}}

{| class="wikitable" border=1
|[[File:Q8Ea2wxy0119.png|600px]]
|[[File:Q8Ea1wxy0119.png|600px]]
|-
|Figure.21 Energy against time plot (HF + H)
|Figure.20 Energy against time plot (F + H2)
|}

By performing MEP calculation with slight increase and decrease of the rFH to 1.820Å to perform F + H2 → H + HF reaction and to 1.800Å to perform H + HF → F + H2 reaction, the activation energies (Ea) of both reactions can be calculated from the differences in their potential energies (Fig.20 Fig.21)ː

H + HF → F + H2ː Ea = -103.751 - (-133.624) = +29.873 kcal mol-1

F + H2 → H + HFː Ea = -103.751 - (-103.972) = + 0.221 kcal mol-1

=== Reaction Dynamics ===

{{fontcolor1|red|Question 9ː In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

Reactionː F + H2 → H + HF

Initial condition setː rHF = 2Å rHH = 0.74Å pHF = -0.5 pHH = 0.1

{| class="wikitable" border=1
|[[File:Q9plot1wxy0119.png|390px]]
|[[File:Q9plot2wxy0119.png|390px]]
|[[File:Q9plot3wxy0119.png|390px]]
|-
|Figure.23 Contour plot
|Figure.24 Surface plot
|Figure.25 Internuclear momentum against time plot
|-
|[[File:Q9ani1wxy0119.png|390px]]
|[[File:Q9ani2wxy0119.png|390px]]
|-
|Figure.26 Animation figure (at the start of the reaction)
|Figure.27 Animation figure (at the end of the reaction)
|}

As shown in the plots and animation figures, the H-H bond in the reactants is broken and H-F bond in the products is formed. There are small momenta between both H and F and H-H bond that provide the initial kinetic energy of the reactants and the initial vibrational energy of the H-H bond in the reactants. The initial energy is required to break the H-H bond and reduce the distance of H and F for bond forming and potential energy is released when H-F bond is formed. Since energy is conserved, the reaction energy released is converted to the vibrational energy of the H-F bond and the translational kinetic energies of the products. This is shown in the internuclear momentum against time graph as the initial vibrations of B-C (H-H bond) becomes flat as two H atoms are separated in the product and the A-B (H-F bond) formed in the product vibrates with a large amplitude. (Fig.25) The contour plot and the surface plot also clearly show there is a significant increase in the amplitude of the oscillation of H-F bond which is an indication of the large vibrational energy of H-F bond. (Fig.23 Fig.24)

The predictions can be proved experimentally. The vibrational energy of H-F bond can be determined by measuring the frequency and the intensity of the absorption band of the H-F bond with infrared spectroscopy. The conversion to the kinetic energies of the products can be observed by measuring the temperature of the reaction as the kinetic energy can be converted to thermal energy as products collide with solvent molecules. The temperature is expected to increase as this is an exothermic reaction.

=== Energy Distribution and Reactivity ===

{{fontcolor|red|Question 10ː 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.}}
{| class="wikitable" border=1
| Condition
| rFH /Å
| rHH /Å
| pFH
| pHH
|-
|1.
| 2
| 0.74
| -0.5
| -3
|-
|2.
| 2
| 0.74
| -0.5
| 1.9
|-
|3.
| 2.1
| 0.74
| -0.5
| 0.1
|-
|4.
| 2
| 0.74
| -0.8
| 0.1
|-
|5.
| 0.91
| 2
| 0.05
| -20
|-
|6.
| 0.91
| 2
| 0.8
| -7.5
|}
{| class="wikitable" border=1
|Condition
|Contour plot
|Observation
|-
|1.
|[[File:Q10plot1wxy0119.png|600px]]
| The products are not formed in the reaction between F and H2 even pHH is very large. Barrier recrossing occurs and the H-H bond oscillates significantly with a large amplitude in the diagram. It should be noted that while keeping pFH to be -0.5 and changing pHH from -3 to 3, the trend is not very obvious as there are many anomalous values of momentum that render the reaction unreactive while slight increase and decrease of momentum make the reaction reactive again. For example, pHH = -0.9 is reactive but pHH = -1.1 is unreactive.
|-
|2.
|[[File:Q10plot2wxy0119.png|600px]]
| There is reaction between F and H2 and products are formed. The F-H bond formed vibrates significantly along BC trajectory.
|-
|3.
|[[File:Q10plot9wxy0119.png|600px]]
| There is no reaction under this condition between F and H2. There is not enough energy to cross over the transition state point. Although investigations have shown that there is reaction when rFH is equal or less than 2Å, but there is no reaction when H and F atoms are further apart.
|-
|4.
|[[File:Q10plot8wxy0119.png|600px]]
| There is reaction between F and H2. By comparison between condition 4 and condition 3, a slight increase in the momentum of F and H which indicates a slight increase in kinetic translational energy result in the formation of products. Investigations have shown that even when rFH is slightly larger than 2Å, the products are still formed. Thus the change in reaction conditions illustrate that the changes in translational energy affect the reactivity the reaction more than the vibrational energy.
|-
|5.
|[[File:Q10plot5wxy0119.png|600px]]
| There is no reaction between H and HF even though the kinetic translational energy is very high and much higher than the activation energy. (pHF = -20) The trajectory reverts back before reaching the transition state structure.
|-
|6.
|[[File:Q10plot4wxy0119.png|600px]]
| There is reaction between H and HF. Compared to condition 5, the translational energy is reduced significantly while vibrational energy increases slightly. Thus, the vibrational energy is more efficient in promoting the reaction than the translational energy.
|}

The transition state illustrated in Question 7 is closer to F and H2 than H and HF. From the above illustrations, it can be shown that the translational energy is more efficient in promoting the reaction between F and H2 which is exothermic, has a lower activation energy and has an earlier transition state. On the other hand, the vibrational energy is more efficient in the reactivity of the reaction between H and HF which is endothermic, has a higher activation energy and has a late transition state. These observations agree with the Polanyi's empirical rules.<ref name="Po Rule" />

== References ==

<references>
<ref name="TS">Eyring, H. (1935). "The Activated Complex in Chemical Reactions". J. Chem. Phys. 3 (2): 107–115.</ref>
</references>

<references>
<ref name="Po Rule">Polanyi J C. Concepts in reaction dynamics[J]. Accounts of Chemical Research, 1972, 5(5): 161-168.</ref>
</references>

MRD:WXY0119

2018-05-18T16:51:10Z

Xw2816: /* Energy Distribution and Reactivity */

= Molecular Reaction Dynamics Report =

== EXERCISE 1ː H + H2 SYSTEM ==

[[File:HandH2wxy0119.png]]

In the H + H2 system, the distance between the first two H atoms (m1, m2) is r1 and the distance between the two H atoms in the H2 molecule (m2, m3) is r2.

=== Dynamics from the transition state region ===

{{fontcolor|red|Question 1 ː What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.}}

{| class="wikitable" border=1
|[[File:Q1TStransitionwxy0119.png|390px]]
|[[File:Q1TS2wxy0119.png|390px]]
|[[File:Q1TS3wxy0119.png|390px]]
|-
|Figure.1 Surface plot of reaction trajectory
|Figure.2 Transition state surface plot
|Figure.3 Transition state surface plot (different angle)
|}

In the reaction trajectory surface plot (Fig.1), AB is the distance r1 and BC is the distance r2. The gradient of the potential energy with regard to r1 and r2 are ∂V(r1)/∂r1 and ∂V(r2)/∂r2 respectively. The second derivatives of the potential energy with regard to r1 and r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 respectively.

At the two minimum structures which are at the two ends of the reaction pathway, ∂V(r1)/∂r1 = 0 and ∂V(r2)/∂r2 = 0 respectively. Since the two minimum structures are at two local minimum point , the second derivatives r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 are both greater than zero.

On the other hand, the components at the transition state structure (Fig.2) are different from that at minimum points. The gradient ∂V(r1)/∂r1 and ∂V(r2)/∂r2 are both equal to zero at the transition structure. However, r2 are ∂2V(r1)/∂r12 >0 and ∂2V(r2)/∂r22 <0. The transition state point is actually a saddle point which is observed more clearly in Fig.3. The Transition state point is thus distinguished from the minimum points as the potential energy surface curves inwards and downwards at the minimum points.

=== Locating the Transition State ===

{{fontcolor|red|Question 2ː 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.}}

{| class="wikitable" border=1
|[[File:TScontourwxy0119.png|390px]]
|[[File:TSsurfaceplwxy0119.png|390px]]
|[[File:TSestimatewxy0119.png|390px]]
|-
|Figure.4 Transition state contour plot
|Figure.5 Transition state surface plot
|Figure.6 Internuclear distance against time plot
|}

The best estimate of the transition state position is rts = 0.9078 Å.

When r1 = r2 = rts and momenta are set to zero, a graph of intermolecular distance against time is plotted. (Fig.6) In this graph, AB and BC lines are the same and overlap, the two lines on the graph are flat and horizontal, indicating that the atomic distances are the same at the position and with no momenta the state is at equilibrium, validating the position is the transition state position. The transition state is shown as a cross on the counter plot (Fig.4) and a dot on the surface plot (Fig.5).

=== Calculating and comparing the reaction path and trajectory ===

{{fontcolor1|red|Question 3ː Comment on how the ''mep'' and the trajectory you just calculated differ.}}

{| class="wikitable" border=1
|[[File:Mep1wxy0119.png|390px]]
|[[File:Mep2wxy0119.png|390px]]
|[[File:Mep3wxy0119.png|390px]]
|-
|Figure.7 mep calculation contour plot
|Figure.8 mep calculation surface plot
|Figure.9 mep calculation internuclear distance against time plot
|-
|[[File:Dynamic1wxy0119.png|390px]]
|[[File:Dynamic2wxy0119.png|390px]]
|[[File:Dynamic3wxy0119.png|390px]]
|-
|Figure.10 Dynamics calculation contour plot
|Figure.11 Dynamics calculation surface plot
|Figure.12 Dynamics calculation internuclear distance against time plot
|}
{| class="wikitable" border=1
|[[File:dynamics4wxy0119.png|600px]]
|[[File:Mep5wxy0119.png|600px]]
|-
|Figure.13 Dynamics calculation internuclear momenta against time plot
|Figure.14 mep calculation internuclear momenta against time plotsurface plot
|}

The mep (minimum energy path) trajectory is a smooth line on contour and surface plot (Fig.7 Fig.8) while the trajectory under dynamics calculation is oscillating along the pathway. (Fig.10 Fig.11) The mep trajectory takes more steps and longer time than the dynamics trajectory as shown in internuclear distance against time plots. (Fig. 9 Fig.12) The momentum is zero all the time under mep calculation (Fig.13) but under dynamics calculation, the momenta lines are oscillating against time.

The differences arise from mep and dynamics calculations can be explained by the the natures of the two calculations. Under mep calculation, the atoms are in extremely slow motion and the momentum is zero for each step, i.e. the atoms are "stop" after every step. The pathway under mep calculation is formed by connecting all the minimum energy points of each step and appears as a smooth line. On the other hand, atoms are in continuous motion and the momentum is accumulated after each step, atoms are then oscillating on the potential energy surface along the pathway. The steps are set to 5000 in the mep calculation but are only 500 in the dynamics calculatoin. Since the steps in mep are small, more time is needed to complete the same length on the trajectories.

=== Reactive and unreactive trajectories ===

{{fontcolor1|red|Question 4ː Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.}}

For the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, run trajectories with the following momenta p1 and p2 combinations:
{| class="wikitable" border=1
| Condition
| p1
| p2
| Total Energy/ kcal mol -1
| Reactivity
|-
|1.
| -1.25
| -2.5
| -99.018
| reactive
|-
|2.
| -1.5
| -2.0
| -100.456
| unreactive
|-
|3.
| -1.5
| -2.5
| -98.956
| reactive
|-
|4.
| -2.5
| -5.0
| -84.956
| unreactive
|-
|5.
| -2.5
| -5.2
| -83.416
| reactive
|}

{| class="wikitable" border=1
| Condition
| Surface Plot
| Contour Plot
| Description
|-
|1.
| [[File:condition1wxy0119.png|350px]]
| [[File:firstwxy0119.png|350px]]
| r2 (BC) decreases when HC approaches bonded HA and HB. The energy processed by the system is sufficient to overcome the activation barrier and crosses the transition state structure to break the HA and HB bond and form a new HB and HC bond. The new bond oscillate as r1 (AB) increases.
|-
|2.
| [[File:condition2wxy0119.png|350px]]
| [[File:secondwxy0119.png|350px]]
| HC approaches HB but the energy is insufficient to reach the transition state point, HC then moves further away from the bonded H2 molecule and no new bond is formed. The oscillation along r2 is a result of an increase in momentum p1.
|-
|3.
| [[File:condition3wxy0119.png|350px]]
| [[File:thirdwxy0119.png|350px]]
| Similarly to condition 1 but with a more negative p1, HA - HB bond oscillates more when r2 decreases. The reaction proceeds and crosses the transition state point to form the products.
|-
|4.
| [[File:condition4wxy0119.png|350px]]
| [[File:fourthwxy0119.png|350px]]
| Both p1 and p2 are more negative than the previous conditions. The HA - HB bond oscillates vigorously with larger amplitudes. The barrier recrossing happens after the system crosses the transition state region but eventually reverts back to the reactants.
|-
|5.
| [[File:condition5wxy0119.png|350px]]
| [[File:fifthwxy0119.png|350px]]
| With a slight more negative value of p1 as compared to that in condition 4, the system now processes appropriate energy to proceed to forming products. The vigorous oscillations are resulting from high momenta and barrier recrossing occurs.
|}

=== Transition State Theory ===

{{fontcolor|red|Question 5ː State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}

The main assumptions of Transition State Theory (TST) are<ref name="TS" />ː

1. The behaviours of the atomic nuclei follow the classical mechanics rules, i.e. the transition structure is formed as long as atoms or molecules collide with enough energy.

2. The intermediates have a long enough lifetime for the energies to be Boltzmann distributed preceding the next step.

3. The lowest energy saddle point on the potential energy surface is passed over in the reaction system.

The TST does not take into account of the quantum tunnelling effect which allows reactions to happen even when the energy carried by the system is lower than the activation energy barrier, especially for reactions with low energy barriers. Nor does the theory consider the intermediates with short lifetimes. When the energies are not fully distributed, the momentum of the reaction trajectory from the reactants to the intermediate can have effects on product selectivity. Moreover, the theory fails at high temperature because higher vibrational energy modes can be populated for a molecule at high temperature. The molecules can have complex motion and their collisions may result in a higher energy transition structure. The experimental results may thus deviate from that predicted by the TST. For example in condition 4, the kinetic energy of the reactants is much higher than the activation energy required. However, the momenta between atoms are very high and higher vibrational modes are populated that result in higher energy transition structures formed and the system does not pass though the lowest transition saddle point as shown in the diagram. Barrier recrossing occurs but the products are not formed as opposed to the predictions by the TST.

== EXERCISE 2: F - H - H SYSTEM ==

=== Reaction Energetics ===

{{fontcolor1|red|Question 6ː 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?}}

{| class="wikitable" border=1
|[[File:Q6FH2wxy0119.png|600px]]
|[[File:Q6HFHwxy0119.png|600px]]
|-
|Figure.15 F + H2 position on surface plot
|Figure.16 HF and H position on surface plot
|}

The surface plots of the potential energy surface of H + HF → F + H2 reaction is shown (Fig.15 Fig.16). Distance AB is the distance between H and F atoms and distance BC is the distance between two H atoms. The two minimum points shown in the graph are positions of F + H2 (Fig.15) and H + HF (Fig.16) respectively. If forward reaction is H + HF → F + H2, then the backward reaction is F + H2 → H + HF. As clearly shown in the diagram F + H2 are at a minimum point of higher potential energy than that of H + HF. Thus, by comparing the potential energy of reactants and products, the signs of changes in enthalpy (ΔH) and the energetics are determined. H + HF → F + H2 is endothermic and F + H2 → H + HF is exothermic.

The energetics of the two reactions are in accordance to the predictions by comparing bond strengths of chemical species. The bond energy of H-F bond (565 kJ/mol) is higher than the bond energy of H-H bond (432 kJ/mol). Thus, energy released by forming the H-H bond is not enough to compensate for energy required for breaking the H-F bond and the reaction H + HF → F + H2 is endothermic. On the contrary, the reaction F + H2 → H + HF is exothermic.

=== Transition State Approximation ===

{{fontcolor1|red|Question 7ː Locate the approximate position of the transition state.}}

{| class="wikitable" border=1
|[[File:Q7TStransitionwxy0119.png|390px]]
|[[File:Q7TS2wxy0119.png|390px]]
|[[File:Q7TS3wxy0119.png|390px]]
|-
|Figure.17 Transition state surface plot
|Figure.18 Transition state contour plot
|Figure.19 Internuclear distance against time plot (at TS point)
|}

The transition state (TS) is shown as a black dot on the potential energy surface plot (Fig.17) and a red cross on the contour plot. (Fig,18) The position of the TS is at the point where the distance between F and H (rFH) is 1.810Å and the distance between two H atoms (rHH) is 0.746Å. The TS is closer to F + H2 and further away from H + HF. At the transition state position, the distance between F and H, between two H atoms are constant with no momentum as shown in the internuclear distance against time plot as horizontal and flat lines, validating the position is the transition state point (Fig.19)

=== Activation Energies ===

{{fontcolor1|red|Question 8ː Report the activation energy for both reactions.}}

{| class="wikitable" border=1
|[[File:Q8Ea2wxy0119.png|600px]]
|[[File:Q8Ea1wxy0119.png|600px]]
|-
|Figure.21 Energy against time plot (HF + H)
|Figure.20 Energy against time plot (F + H2)
|}

By performing MEP calculation with slight increase and decrease of the rFH to 1.820Å to perform F + H2 → H + HF reaction and to 1.800Å to perform H + HF → F + H2 reaction, the activation energies (Ea) of both reactions can be calculated from the differences in their potential energies (Fig.20 Fig.21)ː

H + HF → F + H2ː Ea = -103.751 - (-133.624) = +29.873 kcal mol-1

F + H2 → H + HFː Ea = -103.751 - (-103.972) = + 0.221 kcal mol-1

=== Reaction Dynamics ===

{{fontcolor1|red|Question 9ː In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

Reactionː F + H2 → H + HF

Initial condition setː rHF = 2Å rHH = 0.74Å pHF = -0.5 pHH = 0.1

{| class="wikitable" border=1
|[[File:Q9plot1wxy0119.png|390px]]
|[[File:Q9plot2wxy0119.png|390px]]
|[[File:Q9plot3wxy0119.png|390px]]
|-
|Figure.23 Contour plot
|Figure.24 Surface plot
|Figure.25 Internuclear momentum against time plot
|-
|[[File:Q9ani1wxy0119.png|390px]]
|[[File:Q9ani2wxy0119.png|390px]]
|-
|Figure.26 Animation figure (at the start of the reaction)
|Figure.27 Animation figure (at the end of the reaction)
|}

As shown in the plots and animation figures, the H-H bond in the reactants is broken and H-F bond in the products is formed. There are small momenta between both H and F and H-H bond that provide the initial kinetic energy of the reactants and the initial vibrational energy of the H-H bond in the reactants. The initial energy is required to break the H-H bond and reduce the distance of H and F for bond forming and potential energy is released when H-F bond is formed. Since energy is conserved, the reaction energy released is converted to the vibrational energy of the H-F bond and the translational kinetic energies of the products. This is shown in the internuclear momentum against time graph as the initial vibrations of B-C (H-H bond) becomes flat as two H atoms are separated in the product and the A-B (H-F bond) formed in the product vibrates with a large amplitude. (Fig.25) The contour plot and the surface plot also clearly show there is a significant increase in the amplitude of the oscillation of H-F bond which is an indication of the large vibrational energy of H-F bond. (Fig.23 Fig.24)

The predictions can be proved experimentally. The vibrational energy of H-F bond can be determined by measuring the frequency and the intensity of the absorption band of the H-F bond with infrared spectroscopy. The conversion to the kinetic energies of the products can be observed by measuring the temperature of the reaction as the kinetic energy can be converted to thermal energy as products collide with solvent molecules. The temperature is expected to increase as this is an exothermic reaction.

=== Energy Distribution and Reactivity ===

{{fontcolor|red|Question 10ː 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.}}
{| class="wikitable" border=1
| Condition
| rFH /Å
| rHH /Å
| pFH
| pHH
|-
|1.
| 2
| 0.74
| -0.5
| -3
|-
|2.
| 2
| 0.74
| -0.5
| 1.9
|-
|3.
| 2.1
| 0.74
| -0.5
| 0.1
|-
|4.
| 2
| 0.74
| -0.8
| 0.1
|-
|5.
| 0.91
| 2
| 0.05
| -20
|-
|6.
| 0.91
| 2
| 0.8
| -7.5
|}
{| class="wikitable" border=1
|Condition
|Contour plot
|Observation
|-
|1.
|[[File:Q10plot1wxy0119.png|600px]]
| The products are not formed in the reaction between F and H2 even pHH is very large. Barrier recrossing occurs and the H-H bond oscillates significantly with a large amplitude in the diagram. It should be noted that while keeping pFH to be -0.5 and changing pHH from -3 to 3, the trend is not very obvious as there are many anomalous values of momentum that render the reaction unreactive while slight increase and decrease of momentum make the reaction reactive again. For example, pHH = -0.9 is reactive but pHH = -1.1 is unreactive.
|-
|2.
|[[File:Q10plot2wxy0119.png|600px]]
| There is reaction between F and H2 and products are formed. The F-H bond formed vibrates significantly along BC trajectory.
|-
|3.
|[[File:Q10plot9wxy0119.png|600px]]
| There is no reaction under this condition between F and H2. There is not enough energy to cross over the transition state point. Although investigations have shown that there is reaction when rFH is equal or less than 2Å, but there is no reaction when H and F atoms are further apart.
|-
|4.
|[[File:Q10plot8wxy0119.png|600px]]
| There is reaction between F and H2. By comparison between condition 4 and condition 3, a slight increase in the momentum of F and H which indicates a slight increase in kinetic translational energy result in the formation of products. Investigations have shown that even when rFH is slightly larger than 2Å, the products are still formed. Thus the change in reaction conditions illustrate that the changes in translational energy affect the reactivity the reaction more than the vibrational energy.
|-
|5.
|[[File:Q10plot5wxy0119.png|600px]]
| There is no reaction between H and HF even though the kinetic translational energy is very high and much higher than the activation energy. (pHF = -20) The trajectory reverts back before reaching the transition state structure.
|-
|6.
|[[File:Q10plot4wxy0119.png|600px]]
| There is reaction between H and HF. Compared to condition 5, the translational energy is reduced significantly while vibrational energy increases slightly. Thus, the vibrational energy is more efficient in promoting the reaction than the translational energy.
|}

The transition state illustrated in Question 7 is closer to F and H2 than H and HF. From the above illustrations, it can be shown that the translational energy is more efficient in promoting the reaction between F and H2 which is exothermic, has a lower activation energy and has an earlier transition state. On the other hand, the vibrational energy is more efficient in the reactivity of the reaction between H and HF which is endothermic, has a higher activation energy and has a late transition state. These observations agree with the Polanyi's empirical rules.<ref name="Po Rule" />

== References ==

<references>
<ref name="TS">Eyring, H. (1935). "The Activated Complex in Chemical Reactions". J. Chem. Phys. 3 (2): 107–115.</ref>
</references>

MRD:WXY0119

2018-05-18T16:49:50Z

Xw2816: /* Energy Distribution and Reactivity */

= Molecular Reaction Dynamics Report =

== EXERCISE 1ː H + H2 SYSTEM ==

[[File:HandH2wxy0119.png]]

In the H + H2 system, the distance between the first two H atoms (m1, m2) is r1 and the distance between the two H atoms in the H2 molecule (m2, m3) is r2.

=== Dynamics from the transition state region ===

{{fontcolor|red|Question 1 ː What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.}}

{| class="wikitable" border=1
|[[File:Q1TStransitionwxy0119.png|390px]]
|[[File:Q1TS2wxy0119.png|390px]]
|[[File:Q1TS3wxy0119.png|390px]]
|-
|Figure.1 Surface plot of reaction trajectory
|Figure.2 Transition state surface plot
|Figure.3 Transition state surface plot (different angle)
|}

In the reaction trajectory surface plot (Fig.1), AB is the distance r1 and BC is the distance r2. The gradient of the potential energy with regard to r1 and r2 are ∂V(r1)/∂r1 and ∂V(r2)/∂r2 respectively. The second derivatives of the potential energy with regard to r1 and r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 respectively.

At the two minimum structures which are at the two ends of the reaction pathway, ∂V(r1)/∂r1 = 0 and ∂V(r2)/∂r2 = 0 respectively. Since the two minimum structures are at two local minimum point , the second derivatives r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 are both greater than zero.

On the other hand, the components at the transition state structure (Fig.2) are different from that at minimum points. The gradient ∂V(r1)/∂r1 and ∂V(r2)/∂r2 are both equal to zero at the transition structure. However, r2 are ∂2V(r1)/∂r12 >0 and ∂2V(r2)/∂r22 <0. The transition state point is actually a saddle point which is observed more clearly in Fig.3. The Transition state point is thus distinguished from the minimum points as the potential energy surface curves inwards and downwards at the minimum points.

=== Locating the Transition State ===

{{fontcolor|red|Question 2ː 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.}}

{| class="wikitable" border=1
|[[File:TScontourwxy0119.png|390px]]
|[[File:TSsurfaceplwxy0119.png|390px]]
|[[File:TSestimatewxy0119.png|390px]]
|-
|Figure.4 Transition state contour plot
|Figure.5 Transition state surface plot
|Figure.6 Internuclear distance against time plot
|}

The best estimate of the transition state position is rts = 0.9078 Å.

When r1 = r2 = rts and momenta are set to zero, a graph of intermolecular distance against time is plotted. (Fig.6) In this graph, AB and BC lines are the same and overlap, the two lines on the graph are flat and horizontal, indicating that the atomic distances are the same at the position and with no momenta the state is at equilibrium, validating the position is the transition state position. The transition state is shown as a cross on the counter plot (Fig.4) and a dot on the surface plot (Fig.5).

=== Calculating and comparing the reaction path and trajectory ===

{{fontcolor1|red|Question 3ː Comment on how the ''mep'' and the trajectory you just calculated differ.}}

{| class="wikitable" border=1
|[[File:Mep1wxy0119.png|390px]]
|[[File:Mep2wxy0119.png|390px]]
|[[File:Mep3wxy0119.png|390px]]
|-
|Figure.7 mep calculation contour plot
|Figure.8 mep calculation surface plot
|Figure.9 mep calculation internuclear distance against time plot
|-
|[[File:Dynamic1wxy0119.png|390px]]
|[[File:Dynamic2wxy0119.png|390px]]
|[[File:Dynamic3wxy0119.png|390px]]
|-
|Figure.10 Dynamics calculation contour plot
|Figure.11 Dynamics calculation surface plot
|Figure.12 Dynamics calculation internuclear distance against time plot
|}
{| class="wikitable" border=1
|[[File:dynamics4wxy0119.png|600px]]
|[[File:Mep5wxy0119.png|600px]]
|-
|Figure.13 Dynamics calculation internuclear momenta against time plot
|Figure.14 mep calculation internuclear momenta against time plotsurface plot
|}

The mep (minimum energy path) trajectory is a smooth line on contour and surface plot (Fig.7 Fig.8) while the trajectory under dynamics calculation is oscillating along the pathway. (Fig.10 Fig.11) The mep trajectory takes more steps and longer time than the dynamics trajectory as shown in internuclear distance against time plots. (Fig. 9 Fig.12) The momentum is zero all the time under mep calculation (Fig.13) but under dynamics calculation, the momenta lines are oscillating against time.

The differences arise from mep and dynamics calculations can be explained by the the natures of the two calculations. Under mep calculation, the atoms are in extremely slow motion and the momentum is zero for each step, i.e. the atoms are "stop" after every step. The pathway under mep calculation is formed by connecting all the minimum energy points of each step and appears as a smooth line. On the other hand, atoms are in continuous motion and the momentum is accumulated after each step, atoms are then oscillating on the potential energy surface along the pathway. The steps are set to 5000 in the mep calculation but are only 500 in the dynamics calculatoin. Since the steps in mep are small, more time is needed to complete the same length on the trajectories.

=== Reactive and unreactive trajectories ===

{{fontcolor1|red|Question 4ː Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.}}

For the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, run trajectories with the following momenta p1 and p2 combinations:
{| class="wikitable" border=1
| Condition
| p1
| p2
| Total Energy/ kcal mol -1
| Reactivity
|-
|1.
| -1.25
| -2.5
| -99.018
| reactive
|-
|2.
| -1.5
| -2.0
| -100.456
| unreactive
|-
|3.
| -1.5
| -2.5
| -98.956
| reactive
|-
|4.
| -2.5
| -5.0
| -84.956
| unreactive
|-
|5.
| -2.5
| -5.2
| -83.416
| reactive
|}

{| class="wikitable" border=1
| Condition
| Surface Plot
| Contour Plot
| Description
|-
|1.
| [[File:condition1wxy0119.png|350px]]
| [[File:firstwxy0119.png|350px]]
| r2 (BC) decreases when HC approaches bonded HA and HB. The energy processed by the system is sufficient to overcome the activation barrier and crosses the transition state structure to break the HA and HB bond and form a new HB and HC bond. The new bond oscillate as r1 (AB) increases.
|-
|2.
| [[File:condition2wxy0119.png|350px]]
| [[File:secondwxy0119.png|350px]]
| HC approaches HB but the energy is insufficient to reach the transition state point, HC then moves further away from the bonded H2 molecule and no new bond is formed. The oscillation along r2 is a result of an increase in momentum p1.
|-
|3.
| [[File:condition3wxy0119.png|350px]]
| [[File:thirdwxy0119.png|350px]]
| Similarly to condition 1 but with a more negative p1, HA - HB bond oscillates more when r2 decreases. The reaction proceeds and crosses the transition state point to form the products.
|-
|4.
| [[File:condition4wxy0119.png|350px]]
| [[File:fourthwxy0119.png|350px]]
| Both p1 and p2 are more negative than the previous conditions. The HA - HB bond oscillates vigorously with larger amplitudes. The barrier recrossing happens after the system crosses the transition state region but eventually reverts back to the reactants.
|-
|5.
| [[File:condition5wxy0119.png|350px]]
| [[File:fifthwxy0119.png|350px]]
| With a slight more negative value of p1 as compared to that in condition 4, the system now processes appropriate energy to proceed to forming products. The vigorous oscillations are resulting from high momenta and barrier recrossing occurs.
|}

=== Transition State Theory ===

{{fontcolor|red|Question 5ː State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}

The main assumptions of Transition State Theory (TST) are<ref name="TS" />ː

1. The behaviours of the atomic nuclei follow the classical mechanics rules, i.e. the transition structure is formed as long as atoms or molecules collide with enough energy.

2. The intermediates have a long enough lifetime for the energies to be Boltzmann distributed preceding the next step.

3. The lowest energy saddle point on the potential energy surface is passed over in the reaction system.

The TST does not take into account of the quantum tunnelling effect which allows reactions to happen even when the energy carried by the system is lower than the activation energy barrier, especially for reactions with low energy barriers. Nor does the theory consider the intermediates with short lifetimes. When the energies are not fully distributed, the momentum of the reaction trajectory from the reactants to the intermediate can have effects on product selectivity. Moreover, the theory fails at high temperature because higher vibrational energy modes can be populated for a molecule at high temperature. The molecules can have complex motion and their collisions may result in a higher energy transition structure. The experimental results may thus deviate from that predicted by the TST. For example in condition 4, the kinetic energy of the reactants is much higher than the activation energy required. However, the momenta between atoms are very high and higher vibrational modes are populated that result in higher energy transition structures formed and the system does not pass though the lowest transition saddle point as shown in the diagram. Barrier recrossing occurs but the products are not formed as opposed to the predictions by the TST.

== EXERCISE 2: F - H - H SYSTEM ==

=== Reaction Energetics ===

{{fontcolor1|red|Question 6ː 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?}}

{| class="wikitable" border=1
|[[File:Q6FH2wxy0119.png|600px]]
|[[File:Q6HFHwxy0119.png|600px]]
|-
|Figure.15 F + H2 position on surface plot
|Figure.16 HF and H position on surface plot
|}

The surface plots of the potential energy surface of H + HF → F + H2 reaction is shown (Fig.15 Fig.16). Distance AB is the distance between H and F atoms and distance BC is the distance between two H atoms. The two minimum points shown in the graph are positions of F + H2 (Fig.15) and H + HF (Fig.16) respectively. If forward reaction is H + HF → F + H2, then the backward reaction is F + H2 → H + HF. As clearly shown in the diagram F + H2 are at a minimum point of higher potential energy than that of H + HF. Thus, by comparing the potential energy of reactants and products, the signs of changes in enthalpy (ΔH) and the energetics are determined. H + HF → F + H2 is endothermic and F + H2 → H + HF is exothermic.

The energetics of the two reactions are in accordance to the predictions by comparing bond strengths of chemical species. The bond energy of H-F bond (565 kJ/mol) is higher than the bond energy of H-H bond (432 kJ/mol). Thus, energy released by forming the H-H bond is not enough to compensate for energy required for breaking the H-F bond and the reaction H + HF → F + H2 is endothermic. On the contrary, the reaction F + H2 → H + HF is exothermic.

=== Transition State Approximation ===

{{fontcolor1|red|Question 7ː Locate the approximate position of the transition state.}}

{| class="wikitable" border=1
|[[File:Q7TStransitionwxy0119.png|390px]]
|[[File:Q7TS2wxy0119.png|390px]]
|[[File:Q7TS3wxy0119.png|390px]]
|-
|Figure.17 Transition state surface plot
|Figure.18 Transition state contour plot
|Figure.19 Internuclear distance against time plot (at TS point)
|}

The transition state (TS) is shown as a black dot on the potential energy surface plot (Fig.17) and a red cross on the contour plot. (Fig,18) The position of the TS is at the point where the distance between F and H (rFH) is 1.810Å and the distance between two H atoms (rHH) is 0.746Å. The TS is closer to F + H2 and further away from H + HF. At the transition state position, the distance between F and H, between two H atoms are constant with no momentum as shown in the internuclear distance against time plot as horizontal and flat lines, validating the position is the transition state point (Fig.19)

=== Activation Energies ===

{{fontcolor1|red|Question 8ː Report the activation energy for both reactions.}}

{| class="wikitable" border=1
|[[File:Q8Ea2wxy0119.png|600px]]
|[[File:Q8Ea1wxy0119.png|600px]]
|-
|Figure.21 Energy against time plot (HF + H)
|Figure.20 Energy against time plot (F + H2)
|}

By performing MEP calculation with slight increase and decrease of the rFH to 1.820Å to perform F + H2 → H + HF reaction and to 1.800Å to perform H + HF → F + H2 reaction, the activation energies (Ea) of both reactions can be calculated from the differences in their potential energies (Fig.20 Fig.21)ː

H + HF → F + H2ː Ea = -103.751 - (-133.624) = +29.873 kcal mol-1

F + H2 → H + HFː Ea = -103.751 - (-103.972) = + 0.221 kcal mol-1

=== Reaction Dynamics ===

{{fontcolor1|red|Question 9ː In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

Reactionː F + H2 → H + HF

Initial condition setː rHF = 2Å rHH = 0.74Å pHF = -0.5 pHH = 0.1

{| class="wikitable" border=1
|[[File:Q9plot1wxy0119.png|390px]]
|[[File:Q9plot2wxy0119.png|390px]]
|[[File:Q9plot3wxy0119.png|390px]]
|-
|Figure.23 Contour plot
|Figure.24 Surface plot
|Figure.25 Internuclear momentum against time plot
|-
|[[File:Q9ani1wxy0119.png|390px]]
|[[File:Q9ani2wxy0119.png|390px]]
|-
|Figure.26 Animation figure (at the start of the reaction)
|Figure.27 Animation figure (at the end of the reaction)
|}

As shown in the plots and animation figures, the H-H bond in the reactants is broken and H-F bond in the products is formed. There are small momenta between both H and F and H-H bond that provide the initial kinetic energy of the reactants and the initial vibrational energy of the H-H bond in the reactants. The initial energy is required to break the H-H bond and reduce the distance of H and F for bond forming and potential energy is released when H-F bond is formed. Since energy is conserved, the reaction energy released is converted to the vibrational energy of the H-F bond and the translational kinetic energies of the products. This is shown in the internuclear momentum against time graph as the initial vibrations of B-C (H-H bond) becomes flat as two H atoms are separated in the product and the A-B (H-F bond) formed in the product vibrates with a large amplitude. (Fig.25) The contour plot and the surface plot also clearly show there is a significant increase in the amplitude of the oscillation of H-F bond which is an indication of the large vibrational energy of H-F bond. (Fig.23 Fig.24)

The predictions can be proved experimentally. The vibrational energy of H-F bond can be determined by measuring the frequency and the intensity of the absorption band of the H-F bond with infrared spectroscopy. The conversion to the kinetic energies of the products can be observed by measuring the temperature of the reaction as the kinetic energy can be converted to thermal energy as products collide with solvent molecules. The temperature is expected to increase as this is an exothermic reaction.

=== Energy Distribution and Reactivity ===

{{fontcolor|red|Question 10ː 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.}}
{| class="wikitable" border=1
| Condition
| rFH /Å
| rHH /Å
| pFH
| pHH
|-
|1.
| 2
| 0.74
| -0.5
| -3
|-
|2.
| 2
| 0.74
| -0.5
| 1.9
|-
|3.
| 2.1
| 0.74
| -0.5
| 0.1
|-
|4.
| 2
| 0.74
| -0.8
| 0.1
|-
|5.
| 0.91
| 2
| 0.05
| -20
|-
|6.
| 0.91
| 2
| 0.8
| -7.5
|}
{| class="wikitable" border=1
|Condition
|Contour plot
|Observation
|-
|1.
|[[File:Q10plot1wxy0119.png|600px]]
| The products are not formed in the reaction between F and H2 even pHH is very large. Barrier recrossing occurs and the H-H bond oscillates significantly with a large amplitude in the diagram. It should be noted that while keeping pFH to be -0.5 and changing pHH from -3 to 3, the trend is not very obvious as there are many anomalous values of momentum that render the reaction unreactive while slight increase and decrease of momentum make the reaction reactive again. For example, pHH = -0.9 is reactive but pHH = -1.1 is unreactive.
|-
|2.
|[[File:Q10plot2wxy0119.png|600px]]
| There is reaction between F and H2 and products are formed. The F-H bond formed vibrates significantly along BC trajectory.
|-
|3.
|[[File:Q10plot9wxy0119.png|600px]]
| There is no reaction under this condition between F and H2. There is not enough energy to cross over the transition state point. Although investigations have shown that there is reaction when rFH is equal or less than 2Å, but there is no reaction when H and F atoms are further apart.
|-
|4.
|[[File:Q10plot8wxy0119.png|600px]]
| There is reaction between F and H2. By comparison between condition 4 and condition 3, a slight increase in the momentum of F and H which indicates a slight increase in kinetic translational energy result in the formation of products. Investigations have shown that even when rFH is slightly larger than 2Å, the products are still formed. Thus the change in reaction conditions illustrate that the changes in translational energy affect the reactivity the reaction more than the vibrational energy.
|-
|5.
|[[File:Q10plot5wxy0119.png|600px]]
| There is no reaction between H and HF even though the kinetic translational energy is very high and much higher than the activation energy. (pHF = -20) The trajectory reverts back before reaching the transition state structure.
|-
|6.
|[[File:Q10plot4wxy0119.png|600px]]
| There is reaction between H and HF. Compared to condition 5, the translational energy is reduced significantly while vibrational energy increases slightly. Thus, the vibrational energy is more efficient in promoting the reaction than the translational energy.
|}

The transition state illustrated in Question 7 is closer to F and H2 than H and HF. From the above illustrations, it can be shown that the translational energy is more efficient in promoting the reaction between F and H2 which is exothermic, has a lower activation energy and has an earlier transition state. On the other hand, the vibrational energy is more efficient in the reactivity of the reaction between H and HF which is endothermic, has a higher activation energy and has a late transition state. These observations agree with the Polanyi's empirical rules.

== References ==

<references>
<ref name="TS">Eyring, H. (1935). "The Activated Complex in Chemical Reactions". J. Chem. Phys. 3 (2): 107–115.</ref>
</references>

MRD:WXY0119

2018-05-18T16:46:48Z

Xw2816: /* Activation Energies */

= Molecular Reaction Dynamics Report =

== EXERCISE 1ː H + H2 SYSTEM ==

[[File:HandH2wxy0119.png]]

In the H + H2 system, the distance between the first two H atoms (m1, m2) is r1 and the distance between the two H atoms in the H2 molecule (m2, m3) is r2.

=== Dynamics from the transition state region ===

{{fontcolor|red|Question 1 ː What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.}}

{| class="wikitable" border=1
|[[File:Q1TStransitionwxy0119.png|390px]]
|[[File:Q1TS2wxy0119.png|390px]]
|[[File:Q1TS3wxy0119.png|390px]]
|-
|Figure.1 Surface plot of reaction trajectory
|Figure.2 Transition state surface plot
|Figure.3 Transition state surface plot (different angle)
|}

In the reaction trajectory surface plot (Fig.1), AB is the distance r1 and BC is the distance r2. The gradient of the potential energy with regard to r1 and r2 are ∂V(r1)/∂r1 and ∂V(r2)/∂r2 respectively. The second derivatives of the potential energy with regard to r1 and r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 respectively.

At the two minimum structures which are at the two ends of the reaction pathway, ∂V(r1)/∂r1 = 0 and ∂V(r2)/∂r2 = 0 respectively. Since the two minimum structures are at two local minimum point , the second derivatives r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 are both greater than zero.

On the other hand, the components at the transition state structure (Fig.2) are different from that at minimum points. The gradient ∂V(r1)/∂r1 and ∂V(r2)/∂r2 are both equal to zero at the transition structure. However, r2 are ∂2V(r1)/∂r12 >0 and ∂2V(r2)/∂r22 <0. The transition state point is actually a saddle point which is observed more clearly in Fig.3. The Transition state point is thus distinguished from the minimum points as the potential energy surface curves inwards and downwards at the minimum points.

=== Locating the Transition State ===

{{fontcolor|red|Question 2ː 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.}}

{| class="wikitable" border=1
|[[File:TScontourwxy0119.png|390px]]
|[[File:TSsurfaceplwxy0119.png|390px]]
|[[File:TSestimatewxy0119.png|390px]]
|-
|Figure.4 Transition state contour plot
|Figure.5 Transition state surface plot
|Figure.6 Internuclear distance against time plot
|}

The best estimate of the transition state position is rts = 0.9078 Å.

When r1 = r2 = rts and momenta are set to zero, a graph of intermolecular distance against time is plotted. (Fig.6) In this graph, AB and BC lines are the same and overlap, the two lines on the graph are flat and horizontal, indicating that the atomic distances are the same at the position and with no momenta the state is at equilibrium, validating the position is the transition state position. The transition state is shown as a cross on the counter plot (Fig.4) and a dot on the surface plot (Fig.5).

=== Calculating and comparing the reaction path and trajectory ===

{{fontcolor1|red|Question 3ː Comment on how the ''mep'' and the trajectory you just calculated differ.}}

{| class="wikitable" border=1
|[[File:Mep1wxy0119.png|390px]]
|[[File:Mep2wxy0119.png|390px]]
|[[File:Mep3wxy0119.png|390px]]
|-
|Figure.7 mep calculation contour plot
|Figure.8 mep calculation surface plot
|Figure.9 mep calculation internuclear distance against time plot
|-
|[[File:Dynamic1wxy0119.png|390px]]
|[[File:Dynamic2wxy0119.png|390px]]
|[[File:Dynamic3wxy0119.png|390px]]
|-
|Figure.10 Dynamics calculation contour plot
|Figure.11 Dynamics calculation surface plot
|Figure.12 Dynamics calculation internuclear distance against time plot
|}
{| class="wikitable" border=1
|[[File:dynamics4wxy0119.png|600px]]
|[[File:Mep5wxy0119.png|600px]]
|-
|Figure.13 Dynamics calculation internuclear momenta against time plot
|Figure.14 mep calculation internuclear momenta against time plotsurface plot
|}

The mep (minimum energy path) trajectory is a smooth line on contour and surface plot (Fig.7 Fig.8) while the trajectory under dynamics calculation is oscillating along the pathway. (Fig.10 Fig.11) The mep trajectory takes more steps and longer time than the dynamics trajectory as shown in internuclear distance against time plots. (Fig. 9 Fig.12) The momentum is zero all the time under mep calculation (Fig.13) but under dynamics calculation, the momenta lines are oscillating against time.

The differences arise from mep and dynamics calculations can be explained by the the natures of the two calculations. Under mep calculation, the atoms are in extremely slow motion and the momentum is zero for each step, i.e. the atoms are "stop" after every step. The pathway under mep calculation is formed by connecting all the minimum energy points of each step and appears as a smooth line. On the other hand, atoms are in continuous motion and the momentum is accumulated after each step, atoms are then oscillating on the potential energy surface along the pathway. The steps are set to 5000 in the mep calculation but are only 500 in the dynamics calculatoin. Since the steps in mep are small, more time is needed to complete the same length on the trajectories.

=== Reactive and unreactive trajectories ===

{{fontcolor1|red|Question 4ː Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.}}

For the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, run trajectories with the following momenta p1 and p2 combinations:
{| class="wikitable" border=1
| Condition
| p1
| p2
| Total Energy/ kcal mol -1
| Reactivity
|-
|1.
| -1.25
| -2.5
| -99.018
| reactive
|-
|2.
| -1.5
| -2.0
| -100.456
| unreactive
|-
|3.
| -1.5
| -2.5
| -98.956
| reactive
|-
|4.
| -2.5
| -5.0
| -84.956
| unreactive
|-
|5.
| -2.5
| -5.2
| -83.416
| reactive
|}

{| class="wikitable" border=1
| Condition
| Surface Plot
| Contour Plot
| Description
|-
|1.
| [[File:condition1wxy0119.png|350px]]
| [[File:firstwxy0119.png|350px]]
| r2 (BC) decreases when HC approaches bonded HA and HB. The energy processed by the system is sufficient to overcome the activation barrier and crosses the transition state structure to break the HA and HB bond and form a new HB and HC bond. The new bond oscillate as r1 (AB) increases.
|-
|2.
| [[File:condition2wxy0119.png|350px]]
| [[File:secondwxy0119.png|350px]]
| HC approaches HB but the energy is insufficient to reach the transition state point, HC then moves further away from the bonded H2 molecule and no new bond is formed. The oscillation along r2 is a result of an increase in momentum p1.
|-
|3.
| [[File:condition3wxy0119.png|350px]]
| [[File:thirdwxy0119.png|350px]]
| Similarly to condition 1 but with a more negative p1, HA - HB bond oscillates more when r2 decreases. The reaction proceeds and crosses the transition state point to form the products.
|-
|4.
| [[File:condition4wxy0119.png|350px]]
| [[File:fourthwxy0119.png|350px]]
| Both p1 and p2 are more negative than the previous conditions. The HA - HB bond oscillates vigorously with larger amplitudes. The barrier recrossing happens after the system crosses the transition state region but eventually reverts back to the reactants.
|-
|5.
| [[File:condition5wxy0119.png|350px]]
| [[File:fifthwxy0119.png|350px]]
| With a slight more negative value of p1 as compared to that in condition 4, the system now processes appropriate energy to proceed to forming products. The vigorous oscillations are resulting from high momenta and barrier recrossing occurs.
|}

=== Transition State Theory ===

{{fontcolor|red|Question 5ː State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}

The main assumptions of Transition State Theory (TST) are<ref name="TS" />ː

1. The behaviours of the atomic nuclei follow the classical mechanics rules, i.e. the transition structure is formed as long as atoms or molecules collide with enough energy.

2. The intermediates have a long enough lifetime for the energies to be Boltzmann distributed preceding the next step.

3. The lowest energy saddle point on the potential energy surface is passed over in the reaction system.

The TST does not take into account of the quantum tunnelling effect which allows reactions to happen even when the energy carried by the system is lower than the activation energy barrier, especially for reactions with low energy barriers. Nor does the theory consider the intermediates with short lifetimes. When the energies are not fully distributed, the momentum of the reaction trajectory from the reactants to the intermediate can have effects on product selectivity. Moreover, the theory fails at high temperature because higher vibrational energy modes can be populated for a molecule at high temperature. The molecules can have complex motion and their collisions may result in a higher energy transition structure. The experimental results may thus deviate from that predicted by the TST. For example in condition 4, the kinetic energy of the reactants is much higher than the activation energy required. However, the momenta between atoms are very high and higher vibrational modes are populated that result in higher energy transition structures formed and the system does not pass though the lowest transition saddle point as shown in the diagram. Barrier recrossing occurs but the products are not formed as opposed to the predictions by the TST.

== EXERCISE 2: F - H - H SYSTEM ==

=== Reaction Energetics ===

{{fontcolor1|red|Question 6ː 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?}}

{| class="wikitable" border=1
|[[File:Q6FH2wxy0119.png|600px]]
|[[File:Q6HFHwxy0119.png|600px]]
|-
|Figure.15 F + H2 position on surface plot
|Figure.16 HF and H position on surface plot
|}

The surface plots of the potential energy surface of H + HF → F + H2 reaction is shown (Fig.15 Fig.16). Distance AB is the distance between H and F atoms and distance BC is the distance between two H atoms. The two minimum points shown in the graph are positions of F + H2 (Fig.15) and H + HF (Fig.16) respectively. If forward reaction is H + HF → F + H2, then the backward reaction is F + H2 → H + HF. As clearly shown in the diagram F + H2 are at a minimum point of higher potential energy than that of H + HF. Thus, by comparing the potential energy of reactants and products, the signs of changes in enthalpy (ΔH) and the energetics are determined. H + HF → F + H2 is endothermic and F + H2 → H + HF is exothermic.

The energetics of the two reactions are in accordance to the predictions by comparing bond strengths of chemical species. The bond energy of H-F bond (565 kJ/mol) is higher than the bond energy of H-H bond (432 kJ/mol). Thus, energy released by forming the H-H bond is not enough to compensate for energy required for breaking the H-F bond and the reaction H + HF → F + H2 is endothermic. On the contrary, the reaction F + H2 → H + HF is exothermic.

=== Transition State Approximation ===

{{fontcolor1|red|Question 7ː Locate the approximate position of the transition state.}}

{| class="wikitable" border=1
|[[File:Q7TStransitionwxy0119.png|390px]]
|[[File:Q7TS2wxy0119.png|390px]]
|[[File:Q7TS3wxy0119.png|390px]]
|-
|Figure.17 Transition state surface plot
|Figure.18 Transition state contour plot
|Figure.19 Internuclear distance against time plot (at TS point)
|}

The transition state (TS) is shown as a black dot on the potential energy surface plot (Fig.17) and a red cross on the contour plot. (Fig,18) The position of the TS is at the point where the distance between F and H (rFH) is 1.810Å and the distance between two H atoms (rHH) is 0.746Å. The TS is closer to F + H2 and further away from H + HF. At the transition state position, the distance between F and H, between two H atoms are constant with no momentum as shown in the internuclear distance against time plot as horizontal and flat lines, validating the position is the transition state point (Fig.19)

=== Activation Energies ===

{{fontcolor1|red|Question 8ː Report the activation energy for both reactions.}}

{| class="wikitable" border=1
|[[File:Q8Ea2wxy0119.png|600px]]
|[[File:Q8Ea1wxy0119.png|600px]]
|-
|Figure.21 Energy against time plot (HF + H)
|Figure.20 Energy against time plot (F + H2)
|}

By performing MEP calculation with slight increase and decrease of the rFH to 1.820Å to perform F + H2 → H + HF reaction and to 1.800Å to perform H + HF → F + H2 reaction, the activation energies (Ea) of both reactions can be calculated from the differences in their potential energies (Fig.20 Fig.21)ː

H + HF → F + H2ː Ea = -103.751 - (-133.624) = +29.873 kcal mol-1

F + H2 → H + HFː Ea = -103.751 - (-103.972) = + 0.221 kcal mol-1

=== Reaction Dynamics ===

{{fontcolor1|red|Question 9ː In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

Reactionː F + H2 → H + HF

Initial condition setː rHF = 2Å rHH = 0.74Å pHF = -0.5 pHH = 0.1

{| class="wikitable" border=1
|[[File:Q9plot1wxy0119.png|390px]]
|[[File:Q9plot2wxy0119.png|390px]]
|[[File:Q9plot3wxy0119.png|390px]]
|-
|Figure.23 Contour plot
|Figure.24 Surface plot
|Figure.25 Internuclear momentum against time plot
|-
|[[File:Q9ani1wxy0119.png|390px]]
|[[File:Q9ani2wxy0119.png|390px]]
|-
|Figure.26 Animation figure (at the start of the reaction)
|Figure.27 Animation figure (at the end of the reaction)
|}

As shown in the plots and animation figures, the H-H bond in the reactants is broken and H-F bond in the products is formed. There are small momenta between both H and F and H-H bond that provide the initial kinetic energy of the reactants and the initial vibrational energy of the H-H bond in the reactants. The initial energy is required to break the H-H bond and reduce the distance of H and F for bond forming and potential energy is released when H-F bond is formed. Since energy is conserved, the reaction energy released is converted to the vibrational energy of the H-F bond and the translational kinetic energies of the products. This is shown in the internuclear momentum against time graph as the initial vibrations of B-C (H-H bond) becomes flat as two H atoms are separated in the product and the A-B (H-F bond) formed in the product vibrates with a large amplitude. (Fig.25) The contour plot and the surface plot also clearly show there is a significant increase in the amplitude of the oscillation of H-F bond which is an indication of the large vibrational energy of H-F bond. (Fig.23 Fig.24)

The predictions can be proved experimentally. The vibrational energy of H-F bond can be determined by measuring the frequency and the intensity of the absorption band of the H-F bond with infrared spectroscopy. The conversion to the kinetic energies of the products can be observed by measuring the temperature of the reaction as the kinetic energy can be converted to thermal energy as products collide with solvent molecules. The temperature is expected to increase as this is an exothermic reaction.

=== Energy Distribution and Reactivity ===

{{fontcolor|red|Question 10ː 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.}}
{| class="wikitable" border=1
| Condition
| rFH /Å
| rHH /Å
| pFH
| pHH
|-
|1.
| 2
| 0.74
| -0.5
| -3
|-
|2.
| 2
| 0.74
| -0.5
| 1.9
|-
|3.
| 2.1
| 0.74
| -0.5
| 0.1
|-
|4.
| 2
| 0.74
| -0.8
| 0.1
|-
|5.
| 0.91
| 2
| 0.05
| -20
|-
|6.
| 0.91
| 2
| 0.8
| -7.5
|}
{| class="wikitable" border=1
|Condition
|Contour plot
|Observation
|-
|1.
|[[File:Q10plot1wxy0119.png|600px]]
| The products are not formed in the reaction between F and H2 even pHH is very large. Barrier recrossing occurs and the H-H bond oscillates significantly with a large amplitude in the diagram. It should be noted that while keeping pFH to be -0.5 and changing pHH from -3 to 3, the trend is not very obvious as there are many anomalous values of momentum that render the reaction unreactive while slight increase and decrease of momentum make the reaction reactive again. For example, pHH = -0.9 is reactive but pHH = -1.1 is unreactive.
|-
|2.
|[[File:Q10plot2wxy0119.png|600px]]
| There is reaction between F and H2 and products are formed. The F-H bond formed vibrates significantly along BC trajectory.
|-
|3.
|[[File:Q10plot9wxy0119.png|600px]]
| There is no reaction under this condition between F and H2. There is not enough energy to cross over the transition state point. Although investigations have shown that there is reaction when rFH is equal or less than 2Å, but there is no reaction when H and F atoms are further apart.
|-
|4.
|[[File:Q10plot8wxy0119.png|600px]]
| There is reaction between F and H2. By comparison between condition 4 and condition 3, a slight increase in the momentum of F and H which indicates a slight increase in kinetic translational energy result in the formation of products. Investigations have shown that even when rFH is slightly larger than 2Å, the products are still formed. Thus the change in reaction conditions illustrate that the changes in translational energy affect the reactivity the reaction more than the vibrational energy.
|-
|5.
|[[File:Q10plot5wxy0119.png|600px]]
| There is no reaction between H and HF even though the kinetic translational energy is very high and much higher than the activation energy. (pHF = -20) The trajectory reverts back before reaching the transition state structure.
|-
|6.
|[[File:Q10plot4wxy0119.png|600px]]
| There is reaction between H and HF. Compared to condition 5, the translational energy is reduced significantly while vibrational energy increases slightly. Thus, the vibrational energy is more efficient in promoting the reaction than the translational energy.
|}

The transition state illustrated in Question 7 is closer to F and H2 than H and HF. From the above illustrations, it can be shown that the translational energy is more efficient in promoting the reaction between F and H2 which is exothermic, has a higher activation energy and has an earlier transition state. On the other hand, the vibrational energy is more efficient in the reactivity of the reaction between H and HF which is endothermic, has a higher activation energy and has a lower activation energy and a late transition state. These observations agree with the Polanyi's empirical rules.

== References ==

<references>
<ref name="TS">Eyring, H. (1935). "The Activated Complex in Chemical Reactions". J. Chem. Phys. 3 (2): 107–115.</ref>
</references>

MRD:WXY0119

2018-05-18T16:45:03Z

Xw2816: /* Energy Distribution and Reactivity */

= Molecular Reaction Dynamics Report =

== EXERCISE 1ː H + H2 SYSTEM ==

[[File:HandH2wxy0119.png]]

In the H + H2 system, the distance between the first two H atoms (m1, m2) is r1 and the distance between the two H atoms in the H2 molecule (m2, m3) is r2.

=== Dynamics from the transition state region ===

{{fontcolor|red|Question 1 ː What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.}}

{| class="wikitable" border=1
|[[File:Q1TStransitionwxy0119.png|390px]]
|[[File:Q1TS2wxy0119.png|390px]]
|[[File:Q1TS3wxy0119.png|390px]]
|-
|Figure.1 Surface plot of reaction trajectory
|Figure.2 Transition state surface plot
|Figure.3 Transition state surface plot (different angle)
|}

In the reaction trajectory surface plot (Fig.1), AB is the distance r1 and BC is the distance r2. The gradient of the potential energy with regard to r1 and r2 are ∂V(r1)/∂r1 and ∂V(r2)/∂r2 respectively. The second derivatives of the potential energy with regard to r1 and r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 respectively.

At the two minimum structures which are at the two ends of the reaction pathway, ∂V(r1)/∂r1 = 0 and ∂V(r2)/∂r2 = 0 respectively. Since the two minimum structures are at two local minimum point , the second derivatives r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 are both greater than zero.

On the other hand, the components at the transition state structure (Fig.2) are different from that at minimum points. The gradient ∂V(r1)/∂r1 and ∂V(r2)/∂r2 are both equal to zero at the transition structure. However, r2 are ∂2V(r1)/∂r12 >0 and ∂2V(r2)/∂r22 <0. The transition state point is actually a saddle point which is observed more clearly in Fig.3. The Transition state point is thus distinguished from the minimum points as the potential energy surface curves inwards and downwards at the minimum points.

=== Locating the Transition State ===

{{fontcolor|red|Question 2ː 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.}}

{| class="wikitable" border=1
|[[File:TScontourwxy0119.png|390px]]
|[[File:TSsurfaceplwxy0119.png|390px]]
|[[File:TSestimatewxy0119.png|390px]]
|-
|Figure.4 Transition state contour plot
|Figure.5 Transition state surface plot
|Figure.6 Internuclear distance against time plot
|}

The best estimate of the transition state position is rts = 0.9078 Å.

When r1 = r2 = rts and momenta are set to zero, a graph of intermolecular distance against time is plotted. (Fig.6) In this graph, AB and BC lines are the same and overlap, the two lines on the graph are flat and horizontal, indicating that the atomic distances are the same at the position and with no momenta the state is at equilibrium, validating the position is the transition state position. The transition state is shown as a cross on the counter plot (Fig.4) and a dot on the surface plot (Fig.5).

=== Calculating and comparing the reaction path and trajectory ===

{{fontcolor1|red|Question 3ː Comment on how the ''mep'' and the trajectory you just calculated differ.}}

{| class="wikitable" border=1
|[[File:Mep1wxy0119.png|390px]]
|[[File:Mep2wxy0119.png|390px]]
|[[File:Mep3wxy0119.png|390px]]
|-
|Figure.7 mep calculation contour plot
|Figure.8 mep calculation surface plot
|Figure.9 mep calculation internuclear distance against time plot
|-
|[[File:Dynamic1wxy0119.png|390px]]
|[[File:Dynamic2wxy0119.png|390px]]
|[[File:Dynamic3wxy0119.png|390px]]
|-
|Figure.10 Dynamics calculation contour plot
|Figure.11 Dynamics calculation surface plot
|Figure.12 Dynamics calculation internuclear distance against time plot
|}
{| class="wikitable" border=1
|[[File:dynamics4wxy0119.png|600px]]
|[[File:Mep5wxy0119.png|600px]]
|-
|Figure.13 Dynamics calculation internuclear momenta against time plot
|Figure.14 mep calculation internuclear momenta against time plotsurface plot
|}

The mep (minimum energy path) trajectory is a smooth line on contour and surface plot (Fig.7 Fig.8) while the trajectory under dynamics calculation is oscillating along the pathway. (Fig.10 Fig.11) The mep trajectory takes more steps and longer time than the dynamics trajectory as shown in internuclear distance against time plots. (Fig. 9 Fig.12) The momentum is zero all the time under mep calculation (Fig.13) but under dynamics calculation, the momenta lines are oscillating against time.

The differences arise from mep and dynamics calculations can be explained by the the natures of the two calculations. Under mep calculation, the atoms are in extremely slow motion and the momentum is zero for each step, i.e. the atoms are "stop" after every step. The pathway under mep calculation is formed by connecting all the minimum energy points of each step and appears as a smooth line. On the other hand, atoms are in continuous motion and the momentum is accumulated after each step, atoms are then oscillating on the potential energy surface along the pathway. The steps are set to 5000 in the mep calculation but are only 500 in the dynamics calculatoin. Since the steps in mep are small, more time is needed to complete the same length on the trajectories.

=== Reactive and unreactive trajectories ===

{{fontcolor1|red|Question 4ː Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.}}

For the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, run trajectories with the following momenta p1 and p2 combinations:
{| class="wikitable" border=1
| Condition
| p1
| p2
| Total Energy/ kcal mol -1
| Reactivity
|-
|1.
| -1.25
| -2.5
| -99.018
| reactive
|-
|2.
| -1.5
| -2.0
| -100.456
| unreactive
|-
|3.
| -1.5
| -2.5
| -98.956
| reactive
|-
|4.
| -2.5
| -5.0
| -84.956
| unreactive
|-
|5.
| -2.5
| -5.2
| -83.416
| reactive
|}

{| class="wikitable" border=1
| Condition
| Surface Plot
| Contour Plot
| Description
|-
|1.
| [[File:condition1wxy0119.png|350px]]
| [[File:firstwxy0119.png|350px]]
| r2 (BC) decreases when HC approaches bonded HA and HB. The energy processed by the system is sufficient to overcome the activation barrier and crosses the transition state structure to break the HA and HB bond and form a new HB and HC bond. The new bond oscillate as r1 (AB) increases.
|-
|2.
| [[File:condition2wxy0119.png|350px]]
| [[File:secondwxy0119.png|350px]]
| HC approaches HB but the energy is insufficient to reach the transition state point, HC then moves further away from the bonded H2 molecule and no new bond is formed. The oscillation along r2 is a result of an increase in momentum p1.
|-
|3.
| [[File:condition3wxy0119.png|350px]]
| [[File:thirdwxy0119.png|350px]]
| Similarly to condition 1 but with a more negative p1, HA - HB bond oscillates more when r2 decreases. The reaction proceeds and crosses the transition state point to form the products.
|-
|4.
| [[File:condition4wxy0119.png|350px]]
| [[File:fourthwxy0119.png|350px]]
| Both p1 and p2 are more negative than the previous conditions. The HA - HB bond oscillates vigorously with larger amplitudes. The barrier recrossing happens after the system crosses the transition state region but eventually reverts back to the reactants.
|-
|5.
| [[File:condition5wxy0119.png|350px]]
| [[File:fifthwxy0119.png|350px]]
| With a slight more negative value of p1 as compared to that in condition 4, the system now processes appropriate energy to proceed to forming products. The vigorous oscillations are resulting from high momenta and barrier recrossing occurs.
|}

=== Transition State Theory ===

{{fontcolor|red|Question 5ː State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}

The main assumptions of Transition State Theory (TST) are<ref name="TS" />ː

1. The behaviours of the atomic nuclei follow the classical mechanics rules, i.e. the transition structure is formed as long as atoms or molecules collide with enough energy.

2. The intermediates have a long enough lifetime for the energies to be Boltzmann distributed preceding the next step.

3. The lowest energy saddle point on the potential energy surface is passed over in the reaction system.

The TST does not take into account of the quantum tunnelling effect which allows reactions to happen even when the energy carried by the system is lower than the activation energy barrier, especially for reactions with low energy barriers. Nor does the theory consider the intermediates with short lifetimes. When the energies are not fully distributed, the momentum of the reaction trajectory from the reactants to the intermediate can have effects on product selectivity. Moreover, the theory fails at high temperature because higher vibrational energy modes can be populated for a molecule at high temperature. The molecules can have complex motion and their collisions may result in a higher energy transition structure. The experimental results may thus deviate from that predicted by the TST. For example in condition 4, the kinetic energy of the reactants is much higher than the activation energy required. However, the momenta between atoms are very high and higher vibrational modes are populated that result in higher energy transition structures formed and the system does not pass though the lowest transition saddle point as shown in the diagram. Barrier recrossing occurs but the products are not formed as opposed to the predictions by the TST.

== EXERCISE 2: F - H - H SYSTEM ==

=== Reaction Energetics ===

{{fontcolor1|red|Question 6ː 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?}}

{| class="wikitable" border=1
|[[File:Q6FH2wxy0119.png|600px]]
|[[File:Q6HFHwxy0119.png|600px]]
|-
|Figure.15 F + H2 position on surface plot
|Figure.16 HF and H position on surface plot
|}

The surface plots of the potential energy surface of H + HF → F + H2 reaction is shown (Fig.15 Fig.16). Distance AB is the distance between H and F atoms and distance BC is the distance between two H atoms. The two minimum points shown in the graph are positions of F + H2 (Fig.15) and H + HF (Fig.16) respectively. If forward reaction is H + HF → F + H2, then the backward reaction is F + H2 → H + HF. As clearly shown in the diagram F + H2 are at a minimum point of higher potential energy than that of H + HF. Thus, by comparing the potential energy of reactants and products, the signs of changes in enthalpy (ΔH) and the energetics are determined. H + HF → F + H2 is endothermic and F + H2 → H + HF is exothermic.

The energetics of the two reactions are in accordance to the predictions by comparing bond strengths of chemical species. The bond energy of H-F bond (565 kJ/mol) is higher than the bond energy of H-H bond (432 kJ/mol). Thus, energy released by forming the H-H bond is not enough to compensate for energy required for breaking the H-F bond and the reaction H + HF → F + H2 is endothermic. On the contrary, the reaction F + H2 → H + HF is exothermic.

=== Transition State Approximation ===

{{fontcolor1|red|Question 7ː Locate the approximate position of the transition state.}}

{| class="wikitable" border=1
|[[File:Q7TStransitionwxy0119.png|390px]]
|[[File:Q7TS2wxy0119.png|390px]]
|[[File:Q7TS3wxy0119.png|390px]]
|-
|Figure.17 Transition state surface plot
|Figure.18 Transition state contour plot
|Figure.19 Internuclear distance against time plot (at TS point)
|}

The transition state (TS) is shown as a black dot on the potential energy surface plot (Fig.17) and a red cross on the contour plot. (Fig,18) The position of the TS is at the point where the distance between F and H (rFH) is 1.810Å and the distance between two H atoms (rHH) is 0.746Å. The TS is closer to F + H2 and further away from H + HF. At the transition state position, the distance between F and H, between two H atoms are constant with no momentum as shown in the internuclear distance against time plot as horizontal and flat lines, validating the position is the transition state point (Fig.19)

=== Activation Energies ===

{{fontcolor1|red|Question 8ː Report the activation energy for both reactions.}}

{| class="wikitable" border=1
|[[File:Q8Ea2wxy0119.png|600px]]
|[[File:Q8Ea1wxy0119.png|600px]]
|-
|Figure.21 Energy against time plot (HF + H)
|Figure.20 Energy against time plot (F + H2)
|}

By performing MEP calculation with slight increase and decrease of the rFH to 1.820Å to perform F + H2 → H + HF reaction and to 1.800Å to perform H + HF → F + H2 reaction, the activation energies (Ea) of both reactions can be calculated from the differences in their potential energies (Fig.20 Fig.21)ː

F + H2 → H + HFː Ea = -103.751 - (-133.624) = +29.873 kcal mol-1

H + HF → F + H2ː Ea = -103.751 - (-103.972) = + 0.221 kcal mol-1

=== Reaction Dynamics ===

{{fontcolor1|red|Question 9ː In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

Reactionː F + H2 → H + HF

Initial condition setː rHF = 2Å rHH = 0.74Å pHF = -0.5 pHH = 0.1

{| class="wikitable" border=1
|[[File:Q9plot1wxy0119.png|390px]]
|[[File:Q9plot2wxy0119.png|390px]]
|[[File:Q9plot3wxy0119.png|390px]]
|-
|Figure.23 Contour plot
|Figure.24 Surface plot
|Figure.25 Internuclear momentum against time plot
|-
|[[File:Q9ani1wxy0119.png|390px]]
|[[File:Q9ani2wxy0119.png|390px]]
|-
|Figure.26 Animation figure (at the start of the reaction)
|Figure.27 Animation figure (at the end of the reaction)
|}

As shown in the plots and animation figures, the H-H bond in the reactants is broken and H-F bond in the products is formed. There are small momenta between both H and F and H-H bond that provide the initial kinetic energy of the reactants and the initial vibrational energy of the H-H bond in the reactants. The initial energy is required to break the H-H bond and reduce the distance of H and F for bond forming and potential energy is released when H-F bond is formed. Since energy is conserved, the reaction energy released is converted to the vibrational energy of the H-F bond and the translational kinetic energies of the products. This is shown in the internuclear momentum against time graph as the initial vibrations of B-C (H-H bond) becomes flat as two H atoms are separated in the product and the A-B (H-F bond) formed in the product vibrates with a large amplitude. (Fig.25) The contour plot and the surface plot also clearly show there is a significant increase in the amplitude of the oscillation of H-F bond which is an indication of the large vibrational energy of H-F bond. (Fig.23 Fig.24)

The predictions can be proved experimentally. The vibrational energy of H-F bond can be determined by measuring the frequency and the intensity of the absorption band of the H-F bond with infrared spectroscopy. The conversion to the kinetic energies of the products can be observed by measuring the temperature of the reaction as the kinetic energy can be converted to thermal energy as products collide with solvent molecules. The temperature is expected to increase as this is an exothermic reaction.

=== Energy Distribution and Reactivity ===

{{fontcolor|red|Question 10ː 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.}}
{| class="wikitable" border=1
| Condition
| rFH /Å
| rHH /Å
| pFH
| pHH
|-
|1.
| 2
| 0.74
| -0.5
| -3
|-
|2.
| 2
| 0.74
| -0.5
| 1.9
|-
|3.
| 2.1
| 0.74
| -0.5
| 0.1
|-
|4.
| 2
| 0.74
| -0.8
| 0.1
|-
|5.
| 0.91
| 2
| 0.05
| -20
|-
|6.
| 0.91
| 2
| 0.8
| -7.5
|}
{| class="wikitable" border=1
|Condition
|Contour plot
|Observation
|-
|1.
|[[File:Q10plot1wxy0119.png|600px]]
| The products are not formed in the reaction between F and H2 even pHH is very large. Barrier recrossing occurs and the H-H bond oscillates significantly with a large amplitude in the diagram. It should be noted that while keeping pFH to be -0.5 and changing pHH from -3 to 3, the trend is not very obvious as there are many anomalous values of momentum that render the reaction unreactive while slight increase and decrease of momentum make the reaction reactive again. For example, pHH = -0.9 is reactive but pHH = -1.1 is unreactive.
|-
|2.
|[[File:Q10plot2wxy0119.png|600px]]
| There is reaction between F and H2 and products are formed. The F-H bond formed vibrates significantly along BC trajectory.
|-
|3.
|[[File:Q10plot9wxy0119.png|600px]]
| There is no reaction under this condition between F and H2. There is not enough energy to cross over the transition state point. Although investigations have shown that there is reaction when rFH is equal or less than 2Å, but there is no reaction when H and F atoms are further apart.
|-
|4.
|[[File:Q10plot8wxy0119.png|600px]]
| There is reaction between F and H2. By comparison between condition 4 and condition 3, a slight increase in the momentum of F and H which indicates a slight increase in kinetic translational energy result in the formation of products. Investigations have shown that even when rFH is slightly larger than 2Å, the products are still formed. Thus the change in reaction conditions illustrate that the changes in translational energy affect the reactivity the reaction more than the vibrational energy.
|-
|5.
|[[File:Q10plot5wxy0119.png|600px]]
| There is no reaction between H and HF even though the kinetic translational energy is very high and much higher than the activation energy. (pHF = -20) The trajectory reverts back before reaching the transition state structure.
|-
|6.
|[[File:Q10plot4wxy0119.png|600px]]
| There is reaction between H and HF. Compared to condition 5, the translational energy is reduced significantly while vibrational energy increases slightly. Thus, the vibrational energy is more efficient in promoting the reaction than the translational energy.
|}

The transition state illustrated in Question 7 is closer to F and H2 than H and HF. From the above illustrations, it can be shown that the translational energy is more efficient in promoting the reaction between F and H2 which is exothermic, has a higher activation energy and has an earlier transition state. On the other hand, the vibrational energy is more efficient in the reactivity of the reaction between H and HF which is endothermic, has a higher activation energy and has a lower activation energy and a late transition state. These observations agree with the Polanyi's empirical rules.

== References ==

<references>
<ref name="TS">Eyring, H. (1935). "The Activated Complex in Chemical Reactions". J. Chem. Phys. 3 (2): 107–115.</ref>
</references>

MRD:WXY0119

2018-05-18T16:40:06Z

Xw2816: /* Transition State Approximation */

= Molecular Reaction Dynamics Report =

== EXERCISE 1ː H + H2 SYSTEM ==

[[File:HandH2wxy0119.png]]

In the H + H2 system, the distance between the first two H atoms (m1, m2) is r1 and the distance between the two H atoms in the H2 molecule (m2, m3) is r2.

=== Dynamics from the transition state region ===

{{fontcolor|red|Question 1 ː What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.}}

{| class="wikitable" border=1
|[[File:Q1TStransitionwxy0119.png|390px]]
|[[File:Q1TS2wxy0119.png|390px]]
|[[File:Q1TS3wxy0119.png|390px]]
|-
|Figure.1 Surface plot of reaction trajectory
|Figure.2 Transition state surface plot
|Figure.3 Transition state surface plot (different angle)
|}

In the reaction trajectory surface plot (Fig.1), AB is the distance r1 and BC is the distance r2. The gradient of the potential energy with regard to r1 and r2 are ∂V(r1)/∂r1 and ∂V(r2)/∂r2 respectively. The second derivatives of the potential energy with regard to r1 and r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 respectively.

At the two minimum structures which are at the two ends of the reaction pathway, ∂V(r1)/∂r1 = 0 and ∂V(r2)/∂r2 = 0 respectively. Since the two minimum structures are at two local minimum point , the second derivatives r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 are both greater than zero.

On the other hand, the components at the transition state structure (Fig.2) are different from that at minimum points. The gradient ∂V(r1)/∂r1 and ∂V(r2)/∂r2 are both equal to zero at the transition structure. However, r2 are ∂2V(r1)/∂r12 >0 and ∂2V(r2)/∂r22 <0. The transition state point is actually a saddle point which is observed more clearly in Fig.3. The Transition state point is thus distinguished from the minimum points as the potential energy surface curves inwards and downwards at the minimum points.

=== Locating the Transition State ===

{{fontcolor|red|Question 2ː 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.}}

{| class="wikitable" border=1
|[[File:TScontourwxy0119.png|390px]]
|[[File:TSsurfaceplwxy0119.png|390px]]
|[[File:TSestimatewxy0119.png|390px]]
|-
|Figure.4 Transition state contour plot
|Figure.5 Transition state surface plot
|Figure.6 Internuclear distance against time plot
|}

The best estimate of the transition state position is rts = 0.9078 Å.

When r1 = r2 = rts and momenta are set to zero, a graph of intermolecular distance against time is plotted. (Fig.6) In this graph, AB and BC lines are the same and overlap, the two lines on the graph are flat and horizontal, indicating that the atomic distances are the same at the position and with no momenta the state is at equilibrium, validating the position is the transition state position. The transition state is shown as a cross on the counter plot (Fig.4) and a dot on the surface plot (Fig.5).

=== Calculating and comparing the reaction path and trajectory ===

{{fontcolor1|red|Question 3ː Comment on how the ''mep'' and the trajectory you just calculated differ.}}

{| class="wikitable" border=1
|[[File:Mep1wxy0119.png|390px]]
|[[File:Mep2wxy0119.png|390px]]
|[[File:Mep3wxy0119.png|390px]]
|-
|Figure.7 mep calculation contour plot
|Figure.8 mep calculation surface plot
|Figure.9 mep calculation internuclear distance against time plot
|-
|[[File:Dynamic1wxy0119.png|390px]]
|[[File:Dynamic2wxy0119.png|390px]]
|[[File:Dynamic3wxy0119.png|390px]]
|-
|Figure.10 Dynamics calculation contour plot
|Figure.11 Dynamics calculation surface plot
|Figure.12 Dynamics calculation internuclear distance against time plot
|}
{| class="wikitable" border=1
|[[File:dynamics4wxy0119.png|600px]]
|[[File:Mep5wxy0119.png|600px]]
|-
|Figure.13 Dynamics calculation internuclear momenta against time plot
|Figure.14 mep calculation internuclear momenta against time plotsurface plot
|}

The mep (minimum energy path) trajectory is a smooth line on contour and surface plot (Fig.7 Fig.8) while the trajectory under dynamics calculation is oscillating along the pathway. (Fig.10 Fig.11) The mep trajectory takes more steps and longer time than the dynamics trajectory as shown in internuclear distance against time plots. (Fig. 9 Fig.12) The momentum is zero all the time under mep calculation (Fig.13) but under dynamics calculation, the momenta lines are oscillating against time.

The differences arise from mep and dynamics calculations can be explained by the the natures of the two calculations. Under mep calculation, the atoms are in extremely slow motion and the momentum is zero for each step, i.e. the atoms are "stop" after every step. The pathway under mep calculation is formed by connecting all the minimum energy points of each step and appears as a smooth line. On the other hand, atoms are in continuous motion and the momentum is accumulated after each step, atoms are then oscillating on the potential energy surface along the pathway. The steps are set to 5000 in the mep calculation but are only 500 in the dynamics calculatoin. Since the steps in mep are small, more time is needed to complete the same length on the trajectories.

=== Reactive and unreactive trajectories ===

{{fontcolor1|red|Question 4ː Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.}}

For the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, run trajectories with the following momenta p1 and p2 combinations:
{| class="wikitable" border=1
| Condition
| p1
| p2
| Total Energy/ kcal mol -1
| Reactivity
|-
|1.
| -1.25
| -2.5
| -99.018
| reactive
|-
|2.
| -1.5
| -2.0
| -100.456
| unreactive
|-
|3.
| -1.5
| -2.5
| -98.956
| reactive
|-
|4.
| -2.5
| -5.0
| -84.956
| unreactive
|-
|5.
| -2.5
| -5.2
| -83.416
| reactive
|}

{| class="wikitable" border=1
| Condition
| Surface Plot
| Contour Plot
| Description
|-
|1.
| [[File:condition1wxy0119.png|350px]]
| [[File:firstwxy0119.png|350px]]
| r2 (BC) decreases when HC approaches bonded HA and HB. The energy processed by the system is sufficient to overcome the activation barrier and crosses the transition state structure to break the HA and HB bond and form a new HB and HC bond. The new bond oscillate as r1 (AB) increases.
|-
|2.
| [[File:condition2wxy0119.png|350px]]
| [[File:secondwxy0119.png|350px]]
| HC approaches HB but the energy is insufficient to reach the transition state point, HC then moves further away from the bonded H2 molecule and no new bond is formed. The oscillation along r2 is a result of an increase in momentum p1.
|-
|3.
| [[File:condition3wxy0119.png|350px]]
| [[File:thirdwxy0119.png|350px]]
| Similarly to condition 1 but with a more negative p1, HA - HB bond oscillates more when r2 decreases. The reaction proceeds and crosses the transition state point to form the products.
|-
|4.
| [[File:condition4wxy0119.png|350px]]
| [[File:fourthwxy0119.png|350px]]
| Both p1 and p2 are more negative than the previous conditions. The HA - HB bond oscillates vigorously with larger amplitudes. The barrier recrossing happens after the system crosses the transition state region but eventually reverts back to the reactants.
|-
|5.
| [[File:condition5wxy0119.png|350px]]
| [[File:fifthwxy0119.png|350px]]
| With a slight more negative value of p1 as compared to that in condition 4, the system now processes appropriate energy to proceed to forming products. The vigorous oscillations are resulting from high momenta and barrier recrossing occurs.
|}

=== Transition State Theory ===

{{fontcolor|red|Question 5ː State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}

The main assumptions of Transition State Theory (TST) are<ref name="TS" />ː

1. The behaviours of the atomic nuclei follow the classical mechanics rules, i.e. the transition structure is formed as long as atoms or molecules collide with enough energy.

2. The intermediates have a long enough lifetime for the energies to be Boltzmann distributed preceding the next step.

3. The lowest energy saddle point on the potential energy surface is passed over in the reaction system.

The TST does not take into account of the quantum tunnelling effect which allows reactions to happen even when the energy carried by the system is lower than the activation energy barrier, especially for reactions with low energy barriers. Nor does the theory consider the intermediates with short lifetimes. When the energies are not fully distributed, the momentum of the reaction trajectory from the reactants to the intermediate can have effects on product selectivity. Moreover, the theory fails at high temperature because higher vibrational energy modes can be populated for a molecule at high temperature. The molecules can have complex motion and their collisions may result in a higher energy transition structure. The experimental results may thus deviate from that predicted by the TST. For example in condition 4, the kinetic energy of the reactants is much higher than the activation energy required. However, the momenta between atoms are very high and higher vibrational modes are populated that result in higher energy transition structures formed and the system does not pass though the lowest transition saddle point as shown in the diagram. Barrier recrossing occurs but the products are not formed as opposed to the predictions by the TST.

== EXERCISE 2: F - H - H SYSTEM ==

=== Reaction Energetics ===

{{fontcolor1|red|Question 6ː 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?}}

{| class="wikitable" border=1
|[[File:Q6FH2wxy0119.png|600px]]
|[[File:Q6HFHwxy0119.png|600px]]
|-
|Figure.15 F + H2 position on surface plot
|Figure.16 HF and H position on surface plot
|}

The surface plots of the potential energy surface of H + HF → F + H2 reaction is shown (Fig.15 Fig.16). Distance AB is the distance between H and F atoms and distance BC is the distance between two H atoms. The two minimum points shown in the graph are positions of F + H2 (Fig.15) and H + HF (Fig.16) respectively. If forward reaction is H + HF → F + H2, then the backward reaction is F + H2 → H + HF. As clearly shown in the diagram F + H2 are at a minimum point of higher potential energy than that of H + HF. Thus, by comparing the potential energy of reactants and products, the signs of changes in enthalpy (ΔH) and the energetics are determined. H + HF → F + H2 is endothermic and F + H2 → H + HF is exothermic.

The energetics of the two reactions are in accordance to the predictions by comparing bond strengths of chemical species. The bond energy of H-F bond (565 kJ/mol) is higher than the bond energy of H-H bond (432 kJ/mol). Thus, energy released by forming the H-H bond is not enough to compensate for energy required for breaking the H-F bond and the reaction H + HF → F + H2 is endothermic. On the contrary, the reaction F + H2 → H + HF is exothermic.

=== Transition State Approximation ===

{{fontcolor1|red|Question 7ː Locate the approximate position of the transition state.}}

{| class="wikitable" border=1
|[[File:Q7TStransitionwxy0119.png|390px]]
|[[File:Q7TS2wxy0119.png|390px]]
|[[File:Q7TS3wxy0119.png|390px]]
|-
|Figure.17 Transition state surface plot
|Figure.18 Transition state contour plot
|Figure.19 Internuclear distance against time plot (at TS point)
|}

The transition state (TS) is shown as a black dot on the potential energy surface plot (Fig.17) and a red cross on the contour plot. (Fig,18) The position of the TS is at the point where the distance between F and H (rFH) is 1.810Å and the distance between two H atoms (rHH) is 0.746Å. The TS is closer to F + H2 and further away from H + HF. At the transition state position, the distance between F and H, between two H atoms are constant with no momentum as shown in the internuclear distance against time plot as horizontal and flat lines, validating the position is the transition state point (Fig.19)

=== Activation Energies ===

{{fontcolor1|red|Question 8ː Report the activation energy for both reactions.}}

{| class="wikitable" border=1
|[[File:Q8Ea2wxy0119.png|600px]]
|[[File:Q8Ea1wxy0119.png|600px]]
|-
|Figure.21 Energy against time plot (HF + H)
|Figure.20 Energy against time plot (F + H2)
|}

By performing MEP calculation with slight increase and decrease of the rFH to 1.820Å to perform F + H2 → H + HF reaction and to 1.800Å to perform H + HF → F + H2 reaction, the activation energies (Ea) of both reactions can be calculated from the differences in their potential energies (Fig.20 Fig.21)ː

F + H2 → H + HFː Ea = -103.751 - (-133.624) = +29.873 kcal mol-1

H + HF → F + H2ː Ea = -103.751 - (-103.972) = + 0.221 kcal mol-1

=== Reaction Dynamics ===

{{fontcolor1|red|Question 9ː In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

Reactionː F + H2 → H + HF

Initial condition setː rHF = 2Å rHH = 0.74Å pHF = -0.5 pHH = 0.1

{| class="wikitable" border=1
|[[File:Q9plot1wxy0119.png|390px]]
|[[File:Q9plot2wxy0119.png|390px]]
|[[File:Q9plot3wxy0119.png|390px]]
|-
|Figure.23 Contour plot
|Figure.24 Surface plot
|Figure.25 Internuclear momentum against time plot
|-
|[[File:Q9ani1wxy0119.png|390px]]
|[[File:Q9ani2wxy0119.png|390px]]
|-
|Figure.26 Animation figure (at the start of the reaction)
|Figure.27 Animation figure (at the end of the reaction)
|}

As shown in the plots and animation figures, the H-H bond in the reactants is broken and H-F bond in the products is formed. There are small momenta between both H and F and H-H bond that provide the initial kinetic energy of the reactants and the initial vibrational energy of the H-H bond in the reactants. The initial energy is required to break the H-H bond and reduce the distance of H and F for bond forming and potential energy is released when H-F bond is formed. Since energy is conserved, the reaction energy released is converted to the vibrational energy of the H-F bond and the translational kinetic energies of the products. This is shown in the internuclear momentum against time graph as the initial vibrations of B-C (H-H bond) becomes flat as two H atoms are separated in the product and the A-B (H-F bond) formed in the product vibrates with a large amplitude. (Fig.25) The contour plot and the surface plot also clearly show there is a significant increase in the amplitude of the oscillation of H-F bond which is an indication of the large vibrational energy of H-F bond. (Fig.23 Fig.24)

The predictions can be proved experimentally. The vibrational energy of H-F bond can be determined by measuring the frequency and the intensity of the absorption band of the H-F bond with infrared spectroscopy. The conversion to the kinetic energies of the products can be observed by measuring the temperature of the reaction as the kinetic energy can be converted to thermal energy as products collide with solvent molecules. The temperature is expected to increase as this is an exothermic reaction.

=== Energy Distribution and Reactivity ===

{{fontcolor|red|Question 10ː 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.}}
{| class="wikitable" border=1
| Condition
| rFH /Å
| rHH /Å
| pFH
| pHH
|-
|1.
| 2
| 0.74
| -0.5
| -3
|-
|2.
| 2
| 0.74
| -0.5
| 1.9
|-
|3.
| 2.1
| 0.74
| -0.5
| 0.1
|-
|4.
| 2
| 0.74
| -0.8
| 0.1
|-
|5.
| 0.91
| 2
| 0.05
| -20
|-
|6.
| 0.91
| 2
| 0.8
| -7.5
|}
{| class="wikitable" border=1
|Condition
|Contour plot
|Observation
|-
|1.
|[[File:Q10plot1wxy0119.png|600px]]
| The products are not formed in the reaction between F and H2 even pHH is very large. Barrier recrossing occurs and the H-H bond oscillates significantly with a large amplitude in the diagram. It should be noted that while keeping pFH to be -0.5 and changing pHH from -3 to 3, the trend is not very obvious as there are many anomalous values of momentum that render the reaction unreactive while slight increase and decrease of momentum make the reaction reactive again. For example, pHH = -0.9 is reactive but pHH = -1.1 is unreactive.
|-
|2.
|[[File:Q10plot2wxy0119.png|600px]]
| There is reaction between F and H2 and products are formed. The F-H bond formed vibrates significantly along BC trajectory.
|-
|3.
|[[File:Q10plot9wxy0119.png|600px]]
| There is no reaction under this condition between F and H2. There is not enough energy to cross over the transition state point. Although investigations have shown that there is reaction when rFH is equal or less than 2Å, but there is no reaction when H and F atoms are further apart.
|-
|4.
|[[File:Q10plot8wxy0119.png|600px]]
| There is reaction between F and H2. By comparison between condition 4 and condition 3, a slight increase in the momentum of F and H which indicates a slight increase in kinetic translational energy result in the formation of products. Investigations have shown that even when rFH is slightly larger than 2Å, the products are still formed. Thus the change in reaction conditions illustrate that the changes in translational energy affect the reactivity the reaction more than the vibrational energy.
|-
|5.
|[[File:Q10plot5wxy0119.png|600px]]
| There is no reaction between H and HF even though the kinetic translational energy is very high and much higher than the activation energy. (pHF = -20) The trajectory reverts back before reaching the transition state structure.
|-
|6.
|[[File:Q10plot4wxy0119.png|600px]]
| There is reaction between H and HF. Compared to condition 5, the translational energy is reduced significantly while vibrational energy increases slightly. Thus, the vibrational energy is more efficient in promoting the reaction than the translational energy.
|}

The transition state illustrated in Question 7 is closer to F and H2 than H and HF. From the above illustrations, it can be shown that the translational energy is more efficient in promoting the reaction between F and H2 which is exothermic and has a higher activation energy and an earlier transition state while the

pHH = -8.0 is still very large. This complies with the high activation energy of this reaction as the products formed are thermodynamically less stable than the reactants. It can be concluded that an increase in pHF, which corresponds to a high vibrational energy in H-F coordinate, is necessary for the reaction to take place. pHH, which corresponds to the translational energy in the H-H coordinate that is always very large, does not contribute too much to affect the reactivity of the reaction.

The transition state is present in the exit valley, hence, a late barrier for this endothermic reaction. The reactivity of endothermic reactions is thus dominated by the vibrational energy of the system from the results above.

For substantial exothermic reactions, such as F + H2, the transition state is located in the entrance valley, corresponds to an early barrier of the PES. For substantially endothermic reactions, such as H + HF, the transition state is in the exit valley, corresponds to a late barrier. The favoured degree of freedom for barrier crossing in exothermic reactions would be translation. This means that the momentum of the approaching atom and the atom which it is going to collide with has a greater impact on the rate of the reaction (eg. F atom in F + H2 reaction). However, reagent vibration, which is related to the momentum of the two bonded atoms (eg. HF in H + HF reaction) in the colliding molecule, would be most effective in enabling endothermic reactions to take place.

One reactive trajectory has the conditions of rFH = 0.91 Å, rHH = 2.0 Å, pHF = -0.5 and pHH = -10. The trajectory recrosses the transition state region twice and eventually moves into exit channel, forming products. By comparing this condition to the initial condition, the momentum pHH, which is the translational kinetic energy of the H atom, is reduced by a half; and the pHF, the vibrational energy of the H-F bond is increased to -0.5. This indicates that the vibrational energy has a greater contribution to the feasibility of the reaction than the translational energy. This is the opposite situation of the H2+F reaction.

The results agree with the Polanyi's empirical rules[2], which state that the vibrational energy is more efficient in promoting late-transition-state reactions and the translational energy is more efficient in promoting early-transition-state reactions.

When the momentum pFH is increased slightly from -0.5 to -0.8 with pHH being only 0.1, the reaction now becomes successful. It indicates that the F-H vibration energy has a larger contribution for the feasibility of this reaction compared to the translational energy which is the kinetic energy of the hydrogen atom which is defined by the momentum pHH

When the pHH has a large value above the activation energy (30.231 kcal/mol), which was set to be -20. The hydrogen atom collides with the HF molecule and breaks the H-F bond. However, two hydrogen atoms moves apart and do not form a H2 molecule. The procedure described is illustrated in the following animation snapshots. After the collision, the large momentum that the reactant hydrogen atom contains is still large.As a result, even if an H-H bond can be formed, it would immediately be broken due to an excess vibrational energy between two atoms.

for investigation, 1st set condition pFH = -0.5 pHH = -3 unreactive; 2nd pHH = -2.45 reactive; 3rd pHH = -1.1 unreactive; 4th pHH = -0.9 reactive; 5th pHH = 1.1 unreactive ; 6th pHH = 1.9 reactive ; 7th pHH = 2.4 to 3 unreactive
; pHH = 2.1 no reaction

The cases studied are an illustration of Polanyi's empirical rules.

For H and HF

failed conditionsː 091, 2, 0.05, -20

initial condition of H HF systemː
HF distance = 0.91 HH distance = 2 pHF = 0.8 pHH = -7.5

== References ==

<references>
<ref name="TS">Eyring, H. (1935). "The Activated Complex in Chemical Reactions". J. Chem. Phys. 3 (2): 107–115.</ref>
</references>

MRD:WXY0119

2018-05-18T16:38:36Z

Xw2816: /* Energy Distribution and Reactivity */

= Molecular Reaction Dynamics Report =

== EXERCISE 1ː H + H2 SYSTEM ==

[[File:HandH2wxy0119.png]]

In the H + H2 system, the distance between the first two H atoms (m1, m2) is r1 and the distance between the two H atoms in the H2 molecule (m2, m3) is r2.

=== Dynamics from the transition state region ===

{{fontcolor|red|Question 1 ː What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.}}

{| class="wikitable" border=1
|[[File:Q1TStransitionwxy0119.png|390px]]
|[[File:Q1TS2wxy0119.png|390px]]
|[[File:Q1TS3wxy0119.png|390px]]
|-
|Figure.1 Surface plot of reaction trajectory
|Figure.2 Transition state surface plot
|Figure.3 Transition state surface plot (different angle)
|}

In the reaction trajectory surface plot (Fig.1), AB is the distance r1 and BC is the distance r2. The gradient of the potential energy with regard to r1 and r2 are ∂V(r1)/∂r1 and ∂V(r2)/∂r2 respectively. The second derivatives of the potential energy with regard to r1 and r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 respectively.

At the two minimum structures which are at the two ends of the reaction pathway, ∂V(r1)/∂r1 = 0 and ∂V(r2)/∂r2 = 0 respectively. Since the two minimum structures are at two local minimum point , the second derivatives r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 are both greater than zero.

On the other hand, the components at the transition state structure (Fig.2) are different from that at minimum points. The gradient ∂V(r1)/∂r1 and ∂V(r2)/∂r2 are both equal to zero at the transition structure. However, r2 are ∂2V(r1)/∂r12 >0 and ∂2V(r2)/∂r22 <0. The transition state point is actually a saddle point which is observed more clearly in Fig.3. The Transition state point is thus distinguished from the minimum points as the potential energy surface curves inwards and downwards at the minimum points.

=== Locating the Transition State ===

{{fontcolor|red|Question 2ː 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.}}

{| class="wikitable" border=1
|[[File:TScontourwxy0119.png|390px]]
|[[File:TSsurfaceplwxy0119.png|390px]]
|[[File:TSestimatewxy0119.png|390px]]
|-
|Figure.4 Transition state contour plot
|Figure.5 Transition state surface plot
|Figure.6 Internuclear distance against time plot
|}

The best estimate of the transition state position is rts = 0.9078 Å.

When r1 = r2 = rts and momenta are set to zero, a graph of intermolecular distance against time is plotted. (Fig.6) In this graph, AB and BC lines are the same and overlap, the two lines on the graph are flat and horizontal, indicating that the atomic distances are the same at the position and with no momenta the state is at equilibrium, validating the position is the transition state position. The transition state is shown as a cross on the counter plot (Fig.4) and a dot on the surface plot (Fig.5).

=== Calculating and comparing the reaction path and trajectory ===

{{fontcolor1|red|Question 3ː Comment on how the ''mep'' and the trajectory you just calculated differ.}}

{| class="wikitable" border=1
|[[File:Mep1wxy0119.png|390px]]
|[[File:Mep2wxy0119.png|390px]]
|[[File:Mep3wxy0119.png|390px]]
|-
|Figure.7 mep calculation contour plot
|Figure.8 mep calculation surface plot
|Figure.9 mep calculation internuclear distance against time plot
|-
|[[File:Dynamic1wxy0119.png|390px]]
|[[File:Dynamic2wxy0119.png|390px]]
|[[File:Dynamic3wxy0119.png|390px]]
|-
|Figure.10 Dynamics calculation contour plot
|Figure.11 Dynamics calculation surface plot
|Figure.12 Dynamics calculation internuclear distance against time plot
|}
{| class="wikitable" border=1
|[[File:dynamics4wxy0119.png|600px]]
|[[File:Mep5wxy0119.png|600px]]
|-
|Figure.13 Dynamics calculation internuclear momenta against time plot
|Figure.14 mep calculation internuclear momenta against time plotsurface plot
|}

The mep (minimum energy path) trajectory is a smooth line on contour and surface plot (Fig.7 Fig.8) while the trajectory under dynamics calculation is oscillating along the pathway. (Fig.10 Fig.11) The mep trajectory takes more steps and longer time than the dynamics trajectory as shown in internuclear distance against time plots. (Fig. 9 Fig.12) The momentum is zero all the time under mep calculation (Fig.13) but under dynamics calculation, the momenta lines are oscillating against time.

The differences arise from mep and dynamics calculations can be explained by the the natures of the two calculations. Under mep calculation, the atoms are in extremely slow motion and the momentum is zero for each step, i.e. the atoms are "stop" after every step. The pathway under mep calculation is formed by connecting all the minimum energy points of each step and appears as a smooth line. On the other hand, atoms are in continuous motion and the momentum is accumulated after each step, atoms are then oscillating on the potential energy surface along the pathway. The steps are set to 5000 in the mep calculation but are only 500 in the dynamics calculatoin. Since the steps in mep are small, more time is needed to complete the same length on the trajectories.

=== Reactive and unreactive trajectories ===

{{fontcolor1|red|Question 4ː Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.}}

For the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, run trajectories with the following momenta p1 and p2 combinations:
{| class="wikitable" border=1
| Condition
| p1
| p2
| Total Energy/ kcal mol -1
| Reactivity
|-
|1.
| -1.25
| -2.5
| -99.018
| reactive
|-
|2.
| -1.5
| -2.0
| -100.456
| unreactive
|-
|3.
| -1.5
| -2.5
| -98.956
| reactive
|-
|4.
| -2.5
| -5.0
| -84.956
| unreactive
|-
|5.
| -2.5
| -5.2
| -83.416
| reactive
|}

{| class="wikitable" border=1
| Condition
| Surface Plot
| Contour Plot
| Description
|-
|1.
| [[File:condition1wxy0119.png|350px]]
| [[File:firstwxy0119.png|350px]]
| r2 (BC) decreases when HC approaches bonded HA and HB. The energy processed by the system is sufficient to overcome the activation barrier and crosses the transition state structure to break the HA and HB bond and form a new HB and HC bond. The new bond oscillate as r1 (AB) increases.
|-
|2.
| [[File:condition2wxy0119.png|350px]]
| [[File:secondwxy0119.png|350px]]
| HC approaches HB but the energy is insufficient to reach the transition state point, HC then moves further away from the bonded H2 molecule and no new bond is formed. The oscillation along r2 is a result of an increase in momentum p1.
|-
|3.
| [[File:condition3wxy0119.png|350px]]
| [[File:thirdwxy0119.png|350px]]
| Similarly to condition 1 but with a more negative p1, HA - HB bond oscillates more when r2 decreases. The reaction proceeds and crosses the transition state point to form the products.
|-
|4.
| [[File:condition4wxy0119.png|350px]]
| [[File:fourthwxy0119.png|350px]]
| Both p1 and p2 are more negative than the previous conditions. The HA - HB bond oscillates vigorously with larger amplitudes. The barrier recrossing happens after the system crosses the transition state region but eventually reverts back to the reactants.
|-
|5.
| [[File:condition5wxy0119.png|350px]]
| [[File:fifthwxy0119.png|350px]]
| With a slight more negative value of p1 as compared to that in condition 4, the system now processes appropriate energy to proceed to forming products. The vigorous oscillations are resulting from high momenta and barrier recrossing occurs.
|}

=== Transition State Theory ===

{{fontcolor|red|Question 5ː State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}

The main assumptions of Transition State Theory (TST) are<ref name="TS" />ː

1. The behaviours of the atomic nuclei follow the classical mechanics rules, i.e. the transition structure is formed as long as atoms or molecules collide with enough energy.

2. The intermediates have a long enough lifetime for the energies to be Boltzmann distributed preceding the next step.

3. The lowest energy saddle point on the potential energy surface is passed over in the reaction system.

The TST does not take into account of the quantum tunnelling effect which allows reactions to happen even when the energy carried by the system is lower than the activation energy barrier, especially for reactions with low energy barriers. Nor does the theory consider the intermediates with short lifetimes. When the energies are not fully distributed, the momentum of the reaction trajectory from the reactants to the intermediate can have effects on product selectivity. Moreover, the theory fails at high temperature because higher vibrational energy modes can be populated for a molecule at high temperature. The molecules can have complex motion and their collisions may result in a higher energy transition structure. The experimental results may thus deviate from that predicted by the TST. For example in condition 4, the kinetic energy of the reactants is much higher than the activation energy required. However, the momenta between atoms are very high and higher vibrational modes are populated that result in higher energy transition structures formed and the system does not pass though the lowest transition saddle point as shown in the diagram. Barrier recrossing occurs but the products are not formed as opposed to the predictions by the TST.

== EXERCISE 2: F - H - H SYSTEM ==

=== Reaction Energetics ===

{{fontcolor1|red|Question 6ː 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?}}

{| class="wikitable" border=1
|[[File:Q6FH2wxy0119.png|600px]]
|[[File:Q6HFHwxy0119.png|600px]]
|-
|Figure.15 F + H2 position on surface plot
|Figure.16 HF and H position on surface plot
|}

The surface plots of the potential energy surface of H + HF → F + H2 reaction is shown (Fig.15 Fig.16). Distance AB is the distance between H and F atoms and distance BC is the distance between two H atoms. The two minimum points shown in the graph are positions of F + H2 (Fig.15) and H + HF (Fig.16) respectively. If forward reaction is H + HF → F + H2, then the backward reaction is F + H2 → H + HF. As clearly shown in the diagram F + H2 are at a minimum point of higher potential energy than that of H + HF. Thus, by comparing the potential energy of reactants and products, the signs of changes in enthalpy (ΔH) and the energetics are determined. H + HF → F + H2 is endothermic and F + H2 → H + HF is exothermic.

The energetics of the two reactions are in accordance to the predictions by comparing bond strengths of chemical species. The bond energy of H-F bond (565 kJ/mol) is higher than the bond energy of H-H bond (432 kJ/mol). Thus, energy released by forming the H-H bond is not enough to compensate for energy required for breaking the H-F bond and the reaction H + HF → F + H2 is endothermic. On the contrary, the reaction F + H2 → H + HF is exothermic.

=== Transition State Approximation ===

{{fontcolor1|red|Question 7ː Locate the approximate position of the transition state.}}

{| class="wikitable" border=1
|[[File:Q7TStransitionwxy0119.png|390px]]
|[[File:Q7TS2wxy0119.png|390px]]
|[[File:Q7TS3wxy0119.png|390px]]
|-
|Figure.17 Transition state surface plot
|Figure.18 Transition state contour plot
|Figure.19 Internuclear distance against time plot (at TS point)
|}

The transition state (TS) is shown as a black dot on the potential energy surface plot (Fig.17) and a red cross on the contour plot. (Fig,18) The position of the TS is at the point where the distance between F and H (rFH) is 1.810Å and the distance between two H atoms (rHH) is 0.746Å. At the transition state position, the distance between F and H, between two H atoms are constant with no momentum as shown in the internuclear distance against time plot as horizontal and flat lines, validating the position is the transition state point (Fig.19)

=== Activation Energies ===

{{fontcolor1|red|Question 8ː Report the activation energy for both reactions.}}

{| class="wikitable" border=1
|[[File:Q8Ea2wxy0119.png|600px]]
|[[File:Q8Ea1wxy0119.png|600px]]
|-
|Figure.21 Energy against time plot (HF + H)
|Figure.20 Energy against time plot (F + H2)
|}

By performing MEP calculation with slight increase and decrease of the rFH to 1.820Å to perform F + H2 → H + HF reaction and to 1.800Å to perform H + HF → F + H2 reaction, the activation energies (Ea) of both reactions can be calculated from the differences in their potential energies (Fig.20 Fig.21)ː

F + H2 → H + HFː Ea = -103.751 - (-133.624) = +29.873 kcal mol-1

H + HF → F + H2ː Ea = -103.751 - (-103.972) = + 0.221 kcal mol-1

=== Reaction Dynamics ===

{{fontcolor1|red|Question 9ː In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

Reactionː F + H2 → H + HF

Initial condition setː rHF = 2Å rHH = 0.74Å pHF = -0.5 pHH = 0.1

{| class="wikitable" border=1
|[[File:Q9plot1wxy0119.png|390px]]
|[[File:Q9plot2wxy0119.png|390px]]
|[[File:Q9plot3wxy0119.png|390px]]
|-
|Figure.23 Contour plot
|Figure.24 Surface plot
|Figure.25 Internuclear momentum against time plot
|-
|[[File:Q9ani1wxy0119.png|390px]]
|[[File:Q9ani2wxy0119.png|390px]]
|-
|Figure.26 Animation figure (at the start of the reaction)
|Figure.27 Animation figure (at the end of the reaction)
|}

As shown in the plots and animation figures, the H-H bond in the reactants is broken and H-F bond in the products is formed. There are small momenta between both H and F and H-H bond that provide the initial kinetic energy of the reactants and the initial vibrational energy of the H-H bond in the reactants. The initial energy is required to break the H-H bond and reduce the distance of H and F for bond forming and potential energy is released when H-F bond is formed. Since energy is conserved, the reaction energy released is converted to the vibrational energy of the H-F bond and the translational kinetic energies of the products. This is shown in the internuclear momentum against time graph as the initial vibrations of B-C (H-H bond) becomes flat as two H atoms are separated in the product and the A-B (H-F bond) formed in the product vibrates with a large amplitude. (Fig.25) The contour plot and the surface plot also clearly show there is a significant increase in the amplitude of the oscillation of H-F bond which is an indication of the large vibrational energy of H-F bond. (Fig.23 Fig.24)

The predictions can be proved experimentally. The vibrational energy of H-F bond can be determined by measuring the frequency and the intensity of the absorption band of the H-F bond with infrared spectroscopy. The conversion to the kinetic energies of the products can be observed by measuring the temperature of the reaction as the kinetic energy can be converted to thermal energy as products collide with solvent molecules. The temperature is expected to increase as this is an exothermic reaction.

=== Energy Distribution and Reactivity ===

{{fontcolor|red|Question 10ː 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.}}
{| class="wikitable" border=1
| Condition
| rFH /Å
| rHH /Å
| pFH
| pHH
|-
|1.
| 2
| 0.74
| -0.5
| -3
|-
|2.
| 2
| 0.74
| -0.5
| 1.9
|-
|3.
| 2.1
| 0.74
| -0.5
| 0.1
|-
|4.
| 2
| 0.74
| -0.8
| 0.1
|-
|5.
| 0.91
| 2
| 0.05
| -20
|-
|6.
| 0.91
| 2
| 0.8
| -7.5
|}
{| class="wikitable" border=1
|Condition
|Contour plot
|Observation
|-
|1.
|[[File:Q10plot1wxy0119.png|600px]]
| The products are not formed in the reaction between F and H2 even pHH is very large. Barrier recrossing occurs and the H-H bond oscillates significantly with a large amplitude in the diagram. It should be noted that while keeping pFH to be -0.5 and changing pHH from -3 to 3, the trend is not very obvious as there are many anomalous values of momentum that render the reaction unreactive while slight increase and decrease of momentum make the reaction reactive again. For example, pHH = -0.9 is reactive but pHH = -1.1 is unreactive.
|-
|2.
|[[File:Q10plot2wxy0119.png|600px]]
| There is reaction between F and H2 and products are formed. The F-H bond formed vibrates significantly along BC trajectory.
|-
|3.
|[[File:Q10plot9wxy0119.png|600px]]
| There is no reaction under this condition between F and H2. There is not enough energy to cross over the transition state point. Although investigations have shown that there is reaction when rFH is equal or less than 2Å, but there is no reaction when H and F atoms are further apart.
|-
|4.
|[[File:Q10plot8wxy0119.png|600px]]
| There is reaction between F and H2. By comparison between condition 4 and condition 3, a slight increase in the momentum of F and H which indicates a slight increase in kinetic translational energy result in the formation of products. Investigations have shown that even when rFH is slightly larger than 2Å, the products are still formed. Thus the change in reaction conditions illustrate that the changes in translational energy affect the reactivity the reaction more than the vibrational energy.
|-
|5.
|[[File:Q10plot5wxy0119.png|600px]]
| There is no reaction between H and HF even though the kinetic translational energy is very high and much higher than the activation energy. (pHF = -20) The trajectory reverts back before reaching the transition state structure.
|-
|6.
|[[File:Q10plot4wxy0119.png|600px]]
| There is reaction between H and HF. Compared to condition 5, the translational energy is reduced significantly while vibrational energy increases slightly. Thus, the vibrational energy is more efficient in promoting the reaction than the translational energy.
|}

The transition state illustrated in Question 7 is closer to F and H2 than H and HF. From the above illustrations, it can be shown that the translational energy is more efficient in promoting the reaction between F and H2 which is exothermic and has a higher activation energy and an earlier transition state while the

pHH = -8.0 is still very large. This complies with the high activation energy of this reaction as the products formed are thermodynamically less stable than the reactants. It can be concluded that an increase in pHF, which corresponds to a high vibrational energy in H-F coordinate, is necessary for the reaction to take place. pHH, which corresponds to the translational energy in the H-H coordinate that is always very large, does not contribute too much to affect the reactivity of the reaction.

The transition state is present in the exit valley, hence, a late barrier for this endothermic reaction. The reactivity of endothermic reactions is thus dominated by the vibrational energy of the system from the results above.

For substantial exothermic reactions, such as F + H2, the transition state is located in the entrance valley, corresponds to an early barrier of the PES. For substantially endothermic reactions, such as H + HF, the transition state is in the exit valley, corresponds to a late barrier. The favoured degree of freedom for barrier crossing in exothermic reactions would be translation. This means that the momentum of the approaching atom and the atom which it is going to collide with has a greater impact on the rate of the reaction (eg. F atom in F + H2 reaction). However, reagent vibration, which is related to the momentum of the two bonded atoms (eg. HF in H + HF reaction) in the colliding molecule, would be most effective in enabling endothermic reactions to take place.

One reactive trajectory has the conditions of rFH = 0.91 Å, rHH = 2.0 Å, pHF = -0.5 and pHH = -10. The trajectory recrosses the transition state region twice and eventually moves into exit channel, forming products. By comparing this condition to the initial condition, the momentum pHH, which is the translational kinetic energy of the H atom, is reduced by a half; and the pHF, the vibrational energy of the H-F bond is increased to -0.5. This indicates that the vibrational energy has a greater contribution to the feasibility of the reaction than the translational energy. This is the opposite situation of the H2+F reaction.

The results agree with the Polanyi's empirical rules[2], which state that the vibrational energy is more efficient in promoting late-transition-state reactions and the translational energy is more efficient in promoting early-transition-state reactions.

When the momentum pFH is increased slightly from -0.5 to -0.8 with pHH being only 0.1, the reaction now becomes successful. It indicates that the F-H vibration energy has a larger contribution for the feasibility of this reaction compared to the translational energy which is the kinetic energy of the hydrogen atom which is defined by the momentum pHH

When the pHH has a large value above the activation energy (30.231 kcal/mol), which was set to be -20. The hydrogen atom collides with the HF molecule and breaks the H-F bond. However, two hydrogen atoms moves apart and do not form a H2 molecule. The procedure described is illustrated in the following animation snapshots. After the collision, the large momentum that the reactant hydrogen atom contains is still large.As a result, even if an H-H bond can be formed, it would immediately be broken due to an excess vibrational energy between two atoms.

for investigation, 1st set condition pFH = -0.5 pHH = -3 unreactive; 2nd pHH = -2.45 reactive; 3rd pHH = -1.1 unreactive; 4th pHH = -0.9 reactive; 5th pHH = 1.1 unreactive ; 6th pHH = 1.9 reactive ; 7th pHH = 2.4 to 3 unreactive
; pHH = 2.1 no reaction

The cases studied are an illustration of Polanyi's empirical rules.

For H and HF

failed conditionsː 091, 2, 0.05, -20

initial condition of H HF systemː
HF distance = 0.91 HH distance = 2 pHF = 0.8 pHH = -7.5

== References ==

<references>
<ref name="TS">Eyring, H. (1935). "The Activated Complex in Chemical Reactions". J. Chem. Phys. 3 (2): 107–115.</ref>
</references>

MRD:WXY0119

2018-05-18T16:36:00Z

Xw2816: /* Energy Distribution and Reactivity */

= Molecular Reaction Dynamics Report =

== EXERCISE 1ː H + H2 SYSTEM ==

[[File:HandH2wxy0119.png]]

In the H + H2 system, the distance between the first two H atoms (m1, m2) is r1 and the distance between the two H atoms in the H2 molecule (m2, m3) is r2.

=== Dynamics from the transition state region ===

{{fontcolor|red|Question 1 ː What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.}}

{| class="wikitable" border=1
|[[File:Q1TStransitionwxy0119.png|390px]]
|[[File:Q1TS2wxy0119.png|390px]]
|[[File:Q1TS3wxy0119.png|390px]]
|-
|Figure.1 Surface plot of reaction trajectory
|Figure.2 Transition state surface plot
|Figure.3 Transition state surface plot (different angle)
|}

In the reaction trajectory surface plot (Fig.1), AB is the distance r1 and BC is the distance r2. The gradient of the potential energy with regard to r1 and r2 are ∂V(r1)/∂r1 and ∂V(r2)/∂r2 respectively. The second derivatives of the potential energy with regard to r1 and r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 respectively.

At the two minimum structures which are at the two ends of the reaction pathway, ∂V(r1)/∂r1 = 0 and ∂V(r2)/∂r2 = 0 respectively. Since the two minimum structures are at two local minimum point , the second derivatives r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 are both greater than zero.

On the other hand, the components at the transition state structure (Fig.2) are different from that at minimum points. The gradient ∂V(r1)/∂r1 and ∂V(r2)/∂r2 are both equal to zero at the transition structure. However, r2 are ∂2V(r1)/∂r12 >0 and ∂2V(r2)/∂r22 <0. The transition state point is actually a saddle point which is observed more clearly in Fig.3. The Transition state point is thus distinguished from the minimum points as the potential energy surface curves inwards and downwards at the minimum points.

=== Locating the Transition State ===

{{fontcolor|red|Question 2ː 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.}}

{| class="wikitable" border=1
|[[File:TScontourwxy0119.png|390px]]
|[[File:TSsurfaceplwxy0119.png|390px]]
|[[File:TSestimatewxy0119.png|390px]]
|-
|Figure.4 Transition state contour plot
|Figure.5 Transition state surface plot
|Figure.6 Internuclear distance against time plot
|}

The best estimate of the transition state position is rts = 0.9078 Å.

When r1 = r2 = rts and momenta are set to zero, a graph of intermolecular distance against time is plotted. (Fig.6) In this graph, AB and BC lines are the same and overlap, the two lines on the graph are flat and horizontal, indicating that the atomic distances are the same at the position and with no momenta the state is at equilibrium, validating the position is the transition state position. The transition state is shown as a cross on the counter plot (Fig.4) and a dot on the surface plot (Fig.5).

=== Calculating and comparing the reaction path and trajectory ===

{{fontcolor1|red|Question 3ː Comment on how the ''mep'' and the trajectory you just calculated differ.}}

{| class="wikitable" border=1
|[[File:Mep1wxy0119.png|390px]]
|[[File:Mep2wxy0119.png|390px]]
|[[File:Mep3wxy0119.png|390px]]
|-
|Figure.7 mep calculation contour plot
|Figure.8 mep calculation surface plot
|Figure.9 mep calculation internuclear distance against time plot
|-
|[[File:Dynamic1wxy0119.png|390px]]
|[[File:Dynamic2wxy0119.png|390px]]
|[[File:Dynamic3wxy0119.png|390px]]
|-
|Figure.10 Dynamics calculation contour plot
|Figure.11 Dynamics calculation surface plot
|Figure.12 Dynamics calculation internuclear distance against time plot
|}
{| class="wikitable" border=1
|[[File:dynamics4wxy0119.png|600px]]
|[[File:Mep5wxy0119.png|600px]]
|-
|Figure.13 Dynamics calculation internuclear momenta against time plot
|Figure.14 mep calculation internuclear momenta against time plotsurface plot
|}

The mep (minimum energy path) trajectory is a smooth line on contour and surface plot (Fig.7 Fig.8) while the trajectory under dynamics calculation is oscillating along the pathway. (Fig.10 Fig.11) The mep trajectory takes more steps and longer time than the dynamics trajectory as shown in internuclear distance against time plots. (Fig. 9 Fig.12) The momentum is zero all the time under mep calculation (Fig.13) but under dynamics calculation, the momenta lines are oscillating against time.

The differences arise from mep and dynamics calculations can be explained by the the natures of the two calculations. Under mep calculation, the atoms are in extremely slow motion and the momentum is zero for each step, i.e. the atoms are "stop" after every step. The pathway under mep calculation is formed by connecting all the minimum energy points of each step and appears as a smooth line. On the other hand, atoms are in continuous motion and the momentum is accumulated after each step, atoms are then oscillating on the potential energy surface along the pathway. The steps are set to 5000 in the mep calculation but are only 500 in the dynamics calculatoin. Since the steps in mep are small, more time is needed to complete the same length on the trajectories.

=== Reactive and unreactive trajectories ===

{{fontcolor1|red|Question 4ː Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.}}

For the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, run trajectories with the following momenta p1 and p2 combinations:
{| class="wikitable" border=1
| Condition
| p1
| p2
| Total Energy/ kcal mol -1
| Reactivity
|-
|1.
| -1.25
| -2.5
| -99.018
| reactive
|-
|2.
| -1.5
| -2.0
| -100.456
| unreactive
|-
|3.
| -1.5
| -2.5
| -98.956
| reactive
|-
|4.
| -2.5
| -5.0
| -84.956
| unreactive
|-
|5.
| -2.5
| -5.2
| -83.416
| reactive
|}

{| class="wikitable" border=1
| Condition
| Surface Plot
| Contour Plot
| Description
|-
|1.
| [[File:condition1wxy0119.png|350px]]
| [[File:firstwxy0119.png|350px]]
| r2 (BC) decreases when HC approaches bonded HA and HB. The energy processed by the system is sufficient to overcome the activation barrier and crosses the transition state structure to break the HA and HB bond and form a new HB and HC bond. The new bond oscillate as r1 (AB) increases.
|-
|2.
| [[File:condition2wxy0119.png|350px]]
| [[File:secondwxy0119.png|350px]]
| HC approaches HB but the energy is insufficient to reach the transition state point, HC then moves further away from the bonded H2 molecule and no new bond is formed. The oscillation along r2 is a result of an increase in momentum p1.
|-
|3.
| [[File:condition3wxy0119.png|350px]]
| [[File:thirdwxy0119.png|350px]]
| Similarly to condition 1 but with a more negative p1, HA - HB bond oscillates more when r2 decreases. The reaction proceeds and crosses the transition state point to form the products.
|-
|4.
| [[File:condition4wxy0119.png|350px]]
| [[File:fourthwxy0119.png|350px]]
| Both p1 and p2 are more negative than the previous conditions. The HA - HB bond oscillates vigorously with larger amplitudes. The barrier recrossing happens after the system crosses the transition state region but eventually reverts back to the reactants.
|-
|5.
| [[File:condition5wxy0119.png|350px]]
| [[File:fifthwxy0119.png|350px]]
| With a slight more negative value of p1 as compared to that in condition 4, the system now processes appropriate energy to proceed to forming products. The vigorous oscillations are resulting from high momenta and barrier recrossing occurs.
|}

=== Transition State Theory ===

{{fontcolor|red|Question 5ː State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}

The main assumptions of Transition State Theory (TST) are<ref name="TS" />ː

1. The behaviours of the atomic nuclei follow the classical mechanics rules, i.e. the transition structure is formed as long as atoms or molecules collide with enough energy.

2. The intermediates have a long enough lifetime for the energies to be Boltzmann distributed preceding the next step.

3. The lowest energy saddle point on the potential energy surface is passed over in the reaction system.

The TST does not take into account of the quantum tunnelling effect which allows reactions to happen even when the energy carried by the system is lower than the activation energy barrier, especially for reactions with low energy barriers. Nor does the theory consider the intermediates with short lifetimes. When the energies are not fully distributed, the momentum of the reaction trajectory from the reactants to the intermediate can have effects on product selectivity. Moreover, the theory fails at high temperature because higher vibrational energy modes can be populated for a molecule at high temperature. The molecules can have complex motion and their collisions may result in a higher energy transition structure. The experimental results may thus deviate from that predicted by the TST. For example in condition 4, the kinetic energy of the reactants is much higher than the activation energy required. However, the momenta between atoms are very high and higher vibrational modes are populated that result in higher energy transition structures formed and the system does not pass though the lowest transition saddle point as shown in the diagram. Barrier recrossing occurs but the products are not formed as opposed to the predictions by the TST.

== EXERCISE 2: F - H - H SYSTEM ==

=== Reaction Energetics ===

{{fontcolor1|red|Question 6ː 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?}}

{| class="wikitable" border=1
|[[File:Q6FH2wxy0119.png|600px]]
|[[File:Q6HFHwxy0119.png|600px]]
|-
|Figure.15 F + H2 position on surface plot
|Figure.16 HF and H position on surface plot
|}

The surface plots of the potential energy surface of H + HF → F + H2 reaction is shown (Fig.15 Fig.16). Distance AB is the distance between H and F atoms and distance BC is the distance between two H atoms. The two minimum points shown in the graph are positions of F + H2 (Fig.15) and H + HF (Fig.16) respectively. If forward reaction is H + HF → F + H2, then the backward reaction is F + H2 → H + HF. As clearly shown in the diagram F + H2 are at a minimum point of higher potential energy than that of H + HF. Thus, by comparing the potential energy of reactants and products, the signs of changes in enthalpy (ΔH) and the energetics are determined. H + HF → F + H2 is endothermic and F + H2 → H + HF is exothermic.

The energetics of the two reactions are in accordance to the predictions by comparing bond strengths of chemical species. The bond energy of H-F bond (565 kJ/mol) is higher than the bond energy of H-H bond (432 kJ/mol). Thus, energy released by forming the H-H bond is not enough to compensate for energy required for breaking the H-F bond and the reaction H + HF → F + H2 is endothermic. On the contrary, the reaction F + H2 → H + HF is exothermic.

=== Transition State Approximation ===

{{fontcolor1|red|Question 7ː Locate the approximate position of the transition state.}}

{| class="wikitable" border=1
|[[File:Q7TStransitionwxy0119.png|390px]]
|[[File:Q7TS2wxy0119.png|390px]]
|[[File:Q7TS3wxy0119.png|390px]]
|-
|Figure.17 Transition state surface plot
|Figure.18 Transition state contour plot
|Figure.19 Internuclear distance against time plot (at TS point)
|}

The transition state (TS) is shown as a black dot on the potential energy surface plot (Fig.17) and a red cross on the contour plot. (Fig,18) The position of the TS is at the point where the distance between F and H (rFH) is 1.810Å and the distance between two H atoms (rHH) is 0.746Å. At the transition state position, the distance between F and H, between two H atoms are constant with no momentum as shown in the internuclear distance against time plot as horizontal and flat lines, validating the position is the transition state point (Fig.19)

=== Activation Energies ===

{{fontcolor1|red|Question 8ː Report the activation energy for both reactions.}}

{| class="wikitable" border=1
|[[File:Q8Ea2wxy0119.png|600px]]
|[[File:Q8Ea1wxy0119.png|600px]]
|-
|Figure.21 Energy against time plot (HF + H)
|Figure.20 Energy against time plot (F + H2)
|}

By performing MEP calculation with slight increase and decrease of the rFH to 1.820Å to perform F + H2 → H + HF reaction and to 1.800Å to perform H + HF → F + H2 reaction, the activation energies (Ea) of both reactions can be calculated from the differences in their potential energies (Fig.20 Fig.21)ː

F + H2 → H + HFː Ea = -103.751 - (-133.624) = +29.873 kcal mol-1

H + HF → F + H2ː Ea = -103.751 - (-103.972) = + 0.221 kcal mol-1

=== Reaction Dynamics ===

{{fontcolor1|red|Question 9ː In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

Reactionː F + H2 → H + HF

Initial condition setː rHF = 2Å rHH = 0.74Å pHF = -0.5 pHH = 0.1

{| class="wikitable" border=1
|[[File:Q9plot1wxy0119.png|390px]]
|[[File:Q9plot2wxy0119.png|390px]]
|[[File:Q9plot3wxy0119.png|390px]]
|-
|Figure.23 Contour plot
|Figure.24 Surface plot
|Figure.25 Internuclear momentum against time plot
|-
|[[File:Q9ani1wxy0119.png|390px]]
|[[File:Q9ani2wxy0119.png|390px]]
|-
|Figure.26 Animation figure (at the start of the reaction)
|Figure.27 Animation figure (at the end of the reaction)
|}

As shown in the plots and animation figures, the H-H bond in the reactants is broken and H-F bond in the products is formed. There are small momenta between both H and F and H-H bond that provide the initial kinetic energy of the reactants and the initial vibrational energy of the H-H bond in the reactants. The initial energy is required to break the H-H bond and reduce the distance of H and F for bond forming and potential energy is released when H-F bond is formed. Since energy is conserved, the reaction energy released is converted to the vibrational energy of the H-F bond and the translational kinetic energies of the products. This is shown in the internuclear momentum against time graph as the initial vibrations of B-C (H-H bond) becomes flat as two H atoms are separated in the product and the A-B (H-F bond) formed in the product vibrates with a large amplitude. (Fig.25) The contour plot and the surface plot also clearly show there is a significant increase in the amplitude of the oscillation of H-F bond which is an indication of the large vibrational energy of H-F bond. (Fig.23 Fig.24)

The predictions can be proved experimentally. The vibrational energy of H-F bond can be determined by measuring the frequency and the intensity of the absorption band of the H-F bond with infrared spectroscopy. The conversion to the kinetic energies of the products can be observed by measuring the temperature of the reaction as the kinetic energy can be converted to thermal energy as products collide with solvent molecules. The temperature is expected to increase as this is an exothermic reaction.

=== Energy Distribution and Reactivity ===

{{fontcolor|red|Question 10ː 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.}}
{| class="wikitable" border=1
| Condition
| rFH /Å
| rHH /Å
| pFH
| pHH
|-
|1.
| 2
| 0.74
| -0.5
| -3
|-
|2.
| 2
| 0.74
| -0.5
| 1.9
|-
|3.
| 2.1
| 0.74
| -0.5
| 0.1
|-
|4.
| 2
| 0.74
| -0.8
| 0.1
|-
|5.
| 0.91
| 2
| 0.05
| -20
|-
|6.
| 0.91
| 2
| 0.8
| -7.5
|}
{| class="wikitable" border=1
|Condition
|Contour plot
|Observation
|-
|1.
|[[File:Q10plot1wxy0119.png|600px]]
| The products are not formed in the reaction between F and H2 even pHH is very large. Barrier recrossing occurs and the H-H bond oscillates significantly with a large amplitude in the diagram. It should be noted that while keeping pFH to be -0.5 and changing pHH from -3 to 3, the trend is not very obvious as there are many anomalous values of momentum that render the reaction unreactive while slight increase and decrease of momentum make the reaction reactive again. For example, pHH = -0.9 is reactive but pHH = -1.1 is unreactive.
|-
|2.
|[[File:Q10plot2wxy0119.png|600px]]
| There is reaction between F and H2 and products are formed. The F-H bond formed vibrates significantly along BC trajectory.
|-
|3.
|[[File:Q10plot9wxy0119.png|600px]]
| There is no reaction under this condition between F and H2. There is not enough energy to cross over the transition state point. Although investigations have shown that there is reaction when rFH is equal or less than 2Å, but there is no reaction when H and F atoms are further apart.
|-
|4.
|[[File:Q10plot8wxy0119.png|600px]]
| There is reaction between F and H2. By comparison between condition 4 and condition 3, a slight increase in the momentum of F and H which indicates a slight increase in kinetic translational energy result in the formation of products. Investigations have shown that even when rFH is slightly larger than 2Å, the products are still formed. Thus the change in reaction conditions illustrate that the changes in translational energy affect the reactivity the reaction more than the vibrational energy.
|-
|5.
|[[File:Q10plot5wxy0119.png|600px]]
| There is no reaction between H and HF even though the kinetic translational energy is very high and much higher than the activation energy. (pHF = -20) The trajectory reverts back before reaching the transition state structure.
|-
|6.
|[[File:Q10plot4wxy0119.png|600px]]
| There is reaction between H and HF. Compared to condition 5, the translational energy is reduced significantly while vibrational energy increases slightly. Thus, the vibrational energy is more efficient in promoting the reaction than the translational energy.
|}

From the above illustrations, it can be shown that the translational energy is more efficient in promoting the reaction between F and H2 which is exothermic and has a higher activation energy and an earlier transition state while the

pHH = -8.0 is still very large. This complies with the high activation energy of this reaction as the products formed are thermodynamically less stable than the reactants. It can be concluded that an increase in pHF, which corresponds to a high vibrational energy in H-F coordinate, is necessary for the reaction to take place. pHH, which corresponds to the translational energy in the H-H coordinate that is always very large, does not contribute too much to affect the reactivity of the reaction.

The transition state is present in the exit valley, hence, a late barrier for this endothermic reaction. The reactivity of endothermic reactions is thus dominated by the vibrational energy of the system from the results above.

For substantial exothermic reactions, such as F + H2, the transition state is located in the entrance valley, corresponds to an early barrier of the PES. For substantially endothermic reactions, such as H + HF, the transition state is in the exit valley, corresponds to a late barrier. The favoured degree of freedom for barrier crossing in exothermic reactions would be translation. This means that the momentum of the approaching atom and the atom which it is going to collide with has a greater impact on the rate of the reaction (eg. F atom in F + H2 reaction). However, reagent vibration, which is related to the momentum of the two bonded atoms (eg. HF in H + HF reaction) in the colliding molecule, would be most effective in enabling endothermic reactions to take place.

One reactive trajectory has the conditions of rFH = 0.91 Å, rHH = 2.0 Å, pHF = -0.5 and pHH = -10. The trajectory recrosses the transition state region twice and eventually moves into exit channel, forming products. By comparing this condition to the initial condition, the momentum pHH, which is the translational kinetic energy of the H atom, is reduced by a half; and the pHF, the vibrational energy of the H-F bond is increased to -0.5. This indicates that the vibrational energy has a greater contribution to the feasibility of the reaction than the translational energy. This is the opposite situation of the H2+F reaction.

The results agree with the Polanyi's empirical rules[2], which state that the vibrational energy is more efficient in promoting late-transition-state reactions and the translational energy is more efficient in promoting early-transition-state reactions.

When the momentum pFH is increased slightly from -0.5 to -0.8 with pHH being only 0.1, the reaction now becomes successful. It indicates that the F-H vibration energy has a larger contribution for the feasibility of this reaction compared to the translational energy which is the kinetic energy of the hydrogen atom which is defined by the momentum pHH

When the pHH has a large value above the activation energy (30.231 kcal/mol), which was set to be -20. The hydrogen atom collides with the HF molecule and breaks the H-F bond. However, two hydrogen atoms moves apart and do not form a H2 molecule. The procedure described is illustrated in the following animation snapshots. After the collision, the large momentum that the reactant hydrogen atom contains is still large.As a result, even if an H-H bond can be formed, it would immediately be broken due to an excess vibrational energy between two atoms.

for investigation, 1st set condition pFH = -0.5 pHH = -3 unreactive; 2nd pHH = -2.45 reactive; 3rd pHH = -1.1 unreactive; 4th pHH = -0.9 reactive; 5th pHH = 1.1 unreactive ; 6th pHH = 1.9 reactive ; 7th pHH = 2.4 to 3 unreactive
; pHH = 2.1 no reaction

The cases studied are an illustration of Polanyi's empirical rules.

For H and HF

failed conditionsː 091, 2, 0.05, -20

initial condition of H HF systemː
HF distance = 0.91 HH distance = 2 pHF = 0.8 pHH = -7.5

== References ==

<references>
<ref name="TS">Eyring, H. (1935). "The Activated Complex in Chemical Reactions". J. Chem. Phys. 3 (2): 107–115.</ref>
</references>

MRD:WXY0119

2018-05-18T16:28:07Z

Xw2816: /* Energy Distribution and Reactivity */

= Molecular Reaction Dynamics Report =

== EXERCISE 1ː H + H2 SYSTEM ==

[[File:HandH2wxy0119.png]]

In the H + H2 system, the distance between the first two H atoms (m1, m2) is r1 and the distance between the two H atoms in the H2 molecule (m2, m3) is r2.

=== Dynamics from the transition state region ===

{{fontcolor|red|Question 1 ː What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.}}

{| class="wikitable" border=1
|[[File:Q1TStransitionwxy0119.png|390px]]
|[[File:Q1TS2wxy0119.png|390px]]
|[[File:Q1TS3wxy0119.png|390px]]
|-
|Figure.1 Surface plot of reaction trajectory
|Figure.2 Transition state surface plot
|Figure.3 Transition state surface plot (different angle)
|}

In the reaction trajectory surface plot (Fig.1), AB is the distance r1 and BC is the distance r2. The gradient of the potential energy with regard to r1 and r2 are ∂V(r1)/∂r1 and ∂V(r2)/∂r2 respectively. The second derivatives of the potential energy with regard to r1 and r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 respectively.

At the two minimum structures which are at the two ends of the reaction pathway, ∂V(r1)/∂r1 = 0 and ∂V(r2)/∂r2 = 0 respectively. Since the two minimum structures are at two local minimum point , the second derivatives r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 are both greater than zero.

On the other hand, the components at the transition state structure (Fig.2) are different from that at minimum points. The gradient ∂V(r1)/∂r1 and ∂V(r2)/∂r2 are both equal to zero at the transition structure. However, r2 are ∂2V(r1)/∂r12 >0 and ∂2V(r2)/∂r22 <0. The transition state point is actually a saddle point which is observed more clearly in Fig.3. The Transition state point is thus distinguished from the minimum points as the potential energy surface curves inwards and downwards at the minimum points.

=== Locating the Transition State ===

{{fontcolor|red|Question 2ː 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.}}

{| class="wikitable" border=1
|[[File:TScontourwxy0119.png|390px]]
|[[File:TSsurfaceplwxy0119.png|390px]]
|[[File:TSestimatewxy0119.png|390px]]
|-
|Figure.4 Transition state contour plot
|Figure.5 Transition state surface plot
|Figure.6 Internuclear distance against time plot
|}

The best estimate of the transition state position is rts = 0.9078 Å.

When r1 = r2 = rts and momenta are set to zero, a graph of intermolecular distance against time is plotted. (Fig.6) In this graph, AB and BC lines are the same and overlap, the two lines on the graph are flat and horizontal, indicating that the atomic distances are the same at the position and with no momenta the state is at equilibrium, validating the position is the transition state position. The transition state is shown as a cross on the counter plot (Fig.4) and a dot on the surface plot (Fig.5).

=== Calculating and comparing the reaction path and trajectory ===

{{fontcolor1|red|Question 3ː Comment on how the ''mep'' and the trajectory you just calculated differ.}}

{| class="wikitable" border=1
|[[File:Mep1wxy0119.png|390px]]
|[[File:Mep2wxy0119.png|390px]]
|[[File:Mep3wxy0119.png|390px]]
|-
|Figure.7 mep calculation contour plot
|Figure.8 mep calculation surface plot
|Figure.9 mep calculation internuclear distance against time plot
|-
|[[File:Dynamic1wxy0119.png|390px]]
|[[File:Dynamic2wxy0119.png|390px]]
|[[File:Dynamic3wxy0119.png|390px]]
|-
|Figure.10 Dynamics calculation contour plot
|Figure.11 Dynamics calculation surface plot
|Figure.12 Dynamics calculation internuclear distance against time plot
|}
{| class="wikitable" border=1
|[[File:dynamics4wxy0119.png|600px]]
|[[File:Mep5wxy0119.png|600px]]
|-
|Figure.13 Dynamics calculation internuclear momenta against time plot
|Figure.14 mep calculation internuclear momenta against time plotsurface plot
|}

The mep (minimum energy path) trajectory is a smooth line on contour and surface plot (Fig.7 Fig.8) while the trajectory under dynamics calculation is oscillating along the pathway. (Fig.10 Fig.11) The mep trajectory takes more steps and longer time than the dynamics trajectory as shown in internuclear distance against time plots. (Fig. 9 Fig.12) The momentum is zero all the time under mep calculation (Fig.13) but under dynamics calculation, the momenta lines are oscillating against time.

The differences arise from mep and dynamics calculations can be explained by the the natures of the two calculations. Under mep calculation, the atoms are in extremely slow motion and the momentum is zero for each step, i.e. the atoms are "stop" after every step. The pathway under mep calculation is formed by connecting all the minimum energy points of each step and appears as a smooth line. On the other hand, atoms are in continuous motion and the momentum is accumulated after each step, atoms are then oscillating on the potential energy surface along the pathway. The steps are set to 5000 in the mep calculation but are only 500 in the dynamics calculatoin. Since the steps in mep are small, more time is needed to complete the same length on the trajectories.

=== Reactive and unreactive trajectories ===

{{fontcolor1|red|Question 4ː Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.}}

For the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, run trajectories with the following momenta p1 and p2 combinations:
{| class="wikitable" border=1
| Condition
| p1
| p2
| Total Energy/ kcal mol -1
| Reactivity
|-
|1.
| -1.25
| -2.5
| -99.018
| reactive
|-
|2.
| -1.5
| -2.0
| -100.456
| unreactive
|-
|3.
| -1.5
| -2.5
| -98.956
| reactive
|-
|4.
| -2.5
| -5.0
| -84.956
| unreactive
|-
|5.
| -2.5
| -5.2
| -83.416
| reactive
|}

{| class="wikitable" border=1
| Condition
| Surface Plot
| Contour Plot
| Description
|-
|1.
| [[File:condition1wxy0119.png|350px]]
| [[File:firstwxy0119.png|350px]]
| r2 (BC) decreases when HC approaches bonded HA and HB. The energy processed by the system is sufficient to overcome the activation barrier and crosses the transition state structure to break the HA and HB bond and form a new HB and HC bond. The new bond oscillate as r1 (AB) increases.
|-
|2.
| [[File:condition2wxy0119.png|350px]]
| [[File:secondwxy0119.png|350px]]
| HC approaches HB but the energy is insufficient to reach the transition state point, HC then moves further away from the bonded H2 molecule and no new bond is formed. The oscillation along r2 is a result of an increase in momentum p1.
|-
|3.
| [[File:condition3wxy0119.png|350px]]
| [[File:thirdwxy0119.png|350px]]
| Similarly to condition 1 but with a more negative p1, HA - HB bond oscillates more when r2 decreases. The reaction proceeds and crosses the transition state point to form the products.
|-
|4.
| [[File:condition4wxy0119.png|350px]]
| [[File:fourthwxy0119.png|350px]]
| Both p1 and p2 are more negative than the previous conditions. The HA - HB bond oscillates vigorously with larger amplitudes. The barrier recrossing happens after the system crosses the transition state region but eventually reverts back to the reactants.
|-
|5.
| [[File:condition5wxy0119.png|350px]]
| [[File:fifthwxy0119.png|350px]]
| With a slight more negative value of p1 as compared to that in condition 4, the system now processes appropriate energy to proceed to forming products. The vigorous oscillations are resulting from high momenta and barrier recrossing occurs.
|}

=== Transition State Theory ===

{{fontcolor|red|Question 5ː State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}

The main assumptions of Transition State Theory (TST) are<ref name="TS" />ː

1. The behaviours of the atomic nuclei follow the classical mechanics rules, i.e. the transition structure is formed as long as atoms or molecules collide with enough energy.

2. The intermediates have a long enough lifetime for the energies to be Boltzmann distributed preceding the next step.

3. The lowest energy saddle point on the potential energy surface is passed over in the reaction system.

The TST does not take into account of the quantum tunnelling effect which allows reactions to happen even when the energy carried by the system is lower than the activation energy barrier, especially for reactions with low energy barriers. Nor does the theory consider the intermediates with short lifetimes. When the energies are not fully distributed, the momentum of the reaction trajectory from the reactants to the intermediate can have effects on product selectivity. Moreover, the theory fails at high temperature because higher vibrational energy modes can be populated for a molecule at high temperature. The molecules can have complex motion and their collisions may result in a higher energy transition structure. The experimental results may thus deviate from that predicted by the TST. For example in condition 4, the kinetic energy of the reactants is much higher than the activation energy required. However, the momenta between atoms are very high and higher vibrational modes are populated that result in higher energy transition structures formed and the system does not pass though the lowest transition saddle point as shown in the diagram. Barrier recrossing occurs but the products are not formed as opposed to the predictions by the TST.

== EXERCISE 2: F - H - H SYSTEM ==

=== Reaction Energetics ===

{{fontcolor1|red|Question 6ː 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?}}

{| class="wikitable" border=1
|[[File:Q6FH2wxy0119.png|600px]]
|[[File:Q6HFHwxy0119.png|600px]]
|-
|Figure.15 F + H2 position on surface plot
|Figure.16 HF and H position on surface plot
|}

The surface plots of the potential energy surface of H + HF → F + H2 reaction is shown (Fig.15 Fig.16). Distance AB is the distance between H and F atoms and distance BC is the distance between two H atoms. The two minimum points shown in the graph are positions of F + H2 (Fig.15) and H + HF (Fig.16) respectively. If forward reaction is H + HF → F + H2, then the backward reaction is F + H2 → H + HF. As clearly shown in the diagram F + H2 are at a minimum point of higher potential energy than that of H + HF. Thus, by comparing the potential energy of reactants and products, the signs of changes in enthalpy (ΔH) and the energetics are determined. H + HF → F + H2 is endothermic and F + H2 → H + HF is exothermic.

The energetics of the two reactions are in accordance to the predictions by comparing bond strengths of chemical species. The bond energy of H-F bond (565 kJ/mol) is higher than the bond energy of H-H bond (432 kJ/mol). Thus, energy released by forming the H-H bond is not enough to compensate for energy required for breaking the H-F bond and the reaction H + HF → F + H2 is endothermic. On the contrary, the reaction F + H2 → H + HF is exothermic.

=== Transition State Approximation ===

{{fontcolor1|red|Question 7ː Locate the approximate position of the transition state.}}

{| class="wikitable" border=1
|[[File:Q7TStransitionwxy0119.png|390px]]
|[[File:Q7TS2wxy0119.png|390px]]
|[[File:Q7TS3wxy0119.png|390px]]
|-
|Figure.17 Transition state surface plot
|Figure.18 Transition state contour plot
|Figure.19 Internuclear distance against time plot (at TS point)
|}

The transition state (TS) is shown as a black dot on the potential energy surface plot (Fig.17) and a red cross on the contour plot. (Fig,18) The position of the TS is at the point where the distance between F and H (rFH) is 1.810Å and the distance between two H atoms (rHH) is 0.746Å. At the transition state position, the distance between F and H, between two H atoms are constant with no momentum as shown in the internuclear distance against time plot as horizontal and flat lines, validating the position is the transition state point (Fig.19)

=== Activation Energies ===

{{fontcolor1|red|Question 8ː Report the activation energy for both reactions.}}

{| class="wikitable" border=1
|[[File:Q8Ea2wxy0119.png|600px]]
|[[File:Q8Ea1wxy0119.png|600px]]
|-
|Figure.21 Energy against time plot (HF + H)
|Figure.20 Energy against time plot (F + H2)
|}

By performing MEP calculation with slight increase and decrease of the rFH to 1.820Å to perform F + H2 → H + HF reaction and to 1.800Å to perform H + HF → F + H2 reaction, the activation energies (Ea) of both reactions can be calculated from the differences in their potential energies (Fig.20 Fig.21)ː

F + H2 → H + HFː Ea = -103.751 - (-133.624) = +29.873 kcal mol-1

H + HF → F + H2ː Ea = -103.751 - (-103.972) = + 0.221 kcal mol-1

=== Reaction Dynamics ===

{{fontcolor1|red|Question 9ː In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

Reactionː F + H2 → H + HF

Initial condition setː rHF = 2Å rHH = 0.74Å pHF = -0.5 pHH = 0.1

{| class="wikitable" border=1
|[[File:Q9plot1wxy0119.png|390px]]
|[[File:Q9plot2wxy0119.png|390px]]
|[[File:Q9plot3wxy0119.png|390px]]
|-
|Figure.23 Contour plot
|Figure.24 Surface plot
|Figure.25 Internuclear momentum against time plot
|-
|[[File:Q9ani1wxy0119.png|390px]]
|[[File:Q9ani2wxy0119.png|390px]]
|-
|Figure.26 Animation figure (at the start of the reaction)
|Figure.27 Animation figure (at the end of the reaction)
|}

As shown in the plots and animation figures, the H-H bond in the reactants is broken and H-F bond in the products is formed. There are small momenta between both H and F and H-H bond that provide the initial kinetic energy of the reactants and the initial vibrational energy of the H-H bond in the reactants. The initial energy is required to break the H-H bond and reduce the distance of H and F for bond forming and potential energy is released when H-F bond is formed. Since energy is conserved, the reaction energy released is converted to the vibrational energy of the H-F bond and the translational kinetic energies of the products. This is shown in the internuclear momentum against time graph as the initial vibrations of B-C (H-H bond) becomes flat as two H atoms are separated in the product and the A-B (H-F bond) formed in the product vibrates with a large amplitude. (Fig.25) The contour plot and the surface plot also clearly show there is a significant increase in the amplitude of the oscillation of H-F bond which is an indication of the large vibrational energy of H-F bond. (Fig.23 Fig.24)

The predictions can be proved experimentally. The vibrational energy of H-F bond can be determined by measuring the frequency and the intensity of the absorption band of the H-F bond with infrared spectroscopy. The conversion to the kinetic energies of the products can be observed by measuring the temperature of the reaction as the kinetic energy can be converted to thermal energy as products collide with solvent molecules. The temperature is expected to increase as this is an exothermic reaction.

=== Energy Distribution and Reactivity ===

{{fontcolor|red|Question 10ː 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.}}
{| class="wikitable" border=1
| Condition
| rFH /Å
| rHH /Å
| pFH
| pHH
|-
|1.
| 2
| 0.74
| -0.5
| -3
|-
|2.
| 2
| 0.74
| -0.5
| 1.9
|-
|3.
| 2.1
| 0.74
| -0.5
| 0.1
|-
|4.
| 2
| 0.74
| -0.8
| 0.1
|-
|5.
| 0.91
| 2
| 0.05
| -20
|-
|6.
| 0.91
| 2
| 0.8
| -7.5
|}
{| class="wikitable" border=1
|Condition
|Contour plot
|Observation
|-
|1.
|[[File:Q10plot1wxy0119.png|600px]]
| The products are not formed in the reaction between F and H2 even pHH is very large. Barrier recrossing occurs and the H-H bond oscillates significantly with a large amplitude in the diagram.
|-
|2.
|[[File:Q10plot2wxy0119.png|600px]]
| There is reaction between F and H2 and products are formed. The F-H bond formed vibrates significantly along BC trajectory.
|-
|3.
|[[File:Q10plot9wxy0119.png|600px]]
| There is no reaction under this condition between F and H2. There is not enough energy to cross over the transition state point. Although investigations have shown that there is reaction when rFH is equal or less than 2Å, but there is no reaction when H and F atoms are further apart.
|-
|4.
|[[File:Q10plot8wxy0119.png|600px]]
| There is reaction between F and H2. By comparison between condition 4 and condition 3, a slight increase in the momentum of F and H which indicates a slight increase in kinetic translational energy result in the formation of products. Investigations have shown that even when rFH is slightly larger than 2Å, the products are still formed. Thus the change in reaction conditions illustrate that the changes in translational energy affect the reactivity the reaction more than the vibrational energy.
|-
|5.
|[[File:Q10plot5wxy0119.png|600px]]
| There is no reaction between H and HF even though the kinetic translational energy is very high and much higher than the activation energy. (pHF = -20) The trajectory reverts back before reaching the transition state structure.
|-
|6.
|[[File:Q10plot4wxy0119.png|600px]]
|
|}

pHH = -8.0 is still very large. This complies with the high activation energy of this reaction as the products formed are thermodynamically less stable than the reactants. It can be concluded that an increase in pHF, which corresponds to a high vibrational energy in H-F coordinate, is necessary for the reaction to take place. pHH, which corresponds to the translational energy in the H-H coordinate that is always very large, does not contribute too much to affect the reactivity of the reaction.

The transition state is present in the exit valley, hence, a late barrier for this endothermic reaction. The reactivity of endothermic reactions is thus dominated by the vibrational energy of the system from the results above.

For substantial exothermic reactions, such as F + H2, the transition state is located in the entrance valley, corresponds to an early barrier of the PES. For substantially endothermic reactions, such as H + HF, the transition state is in the exit valley, corresponds to a late barrier. The favoured degree of freedom for barrier crossing in exothermic reactions would be translation. This means that the momentum of the approaching atom and the atom which it is going to collide with has a greater impact on the rate of the reaction (eg. F atom in F + H2 reaction). However, reagent vibration, which is related to the momentum of the two bonded atoms (eg. HF in H + HF reaction) in the colliding molecule, would be most effective in enabling endothermic reactions to take place.

One reactive trajectory has the conditions of rFH = 0.91 Å, rHH = 2.0 Å, pHF = -0.5 and pHH = -10. The trajectory recrosses the transition state region twice and eventually moves into exit channel, forming products. By comparing this condition to the initial condition, the momentum pHH, which is the translational kinetic energy of the H atom, is reduced by a half; and the pHF, the vibrational energy of the H-F bond is increased to -0.5. This indicates that the vibrational energy has a greater contribution to the feasibility of the reaction than the translational energy. This is the opposite situation of the H2+F reaction.

The results agree with the Polanyi's empirical rules[2], which state that the vibrational energy is more efficient in promoting late-transition-state reactions and the translational energy is more efficient in promoting early-transition-state reactions.

When the momentum pFH is increased slightly from -0.5 to -0.8 with pHH being only 0.1, the reaction now becomes successful. It indicates that the F-H vibration energy has a larger contribution for the feasibility of this reaction compared to the translational energy which is the kinetic energy of the hydrogen atom which is defined by the momentum pHH

When the pHH has a large value above the activation energy (30.231 kcal/mol), which was set to be -20. The hydrogen atom collides with the HF molecule and breaks the H-F bond. However, two hydrogen atoms moves apart and do not form a H2 molecule. The procedure described is illustrated in the following animation snapshots. After the collision, the large momentum that the reactant hydrogen atom contains is still large.As a result, even if an H-H bond can be formed, it would immediately be broken due to an excess vibrational energy between two atoms.

for investigation, 1st set condition pFH = -0.5 pHH = -3 unreactive; 2nd pHH = -2.45 reactive; 3rd pHH = -1.1 unreactive; 4th pHH = -0.9 reactive; 5th pHH = 1.1 unreactive ; 6th pHH = 1.9 reactive ; 7th pHH = 2.4 to 3 unreactive
; pHH = 2.1 no reaction

The cases studied are an illustration of Polanyi's empirical rules.

For H and HF

failed conditionsː 091, 2, 0.05, -20

initial condition of H HF systemː
HF distance = 0.91 HH distance = 2 pHF = 0.8 pHH = -7.5

== References ==

<references>
<ref name="TS">Eyring, H. (1935). "The Activated Complex in Chemical Reactions". J. Chem. Phys. 3 (2): 107–115.</ref>
</references>

MRD:WXY0119

2018-05-18T16:22:23Z

Xw2816: /* Energy Distribution and Reactivity */

= Molecular Reaction Dynamics Report =

== EXERCISE 1ː H + H2 SYSTEM ==

[[File:HandH2wxy0119.png]]

In the H + H2 system, the distance between the first two H atoms (m1, m2) is r1 and the distance between the two H atoms in the H2 molecule (m2, m3) is r2.

=== Dynamics from the transition state region ===

{{fontcolor|red|Question 1 ː What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.}}

{| class="wikitable" border=1
|[[File:Q1TStransitionwxy0119.png|390px]]
|[[File:Q1TS2wxy0119.png|390px]]
|[[File:Q1TS3wxy0119.png|390px]]
|-
|Figure.1 Surface plot of reaction trajectory
|Figure.2 Transition state surface plot
|Figure.3 Transition state surface plot (different angle)
|}

In the reaction trajectory surface plot (Fig.1), AB is the distance r1 and BC is the distance r2. The gradient of the potential energy with regard to r1 and r2 are ∂V(r1)/∂r1 and ∂V(r2)/∂r2 respectively. The second derivatives of the potential energy with regard to r1 and r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 respectively.

At the two minimum structures which are at the two ends of the reaction pathway, ∂V(r1)/∂r1 = 0 and ∂V(r2)/∂r2 = 0 respectively. Since the two minimum structures are at two local minimum point , the second derivatives r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 are both greater than zero.

On the other hand, the components at the transition state structure (Fig.2) are different from that at minimum points. The gradient ∂V(r1)/∂r1 and ∂V(r2)/∂r2 are both equal to zero at the transition structure. However, r2 are ∂2V(r1)/∂r12 >0 and ∂2V(r2)/∂r22 <0. The transition state point is actually a saddle point which is observed more clearly in Fig.3. The Transition state point is thus distinguished from the minimum points as the potential energy surface curves inwards and downwards at the minimum points.

=== Locating the Transition State ===

{{fontcolor|red|Question 2ː 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.}}

{| class="wikitable" border=1
|[[File:TScontourwxy0119.png|390px]]
|[[File:TSsurfaceplwxy0119.png|390px]]
|[[File:TSestimatewxy0119.png|390px]]
|-
|Figure.4 Transition state contour plot
|Figure.5 Transition state surface plot
|Figure.6 Internuclear distance against time plot
|}

The best estimate of the transition state position is rts = 0.9078 Å.

When r1 = r2 = rts and momenta are set to zero, a graph of intermolecular distance against time is plotted. (Fig.6) In this graph, AB and BC lines are the same and overlap, the two lines on the graph are flat and horizontal, indicating that the atomic distances are the same at the position and with no momenta the state is at equilibrium, validating the position is the transition state position. The transition state is shown as a cross on the counter plot (Fig.4) and a dot on the surface plot (Fig.5).

=== Calculating and comparing the reaction path and trajectory ===

{{fontcolor1|red|Question 3ː Comment on how the ''mep'' and the trajectory you just calculated differ.}}

{| class="wikitable" border=1
|[[File:Mep1wxy0119.png|390px]]
|[[File:Mep2wxy0119.png|390px]]
|[[File:Mep3wxy0119.png|390px]]
|-
|Figure.7 mep calculation contour plot
|Figure.8 mep calculation surface plot
|Figure.9 mep calculation internuclear distance against time plot
|-
|[[File:Dynamic1wxy0119.png|390px]]
|[[File:Dynamic2wxy0119.png|390px]]
|[[File:Dynamic3wxy0119.png|390px]]
|-
|Figure.10 Dynamics calculation contour plot
|Figure.11 Dynamics calculation surface plot
|Figure.12 Dynamics calculation internuclear distance against time plot
|}
{| class="wikitable" border=1
|[[File:dynamics4wxy0119.png|600px]]
|[[File:Mep5wxy0119.png|600px]]
|-
|Figure.13 Dynamics calculation internuclear momenta against time plot
|Figure.14 mep calculation internuclear momenta against time plotsurface plot
|}

The mep (minimum energy path) trajectory is a smooth line on contour and surface plot (Fig.7 Fig.8) while the trajectory under dynamics calculation is oscillating along the pathway. (Fig.10 Fig.11) The mep trajectory takes more steps and longer time than the dynamics trajectory as shown in internuclear distance against time plots. (Fig. 9 Fig.12) The momentum is zero all the time under mep calculation (Fig.13) but under dynamics calculation, the momenta lines are oscillating against time.

The differences arise from mep and dynamics calculations can be explained by the the natures of the two calculations. Under mep calculation, the atoms are in extremely slow motion and the momentum is zero for each step, i.e. the atoms are "stop" after every step. The pathway under mep calculation is formed by connecting all the minimum energy points of each step and appears as a smooth line. On the other hand, atoms are in continuous motion and the momentum is accumulated after each step, atoms are then oscillating on the potential energy surface along the pathway. The steps are set to 5000 in the mep calculation but are only 500 in the dynamics calculatoin. Since the steps in mep are small, more time is needed to complete the same length on the trajectories.

=== Reactive and unreactive trajectories ===

{{fontcolor1|red|Question 4ː Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.}}

For the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, run trajectories with the following momenta p1 and p2 combinations:
{| class="wikitable" border=1
| Condition
| p1
| p2
| Total Energy/ kcal mol -1
| Reactivity
|-
|1.
| -1.25
| -2.5
| -99.018
| reactive
|-
|2.
| -1.5
| -2.0
| -100.456
| unreactive
|-
|3.
| -1.5
| -2.5
| -98.956
| reactive
|-
|4.
| -2.5
| -5.0
| -84.956
| unreactive
|-
|5.
| -2.5
| -5.2
| -83.416
| reactive
|}

{| class="wikitable" border=1
| Condition
| Surface Plot
| Contour Plot
| Description
|-
|1.
| [[File:condition1wxy0119.png|350px]]
| [[File:firstwxy0119.png|350px]]
| r2 (BC) decreases when HC approaches bonded HA and HB. The energy processed by the system is sufficient to overcome the activation barrier and crosses the transition state structure to break the HA and HB bond and form a new HB and HC bond. The new bond oscillate as r1 (AB) increases.
|-
|2.
| [[File:condition2wxy0119.png|350px]]
| [[File:secondwxy0119.png|350px]]
| HC approaches HB but the energy is insufficient to reach the transition state point, HC then moves further away from the bonded H2 molecule and no new bond is formed. The oscillation along r2 is a result of an increase in momentum p1.
|-
|3.
| [[File:condition3wxy0119.png|350px]]
| [[File:thirdwxy0119.png|350px]]
| Similarly to condition 1 but with a more negative p1, HA - HB bond oscillates more when r2 decreases. The reaction proceeds and crosses the transition state point to form the products.
|-
|4.
| [[File:condition4wxy0119.png|350px]]
| [[File:fourthwxy0119.png|350px]]
| Both p1 and p2 are more negative than the previous conditions. The HA - HB bond oscillates vigorously with larger amplitudes. The barrier recrossing happens after the system crosses the transition state region but eventually reverts back to the reactants.
|-
|5.
| [[File:condition5wxy0119.png|350px]]
| [[File:fifthwxy0119.png|350px]]
| With a slight more negative value of p1 as compared to that in condition 4, the system now processes appropriate energy to proceed to forming products. The vigorous oscillations are resulting from high momenta and barrier recrossing occurs.
|}

=== Transition State Theory ===

{{fontcolor|red|Question 5ː State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}

The main assumptions of Transition State Theory (TST) are<ref name="TS" />ː

1. The behaviours of the atomic nuclei follow the classical mechanics rules, i.e. the transition structure is formed as long as atoms or molecules collide with enough energy.

2. The intermediates have a long enough lifetime for the energies to be Boltzmann distributed preceding the next step.

3. The lowest energy saddle point on the potential energy surface is passed over in the reaction system.

The TST does not take into account of the quantum tunnelling effect which allows reactions to happen even when the energy carried by the system is lower than the activation energy barrier, especially for reactions with low energy barriers. Nor does the theory consider the intermediates with short lifetimes. When the energies are not fully distributed, the momentum of the reaction trajectory from the reactants to the intermediate can have effects on product selectivity. Moreover, the theory fails at high temperature because higher vibrational energy modes can be populated for a molecule at high temperature. The molecules can have complex motion and their collisions may result in a higher energy transition structure. The experimental results may thus deviate from that predicted by the TST. For example in condition 4, the kinetic energy of the reactants is much higher than the activation energy required. However, the momenta between atoms are very high and higher vibrational modes are populated that result in higher energy transition structures formed and the system does not pass though the lowest transition saddle point as shown in the diagram. Barrier recrossing occurs but the products are not formed as opposed to the predictions by the TST.

== EXERCISE 2: F - H - H SYSTEM ==

=== Reaction Energetics ===

{{fontcolor1|red|Question 6ː 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?}}

{| class="wikitable" border=1
|[[File:Q6FH2wxy0119.png|600px]]
|[[File:Q6HFHwxy0119.png|600px]]
|-
|Figure.15 F + H2 position on surface plot
|Figure.16 HF and H position on surface plot
|}

The surface plots of the potential energy surface of H + HF → F + H2 reaction is shown (Fig.15 Fig.16). Distance AB is the distance between H and F atoms and distance BC is the distance between two H atoms. The two minimum points shown in the graph are positions of F + H2 (Fig.15) and H + HF (Fig.16) respectively. If forward reaction is H + HF → F + H2, then the backward reaction is F + H2 → H + HF. As clearly shown in the diagram F + H2 are at a minimum point of higher potential energy than that of H + HF. Thus, by comparing the potential energy of reactants and products, the signs of changes in enthalpy (ΔH) and the energetics are determined. H + HF → F + H2 is endothermic and F + H2 → H + HF is exothermic.

The energetics of the two reactions are in accordance to the predictions by comparing bond strengths of chemical species. The bond energy of H-F bond (565 kJ/mol) is higher than the bond energy of H-H bond (432 kJ/mol). Thus, energy released by forming the H-H bond is not enough to compensate for energy required for breaking the H-F bond and the reaction H + HF → F + H2 is endothermic. On the contrary, the reaction F + H2 → H + HF is exothermic.

=== Transition State Approximation ===

{{fontcolor1|red|Question 7ː Locate the approximate position of the transition state.}}

{| class="wikitable" border=1
|[[File:Q7TStransitionwxy0119.png|390px]]
|[[File:Q7TS2wxy0119.png|390px]]
|[[File:Q7TS3wxy0119.png|390px]]
|-
|Figure.17 Transition state surface plot
|Figure.18 Transition state contour plot
|Figure.19 Internuclear distance against time plot (at TS point)
|}

The transition state (TS) is shown as a black dot on the potential energy surface plot (Fig.17) and a red cross on the contour plot. (Fig,18) The position of the TS is at the point where the distance between F and H (rFH) is 1.810Å and the distance between two H atoms (rHH) is 0.746Å. At the transition state position, the distance between F and H, between two H atoms are constant with no momentum as shown in the internuclear distance against time plot as horizontal and flat lines, validating the position is the transition state point (Fig.19)

=== Activation Energies ===

{{fontcolor1|red|Question 8ː Report the activation energy for both reactions.}}

{| class="wikitable" border=1
|[[File:Q8Ea2wxy0119.png|600px]]
|[[File:Q8Ea1wxy0119.png|600px]]
|-
|Figure.21 Energy against time plot (HF + H)
|Figure.20 Energy against time plot (F + H2)
|}

By performing MEP calculation with slight increase and decrease of the rFH to 1.820Å to perform F + H2 → H + HF reaction and to 1.800Å to perform H + HF → F + H2 reaction, the activation energies (Ea) of both reactions can be calculated from the differences in their potential energies (Fig.20 Fig.21)ː

F + H2 → H + HFː Ea = -103.751 - (-133.624) = +29.873 kcal mol-1

H + HF → F + H2ː Ea = -103.751 - (-103.972) = + 0.221 kcal mol-1

=== Reaction Dynamics ===

{{fontcolor1|red|Question 9ː In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

Reactionː F + H2 → H + HF

Initial condition setː rHF = 2Å rHH = 0.74Å pHF = -0.5 pHH = 0.1

{| class="wikitable" border=1
|[[File:Q9plot1wxy0119.png|390px]]
|[[File:Q9plot2wxy0119.png|390px]]
|[[File:Q9plot3wxy0119.png|390px]]
|-
|Figure.23 Contour plot
|Figure.24 Surface plot
|Figure.25 Internuclear momentum against time plot
|-
|[[File:Q9ani1wxy0119.png|390px]]
|[[File:Q9ani2wxy0119.png|390px]]
|-
|Figure.26 Animation figure (at the start of the reaction)
|Figure.27 Animation figure (at the end of the reaction)
|}

As shown in the plots and animation figures, the H-H bond in the reactants is broken and H-F bond in the products is formed. There are small momenta between both H and F and H-H bond that provide the initial kinetic energy of the reactants and the initial vibrational energy of the H-H bond in the reactants. The initial energy is required to break the H-H bond and reduce the distance of H and F for bond forming and potential energy is released when H-F bond is formed. Since energy is conserved, the reaction energy released is converted to the vibrational energy of the H-F bond and the translational kinetic energies of the products. This is shown in the internuclear momentum against time graph as the initial vibrations of B-C (H-H bond) becomes flat as two H atoms are separated in the product and the A-B (H-F bond) formed in the product vibrates with a large amplitude. (Fig.25) The contour plot and the surface plot also clearly show there is a significant increase in the amplitude of the oscillation of H-F bond which is an indication of the large vibrational energy of H-F bond. (Fig.23 Fig.24)

The predictions can be proved experimentally. The vibrational energy of H-F bond can be determined by measuring the frequency and the intensity of the absorption band of the H-F bond with infrared spectroscopy. The conversion to the kinetic energies of the products can be observed by measuring the temperature of the reaction as the kinetic energy can be converted to thermal energy as products collide with solvent molecules. The temperature is expected to increase as this is an exothermic reaction.

=== Energy Distribution and Reactivity ===

{{fontcolor|red|Question 10ː 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.}}
{| class="wikitable" border=1
| Condition
| rFH /Å
| rHH /Å
| pFH
| pHH
|-
|1.
| 2
| 0.74
| -0.5
| -3
|-
|2.
| 2
| 0.74
| -0.5
| 1.9
|-
|3.
| 2.1
| 0.74
| -0.5
| 0.1
|-
|4.
| 2
| 0.74
| -0.8
| 0.1
|-
|5.
| 0.91
| 2
| 0.05
| -20
|-
|6.
| 0.91
| 2
| 0.8
| -7.5
|}
{| class="wikitable" border=1
|Condition
|Contour plot
|Observation
|-
|1.
|[[File:Q10plot1wxy0119.png|600px]]
| The products are not formed in the reaction between F and H2 even pHH is very large. Barrier recrossing occurs and the H-H bond oscillates significantly with a large amplitude in the diagram.
|-
|2.
|[[File:Q10plot2wxy0119.png|600px]]
| There is reaction between F and H2 and products are formed. The F-H bond formed vibrates significantly along BC trajectory.
|-
|3.
|[[File:Q10plot9wxy0119.png|600px]]
| There is no reaction under this condition between F and H2. Although investigations have shown that there is reaction when rFH is equal or less than 2, but there is no reaction when H and F atoms are further apart.
|-
|4.
|[[File:Q10plot8wxy0119.png|600px]]
| There is reaction between F and H2. By comparison between condition 4 and condition 3, a slight increase in the momentum of F and H which indicates a slight increase in kinetic translational energy result in the formation of products. Investigations have shown that even when rFH is slightly larger than 2, the products are still formed. Thus the change in reaction conditions illustrate that the changes in translational energy affect the reactivity the reaction more than the vibrational energy.
|-
|5.
|[[File:Q10plot5wxy0119.png|600px]]
|
|-
|6.
|[[File:Q10plot4wxy0119.png|600px]]
|
|}

pHH = -8.0 is still very large. This complies with the high activation energy of this reaction as the products formed are thermodynamically less stable than the reactants. It can be concluded that an increase in pHF, which corresponds to a high vibrational energy in H-F coordinate, is necessary for the reaction to take place. pHH, which corresponds to the translational energy in the H-H coordinate that is always very large, does not contribute too much to affect the reactivity of the reaction.

The transition state is present in the exit valley, hence, a late barrier for this endothermic reaction. The reactivity of endothermic reactions is thus dominated by the vibrational energy of the system from the results above.

For substantial exothermic reactions, such as F + H2, the transition state is located in the entrance valley, corresponds to an early barrier of the PES. For substantially endothermic reactions, such as H + HF, the transition state is in the exit valley, corresponds to a late barrier. The favoured degree of freedom for barrier crossing in exothermic reactions would be translation. This means that the momentum of the approaching atom and the atom which it is going to collide with has a greater impact on the rate of the reaction (eg. F atom in F + H2 reaction). However, reagent vibration, which is related to the momentum of the two bonded atoms (eg. HF in H + HF reaction) in the colliding molecule, would be most effective in enabling endothermic reactions to take place.

One reactive trajectory has the conditions of rFH = 0.91 Å, rHH = 2.0 Å, pHF = -0.5 and pHH = -10. The trajectory recrosses the transition state region twice and eventually moves into exit channel, forming products. By comparing this condition to the initial condition, the momentum pHH, which is the translational kinetic energy of the H atom, is reduced by a half; and the pHF, the vibrational energy of the H-F bond is increased to -0.5. This indicates that the vibrational energy has a greater contribution to the feasibility of the reaction than the translational energy. This is the opposite situation of the H2+F reaction.

The results agree with the Polanyi's empirical rules[2], which state that the vibrational energy is more efficient in promoting late-transition-state reactions and the translational energy is more efficient in promoting early-transition-state reactions.

When the momentum pFH is increased slightly from -0.5 to -0.8 with pHH being only 0.1, the reaction now becomes successful. It indicates that the F-H vibration energy has a larger contribution for the feasibility of this reaction compared to the translational energy which is the kinetic energy of the hydrogen atom which is defined by the momentum pHH

When the pHH has a large value above the activation energy (30.231 kcal/mol), which was set to be -20. The hydrogen atom collides with the HF molecule and breaks the H-F bond. However, two hydrogen atoms moves apart and do not form a H2 molecule. The procedure described is illustrated in the following animation snapshots. After the collision, the large momentum that the reactant hydrogen atom contains is still large.As a result, even if an H-H bond can be formed, it would immediately be broken due to an excess vibrational energy between two atoms.

for investigation, 1st set condition pFH = -0.5 pHH = -3 unreactive; 2nd pHH = -2.45 reactive; 3rd pHH = -1.1 unreactive; 4th pHH = -0.9 reactive; 5th pHH = 1.1 unreactive ; 6th pHH = 1.9 reactive ; 7th pHH = 2.4 to 3 unreactive
; pHH = 2.1 no reaction

The cases studied are an illustration of Polanyi's empirical rules.

For H and HF

failed conditionsː 091, 2, 0.05, -20

initial condition of H HF systemː
HF distance = 0.91 HH distance = 2 pHF = 0.8 pHH = -7.5

== References ==

<references>
<ref name="TS">Eyring, H. (1935). "The Activated Complex in Chemical Reactions". J. Chem. Phys. 3 (2): 107–115.</ref>
</references>

MRD:WXY0119

2018-05-18T16:21:46Z

Xw2816: /* Energy Distribution and Reactivity */

= Molecular Reaction Dynamics Report =

== EXERCISE 1ː H + H2 SYSTEM ==

[[File:HandH2wxy0119.png]]

In the H + H2 system, the distance between the first two H atoms (m1, m2) is r1 and the distance between the two H atoms in the H2 molecule (m2, m3) is r2.

=== Dynamics from the transition state region ===

{{fontcolor|red|Question 1 ː What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.}}

{| class="wikitable" border=1
|[[File:Q1TStransitionwxy0119.png|390px]]
|[[File:Q1TS2wxy0119.png|390px]]
|[[File:Q1TS3wxy0119.png|390px]]
|-
|Figure.1 Surface plot of reaction trajectory
|Figure.2 Transition state surface plot
|Figure.3 Transition state surface plot (different angle)
|}

In the reaction trajectory surface plot (Fig.1), AB is the distance r1 and BC is the distance r2. The gradient of the potential energy with regard to r1 and r2 are ∂V(r1)/∂r1 and ∂V(r2)/∂r2 respectively. The second derivatives of the potential energy with regard to r1 and r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 respectively.

At the two minimum structures which are at the two ends of the reaction pathway, ∂V(r1)/∂r1 = 0 and ∂V(r2)/∂r2 = 0 respectively. Since the two minimum structures are at two local minimum point , the second derivatives r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 are both greater than zero.

On the other hand, the components at the transition state structure (Fig.2) are different from that at minimum points. The gradient ∂V(r1)/∂r1 and ∂V(r2)/∂r2 are both equal to zero at the transition structure. However, r2 are ∂2V(r1)/∂r12 >0 and ∂2V(r2)/∂r22 <0. The transition state point is actually a saddle point which is observed more clearly in Fig.3. The Transition state point is thus distinguished from the minimum points as the potential energy surface curves inwards and downwards at the minimum points.

=== Locating the Transition State ===

{{fontcolor|red|Question 2ː 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.}}

{| class="wikitable" border=1
|[[File:TScontourwxy0119.png|390px]]
|[[File:TSsurfaceplwxy0119.png|390px]]
|[[File:TSestimatewxy0119.png|390px]]
|-
|Figure.4 Transition state contour plot
|Figure.5 Transition state surface plot
|Figure.6 Internuclear distance against time plot
|}

The best estimate of the transition state position is rts = 0.9078 Å.

When r1 = r2 = rts and momenta are set to zero, a graph of intermolecular distance against time is plotted. (Fig.6) In this graph, AB and BC lines are the same and overlap, the two lines on the graph are flat and horizontal, indicating that the atomic distances are the same at the position and with no momenta the state is at equilibrium, validating the position is the transition state position. The transition state is shown as a cross on the counter plot (Fig.4) and a dot on the surface plot (Fig.5).

=== Calculating and comparing the reaction path and trajectory ===

{{fontcolor1|red|Question 3ː Comment on how the ''mep'' and the trajectory you just calculated differ.}}

{| class="wikitable" border=1
|[[File:Mep1wxy0119.png|390px]]
|[[File:Mep2wxy0119.png|390px]]
|[[File:Mep3wxy0119.png|390px]]
|-
|Figure.7 mep calculation contour plot
|Figure.8 mep calculation surface plot
|Figure.9 mep calculation internuclear distance against time plot
|-
|[[File:Dynamic1wxy0119.png|390px]]
|[[File:Dynamic2wxy0119.png|390px]]
|[[File:Dynamic3wxy0119.png|390px]]
|-
|Figure.10 Dynamics calculation contour plot
|Figure.11 Dynamics calculation surface plot
|Figure.12 Dynamics calculation internuclear distance against time plot
|}
{| class="wikitable" border=1
|[[File:dynamics4wxy0119.png|600px]]
|[[File:Mep5wxy0119.png|600px]]
|-
|Figure.13 Dynamics calculation internuclear momenta against time plot
|Figure.14 mep calculation internuclear momenta against time plotsurface plot
|}

The mep (minimum energy path) trajectory is a smooth line on contour and surface plot (Fig.7 Fig.8) while the trajectory under dynamics calculation is oscillating along the pathway. (Fig.10 Fig.11) The mep trajectory takes more steps and longer time than the dynamics trajectory as shown in internuclear distance against time plots. (Fig. 9 Fig.12) The momentum is zero all the time under mep calculation (Fig.13) but under dynamics calculation, the momenta lines are oscillating against time.

The differences arise from mep and dynamics calculations can be explained by the the natures of the two calculations. Under mep calculation, the atoms are in extremely slow motion and the momentum is zero for each step, i.e. the atoms are "stop" after every step. The pathway under mep calculation is formed by connecting all the minimum energy points of each step and appears as a smooth line. On the other hand, atoms are in continuous motion and the momentum is accumulated after each step, atoms are then oscillating on the potential energy surface along the pathway. The steps are set to 5000 in the mep calculation but are only 500 in the dynamics calculatoin. Since the steps in mep are small, more time is needed to complete the same length on the trajectories.

=== Reactive and unreactive trajectories ===

{{fontcolor1|red|Question 4ː Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.}}

For the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, run trajectories with the following momenta p1 and p2 combinations:
{| class="wikitable" border=1
| Condition
| p1
| p2
| Total Energy/ kcal mol -1
| Reactivity
|-
|1.
| -1.25
| -2.5
| -99.018
| reactive
|-
|2.
| -1.5
| -2.0
| -100.456
| unreactive
|-
|3.
| -1.5
| -2.5
| -98.956
| reactive
|-
|4.
| -2.5
| -5.0
| -84.956
| unreactive
|-
|5.
| -2.5
| -5.2
| -83.416
| reactive
|}

{| class="wikitable" border=1
| Condition
| Surface Plot
| Contour Plot
| Description
|-
|1.
| [[File:condition1wxy0119.png|350px]]
| [[File:firstwxy0119.png|350px]]
| r2 (BC) decreases when HC approaches bonded HA and HB. The energy processed by the system is sufficient to overcome the activation barrier and crosses the transition state structure to break the HA and HB bond and form a new HB and HC bond. The new bond oscillate as r1 (AB) increases.
|-
|2.
| [[File:condition2wxy0119.png|350px]]
| [[File:secondwxy0119.png|350px]]
| HC approaches HB but the energy is insufficient to reach the transition state point, HC then moves further away from the bonded H2 molecule and no new bond is formed. The oscillation along r2 is a result of an increase in momentum p1.
|-
|3.
| [[File:condition3wxy0119.png|350px]]
| [[File:thirdwxy0119.png|350px]]
| Similarly to condition 1 but with a more negative p1, HA - HB bond oscillates more when r2 decreases. The reaction proceeds and crosses the transition state point to form the products.
|-
|4.
| [[File:condition4wxy0119.png|350px]]
| [[File:fourthwxy0119.png|350px]]
| Both p1 and p2 are more negative than the previous conditions. The HA - HB bond oscillates vigorously with larger amplitudes. The barrier recrossing happens after the system crosses the transition state region but eventually reverts back to the reactants.
|-
|5.
| [[File:condition5wxy0119.png|350px]]
| [[File:fifthwxy0119.png|350px]]
| With a slight more negative value of p1 as compared to that in condition 4, the system now processes appropriate energy to proceed to forming products. The vigorous oscillations are resulting from high momenta and barrier recrossing occurs.
|}

=== Transition State Theory ===

{{fontcolor|red|Question 5ː State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}

The main assumptions of Transition State Theory (TST) are<ref name="TS" />ː

1. The behaviours of the atomic nuclei follow the classical mechanics rules, i.e. the transition structure is formed as long as atoms or molecules collide with enough energy.

2. The intermediates have a long enough lifetime for the energies to be Boltzmann distributed preceding the next step.

3. The lowest energy saddle point on the potential energy surface is passed over in the reaction system.

The TST does not take into account of the quantum tunnelling effect which allows reactions to happen even when the energy carried by the system is lower than the activation energy barrier, especially for reactions with low energy barriers. Nor does the theory consider the intermediates with short lifetimes. When the energies are not fully distributed, the momentum of the reaction trajectory from the reactants to the intermediate can have effects on product selectivity. Moreover, the theory fails at high temperature because higher vibrational energy modes can be populated for a molecule at high temperature. The molecules can have complex motion and their collisions may result in a higher energy transition structure. The experimental results may thus deviate from that predicted by the TST. For example in condition 4, the kinetic energy of the reactants is much higher than the activation energy required. However, the momenta between atoms are very high and higher vibrational modes are populated that result in higher energy transition structures formed and the system does not pass though the lowest transition saddle point as shown in the diagram. Barrier recrossing occurs but the products are not formed as opposed to the predictions by the TST.

== EXERCISE 2: F - H - H SYSTEM ==

=== Reaction Energetics ===

{{fontcolor1|red|Question 6ː 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?}}

{| class="wikitable" border=1
|[[File:Q6FH2wxy0119.png|600px]]
|[[File:Q6HFHwxy0119.png|600px]]
|-
|Figure.15 F + H2 position on surface plot
|Figure.16 HF and H position on surface plot
|}

The surface plots of the potential energy surface of H + HF → F + H2 reaction is shown (Fig.15 Fig.16). Distance AB is the distance between H and F atoms and distance BC is the distance between two H atoms. The two minimum points shown in the graph are positions of F + H2 (Fig.15) and H + HF (Fig.16) respectively. If forward reaction is H + HF → F + H2, then the backward reaction is F + H2 → H + HF. As clearly shown in the diagram F + H2 are at a minimum point of higher potential energy than that of H + HF. Thus, by comparing the potential energy of reactants and products, the signs of changes in enthalpy (ΔH) and the energetics are determined. H + HF → F + H2 is endothermic and F + H2 → H + HF is exothermic.

The energetics of the two reactions are in accordance to the predictions by comparing bond strengths of chemical species. The bond energy of H-F bond (565 kJ/mol) is higher than the bond energy of H-H bond (432 kJ/mol). Thus, energy released by forming the H-H bond is not enough to compensate for energy required for breaking the H-F bond and the reaction H + HF → F + H2 is endothermic. On the contrary, the reaction F + H2 → H + HF is exothermic.

=== Transition State Approximation ===

{{fontcolor1|red|Question 7ː Locate the approximate position of the transition state.}}

{| class="wikitable" border=1
|[[File:Q7TStransitionwxy0119.png|390px]]
|[[File:Q7TS2wxy0119.png|390px]]
|[[File:Q7TS3wxy0119.png|390px]]
|-
|Figure.17 Transition state surface plot
|Figure.18 Transition state contour plot
|Figure.19 Internuclear distance against time plot (at TS point)
|}

The transition state (TS) is shown as a black dot on the potential energy surface plot (Fig.17) and a red cross on the contour plot. (Fig,18) The position of the TS is at the point where the distance between F and H (rFH) is 1.810Å and the distance between two H atoms (rHH) is 0.746Å. At the transition state position, the distance between F and H, between two H atoms are constant with no momentum as shown in the internuclear distance against time plot as horizontal and flat lines, validating the position is the transition state point (Fig.19)

=== Activation Energies ===

{{fontcolor1|red|Question 8ː Report the activation energy for both reactions.}}

{| class="wikitable" border=1
|[[File:Q8Ea2wxy0119.png|600px]]
|[[File:Q8Ea1wxy0119.png|600px]]
|-
|Figure.21 Energy against time plot (HF + H)
|Figure.20 Energy against time plot (F + H2)
|}

By performing MEP calculation with slight increase and decrease of the rFH to 1.820Å to perform F + H2 → H + HF reaction and to 1.800Å to perform H + HF → F + H2 reaction, the activation energies (Ea) of both reactions can be calculated from the differences in their potential energies (Fig.20 Fig.21)ː

F + H2 → H + HFː Ea = -103.751 - (-133.624) = +29.873 kcal mol-1

H + HF → F + H2ː Ea = -103.751 - (-103.972) = + 0.221 kcal mol-1

=== Reaction Dynamics ===

{{fontcolor1|red|Question 9ː In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

Reactionː F + H2 → H + HF

Initial condition setː rHF = 2Å rHH = 0.74Å pHF = -0.5 pHH = 0.1

{| class="wikitable" border=1
|[[File:Q9plot1wxy0119.png|390px]]
|[[File:Q9plot2wxy0119.png|390px]]
|[[File:Q9plot3wxy0119.png|390px]]
|-
|Figure.23 Contour plot
|Figure.24 Surface plot
|Figure.25 Internuclear momentum against time plot
|-
|[[File:Q9ani1wxy0119.png|390px]]
|[[File:Q9ani2wxy0119.png|390px]]
|-
|Figure.26 Animation figure (at the start of the reaction)
|Figure.27 Animation figure (at the end of the reaction)
|}

As shown in the plots and animation figures, the H-H bond in the reactants is broken and H-F bond in the products is formed. There are small momenta between both H and F and H-H bond that provide the initial kinetic energy of the reactants and the initial vibrational energy of the H-H bond in the reactants. The initial energy is required to break the H-H bond and reduce the distance of H and F for bond forming and potential energy is released when H-F bond is formed. Since energy is conserved, the reaction energy released is converted to the vibrational energy of the H-F bond and the translational kinetic energies of the products. This is shown in the internuclear momentum against time graph as the initial vibrations of B-C (H-H bond) becomes flat as two H atoms are separated in the product and the A-B (H-F bond) formed in the product vibrates with a large amplitude. (Fig.25) The contour plot and the surface plot also clearly show there is a significant increase in the amplitude of the oscillation of H-F bond which is an indication of the large vibrational energy of H-F bond. (Fig.23 Fig.24)

The predictions can be proved experimentally. The vibrational energy of H-F bond can be determined by measuring the frequency and the intensity of the absorption band of the H-F bond with infrared spectroscopy. The conversion to the kinetic energies of the products can be observed by measuring the temperature of the reaction as the kinetic energy can be converted to thermal energy as products collide with solvent molecules. The temperature is expected to increase as this is an exothermic reaction.

=== Energy Distribution and Reactivity ===

{{fontcolor|red|Question 10ː 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.}}
{| class="wikitable" border=1
| Condition
| rFH /Å
| rHH /Å
| pFH
| pHH
|-
|1.
| 2
| 0.74
| -0.5
| -3
|-
|2.
| 2
| 0.74
| -0.5
| 1.9
|-
|3.
| 2.1
| 0.74
| -0.5
| 0.1
|-
|4.
| 2
| 0.74
| -0.8
| 0.1
|-
|5.
| 0.91
| 2
| 0.05
| -20
|-
|6.
| 0.91
| 2
| 0.8
| -7.5
|}
{| class="wikitable" border=1
|Condition
|Contour plot
|Observation
|-
|1.
|[[File:Q10plot1wxy0119.png|600px]]
| The products are not formed in the reaction between F and H2 even pHH is very large. Barrier recrossing occurs and the H-H bond oscillates significantly with a large amplitude in the diagram.
|-
|2.
|[[File:Q10plot2wxy0119.png|600px]]
| There is reaction and products are formed in the reaction between F and H2. The F-H bond formed vibrates significantly along BC trajectory.
|-
|3.
|[[File:Q10plot9wxy0119.png|600px]]
| There is no reaction under this condition between F and H2. Although investigations have shown that there is reaction when rFH is equal or less than 2, but there is no reaction when H and F atoms are further apart.
|-
|4.
|[[File:Q10plot8wxy0119.png|600px]]
| There is reaction between F and H2. By comparison between condition 4 and condition 3, a slight increase in the momentum of F and H which indicates a slight increase in kinetic translational energy result in the formation of products. Investigations have shown that even when rFH is slightly larger than 2, the products are still formed. Thus the change in reaction conditions illustrate that the changes in translational energy affect the reactivity the reaction more than the vibrational energy.
|-
|5.
|[[File:Q10plot5wxy0119.png|600px]]
|
|-
|6.
|[[File:Q10plot4wxy0119.png|600px]]
|
|}

pHH = -8.0 is still very large. This complies with the high activation energy of this reaction as the products formed are thermodynamically less stable than the reactants. It can be concluded that an increase in pHF, which corresponds to a high vibrational energy in H-F coordinate, is necessary for the reaction to take place. pHH, which corresponds to the translational energy in the H-H coordinate that is always very large, does not contribute too much to affect the reactivity of the reaction.

The transition state is present in the exit valley, hence, a late barrier for this endothermic reaction. The reactivity of endothermic reactions is thus dominated by the vibrational energy of the system from the results above.

For substantial exothermic reactions, such as F + H2, the transition state is located in the entrance valley, corresponds to an early barrier of the PES. For substantially endothermic reactions, such as H + HF, the transition state is in the exit valley, corresponds to a late barrier. The favoured degree of freedom for barrier crossing in exothermic reactions would be translation. This means that the momentum of the approaching atom and the atom which it is going to collide with has a greater impact on the rate of the reaction (eg. F atom in F + H2 reaction). However, reagent vibration, which is related to the momentum of the two bonded atoms (eg. HF in H + HF reaction) in the colliding molecule, would be most effective in enabling endothermic reactions to take place.

One reactive trajectory has the conditions of rFH = 0.91 Å, rHH = 2.0 Å, pHF = -0.5 and pHH = -10. The trajectory recrosses the transition state region twice and eventually moves into exit channel, forming products. By comparing this condition to the initial condition, the momentum pHH, which is the translational kinetic energy of the H atom, is reduced by a half; and the pHF, the vibrational energy of the H-F bond is increased to -0.5. This indicates that the vibrational energy has a greater contribution to the feasibility of the reaction than the translational energy. This is the opposite situation of the H2+F reaction.

The results agree with the Polanyi's empirical rules[2], which state that the vibrational energy is more efficient in promoting late-transition-state reactions and the translational energy is more efficient in promoting early-transition-state reactions.

When the momentum pFH is increased slightly from -0.5 to -0.8 with pHH being only 0.1, the reaction now becomes successful. It indicates that the F-H vibration energy has a larger contribution for the feasibility of this reaction compared to the translational energy which is the kinetic energy of the hydrogen atom which is defined by the momentum pHH

When the pHH has a large value above the activation energy (30.231 kcal/mol), which was set to be -20. The hydrogen atom collides with the HF molecule and breaks the H-F bond. However, two hydrogen atoms moves apart and do not form a H2 molecule. The procedure described is illustrated in the following animation snapshots. After the collision, the large momentum that the reactant hydrogen atom contains is still large.As a result, even if an H-H bond can be formed, it would immediately be broken due to an excess vibrational energy between two atoms.

for investigation, 1st set condition pFH = -0.5 pHH = -3 unreactive; 2nd pHH = -2.45 reactive; 3rd pHH = -1.1 unreactive; 4th pHH = -0.9 reactive; 5th pHH = 1.1 unreactive ; 6th pHH = 1.9 reactive ; 7th pHH = 2.4 to 3 unreactive
; pHH = 2.1 no reaction

The cases studied are an illustration of Polanyi's empirical rules.

For H and HF

failed conditionsː 091, 2, 0.05, -20

initial condition of H HF systemː
HF distance = 0.91 HH distance = 2 pHF = 0.8 pHH = -7.5

== References ==

<references>
<ref name="TS">Eyring, H. (1935). "The Activated Complex in Chemical Reactions". J. Chem. Phys. 3 (2): 107–115.</ref>
</references>

File:Q10plot9wxy0119.png

2018-05-18T16:10:51Z

Xw2816:

MRD:WXY0119

2018-05-18T16:10:10Z

Xw2816: /* Energy Distribution and Reactivity */

= Molecular Reaction Dynamics Report =

== EXERCISE 1ː H + H2 SYSTEM ==

[[File:HandH2wxy0119.png]]

In the H + H2 system, the distance between the first two H atoms (m1, m2) is r1 and the distance between the two H atoms in the H2 molecule (m2, m3) is r2.

=== Dynamics from the transition state region ===

{{fontcolor|red|Question 1 ː What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.}}

{| class="wikitable" border=1
|[[File:Q1TStransitionwxy0119.png|390px]]
|[[File:Q1TS2wxy0119.png|390px]]
|[[File:Q1TS3wxy0119.png|390px]]
|-
|Figure.1 Surface plot of reaction trajectory
|Figure.2 Transition state surface plot
|Figure.3 Transition state surface plot (different angle)
|}

In the reaction trajectory surface plot (Fig.1), AB is the distance r1 and BC is the distance r2. The gradient of the potential energy with regard to r1 and r2 are ∂V(r1)/∂r1 and ∂V(r2)/∂r2 respectively. The second derivatives of the potential energy with regard to r1 and r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 respectively.

At the two minimum structures which are at the two ends of the reaction pathway, ∂V(r1)/∂r1 = 0 and ∂V(r2)/∂r2 = 0 respectively. Since the two minimum structures are at two local minimum point , the second derivatives r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 are both greater than zero.

On the other hand, the components at the transition state structure (Fig.2) are different from that at minimum points. The gradient ∂V(r1)/∂r1 and ∂V(r2)/∂r2 are both equal to zero at the transition structure. However, r2 are ∂2V(r1)/∂r12 >0 and ∂2V(r2)/∂r22 <0. The transition state point is actually a saddle point which is observed more clearly in Fig.3. The Transition state point is thus distinguished from the minimum points as the potential energy surface curves inwards and downwards at the minimum points.

=== Locating the Transition State ===

{{fontcolor|red|Question 2ː 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.}}

{| class="wikitable" border=1
|[[File:TScontourwxy0119.png|390px]]
|[[File:TSsurfaceplwxy0119.png|390px]]
|[[File:TSestimatewxy0119.png|390px]]
|-
|Figure.4 Transition state contour plot
|Figure.5 Transition state surface plot
|Figure.6 Internuclear distance against time plot
|}

The best estimate of the transition state position is rts = 0.9078 Å.

When r1 = r2 = rts and momenta are set to zero, a graph of intermolecular distance against time is plotted. (Fig.6) In this graph, AB and BC lines are the same and overlap, the two lines on the graph are flat and horizontal, indicating that the atomic distances are the same at the position and with no momenta the state is at equilibrium, validating the position is the transition state position. The transition state is shown as a cross on the counter plot (Fig.4) and a dot on the surface plot (Fig.5).

=== Calculating and comparing the reaction path and trajectory ===

{{fontcolor1|red|Question 3ː Comment on how the ''mep'' and the trajectory you just calculated differ.}}

{| class="wikitable" border=1
|[[File:Mep1wxy0119.png|390px]]
|[[File:Mep2wxy0119.png|390px]]
|[[File:Mep3wxy0119.png|390px]]
|-
|Figure.7 mep calculation contour plot
|Figure.8 mep calculation surface plot
|Figure.9 mep calculation internuclear distance against time plot
|-
|[[File:Dynamic1wxy0119.png|390px]]
|[[File:Dynamic2wxy0119.png|390px]]
|[[File:Dynamic3wxy0119.png|390px]]
|-
|Figure.10 Dynamics calculation contour plot
|Figure.11 Dynamics calculation surface plot
|Figure.12 Dynamics calculation internuclear distance against time plot
|}
{| class="wikitable" border=1
|[[File:dynamics4wxy0119.png|600px]]
|[[File:Mep5wxy0119.png|600px]]
|-
|Figure.13 Dynamics calculation internuclear momenta against time plot
|Figure.14 mep calculation internuclear momenta against time plotsurface plot
|}

The mep (minimum energy path) trajectory is a smooth line on contour and surface plot (Fig.7 Fig.8) while the trajectory under dynamics calculation is oscillating along the pathway. (Fig.10 Fig.11) The mep trajectory takes more steps and longer time than the dynamics trajectory as shown in internuclear distance against time plots. (Fig. 9 Fig.12) The momentum is zero all the time under mep calculation (Fig.13) but under dynamics calculation, the momenta lines are oscillating against time.

The differences arise from mep and dynamics calculations can be explained by the the natures of the two calculations. Under mep calculation, the atoms are in extremely slow motion and the momentum is zero for each step, i.e. the atoms are "stop" after every step. The pathway under mep calculation is formed by connecting all the minimum energy points of each step and appears as a smooth line. On the other hand, atoms are in continuous motion and the momentum is accumulated after each step, atoms are then oscillating on the potential energy surface along the pathway. The steps are set to 5000 in the mep calculation but are only 500 in the dynamics calculatoin. Since the steps in mep are small, more time is needed to complete the same length on the trajectories.

=== Reactive and unreactive trajectories ===

{{fontcolor1|red|Question 4ː Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.}}

For the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, run trajectories with the following momenta p1 and p2 combinations:
{| class="wikitable" border=1
| Condition
| p1
| p2
| Total Energy/ kcal mol -1
| Reactivity
|-
|1.
| -1.25
| -2.5
| -99.018
| reactive
|-
|2.
| -1.5
| -2.0
| -100.456
| unreactive
|-
|3.
| -1.5
| -2.5
| -98.956
| reactive
|-
|4.
| -2.5
| -5.0
| -84.956
| unreactive
|-
|5.
| -2.5
| -5.2
| -83.416
| reactive
|}

{| class="wikitable" border=1
| Condition
| Surface Plot
| Contour Plot
| Description
|-
|1.
| [[File:condition1wxy0119.png|350px]]
| [[File:firstwxy0119.png|350px]]
| r2 (BC) decreases when HC approaches bonded HA and HB. The energy processed by the system is sufficient to overcome the activation barrier and crosses the transition state structure to break the HA and HB bond and form a new HB and HC bond. The new bond oscillate as r1 (AB) increases.
|-
|2.
| [[File:condition2wxy0119.png|350px]]
| [[File:secondwxy0119.png|350px]]
| HC approaches HB but the energy is insufficient to reach the transition state point, HC then moves further away from the bonded H2 molecule and no new bond is formed. The oscillation along r2 is a result of an increase in momentum p1.
|-
|3.
| [[File:condition3wxy0119.png|350px]]
| [[File:thirdwxy0119.png|350px]]
| Similarly to condition 1 but with a more negative p1, HA - HB bond oscillates more when r2 decreases. The reaction proceeds and crosses the transition state point to form the products.
|-
|4.
| [[File:condition4wxy0119.png|350px]]
| [[File:fourthwxy0119.png|350px]]
| Both p1 and p2 are more negative than the previous conditions. The HA - HB bond oscillates vigorously with larger amplitudes. The barrier recrossing happens after the system crosses the transition state region but eventually reverts back to the reactants.
|-
|5.
| [[File:condition5wxy0119.png|350px]]
| [[File:fifthwxy0119.png|350px]]
| With a slight more negative value of p1 as compared to that in condition 4, the system now processes appropriate energy to proceed to forming products. The vigorous oscillations are resulting from high momenta and barrier recrossing occurs.
|}

=== Transition State Theory ===

{{fontcolor|red|Question 5ː State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}

The main assumptions of Transition State Theory (TST) are<ref name="TS" />ː

1. The behaviours of the atomic nuclei follow the classical mechanics rules, i.e. the transition structure is formed as long as atoms or molecules collide with enough energy.

2. The intermediates have a long enough lifetime for the energies to be Boltzmann distributed preceding the next step.

3. The lowest energy saddle point on the potential energy surface is passed over in the reaction system.

The TST does not take into account of the quantum tunnelling effect which allows reactions to happen even when the energy carried by the system is lower than the activation energy barrier, especially for reactions with low energy barriers. Nor does the theory consider the intermediates with short lifetimes. When the energies are not fully distributed, the momentum of the reaction trajectory from the reactants to the intermediate can have effects on product selectivity. Moreover, the theory fails at high temperature because higher vibrational energy modes can be populated for a molecule at high temperature. The molecules can have complex motion and their collisions may result in a higher energy transition structure. The experimental results may thus deviate from that predicted by the TST. For example in condition 4, the kinetic energy of the reactants is much higher than the activation energy required. However, the momenta between atoms are very high and higher vibrational modes are populated that result in higher energy transition structures formed and the system does not pass though the lowest transition saddle point as shown in the diagram. Barrier recrossing occurs but the products are not formed as opposed to the predictions by the TST.

== EXERCISE 2: F - H - H SYSTEM ==

=== Reaction Energetics ===

{{fontcolor1|red|Question 6ː 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?}}

{| class="wikitable" border=1
|[[File:Q6FH2wxy0119.png|600px]]
|[[File:Q6HFHwxy0119.png|600px]]
|-
|Figure.15 F + H2 position on surface plot
|Figure.16 HF and H position on surface plot
|}

The surface plots of the potential energy surface of H + HF → F + H2 reaction is shown (Fig.15 Fig.16). Distance AB is the distance between H and F atoms and distance BC is the distance between two H atoms. The two minimum points shown in the graph are positions of F + H2 (Fig.15) and H + HF (Fig.16) respectively. If forward reaction is H + HF → F + H2, then the backward reaction is F + H2 → H + HF. As clearly shown in the diagram F + H2 are at a minimum point of higher potential energy than that of H + HF. Thus, by comparing the potential energy of reactants and products, the signs of changes in enthalpy (ΔH) and the energetics are determined. H + HF → F + H2 is endothermic and F + H2 → H + HF is exothermic.

The energetics of the two reactions are in accordance to the predictions by comparing bond strengths of chemical species. The bond energy of H-F bond (565 kJ/mol) is higher than the bond energy of H-H bond (432 kJ/mol). Thus, energy released by forming the H-H bond is not enough to compensate for energy required for breaking the H-F bond and the reaction H + HF → F + H2 is endothermic. On the contrary, the reaction F + H2 → H + HF is exothermic.

=== Transition State Approximation ===

{{fontcolor1|red|Question 7ː Locate the approximate position of the transition state.}}

{| class="wikitable" border=1
|[[File:Q7TStransitionwxy0119.png|390px]]
|[[File:Q7TS2wxy0119.png|390px]]
|[[File:Q7TS3wxy0119.png|390px]]
|-
|Figure.17 Transition state surface plot
|Figure.18 Transition state contour plot
|Figure.19 Internuclear distance against time plot (at TS point)
|}

The transition state (TS) is shown as a black dot on the potential energy surface plot (Fig.17) and a red cross on the contour plot. (Fig,18) The position of the TS is at the point where the distance between F and H (rFH) is 1.810Å and the distance between two H atoms (rHH) is 0.746Å. At the transition state position, the distance between F and H, between two H atoms are constant with no momentum as shown in the internuclear distance against time plot as horizontal and flat lines, validating the position is the transition state point (Fig.19)

=== Activation Energies ===

{{fontcolor1|red|Question 8ː Report the activation energy for both reactions.}}

{| class="wikitable" border=1
|[[File:Q8Ea2wxy0119.png|600px]]
|[[File:Q8Ea1wxy0119.png|600px]]
|-
|Figure.21 Energy against time plot (HF + H)
|Figure.20 Energy against time plot (F + H2)
|}

By performing MEP calculation with slight increase and decrease of the rFH to 1.820Å to perform F + H2 → H + HF reaction and to 1.800Å to perform H + HF → F + H2 reaction, the activation energies (Ea) of both reactions can be calculated from the differences in their potential energies (Fig.20 Fig.21)ː

F + H2 → H + HFː Ea = -103.751 - (-133.624) = +29.873 kcal mol-1

H + HF → F + H2ː Ea = -103.751 - (-103.972) = + 0.221 kcal mol-1

=== Reaction Dynamics ===

{{fontcolor1|red|Question 9ː In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

Reactionː F + H2 → H + HF

Initial condition setː rHF = 2Å rHH = 0.74Å pHF = -0.5 pHH = 0.1

{| class="wikitable" border=1
|[[File:Q9plot1wxy0119.png|390px]]
|[[File:Q9plot2wxy0119.png|390px]]
|[[File:Q9plot3wxy0119.png|390px]]
|-
|Figure.23 Contour plot
|Figure.24 Surface plot
|Figure.25 Internuclear momentum against time plot
|-
|[[File:Q9ani1wxy0119.png|390px]]
|[[File:Q9ani2wxy0119.png|390px]]
|-
|Figure.26 Animation figure (at the start of the reaction)
|Figure.27 Animation figure (at the end of the reaction)
|}

As shown in the plots and animation figures, the H-H bond in the reactants is broken and H-F bond in the products is formed. There are small momenta between both H and F and H-H bond that provide the initial kinetic energy of the reactants and the initial vibrational energy of the H-H bond in the reactants. The initial energy is required to break the H-H bond and reduce the distance of H and F for bond forming and potential energy is released when H-F bond is formed. Since energy is conserved, the reaction energy released is converted to the vibrational energy of the H-F bond and the translational kinetic energies of the products. This is shown in the internuclear momentum against time graph as the initial vibrations of B-C (H-H bond) becomes flat as two H atoms are separated in the product and the A-B (H-F bond) formed in the product vibrates with a large amplitude. (Fig.25) The contour plot and the surface plot also clearly show there is a significant increase in the amplitude of the oscillation of H-F bond which is an indication of the large vibrational energy of H-F bond. (Fig.23 Fig.24)

The predictions can be proved experimentally. The vibrational energy of H-F bond can be determined by measuring the frequency and the intensity of the absorption band of the H-F bond with infrared spectroscopy. The conversion to the kinetic energies of the products can be observed by measuring the temperature of the reaction as the kinetic energy can be converted to thermal energy as products collide with solvent molecules. The temperature is expected to increase as this is an exothermic reaction.

=== Energy Distribution and Reactivity ===

{{fontcolor|red|Question 10ː 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.}}
{| class="wikitable" border=1
| Condition
| rFH /Å
| rHH /Å
| pFH
| pHH
|-
|1.
| 2
| 0.74
| -0.5
| -3
|-
|2.
| 2
| 0.74
| -0.5
| 1.9
|-
|3.
| 2.1
| 0.74
| -0.5
| 0.1
|-
|4.
| 2
| 0.74
| -0.8
| 0.1
|-
|5.
| 0.91
| 2
| 0.05
| -20
|-
|6.
| 0.91
| 2
| 0.8
| -7.5
|}
{| class="wikitable" border=1
|Condition
|Contour plot
|Observation
|-
|1.
|[[File:Q10plot1wxy0119.png|600px]]
|
|-
|2.
|[[File:Q10plot2wxy0119.png|600px]]
|
|-
|3.
|[[File:Q10plot9wxy0119.png|600px]]
|
|-
|4.
|[[File:Q10plot8wxy0119.png|600px]]
|
|-
|5.
|[[File:Q10plot5wxy0119.png|600px]]
|
|-
|6.
|[[File:Q10plot4wxy0119.png|600px]]
|
|}

pHH = -8.0 is still very large. This complies with the high activation energy of this reaction as the products formed are thermodynamically less stable than the reactants. It can be concluded that an increase in pHF, which corresponds to a high vibrational energy in H-F coordinate, is necessary for the reaction to take place. pHH, which corresponds to the translational energy in the H-H coordinate that is always very large, does not contribute too much to affect the reactivity of the reaction.

The transition state is present in the exit valley, hence, a late barrier for this endothermic reaction. The reactivity of endothermic reactions is thus dominated by the vibrational energy of the system from the results above.

For substantial exothermic reactions, such as F + H2, the transition state is located in the entrance valley, corresponds to an early barrier of the PES. For substantially endothermic reactions, such as H + HF, the transition state is in the exit valley, corresponds to a late barrier. The favoured degree of freedom for barrier crossing in exothermic reactions would be translation. This means that the momentum of the approaching atom and the atom which it is going to collide with has a greater impact on the rate of the reaction (eg. F atom in F + H2 reaction). However, reagent vibration, which is related to the momentum of the two bonded atoms (eg. HF in H + HF reaction) in the colliding molecule, would be most effective in enabling endothermic reactions to take place.

One reactive trajectory has the conditions of rFH = 0.91 Å, rHH = 2.0 Å, pHF = -0.5 and pHH = -10. The trajectory recrosses the transition state region twice and eventually moves into exit channel, forming products. By comparing this condition to the initial condition, the momentum pHH, which is the translational kinetic energy of the H atom, is reduced by a half; and the pHF, the vibrational energy of the H-F bond is increased to -0.5. This indicates that the vibrational energy has a greater contribution to the feasibility of the reaction than the translational energy. This is the opposite situation of the H2+F reaction.

The results agree with the Polanyi's empirical rules[2], which state that the vibrational energy is more efficient in promoting late-transition-state reactions and the translational energy is more efficient in promoting early-transition-state reactions.

When the momentum pFH is increased slightly from -0.5 to -0.8 with pHH being only 0.1, the reaction now becomes successful. It indicates that the F-H vibration energy has a larger contribution for the feasibility of this reaction compared to the translational energy which is the kinetic energy of the hydrogen atom which is defined by the momentum pHH

When the pHH has a large value above the activation energy (30.231 kcal/mol), which was set to be -20. The hydrogen atom collides with the HF molecule and breaks the H-F bond. However, two hydrogen atoms moves apart and do not form a H2 molecule. The procedure described is illustrated in the following animation snapshots. After the collision, the large momentum that the reactant hydrogen atom contains is still large.As a result, even if an H-H bond can be formed, it would immediately be broken due to an excess vibrational energy between two atoms.

for investigation, 1st set condition pFH = -0.5 pHH = -3 unreactive; 2nd pHH = -2.45 reactive; 3rd pHH = -1.1 unreactive; 4th pHH = -0.9 reactive; 5th pHH = 1.1 unreactive ; 6th pHH = 1.9 reactive ; 7th pHH = 2.4 to 3 unreactive
; pHH = 2.1 no reaction

The cases studied are an illustration of Polanyi's empirical rules.

For H and HF

failed conditionsː 091, 2, 0.05, -20

initial condition of H HF systemː
HF distance = 0.91 HH distance = 2 pHF = 0.8 pHH = -7.5

== References ==

<references>
<ref name="TS">Eyring, H. (1935). "The Activated Complex in Chemical Reactions". J. Chem. Phys. 3 (2): 107–115.</ref>
</references>

MRD:WXY0119

2018-05-18T16:03:14Z

Xw2816: /* EXERCISE 1ː H + H2 SYSTEM */

= Molecular Reaction Dynamics Report =

== EXERCISE 1ː H + H2 SYSTEM ==

[[File:HandH2wxy0119.png]]

In the H + H2 system, the distance between the first two H atoms (m1, m2) is r1 and the distance between the two H atoms in the H2 molecule (m2, m3) is r2.

=== Dynamics from the transition state region ===

{{fontcolor|red|Question 1 ː What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.}}

{| class="wikitable" border=1
|[[File:Q1TStransitionwxy0119.png|390px]]
|[[File:Q1TS2wxy0119.png|390px]]
|[[File:Q1TS3wxy0119.png|390px]]
|-
|Figure.1 Surface plot of reaction trajectory
|Figure.2 Transition state surface plot
|Figure.3 Transition state surface plot (different angle)
|}

In the reaction trajectory surface plot (Fig.1), AB is the distance r1 and BC is the distance r2. The gradient of the potential energy with regard to r1 and r2 are ∂V(r1)/∂r1 and ∂V(r2)/∂r2 respectively. The second derivatives of the potential energy with regard to r1 and r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 respectively.

At the two minimum structures which are at the two ends of the reaction pathway, ∂V(r1)/∂r1 = 0 and ∂V(r2)/∂r2 = 0 respectively. Since the two minimum structures are at two local minimum point , the second derivatives r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 are both greater than zero.

On the other hand, the components at the transition state structure (Fig.2) are different from that at minimum points. The gradient ∂V(r1)/∂r1 and ∂V(r2)/∂r2 are both equal to zero at the transition structure. However, r2 are ∂2V(r1)/∂r12 >0 and ∂2V(r2)/∂r22 <0. The transition state point is actually a saddle point which is observed more clearly in Fig.3. The Transition state point is thus distinguished from the minimum points as the potential energy surface curves inwards and downwards at the minimum points.

=== Locating the Transition State ===

{{fontcolor|red|Question 2ː 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.}}

{| class="wikitable" border=1
|[[File:TScontourwxy0119.png|390px]]
|[[File:TSsurfaceplwxy0119.png|390px]]
|[[File:TSestimatewxy0119.png|390px]]
|-
|Figure.4 Transition state contour plot
|Figure.5 Transition state surface plot
|Figure.6 Internuclear distance against time plot
|}

The best estimate of the transition state position is rts = 0.9078 Å.

When r1 = r2 = rts and momenta are set to zero, a graph of intermolecular distance against time is plotted. (Fig.6) In this graph, AB and BC lines are the same and overlap, the two lines on the graph are flat and horizontal, indicating that the atomic distances are the same at the position and with no momenta the state is at equilibrium, validating the position is the transition state position. The transition state is shown as a cross on the counter plot (Fig.4) and a dot on the surface plot (Fig.5).

=== Calculating and comparing the reaction path and trajectory ===

{{fontcolor1|red|Question 3ː Comment on how the ''mep'' and the trajectory you just calculated differ.}}

{| class="wikitable" border=1
|[[File:Mep1wxy0119.png|390px]]
|[[File:Mep2wxy0119.png|390px]]
|[[File:Mep3wxy0119.png|390px]]
|-
|Figure.7 mep calculation contour plot
|Figure.8 mep calculation surface plot
|Figure.9 mep calculation internuclear distance against time plot
|-
|[[File:Dynamic1wxy0119.png|390px]]
|[[File:Dynamic2wxy0119.png|390px]]
|[[File:Dynamic3wxy0119.png|390px]]
|-
|Figure.10 Dynamics calculation contour plot
|Figure.11 Dynamics calculation surface plot
|Figure.12 Dynamics calculation internuclear distance against time plot
|}
{| class="wikitable" border=1
|[[File:dynamics4wxy0119.png|600px]]
|[[File:Mep5wxy0119.png|600px]]
|-
|Figure.13 Dynamics calculation internuclear momenta against time plot
|Figure.14 mep calculation internuclear momenta against time plotsurface plot
|}

The mep (minimum energy path) trajectory is a smooth line on contour and surface plot (Fig.7 Fig.8) while the trajectory under dynamics calculation is oscillating along the pathway. (Fig.10 Fig.11) The mep trajectory takes more steps and longer time than the dynamics trajectory as shown in internuclear distance against time plots. (Fig. 9 Fig.12) The momentum is zero all the time under mep calculation (Fig.13) but under dynamics calculation, the momenta lines are oscillating against time.

The differences arise from mep and dynamics calculations can be explained by the the natures of the two calculations. Under mep calculation, the atoms are in extremely slow motion and the momentum is zero for each step, i.e. the atoms are "stop" after every step. The pathway under mep calculation is formed by connecting all the minimum energy points of each step and appears as a smooth line. On the other hand, atoms are in continuous motion and the momentum is accumulated after each step, atoms are then oscillating on the potential energy surface along the pathway. The steps are set to 5000 in the mep calculation but are only 500 in the dynamics calculatoin. Since the steps in mep are small, more time is needed to complete the same length on the trajectories.

=== Reactive and unreactive trajectories ===

{{fontcolor1|red|Question 4ː Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.}}

For the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, run trajectories with the following momenta p1 and p2 combinations:
{| class="wikitable" border=1
| Condition
| p1
| p2
| Total Energy/ kcal mol -1
| Reactivity
|-
|1.
| -1.25
| -2.5
| -99.018
| reactive
|-
|2.
| -1.5
| -2.0
| -100.456
| unreactive
|-
|3.
| -1.5
| -2.5
| -98.956
| reactive
|-
|4.
| -2.5
| -5.0
| -84.956
| unreactive
|-
|5.
| -2.5
| -5.2
| -83.416
| reactive
|}

{| class="wikitable" border=1
| Condition
| Surface Plot
| Contour Plot
| Description
|-
|1.
| [[File:condition1wxy0119.png|350px]]
| [[File:firstwxy0119.png|350px]]
| r2 (BC) decreases when HC approaches bonded HA and HB. The energy processed by the system is sufficient to overcome the activation barrier and crosses the transition state structure to break the HA and HB bond and form a new HB and HC bond. The new bond oscillate as r1 (AB) increases.
|-
|2.
| [[File:condition2wxy0119.png|350px]]
| [[File:secondwxy0119.png|350px]]
| HC approaches HB but the energy is insufficient to reach the transition state point, HC then moves further away from the bonded H2 molecule and no new bond is formed. The oscillation along r2 is a result of an increase in momentum p1.
|-
|3.
| [[File:condition3wxy0119.png|350px]]
| [[File:thirdwxy0119.png|350px]]
| Similarly to condition 1 but with a more negative p1, HA - HB bond oscillates more when r2 decreases. The reaction proceeds and crosses the transition state point to form the products.
|-
|4.
| [[File:condition4wxy0119.png|350px]]
| [[File:fourthwxy0119.png|350px]]
| Both p1 and p2 are more negative than the previous conditions. The HA - HB bond oscillates vigorously with larger amplitudes. The barrier recrossing happens after the system crosses the transition state region but eventually reverts back to the reactants.
|-
|5.
| [[File:condition5wxy0119.png|350px]]
| [[File:fifthwxy0119.png|350px]]
| With a slight more negative value of p1 as compared to that in condition 4, the system now processes appropriate energy to proceed to forming products. The vigorous oscillations are resulting from high momenta and barrier recrossing occurs.
|}

=== Transition State Theory ===

{{fontcolor|red|Question 5ː State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}

The main assumptions of Transition State Theory (TST) are<ref name="TS" />ː

1. The behaviours of the atomic nuclei follow the classical mechanics rules, i.e. the transition structure is formed as long as atoms or molecules collide with enough energy.

2. The intermediates have a long enough lifetime for the energies to be Boltzmann distributed preceding the next step.

3. The lowest energy saddle point on the potential energy surface is passed over in the reaction system.

The TST does not take into account of the quantum tunnelling effect which allows reactions to happen even when the energy carried by the system is lower than the activation energy barrier, especially for reactions with low energy barriers. Nor does the theory consider the intermediates with short lifetimes. When the energies are not fully distributed, the momentum of the reaction trajectory from the reactants to the intermediate can have effects on product selectivity. Moreover, the theory fails at high temperature because higher vibrational energy modes can be populated for a molecule at high temperature. The molecules can have complex motion and their collisions may result in a higher energy transition structure. The experimental results may thus deviate from that predicted by the TST. For example in condition 4, the kinetic energy of the reactants is much higher than the activation energy required. However, the momenta between atoms are very high and higher vibrational modes are populated that result in higher energy transition structures formed and the system does not pass though the lowest transition saddle point as shown in the diagram. Barrier recrossing occurs but the products are not formed as opposed to the predictions by the TST.

== EXERCISE 2: F - H - H SYSTEM ==

=== Reaction Energetics ===

{{fontcolor1|red|Question 6ː 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?}}

{| class="wikitable" border=1
|[[File:Q6FH2wxy0119.png|600px]]
|[[File:Q6HFHwxy0119.png|600px]]
|-
|Figure.15 F + H2 position on surface plot
|Figure.16 HF and H position on surface plot
|}

The surface plots of the potential energy surface of H + HF → F + H2 reaction is shown (Fig.15 Fig.16). Distance AB is the distance between H and F atoms and distance BC is the distance between two H atoms. The two minimum points shown in the graph are positions of F + H2 (Fig.15) and H + HF (Fig.16) respectively. If forward reaction is H + HF → F + H2, then the backward reaction is F + H2 → H + HF. As clearly shown in the diagram F + H2 are at a minimum point of higher potential energy than that of H + HF. Thus, by comparing the potential energy of reactants and products, the signs of changes in enthalpy (ΔH) and the energetics are determined. H + HF → F + H2 is endothermic and F + H2 → H + HF is exothermic.

The energetics of the two reactions are in accordance to the predictions by comparing bond strengths of chemical species. The bond energy of H-F bond (565 kJ/mol) is higher than the bond energy of H-H bond (432 kJ/mol). Thus, energy released by forming the H-H bond is not enough to compensate for energy required for breaking the H-F bond and the reaction H + HF → F + H2 is endothermic. On the contrary, the reaction F + H2 → H + HF is exothermic.

=== Transition State Approximation ===

{{fontcolor1|red|Question 7ː Locate the approximate position of the transition state.}}

{| class="wikitable" border=1
|[[File:Q7TStransitionwxy0119.png|390px]]
|[[File:Q7TS2wxy0119.png|390px]]
|[[File:Q7TS3wxy0119.png|390px]]
|-
|Figure.17 Transition state surface plot
|Figure.18 Transition state contour plot
|Figure.19 Internuclear distance against time plot (at TS point)
|}

The transition state (TS) is shown as a black dot on the potential energy surface plot (Fig.17) and a red cross on the contour plot. (Fig,18) The position of the TS is at the point where the distance between F and H (rFH) is 1.810Å and the distance between two H atoms (rHH) is 0.746Å. At the transition state position, the distance between F and H, between two H atoms are constant with no momentum as shown in the internuclear distance against time plot as horizontal and flat lines, validating the position is the transition state point (Fig.19)

=== Activation Energies ===

{{fontcolor1|red|Question 8ː Report the activation energy for both reactions.}}

{| class="wikitable" border=1
|[[File:Q8Ea2wxy0119.png|600px]]
|[[File:Q8Ea1wxy0119.png|600px]]
|-
|Figure.21 Energy against time plot (HF + H)
|Figure.20 Energy against time plot (F + H2)
|}

By performing MEP calculation with slight increase and decrease of the rFH to 1.820Å to perform F + H2 → H + HF reaction and to 1.800Å to perform H + HF → F + H2 reaction, the activation energies (Ea) of both reactions can be calculated from the differences in their potential energies (Fig.20 Fig.21)ː

F + H2 → H + HFː Ea = -103.751 - (-133.624) = +29.873 kcal mol-1

H + HF → F + H2ː Ea = -103.751 - (-103.972) = + 0.221 kcal mol-1

=== Reaction Dynamics ===

{{fontcolor1|red|Question 9ː In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

Reactionː F + H2 → H + HF

Initial condition setː rHF = 2Å rHH = 0.74Å pHF = -0.5 pHH = 0.1

{| class="wikitable" border=1
|[[File:Q9plot1wxy0119.png|390px]]
|[[File:Q9plot2wxy0119.png|390px]]
|[[File:Q9plot3wxy0119.png|390px]]
|-
|Figure.23 Contour plot
|Figure.24 Surface plot
|Figure.25 Internuclear momentum against time plot
|-
|[[File:Q9ani1wxy0119.png|390px]]
|[[File:Q9ani2wxy0119.png|390px]]
|-
|Figure.26 Animation figure (at the start of the reaction)
|Figure.27 Animation figure (at the end of the reaction)
|}

As shown in the plots and animation figures, the H-H bond in the reactants is broken and H-F bond in the products is formed. There are small momenta between both H and F and H-H bond that provide the initial kinetic energy of the reactants and the initial vibrational energy of the H-H bond in the reactants. The initial energy is required to break the H-H bond and reduce the distance of H and F for bond forming and potential energy is released when H-F bond is formed. Since energy is conserved, the reaction energy released is converted to the vibrational energy of the H-F bond and the translational kinetic energies of the products. This is shown in the internuclear momentum against time graph as the initial vibrations of B-C (H-H bond) becomes flat as two H atoms are separated in the product and the A-B (H-F bond) formed in the product vibrates with a large amplitude. (Fig.25) The contour plot and the surface plot also clearly show there is a significant increase in the amplitude of the oscillation of H-F bond which is an indication of the large vibrational energy of H-F bond. (Fig.23 Fig.24)

The predictions can be proved experimentally. The vibrational energy of H-F bond can be determined by measuring the frequency and the intensity of the absorption band of the H-F bond with infrared spectroscopy. The conversion to the kinetic energies of the products can be observed by measuring the temperature of the reaction as the kinetic energy can be converted to thermal energy as products collide with solvent molecules. The temperature is expected to increase as this is an exothermic reaction.

=== Energy Distribution and Reactivity ===

{{fontcolor|red|Question 10ː 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.}}
{| class="wikitable" border=1
| Condition
| rFH /Å
| rHH /Å
| pFH
| pHH
|-
|1.
| 2
| 0.74
| -0.5
| -3
|-
|2.
| 2
| 0.74
| -0.5
| 1.9
|-
|3.
| 2
| 0.74
| -0.5
| 0.1
|-
|4.
| 2
| 0.74
| -0.8
| 0.1
|-
|5.
| 0.91
| 2
| 0.05
| -20
|-
|6.
| 0.91
| 2
| 0.8
| -7.5
|}
{| class="wikitable" border=1
|Condition
|Contour plot
|Observation
|-
|1.
|[[File:Q10plot1wxy0119.png|600px]]
|
|-
|2.
|[[File:Q10plot2wxy0119.png|600px]]
|
|-
|3.
|[[File:Q10plot3wxy0119.png|600px]]
|
|-
|4.
|[[File:Q10plot8wxy0119.png|600px]]
|
|-
|5.
|[[File:Q10plot5wxy0119.png|600px]]
|
|-
|6.
|[[File:Q10plot4wxy0119.png|600px]]
|
|}

pHH = -8.0 is still very large. This complies with the high activation energy of this reaction as the products formed are thermodynamically less stable than the reactants. It can be concluded that an increase in pHF, which corresponds to a high vibrational energy in H-F coordinate, is necessary for the reaction to take place. pHH, which corresponds to the translational energy in the H-H coordinate that is always very large, does not contribute too much to affect the reactivity of the reaction.

The transition state is present in the exit valley, hence, a late barrier for this endothermic reaction. The reactivity of endothermic reactions is thus dominated by the vibrational energy of the system from the results above.

For substantial exothermic reactions, such as F + H2, the transition state is located in the entrance valley, corresponds to an early barrier of the PES. For substantially endothermic reactions, such as H + HF, the transition state is in the exit valley, corresponds to a late barrier. The favoured degree of freedom for barrier crossing in exothermic reactions would be translation. This means that the momentum of the approaching atom and the atom which it is going to collide with has a greater impact on the rate of the reaction (eg. F atom in F + H2 reaction). However, reagent vibration, which is related to the momentum of the two bonded atoms (eg. HF in H + HF reaction) in the colliding molecule, would be most effective in enabling endothermic reactions to take place.

One reactive trajectory has the conditions of rFH = 0.91 Å, rHH = 2.0 Å, pHF = -0.5 and pHH = -10. The trajectory recrosses the transition state region twice and eventually moves into exit channel, forming products. By comparing this condition to the initial condition, the momentum pHH, which is the translational kinetic energy of the H atom, is reduced by a half; and the pHF, the vibrational energy of the H-F bond is increased to -0.5. This indicates that the vibrational energy has a greater contribution to the feasibility of the reaction than the translational energy. This is the opposite situation of the H2+F reaction.

The results agree with the Polanyi's empirical rules[2], which state that the vibrational energy is more efficient in promoting late-transition-state reactions and the translational energy is more efficient in promoting early-transition-state reactions.

When the momentum pFH is increased slightly from -0.5 to -0.8 with pHH being only 0.1, the reaction now becomes successful. It indicates that the F-H vibration energy has a larger contribution for the feasibility of this reaction compared to the translational energy which is the kinetic energy of the hydrogen atom which is defined by the momentum pHH

When the pHH has a large value above the activation energy (30.231 kcal/mol), which was set to be -20. The hydrogen atom collides with the HF molecule and breaks the H-F bond. However, two hydrogen atoms moves apart and do not form a H2 molecule. The procedure described is illustrated in the following animation snapshots. After the collision, the large momentum that the reactant hydrogen atom contains is still large.As a result, even if an H-H bond can be formed, it would immediately be broken due to an excess vibrational energy between two atoms.

for investigation, 1st set condition pFH = -0.5 pHH = -3 unreactive; 2nd pHH = -2.45 reactive; 3rd pHH = -1.1 unreactive; 4th pHH = -0.9 reactive; 5th pHH = 1.1 unreactive ; 6th pHH = 1.9 reactive ; 7th pHH = 2.4 to 3 unreactive
; pHH = 2.1 no reaction

The cases studied are an illustration of Polanyi's empirical rules.

For H and HF

failed conditionsː 091, 2, 0.05, -20

initial condition of H HF systemː
HF distance = 0.91 HH distance = 2 pHF = 0.8 pHH = -7.5

== References ==

<references>
<ref name="TS">Eyring, H. (1935). "The Activated Complex in Chemical Reactions". J. Chem. Phys. 3 (2): 107–115.</ref>
</references>

MRD:WXY0119

2018-05-18T16:01:59Z

Xw2816: /* Rreferences */

= Molecular Reaction Dynamics Report =

== EXERCISE 1ː H + H2 SYSTEM ==

[[File:HandH2wxy0119.png]]

In the H + H2 system, the distance between the first two H atoms is r1 and the distance (bond length) between the two H atoms in the H2 molecule is r2.

=== Dynamics from the transition state region ===

{{fontcolor|red|Question 1 ː What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.}}

{| class="wikitable" border=1
|[[File:Q1TStransitionwxy0119.png|390px]]
|[[File:Q1TS2wxy0119.png|390px]]
|[[File:Q1TS3wxy0119.png|390px]]
|-
|Figure.1 Surface plot of reaction trajectory
|Figure.2 Transition state surface plot
|Figure.3 Transition state surface plot (different angle)
|}

In the reaction trajectory surface plot (Fig.1), AB is the distance r1 and BC is the distance r2. The gradient of the potential energy with regard to r1 and r2 are ∂V(r1)/∂r1 and ∂V(r2)/∂r2 respectively. The second derivatives of the potential energy with regard to r1 and r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 respectively.

At the two minimum structures which are at the two ends of the reaction pathway, ∂V(r1)/∂r1 = 0 and ∂V(r2)/∂r2 = 0 respectively. Since the two minimum structures are at two local minimum point , the second derivatives r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 are both greater than zero.

On the other hand, the components at the transition state structure (Fig.2) are different from that at minimum points. The gradient ∂V(r1)/∂r1 and ∂V(r2)/∂r2 are both equal to zero at the transition structure. However, r2 are ∂2V(r1)/∂r12 >0 and ∂2V(r2)/∂r22 <0. The transition state point is actually a saddle point which is observed more clearly in Fig.3. The Transition state point is thus distinguished from the minimum points as the potential energy surface curves inwards and downwards at the minimum points.

=== Locating the Transition State ===

{{fontcolor|red|Question 2ː 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.}}

{| class="wikitable" border=1
|[[File:TScontourwxy0119.png|390px]]
|[[File:TSsurfaceplwxy0119.png|390px]]
|[[File:TSestimatewxy0119.png|390px]]
|-
|Figure.4 Transition state contour plot
|Figure.5 Transition state surface plot
|Figure.6 Internuclear distance against time plot
|}

The best estimate of the transition state position is rts = 0.9078 Å.

When r1 = r2 = rts and momenta are set to zero, a graph of intermolecular distance against time is plotted. (Fig.6) In this graph, AB and BC lines are the same and overlap, the two lines on the graph are flat and horizontal, indicating that the atomic distances are the same at the position and with no momenta the state is at equilibrium, validating the position is the transition state position. The transition state is shown as a cross on the counter plot (Fig.4) and a dot on the surface plot (Fig.5).

=== Calculating and comparing the reaction path and trajectory ===

{{fontcolor1|red|Question 3ː Comment on how the ''mep'' and the trajectory you just calculated differ.}}

{| class="wikitable" border=1
|[[File:Mep1wxy0119.png|390px]]
|[[File:Mep2wxy0119.png|390px]]
|[[File:Mep3wxy0119.png|390px]]
|-
|Figure.7 mep calculation contour plot
|Figure.8 mep calculation surface plot
|Figure.9 mep calculation internuclear distance against time plot
|-
|[[File:Dynamic1wxy0119.png|390px]]
|[[File:Dynamic2wxy0119.png|390px]]
|[[File:Dynamic3wxy0119.png|390px]]
|-
|Figure.10 Dynamics calculation contour plot
|Figure.11 Dynamics calculation surface plot
|Figure.12 Dynamics calculation internuclear distance against time plot
|}
{| class="wikitable" border=1
|[[File:dynamics4wxy0119.png|600px]]
|[[File:Mep5wxy0119.png|600px]]
|-
|Figure.13 Dynamics calculation internuclear momenta against time plot
|Figure.14 mep calculation internuclear momenta against time plotsurface plot
|}

The mep (minimum energy path) trajectory is a smooth line on contour and surface plot (Fig.7 Fig.8) while the trajectory under dynamics calculation is oscillating along the pathway. (Fig.10 Fig.11) The mep trajectory takes more steps and longer time than the dynamics trajectory as shown in internuclear distance against time plots. (Fig. 9 Fig.12) The momentum is zero all the time under mep calculation (Fig.13) but under dynamics calculation, the momenta lines are oscillating against time.

The differences arise from mep and dynamics calculations can be explained by the the natures of the two calculations. Under mep calculation, the atoms are in extremely slow motion and the momentum is zero for each step, i.e. the atoms are "stop" after every step. The pathway under mep calculation is formed by connecting all the minimum energy points of each step and appears as a smooth line. On the other hand, atoms are in continuous motion and the momentum is accumulated after each step, atoms are then oscillating on the potential energy surface along the pathway. The steps are set to 5000 in the mep calculation but are only 500 in the dynamics calculatoin. Since the steps in mep are small, more time is needed to complete the same length on the trajectories.

=== Reactive and unreactive trajectories ===

{{fontcolor1|red|Question 4ː Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.}}

For the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, run trajectories with the following momenta p1 and p2 combinations:
{| class="wikitable" border=1
| Condition
| p1
| p2
| Total Energy/ kcal mol -1
| Reactivity
|-
|1.
| -1.25
| -2.5
| -99.018
| reactive
|-
|2.
| -1.5
| -2.0
| -100.456
| unreactive
|-
|3.
| -1.5
| -2.5
| -98.956
| reactive
|-
|4.
| -2.5
| -5.0
| -84.956
| unreactive
|-
|5.
| -2.5
| -5.2
| -83.416
| reactive
|}

{| class="wikitable" border=1
| Condition
| Surface Plot
| Contour Plot
| Description
|-
|1.
| [[File:condition1wxy0119.png|350px]]
| [[File:firstwxy0119.png|350px]]
| r2 (BC) decreases when HC approaches bonded HA and HB. The energy processed by the system is sufficient to overcome the activation barrier and crosses the transition state structure to break the HA and HB bond and form a new HB and HC bond. The new bond oscillate as r1 (AB) increases.
|-
|2.
| [[File:condition2wxy0119.png|350px]]
| [[File:secondwxy0119.png|350px]]
| HC approaches HB but the energy is insufficient to reach the transition state point, HC then moves further away from the bonded H2 molecule and no new bond is formed. The oscillation along r2 is a result of an increase in momentum p1.
|-
|3.
| [[File:condition3wxy0119.png|350px]]
| [[File:thirdwxy0119.png|350px]]
| Similarly to condition 1 but with a more negative p1, HA - HB bond oscillates more when r2 decreases. The reaction proceeds and crosses the transition state point to form the products.
|-
|4.
| [[File:condition4wxy0119.png|350px]]
| [[File:fourthwxy0119.png|350px]]
| Both p1 and p2 are more negative than the previous conditions. The HA - HB bond oscillates vigorously with larger amplitudes. The barrier recrossing happens after the system crosses the transition state region but eventually reverts back to the reactants.
|-
|5.
| [[File:condition5wxy0119.png|350px]]
| [[File:fifthwxy0119.png|350px]]
| With a slight more negative value of p1 as compared to that in condition 4, the system now processes appropriate energy to proceed to forming products. The vigorous oscillations are resulting from high momenta and barrier recrossing occurs.
|}

=== Transition State Theory ===

{{fontcolor|red|Question 5ː State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}

The main assumptions of Transition State Theory (TST) are<ref name="TS" />ː

1. The behaviours of the atomic nuclei follow the classical mechanics rules, i.e. the transition structure is formed as long as atoms or molecules collide with enough energy.

2. The intermediates have a long enough lifetime for the energies to be Boltzmann distributed preceding the next step.

3. The lowest energy saddle point on the potential energy surface is passed over in the reaction system.

The TST does not take into account of the quantum tunnelling effect which allows reactions to happen even when the energy carried by the system is lower than the activation energy barrier, especially for reactions with low energy barriers. Nor does the theory consider the intermediates with short lifetimes. When the energies are not fully distributed, the momentum of the reaction trajectory from the reactants to the intermediate can have effects on product selectivity. Moreover, the theory fails at high temperature because higher vibrational energy modes can be populated for a molecule at high temperature. The molecules can have complex motion and their collisions may result in a higher energy transition structure. The experimental results may thus deviate from that predicted by the TST. For example in condition 4, the kinetic energy of the reactants is much higher than the activation energy required. However, the momenta between atoms are very high and higher vibrational modes are populated that result in higher energy transition structures formed and the system does not pass though the lowest transition saddle point as shown in the diagram. Barrier recrossing occurs but the products are not formed as opposed to the predictions by the TST.

== EXERCISE 2: F - H - H SYSTEM ==

=== Reaction Energetics ===

{{fontcolor1|red|Question 6ː 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?}}

{| class="wikitable" border=1
|[[File:Q6FH2wxy0119.png|600px]]
|[[File:Q6HFHwxy0119.png|600px]]
|-
|Figure.15 F + H2 position on surface plot
|Figure.16 HF and H position on surface plot
|}

The surface plots of the potential energy surface of H + HF → F + H2 reaction is shown (Fig.15 Fig.16). Distance AB is the distance between H and F atoms and distance BC is the distance between two H atoms. The two minimum points shown in the graph are positions of F + H2 (Fig.15) and H + HF (Fig.16) respectively. If forward reaction is H + HF → F + H2, then the backward reaction is F + H2 → H + HF. As clearly shown in the diagram F + H2 are at a minimum point of higher potential energy than that of H + HF. Thus, by comparing the potential energy of reactants and products, the signs of changes in enthalpy (ΔH) and the energetics are determined. H + HF → F + H2 is endothermic and F + H2 → H + HF is exothermic.

The energetics of the two reactions are in accordance to the predictions by comparing bond strengths of chemical species. The bond energy of H-F bond (565 kJ/mol) is higher than the bond energy of H-H bond (432 kJ/mol). Thus, energy released by forming the H-H bond is not enough to compensate for energy required for breaking the H-F bond and the reaction H + HF → F + H2 is endothermic. On the contrary, the reaction F + H2 → H + HF is exothermic.

=== Transition State Approximation ===

{{fontcolor1|red|Question 7ː Locate the approximate position of the transition state.}}

{| class="wikitable" border=1
|[[File:Q7TStransitionwxy0119.png|390px]]
|[[File:Q7TS2wxy0119.png|390px]]
|[[File:Q7TS3wxy0119.png|390px]]
|-
|Figure.17 Transition state surface plot
|Figure.18 Transition state contour plot
|Figure.19 Internuclear distance against time plot (at TS point)
|}

The transition state (TS) is shown as a black dot on the potential energy surface plot (Fig.17) and a red cross on the contour plot. (Fig,18) The position of the TS is at the point where the distance between F and H (rFH) is 1.810Å and the distance between two H atoms (rHH) is 0.746Å. At the transition state position, the distance between F and H, between two H atoms are constant with no momentum as shown in the internuclear distance against time plot as horizontal and flat lines, validating the position is the transition state point (Fig.19)

=== Activation Energies ===

{{fontcolor1|red|Question 8ː Report the activation energy for both reactions.}}

{| class="wikitable" border=1
|[[File:Q8Ea2wxy0119.png|600px]]
|[[File:Q8Ea1wxy0119.png|600px]]
|-
|Figure.21 Energy against time plot (HF + H)
|Figure.20 Energy against time plot (F + H2)
|}

By performing MEP calculation with slight increase and decrease of the rFH to 1.820Å to perform F + H2 → H + HF reaction and to 1.800Å to perform H + HF → F + H2 reaction, the activation energies (Ea) of both reactions can be calculated from the differences in their potential energies (Fig.20 Fig.21)ː

F + H2 → H + HFː Ea = -103.751 - (-133.624) = +29.873 kcal mol-1

H + HF → F + H2ː Ea = -103.751 - (-103.972) = + 0.221 kcal mol-1

=== Reaction Dynamics ===

{{fontcolor1|red|Question 9ː In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

Reactionː F + H2 → H + HF

Initial condition setː rHF = 2Å rHH = 0.74Å pHF = -0.5 pHH = 0.1

{| class="wikitable" border=1
|[[File:Q9plot1wxy0119.png|390px]]
|[[File:Q9plot2wxy0119.png|390px]]
|[[File:Q9plot3wxy0119.png|390px]]
|-
|Figure.23 Contour plot
|Figure.24 Surface plot
|Figure.25 Internuclear momentum against time plot
|-
|[[File:Q9ani1wxy0119.png|390px]]
|[[File:Q9ani2wxy0119.png|390px]]
|-
|Figure.26 Animation figure (at the start of the reaction)
|Figure.27 Animation figure (at the end of the reaction)
|}

As shown in the plots and animation figures, the H-H bond in the reactants is broken and H-F bond in the products is formed. There are small momenta between both H and F and H-H bond that provide the initial kinetic energy of the reactants and the initial vibrational energy of the H-H bond in the reactants. The initial energy is required to break the H-H bond and reduce the distance of H and F for bond forming and potential energy is released when H-F bond is formed. Since energy is conserved, the reaction energy released is converted to the vibrational energy of the H-F bond and the translational kinetic energies of the products. This is shown in the internuclear momentum against time graph as the initial vibrations of B-C (H-H bond) becomes flat as two H atoms are separated in the product and the A-B (H-F bond) formed in the product vibrates with a large amplitude. (Fig.25) The contour plot and the surface plot also clearly show there is a significant increase in the amplitude of the oscillation of H-F bond which is an indication of the large vibrational energy of H-F bond. (Fig.23 Fig.24)

The predictions can be proved experimentally. The vibrational energy of H-F bond can be determined by measuring the frequency and the intensity of the absorption band of the H-F bond with infrared spectroscopy. The conversion to the kinetic energies of the products can be observed by measuring the temperature of the reaction as the kinetic energy can be converted to thermal energy as products collide with solvent molecules. The temperature is expected to increase as this is an exothermic reaction.

=== Energy Distribution and Reactivity ===

{{fontcolor|red|Question 10ː 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.}}
{| class="wikitable" border=1
| Condition
| rFH /Å
| rHH /Å
| pFH
| pHH
|-
|1.
| 2
| 0.74
| -0.5
| -3
|-
|2.
| 2
| 0.74
| -0.5
| 1.9
|-
|3.
| 2
| 0.74
| -0.5
| 0.1
|-
|4.
| 2
| 0.74
| -0.8
| 0.1
|-
|5.
| 0.91
| 2
| 0.05
| -20
|-
|6.
| 0.91
| 2
| 0.8
| -7.5
|}
{| class="wikitable" border=1
|Condition
|Contour plot
|Observation
|-
|1.
|[[File:Q10plot1wxy0119.png|600px]]
|
|-
|2.
|[[File:Q10plot2wxy0119.png|600px]]
|
|-
|3.
|[[File:Q10plot3wxy0119.png|600px]]
|
|-
|4.
|[[File:Q10plot8wxy0119.png|600px]]
|
|-
|5.
|[[File:Q10plot5wxy0119.png|600px]]
|
|-
|6.
|[[File:Q10plot4wxy0119.png|600px]]
|
|}

pHH = -8.0 is still very large. This complies with the high activation energy of this reaction as the products formed are thermodynamically less stable than the reactants. It can be concluded that an increase in pHF, which corresponds to a high vibrational energy in H-F coordinate, is necessary for the reaction to take place. pHH, which corresponds to the translational energy in the H-H coordinate that is always very large, does not contribute too much to affect the reactivity of the reaction.

The transition state is present in the exit valley, hence, a late barrier for this endothermic reaction. The reactivity of endothermic reactions is thus dominated by the vibrational energy of the system from the results above.

For substantial exothermic reactions, such as F + H2, the transition state is located in the entrance valley, corresponds to an early barrier of the PES. For substantially endothermic reactions, such as H + HF, the transition state is in the exit valley, corresponds to a late barrier. The favoured degree of freedom for barrier crossing in exothermic reactions would be translation. This means that the momentum of the approaching atom and the atom which it is going to collide with has a greater impact on the rate of the reaction (eg. F atom in F + H2 reaction). However, reagent vibration, which is related to the momentum of the two bonded atoms (eg. HF in H + HF reaction) in the colliding molecule, would be most effective in enabling endothermic reactions to take place.

One reactive trajectory has the conditions of rFH = 0.91 Å, rHH = 2.0 Å, pHF = -0.5 and pHH = -10. The trajectory recrosses the transition state region twice and eventually moves into exit channel, forming products. By comparing this condition to the initial condition, the momentum pHH, which is the translational kinetic energy of the H atom, is reduced by a half; and the pHF, the vibrational energy of the H-F bond is increased to -0.5. This indicates that the vibrational energy has a greater contribution to the feasibility of the reaction than the translational energy. This is the opposite situation of the H2+F reaction.

The results agree with the Polanyi's empirical rules[2], which state that the vibrational energy is more efficient in promoting late-transition-state reactions and the translational energy is more efficient in promoting early-transition-state reactions.

When the momentum pFH is increased slightly from -0.5 to -0.8 with pHH being only 0.1, the reaction now becomes successful. It indicates that the F-H vibration energy has a larger contribution for the feasibility of this reaction compared to the translational energy which is the kinetic energy of the hydrogen atom which is defined by the momentum pHH

When the pHH has a large value above the activation energy (30.231 kcal/mol), which was set to be -20. The hydrogen atom collides with the HF molecule and breaks the H-F bond. However, two hydrogen atoms moves apart and do not form a H2 molecule. The procedure described is illustrated in the following animation snapshots. After the collision, the large momentum that the reactant hydrogen atom contains is still large.As a result, even if an H-H bond can be formed, it would immediately be broken due to an excess vibrational energy between two atoms.

for investigation, 1st set condition pFH = -0.5 pHH = -3 unreactive; 2nd pHH = -2.45 reactive; 3rd pHH = -1.1 unreactive; 4th pHH = -0.9 reactive; 5th pHH = 1.1 unreactive ; 6th pHH = 1.9 reactive ; 7th pHH = 2.4 to 3 unreactive
; pHH = 2.1 no reaction

The cases studied are an illustration of Polanyi's empirical rules.

For H and HF

failed conditionsː 091, 2, 0.05, -20

initial condition of H HF systemː
HF distance = 0.91 HH distance = 2 pHF = 0.8 pHH = -7.5

== References ==

<references>
<ref name="TS">Eyring, H. (1935). "The Activated Complex in Chemical Reactions". J. Chem. Phys. 3 (2): 107–115.</ref>
</references>

MRD:WXY0119

2018-05-18T16:01:48Z

Xw2816: /* EXERCISE 2: F - H - H SYSTEM */

= Molecular Reaction Dynamics Report =

== EXERCISE 1ː H + H2 SYSTEM ==

[[File:HandH2wxy0119.png]]

In the H + H2 system, the distance between the first two H atoms is r1 and the distance (bond length) between the two H atoms in the H2 molecule is r2.

=== Dynamics from the transition state region ===

{{fontcolor|red|Question 1 ː What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.}}

{| class="wikitable" border=1
|[[File:Q1TStransitionwxy0119.png|390px]]
|[[File:Q1TS2wxy0119.png|390px]]
|[[File:Q1TS3wxy0119.png|390px]]
|-
|Figure.1 Surface plot of reaction trajectory
|Figure.2 Transition state surface plot
|Figure.3 Transition state surface plot (different angle)
|}

In the reaction trajectory surface plot (Fig.1), AB is the distance r1 and BC is the distance r2. The gradient of the potential energy with regard to r1 and r2 are ∂V(r1)/∂r1 and ∂V(r2)/∂r2 respectively. The second derivatives of the potential energy with regard to r1 and r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 respectively.

At the two minimum structures which are at the two ends of the reaction pathway, ∂V(r1)/∂r1 = 0 and ∂V(r2)/∂r2 = 0 respectively. Since the two minimum structures are at two local minimum point , the second derivatives r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 are both greater than zero.

On the other hand, the components at the transition state structure (Fig.2) are different from that at minimum points. The gradient ∂V(r1)/∂r1 and ∂V(r2)/∂r2 are both equal to zero at the transition structure. However, r2 are ∂2V(r1)/∂r12 >0 and ∂2V(r2)/∂r22 <0. The transition state point is actually a saddle point which is observed more clearly in Fig.3. The Transition state point is thus distinguished from the minimum points as the potential energy surface curves inwards and downwards at the minimum points.

=== Locating the Transition State ===

{{fontcolor|red|Question 2ː 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.}}

{| class="wikitable" border=1
|[[File:TScontourwxy0119.png|390px]]
|[[File:TSsurfaceplwxy0119.png|390px]]
|[[File:TSestimatewxy0119.png|390px]]
|-
|Figure.4 Transition state contour plot
|Figure.5 Transition state surface plot
|Figure.6 Internuclear distance against time plot
|}

The best estimate of the transition state position is rts = 0.9078 Å.

When r1 = r2 = rts and momenta are set to zero, a graph of intermolecular distance against time is plotted. (Fig.6) In this graph, AB and BC lines are the same and overlap, the two lines on the graph are flat and horizontal, indicating that the atomic distances are the same at the position and with no momenta the state is at equilibrium, validating the position is the transition state position. The transition state is shown as a cross on the counter plot (Fig.4) and a dot on the surface plot (Fig.5).

=== Calculating and comparing the reaction path and trajectory ===

{{fontcolor1|red|Question 3ː Comment on how the ''mep'' and the trajectory you just calculated differ.}}

{| class="wikitable" border=1
|[[File:Mep1wxy0119.png|390px]]
|[[File:Mep2wxy0119.png|390px]]
|[[File:Mep3wxy0119.png|390px]]
|-
|Figure.7 mep calculation contour plot
|Figure.8 mep calculation surface plot
|Figure.9 mep calculation internuclear distance against time plot
|-
|[[File:Dynamic1wxy0119.png|390px]]
|[[File:Dynamic2wxy0119.png|390px]]
|[[File:Dynamic3wxy0119.png|390px]]
|-
|Figure.10 Dynamics calculation contour plot
|Figure.11 Dynamics calculation surface plot
|Figure.12 Dynamics calculation internuclear distance against time plot
|}
{| class="wikitable" border=1
|[[File:dynamics4wxy0119.png|600px]]
|[[File:Mep5wxy0119.png|600px]]
|-
|Figure.13 Dynamics calculation internuclear momenta against time plot
|Figure.14 mep calculation internuclear momenta against time plotsurface plot
|}

The mep (minimum energy path) trajectory is a smooth line on contour and surface plot (Fig.7 Fig.8) while the trajectory under dynamics calculation is oscillating along the pathway. (Fig.10 Fig.11) The mep trajectory takes more steps and longer time than the dynamics trajectory as shown in internuclear distance against time plots. (Fig. 9 Fig.12) The momentum is zero all the time under mep calculation (Fig.13) but under dynamics calculation, the momenta lines are oscillating against time.

The differences arise from mep and dynamics calculations can be explained by the the natures of the two calculations. Under mep calculation, the atoms are in extremely slow motion and the momentum is zero for each step, i.e. the atoms are "stop" after every step. The pathway under mep calculation is formed by connecting all the minimum energy points of each step and appears as a smooth line. On the other hand, atoms are in continuous motion and the momentum is accumulated after each step, atoms are then oscillating on the potential energy surface along the pathway. The steps are set to 5000 in the mep calculation but are only 500 in the dynamics calculatoin. Since the steps in mep are small, more time is needed to complete the same length on the trajectories.

=== Reactive and unreactive trajectories ===

{{fontcolor1|red|Question 4ː Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.}}

For the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, run trajectories with the following momenta p1 and p2 combinations:
{| class="wikitable" border=1
| Condition
| p1
| p2
| Total Energy/ kcal mol -1
| Reactivity
|-
|1.
| -1.25
| -2.5
| -99.018
| reactive
|-
|2.
| -1.5
| -2.0
| -100.456
| unreactive
|-
|3.
| -1.5
| -2.5
| -98.956
| reactive
|-
|4.
| -2.5
| -5.0
| -84.956
| unreactive
|-
|5.
| -2.5
| -5.2
| -83.416
| reactive
|}

{| class="wikitable" border=1
| Condition
| Surface Plot
| Contour Plot
| Description
|-
|1.
| [[File:condition1wxy0119.png|350px]]
| [[File:firstwxy0119.png|350px]]
| r2 (BC) decreases when HC approaches bonded HA and HB. The energy processed by the system is sufficient to overcome the activation barrier and crosses the transition state structure to break the HA and HB bond and form a new HB and HC bond. The new bond oscillate as r1 (AB) increases.
|-
|2.
| [[File:condition2wxy0119.png|350px]]
| [[File:secondwxy0119.png|350px]]
| HC approaches HB but the energy is insufficient to reach the transition state point, HC then moves further away from the bonded H2 molecule and no new bond is formed. The oscillation along r2 is a result of an increase in momentum p1.
|-
|3.
| [[File:condition3wxy0119.png|350px]]
| [[File:thirdwxy0119.png|350px]]
| Similarly to condition 1 but with a more negative p1, HA - HB bond oscillates more when r2 decreases. The reaction proceeds and crosses the transition state point to form the products.
|-
|4.
| [[File:condition4wxy0119.png|350px]]
| [[File:fourthwxy0119.png|350px]]
| Both p1 and p2 are more negative than the previous conditions. The HA - HB bond oscillates vigorously with larger amplitudes. The barrier recrossing happens after the system crosses the transition state region but eventually reverts back to the reactants.
|-
|5.
| [[File:condition5wxy0119.png|350px]]
| [[File:fifthwxy0119.png|350px]]
| With a slight more negative value of p1 as compared to that in condition 4, the system now processes appropriate energy to proceed to forming products. The vigorous oscillations are resulting from high momenta and barrier recrossing occurs.
|}

=== Transition State Theory ===

{{fontcolor|red|Question 5ː State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}

The main assumptions of Transition State Theory (TST) are<ref name="TS" />ː

1. The behaviours of the atomic nuclei follow the classical mechanics rules, i.e. the transition structure is formed as long as atoms or molecules collide with enough energy.

2. The intermediates have a long enough lifetime for the energies to be Boltzmann distributed preceding the next step.

3. The lowest energy saddle point on the potential energy surface is passed over in the reaction system.

The TST does not take into account of the quantum tunnelling effect which allows reactions to happen even when the energy carried by the system is lower than the activation energy barrier, especially for reactions with low energy barriers. Nor does the theory consider the intermediates with short lifetimes. When the energies are not fully distributed, the momentum of the reaction trajectory from the reactants to the intermediate can have effects on product selectivity. Moreover, the theory fails at high temperature because higher vibrational energy modes can be populated for a molecule at high temperature. The molecules can have complex motion and their collisions may result in a higher energy transition structure. The experimental results may thus deviate from that predicted by the TST. For example in condition 4, the kinetic energy of the reactants is much higher than the activation energy required. However, the momenta between atoms are very high and higher vibrational modes are populated that result in higher energy transition structures formed and the system does not pass though the lowest transition saddle point as shown in the diagram. Barrier recrossing occurs but the products are not formed as opposed to the predictions by the TST.

== EXERCISE 2: F - H - H SYSTEM ==

=== Reaction Energetics ===

{{fontcolor1|red|Question 6ː 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?}}

{| class="wikitable" border=1
|[[File:Q6FH2wxy0119.png|600px]]
|[[File:Q6HFHwxy0119.png|600px]]
|-
|Figure.15 F + H2 position on surface plot
|Figure.16 HF and H position on surface plot
|}

The surface plots of the potential energy surface of H + HF → F + H2 reaction is shown (Fig.15 Fig.16). Distance AB is the distance between H and F atoms and distance BC is the distance between two H atoms. The two minimum points shown in the graph are positions of F + H2 (Fig.15) and H + HF (Fig.16) respectively. If forward reaction is H + HF → F + H2, then the backward reaction is F + H2 → H + HF. As clearly shown in the diagram F + H2 are at a minimum point of higher potential energy than that of H + HF. Thus, by comparing the potential energy of reactants and products, the signs of changes in enthalpy (ΔH) and the energetics are determined. H + HF → F + H2 is endothermic and F + H2 → H + HF is exothermic.

The energetics of the two reactions are in accordance to the predictions by comparing bond strengths of chemical species. The bond energy of H-F bond (565 kJ/mol) is higher than the bond energy of H-H bond (432 kJ/mol). Thus, energy released by forming the H-H bond is not enough to compensate for energy required for breaking the H-F bond and the reaction H + HF → F + H2 is endothermic. On the contrary, the reaction F + H2 → H + HF is exothermic.

=== Transition State Approximation ===

{{fontcolor1|red|Question 7ː Locate the approximate position of the transition state.}}

{| class="wikitable" border=1
|[[File:Q7TStransitionwxy0119.png|390px]]
|[[File:Q7TS2wxy0119.png|390px]]
|[[File:Q7TS3wxy0119.png|390px]]
|-
|Figure.17 Transition state surface plot
|Figure.18 Transition state contour plot
|Figure.19 Internuclear distance against time plot (at TS point)
|}

The transition state (TS) is shown as a black dot on the potential energy surface plot (Fig.17) and a red cross on the contour plot. (Fig,18) The position of the TS is at the point where the distance between F and H (rFH) is 1.810Å and the distance between two H atoms (rHH) is 0.746Å. At the transition state position, the distance between F and H, between two H atoms are constant with no momentum as shown in the internuclear distance against time plot as horizontal and flat lines, validating the position is the transition state point (Fig.19)

=== Activation Energies ===

{{fontcolor1|red|Question 8ː Report the activation energy for both reactions.}}

{| class="wikitable" border=1
|[[File:Q8Ea2wxy0119.png|600px]]
|[[File:Q8Ea1wxy0119.png|600px]]
|-
|Figure.21 Energy against time plot (HF + H)
|Figure.20 Energy against time plot (F + H2)
|}

By performing MEP calculation with slight increase and decrease of the rFH to 1.820Å to perform F + H2 → H + HF reaction and to 1.800Å to perform H + HF → F + H2 reaction, the activation energies (Ea) of both reactions can be calculated from the differences in their potential energies (Fig.20 Fig.21)ː

F + H2 → H + HFː Ea = -103.751 - (-133.624) = +29.873 kcal mol-1

H + HF → F + H2ː Ea = -103.751 - (-103.972) = + 0.221 kcal mol-1

=== Reaction Dynamics ===

{{fontcolor1|red|Question 9ː In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

Reactionː F + H2 → H + HF

Initial condition setː rHF = 2Å rHH = 0.74Å pHF = -0.5 pHH = 0.1

{| class="wikitable" border=1
|[[File:Q9plot1wxy0119.png|390px]]
|[[File:Q9plot2wxy0119.png|390px]]
|[[File:Q9plot3wxy0119.png|390px]]
|-
|Figure.23 Contour plot
|Figure.24 Surface plot
|Figure.25 Internuclear momentum against time plot
|-
|[[File:Q9ani1wxy0119.png|390px]]
|[[File:Q9ani2wxy0119.png|390px]]
|-
|Figure.26 Animation figure (at the start of the reaction)
|Figure.27 Animation figure (at the end of the reaction)
|}

As shown in the plots and animation figures, the H-H bond in the reactants is broken and H-F bond in the products is formed. There are small momenta between both H and F and H-H bond that provide the initial kinetic energy of the reactants and the initial vibrational energy of the H-H bond in the reactants. The initial energy is required to break the H-H bond and reduce the distance of H and F for bond forming and potential energy is released when H-F bond is formed. Since energy is conserved, the reaction energy released is converted to the vibrational energy of the H-F bond and the translational kinetic energies of the products. This is shown in the internuclear momentum against time graph as the initial vibrations of B-C (H-H bond) becomes flat as two H atoms are separated in the product and the A-B (H-F bond) formed in the product vibrates with a large amplitude. (Fig.25) The contour plot and the surface plot also clearly show there is a significant increase in the amplitude of the oscillation of H-F bond which is an indication of the large vibrational energy of H-F bond. (Fig.23 Fig.24)

The predictions can be proved experimentally. The vibrational energy of H-F bond can be determined by measuring the frequency and the intensity of the absorption band of the H-F bond with infrared spectroscopy. The conversion to the kinetic energies of the products can be observed by measuring the temperature of the reaction as the kinetic energy can be converted to thermal energy as products collide with solvent molecules. The temperature is expected to increase as this is an exothermic reaction.

=== Energy Distribution and Reactivity ===

{{fontcolor|red|Question 10ː 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.}}
{| class="wikitable" border=1
| Condition
| rFH /Å
| rHH /Å
| pFH
| pHH
|-
|1.
| 2
| 0.74
| -0.5
| -3
|-
|2.
| 2
| 0.74
| -0.5
| 1.9
|-
|3.
| 2
| 0.74
| -0.5
| 0.1
|-
|4.
| 2
| 0.74
| -0.8
| 0.1
|-
|5.
| 0.91
| 2
| 0.05
| -20
|-
|6.
| 0.91
| 2
| 0.8
| -7.5
|}
{| class="wikitable" border=1
|Condition
|Contour plot
|Observation
|-
|1.
|[[File:Q10plot1wxy0119.png|600px]]
|
|-
|2.
|[[File:Q10plot2wxy0119.png|600px]]
|
|-
|3.
|[[File:Q10plot3wxy0119.png|600px]]
|
|-
|4.
|[[File:Q10plot8wxy0119.png|600px]]
|
|-
|5.
|[[File:Q10plot5wxy0119.png|600px]]
|
|-
|6.
|[[File:Q10plot4wxy0119.png|600px]]
|
|}

pHH = -8.0 is still very large. This complies with the high activation energy of this reaction as the products formed are thermodynamically less stable than the reactants. It can be concluded that an increase in pHF, which corresponds to a high vibrational energy in H-F coordinate, is necessary for the reaction to take place. pHH, which corresponds to the translational energy in the H-H coordinate that is always very large, does not contribute too much to affect the reactivity of the reaction.

The transition state is present in the exit valley, hence, a late barrier for this endothermic reaction. The reactivity of endothermic reactions is thus dominated by the vibrational energy of the system from the results above.

For substantial exothermic reactions, such as F + H2, the transition state is located in the entrance valley, corresponds to an early barrier of the PES. For substantially endothermic reactions, such as H + HF, the transition state is in the exit valley, corresponds to a late barrier. The favoured degree of freedom for barrier crossing in exothermic reactions would be translation. This means that the momentum of the approaching atom and the atom which it is going to collide with has a greater impact on the rate of the reaction (eg. F atom in F + H2 reaction). However, reagent vibration, which is related to the momentum of the two bonded atoms (eg. HF in H + HF reaction) in the colliding molecule, would be most effective in enabling endothermic reactions to take place.

One reactive trajectory has the conditions of rFH = 0.91 Å, rHH = 2.0 Å, pHF = -0.5 and pHH = -10. The trajectory recrosses the transition state region twice and eventually moves into exit channel, forming products. By comparing this condition to the initial condition, the momentum pHH, which is the translational kinetic energy of the H atom, is reduced by a half; and the pHF, the vibrational energy of the H-F bond is increased to -0.5. This indicates that the vibrational energy has a greater contribution to the feasibility of the reaction than the translational energy. This is the opposite situation of the H2+F reaction.

The results agree with the Polanyi's empirical rules[2], which state that the vibrational energy is more efficient in promoting late-transition-state reactions and the translational energy is more efficient in promoting early-transition-state reactions.

When the momentum pFH is increased slightly from -0.5 to -0.8 with pHH being only 0.1, the reaction now becomes successful. It indicates that the F-H vibration energy has a larger contribution for the feasibility of this reaction compared to the translational energy which is the kinetic energy of the hydrogen atom which is defined by the momentum pHH

When the pHH has a large value above the activation energy (30.231 kcal/mol), which was set to be -20. The hydrogen atom collides with the HF molecule and breaks the H-F bond. However, two hydrogen atoms moves apart and do not form a H2 molecule. The procedure described is illustrated in the following animation snapshots. After the collision, the large momentum that the reactant hydrogen atom contains is still large.As a result, even if an H-H bond can be formed, it would immediately be broken due to an excess vibrational energy between two atoms.

for investigation, 1st set condition pFH = -0.5 pHH = -3 unreactive; 2nd pHH = -2.45 reactive; 3rd pHH = -1.1 unreactive; 4th pHH = -0.9 reactive; 5th pHH = 1.1 unreactive ; 6th pHH = 1.9 reactive ; 7th pHH = 2.4 to 3 unreactive
; pHH = 2.1 no reaction

The cases studied are an illustration of Polanyi's empirical rules.

For H and HF

failed conditionsː 091, 2, 0.05, -20

initial condition of H HF systemː
HF distance = 0.91 HH distance = 2 pHF = 0.8 pHH = -7.5

== Rreferences ==

<references>
<ref name="TS">Eyring, H. (1935). "The Activated Complex in Chemical Reactions". J. Chem. Phys. 3 (2): 107–115.</ref>
</references>

File:Q10plot2wxy0119.png

2018-05-18T16:00:48Z

Xw2816:

File:Q10plot1wxy0119.png

2018-05-18T16:00:11Z

Xw2816:

MRD:WXY0119

2018-05-18T15:59:49Z

Xw2816: /* Energy Distribution and Reactivity */

= Molecular Reaction Dynamics Report =

== EXERCISE 1ː H + H2 SYSTEM ==

[[File:HandH2wxy0119.png]]

In the H + H2 system, the distance between the first two H atoms is r1 and the distance (bond length) between the two H atoms in the H2 molecule is r2.

=== Dynamics from the transition state region ===

{{fontcolor|red|Question 1 ː What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.}}

{| class="wikitable" border=1
|[[File:Q1TStransitionwxy0119.png|390px]]
|[[File:Q1TS2wxy0119.png|390px]]
|[[File:Q1TS3wxy0119.png|390px]]
|-
|Figure.1 Surface plot of reaction trajectory
|Figure.2 Transition state surface plot
|Figure.3 Transition state surface plot (different angle)
|}

In the reaction trajectory surface plot (Fig.1), AB is the distance r1 and BC is the distance r2. The gradient of the potential energy with regard to r1 and r2 are ∂V(r1)/∂r1 and ∂V(r2)/∂r2 respectively. The second derivatives of the potential energy with regard to r1 and r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 respectively.

At the two minimum structures which are at the two ends of the reaction pathway, ∂V(r1)/∂r1 = 0 and ∂V(r2)/∂r2 = 0 respectively. Since the two minimum structures are at two local minimum point , the second derivatives r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 are both greater than zero.

On the other hand, the components at the transition state structure (Fig.2) are different from that at minimum points. The gradient ∂V(r1)/∂r1 and ∂V(r2)/∂r2 are both equal to zero at the transition structure. However, r2 are ∂2V(r1)/∂r12 >0 and ∂2V(r2)/∂r22 <0. The transition state point is actually a saddle point which is observed more clearly in Fig.3. The Transition state point is thus distinguished from the minimum points as the potential energy surface curves inwards and downwards at the minimum points.

=== Locating the Transition State ===

{{fontcolor|red|Question 2ː 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.}}

{| class="wikitable" border=1
|[[File:TScontourwxy0119.png|390px]]
|[[File:TSsurfaceplwxy0119.png|390px]]
|[[File:TSestimatewxy0119.png|390px]]
|-
|Figure.4 Transition state contour plot
|Figure.5 Transition state surface plot
|Figure.6 Internuclear distance against time plot
|}

The best estimate of the transition state position is rts = 0.9078 Å.

When r1 = r2 = rts and momenta are set to zero, a graph of intermolecular distance against time is plotted. (Fig.6) In this graph, AB and BC lines are the same and overlap, the two lines on the graph are flat and horizontal, indicating that the atomic distances are the same at the position and with no momenta the state is at equilibrium, validating the position is the transition state position. The transition state is shown as a cross on the counter plot (Fig.4) and a dot on the surface plot (Fig.5).

=== Calculating and comparing the reaction path and trajectory ===

{{fontcolor1|red|Question 3ː Comment on how the ''mep'' and the trajectory you just calculated differ.}}

{| class="wikitable" border=1
|[[File:Mep1wxy0119.png|390px]]
|[[File:Mep2wxy0119.png|390px]]
|[[File:Mep3wxy0119.png|390px]]
|-
|Figure.7 mep calculation contour plot
|Figure.8 mep calculation surface plot
|Figure.9 mep calculation internuclear distance against time plot
|-
|[[File:Dynamic1wxy0119.png|390px]]
|[[File:Dynamic2wxy0119.png|390px]]
|[[File:Dynamic3wxy0119.png|390px]]
|-
|Figure.10 Dynamics calculation contour plot
|Figure.11 Dynamics calculation surface plot
|Figure.12 Dynamics calculation internuclear distance against time plot
|}
{| class="wikitable" border=1
|[[File:dynamics4wxy0119.png|600px]]
|[[File:Mep5wxy0119.png|600px]]
|-
|Figure.13 Dynamics calculation internuclear momenta against time plot
|Figure.14 mep calculation internuclear momenta against time plotsurface plot
|}

The mep (minimum energy path) trajectory is a smooth line on contour and surface plot (Fig.7 Fig.8) while the trajectory under dynamics calculation is oscillating along the pathway. (Fig.10 Fig.11) The mep trajectory takes more steps and longer time than the dynamics trajectory as shown in internuclear distance against time plots. (Fig. 9 Fig.12) The momentum is zero all the time under mep calculation (Fig.13) but under dynamics calculation, the momenta lines are oscillating against time.

The differences arise from mep and dynamics calculations can be explained by the the natures of the two calculations. Under mep calculation, the atoms are in extremely slow motion and the momentum is zero for each step, i.e. the atoms are "stop" after every step. The pathway under mep calculation is formed by connecting all the minimum energy points of each step and appears as a smooth line. On the other hand, atoms are in continuous motion and the momentum is accumulated after each step, atoms are then oscillating on the potential energy surface along the pathway. The steps are set to 5000 in the mep calculation but are only 500 in the dynamics calculatoin. Since the steps in mep are small, more time is needed to complete the same length on the trajectories.

=== Reactive and unreactive trajectories ===

{{fontcolor1|red|Question 4ː Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.}}

For the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, run trajectories with the following momenta p1 and p2 combinations:
{| class="wikitable" border=1
| Condition
| p1
| p2
| Total Energy/ kcal mol -1
| Reactivity
|-
|1.
| -1.25
| -2.5
| -99.018
| reactive
|-
|2.
| -1.5
| -2.0
| -100.456
| unreactive
|-
|3.
| -1.5
| -2.5
| -98.956
| reactive
|-
|4.
| -2.5
| -5.0
| -84.956
| unreactive
|-
|5.
| -2.5
| -5.2
| -83.416
| reactive
|}

{| class="wikitable" border=1
| Condition
| Surface Plot
| Contour Plot
| Description
|-
|1.
| [[File:condition1wxy0119.png|350px]]
| [[File:firstwxy0119.png|350px]]
| r2 (BC) decreases when HC approaches bonded HA and HB. The energy processed by the system is sufficient to overcome the activation barrier and crosses the transition state structure to break the HA and HB bond and form a new HB and HC bond. The new bond oscillate as r1 (AB) increases.
|-
|2.
| [[File:condition2wxy0119.png|350px]]
| [[File:secondwxy0119.png|350px]]
| HC approaches HB but the energy is insufficient to reach the transition state point, HC then moves further away from the bonded H2 molecule and no new bond is formed. The oscillation along r2 is a result of an increase in momentum p1.
|-
|3.
| [[File:condition3wxy0119.png|350px]]
| [[File:thirdwxy0119.png|350px]]
| Similarly to condition 1 but with a more negative p1, HA - HB bond oscillates more when r2 decreases. The reaction proceeds and crosses the transition state point to form the products.
|-
|4.
| [[File:condition4wxy0119.png|350px]]
| [[File:fourthwxy0119.png|350px]]
| Both p1 and p2 are more negative than the previous conditions. The HA - HB bond oscillates vigorously with larger amplitudes. The barrier recrossing happens after the system crosses the transition state region but eventually reverts back to the reactants.
|-
|5.
| [[File:condition5wxy0119.png|350px]]
| [[File:fifthwxy0119.png|350px]]
| With a slight more negative value of p1 as compared to that in condition 4, the system now processes appropriate energy to proceed to forming products. The vigorous oscillations are resulting from high momenta and barrier recrossing occurs.
|}

=== Transition State Theory ===

{{fontcolor|red|Question 5ː State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}

The main assumptions of Transition State Theory (TST) are<ref name="TS" />ː

1. The behaviours of the atomic nuclei follow the classical mechanics rules, i.e. the transition structure is formed as long as atoms or molecules collide with enough energy.

2. The intermediates have a long enough lifetime for the energies to be Boltzmann distributed preceding the next step.

3. The lowest energy saddle point on the potential energy surface is passed over in the reaction system.

The TST does not take into account of the quantum tunnelling effect which allows reactions to happen even when the energy carried by the system is lower than the activation energy barrier, especially for reactions with low energy barriers. Nor does the theory consider the intermediates with short lifetimes. When the energies are not fully distributed, the momentum of the reaction trajectory from the reactants to the intermediate can have effects on product selectivity. Moreover, the theory fails at high temperature because higher vibrational energy modes can be populated for a molecule at high temperature. The molecules can have complex motion and their collisions may result in a higher energy transition structure. The experimental results may thus deviate from that predicted by the TST. For example in condition 4, the kinetic energy of the reactants is much higher than the activation energy required. However, the momenta between atoms are very high and higher vibrational modes are populated that result in higher energy transition structures formed and the system does not pass though the lowest transition saddle point as shown in the diagram. Barrier recrossing occurs but the products are not formed as opposed to the predictions by the TST.

== EXERCISE 2: F - H - H SYSTEM ==

=== PES inspection ===

=== Reaction Energetics ===

{{fontcolor1|red|Question 6ː 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?}}

{| class="wikitable" border=1
|[[File:Q6FH2wxy0119.png|600px]]
|[[File:Q6HFHwxy0119.png|600px]]
|-
|Figure.15 F + H2 position on surface plot
|Figure.16 HF and H position on surface plot
|}

The surface plots of the potential energy surface of H + HF → F + H2 reaction is shown (Fig.15 Fig.16). Distance AB is the distance between H and F atoms and distance BC is the distance between two H atoms. The two minimum points shown in the graph are positions of F + H2 (Fig.15) and H + HF (Fig.16) respectively. If forward reaction is H + HF → F + H2, then the backward reaction is F + H2 → H + HF. As clearly shown in the diagram F + H2 are at a minimum point of higher potential energy than that of H + HF. Thus, by comparing the potential energy of reactants and products, the signs of changes in enthalpy (ΔH) and the energetics are determined. H + HF → F + H2 is endothermic and F + H2 → H + HF is exothermic.

The energetics of the two reactions are in accordance to the predictions by comparing bond strengths of chemical species. The bond energy of H-F bond (565 kJ/mol) is higher than the bond energy of H-H bond (432 kJ/mol). Thus, energy released by forming the H-H bond is not enough to compensate for energy required for breaking the H-F bond and the reaction H + HF → F + H2 is endothermic. On the contrary, the reaction F + H2 → H + HF is exothermic.

=== Transition State Approximation ===

{{fontcolor1|red|Question 7ː Locate the approximate position of the transition state.}}

{| class="wikitable" border=1
|[[File:Q7TStransitionwxy0119.png|390px]]
|[[File:Q7TS2wxy0119.png|390px]]
|[[File:Q7TS3wxy0119.png|390px]]
|-
|Figure.17 Transition state surface plot
|Figure.18 Transition state contour plot
|Figure.19 Internuclear distance against time plot (at TS point)
|}

The transition state (TS) is shown as a black dot on the potential energy surface plot (Fig.17) and a red cross on the contour plot. (Fig,18) The position of the TS is at the point where the distance between F and H (rFH) is 1.810Å and the distance between two H atoms (rHH) is 0.746Å. At the transition state position, the distance between F and H, between two H atoms are constant with no momentum as shown in the internuclear distance against time plot as horizontal and flat lines, validating the position is the transition state point (Fig.19)

=== Activation Energies ===

{{fontcolor1|red|Question 8ː Report the activation energy for both reactions.}}

{| class="wikitable" border=1
|[[File:Q8Ea2wxy0119.png|600px]]
|[[File:Q8Ea1wxy0119.png|600px]]
|-
|Figure.21 Energy against time plot (HF + H)
|Figure.20 Energy against time plot (F + H2)
|}

By performing MEP calculation with slight increase and decrease of the rFH to 1.820Å to perform F + H2 → H + HF reaction and to 1.800Å to perform H + HF → F + H2 reaction, the activation energies (Ea) of both reactions can be calculated from the differences in their potential energies (Fig.20 Fig.21)ː

F + H2 → H + HFː Ea = -103.751 - (-133.624) = +29.873 kcal mol-1

H + HF → F + H2ː Ea = -103.751 - (-103.972) = + 0.221 kcal mol-1

=== Reaction Dynamics ===

{{fontcolor1|red|Question 9ː In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

Reactionː F + H2 → H + HF

Initial condition setː rHF = 2Å rHH = 0.74Å pHF = -0.5 pHH = 0.1

{| class="wikitable" border=1
|[[File:Q9plot1wxy0119.png|390px]]
|[[File:Q9plot2wxy0119.png|390px]]
|[[File:Q9plot3wxy0119.png|390px]]
|-
|Figure.23 Contour plot
|Figure.24 Surface plot
|Figure.25 Internuclear momentum against time plot
|-
|[[File:Q9ani1wxy0119.png|390px]]
|[[File:Q9ani2wxy0119.png|390px]]
|-
|Figure.26 Animation figure (at the start of the reaction)
|Figure.27 Animation figure (at the end of the reaction)
|}

As shown in the plots and animation figures, the H-H bond in the reactants is broken and H-F bond in the products is formed. There are small momenta between both H and F and H-H bond that provide the initial kinetic energy of the reactants and the initial vibrational energy of the H-H bond in the reactants. The initial energy is required to break the H-H bond and reduce the distance of H and F for bond forming and potential energy is released when H-F bond is formed. Since energy is conserved, the reaction energy released is converted to the vibrational energy of the H-F bond and the translational kinetic energies of the products. This is shown in the internuclear momentum against time graph as the initial vibrations of B-C (H-H bond) becomes flat as two H atoms are separated in the product and the A-B (H-F bond) formed in the product vibrates with a large amplitude. (Fig.25) The contour plot and the surface plot also clearly show there is a significant increase in the amplitude of the oscillation of H-F bond which is an indication of the large vibrational energy of H-F bond. (Fig.23 Fig.24)

The predictions can be proved experimentally. The vibrational energy of H-F bond can be determined by measuring the frequency and the intensity of the absorption band of the H-F bond with infrared spectroscopy. The conversion to the kinetic energies of the products can be observed by measuring the temperature of the reaction as the kinetic energy can be converted to thermal energy as products collide with solvent molecules. The temperature is expected to increase as this is an exothermic reaction.

=== Energy Distribution and Reactivity ===

{{fontcolor|red|Question 10ː 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.}}
{| class="wikitable" border=1
| Condition
| rFH /Å
| rHH /Å
| pFH
| pHH
|-
|1.
| 2
| 0.74
| -0.5
| -3
|-
|2.
| 2
| 0.74
| -0.5
| 1.9
|-
|3.
| 2
| 0.74
| -0.5
| 0.1
|-
|4.
| 2
| 0.74
| -0.8
| 0.1
|-
|5.
| 0.91
| 2
| 0.05
| -20
|-
|6.
| 0.91
| 2
| 0.8
| -7.5
|}
{| class="wikitable" border=1
|Condition
|Contour plot
|Observation
|-
|1.
|[[File:Q10plot1wxy0119.png|600px]]
|
|-
|2.
|[[File:Q10plot2wxy0119.png|600px]]
|
|-
|3.
|[[File:Q10plot3wxy0119.png|600px]]
|
|-
|4.
|[[File:Q10plot8wxy0119.png|600px]]
|
|-
|5.
|[[File:Q10plot5wxy0119.png|600px]]
|
|-
|6.
|[[File:Q10plot4wxy0119.png|600px]]
|
|}

pHH = -8.0 is still very large. This complies with the high activation energy of this reaction as the products formed are thermodynamically less stable than the reactants. It can be concluded that an increase in pHF, which corresponds to a high vibrational energy in H-F coordinate, is necessary for the reaction to take place. pHH, which corresponds to the translational energy in the H-H coordinate that is always very large, does not contribute too much to affect the reactivity of the reaction.

The transition state is present in the exit valley, hence, a late barrier for this endothermic reaction. The reactivity of endothermic reactions is thus dominated by the vibrational energy of the system from the results above.

For substantial exothermic reactions, such as F + H2, the transition state is located in the entrance valley, corresponds to an early barrier of the PES. For substantially endothermic reactions, such as H + HF, the transition state is in the exit valley, corresponds to a late barrier. The favoured degree of freedom for barrier crossing in exothermic reactions would be translation. This means that the momentum of the approaching atom and the atom which it is going to collide with has a greater impact on the rate of the reaction (eg. F atom in F + H2 reaction). However, reagent vibration, which is related to the momentum of the two bonded atoms (eg. HF in H + HF reaction) in the colliding molecule, would be most effective in enabling endothermic reactions to take place.

One reactive trajectory has the conditions of rFH = 0.91 Å, rHH = 2.0 Å, pHF = -0.5 and pHH = -10. The trajectory recrosses the transition state region twice and eventually moves into exit channel, forming products. By comparing this condition to the initial condition, the momentum pHH, which is the translational kinetic energy of the H atom, is reduced by a half; and the pHF, the vibrational energy of the H-F bond is increased to -0.5. This indicates that the vibrational energy has a greater contribution to the feasibility of the reaction than the translational energy. This is the opposite situation of the H2+F reaction.

The results agree with the Polanyi's empirical rules[2], which state that the vibrational energy is more efficient in promoting late-transition-state reactions and the translational energy is more efficient in promoting early-transition-state reactions.

When the momentum pFH is increased slightly from -0.5 to -0.8 with pHH being only 0.1, the reaction now becomes successful. It indicates that the F-H vibration energy has a larger contribution for the feasibility of this reaction compared to the translational energy which is the kinetic energy of the hydrogen atom which is defined by the momentum pHH

When the pHH has a large value above the activation energy (30.231 kcal/mol), which was set to be -20. The hydrogen atom collides with the HF molecule and breaks the H-F bond. However, two hydrogen atoms moves apart and do not form a H2 molecule. The procedure described is illustrated in the following animation snapshots. After the collision, the large momentum that the reactant hydrogen atom contains is still large.As a result, even if an H-H bond can be formed, it would immediately be broken due to an excess vibrational energy between two atoms.

for investigation, 1st set condition pFH = -0.5 pHH = -3 unreactive; 2nd pHH = -2.45 reactive; 3rd pHH = -1.1 unreactive; 4th pHH = -0.9 reactive; 5th pHH = 1.1 unreactive ; 6th pHH = 1.9 reactive ; 7th pHH = 2.4 to 3 unreactive
; pHH = 2.1 no reaction

The cases studied are an illustration of Polanyi's empirical rules.

For H and HF

failed conditionsː 091, 2, 0.05, -20

initial condition of H HF systemː
HF distance = 0.91 HH distance = 2 pHF = 0.8 pHH = -7.5

== Rreferences ==

<references>
<ref name="TS">Eyring, H. (1935). "The Activated Complex in Chemical Reactions". J. Chem. Phys. 3 (2): 107–115.</ref>
</references>

File:Q10plot3wxy0119.png

2018-05-18T15:57:47Z

Xw2816:

File:Q10plot8wxy0119.png

2018-05-18T15:53:14Z

Xw2816:

MRD:WXY0119

2018-05-18T15:52:41Z

Xw2816: /* Energy Distribution and Reactivity */

= Molecular Reaction Dynamics Report =

== EXERCISE 1ː H + H2 SYSTEM ==

[[File:HandH2wxy0119.png]]

In the H + H2 system, the distance between the first two H atoms is r1 and the distance (bond length) between the two H atoms in the H2 molecule is r2.

=== Dynamics from the transition state region ===

{{fontcolor|red|Question 1 ː What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.}}

{| class="wikitable" border=1
|[[File:Q1TStransitionwxy0119.png|390px]]
|[[File:Q1TS2wxy0119.png|390px]]
|[[File:Q1TS3wxy0119.png|390px]]
|-
|Figure.1 Surface plot of reaction trajectory
|Figure.2 Transition state surface plot
|Figure.3 Transition state surface plot (different angle)
|}

In the reaction trajectory surface plot (Fig.1), AB is the distance r1 and BC is the distance r2. The gradient of the potential energy with regard to r1 and r2 are ∂V(r1)/∂r1 and ∂V(r2)/∂r2 respectively. The second derivatives of the potential energy with regard to r1 and r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 respectively.

At the two minimum structures which are at the two ends of the reaction pathway, ∂V(r1)/∂r1 = 0 and ∂V(r2)/∂r2 = 0 respectively. Since the two minimum structures are at two local minimum point , the second derivatives r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 are both greater than zero.

On the other hand, the components at the transition state structure (Fig.2) are different from that at minimum points. The gradient ∂V(r1)/∂r1 and ∂V(r2)/∂r2 are both equal to zero at the transition structure. However, r2 are ∂2V(r1)/∂r12 >0 and ∂2V(r2)/∂r22 <0. The transition state point is actually a saddle point which is observed more clearly in Fig.3. The Transition state point is thus distinguished from the minimum points as the potential energy surface curves inwards and downwards at the minimum points.

=== Locating the Transition State ===

{{fontcolor|red|Question 2ː 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.}}

{| class="wikitable" border=1
|[[File:TScontourwxy0119.png|390px]]
|[[File:TSsurfaceplwxy0119.png|390px]]
|[[File:TSestimatewxy0119.png|390px]]
|-
|Figure.4 Transition state contour plot
|Figure.5 Transition state surface plot
|Figure.6 Internuclear distance against time plot
|}

The best estimate of the transition state position is rts = 0.9078 Å.

When r1 = r2 = rts and momenta are set to zero, a graph of intermolecular distance against time is plotted. (Fig.6) In this graph, AB and BC lines are the same and overlap, the two lines on the graph are flat and horizontal, indicating that the atomic distances are the same at the position and with no momenta the state is at equilibrium, validating the position is the transition state position. The transition state is shown as a cross on the counter plot (Fig.4) and a dot on the surface plot (Fig.5).

=== Calculating and comparing the reaction path and trajectory ===

{{fontcolor1|red|Question 3ː Comment on how the ''mep'' and the trajectory you just calculated differ.}}

{| class="wikitable" border=1
|[[File:Mep1wxy0119.png|390px]]
|[[File:Mep2wxy0119.png|390px]]
|[[File:Mep3wxy0119.png|390px]]
|-
|Figure.7 mep calculation contour plot
|Figure.8 mep calculation surface plot
|Figure.9 mep calculation internuclear distance against time plot
|-
|[[File:Dynamic1wxy0119.png|390px]]
|[[File:Dynamic2wxy0119.png|390px]]
|[[File:Dynamic3wxy0119.png|390px]]
|-
|Figure.10 Dynamics calculation contour plot
|Figure.11 Dynamics calculation surface plot
|Figure.12 Dynamics calculation internuclear distance against time plot
|}
{| class="wikitable" border=1
|[[File:dynamics4wxy0119.png|600px]]
|[[File:Mep5wxy0119.png|600px]]
|-
|Figure.13 Dynamics calculation internuclear momenta against time plot
|Figure.14 mep calculation internuclear momenta against time plotsurface plot
|}

The mep (minimum energy path) trajectory is a smooth line on contour and surface plot (Fig.7 Fig.8) while the trajectory under dynamics calculation is oscillating along the pathway. (Fig.10 Fig.11) The mep trajectory takes more steps and longer time than the dynamics trajectory as shown in internuclear distance against time plots. (Fig. 9 Fig.12) The momentum is zero all the time under mep calculation (Fig.13) but under dynamics calculation, the momenta lines are oscillating against time.

The differences arise from mep and dynamics calculations can be explained by the the natures of the two calculations. Under mep calculation, the atoms are in extremely slow motion and the momentum is zero for each step, i.e. the atoms are "stop" after every step. The pathway under mep calculation is formed by connecting all the minimum energy points of each step and appears as a smooth line. On the other hand, atoms are in continuous motion and the momentum is accumulated after each step, atoms are then oscillating on the potential energy surface along the pathway. The steps are set to 5000 in the mep calculation but are only 500 in the dynamics calculatoin. Since the steps in mep are small, more time is needed to complete the same length on the trajectories.

=== Reactive and unreactive trajectories ===

{{fontcolor1|red|Question 4ː Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.}}

For the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, run trajectories with the following momenta p1 and p2 combinations:
{| class="wikitable" border=1
| Condition
| p1
| p2
| Total Energy/ kcal mol -1
| Reactivity
|-
|1.
| -1.25
| -2.5
| -99.018
| reactive
|-
|2.
| -1.5
| -2.0
| -100.456
| unreactive
|-
|3.
| -1.5
| -2.5
| -98.956
| reactive
|-
|4.
| -2.5
| -5.0
| -84.956
| unreactive
|-
|5.
| -2.5
| -5.2
| -83.416
| reactive
|}

{| class="wikitable" border=1
| Condition
| Surface Plot
| Contour Plot
| Description
|-
|1.
| [[File:condition1wxy0119.png|350px]]
| [[File:firstwxy0119.png|350px]]
| r2 (BC) decreases when HC approaches bonded HA and HB. The energy processed by the system is sufficient to overcome the activation barrier and crosses the transition state structure to break the HA and HB bond and form a new HB and HC bond. The new bond oscillate as r1 (AB) increases.
|-
|2.
| [[File:condition2wxy0119.png|350px]]
| [[File:secondwxy0119.png|350px]]
| HC approaches HB but the energy is insufficient to reach the transition state point, HC then moves further away from the bonded H2 molecule and no new bond is formed. The oscillation along r2 is a result of an increase in momentum p1.
|-
|3.
| [[File:condition3wxy0119.png|350px]]
| [[File:thirdwxy0119.png|350px]]
| Similarly to condition 1 but with a more negative p1, HA - HB bond oscillates more when r2 decreases. The reaction proceeds and crosses the transition state point to form the products.
|-
|4.
| [[File:condition4wxy0119.png|350px]]
| [[File:fourthwxy0119.png|350px]]
| Both p1 and p2 are more negative than the previous conditions. The HA - HB bond oscillates vigorously with larger amplitudes. The barrier recrossing happens after the system crosses the transition state region but eventually reverts back to the reactants.
|-
|5.
| [[File:condition5wxy0119.png|350px]]
| [[File:fifthwxy0119.png|350px]]
| With a slight more negative value of p1 as compared to that in condition 4, the system now processes appropriate energy to proceed to forming products. The vigorous oscillations are resulting from high momenta and barrier recrossing occurs.
|}

=== Transition State Theory ===

{{fontcolor|red|Question 5ː State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}

The main assumptions of Transition State Theory (TST) are<ref name="TS" />ː

1. The behaviours of the atomic nuclei follow the classical mechanics rules, i.e. the transition structure is formed as long as atoms or molecules collide with enough energy.

2. The intermediates have a long enough lifetime for the energies to be Boltzmann distributed preceding the next step.

3. The lowest energy saddle point on the potential energy surface is passed over in the reaction system.

The TST does not take into account of the quantum tunnelling effect which allows reactions to happen even when the energy carried by the system is lower than the activation energy barrier, especially for reactions with low energy barriers. Nor does the theory consider the intermediates with short lifetimes. When the energies are not fully distributed, the momentum of the reaction trajectory from the reactants to the intermediate can have effects on product selectivity. Moreover, the theory fails at high temperature because higher vibrational energy modes can be populated for a molecule at high temperature. The molecules can have complex motion and their collisions may result in a higher energy transition structure. The experimental results may thus deviate from that predicted by the TST. For example in condition 4, the kinetic energy of the reactants is much higher than the activation energy required. However, the momenta between atoms are very high and higher vibrational modes are populated that result in higher energy transition structures formed and the system does not pass though the lowest transition saddle point as shown in the diagram. Barrier recrossing occurs but the products are not formed as opposed to the predictions by the TST.

== EXERCISE 2: F - H - H SYSTEM ==

=== PES inspection ===

=== Reaction Energetics ===

{{fontcolor1|red|Question 6ː 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?}}

{| class="wikitable" border=1
|[[File:Q6FH2wxy0119.png|600px]]
|[[File:Q6HFHwxy0119.png|600px]]
|-
|Figure.15 F + H2 position on surface plot
|Figure.16 HF and H position on surface plot
|}

The surface plots of the potential energy surface of H + HF → F + H2 reaction is shown (Fig.15 Fig.16). Distance AB is the distance between H and F atoms and distance BC is the distance between two H atoms. The two minimum points shown in the graph are positions of F + H2 (Fig.15) and H + HF (Fig.16) respectively. If forward reaction is H + HF → F + H2, then the backward reaction is F + H2 → H + HF. As clearly shown in the diagram F + H2 are at a minimum point of higher potential energy than that of H + HF. Thus, by comparing the potential energy of reactants and products, the signs of changes in enthalpy (ΔH) and the energetics are determined. H + HF → F + H2 is endothermic and F + H2 → H + HF is exothermic.

The energetics of the two reactions are in accordance to the predictions by comparing bond strengths of chemical species. The bond energy of H-F bond (565 kJ/mol) is higher than the bond energy of H-H bond (432 kJ/mol). Thus, energy released by forming the H-H bond is not enough to compensate for energy required for breaking the H-F bond and the reaction H + HF → F + H2 is endothermic. On the contrary, the reaction F + H2 → H + HF is exothermic.

=== Transition State Approximation ===

{{fontcolor1|red|Question 7ː Locate the approximate position of the transition state.}}

{| class="wikitable" border=1
|[[File:Q7TStransitionwxy0119.png|390px]]
|[[File:Q7TS2wxy0119.png|390px]]
|[[File:Q7TS3wxy0119.png|390px]]
|-
|Figure.17 Transition state surface plot
|Figure.18 Transition state contour plot
|Figure.19 Internuclear distance against time plot (at TS point)
|}

The transition state (TS) is shown as a black dot on the potential energy surface plot (Fig.17) and a red cross on the contour plot. (Fig,18) The position of the TS is at the point where the distance between F and H (rFH) is 1.810Å and the distance between two H atoms (rHH) is 0.746Å. At the transition state position, the distance between F and H, between two H atoms are constant with no momentum as shown in the internuclear distance against time plot as horizontal and flat lines, validating the position is the transition state point (Fig.19)

=== Activation Energies ===

{{fontcolor1|red|Question 8ː Report the activation energy for both reactions.}}

{| class="wikitable" border=1
|[[File:Q8Ea2wxy0119.png|600px]]
|[[File:Q8Ea1wxy0119.png|600px]]
|-
|Figure.21 Energy against time plot (HF + H)
|Figure.20 Energy against time plot (F + H2)
|}

By performing MEP calculation with slight increase and decrease of the rFH to 1.820Å to perform F + H2 → H + HF reaction and to 1.800Å to perform H + HF → F + H2 reaction, the activation energies (Ea) of both reactions can be calculated from the differences in their potential energies (Fig.20 Fig.21)ː

F + H2 → H + HFː Ea = -103.751 - (-133.624) = +29.873 kcal mol-1

H + HF → F + H2ː Ea = -103.751 - (-103.972) = + 0.221 kcal mol-1

=== Reaction Dynamics ===

{{fontcolor1|red|Question 9ː In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

Reactionː F + H2 → H + HF

Initial condition setː rHF = 2Å rHH = 0.74Å pHF = -0.5 pHH = 0.1

{| class="wikitable" border=1
|[[File:Q9plot1wxy0119.png|390px]]
|[[File:Q9plot2wxy0119.png|390px]]
|[[File:Q9plot3wxy0119.png|390px]]
|-
|Figure.23 Contour plot
|Figure.24 Surface plot
|Figure.25 Internuclear momentum against time plot
|-
|[[File:Q9ani1wxy0119.png|390px]]
|[[File:Q9ani2wxy0119.png|390px]]
|-
|Figure.26 Animation figure (at the start of the reaction)
|Figure.27 Animation figure (at the end of the reaction)
|}

As shown in the plots and animation figures, the H-H bond in the reactants is broken and H-F bond in the products is formed. There are small momenta between both H and F and H-H bond that provide the initial kinetic energy of the reactants and the initial vibrational energy of the H-H bond in the reactants. The initial energy is required to break the H-H bond and reduce the distance of H and F for bond forming and potential energy is released when H-F bond is formed. Since energy is conserved, the reaction energy released is converted to the vibrational energy of the H-F bond and the translational kinetic energies of the products. This is shown in the internuclear momentum against time graph as the initial vibrations of B-C (H-H bond) becomes flat as two H atoms are separated in the product and the A-B (H-F bond) formed in the product vibrates with a large amplitude. (Fig.25) The contour plot and the surface plot also clearly show there is a significant increase in the amplitude of the oscillation of H-F bond which is an indication of the large vibrational energy of H-F bond. (Fig.23 Fig.24)

The predictions can be proved experimentally. The vibrational energy of H-F bond can be determined by measuring the frequency and the intensity of the absorption band of the H-F bond with infrared spectroscopy. The conversion to the kinetic energies of the products can be observed by measuring the temperature of the reaction as the kinetic energy can be converted to thermal energy as products collide with solvent molecules. The temperature is expected to increase as this is an exothermic reaction.

=== Energy Distribution and Reactivity ===

{{fontcolor|red|Question 10ː 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.}}
{| class="wikitable" border=1
| Condition
| rFH /Å
| rHH /Å
| pFH
| pHH
|-
|1.
| 2
| 0.74
| -0.5
| ?
|-
|2.
| 2
| 0.74
| -0.5
| unreactive
|-
|3.
| 2
| 0.74
| -0.5
| reactive
|-
|4.
| 2
| 0.74
| -0.8
| 0.1
|-
|5.
| 0.91
| 2
| 0.05
| -20
|-
|6.
| 0.91
| 2
| 0.8
| -7.5
|}
{| class="wikitable" border=1
|Condition
|Contour plot
|Observation
|-
|1.
|[[File:Q10plot1wxy0119.png|600px]]
|
|-
|2.
|[[File:Q10plot2wxy0119.png|600px]]
|
|-
|3.
|[[File:Q10plot3wxy0119.png|600px]]
|
|-
|4.
|[[File:Q10plot8wxy0119.png|600px]]
|
|-
|5.
|[[File:Q10plot5wxy0119.png|600px]]
|
|-
|6.
|[[File:Q10plot4wxy0119.png|600px]]
|
|}

pHH = -8.0 is still very large. This complies with the high activation energy of this reaction as the products formed are thermodynamically less stable than the reactants. It can be concluded that an increase in pHF, which corresponds to a high vibrational energy in H-F coordinate, is necessary for the reaction to take place. pHH, which corresponds to the translational energy in the H-H coordinate that is always very large, does not contribute too much to affect the reactivity of the reaction.

The transition state is present in the exit valley, hence, a late barrier for this endothermic reaction. The reactivity of endothermic reactions is thus dominated by the vibrational energy of the system from the results above.

For substantial exothermic reactions, such as F + H2, the transition state is located in the entrance valley, corresponds to an early barrier of the PES. For substantially endothermic reactions, such as H + HF, the transition state is in the exit valley, corresponds to a late barrier. The favoured degree of freedom for barrier crossing in exothermic reactions would be translation. This means that the momentum of the approaching atom and the atom which it is going to collide with has a greater impact on the rate of the reaction (eg. F atom in F + H2 reaction). However, reagent vibration, which is related to the momentum of the two bonded atoms (eg. HF in H + HF reaction) in the colliding molecule, would be most effective in enabling endothermic reactions to take place.

One reactive trajectory has the conditions of rFH = 0.91 Å, rHH = 2.0 Å, pHF = -0.5 and pHH = -10. The trajectory recrosses the transition state region twice and eventually moves into exit channel, forming products. By comparing this condition to the initial condition, the momentum pHH, which is the translational kinetic energy of the H atom, is reduced by a half; and the pHF, the vibrational energy of the H-F bond is increased to -0.5. This indicates that the vibrational energy has a greater contribution to the feasibility of the reaction than the translational energy. This is the opposite situation of the H2+F reaction.

The results agree with the Polanyi's empirical rules[2], which state that the vibrational energy is more efficient in promoting late-transition-state reactions and the translational energy is more efficient in promoting early-transition-state reactions.

When the momentum pFH is increased slightly from -0.5 to -0.8 with pHH being only 0.1, the reaction now becomes successful. It indicates that the F-H vibration energy has a larger contribution for the feasibility of this reaction compared to the translational energy which is the kinetic energy of the hydrogen atom which is defined by the momentum pHH

When the pHH has a large value above the activation energy (30.231 kcal/mol), which was set to be -20. The hydrogen atom collides with the HF molecule and breaks the H-F bond. However, two hydrogen atoms moves apart and do not form a H2 molecule. The procedure described is illustrated in the following animation snapshots. After the collision, the large momentum that the reactant hydrogen atom contains is still large.As a result, even if an H-H bond can be formed, it would immediately be broken due to an excess vibrational energy between two atoms.

for investigation, 1st set condition pFH = -0.5 pHH = -3 unreactive; 2nd pHH = -2.45 reactive; 3rd pHH = -1.1 unreactive; 4th pHH = -0.9 reactive; 5th pHH = 1.1 unreactive ; 6th pHH = 1.9 reactive ; 7th pHH = 2.4 to 3 unreactive
; pHH = 2.1 no reaction

The cases studied are an illustration of Polanyi's empirical rules.

For H and HF

failed conditionsː 091, 2, 0.05, -20

initial condition of H HF systemː
HF distance = 0.91 HH distance = 2 pHF = 0.8 pHH = -7.5

== Rreferences ==

<references>
<ref name="TS">Eyring, H. (1935). "The Activated Complex in Chemical Reactions". J. Chem. Phys. 3 (2): 107–115.</ref>
</references>

MRD:WXY0119

2018-05-18T15:51:53Z

Xw2816: /* Energy Distribution and Reactivity */

= Molecular Reaction Dynamics Report =

== EXERCISE 1ː H + H2 SYSTEM ==

[[File:HandH2wxy0119.png]]

In the H + H2 system, the distance between the first two H atoms is r1 and the distance (bond length) between the two H atoms in the H2 molecule is r2.

=== Dynamics from the transition state region ===

{{fontcolor|red|Question 1 ː What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.}}

{| class="wikitable" border=1
|[[File:Q1TStransitionwxy0119.png|390px]]
|[[File:Q1TS2wxy0119.png|390px]]
|[[File:Q1TS3wxy0119.png|390px]]
|-
|Figure.1 Surface plot of reaction trajectory
|Figure.2 Transition state surface plot
|Figure.3 Transition state surface plot (different angle)
|}

In the reaction trajectory surface plot (Fig.1), AB is the distance r1 and BC is the distance r2. The gradient of the potential energy with regard to r1 and r2 are ∂V(r1)/∂r1 and ∂V(r2)/∂r2 respectively. The second derivatives of the potential energy with regard to r1 and r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 respectively.

At the two minimum structures which are at the two ends of the reaction pathway, ∂V(r1)/∂r1 = 0 and ∂V(r2)/∂r2 = 0 respectively. Since the two minimum structures are at two local minimum point , the second derivatives r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 are both greater than zero.

On the other hand, the components at the transition state structure (Fig.2) are different from that at minimum points. The gradient ∂V(r1)/∂r1 and ∂V(r2)/∂r2 are both equal to zero at the transition structure. However, r2 are ∂2V(r1)/∂r12 >0 and ∂2V(r2)/∂r22 <0. The transition state point is actually a saddle point which is observed more clearly in Fig.3. The Transition state point is thus distinguished from the minimum points as the potential energy surface curves inwards and downwards at the minimum points.

=== Locating the Transition State ===

{{fontcolor|red|Question 2ː 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.}}

{| class="wikitable" border=1
|[[File:TScontourwxy0119.png|390px]]
|[[File:TSsurfaceplwxy0119.png|390px]]
|[[File:TSestimatewxy0119.png|390px]]
|-
|Figure.4 Transition state contour plot
|Figure.5 Transition state surface plot
|Figure.6 Internuclear distance against time plot
|}

The best estimate of the transition state position is rts = 0.9078 Å.

When r1 = r2 = rts and momenta are set to zero, a graph of intermolecular distance against time is plotted. (Fig.6) In this graph, AB and BC lines are the same and overlap, the two lines on the graph are flat and horizontal, indicating that the atomic distances are the same at the position and with no momenta the state is at equilibrium, validating the position is the transition state position. The transition state is shown as a cross on the counter plot (Fig.4) and a dot on the surface plot (Fig.5).

=== Calculating and comparing the reaction path and trajectory ===

{{fontcolor1|red|Question 3ː Comment on how the ''mep'' and the trajectory you just calculated differ.}}

{| class="wikitable" border=1
|[[File:Mep1wxy0119.png|390px]]
|[[File:Mep2wxy0119.png|390px]]
|[[File:Mep3wxy0119.png|390px]]
|-
|Figure.7 mep calculation contour plot
|Figure.8 mep calculation surface plot
|Figure.9 mep calculation internuclear distance against time plot
|-
|[[File:Dynamic1wxy0119.png|390px]]
|[[File:Dynamic2wxy0119.png|390px]]
|[[File:Dynamic3wxy0119.png|390px]]
|-
|Figure.10 Dynamics calculation contour plot
|Figure.11 Dynamics calculation surface plot
|Figure.12 Dynamics calculation internuclear distance against time plot
|}
{| class="wikitable" border=1
|[[File:dynamics4wxy0119.png|600px]]
|[[File:Mep5wxy0119.png|600px]]
|-
|Figure.13 Dynamics calculation internuclear momenta against time plot
|Figure.14 mep calculation internuclear momenta against time plotsurface plot
|}

The mep (minimum energy path) trajectory is a smooth line on contour and surface plot (Fig.7 Fig.8) while the trajectory under dynamics calculation is oscillating along the pathway. (Fig.10 Fig.11) The mep trajectory takes more steps and longer time than the dynamics trajectory as shown in internuclear distance against time plots. (Fig. 9 Fig.12) The momentum is zero all the time under mep calculation (Fig.13) but under dynamics calculation, the momenta lines are oscillating against time.

The differences arise from mep and dynamics calculations can be explained by the the natures of the two calculations. Under mep calculation, the atoms are in extremely slow motion and the momentum is zero for each step, i.e. the atoms are "stop" after every step. The pathway under mep calculation is formed by connecting all the minimum energy points of each step and appears as a smooth line. On the other hand, atoms are in continuous motion and the momentum is accumulated after each step, atoms are then oscillating on the potential energy surface along the pathway. The steps are set to 5000 in the mep calculation but are only 500 in the dynamics calculatoin. Since the steps in mep are small, more time is needed to complete the same length on the trajectories.

=== Reactive and unreactive trajectories ===

{{fontcolor1|red|Question 4ː Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.}}

For the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, run trajectories with the following momenta p1 and p2 combinations:
{| class="wikitable" border=1
| Condition
| p1
| p2
| Total Energy/ kcal mol -1
| Reactivity
|-
|1.
| -1.25
| -2.5
| -99.018
| reactive
|-
|2.
| -1.5
| -2.0
| -100.456
| unreactive
|-
|3.
| -1.5
| -2.5
| -98.956
| reactive
|-
|4.
| -2.5
| -5.0
| -84.956
| unreactive
|-
|5.
| -2.5
| -5.2
| -83.416
| reactive
|}

{| class="wikitable" border=1
| Condition
| Surface Plot
| Contour Plot
| Description
|-
|1.
| [[File:condition1wxy0119.png|350px]]
| [[File:firstwxy0119.png|350px]]
| r2 (BC) decreases when HC approaches bonded HA and HB. The energy processed by the system is sufficient to overcome the activation barrier and crosses the transition state structure to break the HA and HB bond and form a new HB and HC bond. The new bond oscillate as r1 (AB) increases.
|-
|2.
| [[File:condition2wxy0119.png|350px]]
| [[File:secondwxy0119.png|350px]]
| HC approaches HB but the energy is insufficient to reach the transition state point, HC then moves further away from the bonded H2 molecule and no new bond is formed. The oscillation along r2 is a result of an increase in momentum p1.
|-
|3.
| [[File:condition3wxy0119.png|350px]]
| [[File:thirdwxy0119.png|350px]]
| Similarly to condition 1 but with a more negative p1, HA - HB bond oscillates more when r2 decreases. The reaction proceeds and crosses the transition state point to form the products.
|-
|4.
| [[File:condition4wxy0119.png|350px]]
| [[File:fourthwxy0119.png|350px]]
| Both p1 and p2 are more negative than the previous conditions. The HA - HB bond oscillates vigorously with larger amplitudes. The barrier recrossing happens after the system crosses the transition state region but eventually reverts back to the reactants.
|-
|5.
| [[File:condition5wxy0119.png|350px]]
| [[File:fifthwxy0119.png|350px]]
| With a slight more negative value of p1 as compared to that in condition 4, the system now processes appropriate energy to proceed to forming products. The vigorous oscillations are resulting from high momenta and barrier recrossing occurs.
|}

=== Transition State Theory ===

{{fontcolor|red|Question 5ː State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}

The main assumptions of Transition State Theory (TST) are<ref name="TS" />ː

1. The behaviours of the atomic nuclei follow the classical mechanics rules, i.e. the transition structure is formed as long as atoms or molecules collide with enough energy.

2. The intermediates have a long enough lifetime for the energies to be Boltzmann distributed preceding the next step.

3. The lowest energy saddle point on the potential energy surface is passed over in the reaction system.

The TST does not take into account of the quantum tunnelling effect which allows reactions to happen even when the energy carried by the system is lower than the activation energy barrier, especially for reactions with low energy barriers. Nor does the theory consider the intermediates with short lifetimes. When the energies are not fully distributed, the momentum of the reaction trajectory from the reactants to the intermediate can have effects on product selectivity. Moreover, the theory fails at high temperature because higher vibrational energy modes can be populated for a molecule at high temperature. The molecules can have complex motion and their collisions may result in a higher energy transition structure. The experimental results may thus deviate from that predicted by the TST. For example in condition 4, the kinetic energy of the reactants is much higher than the activation energy required. However, the momenta between atoms are very high and higher vibrational modes are populated that result in higher energy transition structures formed and the system does not pass though the lowest transition saddle point as shown in the diagram. Barrier recrossing occurs but the products are not formed as opposed to the predictions by the TST.

== EXERCISE 2: F - H - H SYSTEM ==

=== PES inspection ===

=== Reaction Energetics ===

{{fontcolor1|red|Question 6ː 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?}}

{| class="wikitable" border=1
|[[File:Q6FH2wxy0119.png|600px]]
|[[File:Q6HFHwxy0119.png|600px]]
|-
|Figure.15 F + H2 position on surface plot
|Figure.16 HF and H position on surface plot
|}

The surface plots of the potential energy surface of H + HF → F + H2 reaction is shown (Fig.15 Fig.16). Distance AB is the distance between H and F atoms and distance BC is the distance between two H atoms. The two minimum points shown in the graph are positions of F + H2 (Fig.15) and H + HF (Fig.16) respectively. If forward reaction is H + HF → F + H2, then the backward reaction is F + H2 → H + HF. As clearly shown in the diagram F + H2 are at a minimum point of higher potential energy than that of H + HF. Thus, by comparing the potential energy of reactants and products, the signs of changes in enthalpy (ΔH) and the energetics are determined. H + HF → F + H2 is endothermic and F + H2 → H + HF is exothermic.

The energetics of the two reactions are in accordance to the predictions by comparing bond strengths of chemical species. The bond energy of H-F bond (565 kJ/mol) is higher than the bond energy of H-H bond (432 kJ/mol). Thus, energy released by forming the H-H bond is not enough to compensate for energy required for breaking the H-F bond and the reaction H + HF → F + H2 is endothermic. On the contrary, the reaction F + H2 → H + HF is exothermic.

=== Transition State Approximation ===

{{fontcolor1|red|Question 7ː Locate the approximate position of the transition state.}}

{| class="wikitable" border=1
|[[File:Q7TStransitionwxy0119.png|390px]]
|[[File:Q7TS2wxy0119.png|390px]]
|[[File:Q7TS3wxy0119.png|390px]]
|-
|Figure.17 Transition state surface plot
|Figure.18 Transition state contour plot
|Figure.19 Internuclear distance against time plot (at TS point)
|}

The transition state (TS) is shown as a black dot on the potential energy surface plot (Fig.17) and a red cross on the contour plot. (Fig,18) The position of the TS is at the point where the distance between F and H (rFH) is 1.810Å and the distance between two H atoms (rHH) is 0.746Å. At the transition state position, the distance between F and H, between two H atoms are constant with no momentum as shown in the internuclear distance against time plot as horizontal and flat lines, validating the position is the transition state point (Fig.19)

=== Activation Energies ===

{{fontcolor1|red|Question 8ː Report the activation energy for both reactions.}}

{| class="wikitable" border=1
|[[File:Q8Ea2wxy0119.png|600px]]
|[[File:Q8Ea1wxy0119.png|600px]]
|-
|Figure.21 Energy against time plot (HF + H)
|Figure.20 Energy against time plot (F + H2)
|}

By performing MEP calculation with slight increase and decrease of the rFH to 1.820Å to perform F + H2 → H + HF reaction and to 1.800Å to perform H + HF → F + H2 reaction, the activation energies (Ea) of both reactions can be calculated from the differences in their potential energies (Fig.20 Fig.21)ː

F + H2 → H + HFː Ea = -103.751 - (-133.624) = +29.873 kcal mol-1

H + HF → F + H2ː Ea = -103.751 - (-103.972) = + 0.221 kcal mol-1

=== Reaction Dynamics ===

{{fontcolor1|red|Question 9ː In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

Reactionː F + H2 → H + HF

Initial condition setː rHF = 2Å rHH = 0.74Å pHF = -0.5 pHH = 0.1

{| class="wikitable" border=1
|[[File:Q9plot1wxy0119.png|390px]]
|[[File:Q9plot2wxy0119.png|390px]]
|[[File:Q9plot3wxy0119.png|390px]]
|-
|Figure.23 Contour plot
|Figure.24 Surface plot
|Figure.25 Internuclear momentum against time plot
|-
|[[File:Q9ani1wxy0119.png|390px]]
|[[File:Q9ani2wxy0119.png|390px]]
|-
|Figure.26 Animation figure (at the start of the reaction)
|Figure.27 Animation figure (at the end of the reaction)
|}

As shown in the plots and animation figures, the H-H bond in the reactants is broken and H-F bond in the products is formed. There are small momenta between both H and F and H-H bond that provide the initial kinetic energy of the reactants and the initial vibrational energy of the H-H bond in the reactants. The initial energy is required to break the H-H bond and reduce the distance of H and F for bond forming and potential energy is released when H-F bond is formed. Since energy is conserved, the reaction energy released is converted to the vibrational energy of the H-F bond and the translational kinetic energies of the products. This is shown in the internuclear momentum against time graph as the initial vibrations of B-C (H-H bond) becomes flat as two H atoms are separated in the product and the A-B (H-F bond) formed in the product vibrates with a large amplitude. (Fig.25) The contour plot and the surface plot also clearly show there is a significant increase in the amplitude of the oscillation of H-F bond which is an indication of the large vibrational energy of H-F bond. (Fig.23 Fig.24)

The predictions can be proved experimentally. The vibrational energy of H-F bond can be determined by measuring the frequency and the intensity of the absorption band of the H-F bond with infrared spectroscopy. The conversion to the kinetic energies of the products can be observed by measuring the temperature of the reaction as the kinetic energy can be converted to thermal energy as products collide with solvent molecules. The temperature is expected to increase as this is an exothermic reaction.

=== Energy Distribution and Reactivity ===

{{fontcolor|red|Question 10ː 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.}}
{| class="wikitable" border=1
| Condition
| rFH /Å
| rHH /Å
| pFH
| pHH
|-
|1.
| 2
| 0.74
| -0.5
| ?
|-
|2.
| 2
| 0.74
| -0.5
| unreactive
|-
|3.
| 2
| 0.74
| -0.5
| reactive
|-
|4.
| 2
| 0.74
| -0.8
| 0.1
|-
|5.
| 0.91
| 2
| 0.05
| -20
|-
|6.
| 0.91
| 2
| 0.8
| -7.5
|}
{| class="wikitable" border=1
|Condition
|Contour plot
|Observation
|-
|1.
|[[File:Q10plot1wxy0119.png|600px]]
|
|-
|2.
|[[File:Q10plot2wxy0119.png|600px]]
|
|-
|3.
|[[File:Q10plot3wxy0119.png|600px]]
|
|-
|4.
|[[File:Q10plot8wxy0119.png|600px]]
|
|-
|5.
|[[File:Q10plot5wxy0119.png|600px]]
|
|-
|6.
|[[File:Q10plot7wxy0119.png|600px]]
|
|}

pHH = -8.0 is still very large. This complies with the high activation energy of this reaction as the products formed are thermodynamically less stable than the reactants. It can be concluded that an increase in pHF, which corresponds to a high vibrational energy in H-F coordinate, is necessary for the reaction to take place. pHH, which corresponds to the translational energy in the H-H coordinate that is always very large, does not contribute too much to affect the reactivity of the reaction.

The transition state is present in the exit valley, hence, a late barrier for this endothermic reaction. The reactivity of endothermic reactions is thus dominated by the vibrational energy of the system from the results above.

For substantial exothermic reactions, such as F + H2, the transition state is located in the entrance valley, corresponds to an early barrier of the PES. For substantially endothermic reactions, such as H + HF, the transition state is in the exit valley, corresponds to a late barrier. The favoured degree of freedom for barrier crossing in exothermic reactions would be translation. This means that the momentum of the approaching atom and the atom which it is going to collide with has a greater impact on the rate of the reaction (eg. F atom in F + H2 reaction). However, reagent vibration, which is related to the momentum of the two bonded atoms (eg. HF in H + HF reaction) in the colliding molecule, would be most effective in enabling endothermic reactions to take place.

One reactive trajectory has the conditions of rFH = 0.91 Å, rHH = 2.0 Å, pHF = -0.5 and pHH = -10. The trajectory recrosses the transition state region twice and eventually moves into exit channel, forming products. By comparing this condition to the initial condition, the momentum pHH, which is the translational kinetic energy of the H atom, is reduced by a half; and the pHF, the vibrational energy of the H-F bond is increased to -0.5. This indicates that the vibrational energy has a greater contribution to the feasibility of the reaction than the translational energy. This is the opposite situation of the H2+F reaction.

The results agree with the Polanyi's empirical rules[2], which state that the vibrational energy is more efficient in promoting late-transition-state reactions and the translational energy is more efficient in promoting early-transition-state reactions.

When the momentum pFH is increased slightly from -0.5 to -0.8 with pHH being only 0.1, the reaction now becomes successful. It indicates that the F-H vibration energy has a larger contribution for the feasibility of this reaction compared to the translational energy which is the kinetic energy of the hydrogen atom which is defined by the momentum pHH

When the pHH has a large value above the activation energy (30.231 kcal/mol), which was set to be -20. The hydrogen atom collides with the HF molecule and breaks the H-F bond. However, two hydrogen atoms moves apart and do not form a H2 molecule. The procedure described is illustrated in the following animation snapshots. After the collision, the large momentum that the reactant hydrogen atom contains is still large.As a result, even if an H-H bond can be formed, it would immediately be broken due to an excess vibrational energy between two atoms.

for investigation, 1st set condition pFH = -0.5 pHH = -3 unreactive; 2nd pHH = -2.45 reactive; 3rd pHH = -1.1 unreactive; 4th pHH = -0.9 reactive; 5th pHH = 1.1 unreactive ; 6th pHH = 1.9 reactive ; 7th pHH = 2.4 to 3 unreactive
; pHH = 2.1 no reaction

The cases studied are an illustration of Polanyi's empirical rules.

For H and HF

failed conditionsː 091, 2, 0.05, -20

initial condition of H HF systemː
HF distance = 0.91 HH distance = 2 pHF = 0.8 pHH = -7.5

== Rreferences ==

<references>
<ref name="TS">Eyring, H. (1935). "The Activated Complex in Chemical Reactions". J. Chem. Phys. 3 (2): 107–115.</ref>
</references>

MRD:WXY0119

2018-05-18T15:51:03Z

Xw2816: /* Energy Distribution and Reactivity */

= Molecular Reaction Dynamics Report =

== EXERCISE 1ː H + H2 SYSTEM ==

[[File:HandH2wxy0119.png]]

In the H + H2 system, the distance between the first two H atoms is r1 and the distance (bond length) between the two H atoms in the H2 molecule is r2.

=== Dynamics from the transition state region ===

{{fontcolor|red|Question 1 ː What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.}}

{| class="wikitable" border=1
|[[File:Q1TStransitionwxy0119.png|390px]]
|[[File:Q1TS2wxy0119.png|390px]]
|[[File:Q1TS3wxy0119.png|390px]]
|-
|Figure.1 Surface plot of reaction trajectory
|Figure.2 Transition state surface plot
|Figure.3 Transition state surface plot (different angle)
|}

In the reaction trajectory surface plot (Fig.1), AB is the distance r1 and BC is the distance r2. The gradient of the potential energy with regard to r1 and r2 are ∂V(r1)/∂r1 and ∂V(r2)/∂r2 respectively. The second derivatives of the potential energy with regard to r1 and r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 respectively.

At the two minimum structures which are at the two ends of the reaction pathway, ∂V(r1)/∂r1 = 0 and ∂V(r2)/∂r2 = 0 respectively. Since the two minimum structures are at two local minimum point , the second derivatives r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 are both greater than zero.

On the other hand, the components at the transition state structure (Fig.2) are different from that at minimum points. The gradient ∂V(r1)/∂r1 and ∂V(r2)/∂r2 are both equal to zero at the transition structure. However, r2 are ∂2V(r1)/∂r12 >0 and ∂2V(r2)/∂r22 <0. The transition state point is actually a saddle point which is observed more clearly in Fig.3. The Transition state point is thus distinguished from the minimum points as the potential energy surface curves inwards and downwards at the minimum points.

=== Locating the Transition State ===

{{fontcolor|red|Question 2ː 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.}}

{| class="wikitable" border=1
|[[File:TScontourwxy0119.png|390px]]
|[[File:TSsurfaceplwxy0119.png|390px]]
|[[File:TSestimatewxy0119.png|390px]]
|-
|Figure.4 Transition state contour plot
|Figure.5 Transition state surface plot
|Figure.6 Internuclear distance against time plot
|}

The best estimate of the transition state position is rts = 0.9078 Å.

When r1 = r2 = rts and momenta are set to zero, a graph of intermolecular distance against time is plotted. (Fig.6) In this graph, AB and BC lines are the same and overlap, the two lines on the graph are flat and horizontal, indicating that the atomic distances are the same at the position and with no momenta the state is at equilibrium, validating the position is the transition state position. The transition state is shown as a cross on the counter plot (Fig.4) and a dot on the surface plot (Fig.5).

=== Calculating and comparing the reaction path and trajectory ===

{{fontcolor1|red|Question 3ː Comment on how the ''mep'' and the trajectory you just calculated differ.}}

{| class="wikitable" border=1
|[[File:Mep1wxy0119.png|390px]]
|[[File:Mep2wxy0119.png|390px]]
|[[File:Mep3wxy0119.png|390px]]
|-
|Figure.7 mep calculation contour plot
|Figure.8 mep calculation surface plot
|Figure.9 mep calculation internuclear distance against time plot
|-
|[[File:Dynamic1wxy0119.png|390px]]
|[[File:Dynamic2wxy0119.png|390px]]
|[[File:Dynamic3wxy0119.png|390px]]
|-
|Figure.10 Dynamics calculation contour plot
|Figure.11 Dynamics calculation surface plot
|Figure.12 Dynamics calculation internuclear distance against time plot
|}
{| class="wikitable" border=1
|[[File:dynamics4wxy0119.png|600px]]
|[[File:Mep5wxy0119.png|600px]]
|-
|Figure.13 Dynamics calculation internuclear momenta against time plot
|Figure.14 mep calculation internuclear momenta against time plotsurface plot
|}

The mep (minimum energy path) trajectory is a smooth line on contour and surface plot (Fig.7 Fig.8) while the trajectory under dynamics calculation is oscillating along the pathway. (Fig.10 Fig.11) The mep trajectory takes more steps and longer time than the dynamics trajectory as shown in internuclear distance against time plots. (Fig. 9 Fig.12) The momentum is zero all the time under mep calculation (Fig.13) but under dynamics calculation, the momenta lines are oscillating against time.

The differences arise from mep and dynamics calculations can be explained by the the natures of the two calculations. Under mep calculation, the atoms are in extremely slow motion and the momentum is zero for each step, i.e. the atoms are "stop" after every step. The pathway under mep calculation is formed by connecting all the minimum energy points of each step and appears as a smooth line. On the other hand, atoms are in continuous motion and the momentum is accumulated after each step, atoms are then oscillating on the potential energy surface along the pathway. The steps are set to 5000 in the mep calculation but are only 500 in the dynamics calculatoin. Since the steps in mep are small, more time is needed to complete the same length on the trajectories.

=== Reactive and unreactive trajectories ===

{{fontcolor1|red|Question 4ː Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.}}

For the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, run trajectories with the following momenta p1 and p2 combinations:
{| class="wikitable" border=1
| Condition
| p1
| p2
| Total Energy/ kcal mol -1
| Reactivity
|-
|1.
| -1.25
| -2.5
| -99.018
| reactive
|-
|2.
| -1.5
| -2.0
| -100.456
| unreactive
|-
|3.
| -1.5
| -2.5
| -98.956
| reactive
|-
|4.
| -2.5
| -5.0
| -84.956
| unreactive
|-
|5.
| -2.5
| -5.2
| -83.416
| reactive
|}

{| class="wikitable" border=1
| Condition
| Surface Plot
| Contour Plot
| Description
|-
|1.
| [[File:condition1wxy0119.png|350px]]
| [[File:firstwxy0119.png|350px]]
| r2 (BC) decreases when HC approaches bonded HA and HB. The energy processed by the system is sufficient to overcome the activation barrier and crosses the transition state structure to break the HA and HB bond and form a new HB and HC bond. The new bond oscillate as r1 (AB) increases.
|-
|2.
| [[File:condition2wxy0119.png|350px]]
| [[File:secondwxy0119.png|350px]]
| HC approaches HB but the energy is insufficient to reach the transition state point, HC then moves further away from the bonded H2 molecule and no new bond is formed. The oscillation along r2 is a result of an increase in momentum p1.
|-
|3.
| [[File:condition3wxy0119.png|350px]]
| [[File:thirdwxy0119.png|350px]]
| Similarly to condition 1 but with a more negative p1, HA - HB bond oscillates more when r2 decreases. The reaction proceeds and crosses the transition state point to form the products.
|-
|4.
| [[File:condition4wxy0119.png|350px]]
| [[File:fourthwxy0119.png|350px]]
| Both p1 and p2 are more negative than the previous conditions. The HA - HB bond oscillates vigorously with larger amplitudes. The barrier recrossing happens after the system crosses the transition state region but eventually reverts back to the reactants.
|-
|5.
| [[File:condition5wxy0119.png|350px]]
| [[File:fifthwxy0119.png|350px]]
| With a slight more negative value of p1 as compared to that in condition 4, the system now processes appropriate energy to proceed to forming products. The vigorous oscillations are resulting from high momenta and barrier recrossing occurs.
|}

=== Transition State Theory ===

{{fontcolor|red|Question 5ː State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}

The main assumptions of Transition State Theory (TST) are<ref name="TS" />ː

1. The behaviours of the atomic nuclei follow the classical mechanics rules, i.e. the transition structure is formed as long as atoms or molecules collide with enough energy.

2. The intermediates have a long enough lifetime for the energies to be Boltzmann distributed preceding the next step.

3. The lowest energy saddle point on the potential energy surface is passed over in the reaction system.

The TST does not take into account of the quantum tunnelling effect which allows reactions to happen even when the energy carried by the system is lower than the activation energy barrier, especially for reactions with low energy barriers. Nor does the theory consider the intermediates with short lifetimes. When the energies are not fully distributed, the momentum of the reaction trajectory from the reactants to the intermediate can have effects on product selectivity. Moreover, the theory fails at high temperature because higher vibrational energy modes can be populated for a molecule at high temperature. The molecules can have complex motion and their collisions may result in a higher energy transition structure. The experimental results may thus deviate from that predicted by the TST. For example in condition 4, the kinetic energy of the reactants is much higher than the activation energy required. However, the momenta between atoms are very high and higher vibrational modes are populated that result in higher energy transition structures formed and the system does not pass though the lowest transition saddle point as shown in the diagram. Barrier recrossing occurs but the products are not formed as opposed to the predictions by the TST.

== EXERCISE 2: F - H - H SYSTEM ==

=== PES inspection ===

=== Reaction Energetics ===

{{fontcolor1|red|Question 6ː 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?}}

{| class="wikitable" border=1
|[[File:Q6FH2wxy0119.png|600px]]
|[[File:Q6HFHwxy0119.png|600px]]
|-
|Figure.15 F + H2 position on surface plot
|Figure.16 HF and H position on surface plot
|}

The surface plots of the potential energy surface of H + HF → F + H2 reaction is shown (Fig.15 Fig.16). Distance AB is the distance between H and F atoms and distance BC is the distance between two H atoms. The two minimum points shown in the graph are positions of F + H2 (Fig.15) and H + HF (Fig.16) respectively. If forward reaction is H + HF → F + H2, then the backward reaction is F + H2 → H + HF. As clearly shown in the diagram F + H2 are at a minimum point of higher potential energy than that of H + HF. Thus, by comparing the potential energy of reactants and products, the signs of changes in enthalpy (ΔH) and the energetics are determined. H + HF → F + H2 is endothermic and F + H2 → H + HF is exothermic.

The energetics of the two reactions are in accordance to the predictions by comparing bond strengths of chemical species. The bond energy of H-F bond (565 kJ/mol) is higher than the bond energy of H-H bond (432 kJ/mol). Thus, energy released by forming the H-H bond is not enough to compensate for energy required for breaking the H-F bond and the reaction H + HF → F + H2 is endothermic. On the contrary, the reaction F + H2 → H + HF is exothermic.

=== Transition State Approximation ===

{{fontcolor1|red|Question 7ː Locate the approximate position of the transition state.}}

{| class="wikitable" border=1
|[[File:Q7TStransitionwxy0119.png|390px]]
|[[File:Q7TS2wxy0119.png|390px]]
|[[File:Q7TS3wxy0119.png|390px]]
|-
|Figure.17 Transition state surface plot
|Figure.18 Transition state contour plot
|Figure.19 Internuclear distance against time plot (at TS point)
|}

The transition state (TS) is shown as a black dot on the potential energy surface plot (Fig.17) and a red cross on the contour plot. (Fig,18) The position of the TS is at the point where the distance between F and H (rFH) is 1.810Å and the distance between two H atoms (rHH) is 0.746Å. At the transition state position, the distance between F and H, between two H atoms are constant with no momentum as shown in the internuclear distance against time plot as horizontal and flat lines, validating the position is the transition state point (Fig.19)

=== Activation Energies ===

{{fontcolor1|red|Question 8ː Report the activation energy for both reactions.}}

{| class="wikitable" border=1
|[[File:Q8Ea2wxy0119.png|600px]]
|[[File:Q8Ea1wxy0119.png|600px]]
|-
|Figure.21 Energy against time plot (HF + H)
|Figure.20 Energy against time plot (F + H2)
|}

By performing MEP calculation with slight increase and decrease of the rFH to 1.820Å to perform F + H2 → H + HF reaction and to 1.800Å to perform H + HF → F + H2 reaction, the activation energies (Ea) of both reactions can be calculated from the differences in their potential energies (Fig.20 Fig.21)ː

F + H2 → H + HFː Ea = -103.751 - (-133.624) = +29.873 kcal mol-1

H + HF → F + H2ː Ea = -103.751 - (-103.972) = + 0.221 kcal mol-1

=== Reaction Dynamics ===

{{fontcolor1|red|Question 9ː In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

Reactionː F + H2 → H + HF

Initial condition setː rHF = 2Å rHH = 0.74Å pHF = -0.5 pHH = 0.1

{| class="wikitable" border=1
|[[File:Q9plot1wxy0119.png|390px]]
|[[File:Q9plot2wxy0119.png|390px]]
|[[File:Q9plot3wxy0119.png|390px]]
|-
|Figure.23 Contour plot
|Figure.24 Surface plot
|Figure.25 Internuclear momentum against time plot
|-
|[[File:Q9ani1wxy0119.png|390px]]
|[[File:Q9ani2wxy0119.png|390px]]
|-
|Figure.26 Animation figure (at the start of the reaction)
|Figure.27 Animation figure (at the end of the reaction)
|}

As shown in the plots and animation figures, the H-H bond in the reactants is broken and H-F bond in the products is formed. There are small momenta between both H and F and H-H bond that provide the initial kinetic energy of the reactants and the initial vibrational energy of the H-H bond in the reactants. The initial energy is required to break the H-H bond and reduce the distance of H and F for bond forming and potential energy is released when H-F bond is formed. Since energy is conserved, the reaction energy released is converted to the vibrational energy of the H-F bond and the translational kinetic energies of the products. This is shown in the internuclear momentum against time graph as the initial vibrations of B-C (H-H bond) becomes flat as two H atoms are separated in the product and the A-B (H-F bond) formed in the product vibrates with a large amplitude. (Fig.25) The contour plot and the surface plot also clearly show there is a significant increase in the amplitude of the oscillation of H-F bond which is an indication of the large vibrational energy of H-F bond. (Fig.23 Fig.24)

The predictions can be proved experimentally. The vibrational energy of H-F bond can be determined by measuring the frequency and the intensity of the absorption band of the H-F bond with infrared spectroscopy. The conversion to the kinetic energies of the products can be observed by measuring the temperature of the reaction as the kinetic energy can be converted to thermal energy as products collide with solvent molecules. The temperature is expected to increase as this is an exothermic reaction.

=== Energy Distribution and Reactivity ===

{{fontcolor|red|Question 10ː 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.}}
{| class="wikitable" border=1
| Condition
| rFH /Å
| rHH /Å
| pFH
| pHH
|-
|1.
| 2
| 0.74
| -0.5
| ?
|-
|2.
| 2
| 0.74
| -0.5
| unreactive
|-
|3.
| 2
| 0.74
| -0.5
| reactive
|-
|4.
| 2
| 0.74
| -0.8
| 0.1
|-
|5.
| 0.91
| 2
| 0.05
| -20
|-
|6.
| 0.91
| 2
| 0.8
| -7.5
|}
{| class="wikitable" border=1
|Condition
|Contour plot
|Observation
|-
|1.
|[[File:Q10plot1wxy0119.png|600px]]
|
|-
|2.
|[[File:Q10plot2wxy0119.png|600px]]
|
|-
|3.
|[[File:Q10plot3wxy0119.png|600px]]
|
|-
|4.
|[[File:Q10plot4wxy0119.png|600px]]
|
|-
|5.
|[[File:Q10plot5wxy0119.png|600px]]
|
|-
|6.
|[[File:Q10plot6wxy0119.png|600px]]
|
|}

pHH = -8.0 is still very large. This complies with the high activation energy of this reaction as the products formed are thermodynamically less stable than the reactants. It can be concluded that an increase in pHF, which corresponds to a high vibrational energy in H-F coordinate, is necessary for the reaction to take place. pHH, which corresponds to the translational energy in the H-H coordinate that is always very large, does not contribute too much to affect the reactivity of the reaction.

The transition state is present in the exit valley, hence, a late barrier for this endothermic reaction. The reactivity of endothermic reactions is thus dominated by the vibrational energy of the system from the results above.

For substantial exothermic reactions, such as F + H2, the transition state is located in the entrance valley, corresponds to an early barrier of the PES. For substantially endothermic reactions, such as H + HF, the transition state is in the exit valley, corresponds to a late barrier. The favoured degree of freedom for barrier crossing in exothermic reactions would be translation. This means that the momentum of the approaching atom and the atom which it is going to collide with has a greater impact on the rate of the reaction (eg. F atom in F + H2 reaction). However, reagent vibration, which is related to the momentum of the two bonded atoms (eg. HF in H + HF reaction) in the colliding molecule, would be most effective in enabling endothermic reactions to take place.

One reactive trajectory has the conditions of rFH = 0.91 Å, rHH = 2.0 Å, pHF = -0.5 and pHH = -10. The trajectory recrosses the transition state region twice and eventually moves into exit channel, forming products. By comparing this condition to the initial condition, the momentum pHH, which is the translational kinetic energy of the H atom, is reduced by a half; and the pHF, the vibrational energy of the H-F bond is increased to -0.5. This indicates that the vibrational energy has a greater contribution to the feasibility of the reaction than the translational energy. This is the opposite situation of the H2+F reaction.

The results agree with the Polanyi's empirical rules[2], which state that the vibrational energy is more efficient in promoting late-transition-state reactions and the translational energy is more efficient in promoting early-transition-state reactions.

When the momentum pFH is increased slightly from -0.5 to -0.8 with pHH being only 0.1, the reaction now becomes successful. It indicates that the F-H vibration energy has a larger contribution for the feasibility of this reaction compared to the translational energy which is the kinetic energy of the hydrogen atom which is defined by the momentum pHH

When the pHH has a large value above the activation energy (30.231 kcal/mol), which was set to be -20. The hydrogen atom collides with the HF molecule and breaks the H-F bond. However, two hydrogen atoms moves apart and do not form a H2 molecule. The procedure described is illustrated in the following animation snapshots. After the collision, the large momentum that the reactant hydrogen atom contains is still large.As a result, even if an H-H bond can be formed, it would immediately be broken due to an excess vibrational energy between two atoms.

for investigation, 1st set condition pFH = -0.5 pHH = -3 unreactive; 2nd pHH = -2.45 reactive; 3rd pHH = -1.1 unreactive; 4th pHH = -0.9 reactive; 5th pHH = 1.1 unreactive ; 6th pHH = 1.9 reactive ; 7th pHH = 2.4 to 3 unreactive
; pHH = 2.1 no reaction

The cases studied are an illustration of Polanyi's empirical rules.

For H and HF

failed conditionsː 091, 2, 0.05, -20

initial condition of H HF systemː
HF distance = 0.91 HH distance = 2 pHF = 0.8 pHH = -7.5

== Rreferences ==

<references>
<ref name="TS">Eyring, H. (1935). "The Activated Complex in Chemical Reactions". J. Chem. Phys. 3 (2): 107–115.</ref>
</references>

File:Q10plot4wxy0119.png

2018-05-18T15:50:06Z

Xw2816:

MRD:WXY0119

2018-05-18T15:49:51Z

Xw2816: /* Energy Distribution and Reactivity */

= Molecular Reaction Dynamics Report =

== EXERCISE 1ː H + H2 SYSTEM ==

[[File:HandH2wxy0119.png]]

In the H + H2 system, the distance between the first two H atoms is r1 and the distance (bond length) between the two H atoms in the H2 molecule is r2.

=== Dynamics from the transition state region ===

{{fontcolor|red|Question 1 ː What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.}}

{| class="wikitable" border=1
|[[File:Q1TStransitionwxy0119.png|390px]]
|[[File:Q1TS2wxy0119.png|390px]]
|[[File:Q1TS3wxy0119.png|390px]]
|-
|Figure.1 Surface plot of reaction trajectory
|Figure.2 Transition state surface plot
|Figure.3 Transition state surface plot (different angle)
|}

In the reaction trajectory surface plot (Fig.1), AB is the distance r1 and BC is the distance r2. The gradient of the potential energy with regard to r1 and r2 are ∂V(r1)/∂r1 and ∂V(r2)/∂r2 respectively. The second derivatives of the potential energy with regard to r1 and r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 respectively.

At the two minimum structures which are at the two ends of the reaction pathway, ∂V(r1)/∂r1 = 0 and ∂V(r2)/∂r2 = 0 respectively. Since the two minimum structures are at two local minimum point , the second derivatives r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 are both greater than zero.

On the other hand, the components at the transition state structure (Fig.2) are different from that at minimum points. The gradient ∂V(r1)/∂r1 and ∂V(r2)/∂r2 are both equal to zero at the transition structure. However, r2 are ∂2V(r1)/∂r12 >0 and ∂2V(r2)/∂r22 <0. The transition state point is actually a saddle point which is observed more clearly in Fig.3. The Transition state point is thus distinguished from the minimum points as the potential energy surface curves inwards and downwards at the minimum points.

=== Locating the Transition State ===

{{fontcolor|red|Question 2ː 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.}}

{| class="wikitable" border=1
|[[File:TScontourwxy0119.png|390px]]
|[[File:TSsurfaceplwxy0119.png|390px]]
|[[File:TSestimatewxy0119.png|390px]]
|-
|Figure.4 Transition state contour plot
|Figure.5 Transition state surface plot
|Figure.6 Internuclear distance against time plot
|}

The best estimate of the transition state position is rts = 0.9078 Å.

When r1 = r2 = rts and momenta are set to zero, a graph of intermolecular distance against time is plotted. (Fig.6) In this graph, AB and BC lines are the same and overlap, the two lines on the graph are flat and horizontal, indicating that the atomic distances are the same at the position and with no momenta the state is at equilibrium, validating the position is the transition state position. The transition state is shown as a cross on the counter plot (Fig.4) and a dot on the surface plot (Fig.5).

=== Calculating and comparing the reaction path and trajectory ===

{{fontcolor1|red|Question 3ː Comment on how the ''mep'' and the trajectory you just calculated differ.}}

{| class="wikitable" border=1
|[[File:Mep1wxy0119.png|390px]]
|[[File:Mep2wxy0119.png|390px]]
|[[File:Mep3wxy0119.png|390px]]
|-
|Figure.7 mep calculation contour plot
|Figure.8 mep calculation surface plot
|Figure.9 mep calculation internuclear distance against time plot
|-
|[[File:Dynamic1wxy0119.png|390px]]
|[[File:Dynamic2wxy0119.png|390px]]
|[[File:Dynamic3wxy0119.png|390px]]
|-
|Figure.10 Dynamics calculation contour plot
|Figure.11 Dynamics calculation surface plot
|Figure.12 Dynamics calculation internuclear distance against time plot
|}
{| class="wikitable" border=1
|[[File:dynamics4wxy0119.png|600px]]
|[[File:Mep5wxy0119.png|600px]]
|-
|Figure.13 Dynamics calculation internuclear momenta against time plot
|Figure.14 mep calculation internuclear momenta against time plotsurface plot
|}

The mep (minimum energy path) trajectory is a smooth line on contour and surface plot (Fig.7 Fig.8) while the trajectory under dynamics calculation is oscillating along the pathway. (Fig.10 Fig.11) The mep trajectory takes more steps and longer time than the dynamics trajectory as shown in internuclear distance against time plots. (Fig. 9 Fig.12) The momentum is zero all the time under mep calculation (Fig.13) but under dynamics calculation, the momenta lines are oscillating against time.

The differences arise from mep and dynamics calculations can be explained by the the natures of the two calculations. Under mep calculation, the atoms are in extremely slow motion and the momentum is zero for each step, i.e. the atoms are "stop" after every step. The pathway under mep calculation is formed by connecting all the minimum energy points of each step and appears as a smooth line. On the other hand, atoms are in continuous motion and the momentum is accumulated after each step, atoms are then oscillating on the potential energy surface along the pathway. The steps are set to 5000 in the mep calculation but are only 500 in the dynamics calculatoin. Since the steps in mep are small, more time is needed to complete the same length on the trajectories.

=== Reactive and unreactive trajectories ===

{{fontcolor1|red|Question 4ː Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.}}

For the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, run trajectories with the following momenta p1 and p2 combinations:
{| class="wikitable" border=1
| Condition
| p1
| p2
| Total Energy/ kcal mol -1
| Reactivity
|-
|1.
| -1.25
| -2.5
| -99.018
| reactive
|-
|2.
| -1.5
| -2.0
| -100.456
| unreactive
|-
|3.
| -1.5
| -2.5
| -98.956
| reactive
|-
|4.
| -2.5
| -5.0
| -84.956
| unreactive
|-
|5.
| -2.5
| -5.2
| -83.416
| reactive
|}

{| class="wikitable" border=1
| Condition
| Surface Plot
| Contour Plot
| Description
|-
|1.
| [[File:condition1wxy0119.png|350px]]
| [[File:firstwxy0119.png|350px]]
| r2 (BC) decreases when HC approaches bonded HA and HB. The energy processed by the system is sufficient to overcome the activation barrier and crosses the transition state structure to break the HA and HB bond and form a new HB and HC bond. The new bond oscillate as r1 (AB) increases.
|-
|2.
| [[File:condition2wxy0119.png|350px]]
| [[File:secondwxy0119.png|350px]]
| HC approaches HB but the energy is insufficient to reach the transition state point, HC then moves further away from the bonded H2 molecule and no new bond is formed. The oscillation along r2 is a result of an increase in momentum p1.
|-
|3.
| [[File:condition3wxy0119.png|350px]]
| [[File:thirdwxy0119.png|350px]]
| Similarly to condition 1 but with a more negative p1, HA - HB bond oscillates more when r2 decreases. The reaction proceeds and crosses the transition state point to form the products.
|-
|4.
| [[File:condition4wxy0119.png|350px]]
| [[File:fourthwxy0119.png|350px]]
| Both p1 and p2 are more negative than the previous conditions. The HA - HB bond oscillates vigorously with larger amplitudes. The barrier recrossing happens after the system crosses the transition state region but eventually reverts back to the reactants.
|-
|5.
| [[File:condition5wxy0119.png|350px]]
| [[File:fifthwxy0119.png|350px]]
| With a slight more negative value of p1 as compared to that in condition 4, the system now processes appropriate energy to proceed to forming products. The vigorous oscillations are resulting from high momenta and barrier recrossing occurs.
|}

=== Transition State Theory ===

{{fontcolor|red|Question 5ː State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}

The main assumptions of Transition State Theory (TST) are<ref name="TS" />ː

1. The behaviours of the atomic nuclei follow the classical mechanics rules, i.e. the transition structure is formed as long as atoms or molecules collide with enough energy.

2. The intermediates have a long enough lifetime for the energies to be Boltzmann distributed preceding the next step.

3. The lowest energy saddle point on the potential energy surface is passed over in the reaction system.

The TST does not take into account of the quantum tunnelling effect which allows reactions to happen even when the energy carried by the system is lower than the activation energy barrier, especially for reactions with low energy barriers. Nor does the theory consider the intermediates with short lifetimes. When the energies are not fully distributed, the momentum of the reaction trajectory from the reactants to the intermediate can have effects on product selectivity. Moreover, the theory fails at high temperature because higher vibrational energy modes can be populated for a molecule at high temperature. The molecules can have complex motion and their collisions may result in a higher energy transition structure. The experimental results may thus deviate from that predicted by the TST. For example in condition 4, the kinetic energy of the reactants is much higher than the activation energy required. However, the momenta between atoms are very high and higher vibrational modes are populated that result in higher energy transition structures formed and the system does not pass though the lowest transition saddle point as shown in the diagram. Barrier recrossing occurs but the products are not formed as opposed to the predictions by the TST.

== EXERCISE 2: F - H - H SYSTEM ==

=== PES inspection ===

=== Reaction Energetics ===

{{fontcolor1|red|Question 6ː 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?}}

{| class="wikitable" border=1
|[[File:Q6FH2wxy0119.png|600px]]
|[[File:Q6HFHwxy0119.png|600px]]
|-
|Figure.15 F + H2 position on surface plot
|Figure.16 HF and H position on surface plot
|}

The surface plots of the potential energy surface of H + HF → F + H2 reaction is shown (Fig.15 Fig.16). Distance AB is the distance between H and F atoms and distance BC is the distance between two H atoms. The two minimum points shown in the graph are positions of F + H2 (Fig.15) and H + HF (Fig.16) respectively. If forward reaction is H + HF → F + H2, then the backward reaction is F + H2 → H + HF. As clearly shown in the diagram F + H2 are at a minimum point of higher potential energy than that of H + HF. Thus, by comparing the potential energy of reactants and products, the signs of changes in enthalpy (ΔH) and the energetics are determined. H + HF → F + H2 is endothermic and F + H2 → H + HF is exothermic.

The energetics of the two reactions are in accordance to the predictions by comparing bond strengths of chemical species. The bond energy of H-F bond (565 kJ/mol) is higher than the bond energy of H-H bond (432 kJ/mol). Thus, energy released by forming the H-H bond is not enough to compensate for energy required for breaking the H-F bond and the reaction H + HF → F + H2 is endothermic. On the contrary, the reaction F + H2 → H + HF is exothermic.

=== Transition State Approximation ===

{{fontcolor1|red|Question 7ː Locate the approximate position of the transition state.}}

{| class="wikitable" border=1
|[[File:Q7TStransitionwxy0119.png|390px]]
|[[File:Q7TS2wxy0119.png|390px]]
|[[File:Q7TS3wxy0119.png|390px]]
|-
|Figure.17 Transition state surface plot
|Figure.18 Transition state contour plot
|Figure.19 Internuclear distance against time plot (at TS point)
|}

The transition state (TS) is shown as a black dot on the potential energy surface plot (Fig.17) and a red cross on the contour plot. (Fig,18) The position of the TS is at the point where the distance between F and H (rFH) is 1.810Å and the distance between two H atoms (rHH) is 0.746Å. At the transition state position, the distance between F and H, between two H atoms are constant with no momentum as shown in the internuclear distance against time plot as horizontal and flat lines, validating the position is the transition state point (Fig.19)

=== Activation Energies ===

{{fontcolor1|red|Question 8ː Report the activation energy for both reactions.}}

{| class="wikitable" border=1
|[[File:Q8Ea2wxy0119.png|600px]]
|[[File:Q8Ea1wxy0119.png|600px]]
|-
|Figure.21 Energy against time plot (HF + H)
|Figure.20 Energy against time plot (F + H2)
|}

By performing MEP calculation with slight increase and decrease of the rFH to 1.820Å to perform F + H2 → H + HF reaction and to 1.800Å to perform H + HF → F + H2 reaction, the activation energies (Ea) of both reactions can be calculated from the differences in their potential energies (Fig.20 Fig.21)ː

F + H2 → H + HFː Ea = -103.751 - (-133.624) = +29.873 kcal mol-1

H + HF → F + H2ː Ea = -103.751 - (-103.972) = + 0.221 kcal mol-1

=== Reaction Dynamics ===

{{fontcolor1|red|Question 9ː In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

Reactionː F + H2 → H + HF

Initial condition setː rHF = 2Å rHH = 0.74Å pHF = -0.5 pHH = 0.1

{| class="wikitable" border=1
|[[File:Q9plot1wxy0119.png|390px]]
|[[File:Q9plot2wxy0119.png|390px]]
|[[File:Q9plot3wxy0119.png|390px]]
|-
|Figure.23 Contour plot
|Figure.24 Surface plot
|Figure.25 Internuclear momentum against time plot
|-
|[[File:Q9ani1wxy0119.png|390px]]
|[[File:Q9ani2wxy0119.png|390px]]
|-
|Figure.26 Animation figure (at the start of the reaction)
|Figure.27 Animation figure (at the end of the reaction)
|}

As shown in the plots and animation figures, the H-H bond in the reactants is broken and H-F bond in the products is formed. There are small momenta between both H and F and H-H bond that provide the initial kinetic energy of the reactants and the initial vibrational energy of the H-H bond in the reactants. The initial energy is required to break the H-H bond and reduce the distance of H and F for bond forming and potential energy is released when H-F bond is formed. Since energy is conserved, the reaction energy released is converted to the vibrational energy of the H-F bond and the translational kinetic energies of the products. This is shown in the internuclear momentum against time graph as the initial vibrations of B-C (H-H bond) becomes flat as two H atoms are separated in the product and the A-B (H-F bond) formed in the product vibrates with a large amplitude. (Fig.25) The contour plot and the surface plot also clearly show there is a significant increase in the amplitude of the oscillation of H-F bond which is an indication of the large vibrational energy of H-F bond. (Fig.23 Fig.24)

The predictions can be proved experimentally. The vibrational energy of H-F bond can be determined by measuring the frequency and the intensity of the absorption band of the H-F bond with infrared spectroscopy. The conversion to the kinetic energies of the products can be observed by measuring the temperature of the reaction as the kinetic energy can be converted to thermal energy as products collide with solvent molecules. The temperature is expected to increase as this is an exothermic reaction.

=== Energy Distribution and Reactivity ===

{{fontcolor|red|Question 10ː 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.}}
{| class="wikitable" border=1
| Condition
| rFH /Å
| rHH /Å
| pFH
| pHH
|-
|1.
| 2
| 0.74
| -0.5
| ?
|-
|2.
| 2
| 0.74
| -0.5
| unreactive
|-
|3.
| 2
| 0.74
| -0.5
| reactive
|-
|4.
| 2
| 0.74
| -0.8
| 0.1
|-
|5.
| 0.91
| 2
| 0.05
| -20
|-
|6.
| 0.91
| 2
| 0.8
| -7.5
|}
{| class="wikitable" border=1
|Condition
|Contour plot
|Observation
|-
|1.
|[[File:Q10plot1wxy0119.png|500px]]
|
|-
|2.
|[[File:Q10plot2wxy0119.png|600px]]
|
|-
|3.
|[[File:Q10plot3wxy0119.png|600px]]
|
|-
|4.
|[[File:Q10plot4wxy0119.png|600px]]
|
|-
|5.
|[[File:Q10plot5wxy0119.png|600px]]
|
|-
|6.
|[[File:Q10plot6wxy0119.png|600px]]
|
|}

pHH = -8.0 is still very large. This complies with the high activation energy of this reaction as the products formed are thermodynamically less stable than the reactants. It can be concluded that an increase in pHF, which corresponds to a high vibrational energy in H-F coordinate, is necessary for the reaction to take place. pHH, which corresponds to the translational energy in the H-H coordinate that is always very large, does not contribute too much to affect the reactivity of the reaction.

The transition state is present in the exit valley, hence, a late barrier for this endothermic reaction. The reactivity of endothermic reactions is thus dominated by the vibrational energy of the system from the results above.

For substantial exothermic reactions, such as F + H2, the transition state is located in the entrance valley, corresponds to an early barrier of the PES. For substantially endothermic reactions, such as H + HF, the transition state is in the exit valley, corresponds to a late barrier. The favoured degree of freedom for barrier crossing in exothermic reactions would be translation. This means that the momentum of the approaching atom and the atom which it is going to collide with has a greater impact on the rate of the reaction (eg. F atom in F + H2 reaction). However, reagent vibration, which is related to the momentum of the two bonded atoms (eg. HF in H + HF reaction) in the colliding molecule, would be most effective in enabling endothermic reactions to take place.

One reactive trajectory has the conditions of rFH = 0.91 Å, rHH = 2.0 Å, pHF = -0.5 and pHH = -10. The trajectory recrosses the transition state region twice and eventually moves into exit channel, forming products. By comparing this condition to the initial condition, the momentum pHH, which is the translational kinetic energy of the H atom, is reduced by a half; and the pHF, the vibrational energy of the H-F bond is increased to -0.5. This indicates that the vibrational energy has a greater contribution to the feasibility of the reaction than the translational energy. This is the opposite situation of the H2+F reaction.

The results agree with the Polanyi's empirical rules[2], which state that the vibrational energy is more efficient in promoting late-transition-state reactions and the translational energy is more efficient in promoting early-transition-state reactions.

When the momentum pFH is increased slightly from -0.5 to -0.8 with pHH being only 0.1, the reaction now becomes successful. It indicates that the F-H vibration energy has a larger contribution for the feasibility of this reaction compared to the translational energy which is the kinetic energy of the hydrogen atom which is defined by the momentum pHH

When the pHH has a large value above the activation energy (30.231 kcal/mol), which was set to be -20. The hydrogen atom collides with the HF molecule and breaks the H-F bond. However, two hydrogen atoms moves apart and do not form a H2 molecule. The procedure described is illustrated in the following animation snapshots. After the collision, the large momentum that the reactant hydrogen atom contains is still large.As a result, even if an H-H bond can be formed, it would immediately be broken due to an excess vibrational energy between two atoms.

for investigation, 1st set condition pFH = -0.5 pHH = -3 unreactive; 2nd pHH = -2.45 reactive; 3rd pHH = -1.1 unreactive; 4th pHH = -0.9 reactive; 5th pHH = 1.1 unreactive ; 6th pHH = 1.9 reactive ; 7th pHH = 2.4 to 3 unreactive
; pHH = 2.1 no reaction

The cases studied are an illustration of Polanyi's empirical rules.

For H and HF

failed conditionsː 091, 2, 0.05, -20

initial condition of H HF systemː
HF distance = 0.91 HH distance = 2 pHF = 0.8 pHH = -7.5

== Rreferences ==

<references>
<ref name="TS">Eyring, H. (1935). "The Activated Complex in Chemical Reactions". J. Chem. Phys. 3 (2): 107–115.</ref>
</references>

File:Q10plot6wxy0119.png

2018-05-18T15:48:38Z

Xw2816:

File:Q10plot5wxy0119.png

2018-05-18T15:47:38Z

Xw2816:

MRD:WXY0119

2018-05-18T15:46:48Z

Xw2816: /* Energy Distribution and Reactivity */

= Molecular Reaction Dynamics Report =

== EXERCISE 1ː H + H2 SYSTEM ==

[[File:HandH2wxy0119.png]]

In the H + H2 system, the distance between the first two H atoms is r1 and the distance (bond length) between the two H atoms in the H2 molecule is r2.

=== Dynamics from the transition state region ===

{{fontcolor|red|Question 1 ː What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.}}

{| class="wikitable" border=1
|[[File:Q1TStransitionwxy0119.png|390px]]
|[[File:Q1TS2wxy0119.png|390px]]
|[[File:Q1TS3wxy0119.png|390px]]
|-
|Figure.1 Surface plot of reaction trajectory
|Figure.2 Transition state surface plot
|Figure.3 Transition state surface plot (different angle)
|}

In the reaction trajectory surface plot (Fig.1), AB is the distance r1 and BC is the distance r2. The gradient of the potential energy with regard to r1 and r2 are ∂V(r1)/∂r1 and ∂V(r2)/∂r2 respectively. The second derivatives of the potential energy with regard to r1 and r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 respectively.

At the two minimum structures which are at the two ends of the reaction pathway, ∂V(r1)/∂r1 = 0 and ∂V(r2)/∂r2 = 0 respectively. Since the two minimum structures are at two local minimum point , the second derivatives r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 are both greater than zero.

On the other hand, the components at the transition state structure (Fig.2) are different from that at minimum points. The gradient ∂V(r1)/∂r1 and ∂V(r2)/∂r2 are both equal to zero at the transition structure. However, r2 are ∂2V(r1)/∂r12 >0 and ∂2V(r2)/∂r22 <0. The transition state point is actually a saddle point which is observed more clearly in Fig.3. The Transition state point is thus distinguished from the minimum points as the potential energy surface curves inwards and downwards at the minimum points.

=== Locating the Transition State ===

{{fontcolor|red|Question 2ː 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.}}

{| class="wikitable" border=1
|[[File:TScontourwxy0119.png|390px]]
|[[File:TSsurfaceplwxy0119.png|390px]]
|[[File:TSestimatewxy0119.png|390px]]
|-
|Figure.4 Transition state contour plot
|Figure.5 Transition state surface plot
|Figure.6 Internuclear distance against time plot
|}

The best estimate of the transition state position is rts = 0.9078 Å.

When r1 = r2 = rts and momenta are set to zero, a graph of intermolecular distance against time is plotted. (Fig.6) In this graph, AB and BC lines are the same and overlap, the two lines on the graph are flat and horizontal, indicating that the atomic distances are the same at the position and with no momenta the state is at equilibrium, validating the position is the transition state position. The transition state is shown as a cross on the counter plot (Fig.4) and a dot on the surface plot (Fig.5).

=== Calculating and comparing the reaction path and trajectory ===

{{fontcolor1|red|Question 3ː Comment on how the ''mep'' and the trajectory you just calculated differ.}}

{| class="wikitable" border=1
|[[File:Mep1wxy0119.png|390px]]
|[[File:Mep2wxy0119.png|390px]]
|[[File:Mep3wxy0119.png|390px]]
|-
|Figure.7 mep calculation contour plot
|Figure.8 mep calculation surface plot
|Figure.9 mep calculation internuclear distance against time plot
|-
|[[File:Dynamic1wxy0119.png|390px]]
|[[File:Dynamic2wxy0119.png|390px]]
|[[File:Dynamic3wxy0119.png|390px]]
|-
|Figure.10 Dynamics calculation contour plot
|Figure.11 Dynamics calculation surface plot
|Figure.12 Dynamics calculation internuclear distance against time plot
|}
{| class="wikitable" border=1
|[[File:dynamics4wxy0119.png|600px]]
|[[File:Mep5wxy0119.png|600px]]
|-
|Figure.13 Dynamics calculation internuclear momenta against time plot
|Figure.14 mep calculation internuclear momenta against time plotsurface plot
|}

The mep (minimum energy path) trajectory is a smooth line on contour and surface plot (Fig.7 Fig.8) while the trajectory under dynamics calculation is oscillating along the pathway. (Fig.10 Fig.11) The mep trajectory takes more steps and longer time than the dynamics trajectory as shown in internuclear distance against time plots. (Fig. 9 Fig.12) The momentum is zero all the time under mep calculation (Fig.13) but under dynamics calculation, the momenta lines are oscillating against time.

The differences arise from mep and dynamics calculations can be explained by the the natures of the two calculations. Under mep calculation, the atoms are in extremely slow motion and the momentum is zero for each step, i.e. the atoms are "stop" after every step. The pathway under mep calculation is formed by connecting all the minimum energy points of each step and appears as a smooth line. On the other hand, atoms are in continuous motion and the momentum is accumulated after each step, atoms are then oscillating on the potential energy surface along the pathway. The steps are set to 5000 in the mep calculation but are only 500 in the dynamics calculatoin. Since the steps in mep are small, more time is needed to complete the same length on the trajectories.

=== Reactive and unreactive trajectories ===

{{fontcolor1|red|Question 4ː Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.}}

For the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, run trajectories with the following momenta p1 and p2 combinations:
{| class="wikitable" border=1
| Condition
| p1
| p2
| Total Energy/ kcal mol -1
| Reactivity
|-
|1.
| -1.25
| -2.5
| -99.018
| reactive
|-
|2.
| -1.5
| -2.0
| -100.456
| unreactive
|-
|3.
| -1.5
| -2.5
| -98.956
| reactive
|-
|4.
| -2.5
| -5.0
| -84.956
| unreactive
|-
|5.
| -2.5
| -5.2
| -83.416
| reactive
|}

{| class="wikitable" border=1
| Condition
| Surface Plot
| Contour Plot
| Description
|-
|1.
| [[File:condition1wxy0119.png|350px]]
| [[File:firstwxy0119.png|350px]]
| r2 (BC) decreases when HC approaches bonded HA and HB. The energy processed by the system is sufficient to overcome the activation barrier and crosses the transition state structure to break the HA and HB bond and form a new HB and HC bond. The new bond oscillate as r1 (AB) increases.
|-
|2.
| [[File:condition2wxy0119.png|350px]]
| [[File:secondwxy0119.png|350px]]
| HC approaches HB but the energy is insufficient to reach the transition state point, HC then moves further away from the bonded H2 molecule and no new bond is formed. The oscillation along r2 is a result of an increase in momentum p1.
|-
|3.
| [[File:condition3wxy0119.png|350px]]
| [[File:thirdwxy0119.png|350px]]
| Similarly to condition 1 but with a more negative p1, HA - HB bond oscillates more when r2 decreases. The reaction proceeds and crosses the transition state point to form the products.
|-
|4.
| [[File:condition4wxy0119.png|350px]]
| [[File:fourthwxy0119.png|350px]]
| Both p1 and p2 are more negative than the previous conditions. The HA - HB bond oscillates vigorously with larger amplitudes. The barrier recrossing happens after the system crosses the transition state region but eventually reverts back to the reactants.
|-
|5.
| [[File:condition5wxy0119.png|350px]]
| [[File:fifthwxy0119.png|350px]]
| With a slight more negative value of p1 as compared to that in condition 4, the system now processes appropriate energy to proceed to forming products. The vigorous oscillations are resulting from high momenta and barrier recrossing occurs.
|}

=== Transition State Theory ===

{{fontcolor|red|Question 5ː State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}

The main assumptions of Transition State Theory (TST) are<ref name="TS" />ː

1. The behaviours of the atomic nuclei follow the classical mechanics rules, i.e. the transition structure is formed as long as atoms or molecules collide with enough energy.

2. The intermediates have a long enough lifetime for the energies to be Boltzmann distributed preceding the next step.

3. The lowest energy saddle point on the potential energy surface is passed over in the reaction system.

The TST does not take into account of the quantum tunnelling effect which allows reactions to happen even when the energy carried by the system is lower than the activation energy barrier, especially for reactions with low energy barriers. Nor does the theory consider the intermediates with short lifetimes. When the energies are not fully distributed, the momentum of the reaction trajectory from the reactants to the intermediate can have effects on product selectivity. Moreover, the theory fails at high temperature because higher vibrational energy modes can be populated for a molecule at high temperature. The molecules can have complex motion and their collisions may result in a higher energy transition structure. The experimental results may thus deviate from that predicted by the TST. For example in condition 4, the kinetic energy of the reactants is much higher than the activation energy required. However, the momenta between atoms are very high and higher vibrational modes are populated that result in higher energy transition structures formed and the system does not pass though the lowest transition saddle point as shown in the diagram. Barrier recrossing occurs but the products are not formed as opposed to the predictions by the TST.

== EXERCISE 2: F - H - H SYSTEM ==

=== PES inspection ===

=== Reaction Energetics ===

{{fontcolor1|red|Question 6ː 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?}}

{| class="wikitable" border=1
|[[File:Q6FH2wxy0119.png|600px]]
|[[File:Q6HFHwxy0119.png|600px]]
|-
|Figure.15 F + H2 position on surface plot
|Figure.16 HF and H position on surface plot
|}

The surface plots of the potential energy surface of H + HF → F + H2 reaction is shown (Fig.15 Fig.16). Distance AB is the distance between H and F atoms and distance BC is the distance between two H atoms. The two minimum points shown in the graph are positions of F + H2 (Fig.15) and H + HF (Fig.16) respectively. If forward reaction is H + HF → F + H2, then the backward reaction is F + H2 → H + HF. As clearly shown in the diagram F + H2 are at a minimum point of higher potential energy than that of H + HF. Thus, by comparing the potential energy of reactants and products, the signs of changes in enthalpy (ΔH) and the energetics are determined. H + HF → F + H2 is endothermic and F + H2 → H + HF is exothermic.

The energetics of the two reactions are in accordance to the predictions by comparing bond strengths of chemical species. The bond energy of H-F bond (565 kJ/mol) is higher than the bond energy of H-H bond (432 kJ/mol). Thus, energy released by forming the H-H bond is not enough to compensate for energy required for breaking the H-F bond and the reaction H + HF → F + H2 is endothermic. On the contrary, the reaction F + H2 → H + HF is exothermic.

=== Transition State Approximation ===

{{fontcolor1|red|Question 7ː Locate the approximate position of the transition state.}}

{| class="wikitable" border=1
|[[File:Q7TStransitionwxy0119.png|390px]]
|[[File:Q7TS2wxy0119.png|390px]]
|[[File:Q7TS3wxy0119.png|390px]]
|-
|Figure.17 Transition state surface plot
|Figure.18 Transition state contour plot
|Figure.19 Internuclear distance against time plot (at TS point)
|}

The transition state (TS) is shown as a black dot on the potential energy surface plot (Fig.17) and a red cross on the contour plot. (Fig,18) The position of the TS is at the point where the distance between F and H (rFH) is 1.810Å and the distance between two H atoms (rHH) is 0.746Å. At the transition state position, the distance between F and H, between two H atoms are constant with no momentum as shown in the internuclear distance against time plot as horizontal and flat lines, validating the position is the transition state point (Fig.19)

=== Activation Energies ===

{{fontcolor1|red|Question 8ː Report the activation energy for both reactions.}}

{| class="wikitable" border=1
|[[File:Q8Ea2wxy0119.png|600px]]
|[[File:Q8Ea1wxy0119.png|600px]]
|-
|Figure.21 Energy against time plot (HF + H)
|Figure.20 Energy against time plot (F + H2)
|}

By performing MEP calculation with slight increase and decrease of the rFH to 1.820Å to perform F + H2 → H + HF reaction and to 1.800Å to perform H + HF → F + H2 reaction, the activation energies (Ea) of both reactions can be calculated from the differences in their potential energies (Fig.20 Fig.21)ː

F + H2 → H + HFː Ea = -103.751 - (-133.624) = +29.873 kcal mol-1

H + HF → F + H2ː Ea = -103.751 - (-103.972) = + 0.221 kcal mol-1

=== Reaction Dynamics ===

{{fontcolor1|red|Question 9ː In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

Reactionː F + H2 → H + HF

Initial condition setː rHF = 2Å rHH = 0.74Å pHF = -0.5 pHH = 0.1

{| class="wikitable" border=1
|[[File:Q9plot1wxy0119.png|390px]]
|[[File:Q9plot2wxy0119.png|390px]]
|[[File:Q9plot3wxy0119.png|390px]]
|-
|Figure.23 Contour plot
|Figure.24 Surface plot
|Figure.25 Internuclear momentum against time plot
|-
|[[File:Q9ani1wxy0119.png|390px]]
|[[File:Q9ani2wxy0119.png|390px]]
|-
|Figure.26 Animation figure (at the start of the reaction)
|Figure.27 Animation figure (at the end of the reaction)
|}

As shown in the plots and animation figures, the H-H bond in the reactants is broken and H-F bond in the products is formed. There are small momenta between both H and F and H-H bond that provide the initial kinetic energy of the reactants and the initial vibrational energy of the H-H bond in the reactants. The initial energy is required to break the H-H bond and reduce the distance of H and F for bond forming and potential energy is released when H-F bond is formed. Since energy is conserved, the reaction energy released is converted to the vibrational energy of the H-F bond and the translational kinetic energies of the products. This is shown in the internuclear momentum against time graph as the initial vibrations of B-C (H-H bond) becomes flat as two H atoms are separated in the product and the A-B (H-F bond) formed in the product vibrates with a large amplitude. (Fig.25) The contour plot and the surface plot also clearly show there is a significant increase in the amplitude of the oscillation of H-F bond which is an indication of the large vibrational energy of H-F bond. (Fig.23 Fig.24)

The predictions can be proved experimentally. The vibrational energy of H-F bond can be determined by measuring the frequency and the intensity of the absorption band of the H-F bond with infrared spectroscopy. The conversion to the kinetic energies of the products can be observed by measuring the temperature of the reaction as the kinetic energy can be converted to thermal energy as products collide with solvent molecules. The temperature is expected to increase as this is an exothermic reaction.

=== Energy Distribution and Reactivity ===

{{fontcolor|red|Question 10ː 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.}}
{| class="wikitable" border=1
| Condition
| rFH /Å
| rHH /Å
| pFH
| pHH
|-
|1.
| 2
| 0.74
| -0.5
| ?
|-
|2.
| 2
| 0.74
| -0.5
| unreactive
|-
|3.
| 2
| 0.74
| -0.5
| reactive
|-
|4.
| 2
| 0.74
| -0.8
| 0.1
|-
|5.
| 0.91
| 2
| 0.05
| -20
|-
|6.
| 0.91
| 2
| 0.8
| -7.5
|}
{| class="wikitable" border=1
|Condition
|Contour plot
|Observation
|-
|1.
|[[File:Q10plot1wxy0119.png|500px]]
|
|-
|2.
|[[File:Q10plot2wxy0119.png|500px]]
|
|-
|3.
|[[File:Q10plot3wxy0119.png|500px]]
|
|-
|4.
|[[File:Q10plot4wxy0119.png|500px]]
|
|-
|5.
|[[File:Q10plot5wxy0119.png|500px]]
|
|-
|6.
|[[File:Q10plot6wxy0119.png|500px]]
|
|}

pHH = -8.0 is still very large. This complies with the high activation energy of this reaction as the products formed are thermodynamically less stable than the reactants. It can be concluded that an increase in pHF, which corresponds to a high vibrational energy in H-F coordinate, is necessary for the reaction to take place. pHH, which corresponds to the translational energy in the H-H coordinate that is always very large, does not contribute too much to affect the reactivity of the reaction.

The transition state is present in the exit valley, hence, a late barrier for this endothermic reaction. The reactivity of endothermic reactions is thus dominated by the vibrational energy of the system from the results above.

For substantial exothermic reactions, such as F + H2, the transition state is located in the entrance valley, corresponds to an early barrier of the PES. For substantially endothermic reactions, such as H + HF, the transition state is in the exit valley, corresponds to a late barrier. The favoured degree of freedom for barrier crossing in exothermic reactions would be translation. This means that the momentum of the approaching atom and the atom which it is going to collide with has a greater impact on the rate of the reaction (eg. F atom in F + H2 reaction). However, reagent vibration, which is related to the momentum of the two bonded atoms (eg. HF in H + HF reaction) in the colliding molecule, would be most effective in enabling endothermic reactions to take place.

One reactive trajectory has the conditions of rFH = 0.91 Å, rHH = 2.0 Å, pHF = -0.5 and pHH = -10. The trajectory recrosses the transition state region twice and eventually moves into exit channel, forming products. By comparing this condition to the initial condition, the momentum pHH, which is the translational kinetic energy of the H atom, is reduced by a half; and the pHF, the vibrational energy of the H-F bond is increased to -0.5. This indicates that the vibrational energy has a greater contribution to the feasibility of the reaction than the translational energy. This is the opposite situation of the H2+F reaction.

The results agree with the Polanyi's empirical rules[2], which state that the vibrational energy is more efficient in promoting late-transition-state reactions and the translational energy is more efficient in promoting early-transition-state reactions.

When the momentum pFH is increased slightly from -0.5 to -0.8 with pHH being only 0.1, the reaction now becomes successful. It indicates that the F-H vibration energy has a larger contribution for the feasibility of this reaction compared to the translational energy which is the kinetic energy of the hydrogen atom which is defined by the momentum pHH

When the pHH has a large value above the activation energy (30.231 kcal/mol), which was set to be -20. The hydrogen atom collides with the HF molecule and breaks the H-F bond. However, two hydrogen atoms moves apart and do not form a H2 molecule. The procedure described is illustrated in the following animation snapshots. After the collision, the large momentum that the reactant hydrogen atom contains is still large.As a result, even if an H-H bond can be formed, it would immediately be broken due to an excess vibrational energy between two atoms.

for investigation, 1st set condition pFH = -0.5 pHH = -3 unreactive; 2nd pHH = -2.45 reactive; 3rd pHH = -1.1 unreactive; 4th pHH = -0.9 reactive; 5th pHH = 1.1 unreactive ; 6th pHH = 1.9 reactive ; 7th pHH = 2.4 to 3 unreactive
; pHH = 2.1 no reaction

The cases studied are an illustration of Polanyi's empirical rules.

For H and HF

failed conditionsː 091, 2, 0.05, -20

initial condition of H HF systemː
HF distance = 0.91 HH distance = 2 pHF = 0.8 pHH = -7.5

== Rreferences ==

<references>
<ref name="TS">Eyring, H. (1935). "The Activated Complex in Chemical Reactions". J. Chem. Phys. 3 (2): 107–115.</ref>
</references>

MRD:WXY0119

2018-05-18T15:46:14Z

Xw2816: /* Energy Distribution and Reactivity */

= Molecular Reaction Dynamics Report =

== EXERCISE 1ː H + H2 SYSTEM ==

[[File:HandH2wxy0119.png]]

In the H + H2 system, the distance between the first two H atoms is r1 and the distance (bond length) between the two H atoms in the H2 molecule is r2.

=== Dynamics from the transition state region ===

{{fontcolor|red|Question 1 ː What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.}}

{| class="wikitable" border=1
|[[File:Q1TStransitionwxy0119.png|390px]]
|[[File:Q1TS2wxy0119.png|390px]]
|[[File:Q1TS3wxy0119.png|390px]]
|-
|Figure.1 Surface plot of reaction trajectory
|Figure.2 Transition state surface plot
|Figure.3 Transition state surface plot (different angle)
|}

In the reaction trajectory surface plot (Fig.1), AB is the distance r1 and BC is the distance r2. The gradient of the potential energy with regard to r1 and r2 are ∂V(r1)/∂r1 and ∂V(r2)/∂r2 respectively. The second derivatives of the potential energy with regard to r1 and r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 respectively.

At the two minimum structures which are at the two ends of the reaction pathway, ∂V(r1)/∂r1 = 0 and ∂V(r2)/∂r2 = 0 respectively. Since the two minimum structures are at two local minimum point , the second derivatives r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 are both greater than zero.

On the other hand, the components at the transition state structure (Fig.2) are different from that at minimum points. The gradient ∂V(r1)/∂r1 and ∂V(r2)/∂r2 are both equal to zero at the transition structure. However, r2 are ∂2V(r1)/∂r12 >0 and ∂2V(r2)/∂r22 <0. The transition state point is actually a saddle point which is observed more clearly in Fig.3. The Transition state point is thus distinguished from the minimum points as the potential energy surface curves inwards and downwards at the minimum points.

=== Locating the Transition State ===

{{fontcolor|red|Question 2ː 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.}}

{| class="wikitable" border=1
|[[File:TScontourwxy0119.png|390px]]
|[[File:TSsurfaceplwxy0119.png|390px]]
|[[File:TSestimatewxy0119.png|390px]]
|-
|Figure.4 Transition state contour plot
|Figure.5 Transition state surface plot
|Figure.6 Internuclear distance against time plot
|}

The best estimate of the transition state position is rts = 0.9078 Å.

When r1 = r2 = rts and momenta are set to zero, a graph of intermolecular distance against time is plotted. (Fig.6) In this graph, AB and BC lines are the same and overlap, the two lines on the graph are flat and horizontal, indicating that the atomic distances are the same at the position and with no momenta the state is at equilibrium, validating the position is the transition state position. The transition state is shown as a cross on the counter plot (Fig.4) and a dot on the surface plot (Fig.5).

=== Calculating and comparing the reaction path and trajectory ===

{{fontcolor1|red|Question 3ː Comment on how the ''mep'' and the trajectory you just calculated differ.}}

{| class="wikitable" border=1
|[[File:Mep1wxy0119.png|390px]]
|[[File:Mep2wxy0119.png|390px]]
|[[File:Mep3wxy0119.png|390px]]
|-
|Figure.7 mep calculation contour plot
|Figure.8 mep calculation surface plot
|Figure.9 mep calculation internuclear distance against time plot
|-
|[[File:Dynamic1wxy0119.png|390px]]
|[[File:Dynamic2wxy0119.png|390px]]
|[[File:Dynamic3wxy0119.png|390px]]
|-
|Figure.10 Dynamics calculation contour plot
|Figure.11 Dynamics calculation surface plot
|Figure.12 Dynamics calculation internuclear distance against time plot
|}
{| class="wikitable" border=1
|[[File:dynamics4wxy0119.png|600px]]
|[[File:Mep5wxy0119.png|600px]]
|-
|Figure.13 Dynamics calculation internuclear momenta against time plot
|Figure.14 mep calculation internuclear momenta against time plotsurface plot
|}

The mep (minimum energy path) trajectory is a smooth line on contour and surface plot (Fig.7 Fig.8) while the trajectory under dynamics calculation is oscillating along the pathway. (Fig.10 Fig.11) The mep trajectory takes more steps and longer time than the dynamics trajectory as shown in internuclear distance against time plots. (Fig. 9 Fig.12) The momentum is zero all the time under mep calculation (Fig.13) but under dynamics calculation, the momenta lines are oscillating against time.

The differences arise from mep and dynamics calculations can be explained by the the natures of the two calculations. Under mep calculation, the atoms are in extremely slow motion and the momentum is zero for each step, i.e. the atoms are "stop" after every step. The pathway under mep calculation is formed by connecting all the minimum energy points of each step and appears as a smooth line. On the other hand, atoms are in continuous motion and the momentum is accumulated after each step, atoms are then oscillating on the potential energy surface along the pathway. The steps are set to 5000 in the mep calculation but are only 500 in the dynamics calculatoin. Since the steps in mep are small, more time is needed to complete the same length on the trajectories.

=== Reactive and unreactive trajectories ===

{{fontcolor1|red|Question 4ː Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.}}

For the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, run trajectories with the following momenta p1 and p2 combinations:
{| class="wikitable" border=1
| Condition
| p1
| p2
| Total Energy/ kcal mol -1
| Reactivity
|-
|1.
| -1.25
| -2.5
| -99.018
| reactive
|-
|2.
| -1.5
| -2.0
| -100.456
| unreactive
|-
|3.
| -1.5
| -2.5
| -98.956
| reactive
|-
|4.
| -2.5
| -5.0
| -84.956
| unreactive
|-
|5.
| -2.5
| -5.2
| -83.416
| reactive
|}

{| class="wikitable" border=1
| Condition
| Surface Plot
| Contour Plot
| Description
|-
|1.
| [[File:condition1wxy0119.png|350px]]
| [[File:firstwxy0119.png|350px]]
| r2 (BC) decreases when HC approaches bonded HA and HB. The energy processed by the system is sufficient to overcome the activation barrier and crosses the transition state structure to break the HA and HB bond and form a new HB and HC bond. The new bond oscillate as r1 (AB) increases.
|-
|2.
| [[File:condition2wxy0119.png|350px]]
| [[File:secondwxy0119.png|350px]]
| HC approaches HB but the energy is insufficient to reach the transition state point, HC then moves further away from the bonded H2 molecule and no new bond is formed. The oscillation along r2 is a result of an increase in momentum p1.
|-
|3.
| [[File:condition3wxy0119.png|350px]]
| [[File:thirdwxy0119.png|350px]]
| Similarly to condition 1 but with a more negative p1, HA - HB bond oscillates more when r2 decreases. The reaction proceeds and crosses the transition state point to form the products.
|-
|4.
| [[File:condition4wxy0119.png|350px]]
| [[File:fourthwxy0119.png|350px]]
| Both p1 and p2 are more negative than the previous conditions. The HA - HB bond oscillates vigorously with larger amplitudes. The barrier recrossing happens after the system crosses the transition state region but eventually reverts back to the reactants.
|-
|5.
| [[File:condition5wxy0119.png|350px]]
| [[File:fifthwxy0119.png|350px]]
| With a slight more negative value of p1 as compared to that in condition 4, the system now processes appropriate energy to proceed to forming products. The vigorous oscillations are resulting from high momenta and barrier recrossing occurs.
|}

=== Transition State Theory ===

{{fontcolor|red|Question 5ː State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}

The main assumptions of Transition State Theory (TST) are<ref name="TS" />ː

1. The behaviours of the atomic nuclei follow the classical mechanics rules, i.e. the transition structure is formed as long as atoms or molecules collide with enough energy.

2. The intermediates have a long enough lifetime for the energies to be Boltzmann distributed preceding the next step.

3. The lowest energy saddle point on the potential energy surface is passed over in the reaction system.

The TST does not take into account of the quantum tunnelling effect which allows reactions to happen even when the energy carried by the system is lower than the activation energy barrier, especially for reactions with low energy barriers. Nor does the theory consider the intermediates with short lifetimes. When the energies are not fully distributed, the momentum of the reaction trajectory from the reactants to the intermediate can have effects on product selectivity. Moreover, the theory fails at high temperature because higher vibrational energy modes can be populated for a molecule at high temperature. The molecules can have complex motion and their collisions may result in a higher energy transition structure. The experimental results may thus deviate from that predicted by the TST. For example in condition 4, the kinetic energy of the reactants is much higher than the activation energy required. However, the momenta between atoms are very high and higher vibrational modes are populated that result in higher energy transition structures formed and the system does not pass though the lowest transition saddle point as shown in the diagram. Barrier recrossing occurs but the products are not formed as opposed to the predictions by the TST.

== EXERCISE 2: F - H - H SYSTEM ==

=== PES inspection ===

=== Reaction Energetics ===

{{fontcolor1|red|Question 6ː 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?}}

{| class="wikitable" border=1
|[[File:Q6FH2wxy0119.png|600px]]
|[[File:Q6HFHwxy0119.png|600px]]
|-
|Figure.15 F + H2 position on surface plot
|Figure.16 HF and H position on surface plot
|}

The surface plots of the potential energy surface of H + HF → F + H2 reaction is shown (Fig.15 Fig.16). Distance AB is the distance between H and F atoms and distance BC is the distance between two H atoms. The two minimum points shown in the graph are positions of F + H2 (Fig.15) and H + HF (Fig.16) respectively. If forward reaction is H + HF → F + H2, then the backward reaction is F + H2 → H + HF. As clearly shown in the diagram F + H2 are at a minimum point of higher potential energy than that of H + HF. Thus, by comparing the potential energy of reactants and products, the signs of changes in enthalpy (ΔH) and the energetics are determined. H + HF → F + H2 is endothermic and F + H2 → H + HF is exothermic.

The energetics of the two reactions are in accordance to the predictions by comparing bond strengths of chemical species. The bond energy of H-F bond (565 kJ/mol) is higher than the bond energy of H-H bond (432 kJ/mol). Thus, energy released by forming the H-H bond is not enough to compensate for energy required for breaking the H-F bond and the reaction H + HF → F + H2 is endothermic. On the contrary, the reaction F + H2 → H + HF is exothermic.

=== Transition State Approximation ===

{{fontcolor1|red|Question 7ː Locate the approximate position of the transition state.}}

{| class="wikitable" border=1
|[[File:Q7TStransitionwxy0119.png|390px]]
|[[File:Q7TS2wxy0119.png|390px]]
|[[File:Q7TS3wxy0119.png|390px]]
|-
|Figure.17 Transition state surface plot
|Figure.18 Transition state contour plot
|Figure.19 Internuclear distance against time plot (at TS point)
|}

The transition state (TS) is shown as a black dot on the potential energy surface plot (Fig.17) and a red cross on the contour plot. (Fig,18) The position of the TS is at the point where the distance between F and H (rFH) is 1.810Å and the distance between two H atoms (rHH) is 0.746Å. At the transition state position, the distance between F and H, between two H atoms are constant with no momentum as shown in the internuclear distance against time plot as horizontal and flat lines, validating the position is the transition state point (Fig.19)

=== Activation Energies ===

{{fontcolor1|red|Question 8ː Report the activation energy for both reactions.}}

{| class="wikitable" border=1
|[[File:Q8Ea2wxy0119.png|600px]]
|[[File:Q8Ea1wxy0119.png|600px]]
|-
|Figure.21 Energy against time plot (HF + H)
|Figure.20 Energy against time plot (F + H2)
|}

By performing MEP calculation with slight increase and decrease of the rFH to 1.820Å to perform F + H2 → H + HF reaction and to 1.800Å to perform H + HF → F + H2 reaction, the activation energies (Ea) of both reactions can be calculated from the differences in their potential energies (Fig.20 Fig.21)ː

F + H2 → H + HFː Ea = -103.751 - (-133.624) = +29.873 kcal mol-1

H + HF → F + H2ː Ea = -103.751 - (-103.972) = + 0.221 kcal mol-1

=== Reaction Dynamics ===

{{fontcolor1|red|Question 9ː In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

Reactionː F + H2 → H + HF

Initial condition setː rHF = 2Å rHH = 0.74Å pHF = -0.5 pHH = 0.1

{| class="wikitable" border=1
|[[File:Q9plot1wxy0119.png|390px]]
|[[File:Q9plot2wxy0119.png|390px]]
|[[File:Q9plot3wxy0119.png|390px]]
|-
|Figure.23 Contour plot
|Figure.24 Surface plot
|Figure.25 Internuclear momentum against time plot
|-
|[[File:Q9ani1wxy0119.png|390px]]
|[[File:Q9ani2wxy0119.png|390px]]
|-
|Figure.26 Animation figure (at the start of the reaction)
|Figure.27 Animation figure (at the end of the reaction)
|}

As shown in the plots and animation figures, the H-H bond in the reactants is broken and H-F bond in the products is formed. There are small momenta between both H and F and H-H bond that provide the initial kinetic energy of the reactants and the initial vibrational energy of the H-H bond in the reactants. The initial energy is required to break the H-H bond and reduce the distance of H and F for bond forming and potential energy is released when H-F bond is formed. Since energy is conserved, the reaction energy released is converted to the vibrational energy of the H-F bond and the translational kinetic energies of the products. This is shown in the internuclear momentum against time graph as the initial vibrations of B-C (H-H bond) becomes flat as two H atoms are separated in the product and the A-B (H-F bond) formed in the product vibrates with a large amplitude. (Fig.25) The contour plot and the surface plot also clearly show there is a significant increase in the amplitude of the oscillation of H-F bond which is an indication of the large vibrational energy of H-F bond. (Fig.23 Fig.24)

The predictions can be proved experimentally. The vibrational energy of H-F bond can be determined by measuring the frequency and the intensity of the absorption band of the H-F bond with infrared spectroscopy. The conversion to the kinetic energies of the products can be observed by measuring the temperature of the reaction as the kinetic energy can be converted to thermal energy as products collide with solvent molecules. The temperature is expected to increase as this is an exothermic reaction.

=== Energy Distribution and Reactivity ===

{{fontcolor|red|Question 10ː 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.}}
{| class="wikitable" border=1
| Condition
| rFH /Å
| rHH /Å
| pFH
| pHH
|-
|1.
| 2
| 0.74
| -0.5
| ?
|-
|2.
| 2
| 0.74
| -0.5
| unreactive
|-
|3.
| 2
| 0.74
| -0.5
| reactive
|-
|4.
| 2
| 0.74
| -0.8
| 0.1
|-
|5.
| 0.91
| 2
| 0.05
| -20
|-
|6.
| 0.91
| 2
| 0.8
| -7.5
|}
{| class="wikitable" border=1
|Condition
|Contour plot
|Observation
|-
|1.
|[[File:Q10plot1wxy0119.png|500px]]
|
|-
|2.
|[[File:Q10plot1wxy0119.png|500px]]
|
|-
|3.
|[[File:Q10plot1wxy0119.png|500px]]
|
|-
|4.
|[[File:Q10plot1wxy0119.png|500px]]
|
|-
|5.
|[[File:Q10plot1wxy0119.png|500px]]
|
|-
|6.
|[[File:Q10plot1wxy0119.png|500px]]
|
|}

pHH = -8.0 is still very large. This complies with the high activation energy of this reaction as the products formed are thermodynamically less stable than the reactants. It can be concluded that an increase in pHF, which corresponds to a high vibrational energy in H-F coordinate, is necessary for the reaction to take place. pHH, which corresponds to the translational energy in the H-H coordinate that is always very large, does not contribute too much to affect the reactivity of the reaction.

The transition state is present in the exit valley, hence, a late barrier for this endothermic reaction. The reactivity of endothermic reactions is thus dominated by the vibrational energy of the system from the results above.

For substantial exothermic reactions, such as F + H2, the transition state is located in the entrance valley, corresponds to an early barrier of the PES. For substantially endothermic reactions, such as H + HF, the transition state is in the exit valley, corresponds to a late barrier. The favoured degree of freedom for barrier crossing in exothermic reactions would be translation. This means that the momentum of the approaching atom and the atom which it is going to collide with has a greater impact on the rate of the reaction (eg. F atom in F + H2 reaction). However, reagent vibration, which is related to the momentum of the two bonded atoms (eg. HF in H + HF reaction) in the colliding molecule, would be most effective in enabling endothermic reactions to take place.

One reactive trajectory has the conditions of rFH = 0.91 Å, rHH = 2.0 Å, pHF = -0.5 and pHH = -10. The trajectory recrosses the transition state region twice and eventually moves into exit channel, forming products. By comparing this condition to the initial condition, the momentum pHH, which is the translational kinetic energy of the H atom, is reduced by a half; and the pHF, the vibrational energy of the H-F bond is increased to -0.5. This indicates that the vibrational energy has a greater contribution to the feasibility of the reaction than the translational energy. This is the opposite situation of the H2+F reaction.

The results agree with the Polanyi's empirical rules[2], which state that the vibrational energy is more efficient in promoting late-transition-state reactions and the translational energy is more efficient in promoting early-transition-state reactions.

When the momentum pFH is increased slightly from -0.5 to -0.8 with pHH being only 0.1, the reaction now becomes successful. It indicates that the F-H vibration energy has a larger contribution for the feasibility of this reaction compared to the translational energy which is the kinetic energy of the hydrogen atom which is defined by the momentum pHH

When the pHH has a large value above the activation energy (30.231 kcal/mol), which was set to be -20. The hydrogen atom collides with the HF molecule and breaks the H-F bond. However, two hydrogen atoms moves apart and do not form a H2 molecule. The procedure described is illustrated in the following animation snapshots. After the collision, the large momentum that the reactant hydrogen atom contains is still large.As a result, even if an H-H bond can be formed, it would immediately be broken due to an excess vibrational energy between two atoms.

for investigation, 1st set condition pFH = -0.5 pHH = -3 unreactive; 2nd pHH = -2.45 reactive; 3rd pHH = -1.1 unreactive; 4th pHH = -0.9 reactive; 5th pHH = 1.1 unreactive ; 6th pHH = 1.9 reactive ; 7th pHH = 2.4 to 3 unreactive
; pHH = 2.1 no reaction

The cases studied are an illustration of Polanyi's empirical rules.

For H and HF

failed conditionsː 091, 2, 0.05, -20

initial condition of H HF systemː
HF distance = 0.91 HH distance = 2 pHF = 0.8 pHH = -7.5

== Rreferences ==

<references>
<ref name="TS">Eyring, H. (1935). "The Activated Complex in Chemical Reactions". J. Chem. Phys. 3 (2): 107–115.</ref>
</references>

MRD:WXY0119

2018-05-18T15:39:30Z

Xw2816: /* Energy Distribution and Reactivity */

= Molecular Reaction Dynamics Report =

== EXERCISE 1ː H + H2 SYSTEM ==

[[File:HandH2wxy0119.png]]

In the H + H2 system, the distance between the first two H atoms is r1 and the distance (bond length) between the two H atoms in the H2 molecule is r2.

=== Dynamics from the transition state region ===

{{fontcolor|red|Question 1 ː What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.}}

{| class="wikitable" border=1
|[[File:Q1TStransitionwxy0119.png|390px]]
|[[File:Q1TS2wxy0119.png|390px]]
|[[File:Q1TS3wxy0119.png|390px]]
|-
|Figure.1 Surface plot of reaction trajectory
|Figure.2 Transition state surface plot
|Figure.3 Transition state surface plot (different angle)
|}

In the reaction trajectory surface plot (Fig.1), AB is the distance r1 and BC is the distance r2. The gradient of the potential energy with regard to r1 and r2 are ∂V(r1)/∂r1 and ∂V(r2)/∂r2 respectively. The second derivatives of the potential energy with regard to r1 and r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 respectively.

At the two minimum structures which are at the two ends of the reaction pathway, ∂V(r1)/∂r1 = 0 and ∂V(r2)/∂r2 = 0 respectively. Since the two minimum structures are at two local minimum point , the second derivatives r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 are both greater than zero.

On the other hand, the components at the transition state structure (Fig.2) are different from that at minimum points. The gradient ∂V(r1)/∂r1 and ∂V(r2)/∂r2 are both equal to zero at the transition structure. However, r2 are ∂2V(r1)/∂r12 >0 and ∂2V(r2)/∂r22 <0. The transition state point is actually a saddle point which is observed more clearly in Fig.3. The Transition state point is thus distinguished from the minimum points as the potential energy surface curves inwards and downwards at the minimum points.

=== Locating the Transition State ===

{{fontcolor|red|Question 2ː 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.}}

{| class="wikitable" border=1
|[[File:TScontourwxy0119.png|390px]]
|[[File:TSsurfaceplwxy0119.png|390px]]
|[[File:TSestimatewxy0119.png|390px]]
|-
|Figure.4 Transition state contour plot
|Figure.5 Transition state surface plot
|Figure.6 Internuclear distance against time plot
|}

The best estimate of the transition state position is rts = 0.9078 Å.

When r1 = r2 = rts and momenta are set to zero, a graph of intermolecular distance against time is plotted. (Fig.6) In this graph, AB and BC lines are the same and overlap, the two lines on the graph are flat and horizontal, indicating that the atomic distances are the same at the position and with no momenta the state is at equilibrium, validating the position is the transition state position. The transition state is shown as a cross on the counter plot (Fig.4) and a dot on the surface plot (Fig.5).

=== Calculating and comparing the reaction path and trajectory ===

{{fontcolor1|red|Question 3ː Comment on how the ''mep'' and the trajectory you just calculated differ.}}

{| class="wikitable" border=1
|[[File:Mep1wxy0119.png|390px]]
|[[File:Mep2wxy0119.png|390px]]
|[[File:Mep3wxy0119.png|390px]]
|-
|Figure.7 mep calculation contour plot
|Figure.8 mep calculation surface plot
|Figure.9 mep calculation internuclear distance against time plot
|-
|[[File:Dynamic1wxy0119.png|390px]]
|[[File:Dynamic2wxy0119.png|390px]]
|[[File:Dynamic3wxy0119.png|390px]]
|-
|Figure.10 Dynamics calculation contour plot
|Figure.11 Dynamics calculation surface plot
|Figure.12 Dynamics calculation internuclear distance against time plot
|}
{| class="wikitable" border=1
|[[File:dynamics4wxy0119.png|600px]]
|[[File:Mep5wxy0119.png|600px]]
|-
|Figure.13 Dynamics calculation internuclear momenta against time plot
|Figure.14 mep calculation internuclear momenta against time plotsurface plot
|}

The mep (minimum energy path) trajectory is a smooth line on contour and surface plot (Fig.7 Fig.8) while the trajectory under dynamics calculation is oscillating along the pathway. (Fig.10 Fig.11) The mep trajectory takes more steps and longer time than the dynamics trajectory as shown in internuclear distance against time plots. (Fig. 9 Fig.12) The momentum is zero all the time under mep calculation (Fig.13) but under dynamics calculation, the momenta lines are oscillating against time.

The differences arise from mep and dynamics calculations can be explained by the the natures of the two calculations. Under mep calculation, the atoms are in extremely slow motion and the momentum is zero for each step, i.e. the atoms are "stop" after every step. The pathway under mep calculation is formed by connecting all the minimum energy points of each step and appears as a smooth line. On the other hand, atoms are in continuous motion and the momentum is accumulated after each step, atoms are then oscillating on the potential energy surface along the pathway. The steps are set to 5000 in the mep calculation but are only 500 in the dynamics calculatoin. Since the steps in mep are small, more time is needed to complete the same length on the trajectories.

=== Reactive and unreactive trajectories ===

{{fontcolor1|red|Question 4ː Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.}}

For the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, run trajectories with the following momenta p1 and p2 combinations:
{| class="wikitable" border=1
| Condition
| p1
| p2
| Total Energy/ kcal mol -1
| Reactivity
|-
|1.
| -1.25
| -2.5
| -99.018
| reactive
|-
|2.
| -1.5
| -2.0
| -100.456
| unreactive
|-
|3.
| -1.5
| -2.5
| -98.956
| reactive
|-
|4.
| -2.5
| -5.0
| -84.956
| unreactive
|-
|5.
| -2.5
| -5.2
| -83.416
| reactive
|}

{| class="wikitable" border=1
| Condition
| Surface Plot
| Contour Plot
| Description
|-
|1.
| [[File:condition1wxy0119.png|350px]]
| [[File:firstwxy0119.png|350px]]
| r2 (BC) decreases when HC approaches bonded HA and HB. The energy processed by the system is sufficient to overcome the activation barrier and crosses the transition state structure to break the HA and HB bond and form a new HB and HC bond. The new bond oscillate as r1 (AB) increases.
|-
|2.
| [[File:condition2wxy0119.png|350px]]
| [[File:secondwxy0119.png|350px]]
| HC approaches HB but the energy is insufficient to reach the transition state point, HC then moves further away from the bonded H2 molecule and no new bond is formed. The oscillation along r2 is a result of an increase in momentum p1.
|-
|3.
| [[File:condition3wxy0119.png|350px]]
| [[File:thirdwxy0119.png|350px]]
| Similarly to condition 1 but with a more negative p1, HA - HB bond oscillates more when r2 decreases. The reaction proceeds and crosses the transition state point to form the products.
|-
|4.
| [[File:condition4wxy0119.png|350px]]
| [[File:fourthwxy0119.png|350px]]
| Both p1 and p2 are more negative than the previous conditions. The HA - HB bond oscillates vigorously with larger amplitudes. The barrier recrossing happens after the system crosses the transition state region but eventually reverts back to the reactants.
|-
|5.
| [[File:condition5wxy0119.png|350px]]
| [[File:fifthwxy0119.png|350px]]
| With a slight more negative value of p1 as compared to that in condition 4, the system now processes appropriate energy to proceed to forming products. The vigorous oscillations are resulting from high momenta and barrier recrossing occurs.
|}

=== Transition State Theory ===

{{fontcolor|red|Question 5ː State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}

The main assumptions of Transition State Theory (TST) are<ref name="TS" />ː

1. The behaviours of the atomic nuclei follow the classical mechanics rules, i.e. the transition structure is formed as long as atoms or molecules collide with enough energy.

2. The intermediates have a long enough lifetime for the energies to be Boltzmann distributed preceding the next step.

3. The lowest energy saddle point on the potential energy surface is passed over in the reaction system.

The TST does not take into account of the quantum tunnelling effect which allows reactions to happen even when the energy carried by the system is lower than the activation energy barrier, especially for reactions with low energy barriers. Nor does the theory consider the intermediates with short lifetimes. When the energies are not fully distributed, the momentum of the reaction trajectory from the reactants to the intermediate can have effects on product selectivity. Moreover, the theory fails at high temperature because higher vibrational energy modes can be populated for a molecule at high temperature. The molecules can have complex motion and their collisions may result in a higher energy transition structure. The experimental results may thus deviate from that predicted by the TST. For example in condition 4, the kinetic energy of the reactants is much higher than the activation energy required. However, the momenta between atoms are very high and higher vibrational modes are populated that result in higher energy transition structures formed and the system does not pass though the lowest transition saddle point as shown in the diagram. Barrier recrossing occurs but the products are not formed as opposed to the predictions by the TST.

== EXERCISE 2: F - H - H SYSTEM ==

=== PES inspection ===

=== Reaction Energetics ===

{{fontcolor1|red|Question 6ː 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?}}

{| class="wikitable" border=1
|[[File:Q6FH2wxy0119.png|600px]]
|[[File:Q6HFHwxy0119.png|600px]]
|-
|Figure.15 F + H2 position on surface plot
|Figure.16 HF and H position on surface plot
|}

The surface plots of the potential energy surface of H + HF → F + H2 reaction is shown (Fig.15 Fig.16). Distance AB is the distance between H and F atoms and distance BC is the distance between two H atoms. The two minimum points shown in the graph are positions of F + H2 (Fig.15) and H + HF (Fig.16) respectively. If forward reaction is H + HF → F + H2, then the backward reaction is F + H2 → H + HF. As clearly shown in the diagram F + H2 are at a minimum point of higher potential energy than that of H + HF. Thus, by comparing the potential energy of reactants and products, the signs of changes in enthalpy (ΔH) and the energetics are determined. H + HF → F + H2 is endothermic and F + H2 → H + HF is exothermic.

The energetics of the two reactions are in accordance to the predictions by comparing bond strengths of chemical species. The bond energy of H-F bond (565 kJ/mol) is higher than the bond energy of H-H bond (432 kJ/mol). Thus, energy released by forming the H-H bond is not enough to compensate for energy required for breaking the H-F bond and the reaction H + HF → F + H2 is endothermic. On the contrary, the reaction F + H2 → H + HF is exothermic.

=== Transition State Approximation ===

{{fontcolor1|red|Question 7ː Locate the approximate position of the transition state.}}

{| class="wikitable" border=1
|[[File:Q7TStransitionwxy0119.png|390px]]
|[[File:Q7TS2wxy0119.png|390px]]
|[[File:Q7TS3wxy0119.png|390px]]
|-
|Figure.17 Transition state surface plot
|Figure.18 Transition state contour plot
|Figure.19 Internuclear distance against time plot (at TS point)
|}

The transition state (TS) is shown as a black dot on the potential energy surface plot (Fig.17) and a red cross on the contour plot. (Fig,18) The position of the TS is at the point where the distance between F and H (rFH) is 1.810Å and the distance between two H atoms (rHH) is 0.746Å. At the transition state position, the distance between F and H, between two H atoms are constant with no momentum as shown in the internuclear distance against time plot as horizontal and flat lines, validating the position is the transition state point (Fig.19)

=== Activation Energies ===

{{fontcolor1|red|Question 8ː Report the activation energy for both reactions.}}

{| class="wikitable" border=1
|[[File:Q8Ea2wxy0119.png|600px]]
|[[File:Q8Ea1wxy0119.png|600px]]
|-
|Figure.21 Energy against time plot (HF + H)
|Figure.20 Energy against time plot (F + H2)
|}

By performing MEP calculation with slight increase and decrease of the rFH to 1.820Å to perform F + H2 → H + HF reaction and to 1.800Å to perform H + HF → F + H2 reaction, the activation energies (Ea) of both reactions can be calculated from the differences in their potential energies (Fig.20 Fig.21)ː

F + H2 → H + HFː Ea = -103.751 - (-133.624) = +29.873 kcal mol-1

H + HF → F + H2ː Ea = -103.751 - (-103.972) = + 0.221 kcal mol-1

=== Reaction Dynamics ===

{{fontcolor1|red|Question 9ː In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

Reactionː F + H2 → H + HF

Initial condition setː rHF = 2Å rHH = 0.74Å pHF = -0.5 pHH = 0.1

{| class="wikitable" border=1
|[[File:Q9plot1wxy0119.png|390px]]
|[[File:Q9plot2wxy0119.png|390px]]
|[[File:Q9plot3wxy0119.png|390px]]
|-
|Figure.23 Contour plot
|Figure.24 Surface plot
|Figure.25 Internuclear momentum against time plot
|-
|[[File:Q9ani1wxy0119.png|390px]]
|[[File:Q9ani2wxy0119.png|390px]]
|-
|Figure.26 Animation figure (at the start of the reaction)
|Figure.27 Animation figure (at the end of the reaction)
|}

As shown in the plots and animation figures, the H-H bond in the reactants is broken and H-F bond in the products is formed. There are small momenta between both H and F and H-H bond that provide the initial kinetic energy of the reactants and the initial vibrational energy of the H-H bond in the reactants. The initial energy is required to break the H-H bond and reduce the distance of H and F for bond forming and potential energy is released when H-F bond is formed. Since energy is conserved, the reaction energy released is converted to the vibrational energy of the H-F bond and the translational kinetic energies of the products. This is shown in the internuclear momentum against time graph as the initial vibrations of B-C (H-H bond) becomes flat as two H atoms are separated in the product and the A-B (H-F bond) formed in the product vibrates with a large amplitude. (Fig.25) The contour plot and the surface plot also clearly show there is a significant increase in the amplitude of the oscillation of H-F bond which is an indication of the large vibrational energy of H-F bond. (Fig.23 Fig.24)

The predictions can be proved experimentally. The vibrational energy of H-F bond can be determined by measuring the frequency and the intensity of the absorption band of the H-F bond with infrared spectroscopy. The conversion to the kinetic energies of the products can be observed by measuring the temperature of the reaction as the kinetic energy can be converted to thermal energy as products collide with solvent molecules. The temperature is expected to increase as this is an exothermic reaction.

=== Energy Distribution and Reactivity ===

{{fontcolor|red|Question 10ː 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.}}
{| class="wikitable" border=1
| Condition
| rFH
| rHH
| pFH
| pHH
|-
|1.
| -1.25
| -2.5
| -99.018
| reactive
|-
|2.
| -1.5
| -2.0
| -100.456
| unreactive
|-
|3.
| -1.5
| -2.5
| -98.956
| reactive
|-
|4.
| -2.5
| -5.0
| -84.956
| unreactive
|-
|5.
| -2.5
| -5.2
| -83.416
| reactive
|}
{| class="wikitable" border=1
|Condition
|Contour plot
|Observation
|-
|1.
|[[File:Q9ani2wxy0119.png|390px]]
|
|-
|2.
|Figure.27 Animation figure (at the end of the reaction)
|
|-
|3.
|[[File:Q9ani2wxy0119.png|390px]]
|
|-
|4.
|[[File:Q9ani2wxy0119.png|390px]]
|
|-
|5.
|[[File:Q9ani2wxy0119.png|390px]]
|
|-
|6.
|[[File:Q9ani2wxy0119.png|390px]]
|
|}

pHH = -8.0 is still very large. This complies with the high activation energy of this reaction as the products formed are thermodynamically less stable than the reactants. It can be concluded that an increase in pHF, which corresponds to a high vibrational energy in H-F coordinate, is necessary for the reaction to take place. pHH, which corresponds to the translational energy in the H-H coordinate that is always very large, does not contribute too much to affect the reactivity of the reaction.

The transition state is present in the exit valley, hence, a late barrier for this endothermic reaction. The reactivity of endothermic reactions is thus dominated by the vibrational energy of the system from the results above.

For substantial exothermic reactions, such as F + H2, the transition state is located in the entrance valley, corresponds to an early barrier of the PES. For substantially endothermic reactions, such as H + HF, the transition state is in the exit valley, corresponds to a late barrier. The favoured degree of freedom for barrier crossing in exothermic reactions would be translation. This means that the momentum of the approaching atom and the atom which it is going to collide with has a greater impact on the rate of the reaction (eg. F atom in F + H2 reaction). However, reagent vibration, which is related to the momentum of the two bonded atoms (eg. HF in H + HF reaction) in the colliding molecule, would be most effective in enabling endothermic reactions to take place.

One reactive trajectory has the conditions of rFH = 0.91 Å, rHH = 2.0 Å, pHF = -0.5 and pHH = -10. The trajectory recrosses the transition state region twice and eventually moves into exit channel, forming products. By comparing this condition to the initial condition, the momentum pHH, which is the translational kinetic energy of the H atom, is reduced by a half; and the pHF, the vibrational energy of the H-F bond is increased to -0.5. This indicates that the vibrational energy has a greater contribution to the feasibility of the reaction than the translational energy. This is the opposite situation of the H2+F reaction.

The results agree with the Polanyi's empirical rules[2], which state that the vibrational energy is more efficient in promoting late-transition-state reactions and the translational energy is more efficient in promoting early-transition-state reactions.

When the momentum pFH is increased slightly from -0.5 to -0.8 with pHH being only 0.1, the reaction now becomes successful. It indicates that the F-H vibration energy has a larger contribution for the feasibility of this reaction compared to the translational energy which is the kinetic energy of the hydrogen atom which is defined by the momentum pHH

When the pHH has a large value above the activation energy (30.231 kcal/mol), which was set to be -20. The hydrogen atom collides with the HF molecule and breaks the H-F bond. However, two hydrogen atoms moves apart and do not form a H2 molecule. The procedure described is illustrated in the following animation snapshots. After the collision, the large momentum that the reactant hydrogen atom contains is still large.As a result, even if an H-H bond can be formed, it would immediately be broken due to an excess vibrational energy between two atoms.

for investigation, 1st set condition pFH = -0.5 pHH = -3 unreactive; 2nd pHH = -2.45 reactive; 3rd pHH = -1.1 unreactive; 4th pHH = -0.9 reactive; 5th pHH = 1.1 unreactive ; 6th pHH = 1.9 reactive ; 7th pHH = 2.4 to 3 unreactive
; pHH = 2.1 no reaction

The cases studied are an illustration of Polanyi's empirical rules.

For H and HF

failed conditionsː 091, 2, 0.05, -20

initial condition of H HF systemː
HF distance = 0.91 HH distance = 2 pHF = 0.8 pHH = -7.5

== Rreferences ==

<references>
<ref name="TS">Eyring, H. (1935). "The Activated Complex in Chemical Reactions". J. Chem. Phys. 3 (2): 107–115.</ref>
</references>

MRD:WXY0119

2018-05-18T15:34:03Z

Xw2816: /* Energy Distribution and Reactivity */

= Molecular Reaction Dynamics Report =

== EXERCISE 1ː H + H2 SYSTEM ==

[[File:HandH2wxy0119.png]]

In the H + H2 system, the distance between the first two H atoms is r1 and the distance (bond length) between the two H atoms in the H2 molecule is r2.

=== Dynamics from the transition state region ===

{{fontcolor|red|Question 1 ː What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.}}

{| class="wikitable" border=1
|[[File:Q1TStransitionwxy0119.png|390px]]
|[[File:Q1TS2wxy0119.png|390px]]
|[[File:Q1TS3wxy0119.png|390px]]
|-
|Figure.1 Surface plot of reaction trajectory
|Figure.2 Transition state surface plot
|Figure.3 Transition state surface plot (different angle)
|}

In the reaction trajectory surface plot (Fig.1), AB is the distance r1 and BC is the distance r2. The gradient of the potential energy with regard to r1 and r2 are ∂V(r1)/∂r1 and ∂V(r2)/∂r2 respectively. The second derivatives of the potential energy with regard to r1 and r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 respectively.

At the two minimum structures which are at the two ends of the reaction pathway, ∂V(r1)/∂r1 = 0 and ∂V(r2)/∂r2 = 0 respectively. Since the two minimum structures are at two local minimum point , the second derivatives r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 are both greater than zero.

On the other hand, the components at the transition state structure (Fig.2) are different from that at minimum points. The gradient ∂V(r1)/∂r1 and ∂V(r2)/∂r2 are both equal to zero at the transition structure. However, r2 are ∂2V(r1)/∂r12 >0 and ∂2V(r2)/∂r22 <0. The transition state point is actually a saddle point which is observed more clearly in Fig.3. The Transition state point is thus distinguished from the minimum points as the potential energy surface curves inwards and downwards at the minimum points.

=== Locating the Transition State ===

{{fontcolor|red|Question 2ː 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.}}

{| class="wikitable" border=1
|[[File:TScontourwxy0119.png|390px]]
|[[File:TSsurfaceplwxy0119.png|390px]]
|[[File:TSestimatewxy0119.png|390px]]
|-
|Figure.4 Transition state contour plot
|Figure.5 Transition state surface plot
|Figure.6 Internuclear distance against time plot
|}

The best estimate of the transition state position is rts = 0.9078 Å.

When r1 = r2 = rts and momenta are set to zero, a graph of intermolecular distance against time is plotted. (Fig.6) In this graph, AB and BC lines are the same and overlap, the two lines on the graph are flat and horizontal, indicating that the atomic distances are the same at the position and with no momenta the state is at equilibrium, validating the position is the transition state position. The transition state is shown as a cross on the counter plot (Fig.4) and a dot on the surface plot (Fig.5).

=== Calculating and comparing the reaction path and trajectory ===

{{fontcolor1|red|Question 3ː Comment on how the ''mep'' and the trajectory you just calculated differ.}}

{| class="wikitable" border=1
|[[File:Mep1wxy0119.png|390px]]
|[[File:Mep2wxy0119.png|390px]]
|[[File:Mep3wxy0119.png|390px]]
|-
|Figure.7 mep calculation contour plot
|Figure.8 mep calculation surface plot
|Figure.9 mep calculation internuclear distance against time plot
|-
|[[File:Dynamic1wxy0119.png|390px]]
|[[File:Dynamic2wxy0119.png|390px]]
|[[File:Dynamic3wxy0119.png|390px]]
|-
|Figure.10 Dynamics calculation contour plot
|Figure.11 Dynamics calculation surface plot
|Figure.12 Dynamics calculation internuclear distance against time plot
|}
{| class="wikitable" border=1
|[[File:dynamics4wxy0119.png|600px]]
|[[File:Mep5wxy0119.png|600px]]
|-
|Figure.13 Dynamics calculation internuclear momenta against time plot
|Figure.14 mep calculation internuclear momenta against time plotsurface plot
|}

The mep (minimum energy path) trajectory is a smooth line on contour and surface plot (Fig.7 Fig.8) while the trajectory under dynamics calculation is oscillating along the pathway. (Fig.10 Fig.11) The mep trajectory takes more steps and longer time than the dynamics trajectory as shown in internuclear distance against time plots. (Fig. 9 Fig.12) The momentum is zero all the time under mep calculation (Fig.13) but under dynamics calculation, the momenta lines are oscillating against time.

The differences arise from mep and dynamics calculations can be explained by the the natures of the two calculations. Under mep calculation, the atoms are in extremely slow motion and the momentum is zero for each step, i.e. the atoms are "stop" after every step. The pathway under mep calculation is formed by connecting all the minimum energy points of each step and appears as a smooth line. On the other hand, atoms are in continuous motion and the momentum is accumulated after each step, atoms are then oscillating on the potential energy surface along the pathway. The steps are set to 5000 in the mep calculation but are only 500 in the dynamics calculatoin. Since the steps in mep are small, more time is needed to complete the same length on the trajectories.

=== Reactive and unreactive trajectories ===

{{fontcolor1|red|Question 4ː Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.}}

For the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, run trajectories with the following momenta p1 and p2 combinations:
{| class="wikitable" border=1
| Condition
| p1
| p2
| Total Energy/ kcal mol -1
| Reactivity
|-
|1.
| -1.25
| -2.5
| -99.018
| reactive
|-
|2.
| -1.5
| -2.0
| -100.456
| unreactive
|-
|3.
| -1.5
| -2.5
| -98.956
| reactive
|-
|4.
| -2.5
| -5.0
| -84.956
| unreactive
|-
|5.
| -2.5
| -5.2
| -83.416
| reactive
|}

{| class="wikitable" border=1
| Condition
| Surface Plot
| Contour Plot
| Description
|-
|1.
| [[File:condition1wxy0119.png|350px]]
| [[File:firstwxy0119.png|350px]]
| r2 (BC) decreases when HC approaches bonded HA and HB. The energy processed by the system is sufficient to overcome the activation barrier and crosses the transition state structure to break the HA and HB bond and form a new HB and HC bond. The new bond oscillate as r1 (AB) increases.
|-
|2.
| [[File:condition2wxy0119.png|350px]]
| [[File:secondwxy0119.png|350px]]
| HC approaches HB but the energy is insufficient to reach the transition state point, HC then moves further away from the bonded H2 molecule and no new bond is formed. The oscillation along r2 is a result of an increase in momentum p1.
|-
|3.
| [[File:condition3wxy0119.png|350px]]
| [[File:thirdwxy0119.png|350px]]
| Similarly to condition 1 but with a more negative p1, HA - HB bond oscillates more when r2 decreases. The reaction proceeds and crosses the transition state point to form the products.
|-
|4.
| [[File:condition4wxy0119.png|350px]]
| [[File:fourthwxy0119.png|350px]]
| Both p1 and p2 are more negative than the previous conditions. The HA - HB bond oscillates vigorously with larger amplitudes. The barrier recrossing happens after the system crosses the transition state region but eventually reverts back to the reactants.
|-
|5.
| [[File:condition5wxy0119.png|350px]]
| [[File:fifthwxy0119.png|350px]]
| With a slight more negative value of p1 as compared to that in condition 4, the system now processes appropriate energy to proceed to forming products. The vigorous oscillations are resulting from high momenta and barrier recrossing occurs.
|}

=== Transition State Theory ===

{{fontcolor|red|Question 5ː State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}

The main assumptions of Transition State Theory (TST) are<ref name="TS" />ː

1. The behaviours of the atomic nuclei follow the classical mechanics rules, i.e. the transition structure is formed as long as atoms or molecules collide with enough energy.

2. The intermediates have a long enough lifetime for the energies to be Boltzmann distributed preceding the next step.

3. The lowest energy saddle point on the potential energy surface is passed over in the reaction system.

The TST does not take into account of the quantum tunnelling effect which allows reactions to happen even when the energy carried by the system is lower than the activation energy barrier, especially for reactions with low energy barriers. Nor does the theory consider the intermediates with short lifetimes. When the energies are not fully distributed, the momentum of the reaction trajectory from the reactants to the intermediate can have effects on product selectivity. Moreover, the theory fails at high temperature because higher vibrational energy modes can be populated for a molecule at high temperature. The molecules can have complex motion and their collisions may result in a higher energy transition structure. The experimental results may thus deviate from that predicted by the TST. For example in condition 4, the kinetic energy of the reactants is much higher than the activation energy required. However, the momenta between atoms are very high and higher vibrational modes are populated that result in higher energy transition structures formed and the system does not pass though the lowest transition saddle point as shown in the diagram. Barrier recrossing occurs but the products are not formed as opposed to the predictions by the TST.

== EXERCISE 2: F - H - H SYSTEM ==

=== PES inspection ===

=== Reaction Energetics ===

{{fontcolor1|red|Question 6ː 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?}}

{| class="wikitable" border=1
|[[File:Q6FH2wxy0119.png|600px]]
|[[File:Q6HFHwxy0119.png|600px]]
|-
|Figure.15 F + H2 position on surface plot
|Figure.16 HF and H position on surface plot
|}

The surface plots of the potential energy surface of H + HF → F + H2 reaction is shown (Fig.15 Fig.16). Distance AB is the distance between H and F atoms and distance BC is the distance between two H atoms. The two minimum points shown in the graph are positions of F + H2 (Fig.15) and H + HF (Fig.16) respectively. If forward reaction is H + HF → F + H2, then the backward reaction is F + H2 → H + HF. As clearly shown in the diagram F + H2 are at a minimum point of higher potential energy than that of H + HF. Thus, by comparing the potential energy of reactants and products, the signs of changes in enthalpy (ΔH) and the energetics are determined. H + HF → F + H2 is endothermic and F + H2 → H + HF is exothermic.

The energetics of the two reactions are in accordance to the predictions by comparing bond strengths of chemical species. The bond energy of H-F bond (565 kJ/mol) is higher than the bond energy of H-H bond (432 kJ/mol). Thus, energy released by forming the H-H bond is not enough to compensate for energy required for breaking the H-F bond and the reaction H + HF → F + H2 is endothermic. On the contrary, the reaction F + H2 → H + HF is exothermic.

=== Transition State Approximation ===

{{fontcolor1|red|Question 7ː Locate the approximate position of the transition state.}}

{| class="wikitable" border=1
|[[File:Q7TStransitionwxy0119.png|390px]]
|[[File:Q7TS2wxy0119.png|390px]]
|[[File:Q7TS3wxy0119.png|390px]]
|-
|Figure.17 Transition state surface plot
|Figure.18 Transition state contour plot
|Figure.19 Internuclear distance against time plot (at TS point)
|}

The transition state (TS) is shown as a black dot on the potential energy surface plot (Fig.17) and a red cross on the contour plot. (Fig,18) The position of the TS is at the point where the distance between F and H (rFH) is 1.810Å and the distance between two H atoms (rHH) is 0.746Å. At the transition state position, the distance between F and H, between two H atoms are constant with no momentum as shown in the internuclear distance against time plot as horizontal and flat lines, validating the position is the transition state point (Fig.19)

=== Activation Energies ===

{{fontcolor1|red|Question 8ː Report the activation energy for both reactions.}}

{| class="wikitable" border=1
|[[File:Q8Ea2wxy0119.png|600px]]
|[[File:Q8Ea1wxy0119.png|600px]]
|-
|Figure.21 Energy against time plot (HF + H)
|Figure.20 Energy against time plot (F + H2)
|}

By performing MEP calculation with slight increase and decrease of the rFH to 1.820Å to perform F + H2 → H + HF reaction and to 1.800Å to perform H + HF → F + H2 reaction, the activation energies (Ea) of both reactions can be calculated from the differences in their potential energies (Fig.20 Fig.21)ː

F + H2 → H + HFː Ea = -103.751 - (-133.624) = +29.873 kcal mol-1

H + HF → F + H2ː Ea = -103.751 - (-103.972) = + 0.221 kcal mol-1

=== Reaction Dynamics ===

{{fontcolor1|red|Question 9ː In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

Reactionː F + H2 → H + HF

Initial condition setː rHF = 2Å rHH = 0.74Å pHF = -0.5 pHH = 0.1

{| class="wikitable" border=1
|[[File:Q9plot1wxy0119.png|390px]]
|[[File:Q9plot2wxy0119.png|390px]]
|[[File:Q9plot3wxy0119.png|390px]]
|-
|Figure.23 Contour plot
|Figure.24 Surface plot
|Figure.25 Internuclear momentum against time plot
|-
|[[File:Q9ani1wxy0119.png|390px]]
|[[File:Q9ani2wxy0119.png|390px]]
|-
|Figure.26 Animation figure (at the start of the reaction)
|Figure.27 Animation figure (at the end of the reaction)
|}

As shown in the plots and animation figures, the H-H bond in the reactants is broken and H-F bond in the products is formed. There are small momenta between both H and F and H-H bond that provide the initial kinetic energy of the reactants and the initial vibrational energy of the H-H bond in the reactants. The initial energy is required to break the H-H bond and reduce the distance of H and F for bond forming and potential energy is released when H-F bond is formed. Since energy is conserved, the reaction energy released is converted to the vibrational energy of the H-F bond and the translational kinetic energies of the products. This is shown in the internuclear momentum against time graph as the initial vibrations of B-C (H-H bond) becomes flat as two H atoms are separated in the product and the A-B (H-F bond) formed in the product vibrates with a large amplitude. (Fig.25) The contour plot and the surface plot also clearly show there is a significant increase in the amplitude of the oscillation of H-F bond which is an indication of the large vibrational energy of H-F bond. (Fig.23 Fig.24)

The predictions can be proved experimentally. The vibrational energy of H-F bond can be determined by measuring the frequency and the intensity of the absorption band of the H-F bond with infrared spectroscopy. The conversion to the kinetic energies of the products can be observed by measuring the temperature of the reaction as the kinetic energy can be converted to thermal energy as products collide with solvent molecules. The temperature is expected to increase as this is an exothermic reaction.

=== Energy Distribution and Reactivity ===

{{fontcolor|red|Question 10ː 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.}}
{| class="wikitable" border=1
| Condition
| rFH
| rHH
| pFH
| pHH
|-
|1.
| -1.25
| -2.5
| -99.018
| reactive
|-
|2.
| -1.5
| -2.0
| -100.456
| unreactive
|-
|3.
| -1.5
| -2.5
| -98.956
| reactive
|-
|4.
| -2.5
| -5.0
| -84.956
| unreactive
|-
|5.
| -2.5
| -5.2
| -83.416
| reactive
|}
{| class="wikitable" border=1
|Condition
|Contour plot
|Observation
|-
|1.
|[[File:Q9ani2wxy0119.png|390px]]
|
|-
|2.
|Figure.27 Animation figure (at the end of the reaction)
|
|-
|3.
|[[File:Q9ani2wxy0119.png|390px]]
|
|-
|4.
|[[File:Q9ani2wxy0119.png|390px]]
|
|-
|5.
|[[File:Q9ani2wxy0119.png|390px]]
|
|-
|6.
|[[File:Q9ani2wxy0119.png|390px]]
|
|}

for investigation, 1st set condition pFH = -0.5 pHH = -3 unreactive; 2nd pHH = -2.45 reactive; 3rd pHH = -1.1 unreactive; 4th pHH = -0.9 reactive; 5th pHH = 1.1 unreactive ; 6th pHH = 1.9 reactive ; 7th pHH = 2.4 to 3 unreactive
; pHH = 2.1 no reaction

The cases studied are an illustration of Polanyi's empirical rules.

For H and HF

failed conditionsː 091, 2, 0.05, -20

initial condition of H HF systemː
HF distance = 0.91 HH distance = 2 pHF = 0.8 pHH = -7.5

== Rreferences ==

<references>
<ref name="TS">Eyring, H. (1935). "The Activated Complex in Chemical Reactions". J. Chem. Phys. 3 (2): 107–115.</ref>
</references>

MRD:WXY0119

2018-05-18T15:33:44Z

Xw2816: /* Energy Distribution and Reactivity */

= Molecular Reaction Dynamics Report =

== EXERCISE 1ː H + H2 SYSTEM ==

[[File:HandH2wxy0119.png]]

In the H + H2 system, the distance between the first two H atoms is r1 and the distance (bond length) between the two H atoms in the H2 molecule is r2.

=== Dynamics from the transition state region ===

{{fontcolor|red|Question 1 ː What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.}}

{| class="wikitable" border=1
|[[File:Q1TStransitionwxy0119.png|390px]]
|[[File:Q1TS2wxy0119.png|390px]]
|[[File:Q1TS3wxy0119.png|390px]]
|-
|Figure.1 Surface plot of reaction trajectory
|Figure.2 Transition state surface plot
|Figure.3 Transition state surface plot (different angle)
|}

In the reaction trajectory surface plot (Fig.1), AB is the distance r1 and BC is the distance r2. The gradient of the potential energy with regard to r1 and r2 are ∂V(r1)/∂r1 and ∂V(r2)/∂r2 respectively. The second derivatives of the potential energy with regard to r1 and r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 respectively.

At the two minimum structures which are at the two ends of the reaction pathway, ∂V(r1)/∂r1 = 0 and ∂V(r2)/∂r2 = 0 respectively. Since the two minimum structures are at two local minimum point , the second derivatives r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 are both greater than zero.

On the other hand, the components at the transition state structure (Fig.2) are different from that at minimum points. The gradient ∂V(r1)/∂r1 and ∂V(r2)/∂r2 are both equal to zero at the transition structure. However, r2 are ∂2V(r1)/∂r12 >0 and ∂2V(r2)/∂r22 <0. The transition state point is actually a saddle point which is observed more clearly in Fig.3. The Transition state point is thus distinguished from the minimum points as the potential energy surface curves inwards and downwards at the minimum points.

=== Locating the Transition State ===

{{fontcolor|red|Question 2ː 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.}}

{| class="wikitable" border=1
|[[File:TScontourwxy0119.png|390px]]
|[[File:TSsurfaceplwxy0119.png|390px]]
|[[File:TSestimatewxy0119.png|390px]]
|-
|Figure.4 Transition state contour plot
|Figure.5 Transition state surface plot
|Figure.6 Internuclear distance against time plot
|}

The best estimate of the transition state position is rts = 0.9078 Å.

When r1 = r2 = rts and momenta are set to zero, a graph of intermolecular distance against time is plotted. (Fig.6) In this graph, AB and BC lines are the same and overlap, the two lines on the graph are flat and horizontal, indicating that the atomic distances are the same at the position and with no momenta the state is at equilibrium, validating the position is the transition state position. The transition state is shown as a cross on the counter plot (Fig.4) and a dot on the surface plot (Fig.5).

=== Calculating and comparing the reaction path and trajectory ===

{{fontcolor1|red|Question 3ː Comment on how the ''mep'' and the trajectory you just calculated differ.}}

{| class="wikitable" border=1
|[[File:Mep1wxy0119.png|390px]]
|[[File:Mep2wxy0119.png|390px]]
|[[File:Mep3wxy0119.png|390px]]
|-
|Figure.7 mep calculation contour plot
|Figure.8 mep calculation surface plot
|Figure.9 mep calculation internuclear distance against time plot
|-
|[[File:Dynamic1wxy0119.png|390px]]
|[[File:Dynamic2wxy0119.png|390px]]
|[[File:Dynamic3wxy0119.png|390px]]
|-
|Figure.10 Dynamics calculation contour plot
|Figure.11 Dynamics calculation surface plot
|Figure.12 Dynamics calculation internuclear distance against time plot
|}
{| class="wikitable" border=1
|[[File:dynamics4wxy0119.png|600px]]
|[[File:Mep5wxy0119.png|600px]]
|-
|Figure.13 Dynamics calculation internuclear momenta against time plot
|Figure.14 mep calculation internuclear momenta against time plotsurface plot
|}

The mep (minimum energy path) trajectory is a smooth line on contour and surface plot (Fig.7 Fig.8) while the trajectory under dynamics calculation is oscillating along the pathway. (Fig.10 Fig.11) The mep trajectory takes more steps and longer time than the dynamics trajectory as shown in internuclear distance against time plots. (Fig. 9 Fig.12) The momentum is zero all the time under mep calculation (Fig.13) but under dynamics calculation, the momenta lines are oscillating against time.

The differences arise from mep and dynamics calculations can be explained by the the natures of the two calculations. Under mep calculation, the atoms are in extremely slow motion and the momentum is zero for each step, i.e. the atoms are "stop" after every step. The pathway under mep calculation is formed by connecting all the minimum energy points of each step and appears as a smooth line. On the other hand, atoms are in continuous motion and the momentum is accumulated after each step, atoms are then oscillating on the potential energy surface along the pathway. The steps are set to 5000 in the mep calculation but are only 500 in the dynamics calculatoin. Since the steps in mep are small, more time is needed to complete the same length on the trajectories.

=== Reactive and unreactive trajectories ===

{{fontcolor1|red|Question 4ː Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.}}

For the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, run trajectories with the following momenta p1 and p2 combinations:
{| class="wikitable" border=1
| Condition
| p1
| p2
| Total Energy/ kcal mol -1
| Reactivity
|-
|1.
| -1.25
| -2.5
| -99.018
| reactive
|-
|2.
| -1.5
| -2.0
| -100.456
| unreactive
|-
|3.
| -1.5
| -2.5
| -98.956
| reactive
|-
|4.
| -2.5
| -5.0
| -84.956
| unreactive
|-
|5.
| -2.5
| -5.2
| -83.416
| reactive
|}

{| class="wikitable" border=1
| Condition
| Surface Plot
| Contour Plot
| Description
|-
|1.
| [[File:condition1wxy0119.png|350px]]
| [[File:firstwxy0119.png|350px]]
| r2 (BC) decreases when HC approaches bonded HA and HB. The energy processed by the system is sufficient to overcome the activation barrier and crosses the transition state structure to break the HA and HB bond and form a new HB and HC bond. The new bond oscillate as r1 (AB) increases.
|-
|2.
| [[File:condition2wxy0119.png|350px]]
| [[File:secondwxy0119.png|350px]]
| HC approaches HB but the energy is insufficient to reach the transition state point, HC then moves further away from the bonded H2 molecule and no new bond is formed. The oscillation along r2 is a result of an increase in momentum p1.
|-
|3.
| [[File:condition3wxy0119.png|350px]]
| [[File:thirdwxy0119.png|350px]]
| Similarly to condition 1 but with a more negative p1, HA - HB bond oscillates more when r2 decreases. The reaction proceeds and crosses the transition state point to form the products.
|-
|4.
| [[File:condition4wxy0119.png|350px]]
| [[File:fourthwxy0119.png|350px]]
| Both p1 and p2 are more negative than the previous conditions. The HA - HB bond oscillates vigorously with larger amplitudes. The barrier recrossing happens after the system crosses the transition state region but eventually reverts back to the reactants.
|-
|5.
| [[File:condition5wxy0119.png|350px]]
| [[File:fifthwxy0119.png|350px]]
| With a slight more negative value of p1 as compared to that in condition 4, the system now processes appropriate energy to proceed to forming products. The vigorous oscillations are resulting from high momenta and barrier recrossing occurs.
|}

=== Transition State Theory ===

{{fontcolor|red|Question 5ː State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}

The main assumptions of Transition State Theory (TST) are<ref name="TS" />ː

1. The behaviours of the atomic nuclei follow the classical mechanics rules, i.e. the transition structure is formed as long as atoms or molecules collide with enough energy.

2. The intermediates have a long enough lifetime for the energies to be Boltzmann distributed preceding the next step.

3. The lowest energy saddle point on the potential energy surface is passed over in the reaction system.

The TST does not take into account of the quantum tunnelling effect which allows reactions to happen even when the energy carried by the system is lower than the activation energy barrier, especially for reactions with low energy barriers. Nor does the theory consider the intermediates with short lifetimes. When the energies are not fully distributed, the momentum of the reaction trajectory from the reactants to the intermediate can have effects on product selectivity. Moreover, the theory fails at high temperature because higher vibrational energy modes can be populated for a molecule at high temperature. The molecules can have complex motion and their collisions may result in a higher energy transition structure. The experimental results may thus deviate from that predicted by the TST. For example in condition 4, the kinetic energy of the reactants is much higher than the activation energy required. However, the momenta between atoms are very high and higher vibrational modes are populated that result in higher energy transition structures formed and the system does not pass though the lowest transition saddle point as shown in the diagram. Barrier recrossing occurs but the products are not formed as opposed to the predictions by the TST.

== EXERCISE 2: F - H - H SYSTEM ==

=== PES inspection ===

=== Reaction Energetics ===

{{fontcolor1|red|Question 6ː 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?}}

{| class="wikitable" border=1
|[[File:Q6FH2wxy0119.png|600px]]
|[[File:Q6HFHwxy0119.png|600px]]
|-
|Figure.15 F + H2 position on surface plot
|Figure.16 HF and H position on surface plot
|}

The surface plots of the potential energy surface of H + HF → F + H2 reaction is shown (Fig.15 Fig.16). Distance AB is the distance between H and F atoms and distance BC is the distance between two H atoms. The two minimum points shown in the graph are positions of F + H2 (Fig.15) and H + HF (Fig.16) respectively. If forward reaction is H + HF → F + H2, then the backward reaction is F + H2 → H + HF. As clearly shown in the diagram F + H2 are at a minimum point of higher potential energy than that of H + HF. Thus, by comparing the potential energy of reactants and products, the signs of changes in enthalpy (ΔH) and the energetics are determined. H + HF → F + H2 is endothermic and F + H2 → H + HF is exothermic.

The energetics of the two reactions are in accordance to the predictions by comparing bond strengths of chemical species. The bond energy of H-F bond (565 kJ/mol) is higher than the bond energy of H-H bond (432 kJ/mol). Thus, energy released by forming the H-H bond is not enough to compensate for energy required for breaking the H-F bond and the reaction H + HF → F + H2 is endothermic. On the contrary, the reaction F + H2 → H + HF is exothermic.

=== Transition State Approximation ===

{{fontcolor1|red|Question 7ː Locate the approximate position of the transition state.}}

{| class="wikitable" border=1
|[[File:Q7TStransitionwxy0119.png|390px]]
|[[File:Q7TS2wxy0119.png|390px]]
|[[File:Q7TS3wxy0119.png|390px]]
|-
|Figure.17 Transition state surface plot
|Figure.18 Transition state contour plot
|Figure.19 Internuclear distance against time plot (at TS point)
|}

The transition state (TS) is shown as a black dot on the potential energy surface plot (Fig.17) and a red cross on the contour plot. (Fig,18) The position of the TS is at the point where the distance between F and H (rFH) is 1.810Å and the distance between two H atoms (rHH) is 0.746Å. At the transition state position, the distance between F and H, between two H atoms are constant with no momentum as shown in the internuclear distance against time plot as horizontal and flat lines, validating the position is the transition state point (Fig.19)

=== Activation Energies ===

{{fontcolor1|red|Question 8ː Report the activation energy for both reactions.}}

{| class="wikitable" border=1
|[[File:Q8Ea2wxy0119.png|600px]]
|[[File:Q8Ea1wxy0119.png|600px]]
|-
|Figure.21 Energy against time plot (HF + H)
|Figure.20 Energy against time plot (F + H2)
|}

By performing MEP calculation with slight increase and decrease of the rFH to 1.820Å to perform F + H2 → H + HF reaction and to 1.800Å to perform H + HF → F + H2 reaction, the activation energies (Ea) of both reactions can be calculated from the differences in their potential energies (Fig.20 Fig.21)ː

F + H2 → H + HFː Ea = -103.751 - (-133.624) = +29.873 kcal mol-1

H + HF → F + H2ː Ea = -103.751 - (-103.972) = + 0.221 kcal mol-1

=== Reaction Dynamics ===

{{fontcolor1|red|Question 9ː In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

Reactionː F + H2 → H + HF

Initial condition setː rHF = 2Å rHH = 0.74Å pHF = -0.5 pHH = 0.1

{| class="wikitable" border=1
|[[File:Q9plot1wxy0119.png|390px]]
|[[File:Q9plot2wxy0119.png|390px]]
|[[File:Q9plot3wxy0119.png|390px]]
|-
|Figure.23 Contour plot
|Figure.24 Surface plot
|Figure.25 Internuclear momentum against time plot
|-
|[[File:Q9ani1wxy0119.png|390px]]
|[[File:Q9ani2wxy0119.png|390px]]
|-
|Figure.26 Animation figure (at the start of the reaction)
|Figure.27 Animation figure (at the end of the reaction)
|}

As shown in the plots and animation figures, the H-H bond in the reactants is broken and H-F bond in the products is formed. There are small momenta between both H and F and H-H bond that provide the initial kinetic energy of the reactants and the initial vibrational energy of the H-H bond in the reactants. The initial energy is required to break the H-H bond and reduce the distance of H and F for bond forming and potential energy is released when H-F bond is formed. Since energy is conserved, the reaction energy released is converted to the vibrational energy of the H-F bond and the translational kinetic energies of the products. This is shown in the internuclear momentum against time graph as the initial vibrations of B-C (H-H bond) becomes flat as two H atoms are separated in the product and the A-B (H-F bond) formed in the product vibrates with a large amplitude. (Fig.25) The contour plot and the surface plot also clearly show there is a significant increase in the amplitude of the oscillation of H-F bond which is an indication of the large vibrational energy of H-F bond. (Fig.23 Fig.24)

The predictions can be proved experimentally. The vibrational energy of H-F bond can be determined by measuring the frequency and the intensity of the absorption band of the H-F bond with infrared spectroscopy. The conversion to the kinetic energies of the products can be observed by measuring the temperature of the reaction as the kinetic energy can be converted to thermal energy as products collide with solvent molecules. The temperature is expected to increase as this is an exothermic reaction.

=== Energy Distribution and Reactivity ===

{{fontcolor|red|Question 10ː 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.}}

{| class="wikitable" border=1
| Condition
| rFH
| rHH
| pFH
| pHH
|-
|1.
| -1.25
| -2.5
| -99.018
| reactive
|-
|2.
| -1.5
| -2.0
| -100.456
| unreactive
|-
|3.
| -1.5
| -2.5
| -98.956
| reactive
|-
|4.
| -2.5
| -5.0
| -84.956
| unreactive
|-
|5.
| -2.5
| -5.2
| -83.416
| reactive
|}

{| class="wikitable" border=1
|Condition
|Contour plot
|Observation
|-
|1.
|[[File:Q9ani2wxy0119.png|390px]]
|
|-
|2.
|Figure.27 Animation figure (at the end of the reaction)
|
|-
|3.
|[[File:Q9ani2wxy0119.png|390px]]
|
|-
|4.
|[[File:Q9ani2wxy0119.png|390px]]
|
|-
|5.
|[[File:Q9ani2wxy0119.png|390px]]
|
|-
|6.
|[[File:Q9ani2wxy0119.png|390px]]
|
|}

for investigation, 1st set condition pFH = -0.5 pHH = -3 unreactive; 2nd pHH = -2.45 reactive; 3rd pHH = -1.1 unreactive; 4th pHH = -0.9 reactive; 5th pHH = 1.1 unreactive ; 6th pHH = 1.9 reactive ; 7th pHH = 2.4 to 3 unreactive
; pHH = 2.1 no reaction

The cases studied are an illustration of Polanyi's empirical rules.

For H and HF

failed conditionsː 091, 2, 0.05, -20

initial condition of H HF systemː
HF distance = 0.91 HH distance = 2 pHF = 0.8 pHH = -7.5

== Rreferences ==

<references>
<ref name="TS">Eyring, H. (1935). "The Activated Complex in Chemical Reactions". J. Chem. Phys. 3 (2): 107–115.</ref>
</references>

MRD:WXY0119

2018-05-18T15:31:31Z

Xw2816: /* Energy Distribution and Reactivity */

= Molecular Reaction Dynamics Report =

== EXERCISE 1ː H + H2 SYSTEM ==

[[File:HandH2wxy0119.png]]

In the H + H2 system, the distance between the first two H atoms is r1 and the distance (bond length) between the two H atoms in the H2 molecule is r2.

=== Dynamics from the transition state region ===

{{fontcolor|red|Question 1 ː What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.}}

{| class="wikitable" border=1
|[[File:Q1TStransitionwxy0119.png|390px]]
|[[File:Q1TS2wxy0119.png|390px]]
|[[File:Q1TS3wxy0119.png|390px]]
|-
|Figure.1 Surface plot of reaction trajectory
|Figure.2 Transition state surface plot
|Figure.3 Transition state surface plot (different angle)
|}

In the reaction trajectory surface plot (Fig.1), AB is the distance r1 and BC is the distance r2. The gradient of the potential energy with regard to r1 and r2 are ∂V(r1)/∂r1 and ∂V(r2)/∂r2 respectively. The second derivatives of the potential energy with regard to r1 and r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 respectively.

At the two minimum structures which are at the two ends of the reaction pathway, ∂V(r1)/∂r1 = 0 and ∂V(r2)/∂r2 = 0 respectively. Since the two minimum structures are at two local minimum point , the second derivatives r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 are both greater than zero.

On the other hand, the components at the transition state structure (Fig.2) are different from that at minimum points. The gradient ∂V(r1)/∂r1 and ∂V(r2)/∂r2 are both equal to zero at the transition structure. However, r2 are ∂2V(r1)/∂r12 >0 and ∂2V(r2)/∂r22 <0. The transition state point is actually a saddle point which is observed more clearly in Fig.3. The Transition state point is thus distinguished from the minimum points as the potential energy surface curves inwards and downwards at the minimum points.

=== Locating the Transition State ===

{{fontcolor|red|Question 2ː 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.}}

{| class="wikitable" border=1
|[[File:TScontourwxy0119.png|390px]]
|[[File:TSsurfaceplwxy0119.png|390px]]
|[[File:TSestimatewxy0119.png|390px]]
|-
|Figure.4 Transition state contour plot
|Figure.5 Transition state surface plot
|Figure.6 Internuclear distance against time plot
|}

The best estimate of the transition state position is rts = 0.9078 Å.

When r1 = r2 = rts and momenta are set to zero, a graph of intermolecular distance against time is plotted. (Fig.6) In this graph, AB and BC lines are the same and overlap, the two lines on the graph are flat and horizontal, indicating that the atomic distances are the same at the position and with no momenta the state is at equilibrium, validating the position is the transition state position. The transition state is shown as a cross on the counter plot (Fig.4) and a dot on the surface plot (Fig.5).

=== Calculating and comparing the reaction path and trajectory ===

{{fontcolor1|red|Question 3ː Comment on how the ''mep'' and the trajectory you just calculated differ.}}

{| class="wikitable" border=1
|[[File:Mep1wxy0119.png|390px]]
|[[File:Mep2wxy0119.png|390px]]
|[[File:Mep3wxy0119.png|390px]]
|-
|Figure.7 mep calculation contour plot
|Figure.8 mep calculation surface plot
|Figure.9 mep calculation internuclear distance against time plot
|-
|[[File:Dynamic1wxy0119.png|390px]]
|[[File:Dynamic2wxy0119.png|390px]]
|[[File:Dynamic3wxy0119.png|390px]]
|-
|Figure.10 Dynamics calculation contour plot
|Figure.11 Dynamics calculation surface plot
|Figure.12 Dynamics calculation internuclear distance against time plot
|}
{| class="wikitable" border=1
|[[File:dynamics4wxy0119.png|600px]]
|[[File:Mep5wxy0119.png|600px]]
|-
|Figure.13 Dynamics calculation internuclear momenta against time plot
|Figure.14 mep calculation internuclear momenta against time plotsurface plot
|}

The mep (minimum energy path) trajectory is a smooth line on contour and surface plot (Fig.7 Fig.8) while the trajectory under dynamics calculation is oscillating along the pathway. (Fig.10 Fig.11) The mep trajectory takes more steps and longer time than the dynamics trajectory as shown in internuclear distance against time plots. (Fig. 9 Fig.12) The momentum is zero all the time under mep calculation (Fig.13) but under dynamics calculation, the momenta lines are oscillating against time.

The differences arise from mep and dynamics calculations can be explained by the the natures of the two calculations. Under mep calculation, the atoms are in extremely slow motion and the momentum is zero for each step, i.e. the atoms are "stop" after every step. The pathway under mep calculation is formed by connecting all the minimum energy points of each step and appears as a smooth line. On the other hand, atoms are in continuous motion and the momentum is accumulated after each step, atoms are then oscillating on the potential energy surface along the pathway. The steps are set to 5000 in the mep calculation but are only 500 in the dynamics calculatoin. Since the steps in mep are small, more time is needed to complete the same length on the trajectories.

=== Reactive and unreactive trajectories ===

{{fontcolor1|red|Question 4ː Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.}}

For the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, run trajectories with the following momenta p1 and p2 combinations:
{| class="wikitable" border=1
| Condition
| p1
| p2
| Total Energy/ kcal mol -1
| Reactivity
|-
|1.
| -1.25
| -2.5
| -99.018
| reactive
|-
|2.
| -1.5
| -2.0
| -100.456
| unreactive
|-
|3.
| -1.5
| -2.5
| -98.956
| reactive
|-
|4.
| -2.5
| -5.0
| -84.956
| unreactive
|-
|5.
| -2.5
| -5.2
| -83.416
| reactive
|}

{| class="wikitable" border=1
| Condition
| Surface Plot
| Contour Plot
| Description
|-
|1.
| [[File:condition1wxy0119.png|350px]]
| [[File:firstwxy0119.png|350px]]
| r2 (BC) decreases when HC approaches bonded HA and HB. The energy processed by the system is sufficient to overcome the activation barrier and crosses the transition state structure to break the HA and HB bond and form a new HB and HC bond. The new bond oscillate as r1 (AB) increases.
|-
|2.
| [[File:condition2wxy0119.png|350px]]
| [[File:secondwxy0119.png|350px]]
| HC approaches HB but the energy is insufficient to reach the transition state point, HC then moves further away from the bonded H2 molecule and no new bond is formed. The oscillation along r2 is a result of an increase in momentum p1.
|-
|3.
| [[File:condition3wxy0119.png|350px]]
| [[File:thirdwxy0119.png|350px]]
| Similarly to condition 1 but with a more negative p1, HA - HB bond oscillates more when r2 decreases. The reaction proceeds and crosses the transition state point to form the products.
|-
|4.
| [[File:condition4wxy0119.png|350px]]
| [[File:fourthwxy0119.png|350px]]
| Both p1 and p2 are more negative than the previous conditions. The HA - HB bond oscillates vigorously with larger amplitudes. The barrier recrossing happens after the system crosses the transition state region but eventually reverts back to the reactants.
|-
|5.
| [[File:condition5wxy0119.png|350px]]
| [[File:fifthwxy0119.png|350px]]
| With a slight more negative value of p1 as compared to that in condition 4, the system now processes appropriate energy to proceed to forming products. The vigorous oscillations are resulting from high momenta and barrier recrossing occurs.
|}

=== Transition State Theory ===

{{fontcolor|red|Question 5ː State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}

The main assumptions of Transition State Theory (TST) are<ref name="TS" />ː

1. The behaviours of the atomic nuclei follow the classical mechanics rules, i.e. the transition structure is formed as long as atoms or molecules collide with enough energy.

2. The intermediates have a long enough lifetime for the energies to be Boltzmann distributed preceding the next step.

3. The lowest energy saddle point on the potential energy surface is passed over in the reaction system.

The TST does not take into account of the quantum tunnelling effect which allows reactions to happen even when the energy carried by the system is lower than the activation energy barrier, especially for reactions with low energy barriers. Nor does the theory consider the intermediates with short lifetimes. When the energies are not fully distributed, the momentum of the reaction trajectory from the reactants to the intermediate can have effects on product selectivity. Moreover, the theory fails at high temperature because higher vibrational energy modes can be populated for a molecule at high temperature. The molecules can have complex motion and their collisions may result in a higher energy transition structure. The experimental results may thus deviate from that predicted by the TST. For example in condition 4, the kinetic energy of the reactants is much higher than the activation energy required. However, the momenta between atoms are very high and higher vibrational modes are populated that result in higher energy transition structures formed and the system does not pass though the lowest transition saddle point as shown in the diagram. Barrier recrossing occurs but the products are not formed as opposed to the predictions by the TST.

== EXERCISE 2: F - H - H SYSTEM ==

=== PES inspection ===

=== Reaction Energetics ===

{{fontcolor1|red|Question 6ː 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?}}

{| class="wikitable" border=1
|[[File:Q6FH2wxy0119.png|600px]]
|[[File:Q6HFHwxy0119.png|600px]]
|-
|Figure.15 F + H2 position on surface plot
|Figure.16 HF and H position on surface plot
|}

The surface plots of the potential energy surface of H + HF → F + H2 reaction is shown (Fig.15 Fig.16). Distance AB is the distance between H and F atoms and distance BC is the distance between two H atoms. The two minimum points shown in the graph are positions of F + H2 (Fig.15) and H + HF (Fig.16) respectively. If forward reaction is H + HF → F + H2, then the backward reaction is F + H2 → H + HF. As clearly shown in the diagram F + H2 are at a minimum point of higher potential energy than that of H + HF. Thus, by comparing the potential energy of reactants and products, the signs of changes in enthalpy (ΔH) and the energetics are determined. H + HF → F + H2 is endothermic and F + H2 → H + HF is exothermic.

The energetics of the two reactions are in accordance to the predictions by comparing bond strengths of chemical species. The bond energy of H-F bond (565 kJ/mol) is higher than the bond energy of H-H bond (432 kJ/mol). Thus, energy released by forming the H-H bond is not enough to compensate for energy required for breaking the H-F bond and the reaction H + HF → F + H2 is endothermic. On the contrary, the reaction F + H2 → H + HF is exothermic.

=== Transition State Approximation ===

{{fontcolor1|red|Question 7ː Locate the approximate position of the transition state.}}

{| class="wikitable" border=1
|[[File:Q7TStransitionwxy0119.png|390px]]
|[[File:Q7TS2wxy0119.png|390px]]
|[[File:Q7TS3wxy0119.png|390px]]
|-
|Figure.17 Transition state surface plot
|Figure.18 Transition state contour plot
|Figure.19 Internuclear distance against time plot (at TS point)
|}

The transition state (TS) is shown as a black dot on the potential energy surface plot (Fig.17) and a red cross on the contour plot. (Fig,18) The position of the TS is at the point where the distance between F and H (rFH) is 1.810Å and the distance between two H atoms (rHH) is 0.746Å. At the transition state position, the distance between F and H, between two H atoms are constant with no momentum as shown in the internuclear distance against time plot as horizontal and flat lines, validating the position is the transition state point (Fig.19)

=== Activation Energies ===

{{fontcolor1|red|Question 8ː Report the activation energy for both reactions.}}

{| class="wikitable" border=1
|[[File:Q8Ea2wxy0119.png|600px]]
|[[File:Q8Ea1wxy0119.png|600px]]
|-
|Figure.21 Energy against time plot (HF + H)
|Figure.20 Energy against time plot (F + H2)
|}

By performing MEP calculation with slight increase and decrease of the rFH to 1.820Å to perform F + H2 → H + HF reaction and to 1.800Å to perform H + HF → F + H2 reaction, the activation energies (Ea) of both reactions can be calculated from the differences in their potential energies (Fig.20 Fig.21)ː

F + H2 → H + HFː Ea = -103.751 - (-133.624) = +29.873 kcal mol-1

H + HF → F + H2ː Ea = -103.751 - (-103.972) = + 0.221 kcal mol-1

=== Reaction Dynamics ===

{{fontcolor1|red|Question 9ː In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

Reactionː F + H2 → H + HF

Initial condition setː rHF = 2Å rHH = 0.74Å pHF = -0.5 pHH = 0.1

{| class="wikitable" border=1
|[[File:Q9plot1wxy0119.png|390px]]
|[[File:Q9plot2wxy0119.png|390px]]
|[[File:Q9plot3wxy0119.png|390px]]
|-
|Figure.23 Contour plot
|Figure.24 Surface plot
|Figure.25 Internuclear momentum against time plot
|-
|[[File:Q9ani1wxy0119.png|390px]]
|[[File:Q9ani2wxy0119.png|390px]]
|-
|Figure.26 Animation figure (at the start of the reaction)
|Figure.27 Animation figure (at the end of the reaction)
|}

As shown in the plots and animation figures, the H-H bond in the reactants is broken and H-F bond in the products is formed. There are small momenta between both H and F and H-H bond that provide the initial kinetic energy of the reactants and the initial vibrational energy of the H-H bond in the reactants. The initial energy is required to break the H-H bond and reduce the distance of H and F for bond forming and potential energy is released when H-F bond is formed. Since energy is conserved, the reaction energy released is converted to the vibrational energy of the H-F bond and the translational kinetic energies of the products. This is shown in the internuclear momentum against time graph as the initial vibrations of B-C (H-H bond) becomes flat as two H atoms are separated in the product and the A-B (H-F bond) formed in the product vibrates with a large amplitude. (Fig.25) The contour plot and the surface plot also clearly show there is a significant increase in the amplitude of the oscillation of H-F bond which is an indication of the large vibrational energy of H-F bond. (Fig.23 Fig.24)

The predictions can be proved experimentally. The vibrational energy of H-F bond can be determined by measuring the frequency and the intensity of the absorption band of the H-F bond with infrared spectroscopy. The conversion to the kinetic energies of the products can be observed by measuring the temperature of the reaction as the kinetic energy can be converted to thermal energy as products collide with solvent molecules. The temperature is expected to increase as this is an exothermic reaction.

=== Energy Distribution and Reactivity ===

{{fontcolor|red|Question 10ː 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.}}

{| class="wikitable" border=1
| Condition
| rFH
| rHH
| pFH
| pHH
|-
|1.
| -1.25
| -2.5
| -99.018
| reactive
|-
|2.
| -1.5
| -2.0
| -100.456
| unreactive
|-
|3.
| -1.5
| -2.5
| -98.956
| reactive
|-
|4.
| -2.5
| -5.0
| -84.956
| unreactive
|-
|5.
| -2.5
| -5.2
| -83.416
| reactive
|}

{| class="wikitable" border=1
|Condition
|Contour plot
|Observation
|-
|[[File:Q9ani1wxy0119.png|390px]]
|[[File:Q9ani2wxy0119.png|390px]]
|-
|Figure.26 Animation figure (at the start of the reaction)
|Figure.27 Animation figure (at the end of the reaction)
|}

for investigation, 1st set condition pFH = -0.5 pHH = -3 unreactive; 2nd pHH = -2.45 reactive; 3rd pHH = -1.1 unreactive; 4th pHH = -0.9 reactive; 5th pHH = 1.1 unreactive ; 6th pHH = 1.9 reactive ; 7th pHH = 2.4 to 3 unreactive
; pHH = 2.1 no reaction

The cases studied are an illustration of Polanyi's empirical rules.

For H and HF

failed conditionsː 091, 2, 0.05, -20

initial condition of H HF systemː
HF distance = 0.91 HH distance = 2 pHF = 0.8 pHH = -7.5

== Rreferences ==

<references>
<ref name="TS">Eyring, H. (1935). "The Activated Complex in Chemical Reactions". J. Chem. Phys. 3 (2): 107–115.</ref>
</references>

MRD:WXY0119

2018-05-18T15:26:22Z

Xw2816: /* Energy Distribution and Reactivity */

= Molecular Reaction Dynamics Report =

== EXERCISE 1ː H + H2 SYSTEM ==

[[File:HandH2wxy0119.png]]

In the H + H2 system, the distance between the first two H atoms is r1 and the distance (bond length) between the two H atoms in the H2 molecule is r2.

=== Dynamics from the transition state region ===

{{fontcolor|red|Question 1 ː What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.}}

{| class="wikitable" border=1
|[[File:Q1TStransitionwxy0119.png|390px]]
|[[File:Q1TS2wxy0119.png|390px]]
|[[File:Q1TS3wxy0119.png|390px]]
|-
|Figure.1 Surface plot of reaction trajectory
|Figure.2 Transition state surface plot
|Figure.3 Transition state surface plot (different angle)
|}

In the reaction trajectory surface plot (Fig.1), AB is the distance r1 and BC is the distance r2. The gradient of the potential energy with regard to r1 and r2 are ∂V(r1)/∂r1 and ∂V(r2)/∂r2 respectively. The second derivatives of the potential energy with regard to r1 and r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 respectively.

At the two minimum structures which are at the two ends of the reaction pathway, ∂V(r1)/∂r1 = 0 and ∂V(r2)/∂r2 = 0 respectively. Since the two minimum structures are at two local minimum point , the second derivatives r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 are both greater than zero.

On the other hand, the components at the transition state structure (Fig.2) are different from that at minimum points. The gradient ∂V(r1)/∂r1 and ∂V(r2)/∂r2 are both equal to zero at the transition structure. However, r2 are ∂2V(r1)/∂r12 >0 and ∂2V(r2)/∂r22 <0. The transition state point is actually a saddle point which is observed more clearly in Fig.3. The Transition state point is thus distinguished from the minimum points as the potential energy surface curves inwards and downwards at the minimum points.

=== Locating the Transition State ===

{{fontcolor|red|Question 2ː 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.}}

{| class="wikitable" border=1
|[[File:TScontourwxy0119.png|390px]]
|[[File:TSsurfaceplwxy0119.png|390px]]
|[[File:TSestimatewxy0119.png|390px]]
|-
|Figure.4 Transition state contour plot
|Figure.5 Transition state surface plot
|Figure.6 Internuclear distance against time plot
|}

The best estimate of the transition state position is rts = 0.9078 Å.

When r1 = r2 = rts and momenta are set to zero, a graph of intermolecular distance against time is plotted. (Fig.6) In this graph, AB and BC lines are the same and overlap, the two lines on the graph are flat and horizontal, indicating that the atomic distances are the same at the position and with no momenta the state is at equilibrium, validating the position is the transition state position. The transition state is shown as a cross on the counter plot (Fig.4) and a dot on the surface plot (Fig.5).

=== Calculating and comparing the reaction path and trajectory ===

{{fontcolor1|red|Question 3ː Comment on how the ''mep'' and the trajectory you just calculated differ.}}

{| class="wikitable" border=1
|[[File:Mep1wxy0119.png|390px]]
|[[File:Mep2wxy0119.png|390px]]
|[[File:Mep3wxy0119.png|390px]]
|-
|Figure.7 mep calculation contour plot
|Figure.8 mep calculation surface plot
|Figure.9 mep calculation internuclear distance against time plot
|-
|[[File:Dynamic1wxy0119.png|390px]]
|[[File:Dynamic2wxy0119.png|390px]]
|[[File:Dynamic3wxy0119.png|390px]]
|-
|Figure.10 Dynamics calculation contour plot
|Figure.11 Dynamics calculation surface plot
|Figure.12 Dynamics calculation internuclear distance against time plot
|}
{| class="wikitable" border=1
|[[File:dynamics4wxy0119.png|600px]]
|[[File:Mep5wxy0119.png|600px]]
|-
|Figure.13 Dynamics calculation internuclear momenta against time plot
|Figure.14 mep calculation internuclear momenta against time plotsurface plot
|}

The mep (minimum energy path) trajectory is a smooth line on contour and surface plot (Fig.7 Fig.8) while the trajectory under dynamics calculation is oscillating along the pathway. (Fig.10 Fig.11) The mep trajectory takes more steps and longer time than the dynamics trajectory as shown in internuclear distance against time plots. (Fig. 9 Fig.12) The momentum is zero all the time under mep calculation (Fig.13) but under dynamics calculation, the momenta lines are oscillating against time.

The differences arise from mep and dynamics calculations can be explained by the the natures of the two calculations. Under mep calculation, the atoms are in extremely slow motion and the momentum is zero for each step, i.e. the atoms are "stop" after every step. The pathway under mep calculation is formed by connecting all the minimum energy points of each step and appears as a smooth line. On the other hand, atoms are in continuous motion and the momentum is accumulated after each step, atoms are then oscillating on the potential energy surface along the pathway. The steps are set to 5000 in the mep calculation but are only 500 in the dynamics calculatoin. Since the steps in mep are small, more time is needed to complete the same length on the trajectories.

=== Reactive and unreactive trajectories ===

{{fontcolor1|red|Question 4ː Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.}}

For the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, run trajectories with the following momenta p1 and p2 combinations:
{| class="wikitable" border=1
| Condition
| p1
| p2
| Total Energy/ kcal mol -1
| Reactivity
|-
|1.
| -1.25
| -2.5
| -99.018
| reactive
|-
|2.
| -1.5
| -2.0
| -100.456
| unreactive
|-
|3.
| -1.5
| -2.5
| -98.956
| reactive
|-
|4.
| -2.5
| -5.0
| -84.956
| unreactive
|-
|5.
| -2.5
| -5.2
| -83.416
| reactive
|}

{| class="wikitable" border=1
| Condition
| Surface Plot
| Contour Plot
| Description
|-
|1.
| [[File:condition1wxy0119.png|350px]]
| [[File:firstwxy0119.png|350px]]
| r2 (BC) decreases when HC approaches bonded HA and HB. The energy processed by the system is sufficient to overcome the activation barrier and crosses the transition state structure to break the HA and HB bond and form a new HB and HC bond. The new bond oscillate as r1 (AB) increases.
|-
|2.
| [[File:condition2wxy0119.png|350px]]
| [[File:secondwxy0119.png|350px]]
| HC approaches HB but the energy is insufficient to reach the transition state point, HC then moves further away from the bonded H2 molecule and no new bond is formed. The oscillation along r2 is a result of an increase in momentum p1.
|-
|3.
| [[File:condition3wxy0119.png|350px]]
| [[File:thirdwxy0119.png|350px]]
| Similarly to condition 1 but with a more negative p1, HA - HB bond oscillates more when r2 decreases. The reaction proceeds and crosses the transition state point to form the products.
|-
|4.
| [[File:condition4wxy0119.png|350px]]
| [[File:fourthwxy0119.png|350px]]
| Both p1 and p2 are more negative than the previous conditions. The HA - HB bond oscillates vigorously with larger amplitudes. The barrier recrossing happens after the system crosses the transition state region but eventually reverts back to the reactants.
|-
|5.
| [[File:condition5wxy0119.png|350px]]
| [[File:fifthwxy0119.png|350px]]
| With a slight more negative value of p1 as compared to that in condition 4, the system now processes appropriate energy to proceed to forming products. The vigorous oscillations are resulting from high momenta and barrier recrossing occurs.
|}

=== Transition State Theory ===

{{fontcolor|red|Question 5ː State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}

The main assumptions of Transition State Theory (TST) are<ref name="TS" />ː

1. The behaviours of the atomic nuclei follow the classical mechanics rules, i.e. the transition structure is formed as long as atoms or molecules collide with enough energy.

2. The intermediates have a long enough lifetime for the energies to be Boltzmann distributed preceding the next step.

3. The lowest energy saddle point on the potential energy surface is passed over in the reaction system.

The TST does not take into account of the quantum tunnelling effect which allows reactions to happen even when the energy carried by the system is lower than the activation energy barrier, especially for reactions with low energy barriers. Nor does the theory consider the intermediates with short lifetimes. When the energies are not fully distributed, the momentum of the reaction trajectory from the reactants to the intermediate can have effects on product selectivity. Moreover, the theory fails at high temperature because higher vibrational energy modes can be populated for a molecule at high temperature. The molecules can have complex motion and their collisions may result in a higher energy transition structure. The experimental results may thus deviate from that predicted by the TST. For example in condition 4, the kinetic energy of the reactants is much higher than the activation energy required. However, the momenta between atoms are very high and higher vibrational modes are populated that result in higher energy transition structures formed and the system does not pass though the lowest transition saddle point as shown in the diagram. Barrier recrossing occurs but the products are not formed as opposed to the predictions by the TST.

== EXERCISE 2: F - H - H SYSTEM ==

=== PES inspection ===

=== Reaction Energetics ===

{{fontcolor1|red|Question 6ː 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?}}

{| class="wikitable" border=1
|[[File:Q6FH2wxy0119.png|600px]]
|[[File:Q6HFHwxy0119.png|600px]]
|-
|Figure.15 F + H2 position on surface plot
|Figure.16 HF and H position on surface plot
|}

The surface plots of the potential energy surface of H + HF → F + H2 reaction is shown (Fig.15 Fig.16). Distance AB is the distance between H and F atoms and distance BC is the distance between two H atoms. The two minimum points shown in the graph are positions of F + H2 (Fig.15) and H + HF (Fig.16) respectively. If forward reaction is H + HF → F + H2, then the backward reaction is F + H2 → H + HF. As clearly shown in the diagram F + H2 are at a minimum point of higher potential energy than that of H + HF. Thus, by comparing the potential energy of reactants and products, the signs of changes in enthalpy (ΔH) and the energetics are determined. H + HF → F + H2 is endothermic and F + H2 → H + HF is exothermic.

The energetics of the two reactions are in accordance to the predictions by comparing bond strengths of chemical species. The bond energy of H-F bond (565 kJ/mol) is higher than the bond energy of H-H bond (432 kJ/mol). Thus, energy released by forming the H-H bond is not enough to compensate for energy required for breaking the H-F bond and the reaction H + HF → F + H2 is endothermic. On the contrary, the reaction F + H2 → H + HF is exothermic.

=== Transition State Approximation ===

{{fontcolor1|red|Question 7ː Locate the approximate position of the transition state.}}

{| class="wikitable" border=1
|[[File:Q7TStransitionwxy0119.png|390px]]
|[[File:Q7TS2wxy0119.png|390px]]
|[[File:Q7TS3wxy0119.png|390px]]
|-
|Figure.17 Transition state surface plot
|Figure.18 Transition state contour plot
|Figure.19 Internuclear distance against time plot (at TS point)
|}

The transition state (TS) is shown as a black dot on the potential energy surface plot (Fig.17) and a red cross on the contour plot. (Fig,18) The position of the TS is at the point where the distance between F and H (rFH) is 1.810Å and the distance between two H atoms (rHH) is 0.746Å. At the transition state position, the distance between F and H, between two H atoms are constant with no momentum as shown in the internuclear distance against time plot as horizontal and flat lines, validating the position is the transition state point (Fig.19)

=== Activation Energies ===

{{fontcolor1|red|Question 8ː Report the activation energy for both reactions.}}

{| class="wikitable" border=1
|[[File:Q8Ea2wxy0119.png|600px]]
|[[File:Q8Ea1wxy0119.png|600px]]
|-
|Figure.21 Energy against time plot (HF + H)
|Figure.20 Energy against time plot (F + H2)
|}

By performing MEP calculation with slight increase and decrease of the rFH to 1.820Å to perform F + H2 → H + HF reaction and to 1.800Å to perform H + HF → F + H2 reaction, the activation energies (Ea) of both reactions can be calculated from the differences in their potential energies (Fig.20 Fig.21)ː

F + H2 → H + HFː Ea = -103.751 - (-133.624) = +29.873 kcal mol-1

H + HF → F + H2ː Ea = -103.751 - (-103.972) = + 0.221 kcal mol-1

=== Reaction Dynamics ===

{{fontcolor1|red|Question 9ː In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

Reactionː F + H2 → H + HF

Initial condition setː rHF = 2Å rHH = 0.74Å pHF = -0.5 pHH = 0.1

{| class="wikitable" border=1
|[[File:Q9plot1wxy0119.png|390px]]
|[[File:Q9plot2wxy0119.png|390px]]
|[[File:Q9plot3wxy0119.png|390px]]
|-
|Figure.23 Contour plot
|Figure.24 Surface plot
|Figure.25 Internuclear momentum against time plot
|-
|[[File:Q9ani1wxy0119.png|390px]]
|[[File:Q9ani2wxy0119.png|390px]]
|-
|Figure.26 Animation figure (at the start of the reaction)
|Figure.27 Animation figure (at the end of the reaction)
|}

As shown in the plots and animation figures, the H-H bond in the reactants is broken and H-F bond in the products is formed. There are small momenta between both H and F and H-H bond that provide the initial kinetic energy of the reactants and the initial vibrational energy of the H-H bond in the reactants. The initial energy is required to break the H-H bond and reduce the distance of H and F for bond forming and potential energy is released when H-F bond is formed. Since energy is conserved, the reaction energy released is converted to the vibrational energy of the H-F bond and the translational kinetic energies of the products. This is shown in the internuclear momentum against time graph as the initial vibrations of B-C (H-H bond) becomes flat as two H atoms are separated in the product and the A-B (H-F bond) formed in the product vibrates with a large amplitude. (Fig.25) The contour plot and the surface plot also clearly show there is a significant increase in the amplitude of the oscillation of H-F bond which is an indication of the large vibrational energy of H-F bond. (Fig.23 Fig.24)

The predictions can be proved experimentally. The vibrational energy of H-F bond can be determined by measuring the frequency and the intensity of the absorption band of the H-F bond with infrared spectroscopy. The conversion to the kinetic energies of the products can be observed by measuring the temperature of the reaction as the kinetic energy can be converted to thermal energy as products collide with solvent molecules. The temperature is expected to increase as this is an exothermic reaction.

=== Energy Distribution and Reactivity ===

{{fontcolor|red|Question 10ː 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.}}

{| class="wikitable" border=1
|
|
|
|-
|
|Figure.24 Surface plot
|Figure.25 Internuclear momentum against time plot
|-
|[[File:Q9ani1wxy0119.png|390px]]
|[[File:Q9ani2wxy0119.png|390px]]
|-
|Figure.26 Animation figure (at the start of the reaction)
|Figure.27 Animation figure (at the end of the reaction)
|}

{| class="wikitable" border=1
|[[File:Q9plot1wxy0119.png|390px]]
|[[File:Q9plot2wxy0119.png|390px]]
|[[File:Q9plot3wxy0119.png|390px]]
|-
|Figure.23 Contour plot
|Figure.24 Surface plot
|Figure.25 Internuclear momentum against time plot
|-
|[[File:Q9ani1wxy0119.png|390px]]
|[[File:Q9ani2wxy0119.png|390px]]
|-
|Figure.26 Animation figure (at the start of the reaction)
|Figure.27 Animation figure (at the end of the reaction)
|}

for investigation, 1st set condition pFH = -0.5 pHH = -3 unreactive; 2nd pHH = -2.45 reactive; 3rd pHH = -1.1 unreactive; 4th pHH = -0.9 reactive; 5th pHH = 1.1 unreactive ; 6th pHH = 1.9 reactive ; 7th pHH = 2.4 to 3 unreactive
; pHH = 2.1 no reaction

The cases studied are an illustration of Polanyi's empirical rules.

For H and HF

failed conditionsː 091, 2, 0.05, -20

initial condition of H HF systemː
HF distance = 0.91 HH distance = 2 pHF = 0.8 pHH = -7.5

== Rreferences ==

<references>
<ref name="TS">Eyring, H. (1935). "The Activated Complex in Chemical Reactions". J. Chem. Phys. 3 (2): 107–115.</ref>
</references>

MRD:WXY0119

2018-05-18T15:25:32Z

Xw2816: /* Energy Distribution and Reactivity */

= Molecular Reaction Dynamics Report =

== EXERCISE 1ː H + H2 SYSTEM ==

[[File:HandH2wxy0119.png]]

In the H + H2 system, the distance between the first two H atoms is r1 and the distance (bond length) between the two H atoms in the H2 molecule is r2.

=== Dynamics from the transition state region ===

{{fontcolor|red|Question 1 ː What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.}}

{| class="wikitable" border=1
|[[File:Q1TStransitionwxy0119.png|390px]]
|[[File:Q1TS2wxy0119.png|390px]]
|[[File:Q1TS3wxy0119.png|390px]]
|-
|Figure.1 Surface plot of reaction trajectory
|Figure.2 Transition state surface plot
|Figure.3 Transition state surface plot (different angle)
|}

In the reaction trajectory surface plot (Fig.1), AB is the distance r1 and BC is the distance r2. The gradient of the potential energy with regard to r1 and r2 are ∂V(r1)/∂r1 and ∂V(r2)/∂r2 respectively. The second derivatives of the potential energy with regard to r1 and r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 respectively.

At the two minimum structures which are at the two ends of the reaction pathway, ∂V(r1)/∂r1 = 0 and ∂V(r2)/∂r2 = 0 respectively. Since the two minimum structures are at two local minimum point , the second derivatives r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 are both greater than zero.

On the other hand, the components at the transition state structure (Fig.2) are different from that at minimum points. The gradient ∂V(r1)/∂r1 and ∂V(r2)/∂r2 are both equal to zero at the transition structure. However, r2 are ∂2V(r1)/∂r12 >0 and ∂2V(r2)/∂r22 <0. The transition state point is actually a saddle point which is observed more clearly in Fig.3. The Transition state point is thus distinguished from the minimum points as the potential energy surface curves inwards and downwards at the minimum points.

=== Locating the Transition State ===

{{fontcolor|red|Question 2ː 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.}}

{| class="wikitable" border=1
|[[File:TScontourwxy0119.png|390px]]
|[[File:TSsurfaceplwxy0119.png|390px]]
|[[File:TSestimatewxy0119.png|390px]]
|-
|Figure.4 Transition state contour plot
|Figure.5 Transition state surface plot
|Figure.6 Internuclear distance against time plot
|}

The best estimate of the transition state position is rts = 0.9078 Å.

When r1 = r2 = rts and momenta are set to zero, a graph of intermolecular distance against time is plotted. (Fig.6) In this graph, AB and BC lines are the same and overlap, the two lines on the graph are flat and horizontal, indicating that the atomic distances are the same at the position and with no momenta the state is at equilibrium, validating the position is the transition state position. The transition state is shown as a cross on the counter plot (Fig.4) and a dot on the surface plot (Fig.5).

=== Calculating and comparing the reaction path and trajectory ===

{{fontcolor1|red|Question 3ː Comment on how the ''mep'' and the trajectory you just calculated differ.}}

{| class="wikitable" border=1
|[[File:Mep1wxy0119.png|390px]]
|[[File:Mep2wxy0119.png|390px]]
|[[File:Mep3wxy0119.png|390px]]
|-
|Figure.7 mep calculation contour plot
|Figure.8 mep calculation surface plot
|Figure.9 mep calculation internuclear distance against time plot
|-
|[[File:Dynamic1wxy0119.png|390px]]
|[[File:Dynamic2wxy0119.png|390px]]
|[[File:Dynamic3wxy0119.png|390px]]
|-
|Figure.10 Dynamics calculation contour plot
|Figure.11 Dynamics calculation surface plot
|Figure.12 Dynamics calculation internuclear distance against time plot
|}
{| class="wikitable" border=1
|[[File:dynamics4wxy0119.png|600px]]
|[[File:Mep5wxy0119.png|600px]]
|-
|Figure.13 Dynamics calculation internuclear momenta against time plot
|Figure.14 mep calculation internuclear momenta against time plotsurface plot
|}

The mep (minimum energy path) trajectory is a smooth line on contour and surface plot (Fig.7 Fig.8) while the trajectory under dynamics calculation is oscillating along the pathway. (Fig.10 Fig.11) The mep trajectory takes more steps and longer time than the dynamics trajectory as shown in internuclear distance against time plots. (Fig. 9 Fig.12) The momentum is zero all the time under mep calculation (Fig.13) but under dynamics calculation, the momenta lines are oscillating against time.

The differences arise from mep and dynamics calculations can be explained by the the natures of the two calculations. Under mep calculation, the atoms are in extremely slow motion and the momentum is zero for each step, i.e. the atoms are "stop" after every step. The pathway under mep calculation is formed by connecting all the minimum energy points of each step and appears as a smooth line. On the other hand, atoms are in continuous motion and the momentum is accumulated after each step, atoms are then oscillating on the potential energy surface along the pathway. The steps are set to 5000 in the mep calculation but are only 500 in the dynamics calculatoin. Since the steps in mep are small, more time is needed to complete the same length on the trajectories.

=== Reactive and unreactive trajectories ===

{{fontcolor1|red|Question 4ː Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.}}

For the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, run trajectories with the following momenta p1 and p2 combinations:
{| class="wikitable" border=1
| Condition
| p1
| p2
| Total Energy/ kcal mol -1
| Reactivity
|-
|1.
| -1.25
| -2.5
| -99.018
| reactive
|-
|2.
| -1.5
| -2.0
| -100.456
| unreactive
|-
|3.
| -1.5
| -2.5
| -98.956
| reactive
|-
|4.
| -2.5
| -5.0
| -84.956
| unreactive
|-
|5.
| -2.5
| -5.2
| -83.416
| reactive
|}

{| class="wikitable" border=1
| Condition
| Surface Plot
| Contour Plot
| Description
|-
|1.
| [[File:condition1wxy0119.png|350px]]
| [[File:firstwxy0119.png|350px]]
| r2 (BC) decreases when HC approaches bonded HA and HB. The energy processed by the system is sufficient to overcome the activation barrier and crosses the transition state structure to break the HA and HB bond and form a new HB and HC bond. The new bond oscillate as r1 (AB) increases.
|-
|2.
| [[File:condition2wxy0119.png|350px]]
| [[File:secondwxy0119.png|350px]]
| HC approaches HB but the energy is insufficient to reach the transition state point, HC then moves further away from the bonded H2 molecule and no new bond is formed. The oscillation along r2 is a result of an increase in momentum p1.
|-
|3.
| [[File:condition3wxy0119.png|350px]]
| [[File:thirdwxy0119.png|350px]]
| Similarly to condition 1 but with a more negative p1, HA - HB bond oscillates more when r2 decreases. The reaction proceeds and crosses the transition state point to form the products.
|-
|4.
| [[File:condition4wxy0119.png|350px]]
| [[File:fourthwxy0119.png|350px]]
| Both p1 and p2 are more negative than the previous conditions. The HA - HB bond oscillates vigorously with larger amplitudes. The barrier recrossing happens after the system crosses the transition state region but eventually reverts back to the reactants.
|-
|5.
| [[File:condition5wxy0119.png|350px]]
| [[File:fifthwxy0119.png|350px]]
| With a slight more negative value of p1 as compared to that in condition 4, the system now processes appropriate energy to proceed to forming products. The vigorous oscillations are resulting from high momenta and barrier recrossing occurs.
|}

=== Transition State Theory ===

{{fontcolor|red|Question 5ː State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}

The main assumptions of Transition State Theory (TST) are<ref name="TS" />ː

1. The behaviours of the atomic nuclei follow the classical mechanics rules, i.e. the transition structure is formed as long as atoms or molecules collide with enough energy.

2. The intermediates have a long enough lifetime for the energies to be Boltzmann distributed preceding the next step.

3. The lowest energy saddle point on the potential energy surface is passed over in the reaction system.

The TST does not take into account of the quantum tunnelling effect which allows reactions to happen even when the energy carried by the system is lower than the activation energy barrier, especially for reactions with low energy barriers. Nor does the theory consider the intermediates with short lifetimes. When the energies are not fully distributed, the momentum of the reaction trajectory from the reactants to the intermediate can have effects on product selectivity. Moreover, the theory fails at high temperature because higher vibrational energy modes can be populated for a molecule at high temperature. The molecules can have complex motion and their collisions may result in a higher energy transition structure. The experimental results may thus deviate from that predicted by the TST. For example in condition 4, the kinetic energy of the reactants is much higher than the activation energy required. However, the momenta between atoms are very high and higher vibrational modes are populated that result in higher energy transition structures formed and the system does not pass though the lowest transition saddle point as shown in the diagram. Barrier recrossing occurs but the products are not formed as opposed to the predictions by the TST.

== EXERCISE 2: F - H - H SYSTEM ==

=== PES inspection ===

=== Reaction Energetics ===

{{fontcolor1|red|Question 6ː 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?}}

{| class="wikitable" border=1
|[[File:Q6FH2wxy0119.png|600px]]
|[[File:Q6HFHwxy0119.png|600px]]
|-
|Figure.15 F + H2 position on surface plot
|Figure.16 HF and H position on surface plot
|}

The surface plots of the potential energy surface of H + HF → F + H2 reaction is shown (Fig.15 Fig.16). Distance AB is the distance between H and F atoms and distance BC is the distance between two H atoms. The two minimum points shown in the graph are positions of F + H2 (Fig.15) and H + HF (Fig.16) respectively. If forward reaction is H + HF → F + H2, then the backward reaction is F + H2 → H + HF. As clearly shown in the diagram F + H2 are at a minimum point of higher potential energy than that of H + HF. Thus, by comparing the potential energy of reactants and products, the signs of changes in enthalpy (ΔH) and the energetics are determined. H + HF → F + H2 is endothermic and F + H2 → H + HF is exothermic.

The energetics of the two reactions are in accordance to the predictions by comparing bond strengths of chemical species. The bond energy of H-F bond (565 kJ/mol) is higher than the bond energy of H-H bond (432 kJ/mol). Thus, energy released by forming the H-H bond is not enough to compensate for energy required for breaking the H-F bond and the reaction H + HF → F + H2 is endothermic. On the contrary, the reaction F + H2 → H + HF is exothermic.

=== Transition State Approximation ===

{{fontcolor1|red|Question 7ː Locate the approximate position of the transition state.}}

{| class="wikitable" border=1
|[[File:Q7TStransitionwxy0119.png|390px]]
|[[File:Q7TS2wxy0119.png|390px]]
|[[File:Q7TS3wxy0119.png|390px]]
|-
|Figure.17 Transition state surface plot
|Figure.18 Transition state contour plot
|Figure.19 Internuclear distance against time plot (at TS point)
|}

The transition state (TS) is shown as a black dot on the potential energy surface plot (Fig.17) and a red cross on the contour plot. (Fig,18) The position of the TS is at the point where the distance between F and H (rFH) is 1.810Å and the distance between two H atoms (rHH) is 0.746Å. At the transition state position, the distance between F and H, between two H atoms are constant with no momentum as shown in the internuclear distance against time plot as horizontal and flat lines, validating the position is the transition state point (Fig.19)

=== Activation Energies ===

{{fontcolor1|red|Question 8ː Report the activation energy for both reactions.}}

{| class="wikitable" border=1
|[[File:Q8Ea2wxy0119.png|600px]]
|[[File:Q8Ea1wxy0119.png|600px]]
|-
|Figure.21 Energy against time plot (HF + H)
|Figure.20 Energy against time plot (F + H2)
|}

By performing MEP calculation with slight increase and decrease of the rFH to 1.820Å to perform F + H2 → H + HF reaction and to 1.800Å to perform H + HF → F + H2 reaction, the activation energies (Ea) of both reactions can be calculated from the differences in their potential energies (Fig.20 Fig.21)ː

F + H2 → H + HFː Ea = -103.751 - (-133.624) = +29.873 kcal mol-1

H + HF → F + H2ː Ea = -103.751 - (-103.972) = + 0.221 kcal mol-1

=== Reaction Dynamics ===

{{fontcolor1|red|Question 9ː In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

Reactionː F + H2 → H + HF

Initial condition setː rHF = 2Å rHH = 0.74Å pHF = -0.5 pHH = 0.1

{| class="wikitable" border=1
|[[File:Q9plot1wxy0119.png|390px]]
|[[File:Q9plot2wxy0119.png|390px]]
|[[File:Q9plot3wxy0119.png|390px]]
|-
|Figure.23 Contour plot
|Figure.24 Surface plot
|Figure.25 Internuclear momentum against time plot
|-
|[[File:Q9ani1wxy0119.png|390px]]
|[[File:Q9ani2wxy0119.png|390px]]
|-
|Figure.26 Animation figure (at the start of the reaction)
|Figure.27 Animation figure (at the end of the reaction)
|}

As shown in the plots and animation figures, the H-H bond in the reactants is broken and H-F bond in the products is formed. There are small momenta between both H and F and H-H bond that provide the initial kinetic energy of the reactants and the initial vibrational energy of the H-H bond in the reactants. The initial energy is required to break the H-H bond and reduce the distance of H and F for bond forming and potential energy is released when H-F bond is formed. Since energy is conserved, the reaction energy released is converted to the vibrational energy of the H-F bond and the translational kinetic energies of the products. This is shown in the internuclear momentum against time graph as the initial vibrations of B-C (H-H bond) becomes flat as two H atoms are separated in the product and the A-B (H-F bond) formed in the product vibrates with a large amplitude. (Fig.25) The contour plot and the surface plot also clearly show there is a significant increase in the amplitude of the oscillation of H-F bond which is an indication of the large vibrational energy of H-F bond. (Fig.23 Fig.24)

The predictions can be proved experimentally. The vibrational energy of H-F bond can be determined by measuring the frequency and the intensity of the absorption band of the H-F bond with infrared spectroscopy. The conversion to the kinetic energies of the products can be observed by measuring the temperature of the reaction as the kinetic energy can be converted to thermal energy as products collide with solvent molecules. The temperature is expected to increase as this is an exothermic reaction.

=== Energy Distribution and Reactivity ===

{{fontcolor|red|Question 10ː 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.}}

{| class="wikitable" border=1
|[[File:Q9plot1wxy0119.png|390px]]
|[[File:Q9plot2wxy0119.png|390px]]
|[[File:Q9plot3wxy0119.png|390px]]
|-
|Figure.23 Contour plot
|Figure.24 Surface plot
|Figure.25 Internuclear momentum against time plot
|-
|[[File:Q9ani1wxy0119.png|390px]]
|[[File:Q9ani2wxy0119.png|390px]]
|-
|Figure.26 Animation figure (at the start of the reaction)
|Figure.27 Animation figure (at the end of the reaction)
|}

for investigation, 1st set condition pFH = -0.5 pHH = -3 unreactive; 2nd pHH = -2.45 reactive; 3rd pHH = -1.1 unreactive; 4th pHH = -0.9 reactive; 5th pHH = 1.1 unreactive ; 6th pHH = 1.9 reactive ; 7th pHH = 2.4 to 3 unreactive
; pHH = 2.1 no reaction

The cases studied are an illustration of Polanyi's empirical rules.

For H and HF

failed conditionsː 091, 2, 0.05, -20

initial condition of H HF systemː
HF distance = 0.91 HH distance = 2 pHF = 0.8 pHH = -7.5

== Rreferences ==

<references>
<ref name="TS">Eyring, H. (1935). "The Activated Complex in Chemical Reactions". J. Chem. Phys. 3 (2): 107–115.</ref>
</references>

MRD:WXY0119

2018-05-18T15:20:21Z

Xw2816: /* Reaction Dynamics */

= Molecular Reaction Dynamics Report =

== EXERCISE 1ː H + H2 SYSTEM ==

[[File:HandH2wxy0119.png]]

In the H + H2 system, the distance between the first two H atoms is r1 and the distance (bond length) between the two H atoms in the H2 molecule is r2.

=== Dynamics from the transition state region ===

{{fontcolor|red|Question 1 ː What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.}}

{| class="wikitable" border=1
|[[File:Q1TStransitionwxy0119.png|390px]]
|[[File:Q1TS2wxy0119.png|390px]]
|[[File:Q1TS3wxy0119.png|390px]]
|-
|Figure.1 Surface plot of reaction trajectory
|Figure.2 Transition state surface plot
|Figure.3 Transition state surface plot (different angle)
|}

In the reaction trajectory surface plot (Fig.1), AB is the distance r1 and BC is the distance r2. The gradient of the potential energy with regard to r1 and r2 are ∂V(r1)/∂r1 and ∂V(r2)/∂r2 respectively. The second derivatives of the potential energy with regard to r1 and r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 respectively.

At the two minimum structures which are at the two ends of the reaction pathway, ∂V(r1)/∂r1 = 0 and ∂V(r2)/∂r2 = 0 respectively. Since the two minimum structures are at two local minimum point , the second derivatives r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 are both greater than zero.

On the other hand, the components at the transition state structure (Fig.2) are different from that at minimum points. The gradient ∂V(r1)/∂r1 and ∂V(r2)/∂r2 are both equal to zero at the transition structure. However, r2 are ∂2V(r1)/∂r12 >0 and ∂2V(r2)/∂r22 <0. The transition state point is actually a saddle point which is observed more clearly in Fig.3. The Transition state point is thus distinguished from the minimum points as the potential energy surface curves inwards and downwards at the minimum points.

=== Locating the Transition State ===

{{fontcolor|red|Question 2ː 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.}}

{| class="wikitable" border=1
|[[File:TScontourwxy0119.png|390px]]
|[[File:TSsurfaceplwxy0119.png|390px]]
|[[File:TSestimatewxy0119.png|390px]]
|-
|Figure.4 Transition state contour plot
|Figure.5 Transition state surface plot
|Figure.6 Internuclear distance against time plot
|}

The best estimate of the transition state position is rts = 0.9078 Å.

When r1 = r2 = rts and momenta are set to zero, a graph of intermolecular distance against time is plotted. (Fig.6) In this graph, AB and BC lines are the same and overlap, the two lines on the graph are flat and horizontal, indicating that the atomic distances are the same at the position and with no momenta the state is at equilibrium, validating the position is the transition state position. The transition state is shown as a cross on the counter plot (Fig.4) and a dot on the surface plot (Fig.5).

=== Calculating and comparing the reaction path and trajectory ===

{{fontcolor1|red|Question 3ː Comment on how the ''mep'' and the trajectory you just calculated differ.}}

{| class="wikitable" border=1
|[[File:Mep1wxy0119.png|390px]]
|[[File:Mep2wxy0119.png|390px]]
|[[File:Mep3wxy0119.png|390px]]
|-
|Figure.7 mep calculation contour plot
|Figure.8 mep calculation surface plot
|Figure.9 mep calculation internuclear distance against time plot
|-
|[[File:Dynamic1wxy0119.png|390px]]
|[[File:Dynamic2wxy0119.png|390px]]
|[[File:Dynamic3wxy0119.png|390px]]
|-
|Figure.10 Dynamics calculation contour plot
|Figure.11 Dynamics calculation surface plot
|Figure.12 Dynamics calculation internuclear distance against time plot
|}
{| class="wikitable" border=1
|[[File:dynamics4wxy0119.png|600px]]
|[[File:Mep5wxy0119.png|600px]]
|-
|Figure.13 Dynamics calculation internuclear momenta against time plot
|Figure.14 mep calculation internuclear momenta against time plotsurface plot
|}

The mep (minimum energy path) trajectory is a smooth line on contour and surface plot (Fig.7 Fig.8) while the trajectory under dynamics calculation is oscillating along the pathway. (Fig.10 Fig.11) The mep trajectory takes more steps and longer time than the dynamics trajectory as shown in internuclear distance against time plots. (Fig. 9 Fig.12) The momentum is zero all the time under mep calculation (Fig.13) but under dynamics calculation, the momenta lines are oscillating against time.

The differences arise from mep and dynamics calculations can be explained by the the natures of the two calculations. Under mep calculation, the atoms are in extremely slow motion and the momentum is zero for each step, i.e. the atoms are "stop" after every step. The pathway under mep calculation is formed by connecting all the minimum energy points of each step and appears as a smooth line. On the other hand, atoms are in continuous motion and the momentum is accumulated after each step, atoms are then oscillating on the potential energy surface along the pathway. The steps are set to 5000 in the mep calculation but are only 500 in the dynamics calculatoin. Since the steps in mep are small, more time is needed to complete the same length on the trajectories.

=== Reactive and unreactive trajectories ===

{{fontcolor1|red|Question 4ː Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.}}

For the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, run trajectories with the following momenta p1 and p2 combinations:
{| class="wikitable" border=1
| Condition
| p1
| p2
| Total Energy/ kcal mol -1
| Reactivity
|-
|1.
| -1.25
| -2.5
| -99.018
| reactive
|-
|2.
| -1.5
| -2.0
| -100.456
| unreactive
|-
|3.
| -1.5
| -2.5
| -98.956
| reactive
|-
|4.
| -2.5
| -5.0
| -84.956
| unreactive
|-
|5.
| -2.5
| -5.2
| -83.416
| reactive
|}

{| class="wikitable" border=1
| Condition
| Surface Plot
| Contour Plot
| Description
|-
|1.
| [[File:condition1wxy0119.png|350px]]
| [[File:firstwxy0119.png|350px]]
| r2 (BC) decreases when HC approaches bonded HA and HB. The energy processed by the system is sufficient to overcome the activation barrier and crosses the transition state structure to break the HA and HB bond and form a new HB and HC bond. The new bond oscillate as r1 (AB) increases.
|-
|2.
| [[File:condition2wxy0119.png|350px]]
| [[File:secondwxy0119.png|350px]]
| HC approaches HB but the energy is insufficient to reach the transition state point, HC then moves further away from the bonded H2 molecule and no new bond is formed. The oscillation along r2 is a result of an increase in momentum p1.
|-
|3.
| [[File:condition3wxy0119.png|350px]]
| [[File:thirdwxy0119.png|350px]]
| Similarly to condition 1 but with a more negative p1, HA - HB bond oscillates more when r2 decreases. The reaction proceeds and crosses the transition state point to form the products.
|-
|4.
| [[File:condition4wxy0119.png|350px]]
| [[File:fourthwxy0119.png|350px]]
| Both p1 and p2 are more negative than the previous conditions. The HA - HB bond oscillates vigorously with larger amplitudes. The barrier recrossing happens after the system crosses the transition state region but eventually reverts back to the reactants.
|-
|5.
| [[File:condition5wxy0119.png|350px]]
| [[File:fifthwxy0119.png|350px]]
| With a slight more negative value of p1 as compared to that in condition 4, the system now processes appropriate energy to proceed to forming products. The vigorous oscillations are resulting from high momenta and barrier recrossing occurs.
|}

=== Transition State Theory ===

{{fontcolor|red|Question 5ː State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}

The main assumptions of Transition State Theory (TST) are<ref name="TS" />ː

1. The behaviours of the atomic nuclei follow the classical mechanics rules, i.e. the transition structure is formed as long as atoms or molecules collide with enough energy.

2. The intermediates have a long enough lifetime for the energies to be Boltzmann distributed preceding the next step.

3. The lowest energy saddle point on the potential energy surface is passed over in the reaction system.

The TST does not take into account of the quantum tunnelling effect which allows reactions to happen even when the energy carried by the system is lower than the activation energy barrier, especially for reactions with low energy barriers. Nor does the theory consider the intermediates with short lifetimes. When the energies are not fully distributed, the momentum of the reaction trajectory from the reactants to the intermediate can have effects on product selectivity. Moreover, the theory fails at high temperature because higher vibrational energy modes can be populated for a molecule at high temperature. The molecules can have complex motion and their collisions may result in a higher energy transition structure. The experimental results may thus deviate from that predicted by the TST. For example in condition 4, the kinetic energy of the reactants is much higher than the activation energy required. However, the momenta between atoms are very high and higher vibrational modes are populated that result in higher energy transition structures formed and the system does not pass though the lowest transition saddle point as shown in the diagram. Barrier recrossing occurs but the products are not formed as opposed to the predictions by the TST.

== EXERCISE 2: F - H - H SYSTEM ==

=== PES inspection ===

=== Reaction Energetics ===

{{fontcolor1|red|Question 6ː 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?}}

{| class="wikitable" border=1
|[[File:Q6FH2wxy0119.png|600px]]
|[[File:Q6HFHwxy0119.png|600px]]
|-
|Figure.15 F + H2 position on surface plot
|Figure.16 HF and H position on surface plot
|}

The surface plots of the potential energy surface of H + HF → F + H2 reaction is shown (Fig.15 Fig.16). Distance AB is the distance between H and F atoms and distance BC is the distance between two H atoms. The two minimum points shown in the graph are positions of F + H2 (Fig.15) and H + HF (Fig.16) respectively. If forward reaction is H + HF → F + H2, then the backward reaction is F + H2 → H + HF. As clearly shown in the diagram F + H2 are at a minimum point of higher potential energy than that of H + HF. Thus, by comparing the potential energy of reactants and products, the signs of changes in enthalpy (ΔH) and the energetics are determined. H + HF → F + H2 is endothermic and F + H2 → H + HF is exothermic.

The energetics of the two reactions are in accordance to the predictions by comparing bond strengths of chemical species. The bond energy of H-F bond (565 kJ/mol) is higher than the bond energy of H-H bond (432 kJ/mol). Thus, energy released by forming the H-H bond is not enough to compensate for energy required for breaking the H-F bond and the reaction H + HF → F + H2 is endothermic. On the contrary, the reaction F + H2 → H + HF is exothermic.

=== Transition State Approximation ===

{{fontcolor1|red|Question 7ː Locate the approximate position of the transition state.}}

{| class="wikitable" border=1
|[[File:Q7TStransitionwxy0119.png|390px]]
|[[File:Q7TS2wxy0119.png|390px]]
|[[File:Q7TS3wxy0119.png|390px]]
|-
|Figure.17 Transition state surface plot
|Figure.18 Transition state contour plot
|Figure.19 Internuclear distance against time plot (at TS point)
|}

The transition state (TS) is shown as a black dot on the potential energy surface plot (Fig.17) and a red cross on the contour plot. (Fig,18) The position of the TS is at the point where the distance between F and H (rFH) is 1.810Å and the distance between two H atoms (rHH) is 0.746Å. At the transition state position, the distance between F and H, between two H atoms are constant with no momentum as shown in the internuclear distance against time plot as horizontal and flat lines, validating the position is the transition state point (Fig.19)

=== Activation Energies ===

{{fontcolor1|red|Question 8ː Report the activation energy for both reactions.}}

{| class="wikitable" border=1
|[[File:Q8Ea2wxy0119.png|600px]]
|[[File:Q8Ea1wxy0119.png|600px]]
|-
|Figure.21 Energy against time plot (HF + H)
|Figure.20 Energy against time plot (F + H2)
|}

By performing MEP calculation with slight increase and decrease of the rFH to 1.820Å to perform F + H2 → H + HF reaction and to 1.800Å to perform H + HF → F + H2 reaction, the activation energies (Ea) of both reactions can be calculated from the differences in their potential energies (Fig.20 Fig.21)ː

F + H2 → H + HFː Ea = -103.751 - (-133.624) = +29.873 kcal mol-1

H + HF → F + H2ː Ea = -103.751 - (-103.972) = + 0.221 kcal mol-1

=== Reaction Dynamics ===

{{fontcolor1|red|Question 9ː In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

Reactionː F + H2 → H + HF

Initial condition setː rHF = 2Å rHH = 0.74Å pHF = -0.5 pHH = 0.1

{| class="wikitable" border=1
|[[File:Q9plot1wxy0119.png|390px]]
|[[File:Q9plot2wxy0119.png|390px]]
|[[File:Q9plot3wxy0119.png|390px]]
|-
|Figure.23 Contour plot
|Figure.24 Surface plot
|Figure.25 Internuclear momentum against time plot
|-
|[[File:Q9ani1wxy0119.png|390px]]
|[[File:Q9ani2wxy0119.png|390px]]
|-
|Figure.26 Animation figure (at the start of the reaction)
|Figure.27 Animation figure (at the end of the reaction)
|}

As shown in the plots and animation figures, the H-H bond in the reactants is broken and H-F bond in the products is formed. There are small momenta between both H and F and H-H bond that provide the initial kinetic energy of the reactants and the initial vibrational energy of the H-H bond in the reactants. The initial energy is required to break the H-H bond and reduce the distance of H and F for bond forming and potential energy is released when H-F bond is formed. Since energy is conserved, the reaction energy released is converted to the vibrational energy of the H-F bond and the translational kinetic energies of the products. This is shown in the internuclear momentum against time graph as the initial vibrations of B-C (H-H bond) becomes flat as two H atoms are separated in the product and the A-B (H-F bond) formed in the product vibrates with a large amplitude. (Fig.25) The contour plot and the surface plot also clearly show there is a significant increase in the amplitude of the oscillation of H-F bond which is an indication of the large vibrational energy of H-F bond. (Fig.23 Fig.24)

The predictions can be proved experimentally. The vibrational energy of H-F bond can be determined by measuring the frequency and the intensity of the absorption band of the H-F bond with infrared spectroscopy. The conversion to the kinetic energies of the products can be observed by measuring the temperature of the reaction as the kinetic energy can be converted to thermal energy as products collide with solvent molecules. The temperature is expected to increase as this is an exothermic reaction.

=== Energy Distribution and Reactivity ===

{{fontcolor|red|Question 10ː 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 investigation, 1st set condition pFH = -0.5 pHH = -3 unreactive; 2nd pHH = -2.45 reactive; 3rd pHH = -1.1 unreactive; 4th pHH = -0.9 reactive; 5th pHH = 1.1 unreactive ; 6th pHH = 1.9 reactive ; 7th pHH = 2.4 to 3 unreactive
; pHH = 2.1 no reaction

The cases studied are an illustration of Polanyi's empirical rules.

For H and HF

failed conditionsː 091, 2, 0.05, -20

initial condition of H HF systemː
HF distance = 0.91 HH distance = 2 pHF = 0.8 pHH = -7.5

== Rreferences ==

<references>
<ref name="TS">Eyring, H. (1935). "The Activated Complex in Chemical Reactions". J. Chem. Phys. 3 (2): 107–115.</ref>
</references>

MRD:WXY0119

2018-05-18T15:18:56Z

Xw2816: Undo revision 723185 by Nw716 (talk)

= Molecular Reaction Dynamics Report =

== EXERCISE 1ː H + H2 SYSTEM ==

[[File:HandH2wxy0119.png]]

In the H + H2 system, the distance between the first two H atoms is r1 and the distance (bond length) between the two H atoms in the H2 molecule is r2.

=== Dynamics from the transition state region ===

{{fontcolor|red|Question 1 ː What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.}}

{| class="wikitable" border=1
|[[File:Q1TStransitionwxy0119.png|390px]]
|[[File:Q1TS2wxy0119.png|390px]]
|[[File:Q1TS3wxy0119.png|390px]]
|-
|Figure.1 Surface plot of reaction trajectory
|Figure.2 Transition state surface plot
|Figure.3 Transition state surface plot (different angle)
|}

In the reaction trajectory surface plot (Fig.1), AB is the distance r1 and BC is the distance r2. The gradient of the potential energy with regard to r1 and r2 are ∂V(r1)/∂r1 and ∂V(r2)/∂r2 respectively. The second derivatives of the potential energy with regard to r1 and r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 respectively.

At the two minimum structures which are at the two ends of the reaction pathway, ∂V(r1)/∂r1 = 0 and ∂V(r2)/∂r2 = 0 respectively. Since the two minimum structures are at two local minimum point , the second derivatives r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 are both greater than zero.

On the other hand, the components at the transition state structure (Fig.2) are different from that at minimum points. The gradient ∂V(r1)/∂r1 and ∂V(r2)/∂r2 are both equal to zero at the transition structure. However, r2 are ∂2V(r1)/∂r12 >0 and ∂2V(r2)/∂r22 <0. The transition state point is actually a saddle point which is observed more clearly in Fig.3. The Transition state point is thus distinguished from the minimum points as the potential energy surface curves inwards and downwards at the minimum points.

=== Locating the Transition State ===

{{fontcolor|red|Question 2ː 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.}}

{| class="wikitable" border=1
|[[File:TScontourwxy0119.png|390px]]
|[[File:TSsurfaceplwxy0119.png|390px]]
|[[File:TSestimatewxy0119.png|390px]]
|-
|Figure.4 Transition state contour plot
|Figure.5 Transition state surface plot
|Figure.6 Internuclear distance against time plot
|}

The best estimate of the transition state position is rts = 0.9078 Å.

When r1 = r2 = rts and momenta are set to zero, a graph of intermolecular distance against time is plotted. (Fig.6) In this graph, AB and BC lines are the same and overlap, the two lines on the graph are flat and horizontal, indicating that the atomic distances are the same at the position and with no momenta the state is at equilibrium, validating the position is the transition state position. The transition state is shown as a cross on the counter plot (Fig.4) and a dot on the surface plot (Fig.5).

=== Calculating and comparing the reaction path and trajectory ===

{{fontcolor1|red|Question 3ː Comment on how the ''mep'' and the trajectory you just calculated differ.}}

{| class="wikitable" border=1
|[[File:Mep1wxy0119.png|390px]]
|[[File:Mep2wxy0119.png|390px]]
|[[File:Mep3wxy0119.png|390px]]
|-
|Figure.7 mep calculation contour plot
|Figure.8 mep calculation surface plot
|Figure.9 mep calculation internuclear distance against time plot
|-
|[[File:Dynamic1wxy0119.png|390px]]
|[[File:Dynamic2wxy0119.png|390px]]
|[[File:Dynamic3wxy0119.png|390px]]
|-
|Figure.10 Dynamics calculation contour plot
|Figure.11 Dynamics calculation surface plot
|Figure.12 Dynamics calculation internuclear distance against time plot
|}
{| class="wikitable" border=1
|[[File:dynamics4wxy0119.png|600px]]
|[[File:Mep5wxy0119.png|600px]]
|-
|Figure.13 Dynamics calculation internuclear momenta against time plot
|Figure.14 mep calculation internuclear momenta against time plotsurface plot
|}

The mep (minimum energy path) trajectory is a smooth line on contour and surface plot (Fig.7 Fig.8) while the trajectory under dynamics calculation is oscillating along the pathway. (Fig.10 Fig.11) The mep trajectory takes more steps and longer time than the dynamics trajectory as shown in internuclear distance against time plots. (Fig. 9 Fig.12) The momentum is zero all the time under mep calculation (Fig.13) but under dynamics calculation, the momenta lines are oscillating against time.

The differences arise from mep and dynamics calculations can be explained by the the natures of the two calculations. Under mep calculation, the atoms are in extremely slow motion and the momentum is zero for each step, i.e. the atoms are "stop" after every step. The pathway under mep calculation is formed by connecting all the minimum energy points of each step and appears as a smooth line. On the other hand, atoms are in continuous motion and the momentum is accumulated after each step, atoms are then oscillating on the potential energy surface along the pathway. The steps are set to 5000 in the mep calculation but are only 500 in the dynamics calculatoin. Since the steps in mep are small, more time is needed to complete the same length on the trajectories.

=== Reactive and unreactive trajectories ===

{{fontcolor1|red|Question 4ː Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.}}

For the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, run trajectories with the following momenta p1 and p2 combinations:
{| class="wikitable" border=1
| Condition
| p1
| p2
| Total Energy/ kcal mol -1
| Reactivity
|-
|1.
| -1.25
| -2.5
| -99.018
| reactive
|-
|2.
| -1.5
| -2.0
| -100.456
| unreactive
|-
|3.
| -1.5
| -2.5
| -98.956
| reactive
|-
|4.
| -2.5
| -5.0
| -84.956
| unreactive
|-
|5.
| -2.5
| -5.2
| -83.416
| reactive
|}

{| class="wikitable" border=1
| Condition
| Surface Plot
| Contour Plot
| Description
|-
|1.
| [[File:condition1wxy0119.png|350px]]
| [[File:firstwxy0119.png|350px]]
| r2 (BC) decreases when HC approaches bonded HA and HB. The energy processed by the system is sufficient to overcome the activation barrier and crosses the transition state structure to break the HA and HB bond and form a new HB and HC bond. The new bond oscillate as r1 (AB) increases.
|-
|2.
| [[File:condition2wxy0119.png|350px]]
| [[File:secondwxy0119.png|350px]]
| HC approaches HB but the energy is insufficient to reach the transition state point, HC then moves further away from the bonded H2 molecule and no new bond is formed. The oscillation along r2 is a result of an increase in momentum p1.
|-
|3.
| [[File:condition3wxy0119.png|350px]]
| [[File:thirdwxy0119.png|350px]]
| Similarly to condition 1 but with a more negative p1, HA - HB bond oscillates more when r2 decreases. The reaction proceeds and crosses the transition state point to form the products.
|-
|4.
| [[File:condition4wxy0119.png|350px]]
| [[File:fourthwxy0119.png|350px]]
| Both p1 and p2 are more negative than the previous conditions. The HA - HB bond oscillates vigorously with larger amplitudes. The barrier recrossing happens after the system crosses the transition state region but eventually reverts back to the reactants.
|-
|5.
| [[File:condition5wxy0119.png|350px]]
| [[File:fifthwxy0119.png|350px]]
| With a slight more negative value of p1 as compared to that in condition 4, the system now processes appropriate energy to proceed to forming products. The vigorous oscillations are resulting from high momenta and barrier recrossing occurs.
|}

=== Transition State Theory ===

{{fontcolor|red|Question 5ː State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}

The main assumptions of Transition State Theory (TST) are<ref name="TS" />ː

1. The behaviours of the atomic nuclei follow the classical mechanics rules, i.e. the transition structure is formed as long as atoms or molecules collide with enough energy.

2. The intermediates have a long enough lifetime for the energies to be Boltzmann distributed preceding the next step.

3. The lowest energy saddle point on the potential energy surface is passed over in the reaction system.

The TST does not take into account of the quantum tunnelling effect which allows reactions to happen even when the energy carried by the system is lower than the activation energy barrier, especially for reactions with low energy barriers. Nor does the theory consider the intermediates with short lifetimes. When the energies are not fully distributed, the momentum of the reaction trajectory from the reactants to the intermediate can have effects on product selectivity. Moreover, the theory fails at high temperature because higher vibrational energy modes can be populated for a molecule at high temperature. The molecules can have complex motion and their collisions may result in a higher energy transition structure. The experimental results may thus deviate from that predicted by the TST. For example in condition 4, the kinetic energy of the reactants is much higher than the activation energy required. However, the momenta between atoms are very high and higher vibrational modes are populated that result in higher energy transition structures formed and the system does not pass though the lowest transition saddle point as shown in the diagram. Barrier recrossing occurs but the products are not formed as opposed to the predictions by the TST.

== EXERCISE 2: F - H - H SYSTEM ==

=== PES inspection ===

=== Reaction Energetics ===

{{fontcolor1|red|Question 6ː 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?}}

{| class="wikitable" border=1
|[[File:Q6FH2wxy0119.png|600px]]
|[[File:Q6HFHwxy0119.png|600px]]
|-
|Figure.15 F + H2 position on surface plot
|Figure.16 HF and H position on surface plot
|}

The surface plots of the potential energy surface of H + HF → F + H2 reaction is shown (Fig.15 Fig.16). Distance AB is the distance between H and F atoms and distance BC is the distance between two H atoms. The two minimum points shown in the graph are positions of F + H2 (Fig.15) and H + HF (Fig.16) respectively. If forward reaction is H + HF → F + H2, then the backward reaction is F + H2 → H + HF. As clearly shown in the diagram F + H2 are at a minimum point of higher potential energy than that of H + HF. Thus, by comparing the potential energy of reactants and products, the signs of changes in enthalpy (ΔH) and the energetics are determined. H + HF → F + H2 is endothermic and F + H2 → H + HF is exothermic.

The energetics of the two reactions are in accordance to the predictions by comparing bond strengths of chemical species. The bond energy of H-F bond (565 kJ/mol) is higher than the bond energy of H-H bond (432 kJ/mol). Thus, energy released by forming the H-H bond is not enough to compensate for energy required for breaking the H-F bond and the reaction H + HF → F + H2 is endothermic. On the contrary, the reaction F + H2 → H + HF is exothermic.

=== Transition State Approximation ===

{{fontcolor1|red|Question 7ː Locate the approximate position of the transition state.}}

{| class="wikitable" border=1
|[[File:Q7TStransitionwxy0119.png|390px]]
|[[File:Q7TS2wxy0119.png|390px]]
|[[File:Q7TS3wxy0119.png|390px]]
|-
|Figure.17 Transition state surface plot
|Figure.18 Transition state contour plot
|Figure.19 Internuclear distance against time plot (at TS point)
|}

The transition state (TS) is shown as a black dot on the potential energy surface plot (Fig.17) and a red cross on the contour plot. (Fig,18) The position of the TS is at the point where the distance between F and H (rFH) is 1.810Å and the distance between two H atoms (rHH) is 0.746Å. At the transition state position, the distance between F and H, between two H atoms are constant with no momentum as shown in the internuclear distance against time plot as horizontal and flat lines, validating the position is the transition state point (Fig.19)

=== Activation Energies ===

{{fontcolor1|red|Question 8ː Report the activation energy for both reactions.}}

{| class="wikitable" border=1
|[[File:Q8Ea2wxy0119.png|600px]]
|[[File:Q8Ea1wxy0119.png|600px]]
|-
|Figure.21 Energy against time plot (HF + H)
|Figure.20 Energy against time plot (F + H2)
|}

By performing MEP calculation with slight increase and decrease of the rFH to 1.820Å to perform F + H2 → H + HF reaction and to 1.800Å to perform H + HF → F + H2 reaction, the activation energies (Ea) of both reactions can be calculated from the differences in their potential energies (Fig.20 Fig.21)ː

F + H2 → H + HFː Ea = -103.751 - (-133.624) = +29.873 kcal mol-1

H + HF → F + H2ː Ea = -103.751 - (-103.972) = + 0.221 kcal mol-1

=== Reaction Dynamics ===

{{fontcolor1|red|Question 9ː In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

Reactionː F + H2 → H + HF

Initial condition setː rHF = 2Å rHH = 0.74Å pHF = -0.5 pHH = 0.1

{| class="wikitable" border=1
|[[File:Q9plot1wxy0119.png|390px]]
|[[File:Q9plot2wxy0119.png|390px]]
|[[File:Q9plot3wxy0119.png|390px]]
|-
|Figure.23 Contour plot
|Figure.24 Surface plot
|Figure.25 Internuclear momentum against time plot
|-
|[[File:Q9ani1wxy0119.png|390px]]
|[[File:Q9ani2wxy0119.png|390px]]
|-
|Figure.26 Animation figure (at the start of the reaction)
|Figure.27 Animation figure (at the end of the reaction)
|}

As shown in the plots and animation figures, the H-H bond in the reactants is broken and H-F bond in the products is formed. There are small momenta between both H and F and H-H bond that provide the initial kinetic energy of the reactants and the initial vibrational energy of the H-H bond in the reactants. The initial energy is required to break the H-H bond and reduce the distance of H and F for bond forming and potential energy is released when H-F bond is formed. Since energy is conserved, the reaction energy released is converted to the vibrational energy of the H-F bond and the translational kinetic energies of the products. This is shown in the internuclear momentum against time graph as the initial vibrations of B-C (H-H bond) becomes flat as two H atoms are separated in the product and the A-B (H-F bond) formed in the product vibrates with a large amplitude. (Fig.25) The contour plot and the surface plot also clearly show there is a significant increase in the amplitude of the oscillation of H-F bond which is an indication of the large vibrational energy of H-F bond. (Fig.23 Fig.24)

The predictions can be proved experimentally. The vibrational energy of H-F bond can be determined by measuring frequency and the intensity of the absorption band of the H-F bond with infrared spectroscopy. The conversion to the kinetic energies of the products can be observed by measuring the temperature of the reaction as the kinetic energy can be converted to thermal energy as products collide with solvent molecules. The temperature is expected to increase as this is an exothermic reaction.

=== Energy Distribution and Reactivity ===

{{fontcolor|red|Question 10ː 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 investigation, 1st set condition pFH = -0.5 pHH = -3 unreactive; 2nd pHH = -2.45 reactive; 3rd pHH = -1.1 unreactive; 4th pHH = -0.9 reactive; 5th pHH = 1.1 unreactive ; 6th pHH = 1.9 reactive ; 7th pHH = 2.4 to 3 unreactive
; pHH = 2.1 no reaction

The cases studied are an illustration of Polanyi's empirical rules.

For H and HF

failed conditionsː 091, 2, 0.05, -20

initial condition of H HF systemː
HF distance = 0.91 HH distance = 2 pHF = 0.8 pHH = -7.5

== Rreferences ==

<references>
<ref name="TS">Eyring, H. (1935). "The Activated Complex in Chemical Reactions". J. Chem. Phys. 3 (2): 107–115.</ref>
</references>

MRD:WXY0119

2018-05-18T15:15:40Z

Xw2816: /* Reaction Dynamics */

= Molecular Reaction Dynamics Report =

== EXERCISE 1ː H + H2 SYSTEM ==

[[File:HandH2wxy0119.png]]

In the H + H2 system, the distance between the first two H atoms is r1 and the distance (bond length) between the two H atoms in the H2 molecule is r2.

=== Dynamics from the transition state region ===

{{fontcolor|red|Question 1 ː What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.}}

{| class="wikitable" border=1
|[[File:Q1TStransitionwxy0119.png|390px]]
|[[File:Q1TS2wxy0119.png|390px]]
|[[File:Q1TS3wxy0119.png|390px]]
|-
|Figure.1 Surface plot of reaction trajectory
|Figure.2 Transition state surface plot
|Figure.3 Transition state surface plot (different angle)
|}

In the reaction trajectory surface plot (Fig.1), AB is the distance r1 and BC is the distance r2. The gradient of the potential energy with regard to r1 and r2 are ∂V(r1)/∂r1 and ∂V(r2)/∂r2 respectively. The second derivatives of the potential energy with regard to r1 and r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 respectively.

At the two minimum structures which are at the two ends of the reaction pathway, ∂V(r1)/∂r1 = 0 and ∂V(r2)/∂r2 = 0 respectively. Since the two minimum structures are at two local minimum point , the second derivatives r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 are both greater than zero.

On the other hand, the components at the transition state structure (Fig.2) are different from that at minimum points. The gradient ∂V(r1)/∂r1 and ∂V(r2)/∂r2 are both equal to zero at the transition structure. However, r2 are ∂2V(r1)/∂r12 >0 and ∂2V(r2)/∂r22 <0. The transition state point is actually a saddle point which is observed more clearly in Fig.3. The Transition state point is thus distinguished from the minimum points as the potential energy surface curves inwards and downwards at the minimum points.

=== Locating the Transition State ===

{{fontcolor|red|Question 2ː 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.}}

{| class="wikitable" border=1
|[[File:TScontourwxy0119.png|390px]]
|[[File:TSsurfaceplwxy0119.png|390px]]
|[[File:TSestimatewxy0119.png|390px]]
|-
|Figure.4 Transition state contour plot
|Figure.5 Transition state surface plot
|Figure.6 Internuclear distance against time plot
|}

The best estimate of the transition state position is rts = 0.9078 Å.

When r1 = r2 = rts and momenta are set to zero, a graph of intermolecular distance against time is plotted. (Fig.6) In this graph, AB and BC lines are the same and overlap, the two lines on the graph are flat and horizontal, indicating that the atomic distances are the same at the position and with no momenta the state is at equilibrium, validating the position is the transition state position. The transition state is shown as a cross on the counter plot (Fig.4) and a dot on the surface plot (Fig.5).

=== Calculating and comparing the reaction path and trajectory ===

{{fontcolor1|red|Question 3ː Comment on how the ''mep'' and the trajectory you just calculated differ.}}

{| class="wikitable" border=1
|[[File:Mep1wxy0119.png|390px]]
|[[File:Mep2wxy0119.png|390px]]
|[[File:Mep3wxy0119.png|390px]]
|-
|Figure.7 mep calculation contour plot
|Figure.8 mep calculation surface plot
|Figure.9 mep calculation internuclear distance against time plot
|-
|[[File:Dynamic1wxy0119.png|390px]]
|[[File:Dynamic2wxy0119.png|390px]]
|[[File:Dynamic3wxy0119.png|390px]]
|-
|Figure.10 Dynamics calculation contour plot
|Figure.11 Dynamics calculation surface plot
|Figure.12 Dynamics calculation internuclear distance against time plot
|}
{| class="wikitable" border=1
|[[File:dynamics4wxy0119.png|600px]]
|[[File:Mep5wxy0119.png|600px]]
|-
|Figure.13 Dynamics calculation internuclear momenta against time plot
|Figure.14 mep calculation internuclear momenta against time plotsurface plot
|}

The mep (minimum energy path) trajectory is a smooth line on contour and surface plot (Fig.7 Fig.8) while the trajectory under dynamics calculation is oscillating along the pathway. (Fig.10 Fig.11) The mep trajectory takes more steps and longer time than the dynamics trajectory as shown in internuclear distance against time plots. (Fig. 9 Fig.12) The momentum is zero all the time under mep calculation (Fig.13) but under dynamics calculation, the momenta lines are oscillating against time.

The differences arise from mep and dynamics calculations can be explained by the the natures of the two calculations. Under mep calculation, the atoms are in extremely slow motion and the momentum is zero for each step, i.e. the atoms are "stop" after every step. The pathway under mep calculation is formed by connecting all the minimum energy points of each step and appears as a smooth line. On the other hand, atoms are in continuous motion and the momentum is accumulated after each step, atoms are then oscillating on the potential energy surface along the pathway. The steps are set to 5000 in the mep calculation but are only 500 in the dynamics calculatoin. Since the steps in mep are small, more time is needed to complete the same length on the trajectories.

=== Reactive and unreactive trajectories ===

{{fontcolor1|red|Question 4ː Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.}}

For the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, run trajectories with the following momenta p1 and p2 combinations:
{| class="wikitable" border=1
| Condition
| p1
| p2
| Total Energy/ kcal mol -1
| Reactivity
|-
|1.
| -1.25
| -2.5
| -99.018
| reactive
|-
|2.
| -1.5
| -2.0
| -100.456
| unreactive
|-
|3.
| -1.5
| -2.5
| -98.956
| reactive
|-
|4.
| -2.5
| -5.0
| -84.956
| unreactive
|-
|5.
| -2.5
| -5.2
| -83.416
| reactive
|}

{| class="wikitable" border=1
| Condition
| Surface Plot
| Contour Plot
| Description
|-
|1.
| [[File:condition1wxy0119.png|350px]]
| [[File:firstwxy0119.png|350px]]
| r2 (BC) decreases when HC approaches bonded HA and HB. The energy processed by the system is sufficient to overcome the activation barrier and crosses the transition state structure to break the HA and HB bond and form a new HB and HC bond. The new bond oscillate as r1 (AB) increases.
|-
|2.
| [[File:condition2wxy0119.png|350px]]
| [[File:secondwxy0119.png|350px]]
| HC approaches HB but the energy is insufficient to reach the transition state point, HC then moves further away from the bonded H2 molecule and no new bond is formed. The oscillation along r2 is a result of an increase in momentum p1.
|-
|3.
| [[File:condition3wxy0119.png|350px]]
| [[File:thirdwxy0119.png|350px]]
| Similarly to condition 1 but with a more negative p1, HA - HB bond oscillates more when r2 decreases. The reaction proceeds and crosses the transition state point to form the products.
|-
|4.
| [[File:condition4wxy0119.png|350px]]
| [[File:fourthwxy0119.png|350px]]
| Both p1 and p2 are more negative than the previous conditions. The HA - HB bond oscillates vigorously with larger amplitudes. The barrier recrossing happens after the system crosses the transition state region but eventually reverts back to the reactants.
|-
|5.
| [[File:condition5wxy0119.png|350px]]
| [[File:fifthwxy0119.png|350px]]
| With a slight more negative value of p1 as compared to that in condition 4, the system now processes appropriate energy to proceed to forming products. The vigorous oscillations are resulting from high momenta and barrier recrossing occurs.
|}

=== Transition State Theory ===

{{fontcolor|red|Question 5ː State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}

The main assumptions of Transition State Theory (TST) are<ref name="TS" />ː

1. The behaviours of the atomic nuclei follow the classical mechanics rules, i.e. the transition structure is formed as long as atoms or molecules collide with enough energy.

2. The intermediates have a long enough lifetime for the energies to be Boltzmann distributed preceding the next step.

3. The lowest energy saddle point on the potential energy surface is passed over in the reaction system.

The TST does not take into account of the quantum tunnelling effect which allows reactions to happen even when the energy carried by the system is lower than the activation energy barrier, especially for reactions with low energy barriers. Nor does the theory consider the intermediates with short lifetimes. When the energies are not fully distributed, the momentum of the reaction trajectory from the reactants to the intermediate can have effects on product selectivity. Moreover, the theory fails at high temperature because higher vibrational energy modes can be populated for a molecule at high temperature. The molecules can have complex motion and their collisions may result in a higher energy transition structure. The experimental results may thus deviate from that predicted by the TST. For example in condition 4, the kinetic energy of the reactants is much higher than the activation energy required. However, the momenta between atoms are very high and higher vibrational modes are populated that result in higher energy transition structures formed and the system does not pass though the lowest transition saddle point as shown in the diagram. Barrier recrossing occurs but the products are not formed as opposed to the predictions by the TST.

== EXERCISE 2: F - H - H SYSTEM ==

=== PES inspection ===

=== Reaction Energetics ===

{{fontcolor1|red|Question 6ː 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?}}

{| class="wikitable" border=1
|[[File:Q6FH2wxy0119.png|600px]]
|[[File:Q6HFHwxy0119.png|600px]]
|-
|Figure.15 F + H2 position on surface plot
|Figure.16 HF and H position on surface plot
|}

The surface plots of the potential energy surface of H + HF → F + H2 reaction is shown (Fig.15 Fig.16). Distance AB is the distance between H and F atoms and distance BC is the distance between two H atoms. The two minimum points shown in the graph are positions of F + H2 (Fig.15) and H + HF (Fig.16) respectively. If forward reaction is H + HF → F + H2, then the backward reaction is F + H2 → H + HF. As clearly shown in the diagram F + H2 are at a minimum point of higher potential energy than that of H + HF. Thus, by comparing the potential energy of reactants and products, the signs of changes in enthalpy (ΔH) and the energetics are determined. H + HF → F + H2 is endothermic and F + H2 → H + HF is exothermic.

The energetics of the two reactions are in accordance to the predictions by comparing bond strengths of chemical species. The bond energy of H-F bond (565 kJ/mol) is higher than the bond energy of H-H bond (432 kJ/mol). Thus, energy released by forming the H-H bond is not enough to compensate for energy required for breaking the H-F bond and the reaction H + HF → F + H2 is endothermic. On the contrary, the reaction F + H2 → H + HF is exothermic.

=== Transition State Approximation ===

{{fontcolor1|red|Question 7ː Locate the approximate position of the transition state.}}

{| class="wikitable" border=1
|[[File:Q7TStransitionwxy0119.png|390px]]
|[[File:Q7TS2wxy0119.png|390px]]
|[[File:Q7TS3wxy0119.png|390px]]
|-
|Figure.17 Transition state surface plot
|Figure.18 Transition state contour plot
|Figure.19 Internuclear distance against time plot (at TS point)
|}

The transition state (TS) is shown as a black dot on the potential energy surface plot (Fig.17) and a red cross on the contour plot. (Fig,18) The position of the TS is at the point where the distance between F and H (rFH) is 1.810Å and the distance between two H atoms (rHH) is 0.746Å. At the transition state position, the distance between F and H, between two H atoms are constant with no momentum as shown in the internuclear distance against time plot as horizontal and flat lines, validating the position is the transition state point (Fig.19)

=== Activation Energies ===

{{fontcolor1|red|Question 8ː Report the activation energy for both reactions.}}

{| class="wikitable" border=1
|[[File:Q8Ea2wxy0119.png|600px]]
|[[File:Q8Ea1wxy0119.png|600px]]
|-
|Figure.21 Energy against time plot (HF + H)
|Figure.20 Energy against time plot (F + H2)
|}

By performing MEP calculation with slight increase and decrease of the rFH to 1.820Å to perform F + H2 → H + HF reaction and to 1.800Å to perform H + HF → F + H2 reaction, the activation energies (Ea) of both reactions can be calculated from the differences in their potential energies (Fig.20 Fig.21)ː

F + H2 → H + HFː Ea = -103.751 - (-133.624) = +29.873 kcal mol-1

H + HF → F + H2ː Ea = -103.751 - (-103.972) = + 0.221 kcal mol-1

=== Reaction Dynamics ===

{{fontcolor1|red|Question 9ː In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

Reactionː F + H2 → H + HF

Initial condition setː rHF = 2Å rHH = 0.74Å pHF = -0.5 pHH = 0.1

{| class="wikitable" border=1
|[[File:Q9plot1wxy0119.png|390px]]
|[[File:Q9plot2wxy0119.png|390px]]
|[[File:Q9plot3wxy0119.png|390px]]
|-
|Figure.23 Contour plot
|Figure.24 Surface plot
|Figure.25 Internuclear momentum against time plot
|-
|[[File:Q9ani1wxy0119.png|390px]]
|[[File:Q9ani2wxy0119.png|390px]]
|-
|Figure.26 Animation figure (at the start of the reaction)
|Figure.27 Animation figure (at the end of the reaction)
|}

As shown in the plots and animation figures, the H-H bond in the reactants is broken and H-F bond in the products is formed. There are small momenta between both H and F and H-H bond that provide the initial kinetic energy of the reactants and the initial vibrational energy of the H-H bond in the reactants. The initial energy is required to break the H-H bond and reduce the distance of H and F for bond forming and potential energy is released when H-F bond is formed. Since energy is conserved, the reaction energy released is converted to the vibrational energy of the H-F bond and the translational kinetic energies of the products. This is shown in the internuclear momentum against time graph as the initial vibrations of B-C (H-H bond) becomes flat as two H atoms are separated in the product and the A-B (H-F bond) formed in the product vibrates with a large amplitude. (Fig.25) The contour plot and the surface plot also clearly show there is a significant increase in the amplitude of the oscillation of H-F bond which is an indication of the large vibrational energy of H-F bond. (Fig.23 Fig.24)

The predictions can be proved experimentally. The vibrational energy of H-F bond can be determined by measuring frequency and the intensity of the absorption band of the H-F bond with infrared spectroscopy. The conversion to the kinetic energies of the products can be observed by measuring the temperature of the reaction as the kinetic energy can be converted to thermal energy as products collide with solvent molecules. The temperature is expected to increase as this is an exothermic reaction.

=== Energy Distribution and Reactivity ===

{{fontcolor|red|Question 10ː 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 investigation, 1st set condition pFH = -0.5 pHH = -3 unreactive; 2nd pHH = -2.45 reactive; 3rd pHH = -1.1 unreactive; 4th pHH = -0.9 reactive; 5th pHH = 1.1 unreactive ; 6th pHH = 1.9 reactive ; 7th pHH = 2.4 to 3 unreactive
; pHH = 2.1 no reaction

The cases studied are an illustration of Polanyi's empirical rules.

For H and HF

failed conditionsː 091, 2, 0.05, -20

initial condition of H HF systemː
HF distance = 0.91 HH distance = 2 pHF = 0.8 pHH = -7.5

== Rreferences ==

<references>
<ref name="TS">Eyring, H. (1935). "The Activated Complex in Chemical Reactions". J. Chem. Phys. 3 (2): 107–115.</ref>
</references>

MRD:WXY0119

2018-05-18T15:15:23Z

Xw2816: /* Reaction Dynamics */

= Molecular Reaction Dynamics Report =

== EXERCISE 1ː H + H2 SYSTEM ==

[[File:HandH2wxy0119.png]]

In the H + H2 system, the distance between the first two H atoms is r1 and the distance (bond length) between the two H atoms in the H2 molecule is r2.

=== Dynamics from the transition state region ===

{{fontcolor|red|Question 1 ː What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.}}

{| class="wikitable" border=1
|[[File:Q1TStransitionwxy0119.png|390px]]
|[[File:Q1TS2wxy0119.png|390px]]
|[[File:Q1TS3wxy0119.png|390px]]
|-
|Figure.1 Surface plot of reaction trajectory
|Figure.2 Transition state surface plot
|Figure.3 Transition state surface plot (different angle)
|}

In the reaction trajectory surface plot (Fig.1), AB is the distance r1 and BC is the distance r2. The gradient of the potential energy with regard to r1 and r2 are ∂V(r1)/∂r1 and ∂V(r2)/∂r2 respectively. The second derivatives of the potential energy with regard to r1 and r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 respectively.

At the two minimum structures which are at the two ends of the reaction pathway, ∂V(r1)/∂r1 = 0 and ∂V(r2)/∂r2 = 0 respectively. Since the two minimum structures are at two local minimum point , the second derivatives r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 are both greater than zero.

On the other hand, the components at the transition state structure (Fig.2) are different from that at minimum points. The gradient ∂V(r1)/∂r1 and ∂V(r2)/∂r2 are both equal to zero at the transition structure. However, r2 are ∂2V(r1)/∂r12 >0 and ∂2V(r2)/∂r22 <0. The transition state point is actually a saddle point which is observed more clearly in Fig.3. The Transition state point is thus distinguished from the minimum points as the potential energy surface curves inwards and downwards at the minimum points.

=== Locating the Transition State ===

{{fontcolor|red|Question 2ː 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.}}

{| class="wikitable" border=1
|[[File:TScontourwxy0119.png|390px]]
|[[File:TSsurfaceplwxy0119.png|390px]]
|[[File:TSestimatewxy0119.png|390px]]
|-
|Figure.4 Transition state contour plot
|Figure.5 Transition state surface plot
|Figure.6 Internuclear distance against time plot
|}

The best estimate of the transition state position is rts = 0.9078 Å.

When r1 = r2 = rts and momenta are set to zero, a graph of intermolecular distance against time is plotted. (Fig.6) In this graph, AB and BC lines are the same and overlap, the two lines on the graph are flat and horizontal, indicating that the atomic distances are the same at the position and with no momenta the state is at equilibrium, validating the position is the transition state position. The transition state is shown as a cross on the counter plot (Fig.4) and a dot on the surface plot (Fig.5).

=== Calculating and comparing the reaction path and trajectory ===

{{fontcolor1|red|Question 3ː Comment on how the ''mep'' and the trajectory you just calculated differ.}}

{| class="wikitable" border=1
|[[File:Mep1wxy0119.png|390px]]
|[[File:Mep2wxy0119.png|390px]]
|[[File:Mep3wxy0119.png|390px]]
|-
|Figure.7 mep calculation contour plot
|Figure.8 mep calculation surface plot
|Figure.9 mep calculation internuclear distance against time plot
|-
|[[File:Dynamic1wxy0119.png|390px]]
|[[File:Dynamic2wxy0119.png|390px]]
|[[File:Dynamic3wxy0119.png|390px]]
|-
|Figure.10 Dynamics calculation contour plot
|Figure.11 Dynamics calculation surface plot
|Figure.12 Dynamics calculation internuclear distance against time plot
|}
{| class="wikitable" border=1
|[[File:dynamics4wxy0119.png|600px]]
|[[File:Mep5wxy0119.png|600px]]
|-
|Figure.13 Dynamics calculation internuclear momenta against time plot
|Figure.14 mep calculation internuclear momenta against time plotsurface plot
|}

The mep (minimum energy path) trajectory is a smooth line on contour and surface plot (Fig.7 Fig.8) while the trajectory under dynamics calculation is oscillating along the pathway. (Fig.10 Fig.11) The mep trajectory takes more steps and longer time than the dynamics trajectory as shown in internuclear distance against time plots. (Fig. 9 Fig.12) The momentum is zero all the time under mep calculation (Fig.13) but under dynamics calculation, the momenta lines are oscillating against time.

The differences arise from mep and dynamics calculations can be explained by the the natures of the two calculations. Under mep calculation, the atoms are in extremely slow motion and the momentum is zero for each step, i.e. the atoms are "stop" after every step. The pathway under mep calculation is formed by connecting all the minimum energy points of each step and appears as a smooth line. On the other hand, atoms are in continuous motion and the momentum is accumulated after each step, atoms are then oscillating on the potential energy surface along the pathway. The steps are set to 5000 in the mep calculation but are only 500 in the dynamics calculatoin. Since the steps in mep are small, more time is needed to complete the same length on the trajectories.

=== Reactive and unreactive trajectories ===

{{fontcolor1|red|Question 4ː Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.}}

For the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, run trajectories with the following momenta p1 and p2 combinations:
{| class="wikitable" border=1
| Condition
| p1
| p2
| Total Energy/ kcal mol -1
| Reactivity
|-
|1.
| -1.25
| -2.5
| -99.018
| reactive
|-
|2.
| -1.5
| -2.0
| -100.456
| unreactive
|-
|3.
| -1.5
| -2.5
| -98.956
| reactive
|-
|4.
| -2.5
| -5.0
| -84.956
| unreactive
|-
|5.
| -2.5
| -5.2
| -83.416
| reactive
|}

{| class="wikitable" border=1
| Condition
| Surface Plot
| Contour Plot
| Description
|-
|1.
| [[File:condition1wxy0119.png|350px]]
| [[File:firstwxy0119.png|350px]]
| r2 (BC) decreases when HC approaches bonded HA and HB. The energy processed by the system is sufficient to overcome the activation barrier and crosses the transition state structure to break the HA and HB bond and form a new HB and HC bond. The new bond oscillate as r1 (AB) increases.
|-
|2.
| [[File:condition2wxy0119.png|350px]]
| [[File:secondwxy0119.png|350px]]
| HC approaches HB but the energy is insufficient to reach the transition state point, HC then moves further away from the bonded H2 molecule and no new bond is formed. The oscillation along r2 is a result of an increase in momentum p1.
|-
|3.
| [[File:condition3wxy0119.png|350px]]
| [[File:thirdwxy0119.png|350px]]
| Similarly to condition 1 but with a more negative p1, HA - HB bond oscillates more when r2 decreases. The reaction proceeds and crosses the transition state point to form the products.
|-
|4.
| [[File:condition4wxy0119.png|350px]]
| [[File:fourthwxy0119.png|350px]]
| Both p1 and p2 are more negative than the previous conditions. The HA - HB bond oscillates vigorously with larger amplitudes. The barrier recrossing happens after the system crosses the transition state region but eventually reverts back to the reactants.
|-
|5.
| [[File:condition5wxy0119.png|350px]]
| [[File:fifthwxy0119.png|350px]]
| With a slight more negative value of p1 as compared to that in condition 4, the system now processes appropriate energy to proceed to forming products. The vigorous oscillations are resulting from high momenta and barrier recrossing occurs.
|}

=== Transition State Theory ===

{{fontcolor|red|Question 5ː State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}

The main assumptions of Transition State Theory (TST) are<ref name="TS" />ː

1. The behaviours of the atomic nuclei follow the classical mechanics rules, i.e. the transition structure is formed as long as atoms or molecules collide with enough energy.

2. The intermediates have a long enough lifetime for the energies to be Boltzmann distributed preceding the next step.

3. The lowest energy saddle point on the potential energy surface is passed over in the reaction system.

The TST does not take into account of the quantum tunnelling effect which allows reactions to happen even when the energy carried by the system is lower than the activation energy barrier, especially for reactions with low energy barriers. Nor does the theory consider the intermediates with short lifetimes. When the energies are not fully distributed, the momentum of the reaction trajectory from the reactants to the intermediate can have effects on product selectivity. Moreover, the theory fails at high temperature because higher vibrational energy modes can be populated for a molecule at high temperature. The molecules can have complex motion and their collisions may result in a higher energy transition structure. The experimental results may thus deviate from that predicted by the TST. For example in condition 4, the kinetic energy of the reactants is much higher than the activation energy required. However, the momenta between atoms are very high and higher vibrational modes are populated that result in higher energy transition structures formed and the system does not pass though the lowest transition saddle point as shown in the diagram. Barrier recrossing occurs but the products are not formed as opposed to the predictions by the TST.

== EXERCISE 2: F - H - H SYSTEM ==

=== PES inspection ===

=== Reaction Energetics ===

{{fontcolor1|red|Question 6ː 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?}}

{| class="wikitable" border=1
|[[File:Q6FH2wxy0119.png|600px]]
|[[File:Q6HFHwxy0119.png|600px]]
|-
|Figure.15 F + H2 position on surface plot
|Figure.16 HF and H position on surface plot
|}

The surface plots of the potential energy surface of H + HF → F + H2 reaction is shown (Fig.15 Fig.16). Distance AB is the distance between H and F atoms and distance BC is the distance between two H atoms. The two minimum points shown in the graph are positions of F + H2 (Fig.15) and H + HF (Fig.16) respectively. If forward reaction is H + HF → F + H2, then the backward reaction is F + H2 → H + HF. As clearly shown in the diagram F + H2 are at a minimum point of higher potential energy than that of H + HF. Thus, by comparing the potential energy of reactants and products, the signs of changes in enthalpy (ΔH) and the energetics are determined. H + HF → F + H2 is endothermic and F + H2 → H + HF is exothermic.

The energetics of the two reactions are in accordance to the predictions by comparing bond strengths of chemical species. The bond energy of H-F bond (565 kJ/mol) is higher than the bond energy of H-H bond (432 kJ/mol). Thus, energy released by forming the H-H bond is not enough to compensate for energy required for breaking the H-F bond and the reaction H + HF → F + H2 is endothermic. On the contrary, the reaction F + H2 → H + HF is exothermic.

=== Transition State Approximation ===

{{fontcolor1|red|Question 7ː Locate the approximate position of the transition state.}}

{| class="wikitable" border=1
|[[File:Q7TStransitionwxy0119.png|390px]]
|[[File:Q7TS2wxy0119.png|390px]]
|[[File:Q7TS3wxy0119.png|390px]]
|-
|Figure.17 Transition state surface plot
|Figure.18 Transition state contour plot
|Figure.19 Internuclear distance against time plot (at TS point)
|}

The transition state (TS) is shown as a black dot on the potential energy surface plot (Fig.17) and a red cross on the contour plot. (Fig,18) The position of the TS is at the point where the distance between F and H (rFH) is 1.810Å and the distance between two H atoms (rHH) is 0.746Å. At the transition state position, the distance between F and H, between two H atoms are constant with no momentum as shown in the internuclear distance against time plot as horizontal and flat lines, validating the position is the transition state point (Fig.19)

=== Activation Energies ===

{{fontcolor1|red|Question 8ː Report the activation energy for both reactions.}}

{| class="wikitable" border=1
|[[File:Q8Ea2wxy0119.png|600px]]
|[[File:Q8Ea1wxy0119.png|600px]]
|-
|Figure.21 Energy against time plot (HF + H)
|Figure.20 Energy against time plot (F + H2)
|}

By performing MEP calculation with slight increase and decrease of the rFH to 1.820Å to perform F + H2 → H + HF reaction and to 1.800Å to perform H + HF → F + H2 reaction, the activation energies (Ea) of both reactions can be calculated from the differences in their potential energies (Fig.20 Fig.21)ː

F + H2 → H + HFː Ea = -103.751 - (-133.624) = +29.873 kcal mol-1

H + HF → F + H2ː Ea = -103.751 - (-103.972) = + 0.221 kcal mol-1

=== Reaction Dynamics ===

{{fontcolor1|red|Question 9ː In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

F + H2 → H + HF

Initial condition setː rHF = 2Å rHH = 0.74Å pHF = -0.5 pHH = 0.1

{| class="wikitable" border=1
|[[File:Q9plot1wxy0119.png|390px]]
|[[File:Q9plot2wxy0119.png|390px]]
|[[File:Q9plot3wxy0119.png|390px]]
|-
|Figure.23 Contour plot
|Figure.24 Surface plot
|Figure.25 Internuclear momentum against time plot
|-
|[[File:Q9ani1wxy0119.png|390px]]
|[[File:Q9ani2wxy0119.png|390px]]
|-
|Figure.26 Animation figure (at the start of the reaction)
|Figure.27 Animation figure (at the end of the reaction)
|}

As shown in the plots and animation figures, the H-H bond in the reactants is broken and H-F bond in the products is formed. There are small momenta between both H and F and H-H bond that provide the initial kinetic energy of the reactants and the initial vibrational energy of the H-H bond in the reactants. The initial energy is required to break the H-H bond and reduce the distance of H and F for bond forming and potential energy is released when H-F bond is formed. Since energy is conserved, the reaction energy released is converted to the vibrational energy of the H-F bond and the translational kinetic energies of the products. This is shown in the internuclear momentum against time graph as the initial vibrations of B-C (H-H bond) becomes flat as two H atoms are separated in the product and the A-B (H-F bond) formed in the product vibrates with a large amplitude. (Fig.25) The contour plot and the surface plot also clearly show there is a significant increase in the amplitude of the oscillation of H-F bond which is an indication of the large vibrational energy of H-F bond. (Fig.23 Fig.24)

The predictions can be proved experimentally. The vibrational energy of H-F bond can be determined by measuring frequency and the intensity of the absorption band of the H-F bond with infrared spectroscopy. The conversion to the kinetic energies of the products can be observed by measuring the temperature of the reaction as the kinetic energy can be converted to thermal energy as products collide with solvent molecules. The temperature is expected to increase as this is an exothermic reaction.

=== Energy Distribution and Reactivity ===

{{fontcolor|red|Question 10ː 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 investigation, 1st set condition pFH = -0.5 pHH = -3 unreactive; 2nd pHH = -2.45 reactive; 3rd pHH = -1.1 unreactive; 4th pHH = -0.9 reactive; 5th pHH = 1.1 unreactive ; 6th pHH = 1.9 reactive ; 7th pHH = 2.4 to 3 unreactive
; pHH = 2.1 no reaction

The cases studied are an illustration of Polanyi's empirical rules.

For H and HF

failed conditionsː 091, 2, 0.05, -20

initial condition of H HF systemː
HF distance = 0.91 HH distance = 2 pHF = 0.8 pHH = -7.5

== Rreferences ==

<references>
<ref name="TS">Eyring, H. (1935). "The Activated Complex in Chemical Reactions". J. Chem. Phys. 3 (2): 107–115.</ref>
</references>

MRD:WXY0119

2018-05-18T15:08:53Z

Xw2816: /* Reaction Dynamics */

= Molecular Reaction Dynamics Report =

== EXERCISE 1ː H + H2 SYSTEM ==

[[File:HandH2wxy0119.png]]

In the H + H2 system, the distance between the first two H atoms is r1 and the distance (bond length) between the two H atoms in the H2 molecule is r2.

=== Dynamics from the transition state region ===

{{fontcolor|red|Question 1 ː What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.}}

{| class="wikitable" border=1
|[[File:Q1TStransitionwxy0119.png|390px]]
|[[File:Q1TS2wxy0119.png|390px]]
|[[File:Q1TS3wxy0119.png|390px]]
|-
|Figure.1 Surface plot of reaction trajectory
|Figure.2 Transition state surface plot
|Figure.3 Transition state surface plot (different angle)
|}

In the reaction trajectory surface plot (Fig.1), AB is the distance r1 and BC is the distance r2. The gradient of the potential energy with regard to r1 and r2 are ∂V(r1)/∂r1 and ∂V(r2)/∂r2 respectively. The second derivatives of the potential energy with regard to r1 and r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 respectively.

At the two minimum structures which are at the two ends of the reaction pathway, ∂V(r1)/∂r1 = 0 and ∂V(r2)/∂r2 = 0 respectively. Since the two minimum structures are at two local minimum point , the second derivatives r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 are both greater than zero.

On the other hand, the components at the transition state structure (Fig.2) are different from that at minimum points. The gradient ∂V(r1)/∂r1 and ∂V(r2)/∂r2 are both equal to zero at the transition structure. However, r2 are ∂2V(r1)/∂r12 >0 and ∂2V(r2)/∂r22 <0. The transition state point is actually a saddle point which is observed more clearly in Fig.3. The Transition state point is thus distinguished from the minimum points as the potential energy surface curves inwards and downwards at the minimum points.

=== Locating the Transition State ===

{{fontcolor|red|Question 2ː 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.}}

{| class="wikitable" border=1
|[[File:TScontourwxy0119.png|390px]]
|[[File:TSsurfaceplwxy0119.png|390px]]
|[[File:TSestimatewxy0119.png|390px]]
|-
|Figure.4 Transition state contour plot
|Figure.5 Transition state surface plot
|Figure.6 Internuclear distance against time plot
|}

The best estimate of the transition state position is rts = 0.9078 Å.

When r1 = r2 = rts and momenta are set to zero, a graph of intermolecular distance against time is plotted. (Fig.6) In this graph, AB and BC lines are the same and overlap, the two lines on the graph are flat and horizontal, indicating that the atomic distances are the same at the position and with no momenta the state is at equilibrium, validating the position is the transition state position. The transition state is shown as a cross on the counter plot (Fig.4) and a dot on the surface plot (Fig.5).

=== Calculating and comparing the reaction path and trajectory ===

{{fontcolor1|red|Question 3ː Comment on how the ''mep'' and the trajectory you just calculated differ.}}

{| class="wikitable" border=1
|[[File:Mep1wxy0119.png|390px]]
|[[File:Mep2wxy0119.png|390px]]
|[[File:Mep3wxy0119.png|390px]]
|-
|Figure.7 mep calculation contour plot
|Figure.8 mep calculation surface plot
|Figure.9 mep calculation internuclear distance against time plot
|-
|[[File:Dynamic1wxy0119.png|390px]]
|[[File:Dynamic2wxy0119.png|390px]]
|[[File:Dynamic3wxy0119.png|390px]]
|-
|Figure.10 Dynamics calculation contour plot
|Figure.11 Dynamics calculation surface plot
|Figure.12 Dynamics calculation internuclear distance against time plot
|}
{| class="wikitable" border=1
|[[File:dynamics4wxy0119.png|600px]]
|[[File:Mep5wxy0119.png|600px]]
|-
|Figure.13 Dynamics calculation internuclear momenta against time plot
|Figure.14 mep calculation internuclear momenta against time plotsurface plot
|}

The mep (minimum energy path) trajectory is a smooth line on contour and surface plot (Fig.7 Fig.8) while the trajectory under dynamics calculation is oscillating along the pathway. (Fig.10 Fig.11) The mep trajectory takes more steps and longer time than the dynamics trajectory as shown in internuclear distance against time plots. (Fig. 9 Fig.12) The momentum is zero all the time under mep calculation (Fig.13) but under dynamics calculation, the momenta lines are oscillating against time.

The differences arise from mep and dynamics calculations can be explained by the the natures of the two calculations. Under mep calculation, the atoms are in extremely slow motion and the momentum is zero for each step, i.e. the atoms are "stop" after every step. The pathway under mep calculation is formed by connecting all the minimum energy points of each step and appears as a smooth line. On the other hand, atoms are in continuous motion and the momentum is accumulated after each step, atoms are then oscillating on the potential energy surface along the pathway. The steps are set to 5000 in the mep calculation but are only 500 in the dynamics calculatoin. Since the steps in mep are small, more time is needed to complete the same length on the trajectories.

=== Reactive and unreactive trajectories ===

{{fontcolor1|red|Question 4ː Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.}}

For the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, run trajectories with the following momenta p1 and p2 combinations:
{| class="wikitable" border=1
| Condition
| p1
| p2
| Total Energy/ kcal mol -1
| Reactivity
|-
|1.
| -1.25
| -2.5
| -99.018
| reactive
|-
|2.
| -1.5
| -2.0
| -100.456
| unreactive
|-
|3.
| -1.5
| -2.5
| -98.956
| reactive
|-
|4.
| -2.5
| -5.0
| -84.956
| unreactive
|-
|5.
| -2.5
| -5.2
| -83.416
| reactive
|}

{| class="wikitable" border=1
| Condition
| Surface Plot
| Contour Plot
| Description
|-
|1.
| [[File:condition1wxy0119.png|350px]]
| [[File:firstwxy0119.png|350px]]
| r2 (BC) decreases when HC approaches bonded HA and HB. The energy processed by the system is sufficient to overcome the activation barrier and crosses the transition state structure to break the HA and HB bond and form a new HB and HC bond. The new bond oscillate as r1 (AB) increases.
|-
|2.
| [[File:condition2wxy0119.png|350px]]
| [[File:secondwxy0119.png|350px]]
| HC approaches HB but the energy is insufficient to reach the transition state point, HC then moves further away from the bonded H2 molecule and no new bond is formed. The oscillation along r2 is a result of an increase in momentum p1.
|-
|3.
| [[File:condition3wxy0119.png|350px]]
| [[File:thirdwxy0119.png|350px]]
| Similarly to condition 1 but with a more negative p1, HA - HB bond oscillates more when r2 decreases. The reaction proceeds and crosses the transition state point to form the products.
|-
|4.
| [[File:condition4wxy0119.png|350px]]
| [[File:fourthwxy0119.png|350px]]
| Both p1 and p2 are more negative than the previous conditions. The HA - HB bond oscillates vigorously with larger amplitudes. The barrier recrossing happens after the system crosses the transition state region but eventually reverts back to the reactants.
|-
|5.
| [[File:condition5wxy0119.png|350px]]
| [[File:fifthwxy0119.png|350px]]
| With a slight more negative value of p1 as compared to that in condition 4, the system now processes appropriate energy to proceed to forming products. The vigorous oscillations are resulting from high momenta and barrier recrossing occurs.
|}

=== Transition State Theory ===

{{fontcolor|red|Question 5ː State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}

The main assumptions of Transition State Theory (TST) are<ref name="TS" />ː

1. The behaviours of the atomic nuclei follow the classical mechanics rules, i.e. the transition structure is formed as long as atoms or molecules collide with enough energy.

2. The intermediates have a long enough lifetime for the energies to be Boltzmann distributed preceding the next step.

3. The lowest energy saddle point on the potential energy surface is passed over in the reaction system.

The TST does not take into account of the quantum tunnelling effect which allows reactions to happen even when the energy carried by the system is lower than the activation energy barrier, especially for reactions with low energy barriers. Nor does the theory consider the intermediates with short lifetimes. When the energies are not fully distributed, the momentum of the reaction trajectory from the reactants to the intermediate can have effects on product selectivity. Moreover, the theory fails at high temperature because higher vibrational energy modes can be populated for a molecule at high temperature. The molecules can have complex motion and their collisions may result in a higher energy transition structure. The experimental results may thus deviate from that predicted by the TST. For example in condition 4, the kinetic energy of the reactants is much higher than the activation energy required. However, the momenta between atoms are very high and higher vibrational modes are populated that result in higher energy transition structures formed and the system does not pass though the lowest transition saddle point as shown in the diagram. Barrier recrossing occurs but the products are not formed as opposed to the predictions by the TST.

== EXERCISE 2: F - H - H SYSTEM ==

=== PES inspection ===

=== Reaction Energetics ===

{{fontcolor1|red|Question 6ː 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?}}

{| class="wikitable" border=1
|[[File:Q6FH2wxy0119.png|600px]]
|[[File:Q6HFHwxy0119.png|600px]]
|-
|Figure.15 F + H2 position on surface plot
|Figure.16 HF and H position on surface plot
|}

The surface plots of the potential energy surface of H + HF → F + H2 reaction is shown (Fig.15 Fig.16). Distance AB is the distance between H and F atoms and distance BC is the distance between two H atoms. The two minimum points shown in the graph are positions of F + H2 (Fig.15) and H + HF (Fig.16) respectively. If forward reaction is H + HF → F + H2, then the backward reaction is F + H2 → H + HF. As clearly shown in the diagram F + H2 are at a minimum point of higher potential energy than that of H + HF. Thus, by comparing the potential energy of reactants and products, the signs of changes in enthalpy (ΔH) and the energetics are determined. H + HF → F + H2 is endothermic and F + H2 → H + HF is exothermic.

The energetics of the two reactions are in accordance to the predictions by comparing bond strengths of chemical species. The bond energy of H-F bond (565 kJ/mol) is higher than the bond energy of H-H bond (432 kJ/mol). Thus, energy released by forming the H-H bond is not enough to compensate for energy required for breaking the H-F bond and the reaction H + HF → F + H2 is endothermic. On the contrary, the reaction F + H2 → H + HF is exothermic.

=== Transition State Approximation ===

{{fontcolor1|red|Question 7ː Locate the approximate position of the transition state.}}

{| class="wikitable" border=1
|[[File:Q7TStransitionwxy0119.png|390px]]
|[[File:Q7TS2wxy0119.png|390px]]
|[[File:Q7TS3wxy0119.png|390px]]
|-
|Figure.17 Transition state surface plot
|Figure.18 Transition state contour plot
|Figure.19 Internuclear distance against time plot (at TS point)
|}

The transition state (TS) is shown as a black dot on the potential energy surface plot (Fig.17) and a red cross on the contour plot. (Fig,18) The position of the TS is at the point where the distance between F and H (rFH) is 1.810Å and the distance between two H atoms (rHH) is 0.746Å. At the transition state position, the distance between F and H, between two H atoms are constant with no momentum as shown in the internuclear distance against time plot as horizontal and flat lines, validating the position is the transition state point (Fig.19)

=== Activation Energies ===

{{fontcolor1|red|Question 8ː Report the activation energy for both reactions.}}

{| class="wikitable" border=1
|[[File:Q8Ea2wxy0119.png|600px]]
|[[File:Q8Ea1wxy0119.png|600px]]
|-
|Figure.21 Energy against time plot (HF + H)
|Figure.20 Energy against time plot (F + H2)
|}

By performing MEP calculation with slight increase and decrease of the rFH to 1.820Å to perform F + H2 → H + HF reaction and to 1.800Å to perform H + HF → F + H2 reaction, the activation energies (Ea) of both reactions can be calculated from the differences in their potential energies (Fig.20 Fig.21)ː

F + H2 → H + HFː Ea = -103.751 - (-133.624) = +29.873 kcal mol-1

H + HF → F + H2ː Ea = -103.751 - (-103.972) = + 0.221 kcal mol-1

=== Reaction Dynamics ===

{{fontcolor1|red|Question 9ː In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

F + H2 → H + HF

Initial condition setː rHF = 2Å rHH = 0.74Å pHF = -0.5 pHH = 0.1

{| class="wikitable" border=1
|[[File:Q9plot1wxy0119.png|390px]]
|[[File:Q9plot2wxy0119.png|390px]]
|[[File:Q9plot3wxy0119.png|390px]]
|-
|Figure.23 Contour plot
|Figure.24 Surface plot
|Figure.25 Internuclear momentum against time plot
|-
|[[File:Q9ani1wxy0119.png|390px]]
|[[File:Q9ani2wxy0119.png|390px]]
|-
|Figure.26 Animation figure (at the start of the reaction)
|Figure.27 Animation figure (at the end of the reaction)
|}

As shown in the plots and animation figures, the H-H bond in the reactants is broken and H-F bond in the products is formed. There are small momenta between both H and F and H-H bond that provide the initial kinetic energy of the reactants and the initial vibrational energy of the H-H bond in the reactants. The initial energy is required to break the H-H bond and reduce the distance of H and F for bond forming and potential energy is released when H-F bond is formed. Since energy is conserved, the reaction energy released is converted to the vibrational energy of the H-F bond and the translational kinetic energies of the products. This is shown in the internuclear momentum against time graph as the initial vibrations of B-C (H-H bond) becomes flat as two H atoms are separated in the product and the A-B (H-F bond) formed in the product vibrates with a large amplitude. (Fig.25) The contour plot and the surface plot also clearly show there is a significant increase in the amplitude of the oscillation of H-F bond which is an indication of the large vibrational energy of H-F bond. (Fig.23 Fig.24)

The predictions can be proved experimentally by measuring the frequency and amplitude of vibration of H-F bond with infrared spectroscopy. The

The initial conditions have been set to be rH-H=0.74 Å , rH-F=2.00 Å and momentumH-F=-10. The potential energy released is converted into the kinetic energy of the hydrogen atom. It can be seen from the Momenta vs Time plot that pHH increases after the collision. The kinetic energy would be further convert into thermal energy as the product hydrogen atoms collide with each other. There would be an temperature increase for this exothermic reaction which can be measured experimentally.

From all of the five conditions above, it can be observed that the final HF molecule contains great vibrational energy, as seen from the oscillation of the reaction paths. Reaction energy released as F approaches H2 becomes the motion in HF, the product vibration, whereas energy released as HF separates from H becomes the motion along the BC distance coordinate, the product translation. The HF vibrational energy can be determined using IR and analyse the frequency of the vibrational band. Translational energy of H atom can be confirmed by measuring the scatter of the products. The energy distribution can be measured by recording the infrared chemiluminescence of the reaction under "arrested relaxation".

=== Energy Distribution and Reactivity ===

{{fontcolor|red|Question 10ː 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 investigation, 1st set condition pFH = -0.5 pHH = -3 unreactive; 2nd pHH = -2.45 reactive; 3rd pHH = -1.1 unreactive; 4th pHH = -0.9 reactive; 5th pHH = 1.1 unreactive ; 6th pHH = 1.9 reactive ; 7th pHH = 2.4 to 3 unreactive
; pHH = 2.1 no reaction

The cases studied are an illustration of Polanyi's empirical rules.

For H and HF

failed conditionsː 091, 2, 0.05, -20

initial condition of H HF systemː
HF distance = 0.91 HH distance = 2 pHF = 0.8 pHH = -7.5

== Rreferences ==

<references>
<ref name="TS">Eyring, H. (1935). "The Activated Complex in Chemical Reactions". J. Chem. Phys. 3 (2): 107–115.</ref>
</references>

MRD:WXY0119

2018-05-18T14:59:13Z

Xw2816: /* Reaction Dynamics */

= Molecular Reaction Dynamics Report =

== EXERCISE 1ː H + H2 SYSTEM ==

[[File:HandH2wxy0119.png]]

In the H + H2 system, the distance between the first two H atoms is r1 and the distance (bond length) between the two H atoms in the H2 molecule is r2.

=== Dynamics from the transition state region ===

{{fontcolor|red|Question 1 ː What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.}}

{| class="wikitable" border=1
|[[File:Q1TStransitionwxy0119.png|390px]]
|[[File:Q1TS2wxy0119.png|390px]]
|[[File:Q1TS3wxy0119.png|390px]]
|-
|Figure.1 Surface plot of reaction trajectory
|Figure.2 Transition state surface plot
|Figure.3 Transition state surface plot (different angle)
|}

In the reaction trajectory surface plot (Fig.1), AB is the distance r1 and BC is the distance r2. The gradient of the potential energy with regard to r1 and r2 are ∂V(r1)/∂r1 and ∂V(r2)/∂r2 respectively. The second derivatives of the potential energy with regard to r1 and r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 respectively.

At the two minimum structures which are at the two ends of the reaction pathway, ∂V(r1)/∂r1 = 0 and ∂V(r2)/∂r2 = 0 respectively. Since the two minimum structures are at two local minimum point , the second derivatives r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 are both greater than zero.

On the other hand, the components at the transition state structure (Fig.2) are different from that at minimum points. The gradient ∂V(r1)/∂r1 and ∂V(r2)/∂r2 are both equal to zero at the transition structure. However, r2 are ∂2V(r1)/∂r12 >0 and ∂2V(r2)/∂r22 <0. The transition state point is actually a saddle point which is observed more clearly in Fig.3. The Transition state point is thus distinguished from the minimum points as the potential energy surface curves inwards and downwards at the minimum points.

=== Locating the Transition State ===

{{fontcolor|red|Question 2ː 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.}}

{| class="wikitable" border=1
|[[File:TScontourwxy0119.png|390px]]
|[[File:TSsurfaceplwxy0119.png|390px]]
|[[File:TSestimatewxy0119.png|390px]]
|-
|Figure.4 Transition state contour plot
|Figure.5 Transition state surface plot
|Figure.6 Internuclear distance against time plot
|}

The best estimate of the transition state position is rts = 0.9078 Å.

When r1 = r2 = rts and momenta are set to zero, a graph of intermolecular distance against time is plotted. (Fig.6) In this graph, AB and BC lines are the same and overlap, the two lines on the graph are flat and horizontal, indicating that the atomic distances are the same at the position and with no momenta the state is at equilibrium, validating the position is the transition state position. The transition state is shown as a cross on the counter plot (Fig.4) and a dot on the surface plot (Fig.5).

=== Calculating and comparing the reaction path and trajectory ===

{{fontcolor1|red|Question 3ː Comment on how the ''mep'' and the trajectory you just calculated differ.}}

{| class="wikitable" border=1
|[[File:Mep1wxy0119.png|390px]]
|[[File:Mep2wxy0119.png|390px]]
|[[File:Mep3wxy0119.png|390px]]
|-
|Figure.7 mep calculation contour plot
|Figure.8 mep calculation surface plot
|Figure.9 mep calculation internuclear distance against time plot
|-
|[[File:Dynamic1wxy0119.png|390px]]
|[[File:Dynamic2wxy0119.png|390px]]
|[[File:Dynamic3wxy0119.png|390px]]
|-
|Figure.10 Dynamics calculation contour plot
|Figure.11 Dynamics calculation surface plot
|Figure.12 Dynamics calculation internuclear distance against time plot
|}
{| class="wikitable" border=1
|[[File:dynamics4wxy0119.png|600px]]
|[[File:Mep5wxy0119.png|600px]]
|-
|Figure.13 Dynamics calculation internuclear momenta against time plot
|Figure.14 mep calculation internuclear momenta against time plotsurface plot
|}

The mep (minimum energy path) trajectory is a smooth line on contour and surface plot (Fig.7 Fig.8) while the trajectory under dynamics calculation is oscillating along the pathway. (Fig.10 Fig.11) The mep trajectory takes more steps and longer time than the dynamics trajectory as shown in internuclear distance against time plots. (Fig. 9 Fig.12) The momentum is zero all the time under mep calculation (Fig.13) but under dynamics calculation, the momenta lines are oscillating against time.

The differences arise from mep and dynamics calculations can be explained by the the natures of the two calculations. Under mep calculation, the atoms are in extremely slow motion and the momentum is zero for each step, i.e. the atoms are "stop" after every step. The pathway under mep calculation is formed by connecting all the minimum energy points of each step and appears as a smooth line. On the other hand, atoms are in continuous motion and the momentum is accumulated after each step, atoms are then oscillating on the potential energy surface along the pathway. The steps are set to 5000 in the mep calculation but are only 500 in the dynamics calculatoin. Since the steps in mep are small, more time is needed to complete the same length on the trajectories.

=== Reactive and unreactive trajectories ===

{{fontcolor1|red|Question 4ː Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.}}

For the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, run trajectories with the following momenta p1 and p2 combinations:
{| class="wikitable" border=1
| Condition
| p1
| p2
| Total Energy/ kcal mol -1
| Reactivity
|-
|1.
| -1.25
| -2.5
| -99.018
| reactive
|-
|2.
| -1.5
| -2.0
| -100.456
| unreactive
|-
|3.
| -1.5
| -2.5
| -98.956
| reactive
|-
|4.
| -2.5
| -5.0
| -84.956
| unreactive
|-
|5.
| -2.5
| -5.2
| -83.416
| reactive
|}

{| class="wikitable" border=1
| Condition
| Surface Plot
| Contour Plot
| Description
|-
|1.
| [[File:condition1wxy0119.png|350px]]
| [[File:firstwxy0119.png|350px]]
| r2 (BC) decreases when HC approaches bonded HA and HB. The energy processed by the system is sufficient to overcome the activation barrier and crosses the transition state structure to break the HA and HB bond and form a new HB and HC bond. The new bond oscillate as r1 (AB) increases.
|-
|2.
| [[File:condition2wxy0119.png|350px]]
| [[File:secondwxy0119.png|350px]]
| HC approaches HB but the energy is insufficient to reach the transition state point, HC then moves further away from the bonded H2 molecule and no new bond is formed. The oscillation along r2 is a result of an increase in momentum p1.
|-
|3.
| [[File:condition3wxy0119.png|350px]]
| [[File:thirdwxy0119.png|350px]]
| Similarly to condition 1 but with a more negative p1, HA - HB bond oscillates more when r2 decreases. The reaction proceeds and crosses the transition state point to form the products.
|-
|4.
| [[File:condition4wxy0119.png|350px]]
| [[File:fourthwxy0119.png|350px]]
| Both p1 and p2 are more negative than the previous conditions. The HA - HB bond oscillates vigorously with larger amplitudes. The barrier recrossing happens after the system crosses the transition state region but eventually reverts back to the reactants.
|-
|5.
| [[File:condition5wxy0119.png|350px]]
| [[File:fifthwxy0119.png|350px]]
| With a slight more negative value of p1 as compared to that in condition 4, the system now processes appropriate energy to proceed to forming products. The vigorous oscillations are resulting from high momenta and barrier recrossing occurs.
|}

=== Transition State Theory ===

{{fontcolor|red|Question 5ː State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}

The main assumptions of Transition State Theory (TST) are<ref name="TS" />ː

1. The behaviours of the atomic nuclei follow the classical mechanics rules, i.e. the transition structure is formed as long as atoms or molecules collide with enough energy.

2. The intermediates have a long enough lifetime for the energies to be Boltzmann distributed preceding the next step.

3. The lowest energy saddle point on the potential energy surface is passed over in the reaction system.

The TST does not take into account of the quantum tunnelling effect which allows reactions to happen even when the energy carried by the system is lower than the activation energy barrier, especially for reactions with low energy barriers. Nor does the theory consider the intermediates with short lifetimes. When the energies are not fully distributed, the momentum of the reaction trajectory from the reactants to the intermediate can have effects on product selectivity. Moreover, the theory fails at high temperature because higher vibrational energy modes can be populated for a molecule at high temperature. The molecules can have complex motion and their collisions may result in a higher energy transition structure. The experimental results may thus deviate from that predicted by the TST. For example in condition 4, the kinetic energy of the reactants is much higher than the activation energy required. However, the momenta between atoms are very high and higher vibrational modes are populated that result in higher energy transition structures formed and the system does not pass though the lowest transition saddle point as shown in the diagram. Barrier recrossing occurs but the products are not formed as opposed to the predictions by the TST.

== EXERCISE 2: F - H - H SYSTEM ==

=== PES inspection ===

=== Reaction Energetics ===

{{fontcolor1|red|Question 6ː 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?}}

{| class="wikitable" border=1
|[[File:Q6FH2wxy0119.png|600px]]
|[[File:Q6HFHwxy0119.png|600px]]
|-
|Figure.15 F + H2 position on surface plot
|Figure.16 HF and H position on surface plot
|}

The surface plots of the potential energy surface of H + HF → F + H2 reaction is shown (Fig.15 Fig.16). Distance AB is the distance between H and F atoms and distance BC is the distance between two H atoms. The two minimum points shown in the graph are positions of F + H2 (Fig.15) and H + HF (Fig.16) respectively. If forward reaction is H + HF → F + H2, then the backward reaction is F + H2 → H + HF. As clearly shown in the diagram F + H2 are at a minimum point of higher potential energy than that of H + HF. Thus, by comparing the potential energy of reactants and products, the signs of changes in enthalpy (ΔH) and the energetics are determined. H + HF → F + H2 is endothermic and F + H2 → H + HF is exothermic.

The energetics of the two reactions are in accordance to the predictions by comparing bond strengths of chemical species. The bond energy of H-F bond (565 kJ/mol) is higher than the bond energy of H-H bond (432 kJ/mol). Thus, energy released by forming the H-H bond is not enough to compensate for energy required for breaking the H-F bond and the reaction H + HF → F + H2 is endothermic. On the contrary, the reaction F + H2 → H + HF is exothermic.

=== Transition State Approximation ===

{{fontcolor1|red|Question 7ː Locate the approximate position of the transition state.}}

{| class="wikitable" border=1
|[[File:Q7TStransitionwxy0119.png|390px]]
|[[File:Q7TS2wxy0119.png|390px]]
|[[File:Q7TS3wxy0119.png|390px]]
|-
|Figure.17 Transition state surface plot
|Figure.18 Transition state contour plot
|Figure.19 Internuclear distance against time plot (at TS point)
|}

The transition state (TS) is shown as a black dot on the potential energy surface plot (Fig.17) and a red cross on the contour plot. (Fig,18) The position of the TS is at the point where the distance between F and H (rFH) is 1.810Å and the distance between two H atoms (rHH) is 0.746Å. At the transition state position, the distance between F and H, between two H atoms are constant with no momentum as shown in the internuclear distance against time plot as horizontal and flat lines, validating the position is the transition state point (Fig.19)

=== Activation Energies ===

{{fontcolor1|red|Question 8ː Report the activation energy for both reactions.}}

{| class="wikitable" border=1
|[[File:Q8Ea2wxy0119.png|600px]]
|[[File:Q8Ea1wxy0119.png|600px]]
|-
|Figure.21 Energy against time plot (HF + H)
|Figure.20 Energy against time plot (F + H2)
|}

By performing MEP calculation with slight increase and decrease of the rFH to 1.820Å to perform F + H2 → H + HF reaction and to 1.800Å to perform H + HF → F + H2 reaction, the activation energies (Ea) of both reactions can be calculated from the differences in their potential energies (Fig.20 Fig.21)ː

F + H2 → H + HFː Ea = -103.751 - (-133.624) = +29.873 kcal mol-1

H + HF → F + H2ː Ea = -103.751 - (-103.972) = + 0.221 kcal mol-1

=== Reaction Dynamics ===

{{fontcolor1|red|Question 9ː In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

F + H2 → H + HF

Initial condition setː rHF = 2Å rHH = 0.74Å pHF = -0.5 pHH = 0.1

{| class="wikitable" border=1
|[[File:Q9plot1wxy0119.png|390px]]
|[[File:Q9plot2wxy0119.png|390px]]
|[[File:Q9plot3wxy0119.png|390px]]
|-
|Figure.23 Contour plot
|Figure.24 Surface plot
|Figure.25 Internuclear momentum against time plot
|-
|[[File:Q9ani1wxy0119.png|390px]]
|[[File:Q9ani2wxy0119.png|390px]]
|-
|Figure.26 Animation figure (at the start of the reaction)
|Figure.27 Animation figure (at the end of the reaction)
|}

As shown in the plots and animation figures, the H-H bond is broken and H-F bond is formed. There are small momenta between both H and F and H-H bond that provide the initial kinetic energy of the reactants and the initial vibrational energy of the H-H bond in the reactants. The initial energy is required to break the H-H bond and reduce the distance of H and F for bond forming and potential energy is released when H-F bond is formed. Since energy is conserved, the reaction energy released is converted to the vibrational energy of the H-F bond and the translational kinetic energies of the products. The
The contour plot and the surface plot clearly shows there is a significant increase in the amplitude of the oscillation of H-F bond which is an indication of the large vibrational energy of H-F bond.

The initial conditions have been set to be rH-H=0.74 Å , rH-F=2.00 Å and momentumH-F=-10. The potential energy released is converted into the kinetic energy of the hydrogen atom. It can be seen from the Momenta vs Time plot that pHH increases after the collision. The kinetic energy would be further convert into thermal energy as the product hydrogen atoms collide with each other. There would be an temperature increase for this exothermic reaction which can be measured experimentally.

From all of the five conditions above, it can be observed that the final HF molecule contains great vibrational energy, as seen from the oscillation of the reaction paths. Reaction energy released as F approaches H2 becomes the motion in HF, the product vibration, whereas energy released as HF separates from H becomes the motion along the BC distance coordinate, the product translation. The HF vibrational energy can be determined using IR and analyse the frequency of the vibrational band. Translational energy of H atom can be confirmed by measuring the scatter of the products. The energy distribution can be measured by recording the infrared chemiluminescence of the reaction under "arrested relaxation".

=== Energy Distribution and Reactivity ===

{{fontcolor|red|Question 10ː 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 investigation, 1st set condition pFH = -0.5 pHH = -3 unreactive; 2nd pHH = -2.45 reactive; 3rd pHH = -1.1 unreactive; 4th pHH = -0.9 reactive; 5th pHH = 1.1 unreactive ; 6th pHH = 1.9 reactive ; 7th pHH = 2.4 to 3 unreactive
; pHH = 2.1 no reaction

The cases studied are an illustration of Polanyi's empirical rules.

For H and HF

failed conditionsː 091, 2, 0.05, -20

initial condition of H HF systemː
HF distance = 0.91 HH distance = 2 pHF = 0.8 pHH = -7.5

== Rreferences ==

<references>
<ref name="TS">Eyring, H. (1935). "The Activated Complex in Chemical Reactions". J. Chem. Phys. 3 (2): 107–115.</ref>
</references>

MRD:WXY0119

2018-05-18T14:46:26Z

Xw2816: /* Reaction Dynamics */

= Molecular Reaction Dynamics Report =

== EXERCISE 1ː H + H2 SYSTEM ==

[[File:HandH2wxy0119.png]]

In the H + H2 system, the distance between the first two H atoms is r1 and the distance (bond length) between the two H atoms in the H2 molecule is r2.

=== Dynamics from the transition state region ===

{{fontcolor|red|Question 1 ː What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.}}

{| class="wikitable" border=1
|[[File:Q1TStransitionwxy0119.png|390px]]
|[[File:Q1TS2wxy0119.png|390px]]
|[[File:Q1TS3wxy0119.png|390px]]
|-
|Figure.1 Surface plot of reaction trajectory
|Figure.2 Transition state surface plot
|Figure.3 Transition state surface plot (different angle)
|}

In the reaction trajectory surface plot (Fig.1), AB is the distance r1 and BC is the distance r2. The gradient of the potential energy with regard to r1 and r2 are ∂V(r1)/∂r1 and ∂V(r2)/∂r2 respectively. The second derivatives of the potential energy with regard to r1 and r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 respectively.

At the two minimum structures which are at the two ends of the reaction pathway, ∂V(r1)/∂r1 = 0 and ∂V(r2)/∂r2 = 0 respectively. Since the two minimum structures are at two local minimum point , the second derivatives r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 are both greater than zero.

On the other hand, the components at the transition state structure (Fig.2) are different from that at minimum points. The gradient ∂V(r1)/∂r1 and ∂V(r2)/∂r2 are both equal to zero at the transition structure. However, r2 are ∂2V(r1)/∂r12 >0 and ∂2V(r2)/∂r22 <0. The transition state point is actually a saddle point which is observed more clearly in Fig.3. The Transition state point is thus distinguished from the minimum points as the potential energy surface curves inwards and downwards at the minimum points.

=== Locating the Transition State ===

{{fontcolor|red|Question 2ː 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.}}

{| class="wikitable" border=1
|[[File:TScontourwxy0119.png|390px]]
|[[File:TSsurfaceplwxy0119.png|390px]]
|[[File:TSestimatewxy0119.png|390px]]
|-
|Figure.4 Transition state contour plot
|Figure.5 Transition state surface plot
|Figure.6 Internuclear distance against time plot
|}

The best estimate of the transition state position is rts = 0.9078 Å.

When r1 = r2 = rts and momenta are set to zero, a graph of intermolecular distance against time is plotted. (Fig.6) In this graph, AB and BC lines are the same and overlap, the two lines on the graph are flat and horizontal, indicating that the atomic distances are the same at the position and with no momenta the state is at equilibrium, validating the position is the transition state position. The transition state is shown as a cross on the counter plot (Fig.4) and a dot on the surface plot (Fig.5).

=== Calculating and comparing the reaction path and trajectory ===

{{fontcolor1|red|Question 3ː Comment on how the ''mep'' and the trajectory you just calculated differ.}}

{| class="wikitable" border=1
|[[File:Mep1wxy0119.png|390px]]
|[[File:Mep2wxy0119.png|390px]]
|[[File:Mep3wxy0119.png|390px]]
|-
|Figure.7 mep calculation contour plot
|Figure.8 mep calculation surface plot
|Figure.9 mep calculation internuclear distance against time plot
|-
|[[File:Dynamic1wxy0119.png|390px]]
|[[File:Dynamic2wxy0119.png|390px]]
|[[File:Dynamic3wxy0119.png|390px]]
|-
|Figure.10 Dynamics calculation contour plot
|Figure.11 Dynamics calculation surface plot
|Figure.12 Dynamics calculation internuclear distance against time plot
|}
{| class="wikitable" border=1
|[[File:dynamics4wxy0119.png|600px]]
|[[File:Mep5wxy0119.png|600px]]
|-
|Figure.13 Dynamics calculation internuclear momenta against time plot
|Figure.14 mep calculation internuclear momenta against time plotsurface plot
|}

The mep (minimum energy path) trajectory is a smooth line on contour and surface plot (Fig.7 Fig.8) while the trajectory under dynamics calculation is oscillating along the pathway. (Fig.10 Fig.11) The mep trajectory takes more steps and longer time than the dynamics trajectory as shown in internuclear distance against time plots. (Fig. 9 Fig.12) The momentum is zero all the time under mep calculation (Fig.13) but under dynamics calculation, the momenta lines are oscillating against time.

The differences arise from mep and dynamics calculations can be explained by the the natures of the two calculations. Under mep calculation, the atoms are in extremely slow motion and the momentum is zero for each step, i.e. the atoms are "stop" after every step. The pathway under mep calculation is formed by connecting all the minimum energy points of each step and appears as a smooth line. On the other hand, atoms are in continuous motion and the momentum is accumulated after each step, atoms are then oscillating on the potential energy surface along the pathway. The steps are set to 5000 in the mep calculation but are only 500 in the dynamics calculatoin. Since the steps in mep are small, more time is needed to complete the same length on the trajectories.

=== Reactive and unreactive trajectories ===

{{fontcolor1|red|Question 4ː Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.}}

For the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, run trajectories with the following momenta p1 and p2 combinations:
{| class="wikitable" border=1
| Condition
| p1
| p2
| Total Energy/ kcal mol -1
| Reactivity
|-
|1.
| -1.25
| -2.5
| -99.018
| reactive
|-
|2.
| -1.5
| -2.0
| -100.456
| unreactive
|-
|3.
| -1.5
| -2.5
| -98.956
| reactive
|-
|4.
| -2.5
| -5.0
| -84.956
| unreactive
|-
|5.
| -2.5
| -5.2
| -83.416
| reactive
|}

{| class="wikitable" border=1
| Condition
| Surface Plot
| Contour Plot
| Description
|-
|1.
| [[File:condition1wxy0119.png|350px]]
| [[File:firstwxy0119.png|350px]]
| r2 (BC) decreases when HC approaches bonded HA and HB. The energy processed by the system is sufficient to overcome the activation barrier and crosses the transition state structure to break the HA and HB bond and form a new HB and HC bond. The new bond oscillate as r1 (AB) increases.
|-
|2.
| [[File:condition2wxy0119.png|350px]]
| [[File:secondwxy0119.png|350px]]
| HC approaches HB but the energy is insufficient to reach the transition state point, HC then moves further away from the bonded H2 molecule and no new bond is formed. The oscillation along r2 is a result of an increase in momentum p1.
|-
|3.
| [[File:condition3wxy0119.png|350px]]
| [[File:thirdwxy0119.png|350px]]
| Similarly to condition 1 but with a more negative p1, HA - HB bond oscillates more when r2 decreases. The reaction proceeds and crosses the transition state point to form the products.
|-
|4.
| [[File:condition4wxy0119.png|350px]]
| [[File:fourthwxy0119.png|350px]]
| Both p1 and p2 are more negative than the previous conditions. The HA - HB bond oscillates vigorously with larger amplitudes. The barrier recrossing happens after the system crosses the transition state region but eventually reverts back to the reactants.
|-
|5.
| [[File:condition5wxy0119.png|350px]]
| [[File:fifthwxy0119.png|350px]]
| With a slight more negative value of p1 as compared to that in condition 4, the system now processes appropriate energy to proceed to forming products. The vigorous oscillations are resulting from high momenta and barrier recrossing occurs.
|}

=== Transition State Theory ===

{{fontcolor|red|Question 5ː State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}

The main assumptions of Transition State Theory (TST) are<ref name="TS" />ː

1. The behaviours of the atomic nuclei follow the classical mechanics rules, i.e. the transition structure is formed as long as atoms or molecules collide with enough energy.

2. The intermediates have a long enough lifetime for the energies to be Boltzmann distributed preceding the next step.

3. The lowest energy saddle point on the potential energy surface is passed over in the reaction system.

The TST does not take into account of the quantum tunnelling effect which allows reactions to happen even when the energy carried by the system is lower than the activation energy barrier, especially for reactions with low energy barriers. Nor does the theory consider the intermediates with short lifetimes. When the energies are not fully distributed, the momentum of the reaction trajectory from the reactants to the intermediate can have effects on product selectivity. Moreover, the theory fails at high temperature because higher vibrational energy modes can be populated for a molecule at high temperature. The molecules can have complex motion and their collisions may result in a higher energy transition structure. The experimental results may thus deviate from that predicted by the TST. For example in condition 4, the kinetic energy of the reactants is much higher than the activation energy required. However, the momenta between atoms are very high and higher vibrational modes are populated that result in higher energy transition structures formed and the system does not pass though the lowest transition saddle point as shown in the diagram. Barrier recrossing occurs but the products are not formed as opposed to the predictions by the TST.

== EXERCISE 2: F - H - H SYSTEM ==

=== PES inspection ===

=== Reaction Energetics ===

{{fontcolor1|red|Question 6ː 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?}}

{| class="wikitable" border=1
|[[File:Q6FH2wxy0119.png|600px]]
|[[File:Q6HFHwxy0119.png|600px]]
|-
|Figure.15 F + H2 position on surface plot
|Figure.16 HF and H position on surface plot
|}

The surface plots of the potential energy surface of H + HF → F + H2 reaction is shown (Fig.15 Fig.16). Distance AB is the distance between H and F atoms and distance BC is the distance between two H atoms. The two minimum points shown in the graph are positions of F + H2 (Fig.15) and H + HF (Fig.16) respectively. If forward reaction is H + HF → F + H2, then the backward reaction is F + H2 → H + HF. As clearly shown in the diagram F + H2 are at a minimum point of higher potential energy than that of H + HF. Thus, by comparing the potential energy of reactants and products, the signs of changes in enthalpy (ΔH) and the energetics are determined. H + HF → F + H2 is endothermic and F + H2 → H + HF is exothermic.

The energetics of the two reactions are in accordance to the predictions by comparing bond strengths of chemical species. The bond energy of H-F bond (565 kJ/mol) is higher than the bond energy of H-H bond (432 kJ/mol). Thus, energy released by forming the H-H bond is not enough to compensate for energy required for breaking the H-F bond and the reaction H + HF → F + H2 is endothermic. On the contrary, the reaction F + H2 → H + HF is exothermic.

=== Transition State Approximation ===

{{fontcolor1|red|Question 7ː Locate the approximate position of the transition state.}}

{| class="wikitable" border=1
|[[File:Q7TStransitionwxy0119.png|390px]]
|[[File:Q7TS2wxy0119.png|390px]]
|[[File:Q7TS3wxy0119.png|390px]]
|-
|Figure.17 Transition state surface plot
|Figure.18 Transition state contour plot
|Figure.19 Internuclear distance against time plot (at TS point)
|}

The transition state (TS) is shown as a black dot on the potential energy surface plot (Fig.17) and a red cross on the contour plot. (Fig,18) The position of the TS is at the point where the distance between F and H (rFH) is 1.810Å and the distance between two H atoms (rHH) is 0.746Å. At the transition state position, the distance between F and H, between two H atoms are constant with no momentum as shown in the internuclear distance against time plot as horizontal and flat lines, validating the position is the transition state point (Fig.19)

=== Activation Energies ===

{{fontcolor1|red|Question 8ː Report the activation energy for both reactions.}}

{| class="wikitable" border=1
|[[File:Q8Ea2wxy0119.png|600px]]
|[[File:Q8Ea1wxy0119.png|600px]]
|-
|Figure.21 Energy against time plot (HF + H)
|Figure.20 Energy against time plot (F + H2)
|}

By performing MEP calculation with slight increase and decrease of the rFH to 1.820Å to perform F + H2 → H + HF reaction and to 1.800Å to perform H + HF → F + H2 reaction, the activation energies (Ea) of both reactions can be calculated from the differences in their potential energies (Fig.20 Fig.21)ː

F + H2 → H + HFː Ea = -103.751 - (-133.624) = +29.873 kcal mol-1

H + HF → F + H2ː Ea = -103.751 - (-103.972) = + 0.221 kcal mol-1

=== Reaction Dynamics ===

{{fontcolor1|red|Question 9ː In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

F + H2 → H + HF

Initial condition setː rHF = 2Å rHH = 0.74Å pHF = -0.5 pHH = 0.1

{| class="wikitable" border=1
|[[File:Q9plot1wxy0119.png|390px]]
|[[File:Q9plot2wxy0119.png|390px]]
|[[File:Q9plot3wxy0119.png|390px]]
|-
|Figure.23 Contour plot
|Figure.24 Surface plot
|Figure.25 Internuclear momentum against time plot
|-
|[[File:Q9ani1wxy0119.png|390px]]
|[[File:Q9ani2wxy0119.png|390px]]
|-
|Figure.26 Animation figure (at the start of the reaction)
|Figure.27 Animation figure (at the end of the reaction)
|}

As shown in the plots and animation figures, the H-H bond is broken and H-F bond is formed. There are small momenta between both H and F and H-H bond, the
The contour plot and the surface plot clearly shows there is a significant increase in the amplitude of the oscillation of H-F bond which is an indication of the large vibrational energy of H-F bond.

The initial conditions have been set to be rH-H=0.74 Å , rH-F=2.00 Å and momentumH-F=-10. The potential energy released is converted into the kinetic energy of the hydrogen atom. It can be seen from the Momenta vs Time plot that pHH increases after the collision. The kinetic energy would be further convert into thermal energy as the product hydrogen atoms collide with each other. There would be an temperature increase for this exothermic reaction which can be measured experimentally.

From all of the five conditions above, it can be observed that the final HF molecule contains great vibrational energy, as seen from the oscillation of the reaction paths. Reaction energy released as F approaches H2 becomes the motion in HF, the product vibration, whereas energy released as HF separates from H becomes the motion along the BC distance coordinate, the product translation. The HF vibrational energy can be determined using IR and analyse the frequency of the vibrational band. Translational energy of H atom can be confirmed by measuring the scatter of the products. The energy distribution can be measured by recording the infrared chemiluminescence of the reaction under "arrested relaxation". [4]

The above five conditions illustrate that a higher p1 (pFH) is always required for the reaction to be reactive, especially in Reaction 1 when p2 (pHH) is zero. Hence, one can assume that pFH, which corresponds to a high translational energy of the reactants, effectively affects the reactivity of the reaction.

=== Energy Distribution and Reactivity ===

{{fontcolor|red|Question 10ː 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 investigation, 1st set condition pFH = -0.5 pHH = -3 unreactive; 2nd pHH = -2.45 reactive; 3rd pHH = -1.1 unreactive; 4th pHH = -0.9 reactive; 5th pHH = 1.1 unreactive ; 6th pHH = 1.9 reactive ; 7th pHH = 2.4 to 3 unreactive
; pHH = 2.1 no reaction

The cases studied are an illustration of Polanyi's empirical rules.

For H and HF

failed conditionsː 091, 2, 0.05, -20

initial condition of H HF systemː
HF distance = 0.91 HH distance = 2 pHF = 0.8 pHH = -7.5

== Rreferences ==

<references>
<ref name="TS">Eyring, H. (1935). "The Activated Complex in Chemical Reactions". J. Chem. Phys. 3 (2): 107–115.</ref>
</references>

MRD:WXY0119

2018-05-18T14:32:19Z

Xw2816: /* Reaction Dynamics */

= Molecular Reaction Dynamics Report =

== EXERCISE 1ː H + H2 SYSTEM ==

[[File:HandH2wxy0119.png]]

In the H + H2 system, the distance between the first two H atoms is r1 and the distance (bond length) between the two H atoms in the H2 molecule is r2.

=== Dynamics from the transition state region ===

{{fontcolor|red|Question 1 ː What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.}}

{| class="wikitable" border=1
|[[File:Q1TStransitionwxy0119.png|390px]]
|[[File:Q1TS2wxy0119.png|390px]]
|[[File:Q1TS3wxy0119.png|390px]]
|-
|Figure.1 Surface plot of reaction trajectory
|Figure.2 Transition state surface plot
|Figure.3 Transition state surface plot (different angle)
|}

In the reaction trajectory surface plot (Fig.1), AB is the distance r1 and BC is the distance r2. The gradient of the potential energy with regard to r1 and r2 are ∂V(r1)/∂r1 and ∂V(r2)/∂r2 respectively. The second derivatives of the potential energy with regard to r1 and r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 respectively.

At the two minimum structures which are at the two ends of the reaction pathway, ∂V(r1)/∂r1 = 0 and ∂V(r2)/∂r2 = 0 respectively. Since the two minimum structures are at two local minimum point , the second derivatives r2 are ∂2V(r1)/∂r12 and ∂2V(r2)/∂r22 are both greater than zero.

On the other hand, the components at the transition state structure (Fig.2) are different from that at minimum points. The gradient ∂V(r1)/∂r1 and ∂V(r2)/∂r2 are both equal to zero at the transition structure. However, r2 are ∂2V(r1)/∂r12 >0 and ∂2V(r2)/∂r22 <0. The transition state point is actually a saddle point which is observed more clearly in Fig.3. The Transition state point is thus distinguished from the minimum points as the potential energy surface curves inwards and downwards at the minimum points.

=== Locating the Transition State ===

{{fontcolor|red|Question 2ː 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.}}

{| class="wikitable" border=1
|[[File:TScontourwxy0119.png|390px]]
|[[File:TSsurfaceplwxy0119.png|390px]]
|[[File:TSestimatewxy0119.png|390px]]
|-
|Figure.4 Transition state contour plot
|Figure.5 Transition state surface plot
|Figure.6 Internuclear distance against time plot
|}

The best estimate of the transition state position is rts = 0.9078 Å.

When r1 = r2 = rts and momenta are set to zero, a graph of intermolecular distance against time is plotted. (Fig.6) In this graph, AB and BC lines are the same and overlap, the two lines on the graph are flat and horizontal, indicating that the atomic distances are the same at the position and with no momenta the state is at equilibrium, validating the position is the transition state position. The transition state is shown as a cross on the counter plot (Fig.4) and a dot on the surface plot (Fig.5).

=== Calculating and comparing the reaction path and trajectory ===

{{fontcolor1|red|Question 3ː Comment on how the ''mep'' and the trajectory you just calculated differ.}}

{| class="wikitable" border=1
|[[File:Mep1wxy0119.png|390px]]
|[[File:Mep2wxy0119.png|390px]]
|[[File:Mep3wxy0119.png|390px]]
|-
|Figure.7 mep calculation contour plot
|Figure.8 mep calculation surface plot
|Figure.9 mep calculation internuclear distance against time plot
|-
|[[File:Dynamic1wxy0119.png|390px]]
|[[File:Dynamic2wxy0119.png|390px]]
|[[File:Dynamic3wxy0119.png|390px]]
|-
|Figure.10 Dynamics calculation contour plot
|Figure.11 Dynamics calculation surface plot
|Figure.12 Dynamics calculation internuclear distance against time plot
|}
{| class="wikitable" border=1
|[[File:dynamics4wxy0119.png|600px]]
|[[File:Mep5wxy0119.png|600px]]
|-
|Figure.13 Dynamics calculation internuclear momenta against time plot
|Figure.14 mep calculation internuclear momenta against time plotsurface plot
|}

The mep (minimum energy path) trajectory is a smooth line on contour and surface plot (Fig.7 Fig.8) while the trajectory under dynamics calculation is oscillating along the pathway. (Fig.10 Fig.11) The mep trajectory takes more steps and longer time than the dynamics trajectory as shown in internuclear distance against time plots. (Fig. 9 Fig.12) The momentum is zero all the time under mep calculation (Fig.13) but under dynamics calculation, the momenta lines are oscillating against time.

The differences arise from mep and dynamics calculations can be explained by the the natures of the two calculations. Under mep calculation, the atoms are in extremely slow motion and the momentum is zero for each step, i.e. the atoms are "stop" after every step. The pathway under mep calculation is formed by connecting all the minimum energy points of each step and appears as a smooth line. On the other hand, atoms are in continuous motion and the momentum is accumulated after each step, atoms are then oscillating on the potential energy surface along the pathway. The steps are set to 5000 in the mep calculation but are only 500 in the dynamics calculatoin. Since the steps in mep are small, more time is needed to complete the same length on the trajectories.

=== Reactive and unreactive trajectories ===

{{fontcolor1|red|Question 4ː Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.}}

For the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, run trajectories with the following momenta p1 and p2 combinations:
{| class="wikitable" border=1
| Condition
| p1
| p2
| Total Energy/ kcal mol -1
| Reactivity
|-
|1.
| -1.25
| -2.5
| -99.018
| reactive
|-
|2.
| -1.5
| -2.0
| -100.456
| unreactive
|-
|3.
| -1.5
| -2.5
| -98.956
| reactive
|-
|4.
| -2.5
| -5.0
| -84.956
| unreactive
|-
|5.
| -2.5
| -5.2
| -83.416
| reactive
|}

{| class="wikitable" border=1
| Condition
| Surface Plot
| Contour Plot
| Description
|-
|1.
| [[File:condition1wxy0119.png|350px]]
| [[File:firstwxy0119.png|350px]]
| r2 (BC) decreases when HC approaches bonded HA and HB. The energy processed by the system is sufficient to overcome the activation barrier and crosses the transition state structure to break the HA and HB bond and form a new HB and HC bond. The new bond oscillate as r1 (AB) increases.
|-
|2.
| [[File:condition2wxy0119.png|350px]]
| [[File:secondwxy0119.png|350px]]
| HC approaches HB but the energy is insufficient to reach the transition state point, HC then moves further away from the bonded H2 molecule and no new bond is formed. The oscillation along r2 is a result of an increase in momentum p1.
|-
|3.
| [[File:condition3wxy0119.png|350px]]
| [[File:thirdwxy0119.png|350px]]
| Similarly to condition 1 but with a more negative p1, HA - HB bond oscillates more when r2 decreases. The reaction proceeds and crosses the transition state point to form the products.
|-
|4.
| [[File:condition4wxy0119.png|350px]]
| [[File:fourthwxy0119.png|350px]]
| Both p1 and p2 are more negative than the previous conditions. The HA - HB bond oscillates vigorously with larger amplitudes. The barrier recrossing happens after the system crosses the transition state region but eventually reverts back to the reactants.
|-
|5.
| [[File:condition5wxy0119.png|350px]]
| [[File:fifthwxy0119.png|350px]]
| With a slight more negative value of p1 as compared to that in condition 4, the system now processes appropriate energy to proceed to forming products. The vigorous oscillations are resulting from high momenta and barrier recrossing occurs.
|}

=== Transition State Theory ===

{{fontcolor|red|Question 5ː State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}

The main assumptions of Transition State Theory (TST) are<ref name="TS" />ː

1. The behaviours of the atomic nuclei follow the classical mechanics rules, i.e. the transition structure is formed as long as atoms or molecules collide with enough energy.

2. The intermediates have a long enough lifetime for the energies to be Boltzmann distributed preceding the next step.

3. The lowest energy saddle point on the potential energy surface is passed over in the reaction system.

The TST does not take into account of the quantum tunnelling effect which allows reactions to happen even when the energy carried by the system is lower than the activation energy barrier, especially for reactions with low energy barriers. Nor does the theory consider the intermediates with short lifetimes. When the energies are not fully distributed, the momentum of the reaction trajectory from the reactants to the intermediate can have effects on product selectivity. Moreover, the theory fails at high temperature because higher vibrational energy modes can be populated for a molecule at high temperature. The molecules can have complex motion and their collisions may result in a higher energy transition structure. The experimental results may thus deviate from that predicted by the TST. For example in condition 4, the kinetic energy of the reactants is much higher than the activation energy required. However, the momenta between atoms are very high and higher vibrational modes are populated that result in higher energy transition structures formed and the system does not pass though the lowest transition saddle point as shown in the diagram. Barrier recrossing occurs but the products are not formed as opposed to the predictions by the TST.

== EXERCISE 2: F - H - H SYSTEM ==

=== PES inspection ===

=== Reaction Energetics ===

{{fontcolor1|red|Question 6ː 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?}}

{| class="wikitable" border=1
|[[File:Q6FH2wxy0119.png|600px]]
|[[File:Q6HFHwxy0119.png|600px]]
|-
|Figure.15 F + H2 position on surface plot
|Figure.16 HF and H position on surface plot
|}

The surface plots of the potential energy surface of H + HF → F + H2 reaction is shown (Fig.15 Fig.16). Distance AB is the distance between H and F atoms and distance BC is the distance between two H atoms. The two minimum points shown in the graph are positions of F + H2 (Fig.15) and H + HF (Fig.16) respectively. If forward reaction is H + HF → F + H2, then the backward reaction is F + H2 → H + HF. As clearly shown in the diagram F + H2 are at a minimum point of higher potential energy than that of H + HF. Thus, by comparing the potential energy of reactants and products, the signs of changes in enthalpy (ΔH) and the energetics are determined. H + HF → F + H2 is endothermic and F + H2 → H + HF is exothermic.

The energetics of the two reactions are in accordance to the predictions by comparing bond strengths of chemical species. The bond energy of H-F bond (565 kJ/mol) is higher than the bond energy of H-H bond (432 kJ/mol). Thus, energy released by forming the H-H bond is not enough to compensate for energy required for breaking the H-F bond and the reaction H + HF → F + H2 is endothermic. On the contrary, the reaction F + H2 → H + HF is exothermic.

=== Transition State Approximation ===

{{fontcolor1|red|Question 7ː Locate the approximate position of the transition state.}}

{| class="wikitable" border=1
|[[File:Q7TStransitionwxy0119.png|390px]]
|[[File:Q7TS2wxy0119.png|390px]]
|[[File:Q7TS3wxy0119.png|390px]]
|-
|Figure.17 Transition state surface plot
|Figure.18 Transition state contour plot
|Figure.19 Internuclear distance against time plot (at TS point)
|}

The transition state (TS) is shown as a black dot on the potential energy surface plot (Fig.17) and a red cross on the contour plot. (Fig,18) The position of the TS is at the point where the distance between F and H (rFH) is 1.810Å and the distance between two H atoms (rHH) is 0.746Å. At the transition state position, the distance between F and H, between two H atoms are constant with no momentum as shown in the internuclear distance against time plot as horizontal and flat lines, validating the position is the transition state point (Fig.19)

=== Activation Energies ===

{{fontcolor1|red|Question 8ː Report the activation energy for both reactions.}}

{| class="wikitable" border=1
|[[File:Q8Ea2wxy0119.png|600px]]
|[[File:Q8Ea1wxy0119.png|600px]]
|-
|Figure.21 Energy against time plot (HF + H)
|Figure.20 Energy against time plot (F + H2)
|}

By performing MEP calculation with slight increase and decrease of the rFH to 1.820Å to perform F + H2 → H + HF reaction and to 1.800Å to perform H + HF → F + H2 reaction, the activation energies (Ea) of both reactions can be calculated from the differences in their potential energies (Fig.20 Fig.21)ː

F + H2 → H + HFː Ea = -103.751 - (-133.624) = +29.873 kcal mol-1

H + HF → F + H2ː Ea = -103.751 - (-103.972) = + 0.221 kcal mol-1

=== Reaction Dynamics ===

{{fontcolor1|red|Question 9ː In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

F + H2 → H + HF
Initial condition setː rHF = 2 rHH = 0.74 pHF = -0.5 pHH = 0.1

{| class="wikitable" border=1
|[[File:Q9plot1wxy0119.png|390px]]
|[[File:Q9plot2wxy0119.png|390px]]
|[[File:Q9plot3wxy0119.png|390px]]
|-
|Figure.23 Contour plot
|Figure.24 Surface plot
|Figure.25 Internuclear momenta against time plot
|-
|[[File:Q9ani1wxy0119.png|390px]]
|[[File:Q9ani2wxy0119.png|390px]]
|-
|Figure.26 Animation plot (at the start of the reaction)
|Figure.27 Animation plot (at the end of the reaction)
|}

=== Energy Distribution and Reactivity ===

{{fontcolor|red|Question 10ː 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 investigation, 1st set condition pFH = -0.5 pHH = -3 unreactive; 2nd pHH = -2.45 reactive; 3rd pHH = -1.1 unreactive; 4th pHH = -0.9 reactive; 5th pHH = 1.1 unreactive ; 6th pHH = 1.9 reactive ; 7th pHH = 2.4 to 3 unreactive
; pHH = 2.1 no reaction

The cases studied are an illustration of Polanyi's empirical rules.

For H and HF

failed conditionsː 091, 2, 0.05, -20

initial condition of H HF systemː
HF distance = 0.91 HH distance = 2 pHF = 0.8 pHH = -7.5

== Rreferences ==

<references>
<ref name="TS">Eyring, H. (1935). "The Activated Complex in Chemical Reactions". J. Chem. Phys. 3 (2): 107–115.</ref>
</references>